Metric Space Topology. Examples, Exercises And Solutions 2023023176, 9789811266973, 9789811266980, 9789811266997

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Series on Analysis, Applications and Computation – Vol. 12 ISAAC

Series on Analysis, Applications and Computation ISSN: 1793-4702 Series Editors:

Heinrich G W Begehr (Freie Univ. Berlin, Germany) Robert Pertsch Gilbert (Univ. Delaware, USA) Tao Qian (Univ. of Macau, China) M W Wong (York Univ., Canada)

Advisory Board Members: Mikhail S Agranovich (Moscow Inst. of Elec. & Math., Russia), Ryuichi Ashino (Osaka Kyoiku Univ., Japan), Alain Bourgeat (Univ. de Lyon, France), Victor Burenkov (Cardiff Univ., UK), Jinyuan Du (Wuhan Univ., China), Antonio Fasano (Univ. di Firenez, Italy), Massimo Lanza de Cristoforis (Univ. di Padova, Italy), Bert-Wolfgang Schulze (Univ. Potsdam, Germany), Masahiro Yamamoto (Univ. of Tokyo, Japan) & Armand Wirgin (CNRS-Marseille, France) Published Vol. 12: Metric Space Topology: Examples, Exercises and Solutions by W-S Cheung Vol. 11: Hardy Operators on Euclidean Spaces and Related Topics by S Lu, Z Fu, F Zhao & S Shi Vol. 10: Fractional Differential Equations and Inclusions: Classical and Advanced Topics by S Abbas, M Benchohra, J E Lazreg, J J Nieto & Y Zhou Vol. 9:

Nonlinear Waves: A Geometrical Approach by P Popivanov & A Slavova

Vol. 8:

The Linearised Dam-Break Problem by D J Needham, S McGovern & J A Leach

Vol. 7:

The “Golden” Non-Euclidean Geometry: Hilbert’s Fourth Problem, “Golden” Dynamical Systems, and the Fine-Structure Constant by A Stakhov & S Aranson, Assisted by S Olsen

Vol. 6:

An Introduction to Pseudo-Differential Operators Third Edition by M W Wong

Vol. 5:

Asymptotic Behavior of Generalized Functions by S Pilipović, B Stanković & J Vindas

More information on this series can be found at http://www.worldscientific.com/series/saac

Series on Analysis, Applications and Computation – Vol. 12 ISAAC

World Scientific

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Names: Cheung, Wing-Sum (Mathematician), author. Title: Metric space topology : examples, exercises and solutions / Wing-Sum Cheung, the University of Hong Kong, Hong Kong. Description: New Jersey : World Scientific, [2024] | Series: Series on analysis, applications and computation, 1793-4702 ; Vol. 12 | Includes bibliographical references and index. Identifiers: LCCN 2023023176 | ISBN 9789811266973 (hardcover) | ISBN 9789811266980 (ebook for institutions) | ISBN 9789811266997 (ebook for individuals) Subjects: LCSH: Metric spaces--Problems, exercises, etc. Classification: LCC QA611.28 .C44 2024 | DDC 514/.325--dc23/eng20230919 LC record available at https://lccn.loc.gov/2023023176

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Copyright © 2024 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

For any available supplementary material, please visit https://www.worldscientific.com/worldscibooks/10.1142/13160#t=suppl Desk Editors: Nimal Koliyat/Lai Fun Kwong Typeset by Stallion Press Email: [email protected] Printed in Singapore

To my parents in heaven

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Preface This is a volume on the Topology of Metric Spaces, which is a subject the author has been teaching in the last couple of decades. It can serve as a textbook or a reference on the subject concerned, mainly for advanced level undergraduates concentrating on Mathematics, Physics, Economics and Finance, etc. A common serious drawback of most existing references/textbooks at this level is the lack of worked examples/exercises with detailed solutions. So the readers are frequently frustrated for not being able to solve the exercises or not being able to tell whether their solutions are valid. To overcome this drawback, it is the purpose of the present volume to provide plenty of worked examples and exercises, including True-or-False type questions and open-ended questions, with detailed solutions. True-or-False type questions and open-ended questions are particularly effective in helping the readers to get a birds eye view of the subject and to master the materials. They are also instrumental in nurturing the mathematical insight and the mathematical maturity of the readers. The emphasis of most standard textbooks or reference books on the subject is the abstract and theoretical aspect, but such an approach may not be readily digestible by most readers. In order to master a new concept, it is most effective to first keep in mind a few simple, concrete and familiar examples, and then when dealing with a problem in the general setting, consider first the problem in such simple concrete settings. In many situations, the treatment of a problem in a simple setting could serve as a clue to tackle the problem in the general setting. On the other hand, pictorization or visualization of abstract problems into simple pictures is very often instrumental to the rigorous treatment of the problems. However, this is rarely provided in most standard

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Metric Space Topology: Examples, Exercises and Solutions

references. In view of this, in this volume, within each section in each chapter, blended with the concise and rigorous treatments on the materials and before tackling problems in the more abstract settings, there will be a number of concrete examples/exercises, supplemented with simple pictorizations as appropriate, to help the readers master the concepts. The learning outcomes of the volume include: • Demonstrate knowledge and understanding of the basic features of mathematical analysis and point set topology (e.g., able to identify objects that are topologically equivalent). • Apply knowledge and skills acquired in mathematical analysis to analyze and handle novel situations in a critical way (e.g., able to determine whether a specific function is uniformly continuous). • Think creatively and laterally to generate innovative examples and solutions to non-standard problems (e.g., able to construct counterexamples to inaccurate mathematical statements). • Acquire sufficient background for further studies in Functional Analysis, Real Analysis, Complex Analysis, Differential and Algebraic Geometry, Probability Theory, Mathematical Physics, Engineering, Economics and Finance, etc. Throughout the 30+ years of teaching experience of the author, the most frequently asked question is “What is the use of the subject?” To be absolutely frank, many branches of pure mathematics do not find any direct application in the real world whatsoever. In fact, except for a small portion of the students who would go on for further studies and take mathematics research as their careers, most theorems or results they have learnt in their mathematics courses will not be seen or used for the rest of their lives. So it is perfectly fine for them to forget all the mathematics contents they have learnt at school or in college. But one thing they should never forget is the study process of and the way of thinking in

Preface

ix

mathematics, that is, the mathematics training they have undergone. It is not the mathematical contents they have learnt that are important but the mathematics training they received. Proper mathematics training nurtures logical arguments, analytical power, critical thinking, and rational thinking. With these attributes, one could excel in any career one may go into. So the target of this volume is, through numerous worked examples and exercises, to guide the readers argue logically, think critically, analyze various possible scenarios, and to nurture a mathematical and logical mind that is able to tackle novel and ill-posed problems they may encounter in the future. With that, the so-called life long self-learning would no longer be a burden but instead, would turn into a basic instinct. The volume is perfect as a textbook or a reference for a onesemester course on Introduction to Metric Space Topology. It is the intention of the author to make it concise instead of stuffing it with unessential materials, as the latter may distract the readers from the natural flow of the development of the subject. Less is more. For advanced undergraduate level students, mastering the central theme of a subject would be much more beneficial than dabbling in a wider range of materials shallowly and prematurely. Finally, the author would like to express his gratitude to the many students who have taken his course on the subject. Their enthusiastic comments and feedback on the teaching materials have provided strong incentive for his determination to write up this book. The professional assistance rendered by the staff of World Scientific, especially Lai Fun Kwong, is also gratefully acknowledged.

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A Note on the Convention There are two different conventions for the notations of “subset” and “proper subset”. In this volume, we will be exclusively adopting the following convention: A is an arbitrary subset of X : A ⊂ X A is a proper subset of X : A  X

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About the Author

Wing-Sum Cheung was a full Professor of the Department of Mathematics of the University of Hong Kong before he retired in 2022. He is currently the Director of Undergraduate Admissions of the Faculty of Science and an Honorary Professor of the Department of Mathematics of the University of Hong Kong. He holds a BSc (with 1st class honors) from the Chinese University of Hong Kong, an MA and a PhD from Harvard University, USA. He has served as Head of Department of Mathematics and Associate Dean of the Faculty of Science of the University of Hong Kong, Vice-President of the Hong Kong Mathematical Society, Council Member of the Southeast Asian Mathematical Society, Council Member of the Hong Kong Institute of Science, Leader of the Hong Kong International Olympiad Team, and Honorary Consultant of the Ministry of Education, Youth and Sports of the Government of Cambodia. He has published over 200 journal articles, conference proceedings and book chapters, in which over 140 appeared in ISI journals. He has been named a top 1% highly cited researcher in the world by Clarivate Analytics’ Essential Science Indicator for 6 times in the last decade. He is on the editorial board of a number of international mathematical journals including Abstract

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Metric Space Topology: Examples, Exercises and Solutions

and Applied Analysis, Asian European Journal of Mathematics, Far East Journal of Mathematical Sciences, Journal of Inequalities and Applications, etc. His current research interests include Differential Geometry, Exterior Differential Systems, Calculus of Variations, Analytic Inequalities, and Differential Equations.

Contents

Preface

vii

A Note on the Convention

xi

About the Author

xiii

1. Metric Spaces

1

1.1

1.2

1.3

1.4

Definitions and Examples . . . . . . . . Exercise 1.1: Part A . . . . . . Exercise 1.1: Part B . . . . . . Topology of Metric Spaces . . . . . . . . Exercise 1.2: Part A . . . . . . Exercise 1.2: Part B . . . . . . Compactness . . . . . . . . . . . . . . . Exercise 1.3: Part A . . . . . . Exercise 1.3: Part B . . . . . . Compactness in the Euclidean Space Rn Exercise 1.4: Part A . . . . . . Exercise 1.4: Part B . . . . . .

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. 1 . 10 . 14 . 36 . 50 . 64 . 85 . 90 . 94 . 108 . 115 . 118

2. Limits and Continuity 2.1

2.2

2.3

129

Convergence in a Metric Space . Exercise 2.1: Part A . . Exercise 2.1: Part B . . Complete Metric Spaces . . . . . Exercise 2.2: Part A . . Exercise 2.2: Part B . . Continuity and Homeomorphism Exercise 2.3: Part A . . Exercise 2.3: Part B . .

3. Connectedness 3.1

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129 134 138 145 150 155 172 193 204 233

Connectedness . . . . . . . . . . . . . . . . . . . . . . . . 233 Exercise 3.1: Part A . . . . . . . . . . . . . . . . 245 Exercise 3.1: Part B . . . . . . . . . . . . . . . . 249

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Metric Space Topology: Examples, Exercises and Solutions

3.2

Path-connectedness . . . . . . . . . . . . . . . . . . . . . . 266 Exercise 3.2: Part A . . . . . . . . . . . . . . . . 278 Exercise 3.2: Part B . . . . . . . . . . . . . . . . 281

4. Uniform Continuity 4.1

4.2

295

Uniform Continuity . . . . . . . . . . . . . . . . . Exercise 4.1: Part A . . . . . . . . . . . Exercise 4.1: Part B . . . . . . . . . . . Contraction and Banach’s Fixed Point Theorem Exercise 4.2: Part A . . . . . . . . . . . Exercise 4.2: Part B . . . . . . . . . . .

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5. Uniform Convergence 5.1

5.2

Sequence of Functions . . . . Exercise 5.1: Part A Exercise 5.1: Part B Series of Functions . . . . . . Exercise 5.2: Part A Exercise 5.2: Part B

296 301 309 322 330 332 349

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349 368 377 389 395 401

Bibliography

421

Index

423

Chapter 1 Metric Spaces In this chapter, the basic concept of metric spaces will be introduced. Naively, they are simply nonempty sets equipped with a structure called metric. For the less matured students, at the beginning, this structure may appear to be a bit abstract and difficult to master. But in practice, this seemingly new concept is nothing more than a tiny little abstractization of the familiar space Rn and so all one needs to do is that whenever one needs to work on a problem in an abstract metric space, one first looks at the problem on Rn , then one would be able to see the clue of how to proceed in the general case. In fact, in general, the most effective way to master a new concept in any branch of mathematics is to keep in mind a couple of typical concrete examples and think of these examples all the time. It is just that easy.

1.1 Definitions and Examples Definition 1.1.1. Let X be a nonempty set. A metric on X is a real-valued function d:X ×X →R satisfying (M1) d(x, y) ≥ 0 and d(x, y) = 0 if and only if x = y, (M2) (symmetry)

d(x, y) = d(y, x),

(M3) (triangle inequality) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X. Given x, y ∈ X, d(x, y) is also known as the distance between x and y with respect to d. The pair (X, d) is called a metric space and elements in X are referred to as points in X. For the sake of convenience, in case there is a clearly defined metric d on X, we shall simply call X a metric space. 1

2

Metric Space Topology: Examples, Exercises and Solutions

Example 1.1.2. (i) Let X := {a, b, c}, consider d1 , d2 , and d3 : X ×X d2 d3 a b c a b c d1 a 0 2 3 a 0 1 2 a 2 0 4 1 0 3 b b b 3 4 0 ; c 2 3 0 ; c c

→ R given by a b c 0 1 2 1 0 4 2 4 0 .

Then d1 and d2 are well-defined metrics on X but d3 is not (why?) (ii) Let X be any nonempty set and d(x, y) := 1 − δxy for any x, y ∈ X, where δxy is the Kronecker delta function defined by  0 if x = y δxy := 1 if x = y . Then d is a well-defined metric called the discrete metric and (X, d) is called a discrete metric space. Notice that in a discrete metric space, all distinct points have the same distance 1, no matter how “far” or “close” they are from each other. (iii) Let X := Rn and de (x, y) := x − y for any x, y ∈ Rn , where n as usual,    stands for the Euclidean norm of R , that is, n  zi2 for any z = (z1 , . . . , zn ) ∈ Rn . Hence z := i=1

  n  de (x, y) =  (xi − yi )2 i=1

for any x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn . Then de is a well-defined metric called the Euclidean metric on Rn and (Rn , de ) is called a Euclidean metric space. In general, when we refer to the “Euclidean space Rn ”, we mean the Euclidean metric space (Rn , de ). In the sequel, unless it is specifically mentioned, all Rn ’s will be equipped with the Euclidean metric de and so sometimes we will be sloppy and write d for de and Rn for (Rn , de ) in case no confusion may arise.

Metric Spaces

3

(iv) Let X := C and dC (x, y) := |x − y| for any x, y ∈ C, where, as usual, |z| := the modulus of z ∈ C. Then dC is a well-defined metric, and we will see that (C, dC ) is “equivalent” to the Euclidean space (R2 , de ). In fact, it is easy to see that under the usual identification C ∼ = R2 , any x, y ∈ C can be considered as points in R2 , and under this identification, dC (x, y) = de (x, y). ˜ y) := 100(x1 − y1 )2 + (x2 − y2 )2 for (v) Let X := R2 and d(x, any x = (x1 , x2 ) and y = (y1 , y2 ) ∈ R2 . Then d˜ is a well-defined metric on X. Note that as real-valued functions of X × X into R, d˜ = de .

(vi) Let X := S 1 ⊂ R2 and d(x, y) := the arc length of the smaller arc in S 1 joining x and y ∈ X. Then d is a well-defined metric on X. Note that d = de as real-valued functions of X × X into R. This is a typical example of “intrinsic metric” that can be defined on a metric space. It is called intrinsic as the distance between two points can be measured without referring to the ambient space R2 . In fact, imagine you are an ant walking on a rubber band. Then for you, the “distance” between two points A, B on the rubber band is d(A, B) instead of de (A, B), because the latter is the Euclidean distance between the two points A, B which is attained by the straight line segment joining the two points, but for you, the ant, unless you can fly, you cannot travel from A to B along the straight line segment joining the two points. (vii) Let X := S 2 ⊂ R3 and x, y ∈ X. Observe that if x = ±y, then there is a unique “great circle” on X passing through x, y. Here, a great circle means a planar circle on S 2 with radius 1. Define d(x, y) := the arc length of the smaller arc on the unique great circle in S 2 joining x and y. If x = y, we define d(x, y) := 0. Finally, if x = −y, then they are antipodal points on S 2 and so they are joined by infinitely many great circles. In this case we define d(x, y) := π = half of the arc length of any such great circles. Then, similar to Example (vi), d is a well-defined intrinsic metric on X.

4

Metric Space Topology: Examples, Exercises and Solutions

(viii) Let X := Rn and dS : X × X → R be defined by dS (x, y) :=

n  i=1

|xi − yi |

for any x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn . Then dS is a well-defined metric on Rn . Note that dS = de . (ix) Let X := Rn and d∞ : X × X → R be defined by d∞ (x, y) := max{|xi − yi | : i = 1, . . . , n} for any x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn . Then d∞ is a well-defined metric on Rn . Note that de = d∞ = dS . (x) Let X := Rn . For any k ∈ N, let dk : X × X → R be defined by dk (x, y) :=



n  i=1

|xi − yi |k

1/k

for any x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn . Then dk is a well-defined metric on Rn . Note that dk = de , dS , nor d∞ . (xi) Let X := C[a, b ]. Clearly, X is a real vector space under pointwise operations. For any f , g ∈ X, define d1 (f, g) :=



b a

|f − g|(x)dx ,

where, as usual, |f − g|(x) := |f (x) − g(x)|

for any x ∈ [a, b ] .

Then d1 is a well-defined metric and so (X, d1 ) is a metric space. [See Exercise 1.1, Part B, Problem #4.]

Metric Spaces

5

(xii) Similar to Example (xi) above, let X := C[a, b ]. For any f , g ∈ X, define d2 (f, g) :=



a

b

|f − g|2 (x)dx

 12

.

Then d2 is a well-defined metric and (X, d2 ) is a metric space. [See Exercise 1.1, Part B, Problem #4.] (xiii) Again, let X := C[a, b ]. For any f, g ∈ X, define d∞ (f, g) := sup{|f (x) − g(x)| : x ∈ [a, b ]} . Then (X, d∞ ) is a metric space. Clearly, d∞ is different from dp , p = 1, 2, and as we will see in the sequel (see, for example, Exercise 1.2, Part B, Problem #19 for details), (X, d∞ ), (X, d1 ), and (X, d2 ) are very different metric spaces. (xiv) Let ∞ be the space of all bounded sequences of real numbers. So an element in ∞ is of the form x = {xn }n∈N with sup{|xn | : n ∈ N} < ∞. For any elements x = {xn }n∈N and y = {yn }n∈N ∈ X, define d(x, y) := sup{|xn − yn | : n ∈ N} . Then it is elementary to verify that (∞ , d) is a metric space. Remark. From Example 1.1.2, we see that on the same nonempty set, different metrics could be defined and they could in turn make the same underlying set into metric spaces with very different properties. Definition 1.1.3. Let (X, d) be a metric space, x ∈ X, and A ⊂ X be a nonempty subset. The distance from x to A is defined as   d(x, A) := inf d(x, a) : a ∈ A

6

Metric Space Topology: Examples, Exercises and Solutions

and the diameter of A is defined as   d(A) := sup d(a1 , a2 ) : a1 , a2 ∈ A .

Note that d(A) can be +∞. For the sake of convenience, we extend the notion of diameter to an empty set by defining d(φ) := −∞ (see Exercise 1.1, Part B, Problem #1.) A is said to be bounded if d(A) = ∞. A mapping from any nonempty set into X is said to be bounded if its image is bounded. Example 1.1.4. In (R3 , d), where as usual, d stands for the Euclidean metric de , we have   d (2, 0, 0), S 2 = 1, d(S 2 ) = 2.

Note that both values are attained in S 2 . That is, there exist a point     p ∈ S 2 (namely, p = (1, 0, 0)) such that d (2, 0, 0), p = d (2, 0, 0), S 2 = 1 and a pair of points r, s ∈ S 2 (any pair of antipodal points in S 2 will do) such that d(r, s) = d(S 2 ) = 2. Example 1.1.5.

In (R, d), if A := (0, 1), we have d(2, A) = 1, d(A) = 1.

Note that neither of these values is attained in A. Definition 1.1.6. Let (X, d) be a metric space, and A, B be nonempty subsets of X. The distance between A and B is d(A, B) := inf{d(a, b) : a ∈ A, b ∈ B}. Clearly, if A ∩ B = φ, then d(A, B) = 0. But the converse is in general not true.

Metric Spaces

7

Example 1.1.7. In (R, d), let A := (0, 1), B := [3, 4], and C := (4, ∞). Then d(A, B) = 2 and d(B, C) = 0. Note, however, that neither of the infima is attained. Example 1.1.8. A good source of metric spaces is the “innerproduct spaces”. In general, an inner-product space is an ordered pair (V, , ), where V is a vector space and , is a function , :V ×V →R called the inner-product in V satisfying (I1) x, x > 0 for all 0 = x ∈ V ,

(I2) x, y = y, x for all x, y ∈ V ,

(I3) , is linear in the first argument, that is, α1 x1 + α2 x2 , y = α1 x1 , y + α2 x2 , y for any x1 , x2 , y ∈ V and any α1 , α2 ∈ R.

Note that by (I2) and (I3), x, 0 = 0, x = 0 for all x ∈ V and , is bilinear. If (V, , ) is an inner-product space, define 

:V →R

by 1

x = x, x 2 . Then 

 satisfies

(N1) x ≥ 0 for all x ∈ V and x > 0 if 0 = x ∈ V ,

(N2) αx = |α| x for all α ∈ R, x ∈ V ,

(N3) x + y ≤ x + y for all x, y ∈ V .

Note that by (N2), we have 0 = 0. In general, any vector space V together with a function   : V → R satisfying (N1)–(N3) is called a normed vector space and the function   is called the norm of V . Hence in particular, every inner-product space is automatically a normed vector space.

8

Metric Space Topology: Examples, Exercises and Solutions

Now if (V,  ) is a normed vector space, define d:V ×V →R by d(x, y) = x − y for any x, y ∈ V. Then it is easily checked that d satisfies (M1) – (M3) and thus (V, d) is a metric space. In general, we have the relations {inner product spaces}  {normed vector spaces}  {metric spaces}.

For instance, each Rn is naturally an inner product space and it gives rise to the usual metric space Rn . Similarly, C[a, b ] is an inner product space with f, g :=



a

b

f (x)g(x)dx

for any f, g ∈ C[a, b ]

and it gives rise to the metric space in Example 1.1.2(xii). Definition 1.1.9. Let (X, d) be a metric space and Y ⊂ X be a nonempty subset. If we denote by d|Y the restriction of d to Y , then d|Y is a well-defined metric on Y and hence (Y, d|Y ) is also a metric space which is known as a metric subspace of X. For the sake of simplicity, unless ambiguity may arise, we normally drop the restriction notation and write d|Y simply as d, and call it the relative metric induced by d on Y . Example 1.1.10. (i) Let Y := (0, ∞) ⊂ R. Then (Y, d) is a metric subspace of R with d = de . In particular, for any x, y ∈ Y , dY (x, y) = de (x, y) = |x − y|.

Metric Spaces

9

(ii) Let Y := N ⊂ R. Then (Y, d) is a metric subspace of R with d = de . In particular, for any m, n ∈ Y, dY (m, n) = de (m, n) = |m − n|.

(iii) Let X := S 1 ⊂ R2 . Using the intrinsic metric d defined in Example 1.1.2 (vi), (X, d) is a well-defined metric space. On the other hand, when considered as a subset of R2 , X inherits the Euclidean metric de and so (X, de ) is also a metric space, or more precisely, a metric subspace of R. Observe that (X, d) is different from (X, de ). For example, if we denote by n the “north pole” and s the “south pole” of S 1 , then d(n, s) = π, while de (n, s) = 2. For the rest of this chapter, unless otherwise specified, X will always stand for a metric space, and S, T , etc., will stand for arbitrary subsets of X.

10

Metric Space Topology: Examples, Exercises and Solutions

Exercise 1.1 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. Let X be a nonempty set and d : X × X → R a function satisfying (i) d(x, y) = 0 ⇐⇒ x = y, (ii) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X. Then d is a metric on X. Answer : False. Example : Let X := {a, b, c}, with d : X × X → R defined by d a b c a 0 1 1 2 0 1 b 1 1 0 , c Then d satisfies (i) and (ii) but it fails to satisfy (M2).

2. Let X be a nonempty set and d : X × X → R a function satisfying (i) d(x, y) = 0 ⇐⇒ x = y, (ii) d(x, y) = d(y, x), (iii) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X. Then d is a metric on X.

Answer : True. Proof . (ii) and (iii) are exactly (M2) and (M3). For (M1), for any x, y ∈ X , by (i), (iii) and (ii), we have 0 = d(x, x) ≤ d(x, y) + d(y, x) = 2d(x, y). Thus (M1) is also satisfied.

3. d(x, y) :=



|x − y| is a metric on R.

Answer : True.

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Proof . (M1) and (M2) are clearly satisfied. For any x, y, z ∈ R, d(x, y)2 = |x − y|

≤ |x − z| + |z − y|

= d(x, z)2 + d(z, y)2  2 ≤ d(x, z) + d(z, y) ,

hence (M3) is also satisfied.

4. The function d : R2 × R2 → R defined by d((x1 , y1 ), (x2 , y2 )) :=



|x1 − x2 | +



2 |y1 − y2 | ,

for any (x1 , y1 ), (x2 , y2 ) ∈ R2 , is a metric on R2 .

Answer : False. Example : Take (x1 , y1 ) = (1, 0) and (x2 , y2 ) = (0, 1). Then   d (x1 , y1 ), (x2 , y2 )     = 4 > 2 = d (x1 , y1 ), (0, 0) + d (0, 0), (x2 , y2 )

and so the triangle inequality is violated.

  5. Let X := Rn . Then d(x, y) := min |xi − yi | : i = 1, . . . , n for any x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) ∈ Rn is a welldefined metric on Rn . Answer : False. Example : Take n = 2, x := (1, 0), y := (−1, 0). Then x = y but d(x, y) = 0. 6. If d1 and d2 are metrics on X, then d := min{d1 , d2 } is a metric on X. Answer : False. Example : Let X := {p, q, r} and d1 , d2 be defined by

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Metric Space Topology: Examples, Exercises and Solutions

d1 p q r p 0 10 1 10 0 10 q 1 10 0 , r It is easily shown that d1 and d2 However,

d2 p q r p 0 10 10 10 0 1 q 10 1 0 . r are well-defined metrics on X.

d(p, q) = min{d1 (p, q), d2 (p, q)} = min{10, 10} = 10 d(p, r) = min{d1 (p, r), d2 (p, r)} = min{1, 10} = 1 d(r, q) = min{d1 (r, q), d2 (r, q)} = min{10, 1} = 1 and so d(p, q) = 10 ≤ 2 = 1 + 1 = d(p, r) + d(r, q) . 7. d(x, S) = 0 if and only if x ∈ S.

Answer : False. Example : Let X := R, S := (0, 1), then 0 ∈ / S but d(0, S) = inf{d(0, s) : s ∈ S} = inf{s : 0 < s < 1} = 0.

8. d(S, T ) = 0 if and only if S ∩ T = φ.

Answer : False. Example : Let X := R, S := (0, 1), T := (1, 2). Then S ∩ T = φ but d(S, T ) = inf{d(s, t) : s ∈ S, t ∈ T } = 0.

9. If S ∩ T = φ, then d(S ∪ T ) ≤ d(S) + d(T ).

Answer : True. Proof . By assumption, there exists x ∈ S ∩ T . For any p, q ∈ S ∪ T , we have 4 cases. Case 1: p, q ∈ S. In this case we have d(p, q) ≤ d(S) ≤ d(S) + d(T ). Case 2: p, q ∈ T . In this case we have d(p, q) ≤ d(T ) ≤ d(S) + d(T ).

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Case 3: p ∈ S, q ∈ T . In this case we have d(p, q) ≤ d(p, x) + d(x, q) ≤ d(S) + d(T ). Case 4: p ∈ T , q ∈ S. In this case we have d(p, q) ≤ d(p, x) + d(x, q) ≤ d(T ) + d(S). Combining, we always have d(p, q) ≤ d(S) + d(T ) and so d(S ∪ T ) ≤ d(S) + d(T ). 10. If S ∩ T = φ, then d(S ∪ T ) ≥ d(S) + d(T ). Answer : False. Example : Let X := R, S := Q ∩ [0, 1], and T := (R \ Q) ∩ [0, 1]. Then S ∩ T = φ, but d(S ∪ T ) = d([0, 1]) = 1 ≥ 2 = 1 + 1 = d(S) + d(T ). 11. d(S ∪ T ) ≤ d(S) + d(T ) + d(S, T ). Answer : True. Proof . For any p, q ∈ S ∪ T , similar to the considerations in Exercise 1.1, Part A, #9, it suffices to consider the case where p ∈ S and q ∈ T . By the definition of infimum, for any ε > 0, there exist s ∈ S , t ∈ T such that

d(S, T ) ≤ d(s, t) < d(S, T ) + ε . Hence

d(p, q) ≤ d(p, s) + d(s, t) + d(t, q) ≤ d(S) + d(S, T ) + d(T ) + ε . Since this is true for all p ∈ S and q ∈ T , we have

d(S ∪ T ) ≤ d(S) + d(T ) + d(S, T ) + ε . Since ε > 0 is arbitrary, we conclude that d(S ∪ T ) ≤ d(S)+ d(T )+

d(S, T ).

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Metric Space Topology: Examples, Exercises and Solutions

Part B: Problems 1. In Definition 1.1.3, we defined d(φ) := −∞. Why? Solution : Recall that we defined d(S) := inf{d(x, y) : x, y ∈ S} for any φ = S ⊂ X . So if we want to the extend the domain of definition of diameter so as to include the empty set, in order to make things defined in a “consistent” or “continuous” manner, we should define

d(φ) := inf{d(x, y) : x, y ∈ φ} = inf φ and so we must first know what inf φ should be. Now observe that for any A, B ⊂ R, if A ⊂ B , we must have inf A ≤ sup B . As φ ⊂ B for all B ⊂ R, we must have inf φ ≤ sup B for all B ⊂ R. Hence this leaves us no choice but to define inf φ := −∞ and so d(φ) must also be defined as −∞. 2. Let (X, d) be a metric space and let M > 0. Define d˜ : X ×X → R by ˜ y) := min(M, d(x, y)) d(x, ˜ a metric space? for any x, y ∈ X. Is (X, d) Answer : Yes. Proof . We need to verify the conditions (M1)-(M3). ˜ y) ≥ 0 for any (M1): Since d(x, y) ≥ 0 and M > 0, we have d(x, x, y ∈ X . Furthermore, for any x, y ∈ X , if x = y , then ˜ y) = min(M, d(x, y)) = min(M, 0) = 0. Conversely, if d(x, ˜ y) = 0, then as M > 0, we must have d(x, y) = 0 and d(x, so x = y . Hence (M1) is satisfied. ˜ y) is symmetric, hence (M2) is satis(M2): It is easily seen that d(x, fied. (M3): For any x. y , z ∈ X ,

˜ z) ≥ M . In this case we have Case 1: d(x, ˜ y) = min(M, d(x, y)) d(x, ˜ z) ≤ d(x, ˜ z) + d(y, ˜ z) . ≤ M ≤ d(x,

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˜ z) ≥ M . Similar to Case 1, we have Case 2: d(y, ˜ y) = min(M, d(x, y)) d(x, ˜ z) ≤ d(x, ˜ z) + d(y, ˜ z) . ≤ M ≤ d(y, ˜ z) < M and d(y, ˜ z) < M . In this case we have Case 3: d(x, ˜ z) = d(x, z) and d(x,

˜ z) = d(y, z) d(y,

and so

˜ y) = min(M, d(x, y)) ≤ d(x, y) d(x, ˜ z) + d(y, ˜ z) . ≤ d(x, z) + d(y, z) = d(x, Combining, we see that (M3) also holds.

3. Determine whether the following functions are metrics on R. (a) d(x, y) := [ |x − y| ]. Here, the Gaussian symbol [ · ] : R → Z is defined as [x] := the largest integer that is less than or equal to x, for any x ∈ R. (b) d(x, y) := (x − y)2 . x . (c) d(x, y) := |f (x) − f (y)|, where f (x) := 1+|x|   −1 −1 (d) d(x, y) := tan x − tan y . Answer :

(a) No.

      Justification . Since d 1, 12 = 1 − 12  = 12 = 0 but 1 = 12 , (M1) is violated.

(b) No.

Justification . Since d(0, 2) = (2 − 0)2 = 4 ≤ 2 = 12 + 12 = d(0, 1) + d(1, 2), (M3) is violated.

(c) Yes.

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Metric Space Topology: Examples, Exercises and Solutions

Proof . It is clear that d(x, y) ≥ 0 and d(x, x) = 0 for all x, y ∈ R. Conversely, if d(x, y) = 0, then y x = f (x) = f (y) = . 1 + |x| 1 + |y|

(*)

Taking absolute values on both sides and simplify, we have |x| =

|y|. Putting it back to (*), we have x = y and so (M1) is

satisfied. (M2) is obvious. (M3) follows immediately from the usual triangle inequality for absolute value:

d(x, y) = |f (x) − f (y)|

≤ |f (x) − f (z)| + |f (z) − f (y)| = d(x, z) + d(z, y) .

(d) Yes.

Proof . (M1) and (M2) are obvious. (M3) follows immediately from the usual triangle inequality for absolute value.

4. Prove Example 1.1.2 (xi) and (xii), that is, on C[a, b ], dp (f, g) :=



b a

|f − g|p (x)dx

 p1

,

p = 1, 2 ,

are well-defined metrics. How about d(f, g) :=



b a

(f − g)2 (x)dx ?

Solution : For p = 1, it is clear that d1 (f, g) ≥ 0 for all f , g ∈ C[a, b ] and d1 (f, g) = 0 for f = g. Furthermore, if d1 (f, g) = 0, then



a

b

|f − g|(x)dx = 0 .

Since |f − g| ≥ 0 and is continuous on [a, b ], this forces f − g = 0 on [a, b ] and so f = g . Hence (M1) is satisfied. Next, it is obvious

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17

that (M2) also holds. Finally, for (M3), let f , g , h ∈ X , we have

d1 (f, g) = ≤ =







1

0 1 0 1 0

  f (x) − g(x)dx

    f (x) − h(x) + h(x) − g(x)dx   f (x) − h(x)dx +

1

0

= d(f, h) + d(h, g) .

  h(x) − g(x)dx

Hence (M3) also holds and so d1 is a metric. Next, for p = 2, it is clear that d2 (f, g) ≥ 0 for all f , g ∈ C[a, b ]

and d2 (f, g) = 0 for f = g . Furthermore, if d2 (f, g) = 0, then



b 2

a

(f − g) (x)dx

 12

=0.

By the continuity of f − g , this forces f − g = 0 on [a, b ] and so

f = g. Hence (M1) is satisfied. Next, it is obvious that (M2) also holds. Finally, for (M3), let f , g , h ∈ X . By the elementary Cauchy–

Schwarz inequality, we have

d22 (f, g)

=



1

(f (x) − g(x))2 dx

0

=



1

(f (x) − h(x) + h(x) − g(x))2 dx

0

=



1 2

(f (x) − h(x)) dx +

0

+2 ≤



1 0



1 0

(h(x) − g(x))2 dx

(f (x) − h(x))(h(x) − g(x))dx

1 (f (x) − h(x))2 dx + (h(x) − g(x))2 dx 0 0  1 1 (f (x) − h(x))2 dx (h(x) − g(x))2 dx +2



1

0

0

2

= (d(f, h) + d(h, g)) .

Hence (M3) also holds. So d2 is a metric.

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Metric Space Topology: Examples, Exercises and Solutions

However,

d(f, g) :=



b a

(f − g)2 (x)dx is not a metric, as triangle

inequality (M3) is violated. For instance, take f :≡ 1, g :≡ 2, and

h :≡ 3 on [0, 1]. Then

d(f, g) + d(g, h) =



1 2

1 dx +

0

0. Hence we must have f (rk ) = g(rk ) for all k ∈ N. That is, f = g on {rk : k ∈ N} which is dense in [0, 1]. Since both f and g are continuous, this forces f = g on [0, 1]. (M2): Trivial. (M3): Observe first the following elementary inequality

min{a, 1} ≤ min{b, 1} + min{c, 1}

(*)

for all a, b, c ≥ 0 with a ≤ b + c. Hence for any f , g , h ∈ X ,

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Metric Space Topology: Examples, Exercises and Solutions

we have

d(f, g) = ≤

∞ 

k=1 ∞  k=1

2−k min{|f (rk ) − g(rk )|, 1} 2−k [min{|f (rk ) − h(rk )|, 1}

+ min{|h(rk ) − g(rk )|, 1}]

=

∞ 

k=1

+

2−k min{|f (rk ) − h(rk )|, 1} ∞ 

k=1

2−k min{|h(rk ) − g(rk )|, 1}

= d(f, h) + d(g, h) . Note that we have made used of the fact that as all terms in the series above are non-negative and so the order of the summation does not matter.

8. If d and d are metrics on a nonempty set X, must D := d · d also be a metric on X? Answer : False. Example : Take X = R and d = d to be the Euclidean metric. Then D fails to be a metric as it violates the triangle inequality (M3). In fact, for x = 1, y = 2, and z = 3, we have D(x, z) = 2 · 2 = 4 > 1 + 1 = D(x, y) + D(y, z) . 9. Determine whether each of the functions d below defines a metric on R2 . In case d is a metric, describe B(0, 1). In case d is not a metric, determine which condition in the definition is violated.  T (a) d(u, v) := (u − v) A(u − v) for any u, v ∈ R2 , where   1 7 −5 . A := −5 7 2

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T

(b) d(u, v) := (u − v) A(u − v) for any u, v ∈ R2 , where   1 −5 7 . A := 7 −5 2 (c) d(u, v) := max{d1 (u, v), d2 (u, v)} for any u, v ∈ R2 , where d1 , d2 are metrics on R2 . (d) d(u, v) := min{d1 (u, v), d2 (u, v)} for any u, v ∈ R2 , where d1 , d2 are metrics on R2 . Answer : (a) True, d is a metric.

Proof . Note that A =

1 P DP T , with 2   1 0 D := 0 6

and

 1 −1 1 1     u1 v1 is an invertible matrix. For any u = and v = ∈ u2 v2 R2 , define s := u1 − v1 , t := u2 − v2 . It follows that P :=

d(u, v) = =





1 2





s+t −s + t

T 

1 0 0 6



s+t −s + t



1 [(s + t)2 + 6(−s + t)2 ] ≥ 0 2

with equality holds if and only if s + t = 0 and s − t = 0,

or equivalently, if and only if s = 0 and t = 0. It follows that

d(u, v) = 0 if and only if u = v . So (M1) holds. On the other hand, as

d(u, v) = =





(u − v)T A(u − v) T

(v − u) A(v − u) = d(v, u) ,

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Metric Space Topology: Examples, Exercises and Solutions

(M2) also holds. Finally, before checking the triangle inequality, we first prove that for any u, v ∈ R2 , we have

√  T  √ u Av  ≤ uT Au v T Av .

In fact, write λ :=

√

uT Au and η :=

√

v T Av . Then

0 ≤ d(ηu, λv)2

= (ηu − λv)T A (ηu − λv)

= η 2 uT Au − 2λ η uT A v + λ2 v T Av

= 2λ2 η 2 − 2λ η uT A v   = 2λ η λ η − uT A v

and hence uT Av ≤ λ η . Similarly, by considering d(η u, −λ v), we have −λ η ≤ uT Av . Thus we have

√  T  √ u Av  ≤ uT Au v T Av

for any u, v ∈ R2 .

Back to our problem. For any u, v , w ∈ R2 , since A is symmetric, we have

d(u, v)2 = (u − v)T A(u − v)   T  = (u − w) + (w − v) A (u − w) + (w − v)

= (u − w)T A(u − w) + 2(w − v)T A(u − w) + (w − v)T A(w − v)

≤ d(u, w)2 + d(w, v)2   T + 2 (w − v) A(w − v) (u − w)T A(u − w)  2 = d(u, w) + d(w, v) .

The triangle inequality follows. Hence d is a metric.

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(b) False, d is not a metric.

    1 0 Justification . In fact, for u = and v = , we have 0 0 −5 = d(u, w) + d(w, v) , 100 5

0 , 0

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Metric Space Topology: Examples, Exercises and Solutions

10. Let X be the space of all sequences of real numbers. So an element in X is of the form x = {xn }n∈N . For any elements x = {xn }n∈N and y = {yn }n∈N ∈ X, define d(x, y) :=

 1 |xn − yn | . 2n 1 + |xn − yn |

n∈N

Show that d is a metric on X. Proof . By Comparison Test, it is elementary to see that the defining series of d is convergent and so d a well-defined real-valued function on X × X . Also, it is evident that (M1) and (M2) are satisfied. For (M3), b a we first observe that if 0 ≤ a ≤ b, then we must have ≤ . 1+a 1+b Hence for any x, y , z ∈ X , we have |xn − zn | + |zn − yn | |xn − yn | ≤ 1 + |xn − yn | 1 + |xn − zn | + |zn − yn | |xn − zn | = 1 + |xn − zn | + |zn − yn | |zn − yn | + 1 + |xn − zn | + |zn − yn | |zn − yn | |xn − zn | + , ≤ 1 + |xn − zn | 1 + |zn − yn | hence

d(x, y) =

 1 |xn − yn | 2n 1 + |xn − yn |

n∈N

≤

 1 |zn − yn |  1 |xn − zn | + n 2 1 + |xn − zn | 2n 1 + |zn − yn |

n∈N

n∈N

= d(x, z) + d(z, y) .

11. Let {(Xn , dn ) : n ∈ N} be a countable collection of metric  spaces and let X := ∞ n=1 Xn . For any x = (xn : n ∈ N) and

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y = (yn : n ∈ N) ∈ X, define d(x, y) :=

∞ 

n=1

2n

Show that d is a metric on X.



dn (xn , yn )  . 1 + dn (xn , yn )

∞ Proof . Note first that the series n=1 n ∈ N and any xn , yn ∈ Xn , we have 0≤

1 2n

is convergent. For any

dn (xn , yn ) ≤1 1 + dn (xn , yn )

and so by comparison test, the series ∞ 

dn (xn , yn ) n (1 + d (x , y )) 2 n n n n=1 is also convergent. Hence d is a well-defined function on X × X and

hence it remains to check conditions (M1) to (M3).

(M1): Since all dn ’s are metrics, it is clear that d(x, y) ≥ 0 for all x,

y ∈ X and d(x, y) = 0 if x = y . Furthermore, if d(x, y) = 0, we must have dn (xn , yn ) = 0 for all n ∈ N and so xn = yn for all n ∈ N, that is, x = y . Hence (M1).

(M2): It is trivial as all dn ’s are metrics.

t is 1+t monotonically increasing for t ≥ 0. In particular, for any n ∈ N and any xn , yn , zn ∈ Xn , by triangle inequality, we have

(M3): By elementary considerations, the function f (t) :=

dn (xn , zn ) ≤ dn (xn , yn ) + dn (yn , zn ) and so

    f dn (xn , zn ) ≤ f dn (xn , yn ) + dn (yn , zn ) ,

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Metric Space Topology: Examples, Exercises and Solutions

that is,

dn (xn , yn ) + dn (yn , zn ) dn (xn , zn )   ≤ 1 + dn (xn , zn ) 1 + dn (xn , yn ) + dn (yn , zn ) dn (xn , yn ) = 1 + dn (xn , yn ) + dn (yn , zn ) dn (yn , zn ) + 1 + dn (xn , yn ) + dn (yn , zn ) dn (yn , zn ) dn (xn , yn ) + . ≤ 1 + dn (xn , yn ) 1 + dn (yn , zn ) It follows that

d(x, z) ∞  d (x , z )   n n n = n 2 (x , z ) 1 + d n n n n=1

≤

∞ 

n=1

2n



∞

 dn (xn , yn ) d (y , z ) +   n n n n 2 (y , z ) 1 + dn (xn , yn ) 1 + d n n n n=1

= d(x, y) + d(y, z) .

This checks the condition (M3). Hence d is a metric on X .

12. Consider the following two metrics on R2 : d1 (x, y) := max |xi − yi | 1≤i≤2

d2 (x, y) :=

2  i=1

|xi − yi |

for any x = (x1 , x2 ), y = (y1 , y2 ) ∈ R2 . Denote the unit ball in (R2 , di ) by Bi (0, 1), = 1, 2, respectively. (a) Show that B2 (0, 1) ⊂ B1 (0, 1). (b) Does there exist a metric d3 on R2 so that B3 (0, 1) ⊂ B2 (0, 1) and B2 (0, 1) ⊂ B3 (0, 1)?

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31

(c) Recall that any straight line in the Euclidean space (R2 , d) can be described by L := {a + tv : t ∈ R} , where a, v ∈ R2 . Let b ∈ R2 \ L. Then there exists a unique t0 ∈ R such that d(b, L) = d(b, a + t0 v) . Is this still valid in (R2 , d1 ) or (R2 , d2 )? (d) Let H be the “upper half plane” H := {(x, y) : y > 0} ⊂ R2 . (i) Show that H is unbounded in (R2 , d1 ). (ii) Is there any metric d4 on R2 such that H is bounded in (R2 , d4 )? Solution : (a) For any (a1 , a2 ) ∈ B2 (0, 1), we have

  |a1 | + |a2 | = d2 (a1 , a2 ), (0, 0) < 1 . 



It follows that |a1 |, |a2 | < 1. Hence d1 (a1 , a2 ), (0, 0)

max{|a1 |, |a2 |} < 1 and so (a1 , a2 ) ∈ B1 (0, 1). (b) Yes. For example, define d3 by d3 (x, y) :=

4 d1 (x, y) . 3

It is obvious that d3 is also a metric, and

4 max{|x1 |, |x2 |} < 1} 3 3 3 = {(x1 , x2 ) : |x1 | < , |x2 | < } . 4 4

B3 (0, 1) = {(x1 , x2 ) :

=

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Metric Space Topology: Examples, Exercises and Solutions

Note that

and

 

 7 , 0 ∈ B2 (0, 1) \ B3 (0, 1) 8

5 5 , 8 8



∈ B3 (0, 1) \ B2 (0, 1) .

(c) It is not valid in either metric space. Let L1 := {(1, t) : t ∈ R}, L2 := {(t, 1 − t) : t ∈ R}, and

b := 0 ∈ R2 . Then in (R2 , d1 ),

  d1 (b, L1 ) = 1 = d1 (0, 0), (1, t)

for any t ∈ R satisfying |t| ≤ 1, while in (R2 , d2 ),

d2 (b, L2 ) = 1 = d2 ((0, 0), (u, v))

(d)

for any u, v ∈ R satisfying u ≥ 0, v ≥ 0, u + v = 1. (i) For any N ∈ N,

d1 ((−N, 1), (N, 1)) = 2N ≥ N . Thus H ⊂ (R2 , d1 ) is unbounded.

(ii) Yes, such d4 exists.

d1 (x, y) for any x, y ∈ R2 . 1 + d1 (x, y) It is easy to verify that d is a well-defined metric on R

Define d4 (x, y) :=

(or it follows directly from Exercise 1.1, Part B, Problem #11). For any x, y ∈ H, we have d4 (x, y) ≤ 1. Thus H is bounded in (R2 , d4 ).

13. For any 1 ≤ p < ∞, let p be the space of sequences x =  |xn |p < ∞. For any elements {xn }n∈N of real numbers with n∈N

x = {xn }n∈N and y = {yn }n∈N ∈ p , define  1/p  p xn − yn  . d(x, y) := n∈N

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Show that d is a metric on p by establishing the following: (a) For any α, β ≥ 0 and 0 < λ < 1, we have αλ β 1−λ ≤ λ α + (1 − λ) β .

(1)

(b) Let p, q ≥ 0 be real numbers satisfying p1 + 1q = 1 (p, q are said to be conjugate pairs). Clearly this forces p > 1 and q > 1. For any x = {xn }n∈N ∈ p and y = {yn }n∈N ∈ q , write zn := xn yn , n ∈ N. Then z := {zn }n∈N ∈ 1 and the following H¨ older’s Inequality holds:  1/p  1/q    p q |xn yn | ≤ |xn | |yn | . (2) n∈N

n∈N

n∈N

(c) For any p ≥ 1 and any elements x = {xn }n∈N and y = {yn }n∈N ∈ p , the following Minkowski Inequality holds:    1/p 1/p 1/p   p   p  p xn + yn      ≤ + . xn yn n∈N

n∈N

n∈N

(3)

Proof . (a) Define ϕ : [0, ∞) → R by

ϕ(t) := (1 − λ) + λ t − tλ

t ∈ [0, ∞) .

It is clear that ϕ is continuous on [0, ∞) and differentiable in

(0, ∞), with ϕ (t) = λ(1 − tλ−1 ) and so

ϕ (t) > 0 for t > 1 , ϕ (t) < 0 for t < 1 . Hence for all t = 1, ϕ(t) > ϕ(1) = 0 and so

(1 − λ) + λ t ≥ tλ ,

(4)

with equality if and only if t = 1. Back to the proof of inequality (1). If β = 0, the inequality is trivial. If β > 0, it follows immediately from (4) by letting

t=

α . β

34

Metric Space Topology: Examples, Exercises and Solutions

(b) Putting

|xn |p α=  , |xk |p

|yn |p β=  , |yk |p

k∈N

λ=

1 , p

1−λ =

1 q

k∈N

into inequality (1), we have,

 =





k∈N



n∈N

|xk



|xn yn |

n∈N 1/p p |



k∈N

|xn |

|xk

|p





k∈N

|yk

1/p · 

|q

1/q



k∈N

|yn | |yk

|q

1/q

1  |xn |p 1  |yn |q   + p p |xk | q |yk |q n∈N n∈N k∈N k∈N   |xn |p |yn |q 1 n∈N 1 n∈N 1 1 =  +  = + =1. p q p |xk | q |yk | p q ≤

k∈N

k∈N

Hence (2). In particular, as x ∈ p and y ∈ q , it follows from inequality (1) that z ∈ 1 .

(c) If p = 1, Minkowski Inequality (3) follows immediately from the usual triangle inequality. So we assume p > 1. By triangle inequality, we have

|xk + yk |p = |xk + yk | |xk + yk |p−1

≤ |xk | |xk + yk |p−1 + |yk | |xk + yk |p−1 .

Metric Spaces

35

Let q > 1 be conjugate to p. Then by H¨ older’s Inequality, we have



n∈N

≤ ≤



n∈N



|xk + yk |p |xn | |xn + yn |p−1 +



|xn |p

+



n∈N



1 p



n∈N



n∈N 1 p

p

|yn |



n∈N

|yn | |xn + yn |p−1 1 q

|xn + yn |(p−1)q





n∈N

1 q

(p−1)q

|xn + yn |

.

By observing that (p − 1)q = p and moving the common factor

on the right hand side to the left hand side, we arrive at Minkowski Inequality (3).

Finally, to show that d is a metric, we need to show (M1), (M2), and (M3). But (M1) and (M2) are clear, and (M3) follows immediately from Minkowski Inequality (3).

36

Metric Space Topology: Examples, Exercises and Solutions

1.2 Topology of Metric Spaces Given just one single structure, namely, metric d, on a nonempty set X , a very rich theory called Topology could be developed. Roughly speaking, Topology is a branch of mathematics which studies the properties of geometric objects which are of the elastic type. That is, objects which are free to bend, twist, stretch, shrink, etc., that is, without closing up existing holes, opening up new holes, tearing, gluing, or passing through itself. Elementary point-set Topology can be considered as a sub-branch of Analysis, as some methods used in Topology are very similar to those used in Analysis. Analysis is a science of approximation. In order to study the properties of an object, we study the properties of its surrounding objects first and then deduce the properties of the whole object under consideration. This makes sense even in real life. One way of understanding a person is to first understand his/her friends. So we will start by studying the so-called local properties of a metric space.

Definition 1.2.1. Let a ∈ X and r > 0. The open ball in X with center a and radius r is   B(a, r) := x ∈ X : d(a, x) < r

and the closed ball in X with center a and radius r is   B(a, r) := x ∈ X : d(a, x) ≤ r . Example 1.2.2. (i) In R, B(0, 1) = (−1, 1). (ii) In R2 , B(0, 1) = {(x1 , x2 ) ∈ R2 : x21 + x22 < 1} = the disc centered at the origin with radius 1 without boundary; and B(0, 1) = {(x1 , x2 ) ∈ R2 : x21 + x22 ≤ 1} = the disc centered at the origin with radius 1 including the boundary. ˜ where d˜ is the metric introduced in Example 1.1.2 (iii) In (R2 , d), (v), B(0, 1) = {(x1 , x2 ) ∈ R2 : 100x21 + x22 < 1}, the region

Metric Spaces

37

strictly inside the ellipse centered at the origin with semi-axes 1 10 and 1, boundary not included; and B(0, 1) = {(x1 , x2 ) ∈ R2 : x21 + 4x22 ≤ 1}, the region enclosed by the same ellipse, boundary included. (iv) Let X := [0, ∞) equipped with the Euclidean metric. Then B(0, 1) = {x ∈ [0, ∞) : d(x, 0) < 1} = [0, 1), while B(0, 1) = {x ∈ [0, ∞) : d(x, 0) ≤ 1} = [0, 1]. (iv) Let X := [0, 1] ∪ (2, 4] equipped with the Euclidean metric. Then B(0, 3) = [0, 1] ∪ (2, 3) and B(0, 3) = [0, 1] ∪ (2, 3]. (vi) At first sight, B(a, r) is only slightly larger than B(a, r). However, in general that may not be the case. For example, if X is a discrete metric space and a ∈ X, then B(a, 1) = {a} is a singleton; while B(a, 1) = X is the entire metric space. Remark. Observe that if φ = S ⊂ X is considered as a metric subspace, the open balls in S are BS (a, r) = {x ∈ S : d(a, x) < r} = BX (a, r) ∩ S. Example 1.2.3.

B[0,∞) (0, 1) = [0, 1) = BR (0, 1) ∩ [0, ∞).

Definition 1.2.4. For any S ⊂ X, a point a ∈ S is called an interior point of S in X if there exists r > 0 such that B(a, r) ⊂ S.

The collection of all interior points of S in X is called the interior of S in X and is denoted by S ◦ or int S. S is said to be open in X if S = S ◦ , closed in X if X \ S is open in X.

38

Metric Space Topology: Examples, Exercises and Solutions

Observe that by definition, S ◦ ⊂ S. Hence S = S ◦ if and only if S ⊂ S ◦ . Hence in order to show that S is open in X, all that is required is to show that every point in S is an interior point of S. Example 1.2.5. (i) If X = R, S = (0, 1), then S ◦ = S, (ii) If X := R, T := [0, 1], then T ◦ = (0, 1), (iii) If X := R, P := [0, 1), then P ◦ = (0, 1). We have now the concepts of open balls, closed balls, open sets, and closed sets. Both “open balls” and “open sets” contain the word “open”, so judging from the terminology it is natural to expect that they are somehow related, although up till this moment we have not investigated whether they have any relations at all. Similar for “closed balls” and “closed sets”. The following example tells how they are related. Example 1.2.6. (i) φ, X ⊂ X are both open and closed (in short, “clopen”) in X. Proof. Since nothing can go beyond X, every point in X is automatically an interior point of X and so X is open in X. Hence by definition, φ = X \ X is closed in X. On the other hand, as S ◦ ⊂ S for any S ⊂ X, we have φ◦ ⊂ φ. But then the only possible subset of φ is φ itself, so φ◦ = φ and so by definition, φ is open in X. Hence X = X \ φ is closed in X.

(ii) Every open ball B(a, r) in X is open in X. Proof. To show that B(a, r) is open in X, we need to show that every point x ∈ B(a, r) is an interior point of B(a, r). So let x ∈ B(a, r). By definition, we have d(x, a) < r. Hence δ := r − d(x, a) > 0 and so B(x, δ) is a well-defined open ball in X. It is not hard to see that B(x, δ) ⊂ B(a, r). In fact, for any y ∈ B(x, δ), we have d(y, x) < δ. Hence d(y, a) ≤ d(y, x) + d(x, a) < δ + d(x, a) = r and so y ∈ B(a, r).

Metric Spaces

39

Since y ∈ B(x, δ) is arbitrary, we conclude that B(x, δ) ⊂ B(a, r) and hence by the definition of interior points, x ∈ B(a, r)◦ . (iii) Every closed ball B(a, r) in X is closed in X. Proof. This can be proven by similar argument as that used in (ii). (iv) If X is a discrete metric space, every subset of X is clopen. Proof. Let S ⊂ X be an arbitrary subset. If S = φ, then it follows from (i) above that S is clopen. So assume that there exists x ∈ S = φ. By Example 1.2.2 (vi), B(x, 1) = {x} ⊂ S and so x ∈ S ◦ . Hence S is open in X. Therefore, all subsets of X are open in X. Since open subsets and closed subsets are complements of each other, all subsets of X are also closed in X. Example 1.2.7. Let X := R, Y := [0, ∞), S := [0, 1). Then for any r > 0, BX (0, r) = (−r, r) ⊂ S and so 0 ∈ S is not an interior point of S in X. Hence in particular, S is not open in X. On the other hand, for any r < 1, BY (0, r) = [0, r) ⊂ S and so 0 ∈ S is an interior point of S in Y . Furthermore, it is easy to see that all points in S other than the point 0 are interior points of S in Y . Hence in particular, S is open in Y . Alternatively, that S is open in Y can also be seen by the fact that S = BY (0, 1).

40

Metric Space Topology: Examples, Exercises and Solutions

Remarks. (i) From Example 1.2.7, we see that a point p ∈ S ⊂ Y ⊂ X is an interior point of S in Y does not necessarily imply that it is an interior point of S in X. Hence in particular, S is open in Y does not necessarily imply S is open in X, etc. So in general, when we talk about openness, closedness, interior, etc., we have to specify which metric space we are referring to. That is why we need to use the seemingly clumsy terminologies “open in X”, “closed in Y ”, “interior point of S in X”, “interior of S in X”, etc. (ii) However, in general, for any S ⊂ Y ⊂ X, it is clear that if S is open in X, then it is also open in Y . Definition 1.2.8. Let φ = S ⊂ X be a nonempty subset of X and a ∈ S. An open neighborhood of a in S is an open set U in S which contains a. More generally, a neighborhood of a in S is a set U in S which contains a in its interior. By Example 1.2.7 above, we see that in general, an open neighborhood of a in S may not be an open neighborhood of a in X. Remarks. (i) By definition, S is open in X if and only if for any x ∈ S, there exists r > 0 such that the open ball B(x, r) is lying entirely in S. Observe that this is equivalent to the condition that for any x ∈ S, there exists an open ball B in X such that x ∈ B ⊂ S, where we do not require that the center of B must be the point x. The point is, if such an open ball B exists, then for sure we can find a (smaller if necessary) open ball B(x, r) centered at x such that B(x, r) ⊂ B, hence B(x, r) ⊂ S. (ii) In the sequel, for the sake of simplicity, the statement “S is open (or closed) in T ” will be abbreviated as “S ⊂ T is open (or closed)”.

Metric Spaces

41

Theorem 1.2.9. Let S ⊂ X be a nonempty subset and U ⊂ S. Then U is open in S if and only if U = V ∩ S for some V ⊂ X which is open in X. Proof. (⇒) If U is open in S, for every x ∈ U , there exists rx > 0 such ! ! that BS (x, rx ) ⊂ U . Then U = x∈U {x} ⊂ x∈U BS (x, rx ) ⊂ U and ! so U = x∈U BS (x, rx ). Observe that BS (x, r) = BX (x, r) ∩ S. So ! this suggests that we set V := x∈U BX (x, r) and so U = V ∩ S. Observe that V must be open in X. In fact, for any y ∈ V , y ∈ BX (x, rx ) for some x ∈ U . By construction, BX (x, rx ) ⊂ V . Hence by definition, V is open in X. (⇐) If U = V ∩ S for some open set V in X, then for any x ∈ U ⊂ V , there exists r > 0 such that BX (x, r) ⊂ V . But then BS (x, r) =  BX (x, r) ∩ S ⊂ V ∩ S = U . Hence U is open in S. Note that from the proof of Theorem 1.2.9, we implicitly showed that the union of any collection of open balls is open. In fact, this will be generalized to general open sets in Theorem 1.2.11 below. But before that, we observe that since closed sets and open sets are complements of each other, whenever we have a result on open sets, we have a corresponding results on closed sets for free. Corollary 1.2.10. Let S ⊂ X be a nonempty subset and D ⊂ S. Then D is closed in S if and only if D = E ∩ S for some E ⊂ X which is closed in X. Proof. This follows immediately from Theorem 1.2.9. In fact, D ⊂ S is closed

⇔ S \ D ⊂ S is open ⇔S\D =V ∩S

for some open set V ⊂ X

⇔ D = S \ (S \ D) = S \ (V ∩ S) = S \ V = S ∩ (X \ V ) = S ∩ E, where E := X \ V ⊂ X is closed.



42

Metric Space Topology: Examples, Exercises and Solutions

Theorem 1.2.11. (a) The union of any collection of open sets is open. (b) The intersection of any finite collection of open sets is open. Proof. ! (a) Let S = λ∈I Sλ , where I is any index set and Sλ is open in X for every λ ∈ I. Then for any a ∈ S, a ∈ Sλ for some λ and so there exists r > 0 such that B(a, r) ⊂ Sλ ⊂ S. Hence S is open in X. "n (b) Let S = i=1 Si where Si is open in X for each i = 1, . . . , n. Let a ∈ S. Then a ∈ Si for all i and so there exists r1 , . . . , rn > 0 such that B(a, ri ) ⊂ Si , i = 1, . . . , n. Let r = min{r1 , . . . , rn }. Then r > 0 and B(a, r) ⊂ B(a, ri ) ⊂ Si for all i and so B(a, r) ⊂ S. Hence S is open in X.  Corollary 1.2.12. (a) The union of finitely many closed sets is closed. (b) The intersection of any collection of closed sets is closed. Proof. It is immediate from Theorem 1.2.11.



Remark. The intersection of infinitely many open sets may not be open. Analogously, the union of infinitely many closed sets may not  "∞  be closed. For example, in the real line R, n=1 − n1 , n1 = {0} is   ! 1 not open in R, while ∞ n=1 0, 1 − n = [0, 1) is not closed in R. Theorem 1.2.13. If A ⊂ X is open and B ⊂ X is closed, then A \ B is open in X and B \ A is closed in X.

Proof. This is obvious because open sets and closed sets are complements of each other. 

Metric Spaces

43

Theorem 1.2.14. φ = U ⊂ X is open if and only if it is a union of open balls. Proof. Suppose U = φ is open in X, then each x ∈ U is contained ! ! in some open ball B(x) ⊂ U , thus U = x∈U {x} ⊂ x∈U B(x) ⊂ U ! ! and so U = x∈U B(x). Conversely, suppose U = α∈Λ Bα , where for each α, Bα is an open ball, then by Theorem 1.2.11, U is open in X.  Theorem 1.2.14 gives a characterization of open sets in a general metric space X. In the special case that X = Rn , we can say more. Theorem 1.2.15 (Lindel¨ of ). If U ⊂ Rn is an open set such that ! U = λ∈Λ Uλ , where Λ is an index set and for each λ ∈ Λ, Uλ is an open set in Rn , then there is a countable subset {Ui }i∈N ⊂ {Uλ }λ∈Λ ! such that U = ∞ i=1 Ui . Proof. First observe that the family

$ #  1 : r ∈ Qn , k ∈ N U = B r, k ! is countable. Now if U ⊂ Rn is open and U = λ∈Λ Uλ , then for any x ∈ U , x ∈ Uλ for some λ and there exists Bx ∈ U such that x ∈ Bx ⊂ Uλ . Since {Bx : x ∈ U } ⊂ U and U is countable, {Bx : x ∈ U } is also countable and so we could rename it as {Bi }i∈N . Note that for each i, there exists Ui ∈ {Uλ }λ∈Λ such that Bi ⊂ Ui . ! ! ! ! Thus U ⊂ i∈N Bi ⊂ i∈N Ui ⊂ λ∈Λ Uλ ⊂ U and so U = i∈N Ui .  Remark. Combining Theorem 1.2.14 and Theorem 1.2.15, we see ! that if U ⊂ Rn is open, then U = ∞ i=1 Bi for some countable family of open balls {Bi }. Furthermore, when n = 1, we can say a bit more.

44

Metric Space Topology: Examples, Exercises and Solutions

Theorem 1.2.16. Every nonempty open subset of R is the union of countably many pairwisely disjoint open intervals in R. Furthermore, such a collection of intervals is unique. Proof. Let U be a nonempty open subset of R. For any x ∈ U , there exists r > 0 such that the open interval B(x, r) ⊂ U . Let % Ix := {B : B is an open interval with x ∈ B ⊂ U }.

Then clearly x ∈ Ix and Ix is an interval in R. Furthermore, since union of open sets is open, we see that Ix is an open interval in R. Repeat this construction for all y ∈ U . Observe that if y ∈ Ix , then Ix ∪ Iy is an open interval containing x, y and is contained in U , thus we must have Iy ⊂ Ix and Ix ⊂ Iy . This forces Ix = Iy . Similarly, if x ∈ Iy , we also have Ix = Iy . Therefore, for any two distinct points x, y ∈ U , we must have either Ix ∩ Iy = φ or Ix = Iy , for if they have a point z in common, then Ix = Iz = Iy . Hence I := {the distinct Ix ’s, x ∈ U }

is a collection of pairwisely disjoint open intervals whose union is U . By Lindel¨ of’s theorem, U equals the union of a countable sub-collection of I. But since elements in I are pairwisely disjoint, no proper sub-collection of I could have union equals U . So this implies that I is itself already countable, and we can write I = {Ii : i ∈ N}. !∞ Finally, for uniqueness. If U = j=1 Jj with the Jj ’s being pairwisely disjoint open intervals, then each Ii must be lying entirely in some unique Jj , and vice versa. Hence the collections {Ii }i∈N and {Jj }j∈N are identical up to order.  Remark. Theorem 1.2.16 is false in Rn . Note however, that we still can show by similar arguments that every open set U ⊂ Rn can be decomposed into a disjoint union of countably many maximal open sets in U called the connected components of U . We will get into that in Chapter 3 below.

Metric Spaces

45

Definition 1.2.17. Let S ⊂ X be a subset. A point x ∈ X is called an adherent point of S if B(x, r) ∩ S = φ for all r > 0. The set of all adherent points of S in X is called the closure of S in X, and is denoted by S. Clearly, S ⊂ S. Example 1.2.18. (i) If X = R, S = (0, 1) ∪ {2}, then S = [0, 1] ∪ {2}.

(ii) If X = a discrete metric space, then S = S for all S ⊂ X.

Theorem 1.2.19. S is the smallest closed subset of X containing S. Proof. For any x ∈ X \S, there exists r > 0 such that B(x, r)∩S = φ. That is, B(x, r) stays away from S. But it is not entirely clear whether it also stays away from S. So the picture may look like this:

We will show that the diagram above is incorrect, that is, B(x, r) actually stays away from S, or B(x, r) ⊂ X \ S. For this, take y ∈    B(x, r). Then B y, r − d(x, y) ⊂ B(x, r) ⊂ X \ S. That is, B y, r −  d(x, y) ∩ S = φ. Hence y ∈ X \ S. Since y ∈ B(x, r) is arbitrary, we have B(x, r) ⊂ X \ S and so X \ S is open in X and so the correct picture should look like this:

46

Metric Space Topology: Examples, Exercises and Solutions

Thus S is closed in X. Next, if S ⊂ C and C is closed in X, then X \ C ⊂ X \ S. For any x ∈ X \ C, since X \ C is open, there exists an open ball B in X such that x ∈ B ⊂ X \ C ⊂ X \ S. Hence B ∩ S = φ. Therefore x ∈ / S or x ∈ X \ S. Since x ∈ X \ C is arbitrary, we have X \ C ⊂ X \ S.  Corollary 1.2.20. For any S ⊂ X, X \ S ◦ = X \ S.

Proof. For any S ⊂ X, as S ◦ ⊂ S, we have X \ S ◦ ⊃ X \ S. But then since X \S ◦ is closed, it contains the smallest closed subset of X containing X \S, i.e., X \S ◦ ⊃ X \ S. Conversely, for any x ∈ X \S ◦ , B(x, r) ⊂ S for any r > 0, i.e., B(x, r) ∩ (X \ S) = φ for any r > 0.  This implies x ∈ X \ S.

Definition 1.2.21. Let S ⊂ T ⊂ X be subsets of X. S is said to be dense in T if S ⊃ T . In particular, S ⊂ X is dense in X if and only if S = X. Geometrically, S is dense in T if arbitrarily close to every point of T , there are points of S. Example 1.2.22. In the metric space R, (i) (0, 1) ∪ (1, 2) is dense in (0, 2).

(ii) Q is dense in R.

(iii) { n1 : n ∈ N} is dense in { n1 : n ∈ N} ∪ {0}.

Metric Spaces

47

Definition 1.2.23. Let S ⊂ X be a subset. A point x ∈ X is called an accumulation point (or, a limit point) of S if it is an adherent point of S \ {x}, that is, if B(x, r) ∩ S \ {x} = φ for all r > 0. Hence x ∈ X is an accumulation point of S ⊂ X if and only if there are points in S which are “arbitrarily close” to x. The set of all accumulation points of S in X is called the derived set of S in X, and is denoted by S  . Clearly we have S  ⊂ S but in general, equality is not expected. Definition 1.2.24. Let S ⊂ X be a subset. A point x ∈ S \ S  is called an isolated point of S. Hence a point x ∈ S is an isolated point of S if and only if there exists r > 0 such that B(x, r) ∩ S \ {x} = φ, or, equivalently, B(x, r) ∩ S = {x}. Example 1.2.25. (i) If X = R, S = [a, b ], then S  = S = S. (ii) If X = R, S = (0, 1) ∪ {2}, then S  = [0, 1].

(iii) If X = R, S = Q, then S  = S = R ⊃ S.

(iv) If X = discrete, then S  = φ for all S ⊂ X.

Theorem 1.2.26. Let S ⊂ X be any subset. Then S = S  ∪ S. Proof. It is straight forward by definition. In fact, it is obvious by definitions that S  ⊂ S and S ⊂ S. Hence S  ∪ S ⊂ S. Conversely, for / S, then B(x, r)∩S \{x} = any x ∈ S. If x ∈ S, then x ∈ S  ∪S. If x ∈  B(x, r) ∩ S = φ for all r > 0. Hence x ∈ S  ⊂ S  ∪ S. Theorem 1.2.27. Let S ⊂ X be any subset and x ∈ X. Then   x ∈ S  if and only if # B(x, r) ∩ S = ∞ for all r > 0, where for any nonempty set A, #(A) denotes the cardinal number of A. Proof. Obvious by definition.



48

Metric Space Topology: Examples, Exercises and Solutions

Remark. By Theorem 1.2.27, a necessary condition for S to have an accumulation point is that it contains infinitely many distinct points. Note that obviously the converse of this is false. For instance, S = N has infinitely many points but it has no accumulation point. Theorem 1.2.28. Let S ⊂ X be any subset. Then the following are equivalent: (a) S is closed in X. (b) S  ⊂ S. (c) S = S.

Proof. (a) ⇔ (c): By Theorem 1.2.19. (b) ⇔ (c): By Theorem 1.2.26.



So far we have, associated to any subset S ⊂ X, studied several types of points, namely, interior points, adherent points, and accumulation points. The readers will see in the sequel that they provide important information for various properties of the subset S. We have yet another type of points: boundary points. At first sight it seems to be so obvious that there is no need to formally define it. But in reality it is not the case. For simple geometric objects it is easy to “see” what the boundary points of a set should be. However, for more complicated sets, and even more so, for subsets of an abstract metric space, it is not entirely clear what boundary $ #1 : n ∈ N ⊂ R and Q ⊂ R, points are. For instance, for the sets n etc., it is not intuitively clear what their boundary points should be. So a formal definition is due. It turns out that it is nothing more than our intuition on simple subsets S ⊂ X. Definition 1.2.29. Let S ⊂ X be a subset. A point x ∈ X is said to be a boundary point of S if  B(x, r) ∩ S = φ B(x, r) ∩ (X \ S) = φ

Metric Spaces

49

for all r > 0. The set of all boundary points of S is called the boundary of S, and is denoted by ∂ S. Note that ∂ S ⊂ S and S ◦ ∩ ∂ S = φ. Example 1.2.30.

  (i) In Rn , ∂ B(a, r) = S(a, r) = x ∈ Rn : x − a = r .

(ii) In R, ∂ (0, ∞) = {0}.

(iii) In R, ∂ Q = R.   (iv) In R, if S := n1 : n ∈ N , then ∂ S = S ∪ {0}.

Remarks. (i) By definition, we have ∂ S = S ∩ X \ S. (ii) Also by definition, we have ∂ S = ∂ (X \ S). (iii) In particular, we see that ∂ S is always closed for any S ⊂ X. Theorem 1.2.31. For any subset S ⊂ X, S = ∂ S ∪· S ◦ , where A ∪· B := A ∪ B with an extra piece of information that A and B are mutually disjoint. Proof. It is rather obvious from definition: For any x ∈ S ◦ , there exists r > 0 such that B(x, r) ⊂ S. Hence B(x, r) ∩ X \ S = φ and so x ∈ / ∂ S. Therefore, S ◦ ∩ ∂ S = φ. On the / S ◦ , then B(x, r) ∩ X \ S = φ for all other hand, for any x ∈ S, if x ∈ r > 0 and so x ∈ ∂ S. 

50

Metric Space Topology: Examples, Exercises and Solutions

Exercise 1.2 Unless otherwise specified, X will stand for a general metric space and S, T , etc., are arbitrary subsets of X. Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. Let (X, d) be a metric space and S, T be subsets of X. (a) S ⊂ T =⇒ S ◦ ⊂ T ◦ ; ◦ (b) (S ◦ ) = S ◦ ; (c) S ◦ ⊂ T ⊂ S =⇒ T ◦ = S ◦ ; (d) S ◦ ∩ T ◦ ⊂ (S ∩ T )◦ ; (e) S ◦ ∩ T ◦ ⊃ (S ∩ T )◦ ; (f) S ◦ ∪ T ◦ ⊂ (S ∪ T )◦ ; (g) S ◦ ∪ T ◦ ⊃ (S ∪ T )◦ . Answer : (a) True.

Proof . Let s ∈ S ◦ . Then there exists r > 0 such that B(s, r) ⊂ S ⊂ T . So s ∈ T ◦ and thus S ◦ ⊂ T ◦ .

(b) True.

Proof . Suffices to show that S ◦ is open. For any x ∈ S ◦ , there exists r > 0 such that B(x, rx ) ⊂ S . Note that actually B(x, rx ) ∈ S ◦ . In fact, for any y ∈ B(x, rx ), since B(x, rx ) is open, there is δ > 0 such that B(y, δ) ⊂ B(x, rx ). Hence B(y, δ) ⊂ S . By definition, y is also an interior point of S , that is, y ∈ S ◦ .

(c) True.

Proof . By Parts (a) and (b), S ◦ ⊂ T ⊂ S ⇒ S ◦ = (S ◦ )◦ ⊂ T ◦ ⊂ S ◦ and so we must have T ◦ = S ◦ .

(d) True.

Metric Spaces

51

Proof . For any x ∈ S ◦ ∩ T ◦ , there exist r1 , r2 > 0 such that B(x, r1 ) ⊂ S and B(x, r2 ) ⊂ T . Let r = min{r1 , r2 } > 0. Then B(x, r) ⊂ S ∩ T . Hence x ∈ (S ∩ T )◦ and so S ◦ ∩ T ◦ ⊂ (S ∩ T )◦ .

(e) True.

Proof . By Part (a), S ∩ T ⊂ S ⇒ (S ∩ T )◦ ⊂ S ◦ . Similarly, S ∩ T ⊂ T ⇒ (S ∩ T )◦ ⊂ T ◦ . Hence (S ∩ T )◦ ⊂ S ◦ ∩ T ◦ .

(f) True.

Proof . In the proof of Part (b), we see that S ◦ and T ◦ are open, so the same is true for their union. Hence S ◦ ∪T ◦ = (S ◦ ∪T ◦ )◦ . Since S ◦ ∪T ◦ ⊂ S∪T , by (a) we have S ◦ ∪T ◦ = (S ◦ ∪T ◦ )◦ ⊂ (S ∪ T )◦ .

(g) False.

Example : Let X := R, S := [0, 1], and T := [1, 2]. Then (S ∪ T )◦ = (0, 2) but S ◦ ∪ T ◦ = (0, 1) ∪ (1, 2). 2. For any φ = S, T ⊂ R, define S · T := {st : s ∈ S, t ∈ T } ,

S + T := {s + t : s ∈ S, t ∈ T } . Then (a) S is open =⇒ S · T is open. (b) S is open =⇒ S + T is open. Answer : (a) False.

Example : Take S := R which is open, and T := {0}. Then S · T = {0} which is not open.

(b) True.

Proof . For any x ∈ S + T , there exist s ∈ S and t ∈ T such that x = s + t. Since S is open, there exists rs > 0 such that B(s, rs ) ⊂ S . Hence for any y ∈ B(x, rs ), |(y − t) − s| = |y − (s + t)| = |y − x| < rs .

52

Metric Space Topology: Examples, Exercises and Solutions

So y − t ∈ B(s, rs ) ⊂ S . As a result,

y = (y − t) + t ∈ S + T . Since y ∈ B(x, rs ) is arbitrary, we conclude that B(x, rs ) ⊂ S + T and we are done.

3. Let (X, d) be a metric space and S, T be subsets of X. (a) S ⊂ T =⇒ S ⊂ T . (b) (S) = S. (c) S ⊂ T ⊂ S =⇒ T = S. (d) S ∪ T ⊂ S ∪ T . (e) S ∪ T ⊃ S ∪ T . (f) S ∩ T ⊂ S ∩ T . (g) S ∩ T ⊃ S ∩ T . Answer : (a) True.

Proof . For any x ∈ S and any r > 0, we have B(x, r) ∩ T ⊃ B(x, r) ∩ S = φ. Hence x ∈ T .

(b) True.

Proof . It follows trivially from the fact that S is closed and so the smallest closed set in X containing S is S itself. Hence (S) = S . (c) True.

Proof . By (a) and (b), S ⊂ T ⊂ S ⇒ S ⊂ T ⊂ (S) = S ⇒ T = S. (d) True.

Proof . As S ⊂ S ∪ T and T ⊂ S ∪ T , we have by (a), S ⊂ S ∪ T , T ⊂ S ∪ T and so S ∪ T ⊂ S ∪ T .

(e) True.

Proof . For any x ∈ S ∪ T and any r > 0, we have (B(x, r)∩ S)∪ (B(x, r)∩ T ) = B(x, r)∩ (S ∪ T ) = φ . (*)

Metric Spaces

53

If B(x, r) ∩ S = φ for all r > 0, then x ∈ S ⊂ S ∪ T .

Otherwise, there exists r0 > 0 such that B(x, r0 ) ∩ S = φ

and so B(x, r) ∩ S = φ for all 0 < r ≤ r0 . But then by (*), we must have B(x, r) ∩ T = φ for all 0 < r < r0 . Thus

B(x, r) ∩ T = φ for all r > 0 and so x ∈ T ⊂ S ∪ T .

(f) False.

Example : Take X := R, S := [0, 1) and T := (1, 2]. Then S ∩ T = {1} but S ∩ T = φ.

(g) True.

Proof . As S ∩T ⊂ S and S ∩T ⊂ T , by (a) we have S ∩ T ⊂ S and S ∩ T ⊂ T . Thus S ∩ T ⊂ S ∩ T . 4. Let (X, d) be a metric space and S, T be subsets of X. (a) S ⊂ T =⇒ S  ⊂ T  . (b) (S  ) ⊂ S  . (c) (S  ) ⊃ S  . (d) S  ∪ T  ⊂ (S ∪ T ) . (e) S  ∪ T  ⊃ (S ∪ T ) . (f) S  ∩ T  ⊂ (S ∩ T ) . (g) S  ∩ T  ⊃ (S ∩ T ) . Answer :

(a) True.

Proof . For any x ∈ S  and any r > 0, by Theorem 1.2.27, we have

#{B(x, r) ∩ T } ⊃ #{B(x, r) ∩ S} = ∞ . Hence by Theorem 1.2.27 again, x ∈ T  .

(b) True.

Proof . Let x ∈ (S  ) and r > 0 be arbitrary. Then there exists y ∈ B(x, r) ∩ S  \ {x}. Put δ := r − d(x, y) > 0. Then B(y, δ) ⊂ B(x, r). As y ∈ S  , #(B(y, δ) ∩ S \ {y}) = ∞. Hence there exists z ∈ B(y, δ)∩S\{x, y} ⊂ B(x, r)∩S\{x}. Therefore x ∈ S  and hence (S  ) ⊂ S  .

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Metric Space Topology: Examples, Exercises and Solutions

(c) False.

Example : Let X := R, S := and (S  ) = φ. (d) True.

1

n

 : n ∈ N . Then S  = {0}

Proof . As S ⊂ S ∪ T and T ⊂ S ∪ T , by (a) we have S  ⊂ (S ∪ T ) , T  ⊂ (S ∪ T ) and so S  ∪ T  ⊂ (S ∪ T ) .

(e) True.

Proof . For any x ∈ (S ∪ T ) and any r > 0, we have (B(x, r) ∩ S \ {x}) ∪ (B(x, r) ∩ T \ {x})

= B(x, r) ∩ (S ∪ T ) \ {x} = φ .

(**)

If B(x, r) ∩ S \ {x} = φ for all r > 0, then x ∈ S  ⊂ S  ∪ T  .

Otherwise, there exists r0 > 0 such that B(x, r0 ) ∩ S \ {x} =

φ and so B(x, r) ∩ S \ {x} = φ for all 0 < r ≤ r0 . But then by (**), we must have B(x, r) ∩ T \ {x} = φ for all 0 < r < r0 . Thus B(x, r) ∩ T \ {x} = φ for all r > 0 and so x ∈ T  ⊂ S ∪ T .

(f) False.

Example : Take X := R, S := [ 0, 1), and T := (1, 2 ]. Then S  ∩ T  = [ 0, 1 ] ∩ [ 1, 2 ] = {1} but (S ∩ T ) = φ.

(g) True.

Proof . As S∩T ⊂ S and S∩T ⊂ T , by (a) we have (S∩T ) ⊂ S  and (S ∩ T ) ⊂ T  and so (S ∩ T ) ⊂ S  ∩ T  . 5. Let (X, d) be a metric space and S, T ⊂ X. (a) S ⊂ T =⇒ ∂S ⊂ ∂T . (b) ∂(∂S) ⊂ ∂S. (c) ∂(∂S) ⊃ ∂S. (d) ∂S ⊂ ∂(S ◦ ). (e) ∂S ⊃ ∂(S ◦ ). (f) ∂S ⊂ ∂(S  ). (g) ∂S ⊃ ∂(S  ). (h) ∂S ⊂ ∂(S).

Metric Spaces

(i) (j) (k) (l) (m) (n) (o) (p) (q)

55

∂S ⊃ ∂(S). ∂(S  ) ⊂ ∂(S ◦ ). ∂(S  ) ⊃ ∂(S ◦ ). ∂(S) ⊂ ∂(S  ). ∂(S) ⊃ ∂(S  ). ∂(S ◦ ) ⊂ ∂(S). ∂(S ◦ ) ⊃ ∂(S). (∂S)◦ = φ. (∂S) = φ.

Answer : (a) False.

Example : Take X := R, S := [0, 3], T := [1, 2]. Then ∂S = {0, 3} while ∂T = {1, 2}.

(b) True.

Proof . Since ∂S is closed, we have ∂(∂S) ⊂ ∂S = ∂S .

(c) False.

Example : Take X := R, S := Q, then ∂S = R while ∂(∂S) = ∂ R = φ. (d) False.

Example : Take X := R, S := (0, 1) ∪ {2}. Then ∂S =   {0, 1, 2} ⊂ {0, 1} = ∂ (0, 1) = ∂(S ◦ ).

(e) True.

Prove . For any x ∈ ∂(S ◦ ) and any r > 0, we have by Corollary 1.2.20,



B(x, r) ∩ S

⊃ B(x, r) ∩ S ◦

= φ

B(x, r) ∩ X \ S = B(x, r) ∩ (X \ S ◦ ) = φ .

(i) (ii)

Observe that from (ii) we actually have B(x, r) ∩ (X \ S) = φ.

In fact, by (ii), there exists y ∈ B(x, r) ∩ X \ S . Let δ > 0 be s.t. B(y, δ) ⊂ B(x, r). As y ∈ X \ S , there is a point

z ∈ B(y, δ) ∩ (X \ S) ⊂ B(x, r) ∩ (X \ S). Thus B(x, r) ∩ (X \ S) = φ. This together with (i) implies x ∈ ∂S .

(f) False.

56

Metric Space Topology: Examples, Exercises and Solutions

Example : Take X := R, S := (0, 1) ∪ {2}. Then ∂S =  {0, 1, 2} ⊂ {0, 1} = ∂ [0, 1]) = ∂(S  ).

(g) False.

  Example : Let X := S := n1 : n ∈ N ∪ {0}. Then ∂S = φ ⊃ {0} = ∂({0}) = ∂(S  ).

(h) False.

Example : Let X := R, S := (0, 1) ∪ (1, 2). Then ∂S =   {0, 1, 2} ⊂ {0, 2} = ∂ [0, 2] = ∂(S).

(i) True.

Proof : ∂S = (S) ∩ X \ S = S ∩ X \ S ⊂ S ∩ X \ S = ∂S .

(j) False.

Example : Let X := R, S := { n1 : n ∈ N}. Then ∂(S  ) = ∂({0}) = {0} ⊂ φ = ∂(φ) = ∂(S ◦ ).

(k) False.

Example : Let X := R, S := (0, 1) ∪ (1, 2). Then ∂(S ◦ ) =   ∂S = {0, 1, 2} ⊂ {0, 2} = ∂ [0, 2] = ∂(S  ).

(l) False.

Example : Let X := R, S := {0}. Then ∂(S) = ∂S = {0} ⊂ φ = ∂φ = ∂(S  ). (m) False.

  Example : Let X := S := n1 : n ∈ N ∪ {0}. Then ∂(S) = ∂(X) = φ ⊃ {0} = ∂{0} = ∂(S  ).

(n) False.

Example : Let X := R, S := (0, 1) ∪ (1, 2). Then ∂(S ◦ ) =   ∂S = {0, 1, 2} ⊂ {0, 2} = ∂ [0, 2] = ∂(S).

(o) False.

Example : Let X := R, S := ∂(φ) = φ ⊃ S = ∂(S).

1

n

 : n ∈ N . Then ∂(S ◦ ) =

(p) False.

Example : Take X := R, S := Q. Then (∂S)◦ = R◦ = R. (q) False.

Example : Take X := R2 , S := {(x, y) : x2 + y 2 ≤ 1}. Then (∂S) = (S 1 ) = S 1 .

Metric Spaces

57

6. Let S, T ⊂ Rn . (a) ∂(S ∪ T ) ⊂ ∂S ∪ ∂T . (b) ∂(S ∪ T ) ⊃ ∂S ∪ ∂T . (c) ∂(S ∩ T ) ⊂ ∂S ∩ ∂T . (d) ∂(S ∩ T ) ⊃ ∂S ∩ ∂T . (e) ∂(S ∩ T ) ⊂ ∂S ∪ ∂T . (f) ∂(S ∩ T ) ⊃ ∂S ∪ ∂T . (g) ∂(S \ T ) ⊂ ∂S ∪ ∂T . (h) ∂(S \ T ) ⊃ ∂S ∪ ∂T . Answer :

(a) True.

Proof . For any x ∈ ∂(S ∪ T ), we have     B(x, r) ∩ S ∪ B(x, r) ∩ T = B(x, r) ∩ (S ∪ T ) = φ for all r > 0 and

  B(x, r) ∩ (X \ S) ∩ (X \ T ) = B(x, r) ∩ X \ (S ∪ T ) = φ for all r > 0. Note that the latter implies that we always have

B(x, r) ∩ (X \ S) = φ and

B(x, r) ∩ (X \ T ) = φ

for all r > 0. If B(x, r) ∩ S = φ for all r > 0, then we have



B(x, r) ∩ S = φ   B(x, r) ∩ X \ S = φ

∀r>0.

Thus x ∈ ∂S .

If there is an r0 > 0 such that B(x, r0 )∩S = φ, then B(x, r)∩

S = φ for all 0 < r < r0 . This forces B(x, r) ∩ T = φ for all 0 < r < r0 and so B(x, r) ∩ T = φ for all 0 < r . Hence  B(x, r) ∩ T = φ   ∀r>0. B(x, r) ∩ X \ T = φ Thus x ∈ ∂T .

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Metric Space Topology: Examples, Exercises and Solutions

(b) False.

Example : Let X := R, S := [ 0, 2 ], T := {1}.     ∂ {1} = ∂S ∪ ∂T .





Then ∂(S∪T ) = ∂ [ 0, 2 ] = {0, 2} ⊃ {0, 1, 2} = ∂ [ 0, 2 ] ∪ (c) False.

Example : Let X := R, S := [ 0, 3 ], T := [ 1, 2 ].   Then ∂(S ∩T ) = ∂ [ 1, 2 ] = {1, 2} ⊂ φ = {0, 3}∩{1, 2} = ∂S ∩ ∂T .

(d) False.

Example : Let X := R, S := [ 0, 1), T := (1, 2 ]. Then ∂(S ∩ T ) = ∂φ = φ ⊃ {1} = {0, 1} ∩ {1, 2 } = ∂S ∩ ∂T .

(e) True.

Proof . For any x ∈ ∂(S ∩ T ), we have B(x, r) ∩ S ∩ T = B(x, r) ∩ (S ∩ T ) = φ for all r > 0 and



   B(x, r) ∩ (X \ S) ∪ B(x, r) ∩ (X \ T )   = B(x, r) ∩ X \ (S ∩ T ) = φ

for all r > 0. Note that the former implies that we always have

B(x, r) ∩ S = φ and

B(x, r) ∩ T = φ

for all r > 0. If B(x, r) ∩ (X \ S) = φ for all r > 0, then we have



B(x, r) ∩ S = φ   B(x, r) ∩ X \ S = φ

∀r>0.

Thus x ∈ ∂S .

If there is an r0 > 0 such that B(x, r0 ) ∩ (X \ S) = φ, then

B(x, r)∩(X \S) = φ for all 0 < r < r0 . This forces B(x, r)∩

Metric Spaces

59

(X \ T ) = φ for all 0 < r < r0 and so B(x, r) ∩ (X \ T ) = φ for all 0 < r . Hence  B(x, r) ∩ T = φ   ∀r>0. B(x, r) ∩ X \ T = φ Thus x ∈ ∂T .

(f) False.

Example : Let X := R, S := [0, 1], T := {2}.

Then ∂(S ∩ T ) = ∂φ = φ ⊃ {0, 1, 2} = ∂S ∪ ∂T .

(g) True.

Proof . Recall that ∂S = ∂(X \ S) for any S ⊂ X . We have

by (e),

  ∂(S \ T ) = ∂ S ∩ (X \ T ) ⊂ ∂S ∪ ∂(X \ T ) = ∂S ∪ ∂T .

(h) False.

Example : Let X := R, S := [0, 1], T := {2}.   Then ∂(S \ T ) = ∂ [0, 1] = {0, 1} ⊃ {0, 1, 2} = ∂S ∪ ∂T .

7. Let (X, d) be a metric space and S, T be subsets of X. (a) (S ∪ T )◦ = S ◦ ∪ T ◦ . (b) (S ∩ T )◦ = S ◦ ∩ T ◦ . (c) S ∪ T = S ∪ T . (d) S ∩ T = S ∩ T . (e) (S ∪ T ) = S  ∪ T  . (f) (S ∩ T ) = S  ∩ T  . (g) S  is closed. (h) S  = φ =⇒ S ∩ S  = φ. (i) S ◦ ⊂ S  (j) S  ⊂ (S  ) . (k) (S  ) ⊂ S  . [This is identical with Exercise 1.2, Part A, Problem #4(b). It is included here to facilitate the readers to make comparison between S  and (S  ) .] (l) S  ⊂ S.

60

Metric Space Topology: Examples, Exercises and Solutions

(m) (n) (o) (p) (q) (r) (s) (t) (u) (v) (w) (x) (y) (z) (aa) (ab)

S  ⊃ S. S  ⊂ S. S  ⊃ S. S ◦ ⊂ S. S ◦ ⊃ S. (S)◦ ⊂ S ◦ . (S)◦ ⊃ S ◦ . (S ◦ ) ⊂ S  . (S ◦ ) ⊃ S  . (S  )◦ ⊂ S ◦ . (S  )◦ ⊃ S ◦ . (S)◦ ⊂ S ◦ . (S)◦ ⊃ S ◦ . (S) ⊂ S  . (S) ⊃ S  . ∂(S  ) ⊂ ∂S ⇐⇒ S  ∩ S ◦ ⊂ (S  )◦ .

Answer : (a) False.

Justification . By Exercise 1.2, Part A, Problem #1(g). (b) True.

Proof . By Exercise 1.2, Part A, Problem #1(d), (e). (c) True.

Proof . By Exercise 1.2, Part A, Problem #3(d), (e). (d) False.

Justification . By Exercise 1.2, Part A, Problem #3(f). (e) True.

Proof . By Exercise 1.2, Part A, Problem #4(d), (e). (f) False.

Justification . By Exercise 1.2, Part A, Problem #4(f). (g) True.

Proof . By Theorem 1.2.26 and Exercise 1.2, Part A, Problem #4(b), S  = (S  ) ∪ S  = S  and so S  is closed.

(h) False.

Metric Spaces

Example : Let X := R, S := φ but S ∩ S  = φ.

61

1

n

 : n ∈ N . Then S  = {0} =

(i) False.

Example : Let X = any discrete metric space and S ⊂ X be nonempty. Then S ◦ = S ⊂ φ = S  .

(j) False.

Example : Let X := R, S := φ = (S  ) .

1

n

 : n ∈ N . Then S  = {0} ⊂

(k) True.

Proof . The readers are referred to the Proof for Exercise 1.2, Part A, Problem #4(b). (l) True.

Proof . By definition, S  ⊂ S . Hence by Exercise 1.2, Part A, Problem #3(a), (b), we have S  ⊂ (S) = S .

(m) False.

Example : Take X := R, S := (0, 1) ∪ {2}. Then S  = [0, 1] ⊃ [0, 1] ∪ {2} = S .

(n) True.

Proof . By definition and Exercise 1.2, Part A, Problem #3(b), S  ⊂ (S) = S .

(o) False.

Example : Take X := R, S := (0, 1) ∪ {2}. Then S  = [0, 1] ⊃ [0, 1] ∪ {2} = S .

(p) True.

Proof . S ◦ ⊂ S =⇒ S ◦ ⊂ S .

(q) False.

Example : Take X := R, S := {0}. Then S ◦ = φ ⊃ {0} = S .

(r) False.

Example : Take X := R, S := Q. Then (S)◦ = R ⊂ φ = S ◦ .

(s) True.

Proof . S ⊂ S =⇒ S ◦ ⊂ (S)◦ .

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Metric Space Topology: Examples, Exercises and Solutions

(t) True.

Proof . For any x ∈ (S ◦ ) and any r > 0, there exists y ∈ B(x, r) ∩ S ◦ \ {x}. As y ∈ S ◦ , there exists 0 < δ < min{d(y, x), r−d(y, x)} small enough such that B(y, δ) ⊂ S . Then B(x, r) ∩ S \ {x} ⊃ B(y, δ) ∩ S \ {x} = B(y, δ) = φ and so x ∈ S  .

(u) False.

Example : Take X := R, S := Q. Then (S ◦ )  = φ ⊃ R = S  .

(v) False.

Example : Take X := R, S := (0, 1) ∪ (1, 2). Then (S  )◦ = (0, 2) ⊂ (0, 1) ∪ (1, 2) = S ◦ .

(w) False.

Example : Let X be discrete and S ⊂ X be any nonempty subset. Then (S  )◦ = φ ⊃ S = S ◦ .

(x) False.

Example : Take X := R, S := Q. Then (S)◦ = R ⊂ φ = S ◦ .

(y) False.

Example : Take X := R, S := (0, 1). Then (S)◦ = (0, 1) ⊃ [0, 1] = S ◦ .

(z) True.

Proof . For any x ∈ (S) and any r > 0, there exists y ∈ B(x, r) ∩ S \ {x}. Let δ := min{d(y, x), r − d(y, x)} > 0. As y ∈ S , we have B(x, r) ∩ S \ {x} ⊃ B(y, δ) ∩ S \ {x} = B(y, δ) ∩ S = φ and hence x ∈ S  .

(aa) True.

Proof . By Exercise 1.2, Part A, Problem #4(a), S ⊂ S =⇒ S  ⊂ (S) .

(ab) True.

Metric Spaces

63

Proof . We have S = S ∩ S

= S  ∩ (∂S ∪· S ◦ )

= (S  ∩ ∂S) ∪· (S  ∩ S ◦ ) .

(*)

On the other hand, as S  is closed, we have ∂(S  ) ⊂ S  and

S  = S  = ∂(S  ) ∪· (S  )◦ .

(**)

Thus

∂(S  ) ⊂ ∂S ⇐⇒ ∂(S  ) ⊂ S  ∩ ∂S

⇐⇒ S  \ ∂(S  ) ⊃ S  \ (S  ∩ ∂S)

⇐⇒ (S  )◦ ⊃ S  ∩ S ◦

by (**) and (*) .

8. Recall that a nonempty set S ⊂ Rn is said to be convex if for any x, y ∈ S and any t ∈ [0, 1], we have (1 − t)x + ty ∈ S. Geometrically, S is convex if the straight line segment joining two points of S lies entirely in S. Suppose S ⊂ Rn . (a) S is convex ⇒ S ◦ is convex. (b) S ◦ is convex ⇒ S is convex. (c) S is convex ⇒ S is convex. (d) S is convex ⇒ S is convex. Answer :

(a) True.

Proof . Let S ⊂ Rn be a convex set, a, b ∈ S ◦ , and t ∈ [0, 1]. It suffices to show that c := (1 − t)a + tb ∈ S ◦ . Since a, b ∈ S ◦ , there exist ra > 0 and rb > 0 such that B(a, ra ) ⊂ S and B(b, rb ) ⊂ S . Define r := min{ra , rb } > 0. Then B(a, r) ⊂ S , B(b, r) ⊂ S . For any p ∈ B(c, r), let x := a+(p−c) ∈ B(a, r) ⊂ S and y := b+(p−c) ∈ B(b, r) ⊂ S . It is easy to verify that p = (1−t)x+ty . Since S is convex and

64

Metric Space Topology: Examples, Exercises and Solutions

x, y ∈ S , this implies p ∈ S . Since p ∈ B(c, r) is arbitrary, we have B(c, r) ⊂ S and so c ∈ S ◦ .

(b) False.

Example : Let X := R, S := (0, 1) ∪ {2}. Then S ◦ = (0, 1) is convex but S is not. (c) True.

Proof . For any a, b ∈ S and any t ∈ [0, 1], it suffices to show that c := (1 − t)a + tb ∈ S . To see this, let r > 0. Then there exist x ∈ B(a, r) ∩ S and y ∈ B(b, r) ∩ S . Write z := (1 − t)x + ty . As S is convex and x, y ∈ S , we have z ∈ S . Furthermore, since     z − c = (1 − t)(x − a) + t(y − b)     ≤ (1 − t)x − a + ty − b ≤ (1 − t)r + tr =r,

we have z ∈ B(c, r) ∩ S and so c ∈ S . Hence S is convex.

(d) False.

Example : Let X := R, S := Q. Then S = R is convex but S is not.

Part B: Problems In what follows, unless otherwise specified, X is a general metric space.

˜ dS 1. Describe the open unit balls B(0, 1) in R2 with metrics d, and d∞ which are defined in Example 1.1.2 (v), (viii) and (ix), respectively. Solution : For d˜:  & (x1 − y1 )2 2 B(0, 1) = (x1 , x2 ) ∈ R2 : + (x − y ) < 1 < 1, 2 2 1 2 ( 10 )

which is the region inside the ellipse centered at the origin with semi

x-axis

1 10

and semi y -axis 1, excluding the boundary.

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65

For dS :

B(0, 1) = {(x1 , x2 ) ∈ R2 : |x1 | + |x2 | < 1}, which is the rectangular region centered at 0 with vertices (±1, 0), (0, ±1), excluding the boundary.

For d∞ :

B(0, 1) = {(x1 , x2 ) ∈ R2 : |x1 | < 1 and |x2 | < 1}, which is the rectangular region centered at 0 with vertices (±1, ±1), excluding the boundary.

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Metric Space Topology: Examples, Exercises and Solutions

2. Compute (a) interior (b) interior (c) interior (d) interior

of of of of

[0, ∞) in R; [0, 1) in R; [0, 1) in [0, ∞); Q in R.

Solution : Recall first that interior points of a set are automatically points in the given set. So to find all interior points of a set, we only need to consider the points in the given set and study whether they are interior points or not. (a) Answer : (0, ∞).

In fact, for any x ∈ (0, ∞), BR (x, x2 ) = ( x2 ,

3x 2 )

⊂ [0, ∞)

and so every point in (0, ∞) is an interior point of [0, ∞) in

R. It remains to consider the point 0 ∈ [0, ∞). As BR (0, r) = (−r, r) ⊂ [0, ∞) for any r > 0, we see that 0 is not an interior point of [0, ∞) in R. Thus the interior of [0, ∞) in R is (0, ∞). (b) Answer : (0, 1). In fact, for any x ∈ (0, 1), if we write r := min{x, 1 − x} > 0, then BR (x, r) ⊂ [0, 1). Hence every point in (0, 1) is an interior point of [0, 1) in R. It remains to consider the point 0 ∈ [0, 1). As BR (0, r) = (−r, r) ⊂ [0, 1) for any r > 0, we see that 0 is not an interior point of [0, 1) in R. Thus the interior of [0, 1) in R is (0, 1). (c) Answer : [0, 1). Similar to part (b) above, for any x ∈ (0, 1), write r := min{x, 1 − x} > 0. We have BR (x, r) ⊂ [0, 1) and so B[0,∞) (x, r) = BR (x, r) ∩ [0, ∞) ⊂ [0, 1) ∩ [0, ∞) = [0, 1). So every point in (0, 1) is an interior point of [0, 1) in [0, ∞). It remains to consider the point 0 ∈ [0, 1). As B[0,∞) (0, r) = BR (0, r) ∩ [0, ∞) = (−r, r) ∩ [0, ∞) = [0, r) ⊂ [0, 1) for any 0 < r < 1, we see that 0 is an also interior point of [0, 1) in [0, ∞). Therefore, the whole set [0, 1) is the interior of [0, 1) in [0, ∞). (d) Answer : φ.

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67

In fact, for any x ∈ Q and any r > 0, BR (x, r) = (x−r, x+r)

is bound to contain irrational numbers. Hence BR (x, r) ⊂ Q.

Therefore, no point in Q can be an interior point of Q in R.

3. Let S = {(−1)n + S.

1 n

: n ∈ N} ⊂ R and T = {1 +

1 2n

: n ∈ N} ⊂

(a) Is S open in R? (b) Is S closed in R? (c) Is T open in R? (d) Is T closed in R? (e) Is T open in S? Solution : (a) False.

  Justification . It is obvious that BR (−1)n + n1 , r ⊂ S for any n ∈ N and any r > 0. Hence no point in S is an interior point of S in R and so S is not open in R.

(b) False. 1 Justification . Consider 1 ∈ R \ S . For any r > 0, 1 + 2r ∈ S ∩ BR (1, r). So 1 ∈ R \ S is not an interior point of R \ S and so R \ S is not open in R. Hence S is not closed in R.

(c) False.

Justification . Similar to (a). Details are left with the readers. (d) False.

Justification . Similar to (b). Details are left with the readers. (e) True.

Proof . Observe first that T contains all the positive elements 1 of S . For any t ∈ T , write t = 1 + 2n , where n ∈ N. Note that  1  1 all elements in BS t, 2n are positive and so BS t, 2n ⊂ T. Hence T is open in S . 4. Show that S ◦ is the largest subset of S which is open in X.

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Metric Space Topology: Examples, Exercises and Solutions

Proof . S ◦ is open: For any x ∈ S ◦ , there exists r > 0 such that B(x, rx ) ⊂ S . Note that actually B(x, rx ) ∈ S ◦ . In fact, for any y ∈ B(x, rx ), since B(x, rx ) is open, there is δ > 0 such that B(y, δ) ⊂ B(x, rx ). Hence B(y, δ) ⊂ S and so y ∈ S ◦ . Since y ∈ B(x, rx ) is arbitrary, we have B(x, rx ) ⊂ S ◦ and so S ◦ is open. S ◦ is the largest open set contained in S : Let U ⊂ S be an open set in X . For any u ∈ U , there exists r > 0 such that B(u, r) ⊂ U . Hence B(u, r) ⊂ S and so u ∈ S ◦ . Since u ∈ U is arbitrary, we have U ⊂ S ◦. 5. Show that S ◦◦ = S ◦ for any subset S of X. Proof . By definition, a subset A ⊂ X is open if and only if A◦ = A. Now by Exercise 1.2, Part B, Problem #4, S ◦ ⊂ X is open, so we have S ◦◦ = S ◦ .   6. Let S = n1 : n ∈ N ⊂ R. What are S, S  , (S  ) , S ∪ {0} , (S ∪ {0}) , and ∂S? Solution :   S = n1 : n ∈ N ∪ {0}, S  = {0}, (S  ) = φ,   S ∪ {0} = n1 : n ∈ N ∪ {0}, (S ∪ {0}) = {0},   ∂S = n1 : n ∈ N ∪ {0}.

7. Let S and T be two subsets of a metric space (X, d). (a) Show that d(x, S) = 0 if and only if x ∈ S. (b) Prove or disprove that if S and T are two disjoint closed subsets of X, then d(S, T ) > 0. (c) Prove or disprove that if S and T are two disjoint closed subsets of X, then there exist two open sets U and V with S ⊂ U , T ⊂ V , and U ∩ V = φ.

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69

Solution : (a) d(x, S) = 0 ⇐⇒ inf d(x, s) = 0 s∈S

⇐⇒ ∀ r > 0, ∃ sr ∈ S s.t. d(x, sr ) < r

⇐⇒ ∀ r > 0, B(x, r) ∩ S = φ ⇐⇒ x ∈ S . (b) False.

Example : In X := R, consider the subsets S := N and T :=   n + 21n : n ∈ N . Clearly, S and T are disjoint closed subsets of R. But as d(n, n + 21n ) = 21n is arbitrarily small as n is arbitrarily large, we have d(S, T ) = 0. (c) True.

Proof . Since S and T are closed and disjoint, we have by (a), d(s, T ) > 0 and d(t, S) > 0 for all s ∈ S and all t ∈ T . Let U :=

%

s∈S

 d(s, T )  , B s, 3

V :=

%

t∈T

 d(t, S)  . B t, 3

Clearly both U and V are open in X , S ⊂ U , and T ⊂ V .

It remains to show that U ∩ V = φ. Suppose not, then there

exists z ∈ U ∩ V . There are s0 ∈ S and t0 ∈ T such that

z ∈ B(s0 , d(s03,T ) )∩ B(t0 , d(t03,S) ). Without loss of generality, assume d(s0 , T ) ≥ d(t0 , S). Then we have d(s0 , T ) ≤ d(s0 , t0 ) ≤ d(s0 , z) + d(z, t0 )
0 be given. For any x ∈ E , by Exercise

1.2, Part B, Problem #7(a), d(x, E) = 0. Hence there exists

e ∈ E such that d(x, e) < ε and so d(a, E) ≤ d(a, e) ≤ d(a, x) + d(x, e) < d(a, x) + ε . Since this is true for all x ∈ E , we have

d(a, E) ≤ d(a, E) + ε . Since ε > 0 is arbitrary, we conclude that d(a, E) ≤ d(a, E) and so the result follows. (b) True.

Proof . Since E ⊂ E , we have {d(x, y) : x, y ∈ E} ⊂ {d(x, y) : x, y ∈ E} and so

d(E) = sup {d(x, y)} ≤ sup {d(x, y)} = d(E) . x,y∈E

x,y∈E

Conversely, let ε > 0 be given. For any x, y ∈ E , there exist

x , y  ∈ E with d(x, x ) < ε and d(y, y  ) < ε. Therefore, d(x, y) ≤ d(x, x ) + d(x , y  ) + d(y  , y) < d(x , y  ) + 2ε ≤ d(E) + 2ε .

Since x, y ∈ E are arbitrary, this implies

d(E) = sup{d(x, y) : x, y ∈ E} ≤ d(E) + 2ε . Finally, since ε > 0 is arbitrary, we conclude that d(E) ≤ d(E) and so the result follows.

Metric Spaces

9.

71

(a) Let S be a nonempty subset of X. Show that S is the intersection of all closed sets of X containing S. !∞ !∞ ◦ (b) Is it always true that ( n=1 An ) = n=1 A◦n ? [Compare with Exercise 1.2, Part A, Problem #1(d), (e) and Problem #7(a).] !∞ !∞ (c) Is it always true that n=1 An = n=1 An ? [Compare with in Exercise 1.2, Part A, Problem #3(d), (e) and Problem #7(c).] !∞ !   (d) Is it always true that ( ∞ n=1 An ) = n=1 An ? [Compare with Exercise 1.2, Part A, Problem #4(d), (e) and Problem #7(e).] "∞ " ◦ ◦ (e) Is it always true that ( ∞ n=1 An ) = n=1 An ? [Compare with Exercise 1.2, Part A, Problem #1(d), (e) and Problem #7(b).] (f) Give examples of a metric space in which B(a, r) = B(a, r), i.e., the closed ball B(a, r) is not the closure of the open ball B(a, r). Solution : (a) First of all, as S ∈ {A ⊂ X : S ⊂ A and A is closed in X}, we have

'

{A ⊂ X : S ⊂ A and A is closed in X} ⊂ S.

On the other hand, since

'

{A ⊂ X : S ⊂ A and A is closed in X}

is closed in X and contains S , and S is the smallest such set, we have

S⊂ (b) No.

'

{A ⊂ X : S ⊂ A and A is closed in X}.

Example : Let X := R and An := [n, n + 1], n ∈ N. Then ! ! ◦ ( ∞ An ) = [1, ∞)◦ = (1, ∞) = ∞ n=1 n=1 (n, n + 1) = !∞ ◦ n=1 An

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Metric Space Topology: Examples, Exercises and Solutions

(c) No.

Example : Let X := R. Since Q is countable, we can write Q = {rn : n ∈ N}. Take An := {rn }, n ∈ N. Then !∞ !∞ n=1 An = Q = R = Q = n=1 An .

(d) No.

!∞

Using the same example as that in (c), we have (

= Q = R = φ = (e) No.

!∞

n=1

An .

n=1

An )



  Example : Let X := R and An := −1, n1 , n ∈ N. Then

∞ '

n=1

An

◦

while

=

∞ '



n=1

A◦n

n=1

(f) Example 1 :

∞  '

=

1 −1, n

∞  '

n=1

◦

1 −1, n

= (−1, 0]◦ = (−1, 0) ,



= (−1, 0] .

Let X be a discrete metric space with #(X) > 1. Let a ∈ X .

Then B(a, 1) = X = {a} = B(a, 1).

Example 2 : Let X := Z with the Euclidean metric. Then B(0, 1) = {−1, 0, 1} = {0} = B(0, 1). Example 3 : Let X := (−∞, 0] ∪ [1, +∞) with the Euclidean metric. Then B(0, 1) = [−1, 0] ∪ {1} = [−1, 0] = B(0, 1).

10. Show that if φ = S ⊂ R is both open and closed, then S = R.

Proof . Suppose to the contrary that S = R. Let T := R \S . Then T is also nonempty, proper, and both open and closed in R. Let α > 0 be such that both (−α, α) ∩ S and (−α, α) ∩ T are nonempty (why exists?). As S is open in R, so is the intersection (−α, α) ∩ S . By Theorem 1.2.16, there is a countable family of pairwisely disjoint open intervals {In } such that

(−α, α) ∩ S =

% n

In .

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73

Write I1 = (a, b). If b = α, by the definition of T , b ∈ T . However,

as T is open, there is 0 < δ < b − a such that (b − δ, b + δ) ⊂ T . But it means that S ∩ T ⊃ (b − δ, b) = φ and contradiction arises.

Therefore b = α. Similarly, one has a = −α. That means the whole

family {In } reduces to one single open interval, namely, (−α, α), i.e.,

(−α, α) ∩ S = (−α, α), or (−α, α) ⊂ S . But that contradicts to T ∩ (−α, α) = φ. Hence S = R.

11. Let X be a nonempty set. A metric d on X is said to be an ultrametric if d(x, y) ≤ max(d(x, z), d(y, z)) for all x, y, z ∈ X. (a) Give an example of ultrametric. (b) Suppose d is an ultrametric. Show that (i) for any a, b, x ∈ X, if d(a, x) < d(a, b), then d(x, b) = d(a, b); (ii) closed balls in (X, d) are open; (iii) open balls in (X, d) are closed; (iv) the sphere {x ∈ X : d(x, a) = r} is clopen for any r > 0. Solution: (a) Example 1 : The discrete metric is an ultrametric. Example 2 : Let p be a prime number. Every non-zero rational number x can be expressed as pk · rs for some k , r ∈ Z and s ∈ N, with neither r nor s is divisible by p. Define |x|p :=



p−k 0

if x = 0

if x = 0,

and d(x, y) := |x − y|p for any x, y ∈ Q. It turns out that

this is a well-defined metric on Q and is known as the “p-adic metric” on Q. In fact, it is not difficult to check that it is non-

negative, symmetric, and d(x, y) = 0 if and only if x = y . Note that the triangle inequality would follow once we have

d(x, y) ≤ max{d(x, z), d(y, z)}. Hence it only remains to show this inequality. To do this, we write |x − z|p = p−n1

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Metric Space Topology: Examples, Exercises and Solutions

and |z − y|p = p−n2 . Without loss of generality, we assume

n1 ≤ n2 . By the definition of | · |p , we have x − z = pn1 · and z − y = pn2 · uv22 with p  u1 , u2 , v1 , v2 , which implies

u1 v1

 u1 n2 −n1 u2 x−y =p +p v1 v2   n2 −n1 u2 v1 u1 v2 + p n1 . =p v1 v2 n1



Since p  v1 v2 and p  u1 v2 , we have

d(x, y) = |x − y|p ≤ p−n1 = max(p−n1 , p−n2 )

= max(|x − z|p , |y − z|p )

= max{d(x, z), d(y, z)} . (b) (i) Since d(a, x) < d(a, b), we have

d(x, b) ≤ max{d(x, a), d(a, b)} = d(a, b)

≤ max{d(a, x), d(x, b)} .

But then again because d(a, x) < d(a, b), the last inequality forces d(x, b) ≥ d(a, b). Combining, we conclude that d(x, b) =

d(a, b).

(ii) For any a ∈ X and any r > 0, let b ∈ B(a, r). For any c ∈ B(b, r), we have

d(c, a) ≤ max{d(c, b), d(b, a)} ≤ r , thus c ∈ B(a, r). Hence B(b, r) ⊂ B(a, r) and so B(a, r) is open.

(iii) For any a ∈ X and any r > 0, let b ∈ X \ B(a, r). If

there is c ∈ B(b, r) ∩ B(a, r), we have

d(a, b) ≤ max{d(a, c), d(c, b)} < r ,

Metric Spaces

75

that is, b ∈ B(a, r), which is absurd. Therefore, B(b, r) ∩

B(a, r) = φ, or equivalently, B(b, r) ⊂ X \ B(a, r). Hence X \ B(a, r) is open.

(iv) Note that

{x ∈ X : d(x, a) = r} = B(a, r) ∩ (X \ B(a, r)). By (ii) and (iii) above, both sets on the right hand side are clopen, hence the sphere is also clopen.

12. Let S ⊂ Rn . Show that there are at most countably many isolated points in S. Hence, show that if S is uncountable then S  is uncountable. Proof . Let Λ ⊂ S be the set of isolated points of S . For any λ ∈ Λ, there is an open set Uλ ⊂ Rn such that Uλ ∩ S = {λ}. So we ! have Λ = λ∈Λ Uλ . By Lindel¨ of’s theorem, there exists a countable ! subcollection {Ui }i∈N ⊂ {Uλ }λ∈Λ such that Λ = i∈N Ui . On the other hand, as each Uλ contains exactly one point in Λ,

no proper subcollection of {Uλ }λ∈Λ could cover Λ. So {Uλ }λ∈Λ = {Ui }i∈N and hence Λ is countable.

Finally, if S is uncountable, S \ Λ is uncountable and so S  ⊃ S \ Λ is uncountable.

13. A collection A of subsets of X is said to have the “countable intersection property” if every countable intersection of elements of A is nonempty. Show that every family of closed nonempty subsets of Rn which has the countable intersection property has a nonempty intersection. Proof . Let A be a family of closed subsets of Rn with the countable " intersection property. Suppose A∈A A = φ. Consider U := {Rn \ A : A ∈ A}. We have   '  %  % U= A = Rn \ φ = Rn . Rn \ A = Rn \ U ∈U

A∈A

A∈A

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Metric Space Topology: Examples, Exercises and Solutions

Hence by Lindel¨ of Theorem, there is a countable subcollection U  ⊂

U such that

%

U = Rn .

U ∈U 

Let A := {Rn \ U : U ∈ U  }. Then A is countable and

φ = Rn \

 %

U ∈U 

  '  ' U = A, Rn \ U = U ∈U 

A∈A

contradicting the countable intersection property of A. Hence

"

A∈A

A = φ.

14. Would Lindel¨ of Theorem remain valid when Rn is equipped with other metrics instead of the Euclidean one? Answer : Not in general. Justification . Let X := R with the discrete metric. Then F :=   {x} : x ∈ X is an infinite collection of open subsets of X which covers X . But then it is obvious that no countable subcollection of F can cover X . 15. Show that a collection of nonempty disjoint open sets in Rn must be countable. What will happen if we replace open sets by closed sets? Solution : Let F be a collection of nonempty disjoint open sets in Rn . Since Qn is countable, we can write Qn = {rk : k ∈ N}. For any S ∈ F , since S is an open in Rn and Qn is dense in Rn , S ∩ Qn = φ. Hence there exists some rS ∈ S ∩ Qn . Do this for every S ∈ F and we form a set {rS : S ∈ F} ⊂ Qn . As Qn is countable, the same is true for the set {rS : S ∈ F}. Observe that if rS = rS  for some S , S  ∈ F , then we have rS ∈ S ∩ Qn and rS ∈ S  ∩ Qn . By the pairwise disjointness of elements in F , this forces S = S  . Hence it follows that F and {rS : S ∈ F} have the same cardinality, and so F is countable.

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77

Finally, the same statement is not true if we replace open sets by closed sets, that is, there are uncountable collections of disjoint closed





sets in Rn . For example, {x} : x ∈ Rn .

16.

(a) If X is a discrete metric space and S ⊂ X, what is ∂S? (b) For any subset S ⊂ X, is ∂S always closed? Solution : (a) ∂S = φ. (b) True.

Proof . Note that ∂S = S \ S ◦ = S ∩ (X \ S ◦ ). Being the intersection of two closed sets, ∂S is always closed. 17. Let X := R, Y := [a, ∞), S =: [a, b), a < b. Compute (a) the boundary of Y in X; (b) the boundary of S in X; (c) the boundary of S in Y . Solution : (a) ∂Y = {a}. (b) ∂S = {a, b}.

(c) The boundary of S in Y is {b}.

18. Let S, T ⊂ X. (a) Show that if S open, then (∂S)◦ = φ. (b) We have seen in Exercise 1.2, Part A, Problem #6(a), (b) that in general, ∂(S ∪ T ) ⊂ ∂S ∪ ∂T but ∂(S ∪ T ) ⊃ ∂S ∪∂T . Show that if S ∩T = φ, then ∂S ∪∂T = ∂(S ∪T ). Proof . (a) For any x ∈ (∂S)◦ , there exists r0 > 0 such that B(x, r0 ) ⊂

∂S . But as x ∈ (∂S)◦ ⊂ ∂S , we have φ = B(x, r0 ) ∩ S ⊂ ∂S ∩ S = ∂S ∩ S ◦ = φ, which is absurd.

(b) It suffices to show that ∂S ∪∂T ⊂ ∂(S ∪T ). Let x ∈ ∂S ∪∂T .

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Metric Space Topology: Examples, Exercises and Solutions

Without loss of generality, assume x ∈ ∂S . Then



B(x, r) ∩ (S ∪ T ) ⊃ B(x, r) ∩ S  φ = B(x, r) ∩ (X \ S) =  φ

∀ r > 0.

On the other hand, as S ∩ T = φ, we have x ∈ ∂S ⊂ S ⊂

X \ T . Hence there exists r0 > 0 such that B(x, r0 ) ⊂ X \ T and so

B(x, r) ⊂ X \ T

∀ 0 < r ≤ r0 .

Hence for all 0 < r ≤ r0 , we have

B(x, r) ∩ (X \ (S ∪ T )) = B(x, r) ∩ (X \ T ) ∩ (X \ S) ⊃ B(x, r) ∩ (X \ T ) ∩ (X \ S)

= B(x, r) ∩ (X \ S) = φ , and so

B(x, r) ∩ (X \ (S ∪ T )) = φ

∀r>0.

Therefore, x ∈ ∂(S ∪ T ).

19. Let X := C[0, 1]. For any f , g ∈ X, define d∞ (f, g) := sup |f (x) − g(x)| , x∈[0,1]

d1 (f, g) := d2 (f, g) :=



1

|f (x) − g(x)| dx ,

0



1 0

2

(f (x) − g(x)) dx

1/2

.

Let S := {f ∈ X : f (0) = 0} and T = {f ∈ S : f (1) = 0}.

Metric Spaces

(a) (b) (c) (d) (e) (f) (g)

79

Show that d∞ , d1 , and d2 are metrics. Determine whether S is closed in (X, d1 ). Determine whether S is closed in (X, d2 ). Determine whether S is closed in (X, d∞ ). Determine whether T is open in (X, d∞ ). Determine whether T is open in (S, d∞ ). Are there two positive real numbers α and β such that α d∞ (f, g) ≤ d2 (f, g) ≤ β d∞ (f, g) for all f , g ∈ X? [In general, if d and d are two metrics on the same underlying set X and there are constants α, β > 0 such that α d(x, y) ≤ d (x, y) ≤ β d(x, y)

∀ x, y ∈ X ,

then the two metrics d and d are said to be equivalent. For example, the metrics d1 , d2 and d∞ on Rn discussed in Exercise 1.1, Part B, Problem #6 are all equivalent. Note that when two metrics d and d on a set X are equivalent, the collections of open sets in (X, d) and (X, d ) are identical. So here the problem is asking whether d∞ and d2 are equivalent.] (h) Show that for all f, g ∈ X with f = g and d2 (f, 0) = d2 (g, 0) = 1, we have d2 (f + g, 0) < 2. What if d2 is replaced by d∞ ? Solution : (a) d∞ is a metric: Clearly, d∞ (f, g) ≥ 0 and d∞ (f, f ) = 0 for all f , g ∈ X . Moreover, if d∞ (f, g) = 0, then for all t ∈ [0, 1], 0 ≤ |f (t) − g(t)| ≤ sup |f (x) − g(x)| = d∞ (f, g) = 0 . x∈[0,1]

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Metric Space Topology: Examples, Exercises and Solutions

It follows that f = g on [0, 1] and hence (M1) holds. Obviously, (M2) also holds. Finally, let f , g , h ∈ X . For all x ∈ [0, 1], we

have

|f (x) − g(x)| ≤ |f (x) − h(x)| + |h(x) − g(x)| ≤ d∞ (f, h) + d∞ (h, g) .

Thus

d∞ (f, g) = sup |f (x) − g(x)| ≤ d∞ (f, h) + d∞ (h, g) x∈[0,1]

and so (M3) also holds.

d1 is a metric: (M1) and (M2) are clear. It remains to verify (M3). For this we let f , g , h ∈ X . For any x ∈ [0, 1], by the usual triangle inequality for the Euclidean metric in R, we have

|f (x) − h(x)| ≤ |f (x) − g(x)| + |g(x) − h(x)| . Since x is arbitrary, we have

d1 (f, h) = sup |f (x) − h(x)| x∈[0,1]

≤ sup

x∈[0,1]



 |f (x) − g(x)| + |g(x) − h(x)|

≤ sup |f (x) − g(x)| + sup |g(x) − h(x)| x∈[0,1]

(why?)

= d1 (f, g) + d1 (g, h) . Hence (M3).

x∈[0,1]

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81

d2 is a metric: (M1) and (M2) are clear. It remains to verify (M3). For this we let f , g , h ∈ X . By Cauchy–Schwarz inequality,

d22 (f, g) 1  2 f (x) − g(x) dx =

= =





0

1

0 1 0

 

+2 ≤



1 0

2 f (x) − h(x) + h(x) − g(x) dx 2 f (x) − h(x) dx +





1

0

0

1

0



2 h(x) − g(x) dx



2 h(x) − g(x) dx

   f (x) − h(x) h(x) − g(x) dx

2 f (x) − h(x) dx +

+2



1



0

1

 2  12 f (x) − h(x) dx

2 = d2 (f, h) + d2 (h, g) .



1

0



2  12 h(x) − g(x) dx

Hence (M3) holds.

(b) Answer : No, S is not closed in (X, d1 ).

Justification . It suffices to show that X \ S is not open in (X, d1 ). This boils down to finding an f ∈ X \ S such that for any r > 0 there is some g ∈ B(f, r) ∩ S . For this we consider f :≡ 1. Then clearly, f ∈ X \ S . For any r > 0, assuming without loss of generality that 0 < r < 1, we define ( x if 0 ≤ x ≤ r r g(x) := if x > r . 1 Then g is continuous and g(0) = 0. Hence, g ∈ S . Note that

d1 (f, g) =



1

|f (x) − g(x)| dx

0

=



0

r



1−

r x dx = < r r 2

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Metric Space Topology: Examples, Exercises and Solutions

and so g ∈ B(f, r). It follows that X \S is not open and hence

S is not closed.

(c) Answer : No, S is not closed in (X, d2 ).

Justification . It suffices to show that X \ S is not open in (X, d2 ). For this, take f :≡ 1 on [0, 1]. Clearly, f ∈ X \ S . For any δ > 0, set fδ (x) :=



1 δx

1

if 0 ≤ x ≤ δ

if δ < x ≤ 1 .

Note that fδ (0) = 0 and hence fδ ∈ S . Furthermore,

d2 (fδ , f ) =



0

δ

 x 2 dx 1− δ

 12

=



δ √ < δ. 3

So for any r > 0, let δ := r 2 . Then we have

d2 (fδ , f )
0,

Bd2 (f, r) ⊂ X \ S . Hence f ∈ X \ S is not an interior point of X \ S and so X \ S is not open in (X, d2 ). (d) Answer : Yes, S is closed in (X, d∞ ).

Proof . It suffices to show that X \ S is open in (X, d∞ ). Let f ∈ X \ S , so f (0) = 0. Then for any g ∈ Bd∞ (f, |f (0)|/2), we have

     |f (0)| − |g(0)|  ≤ |f (0) − g(0)| ≤ sup |f (x) − g(x)| x∈[0,1]

= d∞ (f, g) < |f (0)|/2

and so |g(0)| > |f (0)|/2 > 0. Thus g ∈ X \ S . Since g ∈ Bd∞ (f, |f (0)|/2) is arbitrary, we have Bd∞ (f, |f (0)|/2) ⊂ X \ S and so X \ S is open in (X, d∞ ).

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83

(e) Answer : T is not open in (X, d∞ ).

Justification . Consider the function f (x) := x, x ∈ [0, 1]. Since f (0) = 0 and f (1) = 1 = 0, f ∈ T . For each 0 < δ < 1, set fδ := f + δ2 . Then d∞ (fδ , f ) = sup |fδ (x) − f (x)| = δ/2 < δ , x∈[0,1]

and so fδ ∈ Bd∞ (f, δ). However, as fδ (0) = δ/2 > 0, fδ ∈ / T.

Thus Bd∞ (f, δ) ⊂ T and so T is not open in (X, d∞ ). (f) Answer : T is open in (S, d∞ ).

Proof . In fact, note that T = S ∩ V where V := {f ∈ X : f (1) = 0}. An argument exactly the same as that in (d) shows that

X \ V = {f ∈ X : f (1) = 0} is closed in (X, d∞ ). Hence V is open in (X, d∞ ) and so T = S ∩ V is open in (S, d∞ ). (g) Answer : No, d∞ and d2 are not equivalent.

Justification . Suppose there is a positive real number α such

that

α d∞ (f, g) ≤ d2 (f, g) ∀ f, g ∈ X . (*) # $ n2 −1 1 Take n > max α , 1 , fn (x) := x 2 and g := 0. Then d∞ (fn , g) = sup |fn (x) − 0| = 1 , x∈[0,1]

d2 (fn , g) = By (*), we have α ≤



1 0

1 n,

 n2 −1 2 1/2 1 dx = . x 2 n

which contradicts to the choice of

n. Therefore there is no such α. Similar arguments show that there is no β satisfying d2 (f, g) ≤ β d∞ (f, g) for all f , g ∈ X . Hence we conclude that the metrics d∞ and d2 are not equivalent.

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Metric Space Topology: Examples, Exercises and Solutions

(h) Simple algebra yields

d22 (f + g, 0) + d22 (f − g, 0) = 2[d22 (f, 0) + d22 (g, 0)] = 4 from which the inequality d2 (f +g, 0) < 2 follows immediately. However, the corresponding result for d∞ fails to hold. For example, consider f :≡ 1 and g(x) := x. Then

d∞ (f, 0) = d∞ (g, 0) = 1 but d∞ (f + g, 0) = 2.

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85

1.3 Compactness Definition 1.3.1. Let S be a subset of X. A family F of subsets of X is said to be a cover of S (or, equivalently, F is said to cover S) ! if S ⊂ F ∈F F . If all elements of F are open sets in X, F is called an open cover of S. Example 1.3.2.   (i) X = R, { n1 , n2 }∞ n=1 is an open cover of (0, 1).  ∞ (ii) X = R, (−n, n) n=1 is an open cover of R.   (iii) X = R, (n, n + 2) n∈Z is an open cover of Q.   (iv) X = R, (r, ∞) r>0 is an open cover of (0, ∞). # $ is an open cover of [1, 3). (v) X = [1, ∞), 1, 3 − n1 n∈N

Here, note that the open covers in (i), (ii), (iii) and (v) are countable, while the one in (iv) is uncountable. Definition 1.3.3. A subset S ⊂ X is said to be compact if every open cover of S has a finite sub-cover, that is, for any open cover F = {Uα }α∈Λ of S, where Λ is an index set, there is a finite subcollection ! {U1 , . . . , Un } ⊂ F such that S ⊂ ni=1 Uα . Example 1.3.4.

    (i) In R, F := B x, n1 : x ∈ [0, 1], n ∈ N is an open cover   of [0, 1]. It is clear that B(0, 1), B(1, 1) is a finite subcover of F. But then of course we cannot conclude just with this observation that [0, 1] is compact. The point is, to show that [0, 1] is compact, we need to show that every open cover of X has a finite subcover. Showing that one particular open cover has a finite subcover is not sufficient.

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Metric Space Topology: Examples, Exercises and Solutions

 (n − 1, n + 1) : n ∈ Z is an open cover of R with no finite subcover. Note that since we have exhibited that one specific open cover of R has no finite subcover, we can conclude that R is not compact. $ #  1 1 , 1 − : n ∈ N is an open cover of (0, 1) with no finite (iii) n n subcover. Hence (0, 1) is not compact. (ii)



From these examples, we can imagine that if a metric space is not bounded (like R), or it is not closed (like (0, 1)), then there is no hope that the space is compact. In fact, we have the following necessary condition for compactness. Theorem 1.3.5. Every compact subset of X is closed and bounded. Proof. Let S ⊂ X be compact. If S = φ, the theorem is clear. So   suppose S = φ and so we have a point p ∈ S. Then B(p, n) : n ∈ N is an open cover of S and so it has a finite sub-cover, say, k  B(p, ni ) i=1 . Note that the B(p, ni )’s are concentric open balls. Let !k M := max{n1 , . . . , nk }. Then we have S ⊂ i=1 B(p, ni ) = B(p, M ) and so in particular, S is bounded. Next, for any p$∈ X \ S, since S #   1 is compact, the open cover B x, 2 d(x, p) : x ∈ S of S has a finite   sub-cover, say B(xi , ri ) : i = 1, . . . , n . Let r = min{r1 , . . . , rn }. Then r > 0 and we have B(p, r) ∩ B(xi , ri ) = φ

for all i = 1, . . . , n .

Metric Spaces

Since S ⊂ is closed.

!n

i=1

87

B(xi , ri ), we have B(p, r) ∩ S = φ and therefore, S 

Another useful necessary condition for compactness is the Bolzano-Weierstrass Property. Theorem 1.3.6. Every compact subset S of X has the BolzanoWeierstrass property, that is, every infinite subset of S has an accumulation point in S. Proof. Suppose T ⊂ S is an infinite set with no accumulation point in S. Then every x ∈ S has an open neighborhood B(x) such that   #{B(x)∩T } ≤ 1. Since S is compact, the open cover B(x) : x ∈ S has a finite subcover. In particular, this finite sub-cover also covers T , but this is impossible since T is infinite but each B(x) contains at most 1 point of T . 

Theorem 1.3.7. For any C ⊂ S ⊂ X, C is compact in S if and only if C is compact in X. Proof. (⇒) Let {Uα }α∈Λ be an open cover of C in X. For each α ∈ Λ, Uα is open in X and so Uα ∩ S is open in S. Since %

α∈Λ

(Uα ∩ S) =



%

Uα

α∈Λ



∩S ⊃C ∩S =C ,

{Uα ∩ S}α∈Λ is an open cover of C in S. As C is compact in S, {Uα ∩ S}α∈Λ has a finite subcover, say, {Ui ∩ S}ni=1 . Hence C⊂

n %

(Ui ∩ S) =

i=1



n %

i=1

Ui



∩S ⊂

that is, {Ui }ni=1 is a finite subcover of {Uα }α∈Λ .

n %

i=1

Ui ,

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Metric Space Topology: Examples, Exercises and Solutions

(⇐) Let {Uα }α∈Λ be an open cover of C in S. Then for each α ∈ Λ, Uα = Vα ∩ S for some open set Vα ⊂ X. As C⊂

%

Uα =

α∈Λ

%

α∈Λ

(Vα ∩ S) =



%

α∈Λ

Vα



∩S ⊂

%

Vα ,

α∈Λ

{Vα }α∈Λ is an open cover of C in X. Since C is compact in X, it has a finite subcover, say, {Vi }ni=1 . Hence C⊂



n %

i=1

Vi



∩S =

n %

(Vi ∩ S) =

i=1

n %

Ui ,

i=1

that is, {Ui }ni=1 is a finite subcover of {Uα }α∈Λ .



Remark. By Theorem 1.3.7, we see that unlike openness and closedness, compactness is independent of the ambient space we are working on. So when we talk about compactness, we do not need to specify which metric space it is referred to. A first sufficient condition for compactness is the following. Theorem 1.3.8. Closed subsets of compact sets are compact. More precisely, if S ⊂ X is compact and T ⊂ S is closed in S, then T is compact. Proof. In view of Theorem 1.3.7 and the preceding Remark, we only need to show that T is compact as a subset of S. Let {Uα }α∈Λ be an open cover of T in S. As S \ T is open in S, {Uα , S \ T : α ∈ Λ} is an open cover of S in the metric space S and so it has a finite sub-cover. Without loss of generality, we may assume that the open set S \ T is in this finite subcover and so it takes the form {U1 , . . . , Un , S \ T }. Hence 

n % Ui ∪ (S \ T ) . T ⊂ i=1

Metric Spaces

89

! But then as it is clear that T ∩ (S \ T ) = φ, we have T ⊂ ni=1 Ui and so {Ui }i=1,... ,n is a finite subcover of {Uα }α∈Λ . Hence T is compact.  Remark. It is clear that if T ⊂ S ⊂ X and T is closed in X, then T is also closed in S. Hence the same assertion of Theorem 1.3.8 holds in case S ⊂ X is compact and T ⊂ S is closed in X.

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Metric Space Topology: Examples, Exercises and Solutions

Exercise 1.3 Unless otherwise specified, X will stand for a general metric space and S, T , etc., are arbitrary subsets of X. Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. Every finite subset S ⊂ X is compact. Answer : True. Proof . Let S ⊂ X be a finite set. Write S = {x1 , . . . , xn }. If {Uα }α∈Λ is an open cover of S in X , then for any i = 1, . . . , n, there is an αi ∈ Λ such that xi ∈ Uαi and so {Uαi : i = 1, . . . , n} is a finite subcover of {Uα }α∈Λ . 2. N ⊂ R is compact. Answer : False. Justification . Since N is unbounded. By Theorem 1.3.5, N is not compact.

  n − 12 , n + 12 : n ∈ N is an open cover of N without finite subcover. Hence N is not compact. Alternatively,



3. S := { n1 : n ∈ N} ⊂ R is compact. Answer : False. Justification . Since S = S ∪ {0}  S , S is not closed and so by Theorem 1.3.5, S is not compact. #   $ 1 1 , n(n+1) : n ∈ N is an open cover Alternatively, observe that B n of S . It is easy to see that 1 ∈B k



1 1 , n n(n + 1)



if and only if k = n .

Metric Spaces

Hence for each n ∈ N, B



1 1 n , n(n+1)

91



contains one and only one

point of S and so any finitely many of them could not cover S . Thus

S is not compact. 4. S := { n1 : n ∈ N} ∪ {0} ⊂ R is compact.

Answer : True. Proof . Let U = {Uα }α∈Λ be any open cover of S . Then there is an element, say U0 ∈ U , such that 0 ∈ U0 . Since U0 is open in R, there 1 exists r > 0 such that B(0, r) ⊂ U0 . In particular, n ∈ B(0, r) ⊂ 1 1 U0 for all n > r . On the other hand, for every n ≤ r , there exists Un ∈ U such that n1 ∈ Un . Hence {U0 , Un : n ≤ 1r } is a finite subcover of U .

5. Every discrete metric space is non-compact. Answer : False. The statement is true only when the discrete metric   space is infinite. In fact, in this case, {x} : x ∈ X is an open cover without finite subcover.

Justification . In case the discrete metric space is finite, by Exercise 1.3, Part A, Problem #1, it must be compact.

6. If Ci ⊂ X is compact for every i = 1, . . . , N , then !N (a) i=1 Ci is compact. "N (b) i=1 Ci is compact. Answer :

(a) True.

!N Proof . Let U := {Uα }α∈I be an open cover of i=1 Ci in X . Then for every fixed i = 1, . . . , N , Ci is covered by U and so there is a finite subcover Ui ⊂ U . Collectively, U1 ∪ U2 ∪ · · · ∪ UN ⊂ U is a finite subcover of U .

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Metric Space Topology: Examples, Exercises and Solutions

(b) True.

Proof . For every i, since Ci is compact, it is closed in X . Hence "N "N i=1 Ci is closed in X . In particular, i=1 Ci is a closed subset of the compact set C1 and so by Theorem 1.3.8, it must be

compact.

7. The intersection of any collection of closed sets of X is compact if at least one of them is compact. Answer : True. Proof . Since the intersection of an arbitrary collection of closed sets of X is closed in X , if one of these closed sets is compact, the intersection of them is a closed subset of this compact set. Hence by Theorem 1.3.8, it is compact.

8. The intersection of any collection of compact sets of X is compact. Answer : True. Proof . This follows immediately from Exercise 1.3, Part A, Problem #7.

$ # 1 : x ∈ (0, 1] ⊂ R2 is compact. x, sin x Answer : False. Justification . It suffices to show that A is not closed, or equivalently, R2 \ A is not open. To achieve this, all we need is to show that (0, 0) ∈ R2 \ A is not an interior point of R2 \ A. So let ε > 0 be 1 given. Choose n ∈ N such that n > 2πε . Then

9. A :=

) ) 1 ) 1  ) 1 ) ) ) ) , sin 2nπ − (0, 0) = , 0 0 such that

B(x, r) ⊂ U . Then the set V := B(x, r2 ) will serve the purpose. In fact, we already have {x} ⊂ V ⊂ B(x, r) ⊂ U and V ⊂ V . Hence it only remains to show V ⊂ B(x, r). Let

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Metric Space Topology: Examples, Exercises and Solutions

y ∈ V . Then there is z ∈ B(y, r2 ) ∩ V = B(y, r2 ) ∩ B(x, r2 ). It follows that d(y, x) ≤ d(y, z) + d(z, x) < r2 + r2 = r which means y ∈ B(x, r). Hence V ⊂ B(x, r), as desired.

(b) For each a ∈ S , by (a), there is an open set Va such that

{a} ⊂ Va ⊂ Va ⊂ U . Then the collection {Va : a ∈ S} is an open cover of S . By the compactness of S , {Va : a ∈ S} has a finite subcover, say {Va1 , . . . , Van } for some n ∈ N. Take ! V := ni=1 Vai . Then V is open and satisfies S ⊂ V ⊂ V = !n !n i=1 Vai = i=1 Vai ⊂ U . In case the compactness of S is missing, the assertion will no

longer hold.

Example : Let X := R, S := (0, 1), U := (0, 1). Then S is not compact. Clearly, (0, 1) is the only possible subset of X that contains S and is contained in U , but (0, 1) = [0, 1] ⊂ U .

(c) Since T is closed, X \ T is an open subset of X containing

S and so by (b), there exists an open set US in X such that S ⊂ US ⊂ US ⊂ X \ T . Since US is closed, UT := X \ US is open and contains T . Obviously US ∩ UT = φ.

4. Let S, T be nonempty subsets of X. (a) If S ⊂ X is compact, show that there exists a point x ∈ S such that d(x, T ) = d(S, T ). In this case we say that S has the nearest point property. (b) If S ⊂ X is compact and T ⊂ X is closed such that S ∩ T = φ, show that d(S, T ) > 0. [Compare with Exercise 1.2 Part B, Problem #7(b).]

Solution : (a) If S is finite, then statement is obviously true. So assume S is an infinite set. Write d(S, T ) = a. For any

n ∈ N, there exists xn ∈ S such that a ≤ d(xn , T ) < a + n1 .

If #{xn : n ∈ N} < ∞, then there exists x ∈ {xn : n ∈ N}

Metric Spaces

97

such that

1 n for infinitely many n. Hence d(x, T ) = a. So suppose #{xn : n ∈ N} = ∞. Since S is compact, it a ≤ d(x, T ) < a +

possesses the Bolzano-Weierstrass Property. Hence the infinite

set {xn : n ∈ N} ⊂ S has an accumulation point x ∈ S .

Therefore, for any ε > 0, there is N >

1 ε

such that xN ∈

B(x, ε) and so d(x, T ) ≤ d(x, xN )+d(xN , T ) < ε+a+ N1 < a + 2ε. This forces d(x, T ) = a. (b) Suppose d(S, T ) = 0. By (a), there exists x ∈ S such that d(x, T ) = 0. By Exercise 1.2 Part B, Problem #7(a), x ∈ T = T . So x ∈ S ∩ T , contradicting the assumption that S ∩ T = φ. 5. Let S, T be disjoint subsets of a metric space X, S is compact, and T is closed. Show that there exists δ > 0 such that d(s, t) > δ for any s ∈ S and any t ∈ T . Proof . By Exercise 1.3, Part B, Problem #4(a), there exists s0 ∈ S

such that

d(s0 , T ) = d(S, T ) . Since S ∩T = φ, we have s0 ∈ / T = T and so by Exercise 1.2, Part B, Problem #7, we have

d(S, T ) = d(s0 , T ) > 0 . Hence for any 0 < δ < d(S, T ), we have

d(s, t) ≥ d(S, T ) > δ > 0

for any s ∈ S, t ∈ T .

6. For any φ = A, B ⊂ Rn , define A + B := {a + b : a ∈ A, b ∈ B} . If A is compact and B is closed, show that A + B is closed. What if the compactness of A is missing?

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Metric Space Topology: Examples, Exercises and Solutions

Solution : For any x ∈ A + B , put C = x − B := {x − b : b ∈ B}. As B is closed, it is clear that C is also closed and C ∩ A = φ. By Exercise 1.3, Part B, Problem #5, there exists δ > 0 such that d(a, c) > δ for any a ∈ A and any c ∈ C . We claim that B(x, δ) ∩ (A + B) = φ. In fact, if there exists y ∈ B(x, δ) ∩ (A + B), then y = a0 + b0 for some a0 ∈ A and b0 ∈ B . But then as x − b0 ∈ C ,

we have

δ > d(x, y) = |x − y| = |x − a0 − b0 |

= |(x − b0 ) − a0 | = d(x − b0 , a0 ) > δ ,

which is absurd. Thus B(x, δ) ∩ (A + B) = φ and hence A + B is closed.

In case A is not compact, the assertion no longer holds.

Example : In R2 , consider A : = {(x, tan−1 x) : x ∈ R}

B : = {(x, 0) : x ∈ R} .

It is clear that both A, B are closed in R2 but A is not compact. Observe that A + B = {(x, y) : x ∈ R, −1 < y < 1} is not closed in R2 .

7.

(a) Show that every compact space has a countable dense subset. That is, if X is compact, then there exists a countable subset S ⊂ X such that S = X. (b) Without compactness, is (a) still true? Solution : (a) Since X is compact, for each fixed n ∈ N, the open cover

  1  B x, n : x ∈ X has a finite subcover     &   1 n 1 n 1 n , B x2 , , . . . , B xk n , . B x1 , n n n Write

  Sn := xn1 , xn2 , . . . , xnkn .

Metric Spaces

99

Do this for every n ∈ N and define

S :=

%

Sn .

n∈N

Then S is countable. On the other hand, for any x ∈ X and

  n 1 B x1 , n ,  n 1  n 1  B x2 , n , . . . , B xkn , n covers X , there is j ∈ {1, . . . ,   kn } such that x ∈ B xnj , n1 . But that means any r > 0, let n ∈ N be such that n >

xnj 

and so B x,

1 n





1 ∈ B x, n

1 r.

Since



∩ S = φ. Hence x ∈ S .

(b) No, (a) will not hold any more in case X is not compact.

Example : Let X be a discrete metric space consisting of uncountably many elements. Since S = S for any S ⊂ X , no countable subset of X could be dense in X . Note that in this case X is not compact. 8. Recall (Exercise 1.1, Part B, Problem #13) that 2 is the space  of sequences of real numbers x = {xn }n∈N with n∈N |xn |2 < ∞, and the 2 -metric is defined as d(x, y) :=





n∈N

2

|xn − yn |

1/2

for all x, y ∈ 2 .

Determine whether B(0, 1) is compact. Here as usual, 0 stands for the zero sequence in 2 . Solution : It is non-compact. Denote by en ∈ 2 the sequence defined by en k := δnk . Here we recall that the Kronecker Delta Function δij is defined by δij :=



1 0

if i = j if i = j .

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So en is simply the sequence whose nth term is 1 and all other terms are 0. Clearly, d(en , 0) = 1 and so en ∈ B(0, 1) for all n ∈ N.

√

Furthermore, for any m =$n, d(em , en ) = 2. Now it is clear that # U := B(x, 12 ) : x ∈ 2 is an open cover of 2 . Fix any x ∈ 2 . Note that if there are distinct intergers n = m in N such that both

em , en ∈ B(x, 12 ), then we have √ 1 1 2 = d(em , en ) ≤ d(em , x) + d(en , x) < + = 1 , 2 2 which is absurd. Hence each B(x, 12 ) can contain at most one en . Therefore, no finite subcollection of U could cover all the en ’s, and thus no finite subcollection of U could cover B(0, 1). Hence B(0, 1) is not compact.

9. A metric space X is said to be noetherian if every nonempty family of open subsets has a maximal element. It is known that X is noetherian if and only if it satisfies the ascending chain condition, that is, every increasing sequence of open subsets U1 ⊂ U2 ⊂ · · · of X stabilizes, i.e., there is some N ∈ N such that Un = UN whenever n ≥ N . (a) Show that a space X is noetherian if and only if every open subset is compact. (b) Hence show that a noetherian metric space must be a finite set. Proof . (a) (⇒): Let U ⊂ X be open. Let {Uα }α∈Λ be any open cover of U of nonempty sets, and W be the collection of all finite

unions of the Uα ’s. By definition, the family W has a maximal

element, say W . If W ⊃ U , then there is an element x ∈ U \W .

Since {Uα }α∈Λ covers U , there exists some α0 ∈ Λ such that

x ∈ Uα0 . But then we have

W  W ∪ Uα0 ,

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which contradicts the maximality of W . So W ⊃ U . Note that

W represents a finite subcover of {Uα }α∈Λ for U and so U is

compact.

(⇐): In view of the given equivalent condition for noetherian metric spaces, it suffices to show that X has the ascending chain condition. So let

U1 ⊂ U2 ⊂ · · · be an ascending chain of open sets in X . By assumption, the open set W :=

!

n∈N Un

is compact and so the open cover

{Un }n∈N of W has a finite sub-cover, say, {Uni : 1 ≤ i ≤ m}. Let N := max{ni : 1 ≤ i ≤ m}. Then we have W = UN and hence

UN ⊂ Un ⊂ W = UN

for all n ≥ N ,

which forces Un = UN for all n ≥ N . (b) Suppose X is noetherian. By (a), every open set in X is compact and hence closed. In particular, complements of singletons are closed and so singletons are open. Therefore,

U :=



{x} : x ∈ X



is an open cover of X . Note that X

is open in itself and hence is compact, and so U must have a finite subcover. This can happen only when X is finite.

10. A metric space X is said to be totally bounded if for every ε > 0, there exist k > 0 and x1 , . . . , xk ∈ X such that {B(xi , ε)}ki=1 is an open cover of X. Equivalently, X is totally bounded if for any ε > 0, X can be decomposed into finitely many subsets each of which has diameter less than ε. A subset S ⊂ X is totally bounded if it is totally bounded when considered as a metric space, with metric induced by that of X. (a) Show that if S ⊂ X is totally bounded, then S is bounded. (b) Show that if S ⊂ X is compact, then S is totally bounded.

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(c)

(i) Let X = C[0, 1] with d(f, g) := sup{|f (x) − g(x)| : x ∈ [0, 1]}. Show that B(0, 1) is not totally bounded. (ii) Show that every bounded subset of R is totally bounded. [Hence, together with (a), a subset in R is bounded if and only if it is totally bounded.] (iii) Let X = R with the metric d(x, y) := min{1, |x − y|} .

Find a subset of X that is bounded but not totally bounded. (d) Does closedness + total boundedness imply compactness? Either prove it or give a counterexample. Solution : (a) If S ⊂ X is totally bounded, then S is covered by finitely many open balls of radius 1. Hence S is bounded.

(b) Let S ⊂ X be compact. For any ε > 0, the open cover

{B(s, ε)}s∈S of S has a finite subcover. That is, S is covered by finitely many ε-balls and so it is totally bounded. (c) (i) For any n ∈ N, define fn ∈ X = C[0, 1] by  1 nx if x ∈ [0, n ] fn (x) := 1 if x ∈ [ n , 1] . 1 It is easy to see that fn ∈ B(0, 1) for all n ∈ N and

d(fn , fm ) = m−n for all m > n. Consider the open m cover U := {B(f, 14 ) : f ∈ B(0, 1)} of B(0, 1). Fix an element B(f, 14 ) ∈ U . Suppose B(f, 14 ) constains fk for some k ∈ N. Then for any m = k ,   1 =⇒ d(fm , fk ) ≤ d(fm , f ) + d(fk , f ) fm ∈ B f, 4 1 1 1 < + = . 4 4 2

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In case m > k , this implies

1 m−k < m 2

or

k < m < 2k ;

and in case m < k , this implies

1 k−m < k 2

or

k 0, let N := ε ε-balls   { kε, (k + 2)ε : k = −N − 1, −N, . . . , N − 1} is a finite cover of [−M, M ], hence a finite cover of S .

(iii) Note first that under the given metric d, every subset of R is bounded. So in particular, N ⊂ R is bounded.

However, as B(n, 14 ) ∩ N = {n} for all n ∈ N, any finite

collection of

1 -balls 4

 B(n, 14 ) : n ∈ N cannot cover N.



Hence N is not totally bounded. (d) False.

Example: Let X := (0, 1) with the usual Euclidean metric. Similar to (c)(ii), we see X it is totally bounded. It is also trivial that X is closed in X . However, X is not compact, as   1 the open cover ( n1 , 1 − n ) : n ∈ N has no finite subcover.

11. Let (X, d) be a metric space and U a cover of X. U is said to have a Lebesgue number λ > 0 if every subset of X of diameter less than λ is contained in some element in U , i,e,, for any E ⊂ X, if d(E) < λ, there is some U ∈ U such that E ⊂ U.

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(a) Give an example of a metric space in which an open cover may not have a Lebesgue number. (b) Show that if X is a compact metric space, then every open cover of X has a Lebesgue number. [This is known as Lebesgue Covering Lemma.] Solution : (a) Consider X = (0, 1) with the usual Euclidean metric. Then

{( n1 , 1) : n ∈ N} is an open cover of X without a Lebesgue number. In fact, for any λ > 0, the open set (0, λ2 )∩(0, 1) ⊂ X 1 is of diameter < λ but (0, λ2 ) ∩ (0, 1) ⊂ ( n , 1) for any n ∈ N. (b) Let X be compact and U be an open cover of X . For any x ∈ X , there exists an open set Ux ∈ U such that x ∈ Ux . As Ux is open, there exists rx > 0 such that B(x, rx ) ⊂ Ux . Clearly,

$ #  r  x :x∈X B := B x, 2 is an open cover of X . As X is compact, there is a finite sub-

cover, say

Consider

$ #  r  x B  = B xi , i : i = 1, 2, ..., n . 2 λ := min

#r

xi

2

$ : i = 1, 2, ..., n > 0 .

Then λ is a Lebesgue number for U . To see this, let E be any subset of X with d(E) < λ. Fix p ∈ E . As B  is a cover of

 r  X , there exists k ∈ {1, 2, ..., n} such that p ∈ B xk , x2k . Hence for any q ∈ E , we have rx d(q, xk ) ≤ d(q, p) + d(p, xk ) < λ + k ≤ rxk 2 and so q ∈ B(xk , rxk ) ⊂ Uxk . Thus E ⊂ Uxk .

12. A collection of subsets {Uλ }λ∈Λ of a metric space X is said to have the finite intersection property (f.i.p.) if every finite subcollection of {Uλ }λ∈Λ has nonempty intersection.

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Show that X is compact if and only if every collection of closed subset of X having f.i.p. has nonempty intersection. Proof . (⇒): Let X be compact and {Uλ }λ∈Λ be any collection of closed subsets of X having f.i.p. Assume to the contrary that {Uλ }λ∈Λ has " empty intersection, that is, λ∈Λ Uλ = φ. Then %

λ∈Λ

(X \ Uλ ) = X \

'

Uλ = X

λ∈I

and so {X \ Uλ }λ∈Λ is an open cover of X . Since X is compact,

{X \ Uλ }λ∈Λ has a finite sub-cover, say {X \ Uλi }N i=1 . Hence X=

N %

(X \ Uλi ) = X \

i=1

"N

N '

Uλi

i=1

= φ, contradicting with the f.i.p. of {Uλ : λ ∈ Λ}. (⇐): We prove by contrapositive argument. Suppose X is not compact. Then there exists an open cover {Vλ }λ∈Λ of X which has no finite subcover. For each λ ∈ Λ, Uλ := X \ Vλ is closed in X . Furthermore, for any finite subcollection {Uλi }N i=1 of {Uλ }λ∈Λ , we and so

have

i=1 Uλi

N '

Uλi =

'

Uλ =

i=1

N '

(X \ Vλi ) = X \

'

(X \ Vλ ) = X \

i=1

N %

Vλi = φ

%

Vλ = φ ,

i=1

and so {Uλ }λ∈Λ is a family of closed subsets having f.i.p. However, λ∈Λ

λ∈Λ

λ∈Λ

hence {Uλ }λ∈Λ is a family of closed subsets having f.i.p. but with

empty intersection. Hence the assertion.

13. Let X = C[0, 1] with metric d(f, g) := supx∈[0,1] |f (x) − g(x)|. $ # *1 (a) If S := f ∈ X : 0 f (x) dx = 1 , prove that S is closed and determine whether it is compact.

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(b) If T := {f ∈ X : d(f, 0) ≤ 1, f (0) = 0}, prove that T is closed and bounded. By considering the set {f1 , f2 , . . . }, where  0 if 0 ≤ x ≤ 1 − n1 fn (x) := 1 − n + nx if 1 − n1 < x ≤ 1 , show that T is not compact. Solution : (a) To show the closedness of S , it suffices to prove S  ⊂ S . So let f ∈ S  . For any ε > 0, there is fε ∈ S such that 0
0 is arbitrary, we conclude that i.e., f ∈ S .



1

f (x) dx = 1, 0

However, S is not compact. In fact, for any n ∈ N, define

fn (x) :=



−2n2 x + 2n 0

if 0 ≤ x ≤

1 n

,

otherwise.

It is clear that each fn is continuous and Hence fn ∈ S for all n ∈ N. But then



d(f2n , fn ) ≥ |f2n (0) − fn (0)| = 2n and so S is unbounded, hence not compact.

1

fn (x) dx = 1. 0

for all n ∈ N

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107

(b) Since T ⊂ B(0, 1) ⊂ X , it is bounded. On the other hand, as

T = B(0, 1) ∩ {f ∈ X : f (0) = 0} and both B(0, 1) and {f ∈ X : f (0) = 0} are closed in X , T is closed in X .

However, T is not compact. In fact, consider the subset A := {fn : n ∈ N}, where

fn (x) :=



0 1 − n + nx

if 0 ≤ x ≤ 1 −

if 1 −

1 n

1 n

,

0, there is N ∈ N such that fN ∈ B(f, r), that is, r > d(fN , f ) = sup |fN (x) − f (x)| x∈[0,1]

≥ |fN (1) − f (1)| = |1 − f (1)| .

Since r > 0 is arbitrary, we must have f (1) = 1. On the other hand, since f ∈ T is an accumulation point of A, near f there

are infinitely many elements of A, that is, there are infinitely

many fn ’s. So for any r > 0 and any x ∈ (0, 1), there exists

M ∈ N such that 1 − FM (x) = 0 and so

1 M

> x and fM ∈ B(f, r). Hence

r > d(fn , f ) ≥ |fM (x) − f (x)| = |f (x)| ≥ 0 . Since r > 0 is arbitrary, we have f (x) = 0. Since x ∈ (0, 1)

is arbitrary, we have f = 0 on (0, 1). But then from above we have f (1) = 1 and so f is not continuous at 1, contradicting to the fact that f ∈ T ⊂ C[0, 1]. Hence T is not compact.

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1.4 Compactness in the Euclidean Space Rn As the readers may have already observed by now, in general, compactness in a general metric space is a rather intricate concept to study. In this section, we will restrict our attention to the case where the metric space is the Euclidean space Rn . In fact, we will see that in this setting, neater and more fruitful results on compactness can be obtained. In particular, we shall establish the equivalence among the following conditions on subsets in S ⊂ Rn : (a) S is compact. (b) S is closed and bounded. (c) S has the Bolzano-Weierstrass property.

Observe that by Theorems 1.3.5 and 1.3.6, we already have (a) ⇒ (b)

and (a) ⇒ (c). So in order to establish the equivalence relations, it suffices to show (b) ⇒ (a) and (c) ⇒ (b).

Theorem 1.4.1 (Bolzano-Weierstrass). Every bounded infinite set S ⊂ Rn has an accumulation point in Rn . n Proof. Since S is bounded, S ⊂ J 0 := i=1 [−a, a] for some a > 0. The “mid-planes” xi = 0, i = 1, . . . , n, partition the cube J 0 into 2n identical cubes each side of which has length a. Since S is an infinite set, at least one of these 2n cubes, say J 1 , contains infinitely many points of S. Repeating the same process with J 1 in place of J 0 , we obtain a cube J 2 each side of which has length a2 and contains infinitely many points of S. Inductively, we obtain a decreasing nest k+1 ⊂ J k , with the length of each side of J k of cubes {J k }∞ k=1 , J a being 2k−1 , and for each k, J k contains infinitely many points of S. "∞ Observe that k=1 J k = φ.

Metric Spaces

In fact, for each k ∈ N, if we write k

J =

n /

[aki , bki ] ,

i=1

then for each i = 1, . . . , n, bki − aki =

a 2k−1

109

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Metric Space Topology: Examples, Exercises and Solutions

and so as k → ∞, bki − aki → 0, thus lim aki = lim bki = ti .

k→∞

k→∞

(*)

[Question: why do the limits exist? – see Remark (i) below.] Let t = (t1 , . . . , tn ). Then it is not hard to see that t ∈ J k for every k " k (**) and so t ∈ ∞ k=1 J . [Question: why? – see Remark (ii) below.]

It is then evident that t is an accumulation point of S. In fact, for any r > 0, B(t, r) √will contain the entire cube J k whenever k is   n a < r. Thus B(t, r) ∩ S \ {t} = φ and so large enough so that 2k−1  t ∈ S. Remarks. (i) A common elementary mistake mathematics freshmen would make is “{xk − yk } → 0 as k → ∞

=⇒ {xk } and {yk } converge to the same limit as k → ∞”.

A simple example for this statement to be false is xk := k and yk := k + k1 , as in this case we have {xk − yk } = {− k1 } → 0 as k → ∞, while neither {xk } nor {yk } converges as k → ∞. The more matured readers could tell that the statement above is true only when both {xk } and {yk } are convergent as k → ∞.

Hence to show that (*) is true, we need to show that for each fixed i = 1, . . . , n, both {aki }k∈N and {bki }k∈N are convergent as k → ∞. But these should be immediate once we observe that for every fixed i = 1, . . . , n (that is, at each coordinate), the sequence {aki }k∈N is nondecreasing and bounded above, while {bki }k∈N is nonincreasing and bounded below. Hence both {aki }k∈N and {bki }k∈N are convergent as k → ∞.

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(ii) By the observation in (i), we have, for every fixed i = 1, . . . , n, ≤ · · · ≤ ti ≤ · · · ≤ bk+1 ≤ bki aki ≤ ak+1 i i

for all k ∈ N .

Hence for every i = 1, . . . , n, ti ∈ [aki , bki ] for all k ∈ N and so  t = (t1 , . . . , tn ) ∈ J k = ni=1 [aki , bki ] for every k ∈ N. Hence (**). Theorem 1.4.2 (Cantor Intersection Theorem). Every decreasing nest of closed and bounded nonempty subsets of Rn has nonempty intersection. That is, if {Qk }∞ k=1 is a collection of nonempty closed subsets of Rn satisfying (a) Qk+1 ⊂ Qk for all k, and

(b) Q1 is bounded, " then ∞ k=1 Qk = φ. "∞ Proof. Let S = k=1 Qk . If Qk is finite for some k ∈ N, then there exists r ≥ k such that Qr = Qr+1 = Qr+2 = · · · and hence in this "∞ case k=1 Qk = Qr = φ. So let us assume that Qk is infinite for all k. Let A = {x1 , x2 , . . . }, where for each k, xk ∈ Qk is an arbitrary point distinct from x1 , . . . , xk−1 . This can be done since all Qk ’s are infinite. By BolzanoWeierstrass theorem, A has an accumulation point x ∈ Rn . By Theorem 1.2.27 every open neighborhood U of x in Rn contains infinitely many points of A and thus contains infinitely many points of each Qk . Hence x is an accumulation point of each Qk . Now for each k ∈ N, Qk is closed. Hence we have Qk ⊂ Qk and so x ∈ Qk ⊂ Qk "∞ for all k ∈ N. Therefore, x ∈ k=1 Qk = S. 

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Theorem 1.4.3 (Heine-Borel Theorem). Every closed and bounded subset of Rn is compact. Proof. Let S ⊂ Rn be closed and bounded. Let U = {Uα }α∈Λ be of’s covering theorem (Theorem an open cover of S in Rn . By Lindel¨ 1.2.15), U has a countable sub-cover {Uk }∞ k=1 . For any m ≥ 1, define Sm :=

m %

Uk ,

k=1

Qm := S \ Sm . Then the Sm ’s are open, increasing and hence the Qm ’s are closed and decreasing. Furthermore, since S is bounded, the Qm ’s are also bounded. Now if Qm = φ for all m, by Cantor Intersection Theorem (Theorem 1.4.2), φ =

∞ '

k=1

Qk =

∞  '

k=1



S \ Sk = S \

∞ %

Sk ,

k=1

!∞ which is absurd since S ⊂ k=1 Sk . Thus there exists m ≥ 1 such !m  that Qm = φ and hence S ⊂ Sm = k=1 Uk . Corollary 1.4.4. Let S ⊂ Rn be a subset. Then the following are equivalent: (a) S is compact. (b) S is closed and bounded. (c) S has the Bolzano-Weierstrass property. Proof. By Theorems 1.3.5 and 1.4.3, (a) ⇔ (b). By Theorem 1.3.6, (a) ⇒ (c). It thus remains to show (c) ⇒ (b). So assume that S has the Bolzano-Weierstrass property. Clearly S must be bounded for otherwise there exists a sequence {xk }k∈N in S with xk  ≥ k for all k. But then obviously such a sequence cannot have an accumulation point in S. Next, to show that S is closed, it sufficies to show that

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S  ⊂ S. So let x ∈ S  . Since every open neighborhood of x contains infinitely many points of S, we can choose an infinite sequence of   distinct points {xk : k ∈ N} ⊂ S with xk ∈ B x, k1 for each k ∈ N. Since S has the Bolzano-Weierstrass property, {xk : k ∈ N} has an accumulation point y ∈ S. We claim that y = x. To see this, observe that for any k ∈ N, y − x ≤ y − xk  + xk − x 1 < y − xk  + . k Now certainly the second term k1 on the right is controllable, and is going to be small when k gets large. However, in general the first term y − xk  need not be small for all k ∈ N. In fact, we only know that there are infinitely many k’s for which xk is close to y, but it is not necessarily true that all xk ’s are close to y. But that is already good enough for our purpose: For any r > 0, there are infinitely many k ∈ N such that xk ∈ B(y, r). Hence y − x < r +

1 k

for infinitely many k ∈ N .

This forces y − x ≤ r . But then as r > 0 is arbitrary, this implies x = y ∈ S and so S  ⊂ S. Hence S is closed.  Remark. For a general metric space, Corollary 1.4.4 is only partially true, namely, we have (a) ⇒ (b) (a) ⇒ (c)

by Theorem 1.3.5 , by Theorem 1.3.6 ,

furthermore, it can be proved without much difficulty that (c) ⇒ (a)

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but in general, (b) =⇒ × (a) . For instance, let X := Q with the Euclidean metric, and   S := r ∈ Q : r ∈ (a, b) , where a, b ∈ R \ Q, a < b. Then it is not difficult to see that S is closed and bounded in X but not compact (see Exercise 1.4, Part A, Problem #12 below). Hence for a general metric space, including the case where it is a proper subspace of Rn , we only have (a) ⇔ (c) ⇒ (b) .

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Exercise 1.4 Unless otherwise specified, X will stand for a general metric space and S, T , etc., are arbitrary subsets of X. Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. The circle S 1 := {(x, y) ∈ R2 : x2 + y 2 = 1} ⊂ R2 is compact. Answer : True. Proof . It is closed and bounded in R2 , hence compact. 2. The ellipse

#

(x, y) ∈ R2 :

x2 a2

+

y2 b2

$ = 1 ⊂ R2 is compact.

Answer : True. Proof . It is closed and bounded in R2 , hence compact.

3. The parabola {(x, y) ∈ R2 : y 2 = x} ⊂ R2 is compact. Answer : False. Justification . Although it is closed in R2 , since it is unbounded, it is noncompact.

# 4. The hyperbola (x, y) ∈ R2 :

x2 a2

−

y2 b2

$ = 1 ⊂ R2 is compact.

Answer : False. Justification . Although it is closed in R2 , since it is unbounded, it is

noncompact.

  5. The upper half circle S := (x, y) ∈ R2 : x2 + y 2 = 1 , y > 0 ⊂ R2 is compact. Answer : False. Justification . Although it is bounded, since it is not closed in R2 , it is noncompact.

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6. S := [0, ∞) ⊂ R is compact. Answer : False. Justification . Although S is closed in R, since it is unbounded, it is noncompact. Alternatively, {(−1, n) : n ∈ N} is an open cover of S without finite subcover. Hence S is noncompact.

7. The cone S := {(x, y, z) : z ≥



x2 + y 2 } ⊂ R3 is compact.

Answer : False. Justification . Although it is closed in R3 , since it is not bounded, it is noncompact.

8. The cylinder S := {(x, y, z) : x2 + y 2 = 1} ⊂ R3 is compact. Answer : False. Justification . Although it is closed in R3 , since it is not bounded, it is noncompact.

9. S := (0, 1] ⊂ X = (0, ∞) with Euclidean metric is closed and bounded, hence compact. Answer : The first half is True, the second half is False. Justification. It is clear that S is closed and bounded in X . However, Heine-Borel Theorem does not apply, as the metric space X under con  1 sideration is not the Euclidean space R. In fact, (0, 1 − n ):n∈N is an open cover of S which does not admit any finite subcover. Hence S is not compact.   10. S := n1 : n ∈ N ∪ {0} ⊂ Q is closed and bounded in Q, hence compact. Answer : True. Proof . It is not hard to see that S is closed and bounded in Q. It is also compact. However, the compactness does not follow directly from the closedness and boundedness of S ⊂ Q. Instead, it follows

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from the fact that S when considered as a subset of R, is closed and bounded. Hence by Heine-Borel Theorem, S is compact in R. Finally, by Theorem 1.3.7, S is compact in Q. The compactness of S can also be proved by first principle: Let {Uλ }λ∈Λ

be any open cover of S in X . Then there exists λ0 ∈ Λ such that

0 ∈ Uλ0 . Note that Uλ0 covers all but finitely many points of S . In fact, as Uλ0 is open in Q, there exists an open set V ⊂ R such that Uλ0 = V ∩ Q. Since V ⊂ R is open neighborhood of 0, there exists δ > 0 such that (−δ, δ) ⊂ V . Let N ∈ N be large enough 1 such that N < δ. Then n1 ∈ (−δ, δ) ∩ Q ⊂ V ∩ Q = Uλ0 for all n ≥ N and so the only points of S that are not covered by Uλ0 are { n1 : n = 1, . . . , N − 1}. For each such n, there exists λn ∈ Λ such 1 that n ∈ Uλn . Hence {Uλn : n = 0, 1, 2, . . . , N − 1} is a finite subcover of the open cover {Uλ }λ∈Λ and so S is compact. 11. S := Q ∩ [0, 1] ⊂ R is compact. Answer : False. Justification . S is not closed in R. In fact, (R \ Q) ∩ [0, 1] does not have any interior point, hence (R \ Q) ∩ [0, 1] is not open in [0, 1]   and so S = [0, 1] \ (R \ Q) ∩ [0, 1] is not closed in [0, 1], thus noncompact.

Alternatively, pick any irrational point a ∈ S . Then



1 1 − ,a − n n

  & 1 1 , a − ,1 + n n n∈N

is an open cover of S in R which has no finite subcover. Here, we used the natural convention that (α, β) := φ for α ≥ β .

12. Let a, b ∈ R \ Q, a < b, and S := (a, b) ∩ Q. (a) When considered as a subset of the metric space R with the usual Euclidean metric, S is compact. (b) Consider S as a subset of the metric space Q with the usual Euclidean metric.

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(i) (ii) (iii) (iv)

S S S S

is is is is

bounded. open. closed. compact.

Answer : (a) False.

Justification . As the closure of S in R is S = [a, b ]  S , S is not closed in R, hence not compact. (b)

(i) True.

Proof . It is obvious. (ii) True.

Proof . Since (a, b) is open in R, S = (a, b) ∩ Q is open in Q. (iii) True.

Proof . Since a, b ∈ R \ Q, we have S = (a, b) ∩ Q = [a, b ] ∩ Q. In particular, since [a, b ] is closed in R, S = Q ∩ [a, b ] is closed in Q.

(iv) False.

1 Justification . For any n ∈ N with n > b−a , let Un :=   1 a, b − n ∩ Q. The argument used in Part (ii) shows that Un is open in Q. The sets {Un : n ∈ N} form an open cover of S which obviously has no finite subcover. Hence S ⊂ Q is not compact. Alternatively , by (a), S is not compact in R. Hence by Theorem 1.3.7, it is not compact in Q.

Part B: Problems 1. Without the closedness condition, determine whether Cantor Intersection Theorem is still valid. Answer : Without the closedness condition, Cantor Intersection Theorem is no longer valid.

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Example : Consider Qn := (0, n1 ) ⊂ R, n ∈ N. Then each Qn is "∞ bounded but not closed in R, and n=1 Qn = φ.

2. Without the boundedness condition, determine whether Cantor Intersection Theorem is still valid. Answer : Without the boundedness condition, Cantor Intersection Theorem is no longer valid.

Example : Consider Qn := [n, ∞) ⊂ R, n ∈ N. Then each Qn is "∞ closed but not bounded in R, and n=1 Qn = φ.

3. Determine whether Cantor Intersection Theorem is still valid for any general metric spaces. Answer : Cantor Intersection Theorem is in general not valid for a general metric space.

 n Example : Consider X := (0, 1) ⊂ R. Let Qn := n+1 , 1 for n ∈ N. Each Qn is closed in (0, 1) and is obviously bounded, but "∞ n=1 Qn = φ.

4. Let I0 := [0, 1], I1 := the residue of I0 after the middle open 13 of [0, 1] is removed, and inductively, for any n ∈ N, In+1 be the residue of In after the middle open 13 of each of its subintervals is removed. To be more precise, I0 := [0, 1]       1 2 1 2 , = 0, ∪ ,1 I1 := I0 \ 3 3 3 3   &  7 8 1 2 , ∪ , I2 := I1 \ 9 9 9 9         2 1 2 7 8 1 ∪ , ∪ , ∪ ,1 = 0, 9 9 3 3 9 9 .. .

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or, by abusing notations in the obvious way, we have In+1

In ∪ := 3



2 In + 3 3

&

,

n = 0, 1, 2 . . . .

"∞ Define C := n=0 In . Classically, X is known as the Cantor Set. (a) Show that C = φ. (b) Find C. (c) Find C ◦ . (d) Find C  . (e) Find ∂ C. (f) Determine if C is compact. Solution : (a) As 0 ∈ In for all n, we have 0 ∈ C and so C = φ. (b) Each In is closed in R. Being the intersection of In ’s, C is closed and so C = C . (c) First observe that at step n ∈ N, In consists of 2n closed intervals, each of which has length 31n . So we can denote by Ink , k = 1, . . . , 2n , the subintervals which form In , each of which has length 31n . For any x ∈ C and any δ > 0, let N ∈ N be "∞ large enough such that 31N < 2δ . Then as C := n=0 In , we k have x ∈ IN and so x ∈ IN for some k = 1, . . . , 2N . Note k that when IN +1 is formed, the middle open 13 -interval of IN which is a subset of (x − δ, x + δ) would be removed. As a result, (x − δ, x + δ) ⊂ C . Since this is true for every δ > 0, we conclude that x ∈ / C ◦ and so C ◦ = φ. (d) Observe that every point in x ∈ C \ {0, 1} is a boundary point of some open 13 -interval which has been removed when C is being constructed. In particular, every open interval centered at x would contain infinitely many open

1 3 -intervals

which have

been removed and hence in turn contains infinitely points of C . Hence x ∈ C  . Similarly, 0, 1 ∈ C  and so C ⊂ C  . On the

other hand, by (b), C is closed and so C  ⊂ C . Thus C  = C .

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(e) Since C = C = C ◦ ∪ · ∂ C and C ◦ = φ, we have ∂ C = C .

(f) As C is both closed and bounded in R, by Heine-Borel Theorem,

C is compact. |x − y| . 5. Consider the metric space (R, d), where d(x, y) := 1 + |x − y| √ (a) If d(x, y) < r < 12 , show that |x − y| < r. (b) Show that N is closed and bounded in (R, d). (c) Determine whether N is compact. Solution : (a) Elementary. (b) Since d(x, y) < 1 for any x, y ∈ R, R is bounded and so in particular, N is bounded. On the other hand, consider x ∈

R \ N. If we denote by B(x, r) and Bd (x, r) the open balls at x with radius r > 0 with respect to the Euclidean metric and the metric d, respectively, then there exists δ0 > 0 such that B(x, δ0 ) ∩ N = φ. Observe that by (a), if δ < √1 , we 2 have Bd (x, δ 2 ) ⊂ B(x, δ). Hence for any δ < min{ √1 , δ0 }, 2 we have Bd (x, δ 2 ) ∩ N ⊂ B(x, δ) ∩ N ⊂ B(x, δ0 ) ∩ N = φ. Thus R \ N is open and so N is closed in (R, d).

(c) Although N is closed and bounded in (R, d), Heine-Borel Theorem does not apply here, as the Theorem is valid only when the Euclidean metric is used. In fact, under the metric d defined

here, N is actually not compact. To see this, we observe that for any n ∈ N, Bd (n, 12 ) ∩ N = {n}. Hence the open cover

 Bd (n, 12 ) : n ∈ N does not have any finite subcover. Thus N is not compact. 

6. Determine whether Cantor Intersection Theorem is valid for any compact metric spaces. That is, if X is a compact metric space, is it always true that every decreasing nest of closed nonempty subsets of X (note that subsets of X are automatically bounded) has nonempty intersection.

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Answer : Cantor Intersection Theorem is still valid for arbitrary compact spaces.

Proof . Let X be a compact metric space. Let {Qn : n ∈ N} be a decreasing nest of nonempty closed subsets of X . Assume that "∞ n=1 Qn = φ. For any n ∈ N, let Un := Q1 \ Qn . Then Q1 = Q1 \ φ = Q1 \



∞ '

n=1

Qn



=

∞ %

(Q1 \ Qn ) =

n=1

∞ %

Un .

n=1

Hence {Un : n ∈ N} is an open cover (in Q1 ) for Q1 . Being a closed

subset of the compact space X , Q1 is compact. Thus one can find a

 Unj : j = 1, . . . , J of {Un : n ∈ N} for Q1 . Let m := min{nj : j = 1, . . . , J}. Then Um is the largest one among all the Unj ’s and finite subcover



Q1 =

J %

j=1

Unj = Um = Q1 \ Qm ,

which forces Qm = φ, contradicting to the assumption that Qn = φ for all n. Thus

"∞

n=1

Qn = φ.

7. Determine whether it is true that for every bounded subset S of Rn , we can always find a smallest compact subset of Rn which contains S. Answer : True. Proof . By Heine-Borel Theorem, a subset S ⊂ Rn is compact if and only if it is closed and bounded. Also, the closure S of S is the smallest closed subset of Rn which contains S . So the problem boils down to showing that S of a bounded set S ⊂ Rn is also bounded. To see this, let S ⊂ Rn be bounded. If S = φ, there is nothing to prove. So assume S = φ. Fix x0 ∈ S . As S is bounded, S ⊂ B(x0 , R) for some R > 0. If S is unbounded, then there exists at least one point x1 ∈ S , such that d(x0 , x1 ) > R + 2. So B(x1 , 1) ∩ S = φ. This contradicts with x1 ∈ S . Therefore, S is bounded in X .

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8. Let A ⊂ X be closed, with A◦ = φ. Show that for any nonempty open set U ⊂ X, there exists a nonempty open set V in X such that V ⊂ U and V ∩ A = φ.

Proof . As A◦ = φ and U = φ is open, A ⊃ U . So there exists y ∈ U \ A. As U is open, B(y, ε1 ) ⊂ U for some ε1 > 0. On the other hand, as y ∈ / A and A is closed, B(y, ε2 ) ∩ A = φ for some ε2 > 0. Take ε := min{ε1 /2, ε2 /2} and V := B(y, ). Clearly, V is open, V = φ, V ⊂ B(y, ε1 ) ⊂ U , and V ∩ A ⊂ B(y, ε2 ) ∩ A = φ.

9. X is said to be a Baire space if the union of countably many closed subsets of X, each of which has empty interior, must have empty interior. That is, given any countable collection {An }n∈N of closed subsets of X with A◦n = φ for every n, we ◦ ! = φ. have n∈N An Show that every compact space X is a Baire space. Proof . Let {An }n∈N be a countable collection of closed subsets of X with A◦n = φ for every n. Let V0 ⊂ X be any nonempty open subset. As A◦1 = φ, A1 ⊃ V0 . By Exercise 1.4, Part B, Problem #8, there is a nonempty open set V1 in X such that V1 ⊂ V0 and V1 ∩ A1 = φ. Inductively, we have a sequence of nonempty open sets {Vn }n∈N satisfying Vn ⊂ Vn−1

and

Vn ∩ An = φ for all n ∈ N .

In particular, we have a sequence of nonempty closed sets

· · · ⊂ Vn ⊂ Vn−1 ⊂ · · · ⊂ V2 ⊂ V1 with the finite intersection property. Since X is compact, by Exercise 1.4, Part B, Problem #6 or Exercise 1.3, Part B, Problem #12, we have a point

x∈

'

n∈N

Vn = φ .

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Metric Space Topology: Examples, Exercises and Solutions

In particular, x ∈ V1 ⊂ V0 . Furthermore, since x ∈ Vn and

Vn ∩ An = φ for all n, it follows that x ∈ / An for any n and so ! ! x ∈ / n∈N An . To sum up, we have an x ∈ V0 \ n∈N An , hence ! arbitrary nonempty open subn∈N An ⊃ V0 . Since V0 ⊂ X is any  ! ◦ set of X , we conclude that = φ and thus X is a Baire n∈N An space.

10. Let X be compact and {Un }n∈N be a countable collection of " open sets in X, each of which is dense in X. Show that n∈N Un is also dense in X. Proof . Observe first that S ⊂ X is dense in X if and only if (X \ S)◦ = φ. In fact, S=X ⇔ ∀ x ∈ X, ∀ ε > 0, B(x, ε) ∩ S = φ

⇔ ∀ x ∈ X, ∀ ε > 0, ∃ z ∈ B(x, ε) ∩ S

/ X \S ⇔ ∀ x ∈ X, ∀ ε > 0, ∃ z ∈ B(x, ε) and z ∈

⇔ ∀ x ∈ X, ∀ ε > 0, X \ S ⊃ B(x, ε)

⇔ (X \ S)◦ = φ .

Now if {Un } is a countable collection of open sets each of which is dense in X , then by the observation above, X \ Un is a

countable collection of closed sets, each of which has empty interior. By Exercise 1.4,  Part B, Problem #9, ◦ being compact, X is Baire and ! so in particular, = φ. By the observation above n∈N (X \ Un )

again,

'

n∈N

Un = X \



X\

'

n∈N

Un



=X\

is dense in X .

11. Show that [0, 1] \ Q is dense in [0, 1].



%

n∈N



(X \ Un )

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# $ Proof . [0, 1] \ {r} : r ∈ Q is a countable collection of open sets in [0, 1], each of which is dense in the compact space [0, 1]. By Exercise 1.4, Part B, Problem #10,

[0, 1] \ Q =

 ' [0, 1] \ {r}

r∈Q

is dense in [0, 1].

12. A metric space X is said to be locally compact if every x ∈ X has a compact neighborhood in X. That is, for any x ∈ X, there exists a compact subset C ⊂ X which contains an open neighborhood U of x in X. Show that the following statements are equivalent: (a) X is locally compact. (b) For any x ∈ X, there exists an open neighborhood U of x in X with U compact. (c) For any x ∈ X, there exists r > 0 such that B(x, r) is compact. Proof . (a)⇒(b): By definition of local compactness, for any x ∈ X , there exists a compact set C and an open set U with x ∈ U ⊂ C . Hence we have x ∈ U ⊂ U ⊂ C = C . Note that being a closed subset of the compact set C , U is compact. (b)⇒(c): By assumption, there exists an open neighborhood U of x in X with U compact. Now as U is open, there exists r > 0 such that B(x, r) ⊂ U . Note that B(x, r) ⊂ U and so B(x, r) is compact. (c)⇒(a): By assumption, there exists r > 0 such that B(x, r) is compact. As B(x, r) ⊂ B(x, r) is trivially an open neighborhood of x, (a) follows. 13. Show that (a) R is locally compact. (b) Every discrete metric space is locally compact.

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Proof . (a) It is obvious by Exercise 1.4, Part B, Problem #12, as by HeineBorel Theorem, B(x, r) is compact for every x ∈ R and every

r > 0.

(b) It is obvious since every singleton in a discrete metric space is open and compact.

14.

(a) Show that every compact space is locally compact. (b) Determine whether the converse of (a) is true or false. Solution : (a) If X is compact, then X is trivially open and compact in X , hence X itself is a compact neighborhood of every point in X and so X is locally compact. (b) The converse of (a) is in general not true. Example 1 : By Exercise 1.4, Part B, Problem #13, R is locally compact but not compact.

Example 2 : Let X be a discrete metric space with infinitely many elements. Then X is not compact but by Exercise 1.4, Part B, Problem #13, X is locally compact. 15. Show that every locally compact metric space is a Baire Space. Proof . The proof is similar to that of Exercise 1.4, Part B, Problem #9. Let {An }n∈N be a countable collection of closed subsets of X with A◦n = φ for every n. Let V0 ⊂ X be any nonempty open subset. As A◦1 = φ, A1 ⊃ V0 . By Exercise 1.4, Part B, Problem #8, there is a nonempty open set V1 in X such that V1 ⊂ V0 and V1 ∩ A1 = φ. By the local compactness of X , by choosing a “smaller” subset, we may simply assume that V1 is compact. In fact, pick an arbitrary point z ∈ V1 . By local compactness, we can find an open neighborhood U of z in X such that U is compact. Take W := V1 ∩ U . Then W = V1 ∩ U ⊂ V1 ⊂ V0 , W ∩ A1 ⊂

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V 1 ∩ A1 = φ, and W = V1 ∩ U ⊂ U is compact. So by replacing V1 by W if necessary, we may assume that V1 is compact. Inductively, we have a sequence of nonempty open sets {Vn }n∈N satisfying

Vn ⊂ Vn−1

and

Vn ∩ An = φ for all n ∈ N .

In particular, we have a decreasing nest of nonempty closed sets

· · · ⊂ Vn ⊂ Vn−1 ⊂ · · · ⊂ V2 ⊂ V1 in the compact set V1 . By Exercise 1.4, Part B, Problem #6, we have a point

x∈

'

n∈N

Vn = φ .

In particular, x ∈ V1 ⊂ V0 . Furthermore, since x ∈ Vn and

Vn ∩ An = φ for all n, it follows that x ∈ / An for any n and so ! x∈ / n∈N An . ! ! To sum up, we have an x ∈ V0 \ n∈N An , hence n∈N An ⊃ V0 . Since V0 ⊂ X is any arbitrary nonempty open subset of X , we ! ◦ conclude that = φ and thus X is a Baire space. n∈N An

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Chapter 2 Limits and Continuity Chapter 1 deals with basic topological properties of a single metric space. In this chapter, we shall treat relations of two or more metric spaces. Naturally, by a “relation” between two metric spaces we mean a function, or a mapping, from one metric space to another. In particular, the concepts of limits and continuity in the context of Euclidean spaces will be carried forward to the context of general metric spaces, with the upshot being the study of the “equivalence problem”, namely, the classification of metric spaces into “equivalent classes”. Same as the case for advanced calculus, the basic language here will be the “epsilon-delta” (ε − δ) language. The idea is that in the setting of abstract metric spaces, we have the concept of metrics which leads to the concept of “closeness” among points. Hence the concepts of limits and continuity can easily be mastered once the epsilon-delta language is transformed into the intuition of closeness among points. In particular, for any function f : X → Y of a metric space X into a metric space Y , the limit of f (x) as x approaches a ∈ X is the point in Y for which f (x) is getting arbitrarily close to, when x is getting arbitrarily close to a. Similarly, f is continuous at a ∈ X if points near a are mapped to points near f (a) ∈ Y by f , and two metric spaces are “equivalent” if one can be “continuously deformed into” another, with the process reversible. To certain extent, these intuitive definitions of the concepts of limits and continuity are more important than those using rigorous epsilon-delta language.

2.1 Convergence in a Metric Space Definition 2.1.1. A sequence {xn }n∈N in X is said to be convergent and converge to a ∈ X if for every ε > 0, there exists N ∈ N such 129

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Metric Space Topology: Examples, Exercises and Solutions

that d(xn , a) < ε

whenever

n≥N ,

xn ∈ B(a, ε)

whenever

n≥N .

that is,

In this case we write limn→∞ xn = a, or {xn } → a as n → ∞, or xn → a as n → ∞, or even just xn → a in case no confusion may arise. If {xn }n∈N is not convergent in X, it is said to be divergent. Remarks. (i) Note that {xn } → a in X as n → ∞ if and only if {d(xn , a)} → 0 in R as n → ∞. Hence the convergence of a sequence of points {xn }n∈N in X to a is equivalent to the convergence of the corresponding sequence of real numbers {d(xn , a)}n∈N in R to 0. Proof. {xn } → a as n → ∞

⇐⇒ for any ε > 0, there exists N ∈ N s.t.

|d(xn , a) − 0| = d(xn , a) < ε for all n ≥ N

⇐⇒ {d(xn , a)} → 0 as n → ∞ .

(ii) If a sequence {xn }n∈N in S ⊂ X converges to a point a ∈ S, then as a sequence in X, {xn }n∈N also converges to a. However, in general, the converse need not be true. That is, if a sequence {xn }n∈N in S ⊂ X converges in X to a ∈ X, then as a sequence in S, {xn }n∈N need not be convergent. Proof. The first statement is straight forward, as the metrics in S and X are the same. For the second statement, it suffices to provide a counterexample. This will be done after the following Theorem.

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Theorem 2.1.2. A sequence {xn }n∈N in X can converge to at most one point in X, called the limit of {xn }n∈N . Proof. It is obvious by triangle inequality. In fact, if {xn } → a and {xn } → b as n → ∞ for some a, b ∈ X, then for any ε > 0, there exist N1 , N2 ∈ N such that ε 2 ε d(xn , b) < 2

d(xn , a)
0, there exists N ∈ N such that xn ∈ B(a, r) for all n ≥ N . In particular, B(a, r) ∩ {xn : n ∈ N} = φ and so a ∈ {xn : n ∈ N}. (c) Similar to (b) above, for any r > 0, there exists N ∈ N such that xn ∈ B(a, r) for all n ≥ N . Since #{xn : n ∈ N} = ∞,   # B(a, r) ∩ {xn : n ∈ N} = ∞ and so a ∈ {xn : n ∈ N} . 

A very useful characterization of adherent points and accumulation points is the following. Theorem 2.1.5. Let S ⊂ X and a ∈ X. Then (a) a ∈ S if and only if there is a sequence {xn }n∈N in S such that {xn } → a as n → ∞; (b) a ∈ S  if and only if there is an infinite sequence of distinct points {xn }n∈N in S \ {a} such that {xn } → a as n → ∞. Proof. (a) (⇒) If a ∈ S, then for every n ∈ N, there exists xn ∈ S with d(xn , a) < n1 . Clearly, this sequence {xn }n∈N constructed in S has the desired property that {xn } → a as n → ∞. (⇐) If there is a sequence {xn }n∈N in S converging to a, then for any r > 0, there exists N ∈ N such that d(xn , a) < r for all n ≥ N . Hence B(a, r) ∩ {xn : n ∈ N} = φ and so a ∈ S. (b) (⇒) If a ∈ S  , then near a there are infinitely many points of S. Hence for every n ∈ N, there exists xn ∈ S \ {x1 , . . . , xn−1 } with 0 < d(xn , a) < n1 . By construction, {xn }n∈N is an infinite sequence of distinct points in S \{a} with {xn } → a as n → ∞. (⇐) If there is an infinite sequence of distinct points {xn }n∈N

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133

in S \ {a} converging to a, then for any r > 0, there exists N ∈ N such that 0 < d(xn , a) < r for all n ≥ N . Hence  B(a, r) ∩ {xn : n ∈ N} \ {a} = φ and so a ∈ S  . With Theorem 2.1.5, we have the following very useful necessary and sufficient condition of closedness of a subset S ⊂ X. Theorem 2.1.6. Let S ⊂ X. Then the following are equivalent: (a) S is closed.

(b) If {xn }n∈N is a sequence in S with {xn } → a as n → ∞, then a ∈ S. Proof. It is rather straightforward by Theorem 2.1.5. (a) ⇒ (b): If S is closed and {xn }n∈N is a sequence in S with {xn } → a as n → ∞, then by (a) of Theorem 2.1.5, a ∈ S = S. (b) ⇒ (a): Take any a ∈ S. By (a) of Theorem 2.1.5, there is a sequence {xn }n∈N in S such that {xn } → a as n → ∞. But then by assumption (b), we have a ∈ S. Thus S = S and so S is closed.  To end this section, we have the following trivial result. Theorem 2.1.7. Let {xn }n∈N be a sequence in X. Then {xn } → a as n → ∞ if and only if {xnk } → a as k → ∞ for every subsequence {xnk }k∈N of {xn }n∈N . Proof. Obvious.



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Exercise 2.1 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification.  1. The sequence xn :=

1 n



n∈N

in R is convergent.

Answer : True. Proof . For any ε > 0, let N ∈ N be large enough such that N > 1ε . Then for any n ≥ N , d(xn , 0) = |xn − 0| = |xn | =

1 1 ≤ 0, there exists N ∈ N such that d(xn , a) < ε whenever n ≥ N . Hence

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135

{xn }n∈N → a if and only if there exists ε > 0 such that for any N ∈ N, we can find n ≥ N with d(xn , a) ≥ ε. Back to our problem. We will show that for any real number a ∈ R and any integer N ∈ N, we can find n ≥ N such that d(xn , a) ≥ 12 : Case 1 : a ≤ 0. Then for any even number n ≥ N ,     1 1 1 d(xn , a) = |xn − a| = 1 + − a = 1 + + |a| > . n n 2

Case 2 : a > 0. Then for any odd number n ≥ N ,     1 1 1  d(xn , a) = |xn − a| = −1 + − a = a + 1 − > . n n 2 Hence {xn } → a as n → ∞.

3. (Compare with Theorem 2.1.7) A sequence {xn }n∈N in X is convergent if and only if every subsequence {xnk }k∈N of {xn }n∈N is convergent.

Answer : True. Proof . In view of Theorem 2.1.7, it suffices to show that if all subsequences {xnk }k∈N of {xn }n∈N are convergent, then they

converge to the same limit. So assume that we have two subsequences

{xnk }k∈N and {xmk }k∈N of {xn }n∈N , with lim xnk = a ,

k→∞

lim xmk = b .

k→∞

Then by assumption, a third subsequence {xk }k∈N of {xn }n∈N defined by

xk :=



xnk xm k

if k is odd if k is even

must also be convergent. However, since

x2k+1 = xn2k+1 −→ a as k → ∞ , x2k = xm2k −→ b as k → ∞ ,

we must have a = b.

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Metric Space Topology: Examples, Exercises and Solutions

4. A convergent sequence {xn }n∈N in R with discrete metric is eventually a constant sequence, that is, there exists N ∈ N such that xn = xN for all n ≥ N . Answer : True. Proof . As {xn }n∈N is convergent, it is Cauchy and so there exists N ∈ N such that d(xn , xN ) < 12 for all n ≥ N . However, as d is the discrete metric, it only has values 0 and 1 and so this forces d(xn , xN ) = 0 for all n ≥ N and so xn = xN for all n ≥ N . 5. A convergent sequence {xn }n∈N in Z with the metric induced by the Euclidean metric on R is eventually a constant sequence, that is, there exists N ∈ N such that xn = xN for all n ≥ N . Answer : True. Proof . As {xn }n∈N is convergent, it is Cauchy and so there exists N ∈ N such that d(xn , xN ) < 12 for all n ≥ N . However, as xN and xn ∈ Z for all n, we have d(xn , xN ) ∈ N ∪ {0}. This forces d(xn , xN ) = 0 for all n ≥ N and so xn = xN for all n ≥ N . 6. A convergent sequence {xn }n∈N in a metric space X = X1 ∪ X2 with X1 ∩ X2 = φ such that there are infinitely many distinct xn ’s in each Xi , i = 1, 2, is eventually a constant sequence, that is, there exists N ∈ N such that xn = xN for all n ≥ N . Answer : False. Example : Consider X := X1 ∪ X2 , where X1 := (−∞, 0) and n X2 := [0, ∞). Then X1 ∩ X2 = φ. Let xn := (−1) , n ∈ N. n Then {xn } → 0 ∈ X as n → ∞, and there are infinitely many distinct xn ’s in X1 and in X2 . But the sequence {xn }n∈N is clearly not eventually constant.

7. If X is compact, then every sequence in X has a convergent subsequence.

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137

Answer : True. Proof . Let {xk }k∈N be any infinite sequence in X . If {xk : k ∈ N} ⊂ X is a finite set, then there is at least one K ∈ N s.t. xk = xK for infinitely many k ∈ N. That is, there is a subsequence {xkj }j∈N of {xk }k∈N with xkj = xK for all j ∈ N. In particular, {xkj }j∈N is a convergent subsequence of {xk }. If {xk : k ∈ N} ⊂ X is infinite, by the Bolzano-Weierstrass Property of the compact space X , {xk : k ∈ N} has an accumulation point p ∈ X . In particular, for any j ∈ N, there exists kj ∈ N s.t. d(xkj , p) < 1 j . Without loss of generality, we may assume that kj+1 > kj for all j ∈ N. Therefore, {xkj }j∈N is a subsequence of {xk }k∈N which is convergent (to p). 8. Every bounded sequence in Rn has a convergent subsequence. Answer : True. Proof . Let {xk }k∈N be a bounded sequence in Rn . If {xk : k ∈ N} ⊂ Rn is a finite set, then there is at least one K ∈ N s.t. xk = xK for infinitely many k ∈ N. That is, there is a subsequence {kj }j∈N of {k}k∈N with xkj = xK for all j ∈ N. In particular, {xkj }j∈N is a convergent subsequence of {xk }. If {xk : k ∈ N} ⊂ X is infinite, by Bolzano-Weierstrass Theorem, {xk : k ∈ N} has an accumulation point p ∈ Rn . In particular, for any j ∈ N, there exists kj ∈ N s.t. d(xkj , p) < 1j . Without loss of generality, we may assume that kj+1 > kj for all j ∈ N. Therefore, {xkj }j∈N is a subsequence of {xk }k∈N which is convergent (to p). 9. Let {xn }n∈N be a sequence in X. If every convergent subsequence of {xn }n∈N converges to the same point x ∈ X, then the entire sequence {xn }n∈N is convergent to x. Answer : False. Example : Let X := R and {xn }n∈N be the sequence defined by 1 if n = 2k for some k ∈ N , xn := if n = 2k − 1 for some k ∈ N . k

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It is easy to see that every convergent subsequence of {xn }n∈N converges to 1, but {xn }n∈N itself does not converge.

10. Let {xn }n∈N be a sequence in a compact space X. If every convergent subsequence of {xn }n∈N converges to the same point x ∈ X, then the entire sequence {xn }n∈N is convergent to x. Answer : True. Suppose the contrary, then there exists ε > 0 such that there are infin-

: n ∈ N} not in B(x, ε). By Exercise 2.1, Part A, Problem #7, there is a subsequence {xnk }k∈N of {xn }n∈N such that xnk ∈ / B(x, ε) for all k and {xnk }k∈N converges, ∗ say to some x ∈ X . Clearly x∗ = x, which is a contradiction.

itely many elements of {xn

11. Let X be a metric space and let d1 , d2 be two metrics on X. Let {xn }n∈N be a sequence in X. Then {xn } is convergent in (X, d1 ) if and only if it is convergent in (X, d2 ). Answer : False. Example : Let X = R, d1 be the Euclidean metric, and d2 be the 1 discrete metric. Then the sequence n is convergent in (X, d1 ) n∈N but not in (X, d2 ).

Part B: Problems 1. For each of the following sequences {xn }n∈N , find the derived set {xn : n ∈ N} and the closure {xn : n ∈ N} of the subset {xn : n ∈ N} ⊂ X. (a) xn := (1 + n1 )n ; (b) xn := (−1)n (1 + n1 ); nπ (c) xn := nsin( 2 ) ; ); (d) xn := sin( nπ 3 1/n 2 if n is odd, (e) xn := 1 if n is even.

Limits and Continuity

Answer : (a) {xn (b) {xn (c) {xn (d) {xn (e) {xn

: n ∈ N} : n ∈ N} : n ∈ N} : n ∈ N} : n ∈ N}

139

= {e}; {xn : n ∈ N} = {xn : n ∈ N} ∪ {e}. = {−1, 1}; {xn : n ∈ N} = {−1, 1}. = {0}; {xn : n ∈ N} = {xn : n ∈ N}∪ {0}. = φ; {xn : n ∈ N} = {xn : n ∈ N}. = {1}; {xn : n ∈ N} = {xn : n ∈ N}∪ {1}.

2. Construct a metric on R for which the sequence { n1 }n∈N converges to a point other than 0. Solution : It is clear that if we can construct one such metric, we can construct many. One such examples is as follows: Let f : R → R be

the bijection given by

and define

⎧ ⎪ ⎨ 1 if x = 0 x ∈ R → f (x) := 0 if x = 1 ⎪ ⎩ x otherwise , d(x, y) := |f (x) − f (y)|

for all x, y ∈ R .

It is not hard to show that d such defined is a metric on R. Note that 1 under this metric, { n }n∈N → 1 as n → ∞.

3. If {xn }n∈N and {yn }n∈N are sequences in X, {xn } → x, and {yn } → y as n → ∞, show that {d(xn , yn )} → d(x, y) as n → ∞. Proof . For any ε > 0, there exists N1 , N2 ∈ N such that d(xn , x) < ε for all n ≥ N1 and d(yn , y) < ε for all n ≥ N2 . Take N := max{N1 , N2 }. Then for any n ≥ N , we have, by triangle inequality, |d(xn , yn ) − d(x, y)|

≤ |d(xn , yn ) − d(xn , y)| + |d(xn , y) − d(x, y)|

≤ d(yn , y) + d(xn , x) < 2ε .

Thus {d(xn , yn )} → d(x, y) as n → ∞.

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4. Let X := C[0, 1] with metric d1 (f, g) :=



0

1

  f (x) − g(x) dx.

Determine whether the sequence {fn }n∈N in (X, d1 ) defined by fn (x) := xn for x ∈ [0, 1] is convergent. Answer : Yes, it is convergent in (X, d1 ). Proof . Observe that for any f , g ∈ X , d1 (f, g) is nothing but the area between the graphs of the functions f and g . Observe also that for any fixed x ∈ [0, 1], n

fn (x) = x → f (x) :=



0 if 0 ≤ x < 1

1 if x = 1 .

So the function f such defined may look like a possible candidate for the limit of the sequence {fn }n∈N . However, obviously, f ∈ / X , as

it is discontinuous at x = 1. So it cannot be the limit. However, as

d(f, 0) = the area between the graphs of f and the zero function which is obviously 0, so the zero function which is obviously in X is a good candidate for the limit of the sequence {fn }n∈N . Indeed, we have

d1 (fn , 0) =



0

1

 fn (x) − 0 dx =



0

1

xn dx =

1 →0 n+1

as n → ∞ and so we conclude that {fn }n∈N → 0 in (X, d1 ).

5. Let X := C[0, 1] with metric d∞ (f, g) := sup{|f (x) − g(x)| : x ∈ [0, 1]} . Determine whether the sequence {fn }n∈N in (X, d∞ ) defined by fn (x) := xn for x ∈ [0, 1] is convergent. Answer : No, it is not convergent in (X, d∞ ). Proof . As observed in Exercise 2.1, Part B, Problem #4, n

fn (x) = x → f (x) :=



0 if 0 ≤ x < 1 1 if x = 1 .

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So if {fn }n∈N → g ∈ X in d∞ , then for any ε > 0, there exists

N ∈ N such that for every x ∈ [0, 1],

|fn (x) − g(x)| ≤ d∞ (fn , g) < ε for all n ≥ N . In particular, for all x ∈ [0, 1), we have

|g(x)| = |f (x) − g(x)| = lim |fn (x) − g(x)| ≤ ε . n→∞

As ε > 0 is arbitrary, this shows g(x) = 0 for all x ∈ [0, 1). Now the

only possible function which is continuous on [0, 1] and = 0 on [0, 1) is

the constant zero function. So the problem boils down to determining whether the sequence {fn }n∈N converges to the zero function 0 in

d∞ . Now as

d∞ (fn , 0) = sup{|f (x) − 0| : x ∈ [0, 1]}

= sup{xn : x ∈ [0, 1]} = 1 for all n ∈ N ,

we see that {d∞ (fn , 0)}n∈N → 0 as n → ∞. Thus {fn }n∈N does

not converge in d∞ .

6. A subset A ⊂ X is said to be sequentially compact if every sequence in A has a subsequence which is convergent in A. Prove or disprove the following statements: (a) Every sequentially compact set is closed and bounded. (b) Every closed and bounded subset of Rn is sequentially compact. (c) Every closed and bounded subset of a general metric space X is sequentially compact. Answer : (a) True.

Proof . Let A ⊂ X be sequentially compact. The closedness

of A follows from the definition of sequential compactness. In fact, for any a ∈ A, there is a sequence {xn }n∈N such that

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{xn } → a as n → ∞. Now since A is sequentially compact, there is a subsequence {xnk }k∈N of {xn }n∈N which is convergent to a point b ∈ A. Since all subsequences of {xn }n∈N must converge to a, this forces a = b ∈ A and so A = A. On the other hand, assume that A is not bounded. Fix a ∈ A. There is a sequence {xn }n∈N in A satisfying d(xn , a) > n for all n ∈ N. Since A is sequentially compact, a subsequence {xnk }k∈N of {xn }n∈N must be convergent in A. By construction, {xnk }k∈N is unbounded. But this is impossible as every convergent sequence must be bounded. Hence A must be bounded. (b) True.

Proof . Let A ⊂ Rn be closed and bounded. Let {xk }k∈N be any infinite sequence in A. If {xk : k ∈ N} ⊂ Rn is finite, then there is at least one K ∈ N s.t. xk = xK for infinitely many k ∈ N. That is, there is a subsequence {xkj }j∈N of {xk }k∈N with xkj = xK for all j ∈ N. In particular, {xkj }j∈N is a convergent subsequence of {xk }k∈N . If {xk : k ∈ N} ⊂ Rn is infinite, by Bolzano-Weierstrass Theorem, {xk : k ∈ N} has an accumulation point p ∈ Rn . In particular, for any j ∈ N, there exists kj ∈ N s.t. d(xkj , p) < 1 . Without loss of generality, we may assume that kj+1 > kj j for all j ∈ N. Therefore, {xkj }j∈N is a subsequence of {xk }k∈N which is convergent (to p) and so A is sequentially compact. (c) False.

Example : Consider X := C[0, 1] with d∞ (f, g) := sup{|f (x) − g(x)| : x ∈ [0, 1]} for all f , g ∈ X and

A := B(0, 1) = {f ∈ X : d∞ (f, 0) ≤ 1} ⊂ X . Then trivially, A is bounded. On the other hand, suppose

{fn }n∈N is a sequence in A with {fn } → f ∈ X as n → ∞.

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143

So for any ε > 0, there exists N ∈ N s.t. d∞ (fn , f ) < ε for all n ≥ N . Hence

d∞ (f, 0) ≤ d∞ (f, fn ) + d∞ (fn , 0) < ε + 1 . Since ε > 0 is arbitrary, this forces d∞ (f, 0) ≤ 1 and so

f ∈ B(0, 1) = A. By Theorem 2.1.6, A is closed X . However A is not sequentially compact. To see this, consider the sequence {fn }n∈N in X defined by fn (x) := exp(−nx2 ) for all x ∈ [0, 1] . Clearly, fn ∈ A for all n ∈ N and so {fn }n∈N is a sequence

in A. Suppose {fn }n∈N has a subsequence {fnj }j∈N which is

convergent to some g ∈ A. Then there exists J ∈ N s.t.

sup{|fnj (x) − g(x)| : x ∈ [0, 1]} = d(fnj , g)
n1 such that xn2 ∈ B(p, 12 ) \ {p}. Inductively, there exists nk > nk−1 such that xnk ∈ B(p, k1 ) \ {p} for any k = 2, 3, 4, . . . . Hence in particular, {xnk }k∈N is a subsequence of {xn }n∈N which is convergent (to p). Thus X is sequentially compact. 8. Compactness implies sequential compactness. Proof . It is obvious from Theorem 1.3.6 and Exercise 2.1, Part B, Problem #7.

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2.2 Complete Metric Spaces By elementary calculus, we know that R is complete in the sense that every Cauchy sequence in R is actually convergent in R. In this section, we shall extend the concepts of Cauchyness and completeness to the context of general metric spaces. It turns out, as expected, that Rn is complete for every n ∈ N. Furthermore, as we can imagine, being a very strong condition, compactness implies completeness.

Definition 2.2.1. A sequence {xn }n∈N in X is said to be a Cauchy sequence if for every ε > 0, there exists N ∈ N such that d(xn , xm ) < ε

whenever n, m ≥ N .

Theorem 2.2.2. Every convergent sequence is Cauchy. Proof. It is obvious by triangle inequality. In fact, if {xn } → a ∈ X, then for every ε > 0, there exists N ∈ N such that d(xn , a)
0, there exists N ∈ N such that d(xn , xm ) < ε/2 whenever n, m ≥ N . As   a ∈ {xn : n ∈ N} , there exists m ≥ N such that xm ∈ B a, 2ε . Hence we have d(xn , a) ≤ d(xn , xm ) + d(xm , a)
0, there exists N ∈ N such that d(xn , xm ) < ε/2 whenever n, m ≥ N . As   a ∈ {xn : n ∈ N} , there exists m ≥ N such that xm ∈ B a, 2ε . Hence we have ε ε d(xn , a) ≤ d(xn , xm ) + d(xm , a) < + = ε whenever n ≥ N . 2 2  Thus {xn } → a as n → ∞ and so S is complete. Example 2.2.8. We show that the abstract metric space  1   |f (x) − g(x)| dx, is not complete. C[−1, 1], d , where d(f, g) := −1

Proof. For every n ∈ N, define ⎧  1 ⎪ ⎪ −1 if x ∈ − 1, − ⎪ ⎪ n ⎪ ⎨  1 1 fn (x) := nx if x ∈ − , ⎪ n n ⎪ ⎪ 1  ⎪ ⎪ ⎩1 ,1 . if x ∈ n

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Then {fn }n∈N is a Cauchy sequence in (C[−1, 1], d) which is not convergent.

In fact, suppose {fn } → g ∈ C[−1, 1] in the metric d as n → ∞. For any x0 ∈ (0, 1], if g(x0 ) = 1, there exists a nondegenerate closed interval I with x0 ∈ I ⊂ (0, 1] such that |g(x) − 1| >

|g(x0 ) − 1| >0 2

for all x ∈ I .

Note that we have fn (x) = 1 for all x ∈ I, whenever n >> . Hence d(fn , g) = ≥ =





[−1,1]

I I

>

|g(x) − fn (x)| dx

|g(x) − fn (x)| dx |g(x) − 1| dx

|g(x0 ) − 1| · (I) 2

for all n >> ,

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149

where (I) := the length of I. Since I is nondegenerate, we have (I) > 0. That means {fn } → g in d as n → ∞, which is a contradication. Therefore, we must have g ≡ 1 on (0, 1]. Similarly, we have g ≡ −1 on [−1, 0). But there is no g ∈ C[−1, 1] satisfying g(x) =



1

for x ∈ (0, 1]

−1 for x ∈ [−1, 0) .

Thus {fn }n∈N is not convergent in (C[−1, 1], d) and so (C[−1, 1], d) is not complete.

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Exercise 2.2 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. (0, 1) ⊂ R is complete.

Answer : False. Justification . It is evident that { n1 }∞ n=2 is a Cauchy sequence in (0, 1) which is not convergent there. Alternatively , this follows immediately from Exercise 2.2, Part A,

Problem #8 below.

2. Let X := C[0, 1] with metric d(f, g) := sup{|f (x) − g(x)| : x ∈ [0, 1]} .

The sequence {fn }n∈N in (X, d) given by fn (x) := xn , x ∈ [0, 1], is a Cauchy sequence. Answer : False. Justification . This follows from the observation that   d f2n , fn = sup{|f2n (x) − fn (x)| : x ∈ [0, 1]} ≥ |f2n (2−1/n ) − fn (2−1/n )|

= |(2−1/n )2n − (2−1/n )n | =

1 4

for all n ∈ N .

Hence {fn }n∈N is not Cauchy.

3. Let X := C[0, 1] with metric d(f, g) := sup{|f (x) − g(x)| : x ∈ [0, 1]} . The sequence {fn }n∈N in (X, d) given by nx if 0 ≤ x ≤ 1/n fn (x) := 1 if 1/n < x ≤ 1 is a Cauchy sequence.

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151

Answer : False. Justification . This follows from the observation that   d f2n , fn = sup{|f2n (x) − fn (x)| : x ∈ [0, 1]}     1   1  − fn ≥ f2n 2n 2n  = 1−

1 n = 2n 2

for all n ∈ N .

Hence {fn }n∈N is not Cauchy.

4. Let {xn }n∈N be a sequence in X and Em := {xn : n ≥ m} ⊂ X. Then {xn }n∈N is Cauchy if and only if limm→∞ d(Em ) = 0. Answer : True. Proof . As Em+1 ⊂ Em for all m ∈ N, we have lim d(Em ) = 0

m→∞

⇐⇒ for all ε > 0, there exists N ∈ N s.t. d(EN ) < ε

⇐⇒ for all ε > 0, there exists N ∈ N s.t.

d(xn , xm ) < ε for all n, m ≥ N

⇐⇒ {xn }n∈N is Cauchy .

5. Let {xn }n∈N be a Cauchy sequence in X. Then for any n, m ∈ N large enough, we have d(xn+1 , xm+1 ) ≤ d(xn , xm ) . Answer : False. Example : Let X = R with the Euclidean metric. Define xn :=



0

if n is odd

1 n

if n is even.

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Then {xn } → 0 and hence in particular, it is Cauchy. However, whenever n, m ∈ N are odd and distinct, no matter how large they are, we have

   1 1   − > 0 = d(xn , xm ) . d(xn+1 , xm+1 ) =  n + 1 m + 1 6. Every Cauchy sequence is bounded. Answer : True. Proof . The proof is similar to that of Theorem 2.1.4 (a). In fact, let {xn }n∈N be a Cauchy sequence in X . Then there exists and N ∈ N such that d(xm , xN ) < 1, or equivalently, xm ∈ B(xN , 1) for all m ≥ N . Outside the ball B(xN , 1) there can be at most finitely many points of {xn : n ∈ N}. Hence the whole set {xn : n ∈ N} is lying in a ball centered at xN with finite radius. 7. The “term-wise product” of Cauchy sequences in R is Cauchy, that is, if {xn }n∈N and {yn }n∈N are Cauchy sequences in R, then so is {xn yn }n∈N . Answer : True. Proof . It should be evident from the inequality

|xn yn − xm ym | = |xn yn − xn ym + xn ym − xm ym |

≤ |xn − xm | |ym | + |yn − ym | |xm | .

In fact, by Exercise 2.2, Part A, Problem #6, there exists K1 , K2 > 0 such that |xn | ≤ K1 and |yn | ≤ K2 for all n ∈ N. Let K :=

max{K1 , K2 }. Then we have |xn |, |yn | ≤ K for all n ≥ N . On the other hand, by the definition Cauchyness, for any ε > 0, there exist N1 , N2 ∈ N such that ε 2K ε |yn − ym | < 2K

|xn − xm |
0} and S := (0, ∞) × {0} = {(x, 0) : x > 0} ⊂ X . Then it is clear that S is a closed subset of X . Furthermore, {( n1 , 0)}n∈N is a Cauchy sequence in S which is not convergent in X . Hence S is not complete. 11. Every closed and bounded subset of a metric space is complete. Answer : False.

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√ √ Example : Let X := Q and S := [ 2, 3] ∩ Q ⊂ Q. Then S is closed and bounded in X but it is not complete. In fact, it is easy √ to construct a sequence {xn }n∈N in S which converges to 2 in R. Hence it is a Cauchy sequence in S which is not convergent in S . 12. Every closed subset of a complete metric space is complete. Answer : True. Proof . Let X be complete and Y ⊂ X be closed. Let {xn }n∈N be a Cauchy sequence in Y . As the metric in Y is induced by that in X , when considered as a sequence in X , {xn }n∈N is also Cauchy. Since X is complete, {xn }n∈N converges to some x ∈ X as n → ∞. Note that x ∈ Y = Y . Therefore, {xn }n∈N → x ∈ Y as n → ∞ and so Y is complete. 13. The metric space X := (0, 1] equipped with metric   1 1   x, y ∈ X , d(x, y) :=  −  , x y is a complete metric space.

Answer : True. Proof . Let {xn }n∈N be a Cauchy sequence in (X, d). By definition, for any ε > 0, there exists N ∈ N such that    1 1    xn − xm  = d(xn , xm ) < ε for all n, m ≥ N .   In particular, x1n is a Cauchy sequence in [1, ∞) with the usual n∈N

Euclidean metric, which is, by Exercise  2.2, Part A, Problem #12, a 1 → some y ∈ [1, ∞) in the complete metric space. Hence xn n∈N

usual Euclidean metric, as n → ∞. But then that also means

1 d xn , y



that is, {xn }n∈N →

1 y



   1   = − y  → 0 xn

in d as n → ∞.

as n → ∞ ,

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155

1  14. The metric space N equipped with metric d(m, n) :=  m − n1  is a complete metric space. Answer : False. Justification . For any n ∈ N, let xn := n. It is not hard to see that {xn }n∈N is a Cauchy sequence in (N, d). However, for any n0 ∈ N,   1  1  → 1 = 0 d(xn , xn0 ) = d(n, n0 ) =  − as n → ∞ . n n0  n0 Thus {xn }n∈N is not convergent and so (N, d) is not complete.

15. The metric space N equipped with metric   d(m, n) := e−m − e−n 

m, n ∈ N

is a complete metric space.

Answer : False. Justification . For any n ∈ N, let xn := n. For any ε > 0, let N ∈ N be such that e−N < ε. Then   d(xm , xn ) = d(m, n) = e−m − e−n 

≤ e−m + e−n ≤ 2e−N < 2ε for all m, n ≥ N

and so {xn }n∈N is Cauchy in (N, d). However, for any n0 ∈ N,

  d(xn , n0 ) = d(n, n0 ) = e−n − e−n0  → e−n0 = 0 as n → ∞ .

Thus {xn }n∈N is not convergent and so (N, d) is not complete.

Part B: Problems 1. Let (X, d1 ) be a metric space and d2 (x, y) := min{1, d1 (x, y)} for any x, y ∈ X. It is easily verified that d2 is a well-defined metric on X. Let {xn }n∈N be a sequence in X. Show that it is Cauchy in d1 if and only if it is Cauchy in d2 .

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Proof . (⇒): Suppose {xn }n∈N is Cauchy in d1 . Then for any 1 > ε > 0, there exists N ∈ N such that d1 (xn , xm ) < ε for any n, m ≥ N . It follows that d2 (xn , xm ) = min{1, d1 (xn , xm )} < ε for all n, m ≥ N and so {xn }n∈N is Cauchy in d2 .

(⇐): Suppose {xn }n∈N is Cauchy in d2 . Then for any 1 > ε > 0, there exists N ∈ N such that min{1, d1 (xn , xm )} = d2 (xn , xm ) < ε for all n, m ≥ N . It follows that

d1 (xn , xm ) < ε

for all n, m ≥ N

and so {xn }n∈N is Cauchy in d1 .

2. Recall that a subset S ⊂ X is said to be totally bounded (Exercise 1.3, Part B, Problem #10) if for every ε > 0, there exist k > 0 and x1 , . . . , xk ∈ S such that {B(xi , ε)}ki=1 is an open cover of S. Show that the underlying set of every Cauchy sequence {xn }n∈N in X is totally bounded, that is, the set S := {xn : n ∈ N} ⊂ X is totally bounded. In particular, S is bounded. Proof . Let ε > 0 be given. Since {xn }n∈N is Cauchy, there exists N ∈ N such that d(xn , xm ) < ε for all n, m ≥ N . In particular, xn ∈ B(xN , ε) for all n ≥ N . Hence {B(x1 , ε), . . . , B(xN , ε)} is an open cover of S . Therefore, S is totally bounded. The last statement is obvious.

3. Show that every infinite sequence of distinct points {xn }n∈N in a totally bounded subset S ⊂ X contains a non-constant Cauchy subsequence.

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157

Proof . Since S is totally bounded, it can be decomposed into finitely many subsets each of which has diameter less than 1. One of these finitely many subsets would contain infinitely many xn ’s. Call it S1 . Pick an n1 ∈ N such that xn1 ∈ S1 . Since S1 ⊂ S and S is totally bounded, S1 is also totally bounded. So S1 can be decomposed into finitely many subsets each of which has diameter less than

1 2.

One

of these subsets, say S2 , contains infinitely many xn ’s. Hence there exists n2 ∈ N, n2 > n1 , such that xn2 ∈ S2 . Inductively, we have a decreasing nest of subsets

S ⊃ S1 ⊃ S2 ⊃ S3 ⊃ · · · with d(Sj )
0, let N ∈ N be large enough such that N < ε. Then for any p, q ≥ N , we have xnp ∈ Sp ⊂ SN , xnq ∈ Sq ⊂ SN , and so d(xnp , xnq ) ≤ d(SN )
0, there exists N ∈ N such that d(Sn ) < ε for all n ≥ N . Hence for any n ≥ m ≥ N , d(xn , xm ) ≤ d(Sm ) < ε and so {xn }n∈N is Cauchy.

5. For any sequences {xn }n∈N , {yn }n∈N in X, if {xn }n∈N is Cauchy and limn→∞ d(xn , yn ) = 0, show that {yn }n∈N is also Cauchy.

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Proof . For any ε > 0, there exists N1 ∈ N such that d(xn , yn ) < ε for all n ≥ N1 . On the other hand, there exists N2 ∈ N such that d(xn , xm ) < ε for all n, m ≥ N2 . Take N := max{N1 , N2 }. Then

by triangle inequality,

d(yn , ym ) ≤ d(yn , xn ) + d(xn , xm ) + d(xm , ym ) < 3ε for any n, m ≥ N . Hence {yn }n∈N is Cauchy.

6. For any sequences {xn }n∈N , {yn }n∈N in X, if limn→∞ xn = a and limn→∞ d(xn , yn ) = 0, show that limn→∞ yn = a. Proof . For any ε > 0, there exists N1 ∈ N such that d(xn , yn ) < ε for all n ≥ N1 . On the other hand, there exists N2 ∈ N such that d(xn , a) < ε for all n ≥ N2 . Take N := max{N1 , N2 }. Then by

triangle inequality,

d(yn , a) ≤ d(yn , xn ) + d(xn , a) < 2ε for any n ≥ N . Hence {yn } → a as n → ∞.

7. Consider X := C[0, 1] with metrics d∞ (f, g) := sup |f (x) − g(x)|

and

x∈[0,1]

d1 (f, g) :=



1 0

|f (x) − g(x)| dx .

For each n ∈ N, define fn (x) :=



2n − 2n2 x 0

if 0 ≤ x ≤ 1/n if 1/n < x ≤ 1 .

Discuss whether the sequence {fn }n∈N is Cauchy in (X, d∞ ) and in (X, d1 ).

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Solution : {fn }n∈N is not Cauchy in either (X, d∞ ) or (X, d1 ).

First observe that

d∞ (fn , 0) = sup |2n − 2n2 x| = 2n for all n ∈ N 1 x∈[0, n ]

and so the sequence {fn }n∈N is not bounced, hence in particular, not

Cauchy.

On the other hand, for any n, k ∈ N, elementary calculation shows

d1 (fn+k , fn ) = In particular, d1 (f2n , fn ) = Cauchy.

2 3

2k . 2n + k

for any n ∈ N and so {fn }n∈N is not

8. Suppose {xn }n∈N is a sequence in X such that the underlying set {xn : n ∈ N} ⊂ X is finite. Show that if {xn }n∈N is Cauchy, then it must be convergent. Proof . Under the given assumption, exactly one element of the finite set {xn : n ∈ N} is assumed by the sequence {xn }n∈N infinitely many times. Hence this particular element is the limit of the sequence

{xn }n∈N .

In more details, first observe that if the sequence {xn }n∈N is a constant sequence, then there is nothing to prove. So assume that it is not a constant sequence. Let

r := inf{d(xn , xm ) : n, m ∈ N}. Then as there can be at most finitely many different values of the

xn ’s, we have r > 0. Since the sequence {xn }n∈N is Cauchy, there

exists N ∈ N such that

d(xn , xm )
0, there exists N1 ∈ N such that d(xnk , x)
nk ≥ k ≥ N and so d(xk , p) ≤ d(xk , xnj ) + d(xnj , p) < ε + d(xnj , p) . Letting j → ∞, we have

d(xk , p) ≤ ε

for all k ≥ N

and so {xk } → p as k → ∞. Hence completeness.

12. Show that every totally bounded complete metric space is sequentially compact. Proof . Let X be totally bounded and complete, and {xn }n∈N be any sequence in X . If {xn : n ∈ N} is finite, we have seen before that it

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must have a constant subsequence and so the result is trivial. Hence we only need to consider the case where {xn : n ∈ N} is an infinite

set. For this, we simply assume that all xn ’s are distinct. Since X is

totally bounded, finitely many open balls of radius 1 cover X . One of these open balls, say B(y0 , 1), would cover infinitely many xn ’s. So if we write Λ0 := {n ∈ N : xn ∈ B(y0 , 1)} ⊂ N, we have

#(Λ0 ) = ∞. Similarly, finitely many open balls of radius 12 cover X . One of these balls, say B(y1 , 12 ), would cover infinitely many xn ’s with n ∈ Λ0 . Write Λ1 := {n ∈ Λ0 : xn ∈ B(y1 , 12 )} ⊂ Λ0 . Then we have #(Λ1 ) = ∞. Inductively, for every k ∈ N, we have an open ball B(yk , 21k ) containing infinitely many xn ’s with n ∈ Λk−1 . Write Λk := {n ∈ Λk−1 : xn ∈ B(yk , 21k )} ⊂ Λk−1 . Then #(Λk ) = ∞. In other words, we have constructed a decreasing nest of index sets Λk , each of which is infinite. So we can pick a strictly increasing sequence of natural numbers {nk }k∈N such that nk ∈ Λk for each k ∈ N. Observe that the corresponding sequence {xnk }k∈N is a Cauchy subsequence of {xn }n∈N . In fact, for any k , p ∈ N, we have nk ∈ Λk and nk+p ⊂ Λk+p ⊂ Λk and so xnk , xnk+p ∈ B(yk , 21k ). Therefore,

d(xnk , xnk+p ) ≤ d(xnk , yk ) + d(yk , xnk+p ) 1 1 1 < k + k = k−1 , 2 2 2 from which we see that the subsequence {xnk }k∈N of {xn }n∈N is Cauchy. Since X is complete, {xnk }k∈N is convergent. Hence X is

sequentially compact.

13. Recall that an open cover U of X is said to have a Lebesgue number λ > 0 if every subset of X of diameter less than λ is contained in some element in U . Show that if X is a sequentially compact metric space, then every open cover of X has a Lebesgue number. (Compare with Exercise 1.3, Part B, Problem #11.)

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Proof . Suppose X is sequentially compact. Assume to the contrary that there is an open cover U = {Uα }α∈Λ of X , where Λ is an arbi-

trary index set, without a Lebesgue number. Then for any

n ∈ N, there exists Sn ⊂ X with d(Sn ) < n1 and Sn ⊂ Uα for any α ∈ Λ. Pick a representative element xn ∈ Sn for each n ∈ N. Then as a sequence in the sequentially compact space X , {xn }n∈N has a convergent subsequence {xnk }k∈N , converging to, say, p ∈ X . As U covers X , there exists α0 ∈ Λ such that p ∈ Uα0 . Since Uα0 is open, there exists r > 0 such that p ∈ B(p, r) ⊂ Uα0 . As {xnk } → p, there exists N ∈ N such that d(xnk , p)
N large enough such that |fm (x) − f (x)| < 2ε . Hence |fn (x) − f (x)| ≤ |fn (x) − fm (x)| + |fm (x) − f (x)| ≤ ρ(fn , fm ) + |fm (x) − f (x)| < ε

for all n ≥ N and so ρ(fn , f ) = supx∈X |fn (x) − f (x)| ≤ ε for all n ≥ N . Since ε > 0 is arbitrary, we conclude that

{ρ(fn , f )} → 0 as n → ∞. Hence {fn }n∈N → f in ρ as n → ∞. 17. Let M22 be the space of all 2 × 2 matrices over R, equipped with the metric d(A, B) := max |aij − bij | 1≤i,j≤2

for all A = (aij ), B = (bij ) ∈ M22 . Show that (M22 , d) is complete.

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Proof . It is elementary to verify that d is a well-defined metric on M22 . Let {An = (anij )}n∈N be a Cauchy sequence in (M22 , d). By definition, for any ε > 0, there exists N ∈ N such that max |anij − am ij | = d(An , Am ) < ε whenever n, m ≥ N .

1≤i,j≤2

It follows that for any fixed pair (i, j),

|anij − am ij | < whenever n, m ≥ N. Hence {an ij }n∈N is a Cauchy sequence in R, which must be convergent.

Let aij := limn→∞ an ij . Then A := (aij ) is the limit of An in d. In fact, for any ε > 0, for any i, j = 1, 2, there exists Nij ∈ N such

that

|anij − aij | < ε whenever n ≥ Nij . Take N := max1≤i,j≤2 Nij . Then

d(An , A) = max |anij − aij | < ε whenever n ≥ N . 1≤i,j≤2

18. Determine whether the metric space C([−1, 1], d) with metric d(f, g) := is complete.



1 −1

  f (x) − g(x) dx

for any f, g ∈ C[−1, 1]

Answer : It is not complete. Justification . Observe first that for any f , g ∈ C[−1, 1], d(f, g) = the area of the region bounded by the graphs of f and g . Consider the sequence of functions

⎧ ⎪ ⎨ −1 −1 ≤ x ≤ −1/n, fn (x) := nx −1/n ≤ x ≤ 1/n, ⎪ ⎩ 1 1/n ≤ x ≤ 1 .

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It is clear that fn ∈ C[−1, 1] for all n ∈ N. Furthermore, it is easy



1 to verify that d(fn , fm ) =  n −

large enough such that

1 K



1 m .

< ε. Then

For any ε > 0, let K ∈ N be

     m − n 1 1 < 1 ≤ 1 n ≥ K and so {fn }n∈N is a Cauchy sequence in

C([−1, 1], d).

To show that {fn }n∈N is not convergent, we use contra-positive

argument. Suppose that {fn } → f in C([−1, 1], d) as n → ∞. By intuition, we should have f ≡ 1 on (0, 1]. This can easily be jus-

tified. In fact, suppose f (x0 ) = 1 for some x0 ∈ (0, 1]. Denote by

α := |f (x0 ) − 1| > 0. By the continuity of f , there exists a closed non-degenerate sub-interval [a, b ] ⊂ (0, 1] containing x0 such that |f (x) − 1| > α/2 for all x ∈ [a, b ]. However, as fn ≡ 1 on [a, b ] whenever n ≥ a1 , we have d(fn , f ) = =



1

−1  b a

|fn (x) − f (x)| dx ≥

|f (x) − 1| dx ≥



b a

|fn (x) − f (x)| dx

α 1 (b − a) > 0 for all n ≥ , 2 a

which contradicts to the convergence of {fn } to f in C([−1, 1], d).

Similarly, we also have f ≡ −1 on [−1, 0). However, no f ∈ C[−1, 1] can have such values.

19. Suppose {Fn }n∈N is a decreasing nest of closed nonempty subsets of a complete metric space X, with limn→∞ d(Fn ) = 0. ∞ (a) Show that n=1 Fn = φ. ∞ (b) Determine what n=1 Fn is. (c) If the Fn ’s are not all closed, would (a) still hold? (d) Without the condition limn→∞ d(Fn ) = 0, would (a) still hold?

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(e) If the condition limn→∞ d(Fn ) = 0 is replaced by that all Fn ’s are bounded, would (a) still hold? (f) If X fails to be complete, would (a) still hold? Solution : (a) Since {d(Fn )}n∈N → 0 as n → ∞, for any ε > 0, there

exists N ∈ N such that d(FN ) < ε. For each n ∈ N, pick

xn ∈ Fn . As {Fn }n∈N is decreasing, for any n, m ≥ N , we have xm ∈ Fm ⊂ FN and xn ∈ Fn ⊂ FN . Hence d(xm , xn ) ≤ d(FN ) < ε . Thus {xn }n∈N is a Cauchy sequence in the complete metric

space X and so {xn }n∈N → some x ∈ X as n → ∞. In

particular, for any m ∈ N, {xn }∞ n=m → x as n → ∞ and

so x ∈ {xn : n ≥ m} ⊂ Fm = Fm for all m ∈ N. Hence (b)

x∈ ∞

∞

n=1

Fn .

Fn is a singleton. ∞ To see this, let x, y ∈ n=1 Fn . For any ε > 0, there exists N ∈ N such that d(FN ) < ε. Since x, y ∈ FN , it follows that d(x, y) ≤ d(FN ) < ε. As ε > 0 is arbitrary, we have d(x, y) = 0, which means x = y . Hence the result. n=1

(c) No.

  Example : Consider X := R and Fn := 0, n1 , n ∈ N. Then Fn = φ for all n ∈ N, {Fn } ↓ with limn→∞ d(Fn ) = 0, but ∞ n=1 Fn = φ. Note that Fn ’s are not closed.

(d) No.

Example : Consider X := R and Fn := [n, ∞), n ∈ N. Then ∞ each Fn is closed and nonempty, {Fn } ↓, but n=1 Fn = φ. Note that in this case limn→∞ d(Fn ) = 0.

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(e) No.

Example : Consider X := R with discrete metric and   Fn := 0, n1 , n ∈ N. Then each Fn is closed and nonempty, and {Fn } ↓. Furthermore, since d(Fn ) = 1 for all n ∈ N, all  Fn ’s are bounded. But ∞ n=1 Fn = φ.

(f) No.

Example : Consider X := [−1, 0) ∪ (0, 1] with the usual Euclidean metric. Note that X is not complete. For any n ∈   N, let Fn := − n1 , 0) ∪ (0, n1 . Then each Fn is closed and ∞ nonempty, {Fn } ↓, with limn→∞ d(Fn ) = 0. But n=1 Fn = φ. ∞ 20. Let X be a metric space. If n=1 Fn = φ for any decreasing nest of closed subsets {Fn }n∈N of X with limn→∞ d(Fn ) = 0, show that X is complete. Proof . Let {xn }n∈N be a Cauchy sequence in X . Denote by Em := {xn : n ≥ m}. By Exercise 2.2, Part A, Problem #4, we have limm→∞ d(Em ) = 0. By Exercise 1.2, Part B, Problem #8, we have limm→∞ d(Em ) = 0. Applying the given condition to the decreas∞ ing nest of closed subsets {Em } of X , we have m=1 Em = φ. But then as limm→∞ d(Em ) = 0, by Exercise 2.2, Part B, Problem #19, ∞ m=1 Em must be a singleton, say = {a}. In particular, we have a ∈ Em for all m ∈ N. Now again by limm→∞ d(Em ) = 0, for any ε > 0, there exists N1 ∈ N such that d(xm , a) ≤ d(E m ) < ε/2 for all m ≥ N1 . On the other hand, by the Cauchyness of {xn }n∈N , there exists N2 ∈ N such that d(xn , xm ) < ε/2 for all n, m ≥ N2 . Take N := max{N1 , N2 }. Then for any n, m ≥ N , we have d(xn , a) ≤ d(xn , xm ) + d(xm , a) < ε . Hence limn→∞ xn = a and so X is complete.

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21. For any metric space X, show that every sequence in X has a Cauchy sub-sequence if and only if for every r > 0, there exist n points x1 , . . . , xn ∈ X, such that j=1 B(xj , r) = X. Proof . (⇐): Let {yn }n∈N be any sequence in X . By assumption, N there exist points x1 , . . . , xN ∈ X , such that j=1 B(xj , 12 ) = X . So there is j ∈ {1, . . . , N } such that B(xj , 12 ) contains infinitely many terms of the sequence {yn }n∈N . Hence there is a subsequence ∞ {y1,n }∞ n=1 of {yn }n=1 with d(y1,p , y1,q ) ≤ d(y1,p , xj ) + d(y1,q , xj ) < 1 for all p, q ∈ N . Similarly, since X is a union of finitely many balls of radius 14 , we can pick one of these balls which contains infinitely many terms of {y1,n },

∞ and in turn we can find a subsequence {y2,n }∞ n=1 of {y1,n }n=1 with

d(y2,p , y2,q )
0, there exists δ > 0 such that   d f (x), b < ε

whenever x ∈ S \ {p} and d(x, p) < δ ,

that is,

f (x) ∈ B(b, ε) or

whenever x ∈ B(p, δ) ∩ S \ {p} ,

  f (B(p, δ)) ∩ S \ {p} ⊂ B(b, ε) .

Theorem 2.3.2. Let p ∈ X be an accumulation point of S ⊂ X, b ∈ Y , and f : S → Y be any function. Then lim f (x) = b

x→p

if and only if lim f (xn ) = b

n→∞

for every sequence {xn }n∈N in S \ {p} which converges to p.

Proof. (⇒) As limx→p f (x) = b, by definition, for any ε > 0, there exists δ > 0 such that f (B(p, δ) ∩ S \ {p}) ⊂ B(b, ε) . So if {xn }n∈N is a sequence in S \ {p} such that {xn } → p as n → ∞, then there exists N ∈ N such that xn ∈ B(p, δ) ∩ S \ {p}

whenever

n≥N

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and so

  f (xn ) ∈ f B(p, δ) ∩ S \ {p} ⊂ B(b, ε)

whenever n ≥ N ,

that is, {f (xn )} → b as n → ∞.

(⇐) Suppose f (x) → b as x → p. Then there exists ε > 0 so that for each n ∈ N we can find xn ∈ S \ {p} with 0 < d(xn , p)
0 there exists δ > 0 such that

that is,

  d f (x), f (p) < ε

whenever x ∈ S with d(x, p) < δ ,

f (x) ∈ B(f (p), ε) or

whenever x ∈ B(p, δ) ∩ S ,

    f B(p, δ) ∩ S ⊂ B f (p), ε .

Note that if p is an accumulation point of S, then f is continuous at p if and only if lim f (x) = f (p) , x∈S x→p

and if p is an isolated point of S, that is, if p ∈ S \ S  , the condition is automatically satisfied and so f is automatically continuous at p by default. f is said to be continuous on S if it is continuous at every p ∈ S.

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Theorem 2.3.5. Let S ⊂ X be a subset, p ∈ S, and f : S → Y be a function. Then the following conditions are equivalent: (a) f is continuous at p. (b) limn→∞ f (xn ) = f (p) whenever {xn }n∈N is a sequence in S with limn→∞ {xn } = p. Proof. If p ∈ S  , the Theorem follows immediately from definition  and Theorem 2.3.2. If p ∈ S \ S  , the assertion is trivial. Example 2.3.6. function on Rn .

The Euclidean norm · : Rn → R is a continuous

Proof. This is a simple consequence of Theorem 2.3.3. To see this, let f : Rn → Rn be the identity function f (x) := x, x ∈ Rn . Then for any p ∈ Rn , it is obvious that limx→p f (x) = limx→p x = p and so by Theorem 2.3.3, we have lim x = lim f (x) = p ,

x→p

x→p

which means  ·  is continuous at p. Since p ∈ Rn is arbitrary,  ·  is continuous on Rn . Alternatively, the assertion can also be proved by direct verification from first principle. This will be left to the readers as an exercise. Example 2.3.7. If (X, d) is a metric space, then the metric d is a function d : X × X → R, and we can talk about the continuity of the function d once X × X is equipped with a metric. For example, it is easy to check that   ρ (x1 , y1 ), (x2 , y2 ) := d(x1 , x2 ) + d(y1 , y2 )

is a well-defined metric on X × X. For any ε > 0 and any (x0 , y0 ), (x, y) ∈ X × X, since d(x, y) ≤ d(x, x0 ) + d(x0 , y0 ) + d(y0 , y) ,

d(x0 , y0 ) ≤ d(x0 , x) +

d(x, y) + d(y, y0 ) ,

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we have   |d(x, y) − d(x0 , y0 )| ≤ d(x, x0 ) + d(y, y0 ) < 2ρ (x, y), (x0 , y0 ) .

Hence if we take δ := ε/2, then

  ρ (x, y), (x0 , y0 ) < δ =⇒ |d(x, y) − d(x0 , y0 )| < ε

and so d : X ×X → R is continuous at (x0 , y0 ). Since (x0 , y0 ) ∈ X ×X is arbitrary, d is continuous on (X × X, ρ).

Alternatively, suppose {(xn , yn )}n∈N is a sequence in X × X with {(xn , yn )} → (x, y) in ρ as n → ∞. To show the continuity of d : (X × X, ρ) → R, it suffice to show d(xn , yn ) → d(x, y)

as n → ∞ .

Now observe that (xn , yn ) → (x, y) in ρ as n → ∞   ⇐⇒ d(xn , x) + d(yn , y) = ρ (xn , yn ), (x, y) → 0 as n → ∞ .

Since

d(x, y) ≤ d(x, xn ) + d(xn , yn ) + d(yn , y)

d(xn , yn ) ≤ d(xn , x) + d(x, y) + d(y, yn ) , we have   d(xn , yn ) − d(x, y) ≤ d(x, xn ) + d(y, yn ) → 0 as n → ∞ . Hence d(xn , yn ) → d(x, y) as n → ∞ and so d is continuous.

Remark. Observe that similar arguments can show that d : (X × X, ρ ) → R is also continuous, where the metric ρ is defined by    ρ (x1 , y1 ), (x2 , y2 ) := d(x1 , x2 )2 + d(y1 , y2 )2 .

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Theorem 2.3.8. If f : X → Y is continuous at p ∈ X and g : Y → Z is continuous at f (p) ∈ Y , then g ◦ f : X → Z is continuous at p. Proof. For any ε > 0, since g is continuous at f (p), there exists δ1 > 0 such that     g B(f (p), δ1 ) ⊂ B g ◦ f (p), ε .

On the other hand, since f is continuous at p, there exists δ > 0 such that     f B(p, δ) ⊂ B f (p), δ1 .

Hence       g ◦ f B(p, δ) ⊂ g B(f (p), δ1 ⊂ B g ◦ f (p), ε .



Theorem 2.3.9. (a) If f , g : X → C (or R) are continuous at p, then so are f + g, f − g, f g, and f /g (provided that g(p) = 0).

(b) If f , g : X → Cn (or Rn ) are continuous at p, then so are f + g, λf , f · g, and f  for any λ ∈ C (or R).

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(c) Let fi : X → C or R be functions on X, i = 1, . . . , n. Then f := (f1 , . . . , fn ) : X → Cn (or Rn )

is continuous at p if and only if fi is continuous at p for each i = 1, . . . , n. Proof. It is elementary and will be left as an exercise.



Remark. It is clear that the identity function of C to C (or R to R) is continuous. Hence Theorem 2.3.9 gives rise to plenty of continuous functions, for example, polynomials in x, etc. Remark. By now the readers should be fairly familiar with the interplay between open balls and open neighborhoods. Hence for the sake of simplicity, from this point on, sometimes we will talk about open balls centered at a point p and open neighborhoods of p interchangeably. So for example, f : X → Y is continuous at p ∈ X if for every open neighborhood V of f (p) in Y , there exists an open neighborhood U of p in X such that f (U ) ⊂ V . Same as for other properties, to verify continuity of a function by definition could be tedious and time-consuming, not to mention that such a process in general may not lead to any insightful picture. So as usual, we will try and establish handy necessary and/or sufficient conditions for continuity which could lead to useful characterization and/or properties of continuity. Theorem 2.3.10. f : X → Y is continuous on X if and only if the inverse image f −1 (T ) of any open set T ⊂ Y is open in X.

Proof. (⇒) Let T ⊂ Y be open and p ∈ f −1 (T ). Write q := f (p) ∈ T . Since T is open in Y , there exists ε > 0 such that B(q, ε) ⊂ T .   Since f is continuous at p, there exists δ > 0 such that f B(p, δ) ⊂ B(q, ε). Hence   B(p, δ) ⊂ f −1 B(q, ε) ⊂ f −1 (T )

and so p ∈ f −1 (T )◦ . Hence f −1 (T ) = f −1 (T )◦ is open.

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(⇐) Let p ∈ X and q := f (p). For any ε > 0, since B(q, ε) ⊂ Y   is open, f −1 B(q, ε) ⊂ X is open. Since f (p) = q, we have p ∈   f −1 B(q, ε) and so there exists δ > 0 such that p ∈ B(p, δ) ⊂     f −1 B(q, ε) and thus f B(p, δ) ⊂ B(q, ε). Hence f is continuous at p. Since p ∈ X is arbitrary, f is continuous on X.  Theorem 2.3.11. f : X → Y is continuous on X if and only if the inverse image f −1 (T ) of any closed set T ⊂ Y is closed in X. Proof. This follows from Theorem 2.3.10 and the fact that f −1 (Y \ T ) = X \ f −1 (T ) . In fact, f is continuous ⇔ f −1 (S) ⊂ X is open for any open S ⊂ Y

⇔ f −1 (Y \ T ) ⊂ X is open for any closed T ⊂ Y 

⇔ X \ f −1 (T ) ⊂ X is open for any closed T ⊂ Y .  Remark. In the proof of Theorem 2.3.11, we have made used of the fact that f −1 (Y \ T ) = X \ f −1 (T ) . However, for direct images of sets we only have f (X) \ f (S)  f (X \ S) and the quality holds only when f is 1 − 1. Details will be left as an exercise. Remark. At first sight, the two preceding Theorems on equivalence conditions for continuity may seem to be a little

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unpleasant, and it may look a lot more pleasant in case we have a result saying that continuous functions map open sets to open sets, and closed sets to closed sets, and vice versa. Unfortunately, it is not the case. In general, continuous functions may not map open sets to open sets. For example, the constant function f : R → R given by f (x) := 0 for all x ∈ R is clearly continuous. However, the open set (0, 1) ⊂ R is being mapped to {0} ⊂ R which is not open in R. Similarly, continuous functions may not map closed sets to closed sets. For example, the identity function f : (0, 1) → R given by f (x) := x for all x ∈ R is clearly continuous. However, (0, 1) is closed in (0, 1) while f ((0, 1)) = (0, 1) is not closed in R. In general, a function which maps open sets to open sets is called an open mapping. A function which maps closed sets to closed sets is called a closed mapping. In general, open mappings may not be closed, closed mappings may not be open, open mappings may not be continuous, and closed mappings may not be continuous either. Interested readers are referred to Exercise 2.3, Part A, Problem #26, 28, 29, 30, 31, 32. Having said that, however, openness, closedness, and continuity are not completely unrelated. For example, the following is a nice relation among them. Other examples can be found in Exercise 2.3, Part A, Problem #33 and Exercise 2.3, Part B, Problem #3. Example 2.3.12. If f : X → Y is 1 − 1, then f −1 : f (X) → X exists, and the following statements are equivalent: (a) f is continuous; (b) f −1 is open; (c) f −1 is closed. Proof. First of all, let us set the language straight. In general, for an arbitrary function f : X → Y , f −1 (S) stands for the pre-image of the set S ⊂ Y under f , or more precisely, f −1 (S) := {x ∈ X : f (x) ∈ S} .

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In case f : X → Y is 1 − 1, then f : X → f (X) is bijective and so there is an inverse function mapping f (X) back onto X. As usual, it will be denoted by f −1 : f (X) → X. So if S ⊂ f (X) is a subset of f (X), the inverse function f −1 will map the set S to a set f −1 (S) ⊂ X. So then f −1 (S) is having two meanings. One, it is the pre-image of the set S ⊂ f (X) ⊂ Y under f , and two, it is the image of the set S ⊂ f (X) under the function f −1 . This may seem a bit confusing, but then the good news is, whichever meaning we choose, f −1 (S) represents the same set and so there should indeed be no ambiguity. (a)⇒(b): Let S ⊂ f (X) be open. Then there is an open set U ⊂ Y such that S = U ∩ f (X). Since f is continuous, by Theorem 2.3.10,   f −1 (S) = f −1 (U ∩ f (X)) = f −1 (U ) ∩ f −1 f (X) = f −1 (U ) ∩ X = f −1 (U ) ⊂ X

is open .

Thus f −1 is an open mapping. (b)⇒(c): Let C ⊂ f (X) be closed. Then f (X) \ C ⊂ f (X) is open.   Since f −1 is an open mapping, f −1 f (X) \ C ⊂ X is open and so   f −1 (C) = X \ f −1 f (X) \ C

is closed. Hence f −1 is a closed mapping.

(c)⇒(a): If C ⊂ Y is closed, then C ∩f (X) ⊂ f (X) is closed in f (X). Since f −1 is a closed mapping, it maps closed sets in f (X) to closed sets in X. In particular, f −1 (C) = f −1 (C ∩ f (X)) ⊂ X By Theorem 2.3.11, f is continuous.

is closed in X . 

The interested readers are encouraged to establish the other implications (b)⇒(a), (c)⇒(b), and (a)⇒(c) of Example 2.3.12 directly. These will help achieving a better understanding of the relations among the 3 conditions.

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Theorem 2.3.13. Continuous functions map compact sets to compact sets. Proof. Let f : X → Y be continuous and C ⊂ X be a compact set. Let U = {Uα }α∈Λ be an open cover of f (C) in Y . Since f is   continuous, f −1 (U ) := f −1 (Uα ) α∈Λ is an open cover of C. In fact, f (C) ⊂

=⇒ C ⊂ f



Uα

α∈Λ

−1



 f (C) ⊂ f −1





Uα

α∈Λ



=



f −1 (Uα ) .

α∈Λ

Since C is compact, the open cover f −1 (U ) of C has a finite n  sub-cover, say f −1 (Ui ) i=1 . Then C⊂

n 

f −1 (Ui )

i=1

=⇒ f (C) ⊂ f

n 

f

−1

i=1

n n    −1   (Ui ) = f f (Ui ) ⊂ Ui i=1

and thus f (C) is compact.

i=1



Remark. In the proof of Theorem 2.3.13, we have made used of the facts that   f f −1 (T ) ⊂ T

and

  f −1 f (S) ⊃ S .

Note that in general, equalities may not hold, so this is the most we could say (exercise). Theorem 2.3.14. If f : X → Y is continuous and C ⊂ X is compact, then f is bounded on C. Proof. It is obvious from Theorems 2.3.13 and 1.3.5.



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Theorem 2.3.15. If f : X → R is continuous on a compact set C ⊂ X, then f attains its maximum and minimum in C, that is, there exist p, q ∈ C such that f (p) = sup f (C) = max f (C) , f (q) = inf f (C) = min f (C) . Proof. By Theorems 2.3.13 and 1.3.5, f (C) ⊂ R is compact and hence closed and bounded. Let m = inf f (C) ∈ R. By definition of infimum, we have m ≤ f (c) for all c ∈ C. Observe that m must be an adherent point of f (C). In fact, if not, then there exists r > 0 such that B(m, r)∩f (C) = φ. This implies m < m+ r2 < f (c) for all c ∈ C, violating the assumption that m = inf f (C). Thus m ∈ f (C) = f (C) and so m = f (q) for some q ∈ C. Similarly, f (p) = sup f (C) for some p ∈ C.  Remark. By Theorem 2.3.15, every continuous function f : R → R maps closed and bounded intervals to closed and bounded intervals. In fact, for any [a, b ] ⊂ R, f attains its maximum M and minimum m in [a, b ]. Thus f ([a, b ]) ⊂ [m, M ] with m, M being attained. By Intermediate Value Theorem, f ([a, b ]) = [m, M ]. Theorem 2.3.16. Let f : X → Y be a function. If (a) X is compact,

(b) f is 1 − 1,

(c) f is continuous,

then the inverse function f −1 : f (X) → X is also continuous. Proof. The existence of the inverse function follows from (b). Now let C ⊂ X be closed. By Theorem 1.3.8, C is compact and so by Theorem 2.3.13, (f −1 )−1 (C) = f (C) ⊂ Y is compact. In particular, it is closed in Y . However, since f (C) ⊂ f (X), it follows that (f −1 )−1 (C) = f (C) is also closed in f (X). Therefore,  by Theorem 2.3.11, f −1 is continuous.

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Examples 2.3.17. essential.)

(The compactness of X in Theorem 2.3.16 is

(i) Let f : [1, 2 ] ∪ (3, 4 ] → R be given by x x ∈ [1, 2 ] f (x) = x − 1 x ∈ (3, 4 ] .

Then f is 1 − 1 and continuous. However, lim f −1 (y) = 2 = 3 = lim+ f −1 (y)

y→2−

y→2

and so f −1 is not continuous at 2. This does not violate Theorem 2.3.16 as in this case the domain [1, 2 ] ∪ (3, 4 ] is not compact. (ii) let X = [0, 1) with the usual Euclidean metric and consider f :X →C given by f (x) = e2πix ,

x ∈ [0, 1) .

Clearly f is 1 − 1 and continuous on X with f (X) = S 1 ⊂ C.

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Hence in particular, f −1 : S 1 → X = [0, 1) exists. However, f −1 is not continuous at the point 1 ∈ S 1 ⊂ C. This can be seen by either observing that images of points near 1 ∈ f (X) = S 1 ⊂ C −1 under   f are not necessarily close 0 ∈ X, or noting that we have → f (0) as n → ∞ but f 1 − n1 n∈N

      1 1 −1 = 1− → 0 = f −1 f (0) . f 1− f n n n∈N n∈N

Again, Theorem 2.3.16 is not violated, as in this case the domain X is not compact. Definition 2.3.18. A bijection f : X → Y which is continuous and with continuous inverse f −1 : Y → X is called a homeomorphism or a topological mapping of X onto Y . Any two metric spaces X, Y are said to be homeomorphic, denoted as X ∼ = Y , if there is a homeomorphism of X onto Y . Remark. Clearly, being homeomorphic is an equivalence condition. In fact, it is easy to see that (i) the identity function ι : X → X is a homeomorphism and hence X∼ = X; (ii) if f : X → Y is a homeomorphism, then so f −1 : Y → X is a homeomorphism; (iii) if f : X → Y and g : Y → Z are homeomorphisms, then g ◦ f : X → Z is also a homeomorphism. Remark (more important than customary). Let f : X → Y be a continuous function and imagine that X, Y are elastic geometric objects. Then what f does to X is to move the points in X in a “continuous” manner, that is, to bend, twist, stretch, dilate, contract etc., but there is no tearing. The point is, the action of tearing is not continuous, because points that are originally close to each other will no longer be close to each other after the tearing. So intuitively, a

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homeomorphism of X onto Y is an action that transforms the space X in a “continuous manner” through bending, twisting, stretching, dilating, contracting, etc., but without tearing, onto the space Y in such a way that the process can be reversed. In short, we say that two metric spaces are homeomorphic if one can be “continuously deformed” to another, and back.

Examples 2.3.19. (a) (0, 1) ∼ = (0, 2). (b) [0, 1) ∼ = [0, 2). (c) [0, 1] ∼ = [0, 2]. (d) (0, 1) ∼ = R. (e) S 1 \ {n} ∼ = R, where n is the “north pole” (0, 1) ∈ S 1 ⊂ R2 . (f) S 2 \ {n} ∼ = R2 , where n is the “north pole” (0, 0, 1) ∈ S 2 ⊂ R3 .

Proof. (a) The function f : (0, 1) → (0, 2) given by f (x) := 2x, x ∈ (0, 1), is 1 − 1, onto, continuous, with inverse f −1 (y) = y2 , y ∈ (0, 2), which is clearly also continuous, gives a homeomorphism ∼ = f : (0, 1) −→ (0, 2).

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It is evident that other homeomorphisms of (0, 1) onto (0, 2) can easily be constructed. For example, the function g : (0, 1) → (0, 2) given by g(x) := 2x2 , x ∈ (0, 1), is also 1 − 1, onto,  continuous, with continuous inverse g−1 (y) = y2 , y ∈ (0, 2). Hence it is also a homeomorphism. (b) The same functions f , g as defined in (a) are homeomorphisms ∼ = [0, 1) −→ [0, 2). (c) The same function f as defined in (a) is a homeomorphism ∼ = [0, 1] −→ [0, 2].

(d) It is easily seen that the function f : (0, 1) → (− π2 , π2 ) given by f (x) := (x − 12 )π, x ∈ (0, 1), gives a homeomorphism ∼ =

(0, 1) −→ (− π2 , π2 ). Meanwhile, the function g : (− π2 , π2 ) → R given by g(x) := tan x, x ∈ (− π2 , π2 ), gives a homeomorphism ∼ =

∼ =

(− π2 , π2 ) −→ R. Thus the composition g ◦ f : (0, 1) −→ R is a homeomorphism.

(e) We will prove this descriptively by our geometric insight. The interested readers could try to fill in the detailed computations with explicit formula.

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In the xy-plane, lay the punctured unit circle S 1 \ {n} onto the Rx axis. Then imagine that there is a light source emitting light from the north pole down through the punctured circle onto Rx . In this way, we have a 1 − 1, onto, continuous mapping from the punctured unit circle to the real line Rx with continuous inverse.

That is a homeomorphism of the punctured unit circle onto R, called the stereographic projection. (d) revisited: Based on the result of (e), another homeomorphism ∼ = (0, 1) −→ R can be constructed geometrically by means of the stereographic projection as follows. First of all, by imagining that (0, 1) is an elastic piece of string of unit length, then it can be bent and stretched into a unit circle with the “north pole” removed. That means the line segment (0, 1) is homeomorphic to the punctured circle. By (e), the stereographic projection gives a homeomorphism of the punctured circle onto R. Hence (0, 1) which is homeomorphic to the punctured circle is also homeomorphic to R. (f) Similar to (e), the stereographic projection of the punctured unit ball S 2 \{n} ⊂ R3 onto the plane R2 is a homeomorphism. Details are left for the interested readers. Example 2.3.20. Let f : X → Y be bijective. Then the following conditions are all equivalent:

Limits and Continuity

(a) (b) (c) (d)

f f f f

191

is a homeomorphism. and f −1 are continuous. is continuous and open. is continuous and closed.

Proof. (a)⇔(b) follows from the definition of homeomorphism. The equivalence of (b), (c), and (d) follows immediately from Example 2.3.12, as f −1 is continuous if and only if f = (f −1 )−1 is open if and  only if f = (f −1 )−1 is closed. Definition 2.3.21. A property of a set which remains invariant under (or, is preserved by) topological mappings (or homeomorphisms) is called a topological property or a topological invariant. Example 2.3.22. Determine whether openness, closedness, compactness, boundedness, and the distance between two points are topological invariants. Proof. By Example 2.3.20, openness and closedness are preserved by homeomorphisms, hence they are topological invariants. By Theorem 2.3.13, compactness is preserved by continuous functions, hence by homeomorphisms and so it is also a topological invariant. However, boundedness is not preserved by homeomorphisms. Example 2.3.19 (d) shows that (0, 1) ∼ = R, (0, 1) is bounded, but R is not. Hence boundedness is not a topological invariant. Finally, the distance between two points is not preserved by homeomorphism. For example, from Example 2.3.19 (a), the function f : (0, 1) → (0, 2) x → 2x

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is a homeomorphism. However, it is clear that   d(a, b) = |b − a| = 2|b − a| = d f (a), f (b)

for any a = b in (0, 1). Hence the distance between two points is not a topological invariant. Definition 2.3.23. A mapping f : X → Y is said to be an isometry if it preserves the metric, or preserves the distance, that is,   dY f (p), f (q) = dX (p, q)

for all p, q ∈ X .

Two metric spaces X and Y are said to be isometric if there is an isometry of X onto Y . Observe that surjective isometries are homeomorphisms. Example 2.3.24.

The infinite cylinder

C := {(x, y, z) ∈ R3 : x2 + y 2 = 1} ⊂ R3 equipped with the metric d given by d(p, q) := the shortest path among all paths on C joining the points p, q ∈ C. It is evident that (C, d) is a well-defined metric space. Let S := C \ {a vertical straight line in C} ⊂ S . Then it is intuitively clear that S is isometric to the open infinite paper strip T := {(x, y) ∈ R2 : |x| < π}. The interested readers are encouraged to try and work out the full details.

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193

Exercise 2.3 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. Let (X, d) be a metric space, x ∈ X and φ = A ⊂ X. Recall that the distance between x and A is defined by d(x, A) := inf a∈A d(x, a). We can extend the notion of distance between a point and a set to distance between two nonempty subsets: if A, B ⊂ X, the distance between A and B is d(A, B) := inf{d(a, b) : a ∈ A, b ∈ B} . If A and B are closed and A ∩ B = φ, then d(A, B) > 0.

Answer : False. Example : In X := R2 , take

A := {(x, 0) : x ∈ R}, B :=



1 x, x



 :x>0 .

Then A, B are closed, A ∩ B = φ, but d(A, B) = 0.

2. Let πx , πy : R2 → R be the projection of the plane R2 onto the x- and y- coordinate, respectively, i.e., πx (x, y) := x and

πy (x, y) := y

for all (x, y) ∈ R2 .

Then πx and πy are continuous on R2 . Answer : True. Proof . Suffices to show the continuity of πx , as that for πy can be shown analogously. Fix (a, b) ∈ R2 . For any ε > 0, pick δ := ε. Then   for any (x, y) ∈ BR2 (a, b), δ , we have  |x − a| ≤ |x − a|2 + |y − b|2 < δ = ε ,

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that is, πx (a, b) = x ∈ BR (a, ε) and so πx is continuous at (a, b). Since (a, b) ∈ R2 is arbitrary, πx is continuous on R2 .

3. Let f : X → R be continuous. Then {x ∈ X : f (x) > 0} ⊂ X is open. Answer : True. Proof . Fix a ∈ {x ∈ X : f (x) > 0}. Write ε := f (a). Then by assumption, we have ε > 0. By the continuity of f , there exists δ > 0     such that f BX (a, δ) ⊂ BR f (a), ε = (0, 2ε) ⊂ (0, ∞). That is, BX (a, δ) ⊂ {x ∈ X : f (x) > 0} and so {x ∈ X : f (x) > 0} is open.

Alternatively , since f is continuous and (0, ∞) ⊂ R is open,   {x ∈ X : f (x) > 0} = f −1 (0, ∞) ⊂ X is open.

4. Let f : X → R be continuous. Then {x ∈ X : f (x) ≥ 0} ⊂ X is closed.

Answer : True. Proof . Similar to Exercise 2.3, Part A, Problem #3, it is easy to see that {x ∈ X : f (x) < 0} is open. Hence, being the complement of an open set in X , {x ∈ X : f (x) ≥ 0} is closed. 5. If f : R → R be continuous, then A := {(x, y) ∈ R2 : yf (x) > 0} ⊂ R2 is open. Answer : True. Proof . Let πi : R2 → R be the projection of R2 onto its i-th coordinate, i = 1, 2. By Exercise 2.3, Part A, Problem #2, πi is continuous for i = 1, 2. Define F : R2 → R by F (x, y) := y f (x) = (π2 (x, y)) · (f ◦ π1 (x, y)) .   Then we have A = F −1 (0, ∞) . Since f , π1 and π2 are continuous,   so is F . As (0, ∞) ⊂ R is open, A = F −1 (0, ∞) ⊂ R2 is open.

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6. If f : X → Y maps compact sets to compact sets, then f is continuous. Answer : False. Example : Let X := R, Y := {0, 1}, and f : X → Y be given by f (x) :=



0 1

if x = 0 , if x = 0 .

Then clearly f sends every compact set to a compact set but f is not continuous.

7. Continuous functions map Cauchy sequences to Cauchy sequences. That is, if {xn }n∈N is a Cauchy sequence in X and f : X → Y is continuous, then {f (xn )}n∈N is a Cauchy sequence in Y . Answer : False. Example : Let X = Y := R+ = (0, ∞), xn = 1/n, n ∈ N, and f (x) := x1 for every x ∈ R+ . Then {xn }n∈N is a Cauchy sequence in X and f is continuous on X . But {f (xn )}n∈N = {n}n∈N is not

Cauchy.

8. If f : X → Y maps Cauchy sequences to Cauchy sequences, then f is continuous. Answer : True. Proof . We prove this by contra-positive arguments. Assume that f is discontinuous at some point x ∈ X . Then there exists ε > 0 such 1 that for any n ∈ N, there exists sn ∈ X with 0 < d(sn , x) < n and d(f (sn ), f (x)) ≥ ε. Define, for any n ∈ N, xn :=



sn

if n is even

x

otherwise.

It is clear that the sequence {xn }n∈N is convergent to x, hence in particular is Cauchy, but {f (xn )}n∈N is not Cauchy.

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9. All functions f : X → Y , where X is any metric space and Y is a discrete metric space, are continuous. Answer : False. Example : Let X := R with Euclidean metric and Y := R with discrete metric. Consider the identity function f : X → Y , f (x) := x for all x ∈ X . As {0} ⊂ Y is open but f −1 ({0}) = {0} ⊂ R is not open, f is not continuous. 10. All functions f : X → Y , where X is a discrete metric space and Y is any metric space, are continuous. Answer : True. Proof . For any x ∈ X and any ε > 0, we have f (BX (x, 1)) = f ({x}) = {f (x)} ⊂ BY (f (x), ε) . Hence f is continuous at x. As x ∈ X is arbitrary, f is continuous

on X . Alternatively , since all subsets of a discrete metric space are open, f −1 (V ) ⊂ X is open for any (open) subset V ⊂ Y . Hence f is continuous.

11. If f : X → Y is continuous, then f (S ◦ ) ⊂ f (S)◦ for all S ⊂ X. Answer : False. Example : Let X := R with discrete metric and Y := R with Euclidean metric. Let f : X → Y be the identity mapping f (x) := x for all x ∈ X . As X is discrete, by Exercise 2.3, Part A, Problem #10, f is continuous. Take S := Q ⊂ X . Notice that S ◦ = S and so f (S ◦ ) = f (S), but f (S)◦ = Q◦ (in R with the Euclidean metric) = φ. Hence f (S ◦ ) ⊂ f (S)◦ . 12. If f : X → Y is such that f (S ◦ ) ⊂ f (S)◦ for all S ⊂ X, then f is continuous. Answer : False.

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Example : Let X := R and Y := {0, 1} with discrete metric. Let f : X → Y be the function f (x) :=



1 x≥0

0 x 0 . x

Then f is bijective, open, but it is not continuous at x = 0.

33. If f : X → Y is bijective and open, and Y is compact, then f is continuous. Answer : True. Prove . Since f is bijective, f −1 : Y → X exists. Since f is open, by Example 2.3.3, f −1 is a continuous function. For any closed subset E ⊂ Y , since Y is compact, E is also compact and so f −1 (E) ⊂ X is compact and hence is closed in X . Hence f is continuous on X . 34. Let {An }∞ subsets of a n=0 be a countable collection of closed  ∞  metric space X and f : n=0 An → Y . If f A is continuous n for each n ∈ N, then f is continuous. Answer : False.

  Example : Let X = Y := R, A0 := {0}, An := n1 for every n ∈  ∞ N. Consider the function f : n=0 An → Y defined by f A0 := 1   and f A := 0 for every n ∈ N. Clearly f A is continuous for each n ∞ n n ∈ N, but f is not continuous at 0 ∈ n=0 An .

35. f : X → Y is continuous on X if and only if f is continuous on every compact subset of X. Answer : True.

Proof . Necessity is clear. For the sufficiency, suppose that f is continuous on every compact subset of X . Fix p ∈ X . If p is an isolated point of X , then f is automatically continuous at p. Otherwise, p is an accumulation point of X . Hence there exists a

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sequence {xn }n∈N of distinct points in X with {xn } → p as n → ∞.

Write S := {p}∪{xn : n ∈ N} ⊂ X . Since every infinite sequence in S has an accumulation point (p to be precise) in S , S is sequentially compact. By Exercise 2.2, Part B, Problem #14, S is compact. By assumption, f is continuous at every point of S , hence in particular, continuous at p. Since p is arbitrary, f is continuous on X .

Part B: Problems 1. In each of the following situations, give an example of a function f : S → T such that it is continuous and onto, or else explain why there cannot be such an f : (a) S = (0, 1), T = (0, 1] (b) S = [0, 1] ∪ [2, 3], T = {0, 1} (c) S = [0, 1] × [0, 1], T = R2 Solution :

  2x x ∈ 0, 12   (a) Example : Define f (x) =: 1 x ∈ 12 , 1 . It is clear that f is continuous on (0, 1) and f ((0, 1)) = (0, 1].

0 x ∈ [0, 1] 1 x ∈ [2, 3] . Note that the only open sets in T are φ, {0}, {1}, {0, 1}. As

(b) Example : Define f (x) :=

f −1 (φ) = φ,

f −1 ({0})=[0, 1],

f −1 ({1}) = [2, 3], f −1 ({0, 1})=[0, 1] ∪ [2, 3] are all open in S , we conclude that f is continuous on S . (c) Since S is compact, T is noncompact, and compactness should be preserved by continuous functions, there cannot be any continuous function mapping S onto T .

2. For any A ⊂ X, show that the function f : X → R given by f (x) := d(x, A) is continuous.

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205

Proof . For any x, y ∈ X and any a ∈ A, we have d(x, a) ≤ d(x, y) + d(y, a) . Thus

d(x, A) = inf d(x, a) ≤ inf [d(x, y) + d(y, a)] a∈A

a∈A

= d(x, y) + inf d(y, a) a∈A

= d(x, y) + d(y, A). Reversing x and y , we have d(y, A) ≤ d(x, y) + d(x, A). Combining

the two inequalities, we have

|f (x) − f (y)| = |d(x, A) − d(y, A)| ≤ d(x, y) for all x, y ∈ X , which gives the continuity of f on X . In fact, for any ε > 0, let δ := ε, then

|f (x) − f (y)| ≤ d(x, y) < δ = ε for all x, y ∈ X with d(x, y) < δ .

3. If f : X → Y is a bijective open mapping and Y is compact, show that f is continuous. Proof . Since f is bijective, f −1 : Y → X exists. By assumption, f is an open mapping and so for any open S ⊂ X , (f −1 )−1 (S) = f (S) is open in Y . Hence f −1 : Y → X is continuous. Let T ⊂ Y be any closed subset of Y . Being a closed subset of the compact space Y , T is compact. By the continuity of f −1 , f −1 (T ) is compact and thus is closed. Therefore, f is continuous. 4. Let A, B be disjoint subsets of X. (a) If A is compact and B is closed, show that d(A, B) > 0. (b) Is (a) still valid if A, B are closed but not compact?

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Proof . (a) By Exercise 2.3, Part B, Problem #2, the function

d( · , B) : X → R

x → d(x, B)

is continuous. As A is compact, by Theorem 2.3.10, there exists

a0 ∈ A such that d(a0 , B) ≤ d(a, B)

for all a ∈ A .

Since B is closed and a0 ∈ / B , by Exercise 1.2, Part B, Problem

#7,

we have d(a0 , B) > 0. Hence

d(a, b) ≥ d(a, B) ≥ d(a0 , B) > 0 for all a ∈ A, b ∈ B . Hence

d(A, B) =

inf

a∈A,b∈B

d(a, b) ≥ d(a0 , B) > 0 .

(b) Without the compactness condition, (a) will no longer hold.

Example : Consider X := R2 , A := {(x, 0) : x ∈ R} ,    1 :x>0 . B := x, x It is elementary to check that both A, B are closed in R2 ,

A ∩ B = φ, and d(A, B) = 0. 5. For any f , g ∈ C[a, b ], define h(x) := min{f (x), g(x)} and H(x) := max{f (x), g(x)} for any x ∈ [a, b ]. Show that H, h ∈ C[a, b ].

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Proof . Since f , g are continuous, for any x0 ∈ [a, b ] and ε > 0, there is δ > 0 such that for any x ∈ [a, b ] with |x − x0 | < δ , we have

f (x0 ) − ε < f (x) < f (x0 ) + ε, g(x0 ) − ε < g(x) < g(x0 ) + ε.

Hence for any x ∈ [a, b ] with |x − x0 | < δ ,

h(x) = min{f (x), g(x)} < min{f (x0 ) + ε, g(x0 ) + ε} = min{f (x0 ), g(x0 )} + ε = h(x0 ) + ε , and

h(x) = min{f (x), g(x)} > min{f (x0 ) − ε, g(x0 ) − ε} = min{f (x0 ), g(x0 )} − ε = h(x0 ) − ε .

Therefore, h is continuous at x0 . Since x0 ∈ [a, b ] is arbitrary, we

have h ∈ C[a, b ]. Analogously, we also have H ∈ C[a, b ].

6. For any sequence {fn }n∈N in C[a, b ], define h(x) := inf {fn (x)} and H(x) := sup{fn (x)} for any x ∈ [a, b ] . n∈N

n∈N

Determine whether H, h ∈ C[a, b ]. Answer . Neither h nor H is necessarily continuous on [a, b ]. Example : On [0, 1], define, for any n ∈ N, fn (x) :=



nx 0 ≤ x ≤

1

1 n

1 n

≤ x ≤ 1.

Clearly, fn ∈ C[0, 1] for all n ∈ N. Since fn (0) = 0 for all n, we

have H(0) = 0. On the other hand, for any x0 ∈ (0, 1], there exists

N ∈ N such that

1 N

< x0 . Hence fN (x0 ) = 1 and so H(x0 ) = 1.

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Since this is true for all x0 ∈ (0, 1], we have H(x) = 1 for all

x ∈ (0, 1] and so in particular, H is not continuous at 0. Finally, by considering the sequence of functions {−fn }n∈N in C[0, 1] and by analogous arguments, it is not hard to see that h ∈ / C[0, 1].

7. Let A be a nonempty subset of X. The characteristic function of A is the function χA : X → R defined by χA (x) =



1 0

if x ∈ A, if x ∈ X \ A.

Identify the points at which χA is discontinuous. Answer : χA is discontinuous on ∂A. Proof . First observe that χA is continuous on X \ ∂A. In fact, since X = A ∪· (X \ A) = A◦ ∪· ∂A ∪· (X \ A), we have X \ ∂A = A◦ ∪· (X \ A). So for any x ∈ X \ ∂A, we have x ∈ A◦ ∪ (X \ A). If x ∈ A◦ , then there exists r > 0 such that B(x, r) ⊂ A◦ and so χA ≡ 1 on B(x, r). If x ∈ X \ A, then there exists r > 0 such that B(x, r) ⊂ X \ A ⊂ X \ A and so χA ≡ 0 on B(x, r). In either case, f is continuous at x and so χA is continuous on X \ ∂A. On the other hand, consider x ∈ ∂A. Take ε = 12 . For any r > 0, there exist y ∈ B(x, r) ∩ A and z ∈ B(x, r) ∩ X \ A. If x ∈ A, then |χA (x) − χA (z)| = |1 − 0| = 1 > ε; if x ∈ X \ A, then |χA (x) − χA (y)| = |0 − 1| = 1 > ε. Combining, we see that for any r > 0, there are points in B(x, r) with images under χA far away from that of x. Hence χA is not continuous at x. 8. Let f : (0, 1) → R be given by ⎧ ⎨0 f (x) := 1 ⎩ q

x ∈ (0, 1) \ Q p x = ∈ (0, 1) ∩ Q , p, q ∈ N, (p, q) = 1. q

Determine the set of points x ∈ (0, 1) at which f is continuous.

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Answer : f is continuous at all irrational points and discontinuous at all rational points in (0, 1). Proof . Consider a ∈ (0, 1) ∩ R \ Q. For any ε > 0, there are at most finitely many q ∈ N with 1q ≥ ε, hence there are at most finitely many rational numbers of the form pq in (0, 1) with p, q ∈ N, (p, q) = 1, such p that 1q ≥ ε. Clearly, each of these finitely many rational numbers q has a positive distance from the irrational number a. Let δ := the minimum of these finitely many distances. Then δ > 0, and points in B(0,1) (a, δ) p are either irrational, or rational but of the form q with p, q ∈ N, (p, q) = 1, and 1q < ε. Hence for any x ∈ B(0,1) (a, δ), we have |f (x) − f (a)| = |f (x)| if x ∈ R \ Q 0 = 1 p if x = q , p, q ∈ N, (p, q) = 1, 1q < ε q 0 and points in B(0,1) (a, δ) are either irrational, or rational but of the form pq with p, q ∈ N, (p, q) = 1, and q > qa . Hence for any x ∈ B(0,1) (a, δ), we have     1     f (x) − f (a) = f (x) −  qa  ⎧   ⎨ 0 − q1a  if x ∈ R \ Q  =  p ⎩  1 − 1  if x = q , p, q ∈ N, (p, q) = 1, q > qa q qa ! 1 if x ∈ R \ Q q = 1a p 1 if x = q , p, q ∈ N, (p, q) = 1, q > qa qa − q 1 1 ≥ − which is a fixed positive number, qa qa + 1 hence f is discontinuous at a.

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9. Let f , g be real valued functions on [0, 1] defined by f (x) :=



0 1

if x ∈ /Q

if x ∈ Q ;

g(x) :=



0 x

if x ∈ /Q

if x ∈ Q .

(a) Show that f is not continuous anywhere in [0, 1]. (b) Show that g is continuous only at x = 0. (c) If h : [0, 1] → R is such that h(x) = c = constant for all x ∈ Q, give a necessary and sufficient condition for h to be continuous on [0, 1]. Justify your answer. Proof . (a) For any x ∈ [0, 1] ∩ Q, take ε :=

1 . 2

For any δ > 0, there

exists y ∈ B[0,1] (x, δ)∩ (R \Q), and we have d(f (x), f (y)) =

|1 − 0| = 1 > ε. So f is not continuous at x. Similarly, for any x ∈ [0, 1] ∩ (R \ Q), take ε := 12 . For any δ > 0, there exists y ∈ B[0,1] (x, δ) ∩ Q, and we have d(f (x), f (y)) = |0 − 1| = 1 > ε. So f is not continuous at x. Combining, we see that f is not continuous anywhere in [0, 1]. (b) For x = 0, for any ε > 0, take δ := ε. Then we have g(B[0,1] (0, δ)) = g([0, δ)) = g([0, ε)) ⊂ [0, ε) ⊂ BR (0, ε) = BR (g(0), ε). Hence g is continuous at x = 0. For any x ∈ (0, 1] ∩ Q, take ε := x2 . For any δ > 0, there exists y ∈ B[0,1] (x, δ) ∩ (R \ Q), and we have d(g(x), g(y)) = |x − 0| = x > ε. So g is not continuous at x. Similarly, for any x ∈ (0, 1] ∩ (R \ Q), take ε := x2 . For any δ ∈ (0, x2 ), there exists y ∈ B[0,1] (x, δ) ∩ Q, and we have d(g(x), g(y)) = |0 − y| = y > x − δ > x2 = ε. So g is not continuous at x. Combining, we conclude that g is continuous only at x = 0. (c) h is continuous if and only if h(x) = c for all x ∈ [0, 1]. In fact, it is easy to verify that the constant function h ≡ c is continuous. Conversely, let S = [0, 1] ∩ Q. Then S = [0, 1] and so for every x ∈ [0, 1], there exists a sequence {xn }n∈N in S such that {xn } → x as n → ∞. By the continuity of

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211

h, we have h(x) = lim h(xn ) = c. Hence h(x) = c for all x ∈ [0, 1].

n→∞

10. Let (X, d) be a metric space. Equip X × X with metric D : (X × X) × (X × X) → R defined by   D (x, y), (x , y  ) := max{d(x, x ), d(y, y  )} . (a) Show that the metric d, when considered as a real-valued function d : X × X → R on the metric space X × X, is continuous. (b) Let f : X → X be a continuous function. Show that G := {(x, f (x)) : x ∈ X} ⊂ X × X is closed.

Proof .

(a) It is elementary to verify that D is a well-defined metric on

X × X . Let {(xn , yn )}n∈N be a sequence in X × X such that {(xn , yn )} → (x, y) as n → ∞. For any ε > 0, there is N ∈ N such that D((xn , yn ), (x, y)) < ε for all n ≥ N . Hence by triangle inequality,

|d(xn , yn ) − d(x, y)|

≤ |d(xn , yn ) − d(xn , y)| + |d(xn , y) − d(x, y)|

≤ d(yn , y) + d(xn , x)   ≤ 2D (xn , yn ), (x, y) < 2 for all n ≥ N .

Therefore, {d(xn , yn )} → d(x, y) as n → ∞ and so d is continuous.

(b) Let {(xn , f (xn ))}n∈N be a convergent sequence in G, say,

{(xn , f (xn ))} → (x, y) as n → ∞. By the definition of D , we see easily that {xn } → x and {f (xn )} → y as n → ∞. As f is continuous, we have {f (xn )} → f (x) as n → ∞ and so f (x) = y . Hence {(xn , f (xn ))} → (x, f (x)) ∈ G as n → ∞. Therefore, G is closed in X × X .

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11. Suppose S ⊂ X and f : S → Y . The function f is said to be Lipschitz continuous if there is M > 0 such that dY (f (s1 ), f (s2 )) ≤ M dX (s1 , s2 ) for all s1 , s2 ∈ S . Let f : S → Y be Lipschitz continuous. (a) Show that f is continuous. (b) Prove or disprove that f maps Cauchy sequences in S to Cauchy sequences in Y . (c) Suppose that Y is complete. Show that there is a continuous function f˜ : S → Y which extends f to S  in the sense that f˜S = f . Proof .

(a) Let s ∈ S . Suffices to show that if {sn }n∈N is a sequence in S

convergent to s, then limn→∞ f (sn ) exists and equals f (s).

For any ε > 0, as {sn } → s as n → ∞, there exists N ∈ N such that

dX (sn , s)
0, there is N ∈ N such that dX (sn , sm )
0, since f is continuous at

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215

x, there exists δ > 0 such that f (B(x, δ)) ⊂ B(y, ε). Since x ∈ S  , B(x, δ) ∩ (S \ {x}) = φ. In particular, there exists x = p ∈ B(x, δ) ∩ S . Since f is injective, f (p) = f (x) = y and so f (p) ∈ B(y, ε) ∩ f (S) \ {y}. Hence y ∈ f (S) . 14. Let f : X → Y be a continuous function and S ⊂ X. (a) Show by an example that in general, f (∂S) ⊂ ∂f (S). (b) Give an additional condition on f which can guarantee that f (∂S) ⊂ ∂f (S).

Solution : (a) Example : Consider f : R → R defined by f (x) := x2 . Then clearly f is continuous. Let S := [−1, 2 ]. Then f (∂S) = f ({−1, 2}) = {1, 4} ⊂ {0, 4} = ∂([0, 4 ]) = ∂f (S) . (b) If f : X → Y is continuous and is injective, then f (∂S) ⊂

∂f (S) always holds. Proof . For any y ∈ f (∂S), there exists a unique x ∈ ∂S = S ∩ (X \ S) such that f (x) = y . Since f is injective, we have     y ∈ f S ∩ X \ S = f (S) ∩ f X \ S .  

By Exercise 2.3, Part A, Problem #16, we have f S ⊂ f (S)

  and f X \ S ⊂ f (X \ S). Since f is injective,

f (X \ S) = f (X) \ f (S) ⊂ Y \ f (S) .

Hence

  y ∈ f (S) ∩ f X \ S ⊂ f (S) ∩ f (X \ S) ⊂ f (S) ∩ Y \ f (S) = ∂f (S) .

15. Let A ⊂ Rn be any subset and f : A → A be a continuous function satisfying d(f (x), f (y)) ≥ d(x, y) for any x, y ∈ A, where d is usual Euclidean metric.

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(a) Prove that f is injective and that the inverse function f −1 : f (A) → A is also continuous. (b) If A is compact, prove that f is surjective. Proof . (a) For any x, y ∈ A, if f (x) = f (y), then 0 = d(f (x), f (y)) ≥

d(x, y). This forces x = y and so f is injective. Hence the inverse function f −1 : f (A) → A exists. Next, for any u, v ∈ f (A), we have d(f −1 (u), f −1 (v)) ≤ d(f (f −1 (u)), f (f −1 (v))) = d(u, v) ,

from which the continuity of f −1 follows. [For any ε > 0, let

δ := ε. Then whenever d(u, v) < δ, we have d(f −1 (u), f −1 (v)) ≤ d(u, v) < δ = ε . Thus f −1 is continuous.] (b) Assume to the contrary that there exists x ∈ A \ f (A). Since

A is compact and f is continuous, f (A) is compact. Hence d(x, f (A)) > 0. Write d := d(x, f (A)). Construct a sequence {xk }k∈N in A by x0 := x;

xk := f (xk−1 ) for any k ∈ N .

Note that xk ∈ f (A) for any k ∈ N. Since A is compact

in Rn , by Exercise 2.1, Part B, Problem #8, it is sequentially compact in Rn . Hence the sequence {xk }k∈N has a convergent

subsequence {xkj }j∈N which converges in A. This implies that

there exists m ∈ N such that

d(xkm+1 , xkm ) < d . On the other hand,

d(xkm+1 , xkm ) = d(f (xkm+1 −1 ), f (xkm −1 )) ≥ d(xkm+1 −1 , xkm −1 ) .

(*)

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Repeating the process indefinitely, we obtain

d(xkm+1 , xkm ) ≥ d(xkm+1 −1 , xkm −1 ) ≥ ···

≥ d(xkm+1 −km , x0 )

= d(xkm+1 −km , x) .

Since xkm+1 −km ∈ f (A), we have

d(xkm+1 , xkm ) ≥ d(xkm+1 −km , x)

≥ d(x, f (A)) = d ,

which contradicts to (*).

16.

(a) Let A, B ⊂ X be disjoint nonempty closed sets of X. Define f : X → R by f (x) :=

d(x, A) d(x, A) + d(x, B)

x∈X .

Show that f is a continuous function on X, 0 ≤ f ≤ 1, f = 0 precisely on A, and f = 1 precisely on B. (b) For any g : X → R, the zero set Z(g) of g is defined as Z(g) := f −1 ({0}). Show that every nonempty closed set A ⊂ X can be written as the zero set of some continuous real-valued function on X. (c) Show that for any pair of disjoint closed sets A, B ⊂ X, there is a pair of disjoint open sets V , W ⊂ X such that A ⊂ V and B ⊂ W .

Proof .

(a) First of all, by Exercise 2.3, Part B, Problem #2, both d(x, A) and d(x, B) are continuous. Furthermore, as A ∩ B = φ,

d(x, A) + d(x, B) > 0 for all x ∈ X and so the function f is well-defined and continuous on X . It is also evident that

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0 ≤ f ≤ 1. Finally, by Exercise 1.2, Part B, Problem #7,

we have

f (x) = 0 ⇐⇒ d(x, A)= 0 ⇐⇒ x ∈ A = A

f (x) = 1 ⇐⇒ d(x, B)= 0 ⇐⇒ x ∈ B = B . (b) If A = X , then A = Z(f ) for f ≡ 0. If A = X , pick

p ∈ X \ A and define f : X → R by f (x) :=

d(x, A) d(x, A) + d(x, {p})

x∈X .

By (a), Z(f ) = A. (c) For any pair of disjoint closed subsets A, B ⊂ X , define

  f : X → [0, 1] by the formula in (a). Set V := f −1 [0, 12 )   and W := f −1 ( 12 , 1 ] . Then  1 =V , A = f ({0}) ⊂ f 0, 2   1  ,1 =W . B = f −1 ({1}) ⊂ f −1 2 −1

−1





Furthermore, since f is continuous, V , W are open disjoint subsets of X .

17. Let π : R2 → R be the natural projection of R2 onto the first coordinate, that is, π(x, y) := x for any (x, y) ∈ R2 . (a) Determine whether π is open on R2 . (b) Determine whether π is closed on R2 . Solution : (a) True.

Proof . For any open set U ⊂ R2 and any x ∈ π(U ) ⊂ R, there exists y ∈ R such that (x, y) ∈ U . Since U ⊂ R2 is open, there exists r > 0 such that BR2 ((x, y), r) ⊂ U . Hence x ∈ (x − r, x + r) = BR (x, r) ⊂ π (BR2 ((x, y), r)) ⊂ π(U ). Therefore, π(U ) is open and so π is open.

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(b) False.

   Example : S := x, x1 : x > 0 ⊂ R2 is closed but π(S) = (0, ∞) ⊂ R is not closed.

18. If f : X → Y is open and continuous, must it be closed? Answer : False. Example : Let X := (0, 1) ⊂ R and Y := R with the usual Euclidean metric. Consider f (x) := x, x ∈ X . As open sets in (0, 1) are open in R, f is open. It is trivial that f is continuous. However, as the image of the closed set (0, 1) ⊂ X equals (0, 1) ⊂ Y which is not closed in Y , f is not closed. 19. If f : X → Y is closed and continuous, must it be open? Answer : False. Example : Let X := {0} ⊂ R and Y := R with the usual Euclidean metric. Consider f (x) := x, x ∈ X . Then it is evident that f is closed and continuous. However, as the image of the open set {0} ⊂ X equals {0} ⊂ Y which is not open in Y , f is not open. 20. If f : X → Y is open and closed, must it be continuous? Answer : False. Example : Let X := R with the usual Euclidean metric and let Y := R with the discrete metric. Consider f (x) := x, x ∈ X . As all subsets of Y are clopen, f is both open and closed. However, as the pre-image of the open set {0} ⊂ Y equals {0} ⊂ X which is not open in X , f is not continuous.

21. Let f : X → Y be a continuous function. Determine whether each the following statements is true or false: (a) If f is injective, then f preserves interior points, i.e., f (S ◦ ) ⊂ f (S)◦ for all S ⊂ X. (b) If f is surjective, then f preserves interior points, i.e., f (S ◦ ) ⊂ f (S)◦ for all S ⊂ X.

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(c) If f is injective, then f preserves accumulation points, i.e., f (S  ) ⊂ f (S) for all S ⊂ X. (d) If f is surjective, then f preserves accumulation points, i.e., f (S  ) ⊂ f (S) for all S ⊂ X. (e) If f is injective, then f preserves isolated points, i.e., f (S \ S  ) ⊂ f (S) \ (f (S)) for all S ⊂ X. (f) If f is surjective, then f preserves isolated points, i.e.,  f (S \ S  ) ⊂ f (S) \ (f (S)) for all S ⊂ X. (g) If f is injective, then f preserves boundary points, i.e., f (∂S) ⊂ ∂f (S) for all S ⊂ X. (h) If f is surjective, then f preserves boundary points, i.e., f (∂S) ⊂ ∂f (S) for all S ⊂ X. Solution : (a) Answer : False. Example : Let X := R with the discrete metric and Y := R with the usual Euclidean metric. Consider f (x) := x, x ∈ X , and S := [0, 1]. It is clear that f is continuous and injective, but f (S ◦ ) = f ([0, 1]) = [0, 1] ⊂ (0, 1) = f (S)◦ . (b) Answer : False. Example : The same Example as that in (a) applies (note that the function f is also surjective). (c) Answer : True. Proof . Let a ∈ S  . Then there is an infinite sequence of distinct points {xn }n∈N in S such that {xn } → a as n → ∞. By the injectivity of f , {f (xn )}n∈N is an infinite sequence of distinct points in f (S). By the continuity of f , we have {f (xn )} → f (a) as n → ∞. Hence f (a) ∈ f (S) . (d) Answer : False. Example : Let X := R with the usual Euclidean metric and Y := {0}. Consider f :≡ 0 and S := [0, 1]. Then f is continuous and surjective, but f (S  ) = f ([0, 1]) = {0} ⊂ φ = {0} = f (S) . (e) Answer : False.

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221

Example : Let X := R with the discrete metric and Y := R with the usual Euclidean metric. Consider f (x) := x, x ∈ X , and S := X . Then it is clear that f is continuous and surjective, but f (S \ S  ) = f (S \ φ) = f (S) = R ⊂ φ = R \ R = f (S) \ (f (S)) . (f) Answer : False. Example : The same Example as that in (e) applies (note that the function f is also surjective). (g) Answer : True. Proof . Since f is injective, we have f (X \ S) = f (X) \ f (S) ⊂ Y \ f (S) . As ∂S = S ∩ X \ S , by Exercise 2.3, Part A, Problem#16, we

have

f (∂S) = f (S ∩ X \ S) = f (S) ∩ f (X \ S) ⊂ f (S) ∩ f (X \ S) ⊂ f (S) ∩ Y \ f (S) = ∂f (S) . (h) Answer : False.

Example : Let X := R with the usual Euclidean metric and Y := {0}. Consider f :≡ 0 and S := [0, 1]. Then f is continuous and surjective, but f (∂S) = f ({0, 1}) = {0} ⊂ φ = ∂f (S). 22. Let X = S 2 \ {(0, 0, 1)}, where as usual, S 2 is the unit sphere in R3 . By considering the function f : X → R2 defined by f (x, y, z) :=



y x , 1−z 1−z



or otherwise, show that X ∼ = R2 .

for all (x, y, z) ∈ X ,

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Proof . Write f (x, y, z) = (u, v). Observe that u2 + v 2 =

x2 + y 2 1 − z2 1+z = = . 2 2 (1 − z) (1 − z) 1−z

Solving for z , we get

z=

u2 + v 2 − 1 1 + u2 + v 2

and hence

2u , 1 + u2 + v 2 2v . y = v(1 − z) = 1 + u2 + v 2

x = u(1 − z) =

This suggests that we set, for any (u, v) ∈ R2 ,

g(u, v) :=



2u 2v u2 + v 2 − 1 , , 1 + u2 + v 2 1 + u2 + v 2 1 + u2 + v 2



.

It is a routine computation to check that (f ◦ g)(u, v) = (u, v) for

all (u, v) ∈ R2 and (g ◦ f )(x, y, z) = (x, y, z) for all (x, y, z) ∈ X .

Thus f is a bijection from X onto R2 with inverse function g . Clearly,

both f and g are continuous and so the two spaces are homeomorphic.

23. For each of the following statements, determine whether it is true or false. (a) The spaces (0, 1) and R with Euclidean metric are homeomorphic. (b) The unit circle S 1 ⊂ R2 with Euclidean metric is homeomorphic to R with Euclidean metric. (c) The spaces (0, 1) ∪ {2} and (0, 1 ] with Euclidean metric are homeomorphic. Solution : (a) Answer : True. Proof . Consider the function f : (0, 1) → R given by    1 , x ∈ (0, 1) . f (x) := tan π x − 2

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223

It is evident that f is 1 − 1, onto, continuous, with continuous

inverse f −1 : R → (0, 1) given by

f −1 (y) :=

1 1 tan−1 y + , π 2

y∈R.

Thus (0, 1) ∼ = R. (b) Answer : False.

Justification . Being a closed and bounded subset of R2 , S 1 is compact. Since compactness is a topological property, S 1 cannot be homeomorphic to the non-compact space R. (c) Answer : False.

Justification . Write X = (0, 1) ∪ {2}. Suppose there were a homeomorphism f : X → (0, 1 ]. Then as (0, 1) ⊂ X is closed,   f (0, 1) ⊂ (0, 1 ] should also be closed. However, observe that   f (0, 1) = (0, 1 ] \ {f (2)} (0, 1) = (0, a) ∪ (a, 1 ] 



if f (2) = 1 if f (2) = a ∈ (0, 1) .

In either case, f (0, 1) is not closed in (0, 1 ].

24. Determine which of the following spaces are homeomorphic: (i) [0, 1 ], (ii) (−1, 1), (iii) [−1, 1 ], and (iv) R. All are equipped with the Euclidean metric. Solution : By Exercise 2.3, Part B, Problem #23, (0, 1) ∼ = R. But ∼ then it is clear that (−1, 1) = (0, 1) via the homeomorphism f (t) := t+1 , t ∈ (−1, 1), we have (−1, 1) ∼ = R. On the other hand, [0, 1 ] ∼ = 2

[−1, 1] via the homeomorphism g(t) := 2t − 1, t ∈ [0, 1 ]. Finally, (−1, 1) ∼ = [−1, 1 ]. In fact, if there was a homeomorphism h : [−1, 1 ] → (−1, 1), then (−1, 1) = h[−1, 1 ] has to be com-

pact, which is absurd. Hence we have

[0, 1 ] ∼  (−1, 1) ∼ = [−1, 1 ] ∼ = =R.

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Metric Space Topology: Examples, Exercises and Solutions

(a) If f : [a, b ] → R is one to one and continuous, f (a) = c = d = f (b), describe the image set f ([a, b ]). (b) Determine whether the intervals (−1, 1) and (−1, 1 ] are homeomorphic to each other. Solution : (a) Suppose first that c < d. As c and d are in the range of f , by Intermediate value Theorem, all points in [c, d ] are in the range of f , that is, f ([a, b ]) ⊃ [c, d ]. Suppose there exists t ∈

/ [c, d], say f (t) = p > d. By Intermediate[a, b ] with f (t) ∈

value Theorem, for any s ∈ (d, p), there exist u ∈ (a, t) and

v ∈ (t, b) such that f (u) = s = f (v), which contradicts to the injectivity of f . Therefore f ([a, b ]) ⊂ [c, d]. Hence f ([a, b ]) = [c, d ]. In case d < c, by analogous arguments, we get f ([a, b ]) =

[d, c ]. (b) Suppose there exists a homeomorphism f : (−1, 1) → (−1, 1 ].

Let c ∈ (−1, 1) be the unique point such that f (c) = 1.

Pick any x0 ∈ (−1, c) and write y0 := f (x0 ). Note that

y0 = 1 and so y0 ∈ (−1, 1). Then f : [x0 , c] → R is one to one, continuous, and with f (x0 ) = y0 < 1 = f (c). By (a), f ([x0 , c]) = [y0 , 1]. On the other hand, if we choose x0 ∈ (c, 1), we would get f ([c, x0 ]) = [y0 , 1] = f ([x0 , c]), which contradicts to the injectivity of f .

26. For each of the following statements, determine if it is true or false. If it is true, prove it; if not, give a counterexample. (a) Homeomorphisms preserve isolated points. (b) Homeomorphisms preserve boundedness. (c) Homeomorphisms preserve denseness. (d) Homeomorphisms preserve completeness. Solution : (a) Answer : True.

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∼ =

Proof . Let f : X −→ Y be a homeomorphism and x ∈ X be an isolated point of X . Then there exists r > 0 such that BX (x, r) = {x}. Since every homeomorphism is an open mapping, {f (x)} = f ({x}) = f (BX (x, r)) ⊂ Y is an open set in Y . Hence f (x) is an isolated point of Y . (b) Answer : False. Example : By Exercise 2.3, Part B, Problem #23, (0, 1) ∼ = R. (0, 1) is bounded but R is not. (c) Answer : True. Proof . Let U ⊂ X be a dense subset of X and f : X → Y be a homeomorphism. For any nonempty open set V ⊂ Y , since f is continuous, f −1 (V ) open in X . Since f is onto, f −1 (V ) = φ. Since U ⊂ X is dense, we have U = X and so U ∩ f −1 (V ) = φ. Since f is bijective, we have f (U ) ∩ V = f (U ∩ f −1 (V )) = φ. This shows f (U ) intersects every nonempty open subset of Y and so f (U ) is dense in Y . (d) Answer : False. Example : By Exercise 2.3, Part B, Problem #23, (0, 1) ∼ = R. R is complete but (0, 1) is not. 27. Let n ∈ N. Consider the open unit ball B n := {x ∈ Rn : x < 1} ⊂ Rn and the function f : B n → Rn defined by f (x) :=

1 x. 1 − x

(a) Show that f is a continuous bijection. (b) Determine whether B n and Rn are homeomorphic. Solution :

(a) By Example 2.3.6, f is continuous. Suppose there exist x, y ∈

B n with f (x) = f (y). Observe that y x = f (x) = f (y) =⇒ 1 − x 1 − y y x = =⇒ x = y . =⇒ 1 − x 1 − y

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Combining x = y and f (x) = f (y) yields x = y , as desired. Finally, for any z ∈ Rn , the point

x :=

z ∈ B n (how to come up with that?) 1 + z

will satisfy f (x) = z . (b) As f is bijective, the inverse function f −1 : Rn → B n exists.

z . By Example 1 + z is also continuous. Hence B n ∼ = Rn .

By our consideration in (a), f −1 (z) = 2.3.6 again, f −1

28. Show that every metric space is homeomorphic to a bounded metric space. Proof . Let (X, d) be a metric space. It is easily verified that d1 := min{1, d} is also a well-defined metric on X . We claim that the identity function f : (X, d) → (X, d1 ) is a homeomorphism. In fact, it is obvious that f is a bijection. On the other hand, since d1 ≤ d, whenever {xn }n∈N is a sequence in X which is convergent in d to a point x ∈ X , the image sequence {f (xn )}n∈N = {xn }n∈N must be convergent in d1 to x, which is just f (x). Hence f is continuous. Conversely, if {xn }n∈N is a sequence in X with {xn } → x in d1 as n → ∞, then for any 1 > ε > 0, there exists N ∈ N such that d1 (xn , x) < ε < 1 for all n ≥ N . But then as d1 (xn , x) = min{1, d(xn , x)}, the last inequality implies d(xn , x) = d1 (xn , x) < ε for all n ≥ N and so {xn } → x

in d as n → ∞. Therefore, f −1 is also continuous.

29. Recall the definitions of metrics d1 and d∞ on C[0, 1] as defined in Example 1.1.2. Determine whether the identity function ι : (C[0, 1], d1 ) → (C([0, 1], d∞ )) defined by ι(f ) := f is a homeomorphism. Answer : No.

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Justification . It is clear that ι is bijective. However, ι is not continuous. In fact, consider, for any n ∈ N, fn : [0, 1] → R defined by fn (x) := xn for any x ∈ [0, 1]. Clearly, fn ∈ C[0, 1] for all n ∈ N. Observe that  1 1 d1 (fn , 0) = xn dx = for all n ∈ N , n+1 0   for all n ∈ N . d∞ (fn , 0) = sup |xn | : x ∈ [0, 1] = 1 Thus {fn } → 0 in d1 as n → ∞ but {ι(fn )} = {fn } → 0 = ι(0) in

d∞ . Therefore, by Theorem 2.3.5, we conclude that ι is not continuous.

[Note, however, that the inverse ι−1 : (C[0, 1], d∞ ) → (C[0, 1], d1 ) is continuous. In fact, if {fn } → f in d∞ as n → ∞, then for any

ε > 0, there exists N ∈ N such that

|fn (x) − f (x)| ≤ d∞ (fn , f ) < ε for all x ∈ [0, 1] and all n ≥ N and so

d1 (fn , f ) =



1 0

|fn (x)−f (x)| dx
0, if z ∈ BX (x, ε), we have

  dY g(z), g(x) = dX (z, x) < ε 



and so g(z) ∈ Bg(X) g(x), ε . Thus

    g BX (x, ε) ⊂ Bg(X) g(x), ε .

In particular, this shows g : X → g(X) is continuous.

Conversely, by analogous arguments we have

    Bg(X) g(x), ε ⊂ g BX (x, ε)

and so g −1 : g(X) → X is also continuous. Hence the assertion.

Remark . In fact, it is not hard to show that actually we have     g BX (x, ε) = Bg(X) g(x), ε and     Bg(X) g(x), ε = g BX (x, ε) ,

although we do not require these in the proof.

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33. Let (X, d) be an arbitrary metric space, and B(X, R) the set of all bounded functions of X into R as defined in Exercise 2.3, Part B, Problem #31. (a) Fix x0 ∈ X. For any a ∈ X, consider the function φa : X → R defined by φa (x) := d(x, a) − d(x, x0 )

for any x ∈ X .

Show that φa ∈ B(X, R). (b) Show that X can be isometrically imbedded into B(X, R) (cf. Exercise 2.3, Part B, Problem #32.) Proof . (a) For any x ∈ X , we have

d(x, a) ≤ d(x, x0 ) + d(x0 , a) ,

d(x, x0 ) ≤ d(x, a) + d(a, x0 ) , thus

  |φa (x)| = d(x, a) − d(x, x0 ) ≤ d(x0 , a) .

Hence φa is bounded on X and so φa ∈ B(X, R).

(b) Define Φ : X → B(X, R) by

Φ(a) := φa

for any a ∈ X .

Using the metric ρ on B(X, R) defined in Exercise 2.3, Part B, Problem #31, namely,

ρ(f, g) := sup{|f (x) − g(x)| : x ∈ X},

f, g ∈ B(X, R) ,

we have, by triangle inequality,

ρ(Φ(a), Φ(b)) = ρ(φa , φb ) = sup{|φa (x) − φb (x)| : x ∈ X}

= sup{|d(x, a) − d(x, b)| : x ∈ X} ≤ d(a, b) .

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On the other hand, we have trivially

d(a, b) = |d(a, a) − d(a, b)|

≤ sup{|d(x, a) − d(x, b)| : x ∈ X} = sup{|φa (x) − φb (x)| : x ∈ X}

= ρ(φa , φb ) = ρ(Φ(a), Φ(b)) .

Combining, we see that d(a, b) = ρ(Φ(a), Φ(b)) for all a, b ∈

X . That is, Φ is metric-preserving.

34. Prove that every metric space X can be isometrically imbedded into a complete metric space X ∗ in which X is dense. Proof . By Exercise 2.3, Part B, Problem #31, (B(X, R), ρ) is complete. By Exercise 2.3, Part B, Problem #33, the mapping

Φ : X → B(X, R) defined there is metric-preserving. As a closed subset of the complete space (B(X, R), ρ), Φ(X) is complete. So Φ : X → Φ(X) ⊂ Φ(X) is an isometric imbedding of X into the complete space X ∗ := Φ(X) in which Φ(X) is dense.

35.

(a) Consider two metric subspaces X and Y of R given by X :=

∞      (3n, 3n+1)∪{3n+2} ; Y := X \{2} ∪{1} .

n=0



1 if x = 2 Define f : X → Y by f (x) := x if x = 2 , ⎧ y if y ∈ (0, 1] ⎪ ⎨ 2 y−2 and g : Y → X by g(y) := if y ∈ (3, 4) 2 ⎪ ⎩ y − 3 if y ∈ Y \ (0, 4) . Show that both f and g are continuous bijections. (b) If two metric spaces each of which is the image of the other under a bijective continuous function, must they be homeomorphic?

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Proof . (a) It is elementary to check that both f and g are bijections. Note that f |(3n,3n+1) and f |{3n+2} are continuous for all n ≥ 0,

so f is continuous on X . Similarly, g|(0,1] , g|(3n,3n+1) , and

g|{3n+2} are continuous for all n ≥ 1, so g is continuous on X .

(b) The statement is false.

Example : Consider the example in (a). So we know that f : X → Y and g : Y → X are bijective and continuous. However, X and Y are not homeomorphic. In fact, suppose ∼ = there were a homeomorphism h : X −→ Y . Since (0, 1] is   a connected component of Y , h−1 (0, 1] is a connected set   in X . Furthermore, since h is a bijection, h−1 (0, 1] is not a   singleton and so h−1 (0, 1] should lie in an interval of the form (3k, 3k + 1), k ≥ 0. In particular, the point z := h−1 (1) ∈ (3k, 3k + 1). Since h is a homeomorphism, it maps intervals   to intervals (why?). Hence h (3k, z] is some interval in (0, 1]   containing 1, and h [z, 3k + 1) is also some interval of (0, 1] containing 1. But then as h is 1 − 1, this is simply impossible.

Chapter 3 Connectedness In this chapter, we shall study another important concept of a metric space: connectedness. Naively, we all seem to understand what we mean by saying that two objects are connected or separated. For example, it is clear that the intervals [ 0, 1 ] and [ 2, 3 ] are “separated” or, equivalently, “not connected”. However, for more subtle situations, the concept of “separatedness” or “connectedness” may not be as clear. For example, the intervals [ 0, 1 ] and ( 1, 2 ] are really disjoint. But they are back-to-back close to each other. In fact, the two of them together forms a single interval [ 0, 2 ] and they are just an artificial splitting of the interval. In that case, are they separated or connected? How about the intervals [ 0, 1) and    ( 1, 2 ]? How about the two sets x, sin x1 ∈ R2 : 0 < x ≤ 1 and   ( 0, 0 ) ? In order to answer such questions, like what we have to do with other seemingly obvious but in reality subtle concepts, we have to rigorously formulate the concept of “connectedness”. It turns out that it would lead to some beautiful results which in turn have some important feedback to the equivalence problem.

3.1 Connectedness Unlike other terminologies, it is rather clumsy to define the concept of connectedness in a direct manner. Instead, we shall first define the concept of disconnectedness or separatedness, and then define connectedness as the state of being not disconnected. Definition 3.1.1. A metric space X is said to be disconnected (or separated) if X = A ∪ B, where A = φ, B = φ, A ∩ B = φ, and both A, B are open in X. In this case, {A, B} is called a separation 233

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of X. X is said to be connected if it is not disconnected. A subset S ⊂ X is said to be connected if, when considered as a metric space itself under the induced metric, is connected. Examples 3.1.2. (i) X := [ 0, 1 ] ∪ [ 2, 3 ] is disconnected. Proof . As [ 0, 1 ] and [ 2, 3 ] are nonempty disjoint open subsets of X whose union is X, they form a separation of X. (ii) X := [ 0, 1 ) ∪ ( 1, 2 ] is disconnected. Proof . As [ 0, 1 ) and ( 1, 2 ] are nonempty disjoint open subsets of X whose union is X, they form a separation of X. (iii) X := R \ {0} is disconnected. Proof . As (−∞, 0) and (0, ∞) are nonempty disjoint open subsets of X whose union is X, they form a separation of X. (iv) Q is disconnected.   Proof . Take any x ∈ R \ Q. Then (−∞, x) ∩ Q, (x, ∞) ∩ Q is a separation of Q. (v) Every discrete metric space with more than one element is disconnected. Proof . Let X be a discrete metric space with #(X) > 1 and   a ∈ X. Then X \ {a} = φ and so {a}, X \ {a} is a separation of X. (vi) Every metric space contains connected subsets. Proof . Since a singleton cannot be split into two nonempty subsets, it is clear that every singleton is connected. In particular, every metric space contains connected subsets. (vii) S 1 \ {two distinct points on S 1 } ⊂ R2 is disconnected. Proof . Without loss of generality, we consider S 1 \{n, s}, where n := (0, 1) denotes the “north pole” and s := (0, −1) the “south pole” of the unit circle S 1 . Then      S 1 ∩ (x, y) ∈ R2 : x > 0 , S 1 ∩ (x, y) ∈ R2 : x < 0 is a separation of S 1 \ {n, s}.

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235

(viii) S 2 \ S 1 ⊂ R3 is disconnected, here by S 1 we mean any great circle (i.e., circle of unit length) on the unit sphere S 2 . Proof . Without loss of generality, we take S 1 as the unit circle   (x, y, 0) : x2 + y 2 = 1 in the xy-plane in R3 . Then 

    S 2 ∩ (x, y, z) ∈ R3 : z > 0 , {S 2 ∩ (x, y, z) ∈ R3 : z < 0

is a separation of S 2 \ S 1 .

Remark. By taking a careful look into Example 3.1.2 (iv), (v), (vi), (vii), (viii), it is that evident that in general, if a metric space X is disconnected, there could be more than one separation of X. Corollary 3.1.3. The following statements are all equivalent: (a) X is disconnected, i.e., X = A∪B, where A, B = φ, A∩B = φ, A ∪ B = X, and both A, B are open. (b) X = A ∪ B, where A, B = φ, A ∩ B = φ, A ∪ B = X, and both A, B are closed. (c) There exists a proper nonempty clopen subset of X. Proof. (a) ⇒ (b) : If X = A ∪ B, where A, B = φ, A ∩ B = φ, A ∪ B = X, and both A, B are open, then as complements of each other, B = X \ A and A = X \ B are closed. Hence (b). (b) ⇒ (c) : If X = A ∪ B, where A, B = φ, A ∩ B = φ, A ∪ B = X, and both A, B are closed, then A = X \ B is open. As B = φ, A = X \ B  X and so A is a proper nonempty clopen subset of X.

(c) ⇒ (a): If there exists a proper nonempty clopen subset A ⊂ X, define B := X \ A. As A is proper, B =  φ. As A is closed, B = X \ A is open. It is also clear that A ∩ B = φ and X = A ∪ B. Hence (a). 

In general, it is not difficult to prove that a metric space is disconnected, as all that is required is to produce one single separation of it. As expected, it is more difficult to show that a

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metric space is connected, as we need to show that there cannot be any separation of it, that is, we need to exhaust all possible splittings of the metric space into two portions and show that none could constitute a separation. To this end, the use of a continuous function with only two possible values would simplify the arguments enormously. Definition 3.1.4. Any continuous function from X into the space {0, 1} with discrete metric is called a 2-valued function on X. Theorem 3.1.5. X is connected if and only if the only possible 2-valued functions on X are constant functions. Proof. (⇒): Let f : X → {0, 1} be a 2-valued function. Since f is continuous, both A := f −1 (0) and B := f −1 (1) are open in X, A ∪ B = X and A ∩ B = φ. But since X is connected, this can happen only when one of A, B is empty. If A = φ, then X = B and so f ≡ 1. If B = φ, then X = A and so f ≡ 0. (⇐): We use contra-positive argument. If X is disconnected, then X = A ∪ B, A = φ, B = φ, A ∩ B = φ, and both A, B are open in X. Define f : X → {0, 1} by f (x) :=



0 x∈A 1 x∈B .

As the only possible open sets in {0, 1} are φ, {0}, {1}, and {0, 1}, with inverse images φ, A, B, and X, respectively, which are all open in X, the function f is continuous. Since A = φ and B = φ, f is a non-constant 2-valued function on X.  Theorem 3.1.6. Intervals in R are connected and they are all connected subsets of R, that is, I ⊂ R is an interval if and only if it is connected.

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237

Proof. (⇒): Let I ⊂ R be an interval. If I is a singleton, there is nothing to prove. So assume that I is a nondegenerate interval, that is, #(I) > 1. Let f : I → {0, 1} be a 2-valued function on I. Suppose there are points a < b in I such that f (a) = f (b), say f (a) = 0, f (b) = 1. Let t := inf{x ∈ [ a, b ] : f (x) = 1}. Then t ∈ (a, b] ⊂ I and f (x) = 0 for all x < t, hence by the continuity of f , we have f (t) = limx→t− f (x) = 0. On the other hand, by the definition  of infimum, for any n ∈ N, there exists xn ∈ t, t + n1 ∩ I such that f (xn ) = 1. Note that {xn }n∈N is a sequence in I converging to t as n → ∞. So by the continuity of f , we must have f (t) = limx→t f (x) = limn→∞ f (xn ) = 1, which is a contradiction. Hence f = constant and so by Theorem 3.1.5, I is connected. (⇐): If I is not an interval, there exist a < c < b with a, b ∈ I and   c∈ / I. Then I ∩ (−∞, c), I ∩ (c, ∞) is a separation of I and so I is disconnected.  Example 3.1.7. S := [ 0, 1 ] ∪ ( 1, 2 ] ⊂ R is connected. Proof. Although S is artificially split into two disjoint portions, it is actually the single interval [ 0, 2 ] and so in particular, by Theorem 3.1.6, S is connected.

Definition 3.1.8. Two subsets S, T of a metric space X are said to be separated if S ∩ T = φ and S ∩ T = φ. Example 3.1.9. (a) [ 0, 1 ] and {2 } are separated subsets of R. (b) [ 0, 1) and (1, 2 ] are separated subsets of R. (c) [ 0, 1 ] and (1, 2 ] are not separated subsets of R. Theorem 3.1.10. If S, T ⊂ X are connected subsets which are not separated, then S ∪ T is connected.

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Metric Space Topology: Examples, Exercises and Solutions

Proof. As S and T are not separated, without loss of generality, we may assume that S ∩ T = φ. Let f be a 2-valued function on S ∪ T . As S and T are connected, being 2-valued functions on connected sets, f |S and f |T are constant functions. Take a ∈ S ∩ T . There is a sequence {xn }n∈N in S such that {xn } → a as n → ∞. As f |S is constant, say f ≡ 0 on S, we have f (xn ) = 0 for all n. By the continuity of f , we have f (a) = limn→∞ f (xn ) = 0. But then as a ∈ T and f is constant on T , we also have f |T ≡ f (a) = 0 and so f ≡ 0 on S ∪ T . By Theorem 3.1.5, S ∪ T is connected.  Example 3.1.11. We re-visit Example 3.1.7. In view of Example 3.1.9 (c), [ 0, 1 ] and (1, 2 ] are connected subsets of R which are not separated, so by Theorem 3.1.10, [ 0, 1 ] ∪ ( 1, 2 ] is connected. Theorem 3.1.12. If S ⊂ X is connected, then every T ⊂ X with S ⊂ T ⊂ S is connected. Proof. If f : T → {0, 1} is a 2-valued function on T , f |S is a 2valued function on S. Since S is connected, by Theorem 3.1.5, f |S = constant, say 0. For any x ∈ T ⊂ S, there exists a sequence {xn }n∈N in S such that {xn } → x as n → ∞. Since f is continuous, we have f (x) = limn→∞ f (xn ) = 0. Thus f ≡ 0 on T and so T is connected.  Theorem 3.1.13. Connectedness is preserved by continuous functions, hence in particular, connectedness is a topological property. Proof. Let f : X → Y be continuous and S ⊂ X a connected set. Let g : f (S) → {0, 1} be any 2-valued function on f (S). Then g ◦ f |S : S → {0, 1} is a 2-valued function on S and hence must be constant on S. Therefore, g is constant on f (S) and thus f (S) is connected. 

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239

Examples 3.1.14. (i) S 1 is connected. Proof . Consider the function f : [ 0, 2π] → C given by f (x) := eix , x ∈ [ 0, 2π]. It is clear that f is continuous and f ([ 0, 2π]) = S 1 . By Theorem 3.1.6, [ 0, 2π] is connected. Hence, by Theorem   3.1.13, S 1 = f [ 0, 2π] is connected. (ii) S 1 \ {a point on S 1 } is connected. Proof . Without loss of generality, we work on S 1 \{(1, 0)} ⊂ C. Consider again the function f : (0, 2π) → C given by f (x) :=   eix , x ∈ (0, 2π). As above, f is continuous and f (0, 2π) = S 1 \ {(1, 0)} [note that it is indeed a homeomorphism of (0, 2π) onto S 1 \ {(1, 0)}]. By Theorem 3.1.6, (0, 2π) is connected. Hence by   Theorem 3.1.13, S 1 \ {(1, 0)} = f (0, 2π) is connected. (iii) Continuous functions map intervals onto intervals. Proof . Let f : R → R be continuous and I ⊂ R be an interval. By Theorem 3.1.6, I is connected. By Theorem 3.1.13, f (I) ⊂ R is also connected. By Theorem 3.1.6 again, f (I) is an interval. Hence continuous functions map intervals onto intervals. Theorem 3.1.15. The graph of a continuous function f : I → R over an interval I is connected. Proof. Equip I × R with the Euclidean metric on R2 . It is easy to verify that the function F : I → I × R given by   F (x) := x, f (x) , x ∈ I ,

is continuous. Note that the graph of f is precisely the image of I under the continuous function F and hence is connected by Theorem 3.1.13.  Example 3.1.16. The set    

1 : 0 < x ≤ 1 ∪ ( 0, 0) ⊂ R2 T := x, sin x is connected.

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Proof. Observe first that the set  

1 : 0 < x ≤ 1 ⊂ R2 S := x, sin x

is the graph of the continuous function f : (0, 1] → R defined by f (x) := sin x1 . By Theorem 3.1.15, S is connected. Note that  

1 : 0 < x ≤ 1 ∪ {(0, y) : −1 ≤ y ≤ 1} S= x, sin x

and S ⊂ T ⊂ S. By Theorem 3.1.12, T is connected.

A useful result in elementary Calculus is the Intermediate Value Theorem, which states that if a continuous real-valued function assumes two values a, b in an interval I, then it will assume all values between a and b in I. This result can easily be extended to the context of continuous real-valued functions on a general metric space. Theorem 3.1.17 (Intermediate Value Theorem). Let f : X → R be a continuous function on a connected metric space X. If f takes on two different values a, b in X, then f takes on every real number between a and b in X. Proof. It is obvious in view of Theorem 3.1.13 and Theorem 3.1.6. In fact, as f is continuous and X is connected, f (X) ⊂ R is connected

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and hence it is an interval. So if a, b ∈ f (X), say a < b, then [ a, b ] ⊂ f (X).  Theorem 3.1.18. The union of any collection of connected subsets of X with nonempty intersection is connected. Proof. Let {Uα }α∈Λ be a collection of connected subsets of X with

α∈Λ Uα = φ. Fix t ∈ α∈Λ Uα . Let f : α∈Λ Uα → {0, 1} be a 2valued function. Then for any β ∈ Λ, f |Uβ : Uβ → {0, 1} is a 2-valued function on Uβ . Since Uβ is connected, by Theorem 3.1.5, f |Uβ must be constant. In particular, f (x) = f (t) for all x ∈ Uβ . Since this is

true for all β ∈ Λ, f (x) = f (t) for all x ∈ α∈Λ Uα . Hence f is a  constant and so by Theorem 3.1.5, α∈Λ Uα is connected.

Remark. It is evident that the condition in Theorem 3.1.18 is excessively strong. For example, if there are three connected sets C1 , C2 , C3 ⊂ X such that C1 ∩ C2 = φ and C2 ∩ C3 = φ, then even if C1 ∩ C2 ∩ C3 = φ, we can still conclude that C1 ∪ C2 ∪ C3 is connected. So the condition for the Theorem could be suitably relaxed (see Exercise 3.1, Part B, #2, 3). However, as any relaxation of the condition would lead to more tedious and technical statements which may blur the vision, it is best to leave it as it is, as the readers could readily apply it to handle various situations. Examples 3.1.19. (i) We have seen in Example 3.1.14 (i) that S 1 is connected. An alternative proof goes as follows: By Example 3.1.14 (ii), the subsets S 1 \ {(0, 1)} and S 1 \ {(0, −1)} are connected. Since     1 S \ {(0, 1)} ∩ S 1 \ {(0, −1)} = φ, by Theorem 3.1.18, S 1 =     1 S \ {(0, 1)} ∪ S 1 \ {(0, −1)} is connected.

(ii) B(a, r) and B(a, r) ⊂ Rn are connected. Proof . Every radius of B(a, r) is clearly homeomorphic to the interval [0, r) ⊂ R, which is connected. Hence all radii of B(a, r) are connected. But then as the intersection of all radii of B(a, r)

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is clearly nonempty, actually equals the singleton {0} ⊂ Rn to be precise, by Theorem 3.1.18, the union of all radii of B(a, r), which is the entire B(a, r), is connected. The connectedness of B(a, r) can be derived by similar arguments, or by the observation that since we are working in Rn , we have B(a, r) = B(a, r) and so by Theorem 3.1.12, B(a, r) is connected. (iii) In R2 , let  1 , A := (x, y) ∈ R : 0 < x ≤ 1 , y = sin x   B := (x, y) ∈ R2 : −1 ≤ x ≤ 0 , y = 0 , 

2

X := A ∪ B . Then X is connected.

Proof . Let f : X → {0, 1} be a 2-valued function. Then f |A and f |B are 2-valued functions on A and B, respectively. Since A, B are connected, f |A and f |B are constants. Say, f ≡ 0   on B. In particular, f (0, 0) = 0. Since f is continuous on X, f ≡ 0 in a neighborhood of (0, 0) in X. But then as each neighborhood of (0, 0) in X contains points in A, so f = 0 at these points of A. Since f |A = constant, we have f ≡ 0 in A and so f ≡ 0 on X. Thus X is connected.

Alternatively, in view of Example 3.1.16, the set A∪{(0, 0)} is connected. On the other hand, as B is connected and

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(A ∪ {(0, 0)}) ∩ B = φ, by Theorem 3.1.18, X = A ∪ B = (A ∪ {(0, 0)}) ∪ B is connected. Definition 3.1.20. A maximal connected set in a metric space is called a (connected) component of the metric space. Remark. Fix x ∈ X. Clearly, x is contained in some nonempty connected subset of X. By Theorem 3.1.18, the union U (x) of all connected subsets of X which contain x is also connected. Now as no connected subset of X which contains x could be larger than U (x), we see that U (x) is a connected component of X. Observe that connected components of X are either identical or disjoint. Hence every metric space X is decomposed into a disjoint union of connected components of X. Corollary 3.1.21. (a) Every component is closed. (b) Components may not be open. (c) Components of open subsets S ⊂ Rn are open in Rn . Proof . (a) If S ⊂ X is a component, S is connected and so S ⊃ S must also be connected. By the maximality of S, we have S = S.   (b) Example: {0} is a component of X := n1 : n ∈ N ∪ {0} ⊂ R but it is not open in X. (c) Let S ⊂ Rn be open and T ⊂ S be a component of S. Then for any x ∈ T , there is an open ball BRn (x) ⊂ S ⊂ Rn . By Example 3.1.13, BRn (x) is connected. By the maximality of T ,  we must have BRn (x) ⊂ T . Hence T is open. Theorem 3.1.22. Every open set S ⊂ Rn can be expressed in one and only one way as a countable disjoint union of open connected sets, which are precisely the components of S. Proof. By the previous Remark and Corollary 3.1.21, S is a disjoint union of its components which are open sets in Rn . By Lindel¨of’s

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Theorem (Theorem 1.2.15), a countable sub-collection of all components of S forms a cover of S. But then as the collection of components are pairwisely disjoint, no proper sub-collection could cover S. Hence the original collection of components must already be countable. Finally, for uniqueness, suppose S = ∪∞ n=1 Cn , where the Cn ’s are pairwisely disjoint open connected subsets. Observe that being connected, each Cn is lying in some component of S. Since all the Cn ’s are open, for each m ∈ N, Cm = S \ ∪n=m Cn is also closed in S. Hence each Cm is clopen in S and so each Cm must be a component of S, for otherwise it would be a proper clopen subset of the component which contains it, which violates the fact that components are connected. Hence the decomposition of S into components is unique up to order.  Remark. Recall that Theorem 1.2.16 states that every nonempty open subset of R can be expressed as the union of countably many pairwisely disjoint open intervals in R, and that such a collection of intervals is unique. It has been remarked that since these open intervals are pairwisely disjoint, each of them is indeed maximal. Theorem 3.1.22 is the corresponding result in Rn .

Connectedness

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Exercise 3.1 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. If S, T ⊂ X are disjoint, then S ∪ T is disconnected.

Answer : False. Example : Let X := R, S := [0, 1], T := (1, 2). Clearly S and T are disjoint but S ∪ T = [0, 2) is connected.

2. If S ⊂ X is connected, then S ⊂ S  .

Answer : False. Example : Let X := R and S := {0}. Then S is connected but S ⊂ φ = S  .

3. If S ⊂ X is connected, then ∂S is connected.

Answer : False. Example : Let X := R and S := (0, 1). Then S is connected but ∂S = {0, 1} is not.

4. If S ⊂ X is connected, then S = S  .

Answer : False. Example : Let X := R and S := {0}. Then S is connected but S = {0} = φ = S  .

5. If S ⊂ X and S = S  , then S is connected.

Answer : False. Example : Let X := R and S := the Cantor Set. Then by Exercise 1.4, Part B, Problem #4, S = S  . But it is obvious that S is not connected.

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6. If S ⊂ X and S ◦ is connected, then S is connected. Answer : False. Example : Let X := R and S := (0, 1) ∪ {2 }. Then S ◦ = (0, 1) is connected but S is not. 7. If S ⊂ X and S is connected, then S is connected. Answer : False. Example : Let X := R and S := Q. Then S = R is connected but S is not. 8. If S ⊂ X and S  is connected, then S is connected. Answer : False. Example : Let X := R and S := Q. Then S  = R is connected but S is not. 9. If S ⊂ X and the set of isolated points S \S  is connected, then S is connected. Answer : False. Example : Let X := R and S := (0, 1) ∪ {2}. Then S \ S  = {2} is connected but S is not. 10. If S ⊂ X and ∂S is connected, then S is connected. Answer : False. Example : Let X := R and S := Q. Then ∂S = R is connected but S is not. 11. If S ⊂ X is connected, then S ◦ is connected. Answer : False. Example : Let X := R2 and S := {(x, y) : xy ≥ 0} ⊂ X . Then S is connected but S ◦ = {(x, y) : xy > 0} is disconnected, with a   separation {(x, y) : x > 0, y > 0}, {(x, y) : x < 0, y < 0} .

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12. If S, T ⊂ X are connected and S ∩ T = φ, then S ∩ T is connected. Answer : False. Example : Let X := R2 , S := S 1 , and T := {0} × R. Then S , T are connected and S ∩ T = {(0, 1), (0, −1)} = φ, but it is clearly

not connected.

13. If S, T ⊂ X are connected, then S ∪ T is connected. Answer : False. Example : Let X := R, S := (0, 1), and T = (1, 2). Then S and T are connected but S ∪ T = (0, 1) ∪ (1, 2) is not. 14. If {A, B } is a separation of X and S ⊂ X is connected, then either S ⊂ A or S ⊂ B. Answer : True. Proof . As A ∪ B = X , A ∩ B = φ and both A, B are open in X , we have

(S ∩ A) ∪ (S ∩ B) = S ∩ (A ∪ B) = S ∩ X = S , (S ∩ A) ∩ (S ∩ B) = S ∩ (A ∩ B) = S ∩ φ = φ ,

and both S ∩ A and S ∩ B are open in S . Since S is connected, we must have S ∩ A = φ or S ∩ B = φ, for otherwise S ∩ A and S ∩ B

will form a separation of S . Hence S ⊂ B or S ⊂ A.

15. If A ⊂ X is clopen and S ⊂ X is connected, then either S ⊂ A or S ∩ A = φ. Answer : True. Proof . It follows immediately from Exercise 3.1, Part A, Problem #14.

16. For any S ⊂ X, if S ◦ = φ and S = X, then X \ ∂S is disconnected.

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Answer : True. Proof . As X = S ∪· (X \ S) = S ◦ ∪· ∂S ∪· (X \ S), we have (X \ ∂S) = S ◦ ∪· (X \ S). As both S ◦ and X \ S are open and nonempty, they form a separation of X . 17. Every open ball B(a, r) ⊂ X is connected.

Answer : False. Example : Let X := R \ {0}. Then B(1, 2) = (−1, 0) ∪ (0, 3) is disconnected.

   18. Let S := x, sin x1 : 0 < x ≤ 1 ⊂ R2 . Then T := S ∪ {(0, 1)} ⊂ R2 is connected.

Answer : True. Proof . It is readily seen that S = S ∪ {(0, y) : −1 ≤ y ≤ 1}. Being the graph of a continuous function over an interval, S is connected. Since S ⊂ T ⊂ S , by Theorem 3.1.12, T is connected.

19. If f : X → Y is continuous and T ⊂ Y is connected, then f −1 (T ) is connected. Answer : False. Example : Consider f : S 1 → R defined by f (x, y) := x, which   is clearly continuous on S 1 . The interval T := − 12 , 12 ⊂ R is   connected but f −1 (− 12 , 12 ) is separated into two disjoint open arcs on the unit circle and is hence disconnected.

20. Every finite metric space with at least two elements is disconnected. Answer : True. Proof . Write X = {x1 , . . . , xn }, n ≥ 2. Let r := min2≤i≤n d(xi , x1 ). Then r > 0 and xi ∈ B(x1 , r) for every i = 2, . . . , n. That means {x1 } = B(x1 , r) ⊂ X is open in X . But then being a singleton, {x1 } is also closed in X . Hence it is a proper clopen nonempty subset of X and so X must be disconnected.

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Alternatively , consider the function: f : X → {0, 1} defined by f (x1 ) = 0 and f (xi ) = 1 for i = 2, . . . , n. Similar to the first proof, it is evident that {xi } is open in X for every i = 1, . . . , n and so in particular, the inverse image of any open sets in {0, 1} under f are open in X . That is, f is continuous. Hence X admits a non-constant 2-valued function and so it is disconnected. 21. Every connected metric space with at least 2 elements is uncountable. Answer : True. Proof . Let a, b ∈ X be two distinct elements in X . Let f : X → R be defined by

f (x) := d(x, a)

for all x ∈ X .

Then f is continuous. Since X is connected, by Theorem 3.1.13,

f (X) ⊂ R is also connected, hence an interval. Since f (a) = 0 and f (b) = d(b, a) > 0, by Theorem 3.1.17, f (X) is a nondegenerate interval containing the interval [0, d(b, a)]. In particular, f (X) is uncountable and hence X is uncountable.

Part B: Problems 1. Show that  the following sets in R2are connected: x2 x2 (a) S := (x1 , x2 ) : 41 + 92 = 1 ⊂ R2 ;   (b) T := xT Ax : x ∈ R2 , xT x = 1 , where A is a 2 × 2 real matrix over R. Here, as usual, x ∈ R2 is written in Cartesian coordinates as a column vector, and xT stands for its transpose. Proof . (a) Let f : R → R2 be defined by f (x) := (2 cos x, 3 sin x),

x ∈ R. It is clear that f is continuous on R and f (R) = S . Since R is connected, f (R) is connected. Hence S is connected.

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(b) Let g : R → R2 be defined by g(x) := (cos x, sin x), x ∈ R. It is clear that g is continuous on R and f (R) = S 1 =

 x ∈ R2 : xT x = 1 . Since R is connected, f (R) is connected and so S 1 is connected. On the other hand, the function h : S 1 → R given by h(x) := xT Ax, x ∈ S 1 , is also readily seen to be continuous on S 1 and h(S 1 ) = T . Since S 1 is connected, T = h(S 1 ) is connected. 

2.

(a) Let A1 , . . . , An be connected subsets of X such that

n Ai ∩ Ai+1 = φ for all i = 1, . . . , n − 1. Show that i=1 Ai is connected. (b) Does (a) remain true for a countable family of connected sets {Ai : i ∈ N}? Solution : (a) Write Bm :=

m

Ai for any m = 1, . . . , n. Then B1 = A1 is connected. Assume that Bm is connected for some m < n. Then Bm ∩ Am+1 ⊃ Am ∩ Am+1 = φ. As Bm and Am+1 are connected, by Theorem 3.1.18, Bm+1 = Bm ∪ Am+1 is

n connected. Hence by induction, i=1 Ai = Bn is connected. i=1

(b) Answer : Yes, (a) remains valid.

Proof . Write A :=

∞

i=1 Ai .

Suppose to the contrary that A

is disconnected. Then A has a proper clopen nonempty subset

U . Observe that for any i ∈ N, U ∩ Ai is a clopen subset of Ai .

Since Ai is connected, we have U ∩ Ai = φ or U ∩ Ai = Ai ,

i.e., U ∩ Ai = φ or U ⊃ Ai . Let k ∈ N be the smallest integer

such that Ak ⊂ U . Then we have U ∩Ak+1 ⊃ Ak ∩Ak+1 = φ and so U ⊃ Ak+1 . Inductively, we have U ⊃ Ai for all i ≥ k .

If k = 1, then U ⊃ Ai for all i ∈ N. This implies that A =

∞

i=1

Ai ⊂ U  A, which is absurd. If k > 1, then Ak ⊂ U

and Ak−1 ∩U = φ. But then this implies that Ak ∩Ak−1 = φ,

which is also absurd. Hence A is connected.

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3. Let Aα , α ∈ Λ, be connected subsets of X, where Λ is an arbitrary index set, such that Aγ and Aδ are not separated for

any γ, δ ∈ Λ. Show that A := α∈Λ Aα is connected.

Proof . Suppose A is disconnected. Then A = S ∪ T for some open nonempty subsets S , T ⊂ A such that S ∩ T = φ. Fix α0 ∈ Λ. Then Aα0 ⊂ A and so by Exercise 3.1, Part A, Problem #14, either Aα0 ⊂ S or Aα0 ⊂ T . Without loss of generality, suppose Aα0 ⊂ S . For any β ∈ Λ, by assumption, Aα0 and Aβ are not separated. By Theorem 3.1.10, Aα0 ∪Aβ is a connected subset of A. By Exercise 3.1, Part A, Problem #14 again, either Aα0 ∪ Aβ ⊂ S or Aα0 ∪ Aβ ⊂ T . As we know Aα0 ⊂ S , this forces Aα0 ∪ Aβ ⊂ S and so in particular,

Aβ ⊂ S . Since β ∈ Λ is arbitrary, we conclude that A = β∈Λ Aβ ⊂ S and so T = φ, which is absurd.

4. If h : [ a, b ] → R is continuous and 1 − 1, show that h is monotone. In particular, h([ a, b ]) = [ h(a), h(b) ] if h is monotonic increasing, and h([ a, b ]) = [ h(b), h(a) ] if h is monotonic decreasing. Proof . Let D := {(s, t) : a ≤ s < t ≤ b} ⊂ R2 . Geometrically, D is the triangular region in R2 with vertices (a, a), (a, b), and (b, b), excluding the hypotenuse. Hence in particular, D is connected. Define a function H : D → R by H(s, t) := h(s)−h(t). As h is continuous, H is continuous and so H(D) ⊂ R is connected, hence an interval. Since h is 1 − 1, H is never zero and so H(D) ⊂ R \ {0}. So either H(D) ⊂ (−∞, 0) in which case h is monotonic increasing, or H(D) ⊂ (0, ∞) in which case h is monotonic decreasing. 5. A set S ⊂ X is said to be totally disconnected if all connected components of S are singletons, that is, every subset T ⊂ S with #(T ) > 1 is disconnected. Show that Q ⊂ R is totally disconnected.

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Proof . It is obvious that singletons of Q are connected. Consider any subset A ⊂ Q with at least two elements, say, a < b. Pick any   c ∈ R \ Q such that a < c < b. Then (−∞, c) ∩ A, (c, ∞) ∩ A is a separation of A and so A is disconnected. Hence Q is totally

disconnected.

6. If S ⊂ X is totally disconnected, determine whether X \ S is connected. Answer : No, X \ S need not be connected.

Example : By Exercise 3.1, Part B, #5, Q ⊂ R is totally disconnected but R \ Q is clearly disconnected. 7. If S ⊂ R is totally disconnected, show that S ◦ = φ.  φ, pick a ∈ S ◦ . Then there exists r > 0 such Proof . If S ◦ = that BR (a, r) ⊂ S . But BR (a, r) is connected. Thus S contains a connected set which is not a singleton. But that contradicts the assumption that S is totally disconnected.

8. Two points a, b ∈ X are said to be connected if there is a connected subspace of X containing both a and b. Prove that if every pair of points in X are connected, then X is connected. Proof . Suppose not, then there exists a proper clopen nonempty subset U ⊂ X . Pick a ∈ U , b ∈ X \ U . By assumption, there exists a connected subset C ⊂ X containing a and b. But then U ∩ C is a proper nonempty clopen subset of C , contradicting to the connectedness of C . Alternatively , let f : X → {0, 1} be a 2-valued function on X . Fix a ∈ X . For any x ∈ X , there exists a connected set C ⊂ X  containing a and x. The restriction f C : C → {0, 1} is a 2-valued  function on C , and hence f C is constant. In particular, f (x) = f (a). Since x ∈ X is arbitrary, we conclude that f ≡ f (a) on X and so X is connected.

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9. Let Y ⊂ X. (a) For any A ⊂ Y , show that the closure of A in Y is A ∩ Y , where, as usual, A stands for the closure of A in X. (b) Prove that Y is disconnected if and only if there are nonempty subsets A, B ⊂ Y such that A ∪ B = Y , A ∩ B = φ, and B ∩ A = φ. Proof .

(a) Let B be the closure of A in Y . So B is the smallest closed set in Y containing A. Since A is closed in X , A ∩ Y is closed in

Y and it is obvious that A ⊂ A ∩ Y . Hence B ⊂ A ∩ Y . On the other hand, since B is closed in Y , we can write B = C ∩Y for some C ⊂ X that is closed in X . Since A ⊂ B ⊂ C , we have A ⊂ C = C . Thus, B = C ∩ Y ⊃ A ∩ Y . Combining, we conclude that B = A ∩ Y . (b) (⇒): Suppose Y is disconnected. Then there exist nonempty subsets A, B ⊂ Y , both closed in Y , such that A ∩ B = φ and A ∪ B = Y . By (a), we have A = [the closure of A in Y ] = A ∩ Y . Thus

A ∩ B = A ∩ (Y ∩ B) = (A ∩ Y ) ∩ B = A ∩ B = φ. Similarly, B ∩ A = φ.

(⇐) Suppose there are nonempty subsets A, B ⊂ Y such that

A ∪ B = Y , A ∩ B = φ, and B ∩ A = φ. Then

A = A ∪ φ = A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ Y and so by (a), A = the closure of A in Y , that is, A is closed in

Y . Similar arguments shows that B is closed in Y . Since it is obvious that A ∩ B = φ, we conclude that Y is disconnected. 10. Show that X is connected if and only if every nonempty, proper subset S of X has a nonempty boundary.

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Proof . Note first that for any S ⊂ X , S is clopen if and only if S = S = S ◦ if and only if ∂S = φ. Hence X is disconnected if and only if there exists a nonempty proper clopen subset S ⊂ X if and only if there exists a nonempty proper subset S ⊂ X with ∂S = φ. So by contrapositive, X is connected if and only if no proper subset S of X can have nonempty boundary if and only if every nonempty proper subset S of X must have nonempty boundary. 11. Show that every connected clopen subset of a metric space is a connected component. Proof . Let S ⊂ X be a connected clopen subset. For any T ⊂ X with S  T , as S is clopen in X , S = T ∩ S is clopen in T . Having a proper nonempty clopen subset S , the set T must be disconnected. Hence S is a connected component. 12. Identify all the components of the following metric spaces: (a) X = A ∪ B ⊂ R2 , where A := {(x, 0) : x ∈ R} and    B := x, x1 : x > 0 .    (b) X = P ∪Q∪R ⊂ R2 , where P := x, sin x1 : 0 < x ≤ 1 , Q := {(x, 0) : −1 ≤ x ≤ 0}, R := {(0, y) : −1 ≤ y ≤ 1}. Solution : (a) Observe first that both A and B are connected. Furthermore, it is elementary to see that both A and B are closed and A∩B =

φ, so A = X \ B and B = X \ A are also open in X . Being connected clopen subsets of X , by Exercise 3.1, Part B, Problem #11, both A and B are connected components of X . (b) By Example 3.1.19 (iii), P ∪ Q is connected. Next, being a line segment, R is also connected. Furthermore, since (P ∪Q)∩R = {(0, 0)} = φ, by Theorem 3.1.18, X = P ∪Q∪R is connected. So X has only one connected component, namely, X itself. 13. Let h : S 1 → R be continuous. Show that there exists a pair of antipodal points of S 1 which have the same image under h.

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That is, there exists (x, y) ∈ S 1 such that h(x, y) = h(−x, −y). Proof . If h(1, 0) = h(−1, 0), then we are done. So assume that h(1, 0) = h(−1, 0). Consider the function H : S 1 → S 1 given by H(x, y) := h(x, y) − h(−x, −y) for any (x, y) ∈ S 1 . Clearly we have H(−1, 0) = −H(1, 0) = 0, so exactly one of H(1, 0) and H(−1, 0) is positive, and the other negative. By Intermediate Value Theorem, there exists (x0 , y0 ) ∈ S 1 such that H(x0 , y0 ) = 0. That is, h(x0 , y0 ) = h(−x0 , −y0 ). 14.

(a) Let I = [0, 1] and f : I → I be any continuous function. Show that f has a fixed point in I, that is, there exists t ∈ I such that f (t) = t. (b) Is (a) still valid in case I = (0, 1]? (c) Is (a) still valid in case I = [0, ∞)? Solution : (a) First observe that if f (0) = 0 or f (1) = 1, then there is nothing to prove. So assume f (0) > 0 and f (1) < 1. Consider the function g : I → R given by

g(x) := f (x) − x

for any x ∈ I .

Then g is continuous, g(0) = f (0) − 0 > 0, and g(1) =

f (1) − 1 < 0. By Intermediate Value Theorem, there is a point t ∈ I such that g(t) = 0, or f (t) = t. (b) Answer : No, (a) will no longer be valid. Example : The function f : (0, 1] → (0, 1] given by f (x) := x , x ∈ (0, 1], is continuous on (0, 1] but with no fixed point 2 there. In fact, solving the equation f (x) = x we get x = 0 which is not in (0, 1]. So f has no fixed point in (0, 1]. (c) Answer : No, (a) will no longer be valid. Example : The function f : [0, ∞) → [0, ∞) given by f (x) := x + 1, x ∈ [0, ∞), is continuous on [0, ∞) but with no fixed point there.

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Metric Space Topology: Examples, Exercises and Solutions

15. If X is a metric space such that every continuous function f : X → R has the intermediate value property (i.e., if y1 , y2 ∈ f (X) and y is a real number between y1 and y2 , then there exists an x ∈ X such that f (x) = y), show that X is connected. Proof . Assume X is disconnected. Then there exist nonempty open subsets A, B ∈ X such that A ∪ B = X and A ∩ B = φ. Define f : X → R by f :≡ 0 on A and f ≡ 1 on B . Since the pre-image of any open set in R under f must be one of φ, A, B , or X , which are all open, f is continuous. However, it is evident that f does not have the intermediate value property. Hence X must be connected. Alternatively , Let g : X → {0, 1} be a 2-valued function on X . By assumption, g has the intermediate value property. But this is possible only if g ≡ 0 or g ≡ 1. That is, g must be a constant and so X is connected.

16. Recall that two subsets A, B ⊂ X are separated if each is disjoint from the closure of the other, i.e., if B ∩A = φ = A∩B. (a) Are disjoint subsets always separated? (b) If two sets are separated, must their closures be disjoint? (c) If A, B ⊂ X are separated and C ⊂ A ∪ B is connected, what relation must A, B and C have? Solution : (a) Answer : No.

Example : A = (−∞, 0] and B = (0, ∞) ⊂ R are disjoint, but they are not separated, as A ∩ B = {0} = φ. (b) Answer : No.

Example : A = (−∞, 0) and B = (0, ∞) ⊂ R are separated, but A ∩ B = {0} = φ. (c) Answer : C ⊂ A or C ⊂ B .

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257

Proof . Suppose A, B ⊂ X are separated and C ⊂ A ∪ B is connected. Then

(C ∩ A) ∩ (C ∩ B) ⊂ A ∩ B = φ , (C ∩ A) ∩ (C ∩ B) ⊂ A ∩ B = φ , and so C ∩ A and C ∩ B are separated. Furthermore, it is

obvious that (C ∩ A) ∪ (C ∩ B) = C . Hence if both C ∩ A and

C ∩ B are nonempty, then by Exercise 3.1, Part B, Problem #9, C is disconnected, which is absurd. Therefore, we must have either C ∩ A = φ or C ∩ B = φ, that is, either C ⊂ B or C ⊂ A.

17.

∼ =

(a) If f : X −→ Y is a homeomorphism and x0 ∈ X, then the restriction function f |X\{x0 } : X \{x0 } → Y \{f (x0 )} remains a homeomorphism. (b) Prove (again) that [0, 1) ∼  (0, 1) by using the concept of = connectedness. Proof . (a) It is obvious that the restriction function

f |X\{x0 } : X \ {x0 } → Y \ {f (x0 )} is still 1 − 1, onto, continuous, and with continuous inverse,

hence it remains a homeomorphism.

∼ =

(b) Suppose there were a homeomorphism f : [0, 1) −→ (0, 1). By (a), the restriction

  f

[0,1)\{0}

: [0, 1) \ {0} → (0, 1) \ {f (0)}

remains a homeomorphism. But then [0, 1) \ {0} = (0, 1) is





connected, while its direct image (0, 1) \ {f (0)} = 0, f (0) ∪

  f (0), 1 is disconnected. This contradicts the fact that connectedness is a topological property. So [0, 1) ∼  (0, 1). =

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18. Let f : X → Y be non-constant and continuous. If the direct image f (X) ⊂ Y of X under f has an isolated point, show that X is disconnected. Proof . Suppose y0 ∈ f (X) is an isolated point of f (X). By definition, there exists r > 0 such that BY (y0 , r) ∩ f (X) = {y0 }. Since f is continuous, both U := f −1 (BY (y0 , r))

and

V := f −1 (Y \ {y0 })

are open in X . Moreover, it is clear that U = φ and U ∪ V = X .

Since f is non-constant, we have f (X) = {y0 } and so V = φ. Finally,

since BY (y0 , r) ∩ f (X) = {y0 }, we have

(BY (y0 , r) \ {y0 }) ∩ f (X) = φ and so

U ∩ V = f −1 (BY (y0 , r)) ∩ f −1 (Y \ {y0 }) = f −1 (BY (y0 , r) ∩ (Y \ {y0 })) = f −1 (BY (y0 , r) \ {y0 }) = φ .

So X is the disjoint union of two nonempty open subsets and is thus disconnected.

Alternatively , suppose y0 ∈ f (X) is an isolated point of f (X). Let B := f (X) \ {y0 }. Since y0 is an isolated point, {y0 } and B are disjoint closed subsets of Y . Since f is non-constant, f (X) = {y0 }. Hence f (X) \ {y0 } = φ and B = φ. Define g : X → R by g(x) :=

d (f (x), y0 ) d (f (x), y0 ) + d (f (x), B)

for any x ∈ X .

Note that



f (x) = y0

⇐⇒ d(f (x), y0 ) = 0 ⇐⇒ g(x) = 0

f (x) ∈ f (X) \ {y0 } ⇐⇒ d(f (x), B) = 0 ⇐⇒ g(x) = 1 .

Hence g(X) = {0, 1}. That is, g is a non-constant 2-valued function on X and so X is disconnected.

Connectedness

259

19. Determine whether the following statements are true or false: (a) Every connected subset S ⊂ Rn is convex. (b) Every convex subset S ⊂ Rn is connected. (a) Answer : False.

Example : The unit circle S 1 ⊂ R2 is connected but obviously

not convex.

(b) Answer : True.

Proof . Suppose S is not connected. Then there exist open nonempty subsets U and V of S such that U ∩ V = φ and S = U ∪ V . Fix x ∈ U and y ∈ V . Since S is convex, the line segment joining them is contained in S , i.e., the function f : [0, 1] → S given by f (t) := (1 − t)x + ty is well defined. Moreover, it is evident that f is continuous on [0, 1]. In fact, for any t0 ∈ [0, 1] and any ε > 0, take δ := ε/x − y, then we have

d(f (t), f (t0 )) = f (t) − f (t0 ) = |t − t0 | x − y < ε whenever |t − t0 | < δ . Therefore, f is continuous at t0 .

Hence f −1 (U ) and f −1 (V ) are disjoint nonempty open sets in

[0, 1] whose union is [0, 1], that is, they form a separation of the connected set [0, 1], which is absurd. Alternatively , let f : S → {0, 1} be a 2-valued function on S . Fix a ∈ S . Since S is convex, the line segment ax joining a and  x is lying entirely in S . The restriction of f ax : ax → {0, 1} is a 2-valued function on the connected set ax and so it must be a constant. In particular, f (x) = f (a). Since x ∈ S is arbitrary, this shows f ≡ f (a) on S and so S is connected. 20. Show that homeomorphisms preserve connected components. ∼ =

Proof . Let f : X −→ Y be a homeomorphism and S ⊂ X be a connected component of X . Then f (S) ⊂ Y is connected and so it is contained in a unique connected component of Y . Call that T . Since f −1 is also continuous, f −1 (T ) ⊂ X is connected. But since

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f (S) ⊂ T , we have S ⊂ f −1 (T ). By the maximality of S , this forces S = f −1 (T ) and so f (S) = T is a connected component. 21. If X ∼ = Y , show that there is a 1 − 1 correspondence between connected components of X and those of Y . ∼ =

Proof . Let f : X −→ Y be a homeomorphism. By Exercise 3.1, Part B, Problem #20, connected components of X are mapped by f to connected components of Y . Furthermore, as f is 1 − 1, distinct connected components of X are mapped by f to distinct connected components of Y . Hence there is a 1 − 1 correspondence from the collection of connected components of X into the collection of connected components of Y . On the other hand, applying the same arguments to the homeomorphism f −1 , we see that it maps distinct connected components of Y to distinct connected components of X . Hence the assertion.

22. Determine whether the two subsets of R2 : A := {(x, y) : xy = 0} , B := {(x, y) : xy(x2 − y 2 ) = 0} are homeomorphic. Answer : No. Justification . Observe that A is the union of 2 intersecting straight lines y = 0 and x = 0, and B is the union of 4 concurrent straight lines y = 0, x = 0, and y = ±x. If there were a homeomorphism ∼ = f : B −→ A, then by Exercise 3.1, Part B, Problem #17,   f : B \ {(0, 0)} −→ A \ {f (0, 0)} B\{(0,0)}

remains a homeomorphism. However, observe that B \ {(0, 0)} has

8 components, while A \ {f (0, 0)} has either 4 or 2 components, depending on whether f (0, 0) = (0, 0) or not. That violates Exercise 3.1, Part B, Problem #21.

23. Determine whether each pair of the following subsets of R2 are homeomorphic.

Connectedness

(a) (b) (c) (d) (e) (f)

The The The The The The

261

digit 8 and the letter O. symbols × and ⊥. symbols ∃ and ∀. . objects ∇ and symbols  and ⊗. symbols −→ and =⇒.

Answer : (a) No.

Justification . Removing any point from the letter O we have one connected component, while removing the middle point from figure 8 we have two connected components. In view of Exercise 3.1, Part B, Problem #21, the two objects are not homeomorphic. (b) No.

Justification . Removing the middle point from symbol × we

have four connected components, while removing a point from symbol ⊥ we have either two of three connected components.

Hence the two objects are not homeomorphic. (c) No.

Justification . Removing a point from the symbol ∀ we have one or

two

connected

components,

while

removing

the

middle point on the vertical edge of the symbol ∃ we have three

connected components. Hence the two objects cannot not be homeomorphic. (d) Yes.

Proof . Both objects are homeomorphic to the circle. Details are left with the readers. (e) No.

Justification . Removing the two antipodal points at the end of the diameter from the symbol  we have three connected components, while removing any two points from the symbol ⊗

we have no more than two connected components. By repeated arguments of Exercise 3.1, Part B, Problems #17, and

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Metric Space Topology: Examples, Exercises and Solutions

by Exercise 3.1, Part B, Problems #21, we conclude that the two objects are not homeomorphic. (f) No.

Justification . Removing the two points in the symbol =⇒ which are the intersecting points of the horizontal lines with the slanted lines, we have 5 connected components, while removing any two points from the symbol −→ we are left with no more than four connected components. Hence the two objects are not homeomorphic.

24. Let Λ = be some index set. If V and Vα , α ∈ Λ, are connected subsets of X such that V ∩ Vα = φ for all α ∈ Λ, then V ∪

α∈Λ Vα is connected.

Answer : True. Proof . For every α ∈ Λ, since V , Vα are connected and V ∩ Vα = φ, by Theorem 3.1.18, V ∪ Vα is connected. Since 

  

Vα = φ , V ∪ Vα = V ∩ α∈Λ

α∈Λ

{V ∪ Vα }α∈Λ is a collection of connected sets with nonempty

intersection, by Theorem 3.1.12 again,

V ∪



α∈Λ

  

Vα = V ∪ Vα α∈Λ

is connected.

25. Let (X, dX ) and (Y, dY ) be metric spaces. It is easy to see that d : X × Y → R given by     d (x1 , y1 ), (x2 , y2 ) := max dX (x1 , x2 ), dY (y1 , y2 )

for any (xi , yi ) ∈ X × Y , i = 1, 2, is a well-defined metric, making (X × Y, d) a metric space called the product space of X and Y .

Connectedness

263

(a) For any (x0 , y0 ) ∈ X × Y , prove that {x0 } × Y and X × {y0 } are homeomorphic to Y and X, respectively.

(b) For any φ = A ⊂ X and φ = B ⊂ Y , if A × B is closed in X × Y , show that A is closed in X and B is closed in Y .

(c) Prove that the product space X × Y is connected if and only if both X and Y are connected.

Proof . (a) Observe that {x0 } × Y and X × {y0 } are metric subspaces of

X × Y . The projection maps π1 : X × {y0 } → X and π2 : {x0 } × Y → Y given by π1 (x, y0 ) := x and π2 (x0 , y) := y are isometries, hence homeomorphisms.

(b) Pick x0 ∈ A and y0 ∈ B . Since A × B is closed in X × Y ,

A×{y0 } = (A×B)∩(X ×{y0 }) is closed in X ×{y0 }. By (a), the projection map π1 : X × {y0 } → X is a homeomorphism. Since homeomorphisms preserve closedness, it follows that A = π1 (A × {y0 }) is closed in X . Similarly, B = π2 ({x0 } × B) is closed in Y .

(c) (⇒) Consider the projection maps π1 : X × Y → X and

π2 : X × Y → Y given by π1 (x, y) := x and π2 (x, y) := y for all (x, y) ∈ X × Y . It is elementary to check that both π1 and π2 are continuous on X × Y . So if X × Y is connected, then X = π1 (X × Y ) and Y = π2 (X × Y ) are connected.

(⇐) Suppose X and Y are connected. Fix x0 ∈ X . Write V := {x0 } × Y . For any y ∈ Y , write Uy := X × {y}. By (a), V ∼ = Y and for all y ∈ Y , Uy ∼ = X . In particular, V and all Uy ’s are connected. Moreover, V ∩ Uy = {(x0 , y)} = φ for

every y ∈ Y and X × Y = V ∪ y∈Y Uy . Hence by Exercise 3.1, Part B, Problem #24, X × Y is connected. 26. If a function preserves connectedness, determine whether it is continuous. Answer : No.

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Metric Space Topology: Examples, Exercises and Solutions

Example : Consider the function f : [0, ∞) → R defined by f (x) =



0

x=0

1 sin x

x = 0 .

Then f preserves connectedness. In fact, suppose S ⊂ [0, ∞) is con-

nected. If S is a singleton, f (S) is also a singleton and hence is connected. If S is not a singleton, then it is a nondegenerate interval.

/ S , then f is continuous on S and so f (S) is connected. If 0 ∈

If 0 ∈ S , then S is a nondegenerate interval in [0, ∞) containing the

point 0. No matter how small it is, by the definition of f , we always have f (S) = [−1, 1] which is connected. Combining, we see that f preserves connectedness. However, it is elementary to see that f is discontinuous at 0.

27. Let S be a subset of R3 consisting of the four points (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1) together with the straight line segments joining each pair of these points. If f : S → S is a self-homeomorphism, how many possibilities are there for f (0, 0, 0)? Answer : 4. Prove . By the continuity of f , there exists δ1 > 0 such that f (BS ((0, 0, 0), δ1 )) ⊂ B(f (0, 0, 0), 0.1) . Take δ := min{δ1 , 0.1}. Observe that we have BS ((0, 0, 0), δ) ∼ =

f (BS ((0, 0, 0), δ)) and hence in particular,

    BS (0, 0, 0), δ \ {(0, 0, 0)} ∼ = f BS ((0, 0, 0), δ) \ {f (0, 0, 0)} . 



Note that BS (0, 0, 0), δ \ {(0, 0, 0)} has 3 connected components.

By Exercise 3.1, Part B, Problem #21,

  f BS ((0, 0, 0), δ) \ {f (0, 0, 0)}

Connectedness

265

should also have 3 connected components. Observe that the subset

  f BS ((0, 0, 0), δ) ⊂ S is connected and the only possible scenario of removing a point from a connected subset of S would result in 3 connected components is when the point removed is one of the four given points. Thus any self-homeomorphism of S must permute the four given points and so there are exactly 4 possibilities for f (0, 0, 0).

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Metric Space Topology: Examples, Exercises and Solutions

3.2 Path-connectedness In the last section, we defined connectedness of a metric space as the state of not separated. This approach is rather indirect and causes uneasiness. At least it owes us an intuitive geometrical feeling. In this section we would discuss yet another concept of “connectedness” which is in certain sense stronger and provides a simple intuitive geometrical meaning. That is called

path-connectedness . It turns out that path-connectedness is on the one hand stronger than connectedness, and on the other hand easier to establish, hence facilitating us an efficient way of proving the connectedness of a geometric object.

Definition 3.2.1. A path (or a curve) in a metric space X is a continuous function γ : [0, 1] → X. If we write p := α(0) and q := α(1), we would refer to α as a path in X joining p to q, and the independent variable t ∈ [0, 1] is called the parameter of the path α. Sometimes it would be convenient to identify the direct image of a path with the function itself and regard the parameter t as time. Under such an identification, α (the direct image of the function α in the space X) is the trajectory in X when we travel from the point p at time t = 0 to the point q at time t = 1. Observe also that under this identification, every path in S is a (1-dimensional, although we would not make it clear here) connected subset of X. Let α : [0, 1] → X and β : [0, 1] → X be paths in X with α(0) = p, α(1) = q = β(0), β(1) = r. Then γ : [0, 1] → X defined by

γ(t) =



α(2t) β(2t − 1)

0≤t≤ 1 2

1 2

≤t≤1

is a path in X joining p to r via q through α and β, called the product of the paths α and β, and is denoted as γ = β · α.

Connectedness

267

Geometrically, what γ does is that we travel along the trajectory of α, followed by that of β, but at a speed twice as fast as that of α and β. Examples 3.2.2 (i) Let X := Rn and a, b ∈ X. The image of the function γ : [0, 1] → R defined by γ(t) := (1 − t)a + tb, t ∈ [0, 1], is a straight line segment in Rn joining a to b. It is clear that γ is continuous and so it is a path in Rn joining a and b. (ii) Let X := Rn , and α, β, γ, δ : [0, 1] → X be functions defined by α(t) := (−1 + 2t, −1) β(t) := (1, −1 + 2t) γ(t) := (1 − 2t, 1)

δ(t) := (−1, 1 − 2t)

for all t ∈ [0, 1]. Then α, β, γ, and δ are paths in R2 which form the boundary of the rectangular region D := {(x, y) ∈ R2 : |x| ≤ 1, |y| ≤ 1}. In particular, the boundary ∂D is the image of the single path δ · γ · β · α : [0, 1] → R2 . Remark. The more matured readers would notice from the definition that the requirement that the domain of a path has to be [0, 1] is a bit artificial and not strictly necessary. In fact, for any a < b ∈ R, and any continuous function γ : [a, b] → X, if we define

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Metric Space Topology: Examples, Exercises and Solutions

α : [0, 1] → [a, b] by α(t) := a+(b−a)t for any t ∈ [0, 1], then clearly α is continuous and so the composition function β := γ ◦ α : [0, 1] → X is a continuous function of [0, 1] into X, hence it is a path in X. Furthermore, it is easy to see that the direct image of β in X is exactly the same as that of the given continuous function γ. This shows that any continuous function γ : [a, b] → X can be converted into a path β : [0, 1] → X with identical direct image. In this respect, any continuous function γ : [a, b] → X can be regarded as a path in X in a broad sense, and the process of changing variables of the “path” in the broad sense is called a reparametrization. So in the broad sense, a path in X is simply a continuous function from a closed and bounded interval into X. Having said that, however, in order to standardize things, in the sequel we still require a path in X to have domain [0, 1]. Definition 3.2.3. A metric space X is said to be path-connected (or arcwise-connected) if any two points p, q ∈ X can be joined by a path in X, that is, there is a path γ : [0, 1] → X with γ(0) = p, γ(1) = q. Examples 3.2.4. (i) Let X := Rn and p, q ∈ X. By Example 3.2.2 (i), the straight line segment γ(t) := (1 − t)p + tq, t ∈ [0, 1], is a path in Rn joining p to q. Hence Rn is path-connected. (ii) Every convex set in Rn is path-connected. Proof . Let S ⊂ Rn be convex and a, b ∈ S. Similar to (i) above, the straight line segment γ(t) := (1 − t)a + tb, t ∈ [0, 1], is a path in Rn joining a to b. Since S is convex, the whole line segment joining a, b is lying entirely in S. Hence γ is a path in S joining a to b and so S is path-connected. (iii) All open balls and closed balls in Rn are path-connected. Proof . As open balls and closed balls in Rn are convex, the assertion follows immediately from (ii).

Connectedness

269

(iv) The unit circle S 1 ⊂ R2 is path-connected. Proof . The function γ : [0, 2π] → S 1 given by γ(t) := (cos t, sin t) ,

t ∈ [0, 2π] ,

is clearly continuous and it passes through all points in S 1 . Hence γ is, in the broad sense, a path in S 1 which goes through every point of S 1 , and so S 1 should be path-connected. To put everything into absolute rigor, we argue as follows: For any p, q ∈ S 1 , write p = (cos a, sin a) and q = (cos b, sin b) for some unique a, b ∈ [0, 2π). Without loss of generality, we assume a < b. Define α : [0, 1] → [a, b] by α(t) := a + (b − a)t ,

t ∈ [0, 1] .

Then clearly α is continuous and so the composition function β := γ ◦ α : [0, 1] → S 1 is a path in S 1 with    β(0) = γ α(0) = γ(a) = p   β(1) = γ α(1) = γ(b) = q .

Thus S 1 is path-connected.

(v) A torus with two generating circles removed is path-connected. Proof . Let T be a torus with two generating circles removed and a, b ∈ T . We have seen that T is homeomorphic to an ∼ = open rectangular region R ⊂ R2 . Let f : T −→ R be a homeomorphism. Since R is convex, by (ii), it is path-connected. Hence there is a path γ : [0, 1] → R joining f (a) and f (b) ∈ R. Then f −1 ◦ γ : [0, 1] → T is a path in T joining a and b.

(vi) If S and T ⊂ X are path-connected subsets of X such that S ∩ T = φ, then S ∪ T is also path-connected. Proof . Fix a ∈ S ∩ T . For any p ∈ S and any q ∈ T , since S and T are path-connected, there exist a path α : [0, 1] → S joining p to a, and a path β : [0, 1] → T joining a to q. Then the product β · α : [0, 1] → S ∪ T is a path in S ∪ T joining p to q via a. Hence S ∪ T is path-connected.

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Metric Space Topology: Examples, Exercises and Solutions

(vii) A torus in R3 is path-connected. Proof . Let T be a torus in R3 . Let {C1 , C2 }, {C˜1 , C˜2 } be two different pairs of generating circles of T , and S := T \(C1 ∪C2 ), S˜ := T \ (C˜1 ∪ C˜2 ). By (v) above, both S and S˜ are pathconnected. Furthermore, it is obvious that S ∩ S˜ = φ and so by (vi), S ∪ S˜ is path-connected. On the other hand, by (iv) above, ˜ ∩ C1 = φ, by (vi), C1 and C2 are path-connected. Since (S ∪ S) S ∪ S˜ ∪C1 is path-connected. Finally, since (S ∪ S˜ ∪C1 )∩C2 = φ, by (vi) again, S ∪ S˜ ∪ C1 ∪ C2 is path-connected. Note that S ∪ S˜ ∪ C1 ∪ C2 = T . Theorem 3.2.5. Every path-connected set is connected. Proof. Let S ⊂ X be path-connected, and f : S → {0, 1} a 2-valued function on S. Fix a ∈ S. For any x ∈ S, there is a path γ : [0, 1] → S in S joining a to x. Since γ is continuous and [0, 1] is   connected, γ [0, 1] is connected. The 2-valued function    f γ([0,1]) : γ [0, 1] → {0, 1}

must therefore be a constant and so f (x) = f (γ(1)) = f (γ(0)) = f (a). Since x ∈ S is arbitrary, f ≡ constant on S. That is, the only possible 2-valued function on S must be constant. So by Theorem 3.1.5, S is connected.  Remark. In general, the converse of Theorem 3.2.5 is not valid. So path-connectedness is strictly stronger than connectedness. Example 3.2.6 (Connectedness =⇒ × path-connectedness). In R2 , consider  1 , A := (x, y) ∈ R2 : 0 < x ≤ 1 , y = sin x   B := (x, y) ∈ R2 : −1 ≤ x ≤ 0 , y = 0 , X := A ∪ B .

Connectedness

271

It turns out that X is connected but not path-connected. Proof. In Example 3.1.16, we saw already that X is connected. To show that X is not path-connected, we first observe that both A and B are path-connected. Every point in A can be joined to another point in A by a path in A ⊂ X, and every point in B can be joined to another point in B by a path in B ⊂ X. So we only need to show that points in A cannot be joined to points in B by any path in X. In other words, it suffices to show that any path α : [0, 1] → X which passes through a point in B must stay forever in B. Equivalently, we need to show α([0, 1]) ⊂ B, or α−1 (B) = [0, 1]. Now as α passes through a point in B, we have φ = α−1 (B) ⊂ [0, 1]. Since [0, 1] is connected, it remains to show α−1 (B) ⊂ [0, 1] is clopen. Since B is closed in R2 , it is closed in X. As α is continuous, α−1 (B) is closed in [0, 1]. On the other hand, for any t0 ∈ α−1 (B) ⊂ [0, 1], since α is continuous, there exists δ > 0 s.t.

  1 . α B[0,1] (t0 , δ) ⊂ BX α(t0 ), 2

  Since B[0,1] (t0 , δ) = (t0 − δ, t0 + δ)∩ [0, 1] is connected, α B[0,1] (t0 , δ) is also connected and contains the point α(t0 ), so it is contained in the   unique connected component of BX α(t0 ), 12 which contains α(t0 ).   Now observe that the set BX α(t0 ), 12 splits into a number of parts,   including the horizontal piece BX α(t0 ), 12 ∩ B and the “almost   vertical” pieces of the curve y = sin x1 in BX α(t0 ), 12 .

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Metric Space Topology: Examples, Exercises and Solutions

It is evident that each of the “almost vertical” pieces of the   curve y = sin x1 in BX α(t0 ), 12 is a connected component of     BX α(t0 ), 12 . On the other hand, as BX α(t0 ), 12 ∩ B is an interval, it is connected and we claim that it is indeed another   connected component of BX α(t0 ), 12 . In fact, for any S ⊂     BX α(t0 ), 12 which is a strict superset of BX α(t0 ), 12 ∩ B, S contains at least one point in one of the “almost vertical” pieces of y = sin x1 . Name one of such almost vertical pieces V . It is clear that {S ∩ V, S \ (S ∩ V )} forms a separation of S and so S cannot   be connected. Thus BX α(t0 ), 12 ∩ B is a connected component of   BX α(t0 ), 12 and hence it is indeed the unique connected component   of BX α(t0 ), 12 that contains α(t0 ). Therefore, we have

Hence

  1 ∩B . α B[0,1] (t0 , δ) ⊂ BX α(t0 ), 2 −1

B[0,1] (t0 , δ) ⊂ α



BX



1 ∩B α(t0 ), 2



⊂ α−1 (B)

and so α−1 (B) is open in [0, 1]. Since [0, 1] is connected, this forces   α−1 (B) = [0, 1], or α [0, 1] ⊂ B.

Connectedness

273

Remark. The curious readers should wonder why we chose radius 1 1 2 in the proof. In fact, 2 is not such a magical number. Any number smaller than 1 would do. The point is, if we happen to choose a radius   r > 1, then BX α(t0 ), r will be so large that the sine curve will no longer be cut into those “almost vertical” pieces. Instead, when we are looking at a vicinity near the y-axis, the whole piece of the sine curve is still there and so the problem remains the same. Although in general, connectedness is weaker than pathconnectedness, under some special circumstances, they turn out to be the same. Theorem 3.2.7. Every open connected set in Rn is path-connected. Proof. Let S ⊂ Rn be open and connected. Fix a ∈ S. Let  A := x ∈ S : x and a are joined by a path in S},

B := S \ A .

As a ∈ A, we have A = φ. Note that by construction, A is pathconnected. Furthermore, it is obvious that A ∪ B = S and A ∩ B = φ. So if we can show that both A, B are open in S, then since S is connected and A = φ, we must have B = φ and we will be done, because in this case S = A is path-connected. So the proof boils down to showing the openness of A and B in S. Interestingly, the proofs are based on the same trick. For any x ∈ A, there is a path α : [0, 1] → S joining a to x. Since x ∈ S and S is open in Rn , there is an open ball BRn (x) centered at x with BRn (x) ⊂ S. Since BRn (x) is convex, by Example 3.2.4 (ii), it is path-connected. So for any y ∈ BRn (x), there is a path (for example, a line segment) β : [0, 1] → BRn (x) ⊂ S joining x to y.

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Therefore, β · α : [0, 1] → S is a path in S joining a to y via x. Hence y ∈ A. Since y ∈ BRn (x) is arbitrary, we conclude that BRn (x) ⊂ A and so A is open in Rn , hence open in S. Similarly, for any z ∈ B ⊂ S, there is an open ball BRn (z) centered at z with BRn (z) ⊂ S. For any w ∈ BRn (z), since BRn (z) is convex and hence path-connected, there is a path γ (for example, a line segment) in BRn (z) ⊂ S joining w to z. Observe that w ∈ B. For if w ∈ X \ B = A, then there is a path δ in S joining a to w and so the composition γ · δ would be a path in S joining a to z via w, which means z ∈ A, contradicting to the choice of z. Thus w ∈ B. Since w in BRn (y) is arbitrary, we have BRn (y) ⊂ B and so B is open  in Rn , hence open in S.

Remark. Observe that in the proof of Theorem 3.2.7, the only place where Rn comes into play is that at every point x ∈ S, we can find a small open ball BRn (x) ⊂ S and the only property of such an open ball that we need is that it is path-connected. Hence Theorem 3.2.7 can be relaxed to a more general situation that at each point of S we can find arbitrarily small path-connected open neighborhood. For the details, the readers are referred to Exercise 3.2, Part B, Problem #12. Remark. A path in Rn is said to be polygonal if its image in Rn is the union of a finite number of straight line segments.

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275

Observe that precisely the same arguments used in the proof of Theorem 3.2.7 shows that every open connected set in Rn is polygonally connected, that is, any two points can be joined by a polygonal path lying entirely in the set. Remark. As connectedness and path-connectedness have rather intimate connections, it would lead to speculations that results on connectedness can be carried forward to analogous results on pathconnectedness. Certainly some of them can, but not all.

Example 3.2.8 (A is path-connected =⇒ × A is path-connected). 2 In R , consider  1 , A := (x, y) ∈ R2 : 0 < x ≤ 1 , y = sin x   B := (x, y) ∈ R2 : x = 0 , −1 ≤ y ≤ 1 , X := A ∪ B .

It turns out that A is path-connected but A is not. Proof . That A is path-connected is clear. On the other hand, it is readily seen that A = A ∪ B = X. To show that A = X is not path-

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connected, we use arguments similar to those in Example 3.2.6: It suffices to show that points in A cannot be joined to points in B by a path in X. In other words, we need to show that any path γ : [0, 1] → X which passes through a point in B must stay forever in B, that is, γ should satisfy γ([0, 1]) ⊂ B, or γ −1 (B) = [0, 1]. Now as γ passes through a point in B, we have φ = γ −1 (B) ⊂ [0, 1]. Since [0, 1] is connected, it remains to show γ −1 (B) ⊂ [0, 1] is clopen. Since B is closed in R2 , it is closed in X. As γ is continuous, γ −1 (B) is closed in [0, 1]. On the other hand, for any t0 ∈ γ −1 (B) ⊂ [0, 1], since γ is continuous, there exists δ > 0 s.t.

  1 . γ B[0,1] (t0 , δ) ⊂ BX γ(t0 ), 2

  Since B[0,1] (t0 , δ) = (t0 − δ, t0 + δ)∩ [0, 1] is connected, γ B[0,1] (t0 , δ) is also connected and contains the point γ(t0 ), so it is contained in the   unique connected component of BX γ(t0 ), 12 which contains γ(t0 ).

  Observe that the set BX γ(t0 ), 12 splits into a number of parts,   including the vertical piece BX γ(t0 ), 12 ∩ B and the “almost   vertical” pieces of the curve y = sin x1 in BX γ(t0 ), 12 . It is evident that each of the “almost vertical” pieces of the curve     y = sin x1 in BX γ(t0 ), 12 is a connected component of BX γ(t0 ), 12 .

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  On the other hand, as BX γ(t0 ), 12 ∩ B is an interval, it is connected and we claim that it is indeed another connected     component of BX γ(t0 ), 12 . In fact, for any S ⊂ BX γ(t0 ), 12 which   is a strict superset of BX γ(t0 ), 12 ∩ B, S contains at least one point in one of the “almost vertical” pieces of y = sin x1 . Name one of such almost vertical pieces V . Then it is clear that {S ∩ V, S \ (S ∩ V )} is a   separation of S and so S cannot be connected. Thus BX γ(t0 ), 12 ∩B   is a connected component of BX γ(t0 ), 12 and so it is the unique   connected component of BX γ(t0 ), 12 that contains γ(t0 ). Hence we have

  1 ∩B . γ B[0,1] (t0 , δ) ⊂ BX γ(t0 ), 2 Therefore, B[0,1] (t0 , δ) ⊂ γ

−1



BX



1 ∩B γ(t0 ), 2



⊂ γ −1 (B)

and so γ −1 (B) is open in [0, 1]. Since [0, 1] is connected, this forces   γ −1 (B) = [0, 1], or γ [0, 1] ⊂ B.

Alternatively, by Example 3.2.6, it is evident that {(0, 0)} cannot be joined to any point in A by a path in A∪{(0, 0)}. That is, A∪{(0, 0)} is not path-connected. Similar argument shows that A ∪ {p} for any p ∈ B is not path-connected. Now if A ∪ B were path-connected, then there exists a path γ joining a point in A to a point in B. In particular, there exists a point in B which is a point at which γ “enters” B. In other words, there is a point p ∈ B such that A ∪ {p} is path-connected. That is a contradiction. Hence A ∪ B is not pathconnected.

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Exercise 3.2 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. If S, T ⊂ X are disjoint, then S ∪ T is not path-connected. Answer : False. Example : Let X := R, S := [0, 1], T := (1, 2). Clearly S and T are disjoint but S ∪ T = [0, 2) is path-connected. 2. If S ⊂ X is path-connected, then S ⊂ S  . Answer : False. Example : Let X := R and S := {0}. Then S is path-connected but S ⊂ φ = S  . 3. If S ⊂ X is path-connected, then ∂S is connected. Answer : False. Example : Let X := R and S := (0, 1). Then S is path-connected but ∂S = {0, 1} is not. 4. If S ⊂ X is path-connected, then S = S  . Answer : False. Example : Let X := R and S := {0}. Then S is path-connected but S = {0} = φ = S  . 5. If S ⊂ X and S = S  , then S is path-connected. Answer : False. Example : Let X := R and S := the Cantor Set. Then by Exercise 1.4, Part B, Problem #4, S = S  . But it is obvious that S is not connected, hence by Theorem 3.2.5, not path-connected.

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6. If S ⊂ X and S ◦ is path-connected, then S is path-connected.

Answer : False. Example : Let X := R and S := (0, 1) ∪ {2}. Then S ◦ = (0, 1) is path-connected but S is not.

7. If S ⊂ X and S is path-connected, then S is path-connected.

Answer : False. Example : Let X := R and S := Q. Then S = R is path-connected but S is not.

8. If S ⊂ X and S  is path-connected, then S is path-connected.

Answer : False. Example : Let X := R and S := Q. Then S  = R is path-connected but S is not.

9. If S ⊂ X and the set of isolated points S \S  is path-connected, then S is path-connected. Answer : False. Example : Let X := R and S := (0, 1) ∪ {2}. Then S \ S  = {2} is path-connected but S is not. 10. If S ⊂ X and ∂S is path-connected, then S is path-connected. Answer : False. Example : Let X := R and S := Q. Then ∂S = R is path-connected but S is not.

11. If S ⊂ X is path-connected, then S ◦ is path-connected.

Answer : False. Example : Let X := R2 and S := {(x, y) : xy ≥ 0} ⊂ X . Then S is path-connected but S ◦ = {(x, y) : xy > 0} is not path-connected.

12. If S, T ⊂ X are path-connected and S ∩ T = φ, then S ∩ T is path-connected.

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Answer : False. Example : Let X := R2 , S := S 1 , and T := {0} × R. Then S , T are path-connected and S ∩ T = {(0, 1), (0, −1)} = φ, but it is

clearly not path-connected.

13. If S, T ⊂ X are path-connected, then S ∪ T is path-connected. Answer : False. Example : Let X := R, S := (0, 1), and T = (1, 2). Then S and T are path-connected but S ∪ T = (0, 1) ∪ (1, 2) is not. 14. The union of any collection of path-connected subsets of X with nonempty intersection is path-connected. Answer : True. Proof . Let {Sλ }λ∈Λ be any collection of path-connected subsets of

X with λ∈Λ Sλ = φ, and S := λ∈Λ Sλ . Let a ∈ λ∈Λ Sλ . For any x1 , x2 ∈ S , there exists λ1 , λ2 ∈ Λ such that xi ∈ Sλi , i = 1, 2. For each i = 1, 2, since Sλi is path-connected, there exists a path αi : [0, 1] → Sλi ⊂ S joining xi to a. The composition α−1 2 · α1 is then a path in S joining x1 to x2 . Thus S is path-connected. 15. Every open ball B(a, r) ⊂ X is path-connected. Answer : False. Example : Let X := R \ {0}. Then B(1, 2) = (−1, 0) ∪ (0, 3) is not path-connected.

16. For any n ≥ 2, Rn \{0} is path-connected hence also connected. Answer : True. Proof . For any x = y ∈ S := Rn \{0}, if x = −y , then the function γ : [0, 1] → S given by γ(t) := (1 − t)x + ty , t ∈ [0, 1] is a path in S joining x to y . If x = −y , as n ≥ 2, there is a point z ∈ S \ xy, where, as usual, xy stands for the line segments joining x and y . Then α : [0, 1] → S given by α(t) := (1 − t)x + tz , and β : [0, 1] → S

Connectedness

281

given by β(t) := (1 − t)z + ty , t ∈ [0, 1], are paths in S joining x to z and z to y , respectively. Hence the composition β · α is a path in S joining x to y via z .

17. For each n = 0, 1, 2, . . . , the unit sphere S n := {x ∈ Rn+1 : x = 1} ⊂ Rn+1 is path-connected.

Answer : All S n ’s are path-connected except S 0 . Proof . It is clear that S 0 = {−1, 1} is not path-connected. For any n ≥ 1, for any x, y ∈ S n , there is a “great circle” (a circle of radius 1) on S n passing through x and y . It is evident that either arc on such a great circle determined by x and y is the image of a path on S n . Hence S n is path-connected.

18. Every finite metric space with at least two elements is not pathconnected. Answer : True. Proof . Let X be a finite metric space. As the image of any path in X must be connected, and the only connected subsets of X are singletons, every path in X must be a constant path. Hence if X contains at least two elements, it cannot be path-connected.

19. Every path-connected metric space with at least 2 elements is uncountable. Answer : True. Proof . This follows from Exercise 3.1, Part A, Problem #21, and Theorem 3.2.5.

Part B: Problems 1. Let a, b ∈ Rn be points such that a  < 1 and b  > 1. If α : [0, 1] → Rn is a path joining a to b, show that there exists t ∈ (0, 1) with α(t) = 1.

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Proof . Observe first that the norm function  ·  : Rn → R is continuous. Hence the composition g =  ·  ◦ α : [0, 1] → R is continuous, with g(0) = α(0) < 1, g(1) = α(1) > 1. By the

ordinary Intermediate Value Theorem of real-valued functions, there exists t ∈ (0, 1) such that α(t) = g(t) = 1.

2. Show that continuous functions preserve path-connectedness. That is, let X, Y be metric spaces and f : X → Y be continuous. If S ⊂ X is path-connected, then f (S) is also pathconnected. Proof . Fix y1 , y2 ∈ f (S). Let s1 , s2 ∈ S be such that f (si ) = yi , i = 1, 2. Since S is path-connected, there exists a path α : [0, 1] → S such that α(0) = s1 , α(1) = s2 . As f and α are continuous, the composition f ◦ α : [0, 1] → f (S) is also continuous. Moreover, we have f ◦ α(0) = f (s1 ) = y1 , f ◦ α(1) = f (s2 ) = y2 and so f (S) is path-connected.

3. Determine whether the pre-image of a path-connected set under a continuous function must be path-connected or not. That is, if f : X → Y is continuous and T ⊂ Y is path-connected, determine whether f −1 (T ) is path-connected. Answer : No. Example : Consider f : S 1 → R defined by f (x, y) := x, which   is clearly continuous on S 1 . The interval T := − 12 , 12 ⊂ R is   connected but f −1 (− 12 , 12 ) is separated into two disjoint open arcs on the unit circle and is hence not-path-connected.

4. Show that the graph Γ := {(x, f (x)) : x ∈ I} of a continuous function f : I → R on an interval I is path-connected.

Proof . For any (a, f (a)), (b, f (b)) ∈ Γ, say, a < b. Define γ : [0, 1] → Γ by

  γ(t) := (1 − t)a + b t, f (1 − t)a + b t .

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283

Obviously γ([0, 1]) = Γ, γ(0) = (a, f (a)), γ(1) = (b, f (b)), and γ is continuous. Thus Γ is path-connected.

5. If S, T ⊂ X are path-connected, with S ∩ T = φ or S ∩ T = φ, determine whether S ∪ T is path-connected. [Compare with Theorem 3.1.10.] Answer : False. Example : Consider X := S ∪ T ⊂ R2 , where 

1 , S := (x, y) ∈ R2 : 0 < x ≤ 1 , y = sin x   T := (x, y) ∈ R2 : −1 ≤ x ≤ 0 , y = 0 .

By Exercise 3.2, Part B, Problem #4, S and T are path-connected. Furthermore, it is easy to see that in the metric space X , S = S ∪   (0, 0) ⊂ X . Hence S ∩ T = {(0, 0)} = φ. However, by Example 3.2.6, X = S ∪ T is not path-connected.

6. For any S ⊂ X, if S ◦ = φ and S = X, determine whether X \ ∂S is path-connected. [Compare with Exercise 3.1, Part A, Problem #16.] Answer : No. Example . Consider X := S ∪ T ⊂ R2 , where  1 , S := (x, y) ∈ R2 : 0 < x ≤ 1 , y = sin x   T := (x, y) ∈ R2 : −1 ≤ x ≤ 0 , y = 0 .

It is clear that in the metric space X , S ◦ = S = φ and S = S ∪

{(0, 0)} = X . However, it is elementary to see that ∂S = {(0, 0)}   and so X \ ∂S = S ∪ (x, y) ∈ R2 : −1 ≤ x < 0 , y = 0 which is clearly not path-connected.

7. Prove that the set   A := f ∈ C[0, 1] :

1 0

   f (x)dx = 0 ⊂ C[0, 1], d∞ ,

  where d∞ (f, g) := supx∈[0,1] f (x) − g(x), is path-connected.

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Proof . For any fixed f , g ∈ A, define a function γ : [0, 1] → C[0, 1] by γ(t) := tg + (1 − t)f for all t ∈ [0, 1]. Clearly γ is well-defined. Furthermore, as

     d∞ γ(t), γ(s)) = sup γ(t)(x) − γ(s)(x) x∈[0,1]

    = sup (t − s) g(x) − f (x)  x∈[0,1]

    = |t − s| sup g(x) − f (x) = |t − s| d∞ (g, f ) , x∈[0,1]

γ : [0, 1] → C[0, 1] is continuous and so it is a path in C[0, 1]. Clearly, γ(0) = f and γ(1) = g . Furthermore, for every t ∈ [0, 1],

we have



1 0



 γ(t) (x)dx = t



1 0

g(x)dx + (1 − t)



1

f (x)dx = 0 .

0

So γ(t) ∈ A for all t ∈ [0, 1]. Hence γ is a path in A joining f to g and so A is path-connected.

8. In R2 , consider the “Deleted Comb Space” S := A ∪ B ∪ C, where    1 , y : 0 ≤ y ≤ 1, n ∈ N , A := n   B := (x, 0) : 0 ≤ x ≤ 1 , C := {(0, 1)} .

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285

(a) Determine whether S is connected. (b) Determine whether S is path-connected. Solution : (a) Answer : S is connected. Proof . It is clear that A ∪ B is path-connected and hence connected. Furthermore, as

A ∪ B ⊂ S ⊂ A ∪ B ∪ {(0, y) : 0 ≤ y ≤ 1} = A ∪ B , S is connected. (b) Answer : S is not path-connected.

Justification . It suffices to show that the point P := (0, 1) cannot be joined to any point in A ∪ B by a path in S . Let γ : [0, 1] → S be a path through P . Then φ = γ −1 (C) ⊂ [0, 1]. So the problem boils down to showing γ −1 (C) = [0, 1], for then γ will always stay in C and cannot go to any point in A ∪ B. Observe that as C is closed in R2 , it is closed in S and so γ −1 (C) ⊂ [0, 1] is closed. It thus remains to show that γ −1 (C) ⊂ [0, 1] is also open, for then γ −1 (C) is a nonempty clopen subset of the connected space [0, 1] and so it forces γ −1 (C) = [0, 1].

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Pick any t0 ∈ γ −1 (C). As γ is continuous at t0 , there

exists δ > 0 such that γ(B[0,1] (t0 , δ)) ⊂ BS (P, 12 ). Since

B[0,1] (t0 , δ) = (t0 − δ, t0 + δ) ∩ [0, 1] is connected, γ(B[0,1] (t0 , δ)) is a connected subset of BS (P, 12 ) which contains P , hence it is lying in the connected component of P in BS (P, 12 ). But then it is evident that the connected component of P in BS (P, 12 ) is C = {P }, so we have γ(B[0,1] (t0 , δ)) ⊂ C and hence B[0,1] (t0 , δ) ⊂ γ −1 (C). So γ −1 (C) is open. 9. In R2 , consider the “Deleted Infinite Broom” S := {(1, 0)} ∪

  s  : n ∈ N, 0 ≤ s ≤ 1 . s, n

(a) Determine whether S is connected. (b) Determine whether S is path-connected.

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287

Solution : (a) Answer : S is connected.    Proof . For each n ∈ N, write Ln := s, ns : 0 ≤ s ≤ 1 . Then each Ln is a line segment and so each Ln is connected.

As n∈N Ln = {(0, 0)} = φ, n∈N Ln is connected. On the other hand, as it is evident that

(1, 0) ∈



Ln ,

n∈N

we have



n∈N

    Ln ⊂ S = (1, 0) ∪ Ln ⊂ Ln n∈N

n∈N

and so S is connected. (b) Answer : S is not path-connected.

Justification . As the Ln ’s defined in (a) are line segments, each of them is path-connected and they all pass through the point O := (0, 0). Hence

n∈N

Ln is path-connected. It thus

remains to show that any path γ : [0, 1] → S passing through

P := (1, 0) cannot join any point in any Ln . Let γ : [0, 1] → S be a path passing through P . Without loss of generality, we assume that γ(0) = P . Let C := γ −1 ({P }). It suffices to show that C = [0, 1]. Now since 0 ∈ C , C = φ. So the problem boils down to showing that C is clopen in [0, 1], for then C is a nonempty clopen subset of the connected space [0, 1] and so if forces C = [0, 1]. As γ is continuous and {P } ⊂ R2 is closed, C = γ −1 ({P }) is closed in [0, 1]. On the other hand, pick any t0 ∈ C . By the continuity of γ , there exists δ > 0 such that γ(B[0,1] (t0 , δ)) ⊂ BS (P, 12 ).

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Since B[0,1] (t0 , δ) = (t0 − δ, t0 + δ) ∩ [0, 1] is connected,

γ(B[0,1] (t0 , δ)) is a connected subset of BS (P, 12 ) which passes through P , hence it is lying in the connected component of P in BS (P, 12 ). But then it is clear that the connected component of BS (P, 12 ) is {P }, so we have γ(B[0,1] (t0 , δ)) ⊂ {P } and hence B[0,1] (t0 , δ) ⊂ γ −1 ({P }) = C . So C is open in [0, 1]. 10. If a function preserves path-connectedness, determine whether it is continuous. Answer : False. Example : Consider the function f : [0, ∞) → R defined by  0 x=0 f (x) = 1 x>0. sin x Then f preserves path-connectedness. In fact, suppose S ⊂ [0, ∞) is path-connected. If S is a singleton, f (S) is also a singleton and

hence is path-connected. If S is not a singleton, then since S must be connected, it is a nondegenerate interval. Let x, y ∈ f (S) be two

Connectedness

289

distinct points. Then there exists u, v ∈ S such that f (u) = x and

f (v) = y . Assume without loss of generality that u < v . Case 1 : u > 0.

Since S is an interval, 0 < (1 − t)u + tv ∈ S for any t ∈ [0, 1]. The





function α : [0, 1] → f (S) ⊂ R given by α(t) := f (1 − t)u + tv ,

t ∈ [0, 1], is thus a well-defined continuous function with α(0) = f (u) = x, and α(1) = f (v) = y . That is, it is a path in f (S) joining x to y . Case 2 : u = 0. Note that as S is a nondegenerate interval in [0, ∞) containing the

point u = 0, by the definition of f , we have f (S) = [−1, 1] which is obviously path-connected. Combining, we see that f preserves path-connectedness. However, it is elementary to see that f is discontinuous at 0.

11. For any n ≥ 2, show that Rn \ S is path-connected for any countable subset S ⊂ Rn . Proof . Since Rn is uncountable, Rn \ S is uncountable. Let x, y ∈ Rn \ S be any two distinct points. We need to construct a path in Rn \ S from x to y . Note first that there are uncountably many lines passing through x. Now for any z ∈ S , there is exactly one line passing through x and z . It follows (by Pigeon-hole principle) that there must exists at least one line L passing through x, but not any point of S , that is, L ⊂ Rn \ S . If L passes through y , then we are done. So now suppose L does not pass through y . Similar to considerations above, there are uncountably many lines passing through y , and for any z ∈ S , there is exactly one line passing through y and z . As there are uncountably many points on L, it follows (by Pigeon-hole principle) that there must exist at least one line L passing through y and some point, say u, of L but not any point of S . Thus L ⊂ Rn \S . This gives rise to a polygonal path from x to y via u in Rn \ S . The result follows.

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12. A metric space X is said to be locally path-connected if for each point x ∈ X and for each neighborhood N of x, there is a path-connected open set U ⊂ X such that x ∈ U ⊂ N . If X is connected and locally path-connected, show that X is path-connected. Proof . Fix a ∈ X . It suffices to show that every point in X can be joined by a path in X to a. Imitating the proof of Theorem 3.2.7, let  A := x ∈ X : x and a are joined by a path in X} .

As a ∈ A, we have A = φ. So if we can show that A is clopen in

X , then since X is connected and A = φ, we must have X = A is path-connected.

For any x ∈ A, there is a path α in X joining a to x. Choose a

path-connected open set U ⊂ X containing x. Then for any y ∈ U ,

there is a path β in U joining x to y and so the composition β · α is

a path in X joining a to y via x. Hence y ∈ A and so U ⊂ A. That

is, A is open.

On the other hand, for any z ∈ / A, choose a path-connected open set

V ⊂ X containing z . Then for any w ∈ V , there is a path γ in V joining w to z . Note that w ∈ / A. In fact, if w ∈ A, there would be a path δ in X joining a to w and so the composition γ · δ would be a path in X joining a to z via w, contradicting to the choice of z . Hence we have w ∈ / A and so V ⊂ X \ A, that is, A is closed.

13. Prove or disprove the following statements: (a) Local path-connectedness implies path-connectedness. (b) Path-connectedness implies local path-connectedness. Solution : (a) Answer : False. Example : S := [0, 1] ∪ [2, 3] ⊂ R is locally path-connected but not path-connected.

(b) Answer : False.

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Example : The “Comb Space”  

  1 : n ∈ N × [0, 1] ∪ [0, 1] × {0} C = {0} ∪ n

in R2 is clearly path-connected. But at the point p := (0, 1), the neighborhood BC (p, 12 ) does not contain any path-connected open subset and so C is not locally path-connected.

14. Let X be a metric space and S ⊂ X. If X and ∂S are pathconnected, determine whether each of the following is true: (a) S is path-connected. (b) S is path-connected. Solution : (a) Answer : False. Example : Let X := R2 and S := {(x, y) ∈ R2 : x = 0}. Then X and ∂S = {(x, y) ∈ R2 : x = 0} are path-connected, but S is not. (b) Answer : True. Proof . Let x, y ∈ S . Since X is path-connected, there exists a path γ : [0, 1] → X such that γ(0) = x and γ(1) = y . If γ([0, 1]) ⊂ S , then we are done. Otherwise, there exists some t0 ∈ (0, 1) such that γ(t0 ) ∈ / S . Let T := {t ∈ [0, 1] : γ(t) ∈ X \ S} . Write a := inf T ∈ [0, 1]. Observe that a ∈ / T . In fact, suppose

a ∈ T . Then γ(a) ∈ X \ S . Since γ(0) ∈ S , we have a = 0. Meanwhile, since X \ S is open in X , by the continuity of γ , there exists ε > 0 such that γ(B[0,1] (a, ε)) ⊂ X \ S and so B[0,1] (a, ε) ⊂ T . This contradicts that a = inf T . Hence a∈ / T and so γ(a) ∈ S .

On the other hand, by the definition of infimum, there is a sequence {tn }n∈N in T which converges to a as n → ∞. Hence

γ(a) = lim γ(tn ) ∈ (X \ S) = X \ (S)◦ ⊂ X \ S ◦ . n→∞

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Combining, we have γ(a) ∈ S ∩ (X \ S ◦ ) = S \ S ◦ = ∂S .

Similarly, setting b := sup T , we have γ(b) ∈ ∂S .

Finally, since ∂S is path-connected, there exists a path

σ : [0, 1] → ∂S such that σ(0) = γ(a) and σ(1) = γ(b). So the path τ : [0, 1] → S defined by  γ(t) if t ∈ [0, a] ∪ [b, 1] 

τ (t) := t−a if t ∈ (a, b) σ b−a joins x to y . Hence S is path-connected.

15. Let Mn be the set of all n × n matrices with real entries. For any A, B ∈ Mn , define ⎤1/2 n    aij − bij 2 ⎦ , d(A, B) := ⎣ ⎡

i,j=1

where as usual, aij and bij denote the ij-th entry of A and B, respectively. (a) Show that d is a metric on Mn . (b) Show that the determinant function det : Mn → R is continuous. (c) The orthogonal group O(n) ⊂ Mn is defined as O(n) := {A ∈ Mn : AT A = I} , where AT denotes the transpose of A and I ∈ Mn denotes the n × n identity matrix. Determine whether each of the following is true or false: (i) O(n) is compact. (ii) O(n) is complete. (iii) O(n) is connected. (iv) O(n) is path-connected. (v) When n is odd, there is a matrix A ∈ Mn such that   n i = 0. det i=0 (i + 1)A

Connectedness

293

Solution : 2

(a) Let Φ : Mn → Rn be defined by

Φ(A) := (a11 , a12 , . . . , a1n , a21 , a22 , . . . , a2n , . . . , ann ) . Clearly, Φ is a bijection. Observe that

d(A, B) = Φ(A) − Φ(B) for all A, B ∈ Mn . So d is a well-defined metric on Mn induced by Φ from the 2

Euclidean metric  ·  on Rn . 2





(b) For any x ∈ Rn , the function f (x) := det Φ−1 (x) is a polynomial of n2 variables, which is continuous. It follows that

det = f ◦ Φ is also continuous. (c) (i) Answer : True. Proof . Note that O(n) =

n 

i,j=1

Fij−1 ({δij }) ,

where for any i, j = 1, . . . , n and Fij : Mn → R is defined by

Fij (A) :=

m 

k=1

aik ajk for all A ∈ Mn ,

and δij is the Kronecker delta function. Note that for all

i, j = 1, . . . , n, Fij is continuous and being a singleton, {δij } ⊂ R is closed. Hence O(n) is closed. On the other hand, by definition, for all A ∈ O(n), we have n 

a2ij = 1 for all i = 1, . . . , n .

j=1

Thus if we denote by O ∈ Mn the zero n × n matrix, then

d(A, O) =

n 

i,j=1

a2ij = n for all A ∈ O(n)

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and so O(n) is bounded in Mn . As Φ defined in (a) is by construction an isometry isomorphism, Φ(O(n)) is 2

closed and bounded in Rn . Hence Φ(O(n)) is compact. Since Φ is a homeomorphism, O(n) is also compact. (ii) Answer : True.

Proof . As O(n) is compact, by Theorem 2.2.7, it is complete. (iii) Answer : False.

Justification . For any A ∈ O(n), since AT A = I , 2 we have (det(A)) = 1 and hence det A = ±1. So   det O(n) ⊂ {±1}. On the other hand, it is obvious that I ∈ O(n) with det(I) = 1 and the diagonal matrix J with diagonal entries (−1, 1, 1, . . . , 1) is also in O(n)   but with det(J) = −1. Hence det O(n) = {±1}. That is, the continuous function det maps O(n) onto a disconnected set {±1}. Since continuous functions must preserve connectedness, O(n) must be disconnected. (iv) Answer : False. Justification . By (iii), O(n) is not connected, hence it is not path-connected. (v) Answer : True. 2

Proof . As the function Φ : Mn → Rn defined in (a) 2 is a homeomorphism and Rn is connected, Mn is also connected. Define F : Mn → R by  n   i F (A) := det (i + 1)A , A ∈ Mn . i=0

Since det is continuous, F is continuous. Observe that

F (O) = det (I) = 1 > 0 and  n  !n  n+1 i F (−I) = det (i + 1)(−I) = − 0, we can find a δ > 0 such that   dY f (x), f (a) < ε

whenever

dX (x, a) < δ ,

or equivalently,

    f BX (a, δ) ⊂ BY f (a), ε) .

Obviously, the δ we choose depends on the given ε. In general, if we are given a smaller ε > 0, we would need to find correspondingly a smaller δ. So to be absolutely precise, δ should be written as δ(ε) to indicate its dependence on ε. To avoid unnecessarily complicated symbols, we would simply say that “we can find a δ = δ(ε) > 0 such that ...” to indicate the dependence and then use the symbol δ instead of δ(ε) to simplify the notations afterwards. Globally speaking, f : X → Y is continuous on X if it is continuous at every point a ∈ X. That is, for every given ε > 0, for every a ∈ X, we can find a δ > 0 such that   dY f (x), f (a) < ε whenever

dX (x, a) < δ .

(*)

An important point to note here is that in general, for a specific ε > 0 and a specific a ∈ X, there is a corresponding δ. That is, the δ we choose depends not only on the given ε but also on the point a ∈ X in question. So to be absolutely precise, δ should be written as δ(ε, a) to indicate its dependence on both ε and a. 295

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In this Chapter, we shall investigate what types of continuous functions f : X → Y would have the property that the same δ is good for all points of X, that is, given any ε > 0, there is a “uniform” δ = δ(ε) > 0 such that (∗) holds for every a ∈ X. We shall also investigate what properties such continuous functions would possess.

4.1 Uniform Continuity Example 4.1.1. Let us verify that the function f : (0, ∞) → R given by f (x) := x1 is continuous on (0, ∞). The thinking process goes as follows: Fix a ∈ (0, ∞). For any given ε > 0, we need to find a δ > 0 such that whenever |x − a| < δ, we have    |x − a| 1 1 δ < aε stricting x to |x − a| < δ and so in particular, we have x > a − δ. δ a2 ε , or equivalently, δ < 1+aε , we would have Hence in case a − δ > aε δ and we are done. Hence the formal proof goes as x > a − δ > aε follows: Fix a ∈ (0, ∞). For any given ε > 0, choose a δ > 0 satisfying a2 ε . Then δ < aε(a − δ) and so 0 < δ < 1+aε

  1 1  |x − a| δ aε(a − δ)  < < =ε |f (x) − f (a)| =  −  = x a ax ax a(a − δ)

whenever |x − a| < δ. Hence f is continuous at a. Since a ∈ (0, ∞) is arbitrary, we conclude that f is continuous on (0, ∞). Note, however, that for this particular function f , for any given ε > 0, the δ > 0 we need to choose depends on the position of the point a. In particular, when a is large, we can choose a relatively large δ. But then when a is getting closer and closer to 0, we need

Uniform Continuity

297

to pick a smaller and smaller δ. In the limiting case when a → 0+ , it is easy to see that δ → 0+ and so there does not exist a fixed δ > 0 that is good for every point in (0, ∞). Definition 4.1.2. A function f : X → Y is said to be uniformly continuous on a subset S ⊂ X if for any ε > 0, there exists δ = δ(ε) > 0 such that   d f (x1 ), f (x2 ) < ε

whenever x1 , x2 ∈ S with d(x1 , x2 ) < δ ,

or, equivalently, for any ε > 0, there exists δ = δ(ε) > 0 such that     f BX (x, δ) ∩ S ⊂ BY f (x), ε

for every x ∈ S .

That is to say, f is uniformly continuous on S if f is pointwisely continuous on S and associated to a given ε > 0, there is a “uniform δ” which is good for (i.e., satisfies the definition of continuity of f at) every point in S. Example 4.1.3. If X is discrete and Y is any metric space, then every function f : X → Y is uniformly continuous on X. [Compare with Exercise 2.3, Part A, Problem #10.] Proof. For any ε > 0, let δ := 12 . Then for any x1 , x2 ∈ X with  d(x1 , x2 ) < δ, since X is discrete, we have x1 = x2 and so d f (x1 ),  f (x2 ) = 0 < ε. Hence f is uniformly continuous on X.

Example 4.1.4. f : R → R defined by f (x) := x2 is uniformly continuous on (0, 1]. Proof. For any ε > 0 and any x1 , x2 ∈ (0, 1], we have |f (x1 ) − f (x2 )| = |x21 − x22 | = |x1 + x2 | |x1 − x2 | ≤ 2|x1 − x2 | .

Hence whenever 2|x1 − x2 | < ε, we have |f (x1 ) − f (x2 )| < ε. This suggests that δ := ε/2 is a uniform δ that works everywhere on (0, 1].

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Metric Space Topology: Examples, Exercises and Solutions

So the formal proof goes as follows: For any ε > 0, pick δ := ε/2. Then we have |f (x1 ) − f (x2 )| = |x21 − x22 | = |x1 + x2 | |x1 − x2 | ≤ 2|x1 − x2 | < 2δ = ε for all x1 , x2 ∈ S with |x1 − x2 | < δ. Hence f is uniformly continuous on (0, 1]. Remark. The careful readers should notice that in Example 4.1.4, the only reason why we restrict to the subset (0, 1] ⊂ R is to make sure that all points in there are bounded. Hence by similar arguments, it is easy to see that the function f is actually uniformly continuous on every bounded subset of R. Remark. It is obvious that uniformly continuous functions are automatically continuous, but the converse need not be true. In order to prove from (first principle) that a function f : X → Y is not uniformly continuous on S ⊂ X, we need to establish the negation of Definition 4.1.2, namely, there exists an ε > 0 such that for any δ > 0, the condition   “d f (x1 ), f (x2 ) < ε whenever x1 , x2 ∈ S with d(x1 , x2 ) < δ”

fails to hold, that is, there are points x1 , x2 ∈ S with d(x1 , x2 ) < δ   but d f (x1 ), f (x2 ) ≥ ε.

Example 4.1.5. f : R → R defined by f (x) := x2 is not uniformly continuous on (0, ∞). Proof. It is clear that f is continuous everywhere on R. Also, by the Remark following Example 4.1.4, f is uniformly continuous on every bounded subset of R. However, it is not uniformly continuous on (0, ∞). In fact, in view of the preceding Remark, let us take ε := 1. For any δ > 0, take x1 := 1δ and x2 := 1δ + δ2 . Then x1 , x2 ∈ (0, ∞)

Uniform Continuity

and d(x1 , x2 ) = the assertion.

δ 2

< δ, but d(f (x1 ), f (x2 )) = 1 +

299 δ2 4

> 1 = ε. Hence

Example 4.1.6. f : (0, ∞) → R defined by f (x) := formly continuous on (0, ∞).

1 x

is not uni-

Proof. In Example 4.1.1, we saw that f is continuous on (0, ∞) and we argued that there is no uniform δ for all points in (0, ∞). For the more matured readers, that is accepted to be a rigorous proof that f is not uniformly continuous on (0, ∞). A more concrete proof using the preceding Remark is as follows: Take ε := 1. For any δ > 0, without loss of generality we can assume that 1 > δ > 0, we need to pick points x1 , x2 ∈ (0, ∞) which are close to each other (within a distance < δ) but with images under f far apart from each other (≥ ε). In view of the considerations in Example 4.1.1, we see that the graph of f blows up very quickly as x → 0+ and so this is exactly the place where x1 and x2 should be in. This suggests that we pick x1 := δ and x2 := 2δ . Then both x1 ,  x2 ∈ (0, 1) ⊂ (0, ∞), and we have |x1 − x2 | = 2δ < δ, but f (x1 ) −  f (x2 ) = | 1δ − 2δ | = 1δ > ε. Hence by the preceding Remark, f is not uniformly continuous on (0, ∞). Theorem 4.1.7 (Heine). If f : X → Y is continuous on a compact subset C ⊂ X, then f is uniformly continuous on C. Proof. Let ε > 0 be given. Since f is continuous on C, for any c ∈ C, there exists δc > 0 such that   ε whenever x ∈ B(c, δc ) ∩ C . d f (x), f (c) < 2     Collectively, B c, δ2c : c ∈ C is an open cover of the compact set     C and so it has a finite subcover, say, B ci , δ2i : i = 1, . . . , n . Let   δ := min δ21 , . . . , δ2n . Then δ > 0. For any p, q ∈ C with d(p, q) < δ,

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Metric Space Topology: Examples, Exercises and Solutions

  there is an i ∈ {1, . . . , n} such that p ∈ B ci , δ2i . Observe that by triangle inequality, d(q, ci ) ≤ d(q, p) + d(p, ci ) < δ +

δi δi δi ≤ + = δi . 2 2 2

Hence q ∈ B(ci , δi ) and so both p, q ∈ B(ci , δi ). Therefore, by triangle inequality again, we have    ε ε    d f (p), f (q) ≤ d f (p), f (ci ) + d f (q), f (ci ) < + = ε . 2 2

This completes the proof of the Theorem.



Note that in view of Heine’s Theorem, Example 4.1.4 and its immediate Remark become trivial. Theorem 4.1.8. Composition of uniformly continuous functions is uniformly continuous. Proof. The proof is similar to that for Theorem 2.3.8. Suppose f : X → Y and g : Y → Z are uniformly continuous. For any ε > 0, by the uniform continuity of g on Y , there exists δ1 = δ1 (ε) > 0 such that     g BY (y, δ1 )) ⊂ BZ g(y), ε for all y ∈ Y . By the uniform continuity of f , there exists δ = δ(δ1 (ε)) > 0 such that     for all x ∈ X f BX (x, δ) ⊂ BY f (x), δ1 and so

     g ◦ f BX (x, δ) ⊂ g BY f (x), δ1     ⊂ BZ g(f (x)), ε = BZ g ◦ f (x), ε

for all x ∈ X. Hence g ◦ f is uniformly continuous on X.



Uniform Continuity

301

Exercise 4.1 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. If X is discrete and Y is any metric space, then every function f : X → Y is uniformly continuous. [Compare with Exercise 2.3, Part A, Problem #10.] Answer : True. Proof . For any x ∈ X and any ε > 0, f (BX (x, 1)) = f ({x}) = {f (x)} ⊂ BY (f (x), ε) . Hence f is uniformly continuous on X .

2. f : R → R given by f (x) := ex , x ∈ R, is uniformly continuous on [0, 1]. Answer : True. Proof . First, by elementary Calculus, f is differentiable everywhere in R with derivative f  (x) = ex , x ∈ R. For any x, y ∈ [0, 1], say, x < y , by Mean Value Theorem, there is z ∈ (x, y) such that ey − ex = ez < ey ≤ e. Hence for any ε > 0, take δ := εe , then y−x ey − ex < (y − x)e < δe = ε whenever x < y ∈ [0, 1] with y − x < δ. Hence f is uniformly continuous on [0, 1]. 3. f : R → R given by f (x) := ex , x ∈ R, is uniformly continuous on every bounded subset of R. Answer : True. Proof . Let S ⊂ R be a bounded set. Then there exists M > 0 such that S ⊂ [−M, M ]. Similar to the proof of Exercise 4.1, Part A, Problem #2, it is evident that f is uniformly continuous on [−M, M ], hence uniformly continuous on S .

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4. f : R → R given by f (x) := ex , x ∈ R, is uniformly continuous on R. Answer : False. Justification . By elementary Calculus, f is differentiable everywhere in R. Let ε := 1. For any x, y ∈ R, say, x < y , by Mean Value ey − ex = ez > ex . Hence Theorem, there is z ∈ (x, y) such that y−x for any 1 > δ > 0, the points x := − ln 2δ and y := δ2 − ln δ2 will δ satisfy 0 < y − x = δ2 < δ but ey − ex > (y − x)ex = δ2 e− ln 2 = δ 2 2 · δ = 1 = ε. Hence f is not uniformly continuous on R. 5. The function f (x) := ln x is uniformly continuous on (0, ∞).

Answer : False. Justification . Take ε := ln 2. For all δ > 0, let n be large enough 1 1 1 such that 2n < δ and p := n1 , q := 2n . Then |p − q| = 2n < δ but |f (p) − f (q)| = ln 2 < ε.

6. The function f (x) := ln x is uniformly continuous on (1, 2). Answer : True. Proof . By Mean Value Theorem, for any x = y ∈ (1, 2), we have |f (x) − f (y)| = | ln x − ln y| = 1c |x − y| for some c between x and y . As c ∈ (1, 2), we have 1c < 1 and so |f (x) − f (y)| < |x − y|. It is then evident that f is uniformly continuous on (1, 2). 7. The function f (x) := tan−1 x is uniformly continuous on R. Answer : True. Proof . By mean value theorem, for any x, y ∈ R,

    f (x) − f (y) =  tan−1 x − tan−1 y  = |x − y| ≤ |x − y| (1 + c2 )

for some c between x and y . It is then evident that f is uniformly continuous on R.

8. The function f (x) :=

√ x is uniformly continuous on [0, ∞).

Uniform Continuity

303

Answer : True. Proof . For any ε > 0, take δ := ε2 . For any x, y ∈ [0, ∞) with |x − y| < δ, we have  √ √ √ √ 2 |f (x) − f (y)| = | x − y| =  x − y  √  √     x − √y   x + √y  ≤  √ √  √ √  =  x − y  x + y   √ = x − y  < δ = ε . Hence f is uniformly continuous on [0, ∞).

9. For any n ∈ N, the function f : (0, ∞) → R defined by f (x) := 1 xn , x ∈ (0, ∞), is uniformly continuous on [1, ∞).

Answer : True. Proof . By elementary Calculus, f is differentiable everywhere in (0, ∞). For any x, y ∈ [1, ∞), say, x < y , by Mean Value Theorem, there 1 1 n y n − xn = − n+1 . For any ε > 0, take is z ∈ (x, y) such that y−x z δ := nε . Noting that z > 1, we have     1   1     ·  − n  < |y − x| n < δ n = ε − = y − x  yn xn  z n+1 whenever |y − x| < δ . Hence f is uniformly continuous on [1, ∞).

10. For some n ∈ N, the function f : (0, ∞) → R defined by f (x) := xn , x ∈ (0, ∞), is uniformly continuous on [1, ∞).

Answer : False. f is not uniformly continuous for any n ∈ N. Justification . Fix n ∈ N. Take ε = 1. For any δ > 0, take x ∈ 2 and y := x+ δ2 . Then |y − [1, ∞) be any point satisfying xn−1 > nδ x| = δ2 < δ. By elementary Calculus, f is differentiable everywhere in (0, ∞) and so by Mean Value Theorem, there is z ∈ (x, y) such f (y) − f (x) = nz n−1 > nxn−1 > 2δ and hence we have that y−x

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  f (y) − f (x) > |y − x| · continuous on [1, ∞).

2 δ

= 1 = ε. Thus f is not uniformly

11. For some n ∈ N, the function f : (0, ∞) → R defined by f (x) := x1n , x ∈ (0, ∞), is uniformly continuous on (0, 1].

Answer : False. f is not uniformly continuous for any n ∈ N. Justification . Fix n ∈ N. Take ε := 1. For any 1 > δ > 0, pick x := δ and y := 2δ . Then both x, y ∈ (0, 1) ⊂ (0, ∞), and we have   n n |x − y| = δ2 < δ but f (x) − f (y) = | δ1n − 2δn | = 2 δ−1 > 1 = ε. n Hence f is not uniformly continuous on (0, 1].

12. The function f : R → R given by f (x) := continuous.

1 is uniformly 1 + x4

Answer : True. 2ε Proof . For any ε > 0, let δ := , then for any x, y ∈ R with 3 |x − y| < δ, since geometric mean ≤ algebraic mean, we have 1 (8|x|3 + 12|x|2 |y| + 6|x| |y|2 + |y|3 ) 27 1 (|x|3 + 6|x|2 |y| + 12|x| |y|2 + 8|y|3 ) |xy 2 | ≤ 27 |x2 y| ≤

and so

|x2 y| + |xy 2 | ≤ |x|3 + |y|3 . Hence

      y 4 − x4  f (x) − f (y) =   (1 + x4 )(1 + y 4 )    3 y + y 2 x + yx2 + x3  |y − x| = (1 + x4 )(1 + y 4 )  3   2   2   3  y  + y x + yx  + x  |y − x| ≤ (1 + x4 )(1 + y 4 ) 2|y|3 + 2|x|3 |y − x| . ≤ (1 + x4 )(1 + y 4 )

Uniform Continuity

305

Also because geometric mean ≤ algebraic mean, we have

|y|3 ≤

|y|4 + |y|4 + |y|4 + 1 |x|4 + |x|4 + |x|4 + 1 , |x|3 ≤ , 4 4

and so

  f (x) − f (y) ≤

3 2 + 3y 4 + 3x4 |y − x| < δ = ε . 2(1 + x4 + y 4 + x4 y 4 ) 2

Alternative Proof . It is elementary to check that  3  y + y 2 x + yx2 + x3  ≤



4|x|3 if |y| ≤ |x|

4|y|3 3

if |x| ≤ |y|

≤ 4(|x| + |y|3 ) ,

while 3

|x| ≤



1 4

|x|

if |x| ≤ 1

if |x| ≥ 1

≤ (1 + |x|4 )(1 + |y|4 ) ,

and similarly,

|y|3 ≤ (1 + |x|4 )(1 + |y|4 ) . Hence

    y 4 − x4   |f (x) − f (y)| =  4 4 (1 + x )(1 + y )    3 y + y 2 x + yx2 + x3  |y − x| = (1 + x4 )(1 + y 4 ) ≤8|x − y| . So for any ε > 0, by taking δ := ε/8, the uniform continuity of f is readily seen.

13. f : [a, ∞) → R defined by f (x) := 1/(1 + x) is uniformly continuous for every a > −1.

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Metric Space Topology: Examples, Exercises and Solutions

Answer : True. Proof . For any ε > 0, take δ := ε(1 + a)2 > 0, then |f (x) − f (y)| =

|x − y| 1 |x − y| ≤ 0, and any N ∈ N with N >

|xN − yN | =

1 N

1 , δ

we have

< δ but

2   f (xN ) − f (yN ) = 3N + 2N ≥ 5 > 1 = ε 4 4

and so f is not uniformly continuous.

Alternative Justification . Take ε := 1. For any 1 > δ > 0, take x1 := 1 + δ and x2 := 1 + δ2 . Then   f (x1 ) − f (x2 ) = 3 + δ = 3 + 1 > 4 > ε δ2 δ2 δ

and so f is not uniformly continuous.

15. If f : R → R satisfies the following conditions (i) f is continuous at x = 0, (ii) f (0) = 0, (iii) f is “sub-additive”, that is, f (x1 + x2 ) ≤ f (x1 ) + f (x2 ) for all x1 , x2 ∈ R, then f is uniformly continuous on R.

Uniform Continuity

307

Answer : True. Proof . For any ε > 0, since f is continuous at 0 and f (0) = 0, there exists δ > 0 such that |f (t)| = |f (t) − f (0)| < ε whenever |t| = |t − 0| < δ. So for any x, t ∈ R with |t| < δ, we have, by (iii), f (x + t) − f (x) ≤ f (t) < ε and

f (x) − f (x + t) ≤ f (−t) < ε . Combining, we have |f (x + t) − f (x)| < ε and so f is uniformly

continuous on R.

16. Every differentiable function f continuous.

: R → R is uniformly

Answer : False. Example : The function f : R → R given by f (x) := x2 is differentiable everywhere but by Example 4.1.5, it is not uniformly continuous on (0, ∞) and hence neither is it uniformly continuous

on R.

17. Every uniformly continuous function f : R → R is differentiable. Answer : False. Example : The function f : R → R given by f (x) := |x| is clearly uniformly continuous but it is not differentiable at 0. 18. If f : S ⊂ R → R is uniformly continuous, then so is |f |.

Answer : True. Proof . For any ε > 0, there exists δ > 0 such that |f (y)−f (x)| < ε for all x, y ∈ S with |y − x| < δ . Hence by triangle inequality,              |f |(y) − |f |(x)  =  |f (y)| − |f (x)|  ≤ f (y) − f (x) < ε for all x, y ∈ S with |y − x| < δ . Hence the assertion.

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19. If f : X → Y is continuous on a closed and bounded subset S ⊂ X, then f is uniformly continuous on S. Answer : False. Example : Let X := (0, 1] and f : X → R be given by f (x) := 1/x. Then S := X is closed (in X ) and bounded, f is continuous but not uniformly continuous.

20. The sum of two uniformly continuous functions f , g : S ⊂ R → R is uniformly continuous. Answer : True. Proof . For any ε > 0, there exists δf , δg > 0 such that   f (y) − f (x) < ε 2   ε g(y) − g(x) < 2

for all x, y ∈ S with |y − x| < δf , for all x, y ∈ S with |y − x| < δg .

Take δ := min{δf , δg } > 0. For all x, y ∈ S with |y − x| < δ , we have

    (f + g)(y) − (f + g)(x) = f (y) + g(y) − f (x) − g(x)     ≤ f (y) − f (x) + g(y) − g(x) < ε/2 + ε/2 = ε

and so f + g is uniformly continuous on S .

21. The product of two uniformly continuous functions f , g : S ⊂ R → R is uniformly continuous.

Answer : False. Example : The functions f , g : R → R given by f (x) := g(x) := x, x ∈ R, are clearly uniformly continuous on R. However, as has been shown in Example 4.1.5, the function (f g)(x) = x2 is not.

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Part B: Problems 1. If f : R → R is differentiable with f  bounded on an interval I ⊂ R, show that f is uniformly continuous on I.

Proof . Suppose |f  | ≤ M on I . For any x = y ∈ I , say, x < y , by Mean Value Theorem, there exists z ∈ (x, y) such that |f (x) − f (y)| = |f  (z)(y − x)| = |f  (z)| |y − x| ≤ M |y − x| . For any ε > 0, let δ := ε/M . Then

|f (x) − f (y)| < ε whenever x, y ∈ I with |x − y| < δ and so f is uniformly continuous on I .

2. Show that every polynomial function is uniformly continuous on every bounded subset of R. Proof . Let p be a polynomial and S ⊂ R be a bounded subset. Then

certainly there is a bounded interval I ⊂ R such that S ⊂ I . Observe

that every polynomial is bounded on every bounded subset of R, so

p which is again a polynomial, is bounded on I . Hence by Exercise 4.1, Part B, Problem #1, p is uniformly continuous on I and hence it is uniformly continuous on S . 3. Determine whether it is true that every non-constant polynomial function is uniformly continuous on R. Answer : False. Justification . Let p(x) := an xn + an−1 xn−1 + · · · + a1 x + a0 be a non-constant polynomial. Without loss of generality, we assume that an > 0. We have

p (x) = nan xn−1 + (n − 1)an−1 xn−2 + · · · + a1 .

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As the leading term nan xn−1 is the dominating term of p , there exists R >> such that p (x) >

nan n−1 2 x

for all x ≥ R.

Let ε := 1. For any δ > 0, by taking a larger R if necessary, we may assume that Rn−1 ≥

|y − x| =

δ 2

4 nan δ .

Let x := R and y := R + δ2 . Then

< δ. Since p is differentiable everywhere, by Mean Value Theorem, there exists z ∈ (x, y) such that p(y) − p(x) = (y − x)p (z)

and so

δ  |p (z)| 2 δ nan n−1 δ nan n−1 z R ≥ ≥1=ε. > 2 2 2 2 Hence p is not uniformly continuous on R. |p(y) − p(x)| = |y − x| |p (z)| =

4. Prove that the sine function is uniformly continuous on R. Proof . Since the sine function is differentiable in R and with derivative (the cosine function) bounded on R, by Exercise 4.1, Part B, Problem #1, the sine function is uniformly continuous on R. 5. Prove that the cosine function is uniformly continuous on R. Proof . Since the cosine function is differentiable in R and with derivative (negative of the sine function) bounded on R, by Exercise 4.1, Part A, Problem #1, the cosine function is uniformly continuous on R.

6. Prove or disprove: f : (0, 1) → R given by f (x) := cos x1 , x ∈ (0, 1), is uniformly continuous. Answer : False. Justification . Let ε = 1. For any δ > 0, let n ∈ N be such that 1 1 1 < δ. Then d 2nπ , (2n+1)π < δ but 2n(2n+1)π

  1 1  ,f =2>1=ε. d f 2nπ (2n + 1)π Thus f is not uniformly continuous.

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7. Prove or disprove: f : (0, 1 ] → R given by f (x) := x cos x1 , x ∈ (0, 1 ], is uniformly continuous. Answer : True. Proof . Let f˜ : [0, 1 ] → R be the trivial extension of f to [0, 1 ],

that is,

f˜(x) :=



f (x) = x cos x1 0

for x ∈ (0, 1 ] for x = 0.

It is elementary to check that f˜ is continuous on the compact set [0, 1 ] and so it is uniformly continuous there. In particular, its restriction to (0, 1 ] is also uniformly continuous. That means f is uniformly continuous on (0, 1 ].

8. The product of two uniformly continuous functions f , g : S ⊂ R → R, where g is bounded, is uniformly continuous. [Compare with Exercise 4.1, Part A, Problem #21.] Answer : False. Example : The function f : R → R given by f (x) := x, x ∈ R, is clearly uniformly continuous on R. On the other hand, the function g : R → R given by f (x) := sin x, x ∈ R, is clearly bounded on R and by Exercise 4.1, Part B, Problem #4, uniformly continuous on R. However, (f g)(x) = x sin x is not uniformly continuous on R. In fact, take ε := 1. Then for any 1 > δ > 0, whenever n ∈ N is large, we have 2nπ sin(δ/2) > 1 and so       (f g) 2nπ + δ − (f g)(2nπ) = 2nπ + δ sin δ > 1 = ε .   2 2 2 9. Prove or disprove: Every uniformly continuous function is bounded.

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Answer : False. Example : The identity function f : R → R given by f (x) := x for all x ∈ R is clearly uniformly continuous on R. However, it is obvious that f is unbounded. 10. Prove or disprove: If S ⊂ R is bounded and f : S ⊂ R → R is uniformly continuous, then f is bounded on S. Answer : True. Proof . By the uniform continuity of f on S , for any ε > 0, there exists δ > 0 such that f (z) ∈ BR (f (x), ε) whenever z ∈ BS (x, δ) . By Exercise 1.3, Part B, Problem #10(c)(ii), S ⊂ R is totally bounded

and so it is covered by finitely many open balls of radius δ , say

BS (xi , δ), i = 1, . . . , n. Hence for any x ∈ S , x ∈ BS (xi , δ)   for some i = 1, . . . , n and so f (x) ∈ BR f (xi ), ε . Therefore,   n f (S) ⊂ i=1 B f (xi ), ε and so f is bounded on S .

11. Prove or disprove: The tangent function is uniformly   continuous on − π2 , π2 . Answer : False. Justification . Intuitively, it is evident that the tangent function blows up as x approaches π2 , so there is no hope that there could be a uniform   δ for all x ∈ − π2 , π2 . To prove this in more concrete terms, we let f (x) := tan x for x ∈  π π − 2 , 2 . Let ε := 1. For any 0 < δ < π2 , let x1 := π2 − δ > 0. Then f (x1 ) is a finite positive number. Since limx→ π2 f (x) = ∞, there exists x2 ∈ (x1 , π2 ) such that f (x2 ) > f (x1 ) + 1. Thus x1 , x2 ∈ (0, π2 ) are such that |x1 −x2 | < δ but |f (x1 )−f (x2 )| > 1 = ε. Thus f is not uniformly continuous. Alternatively , here is a simple indirect proof by Exercise 4.1, Part B, Problem #10: In fact, since the tangent function is unbounded on the



bounded set − π2 ,

π 2



, it cannot be uniformly continuous there.

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12. Prove or disprove: Every linear map T : Rn → Rm is uniformly continuous. Answer : True. Proof . By elementary Linear Algebra, the linear map T : Rn → Rm is represented by an m × n matrix A = (aij )m×n of real entries aij , 1 ≤ i ≤ m, 1 ≤ j ≤ n. Then for all x, y ∈ Rn , it is easily verified that

T x − T y = T (x − y) = A(x − y) ≤ M x − y , where M := sup{|aij | : 1 ≤ i ≤ m, 1 ≤ j ≤ n}, from which it is now obvious that T is uniformly continuous.

13. Let A ⊂ X be any subset. Show that the function f : X → R given by f (x) := d(x, A), x ∈ X, is uniformly continuous. [Compare with Exercise 2.3, Part B, Problem #2.] Proof . The proof is exactly the same as that for Exercise 2.3, Part B, Problem #2, noting that the δ chosen is independent of the choice of x ∈ X . 14. Let {xn }n∈N be a fixed sequence in X. Show that the function f : X → R given by f (x) := inf{d(x, xn ) : n ∈ N} is uniformly continuous on X. Proof . For any x, y ∈ X and any n ∈ N, by triangle inequality, we have

|d(x, xn ) − d(y, xn )| ≤ d(x, y) or

d(y, xn ) − d(x, y) ≤ d(x, xn ) ≤ d(y, xn ) + d(x, y) . Thus

f (y) − d(x, y) = inf d(y, xn ) − d(x, y) ≤ inf d(x, xn ) n∈N

= f (x) ≤ inf d(y, xn ) + d(x, y) n∈N

= f (y) + d(x, y)

n∈N

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and so

|f (x) − f (y)| ≤ d(x, y) , from which it is readily seen that f is uniformly continuous on X .

15. A function f : X → Y is said to be Lipschitz if there exists M > 0 called a Lipschitz constant of f such that   d f (x1 ), f (x2 ) ≤ M d(x1 , x2 ) for all x1 , x2 ∈ X .

If f : R → R is differentiable and f  is bounded on R, show that f is Lipschitz. Proof . Suppose f : R → R is differentiable with |f  | ≤ M for some M > 0. Then for any x1 , x2 ∈ R, by Mean Value Theorem, there exists t between x1 and x2 such that |f (x2 )−f (x1 )| = |f  (t)(x2 −x1 )| = |f  (t)| |x2 −x1 | ≤ M |x2 −x1 | . Hence f is Lipschitz.

16. Show that every Lipschitz function is uniformly continuous. Proof . Suppose f : X → Y is Lipschitz with Lipschitz constant M . For any ε > 0, let δ := ε/M . Then for any x1 , x2 ∈ X with d(x1 , x2 ) < δ, we have d(f (x1 ), f (x2 )) ≤ M d(x1 , x2 ) < M δ = ε and so f is uniformly continuous. 17. Let (X, d) be a metric space. Equip the product space X × X with the metric   ρ (x1 , y1 ), (x2 , y2 ) := d(x1 , x2 ) + d(y1 , y2 )

for any xi , yi ∈ X, i = 1, 2. The the metric d on X can be considered as a function d : X × X → R. Show that d is uniformly continuous.

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Proof . By Example 2.3.7, ρ is a well-defined metric on X × X . Furthermore, from the first proof of the continuity of d on X × X in the same Example, it is easy to see that associated to any ε > 0, a uniform δ (= 2ε ) can be chosen. Hence d is indeed uniformly continuous on X × X . 18. Show that Cauchyness is preserved by uniformly continuous functions, that is, uniform continuous functions map Cauchy sequences to Cauchy sequences. [Compare with Exercise 2.3, Part A, Problem #7.] Proof . Suppose f : X → Y is uniformly continuous and {xn }n∈N is a Cauchy sequence in X . Then for any ε > 0, there exists δ = δ(ε) > 0 such that   dY f (x1 ), f (x2 ) < ε whenever dX (x1 , x2 ) < δ .

Since {xn }n∈N is Cauchy, there exists N ∈ N such that dX (xn , xm )
0. Suppose f (S) ⊂ Y is not totally bounded. Then f (S) cannot be covered by any finite collection of open balls in Y of radius r . Pick y1 ∈ f (S). Then B(y1 , r) ⊃ f (S) and so there exists y2 ∈ f (S) \ B(y1 , r). In particular, we have d(y2 , y1 ) ≥ r . Since B(y1 , r) ∪ B(y2 , r) ⊃ f (S), there exists 2 y3 ∈ f (S) \ i=1 B(yi , r). In particular, we have d(yi , yj ) ≥ r for any i = j ∈ {1, 2, 3}. Inductively, we have a sequence {yn }n∈N in f (S) such that d(yi , yj ) ≥ r for any i = j . Let {xn }n∈N be a sequence in S with f (xn ) = yn for all n ∈ N. As the yi ’s are all distinct, the xi ’s must also be all distinct. Since S is totally bounded, by Exercise 2.2, Part B, Problem #3, {xn }n∈N contains a Cauchy subsequence, say, {xnk }k∈N . By Exercise 4.1, Part B, Problem #18, {ynk }k∈N = {f (xnk )}k∈N is a Cauchy sequence, which is impossible. Hence f (S) must be totally bounded. 22. Determine whether the converse of Exercise 4.1, Part B, Problem #21 is valid. That is, whether a function is uniformly continuous if it preserves total boundedness. Answer : No.

Uniform Continuity

317

Example : First of all, it is elementary to verify that every bounded subset of R is totally bounded. Consider the exponential function f : R → R, f (x) := ex . Let S ⊂ R be a totally bounded subset. For any r > 0, S can be covered by finitely many open balls Bi (r) ⊂ R of radius r , i = 1, . . . , n, which are bounded subsets of R and so they are all totally bounded. Being continuous on R, by Theorem 4.1.7, f is uniformly continuous on every compact subset of R. Since every bounded subset of R is contained in a compact subset of R, f is uniformly continuous on every bounded subset of R. So in particular, f is uniformly continuous on each Bi (r). By Exercise 4.1, Part B, Problem #21, f (Bi (r)) is totally n bounded for each i = 1, . . . , n. Therefore, f (S) = i=1 f (Bi (r)) is totally bounded. Hence f preserves total boundedness. However, we have seen in Exercise 4.1, Part A, #4 that it is not uniformly continuous on R.

23. Let f : X → Y be any function and g : Y → Z be 1 − 1 and continuous. Suppose h := g ◦ f : X → Z is uniformly continuous. (a) If Y is compact, show that f is also uniformly continuous. (b) If g is not 1 − 1, is the assertion of (a) still valid? (c) If X and Z are compact but Y is not, is the assertion of (a) still valid? Solution . (a) Since g is 1 − 1 and continuous and Y is compact, by Theorem

2.3.16, g −1 : g(Y ) → Y exists and is continuous. Since Y is compact and g is continuous, g(Y ) is also compact and so by

Theorem 4.1.7, g −1 is uniformly continuous. Hence f = g −1 ◦h is also uniformly continuous by Theorem 4.1.8.

(b) The assertion of (a) will no longer be valid.

Example: Let Y = singleton be any compact space, f : X → Y be any discontinuous function and g : Y → Z be a constant function. Being a constant function, g is continuous but not

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1 − 1. Then h = g ◦ f is a constant function and so it is uniformly continuous. However, f is not uniformly continuous. (c) The assertion of (a) will no longer be valid.

Example: Let X := [0, 2], Y := [0, 1] ∪ (2, 3], Z := [0, 2]. Then X and Z are compact but Y is not. Consider f : X → Y and g : Y → Z given by

x x ∈ [0, 1] f (x) : = x + 1 x ∈ (1, 2],

y y ∈ [0, 1] [12pt]g(y) : = y − 1 y ∈ (2, 3] .

Then g is 1 − 1 and continuous on Y , h(x) = g ◦ f (x) = x is uniformly continuous on X , but f is not continuous on X .

24. Let D ⊂ X be a dense subset of X, Y be a complete metric space, and f : D → Y be uniformly continuous. Show that f can be extended to a uniformly continuous function F : X → Y in the sense that F |D = f , and that such an extension is unique. Proof . Since D is dense in X , for any x ∈ X , we have a sequence {xn }n∈N in D which converges to x as n → ∞. In particular, {xn }n∈N is Cauchy and so by Exercise 4.1, Part B, Problem #18, {f (xn )}n∈N is a Cauchy sequence in the complete metric space Y . Hence {f (xn )} → some y ∈ Y as n → ∞. Define F (x) := y . We claim that F : X → Y such defined is actually well-defined, that is, for any other sequence {tn }n∈N in D which converges to x as n → ∞, we must have {f (tn )} → y as n → ∞. In fact, for any ε > 0, by the uniform continuity of f on D , there exists δ > 0 such that d(f (p), f (q)) < ε for any p, q ∈ D with d(p, q) < δ. Since {xn } → x and {tn } → x as n → ∞, there exists N ∈ N such that d(xn , x) < δ/2 and d(tn , x) < δ/2 for all n ≥ N . By triangle inequality, we have d(xn , tn ) < δ for all n ≥ N . Hence d(f (xn ), f (tn )) < ε for all n ≥ N . As it is known that {f (xn )} → y as n → ∞, this forces {f (tn )} → y as n → ∞ (why?) This proves the claim and F : X → Y is well-defined.

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Next, for any x ∈ D , the constant sequence {xn := x}n∈N clearly

converges to x as n → ∞, hence F (x) = f (x) and so F |D = f .

We move onto show the uniform continuity of F on X . For any

ε > 0, choose δ > 0 as above. For any p, q ∈ X with d(p, q) < δ/3, there exist sequences {pn }n∈N and {qn }n∈N in D such that {pn } → p and {qn } → q as n → ∞. Hence there exists N ∈ N such that d(pn , p) < δ/3 and d(qn , q) < δ/3 for all n ≥ N . By triangle

inequality,

d(pn , qn ) ≤ d(pn , p) + d(p, q) + d(q, qn ) < δ for all n ≥ N . Hence d(f (pn ), f (qn )) < ε for all n ≥ N . On the other hand,

by the definition of F , we have F (p) = limn→∞ f (pn ) and F (q) =

limn→∞ f (qn ). Hence there exists M ∈ N such that d(F (p), f (pn )) < ε and d(F (q), f (qn )) < ε for all n ≥ M . Therefore, d(F (p), F (q)) ≤ d(F (p), F (pn )) + d(F (pn ), F (qn )) + d(F (qn ), F (q))

= d(F (p), f (pn )) + d(f (pn ), f (qn )) + d(f (qn ), F (q)) < 3ε

for all n ≥ max{N, M }. Hence F is uniformly continuous on X .

Finally, uniqueness. If G is another extension of f to X such that

G is (uniformly) continuous, then by construction, F |D = G|D . For any x ∈ X \ D , there is a sequence {xn }n∈N in D which converges to x as n → ∞. Then as both F and G are continuous, we have G(x) = limn→∞ G(xn ) = limn→∞ F (xn ) = F (x) and so G = F . Hence uniqueness.

25. Let f : [0, ∞) → R be a function such that limn→∞ f (n+x) = 0 for every x ∈ [0, 1). (a) If f is uniformly continuous, prove that limx→∞ f (x) = 0. (b) Determine whether (a) remain valid if the continuity of f is only pointwise but not uniform.

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Solution : (a) Since f is uniformly continuous, for all ε > 0, there exists δ > 0 such that

|f (x) − f (y)|
1/δ . Partition

the interval [0, 1] into 0 = x0 < . . . < xK = 1 such that

0 < xi − xi−1 < δ for all 1 ≤ i ≤ K . For each fixed i ∈ {0, . . . , K − 1}, since f (n + xi ) → 0 as n → ∞, there exists Ni ∈ N such that   f (n + xi ) < ε 2

whenever n ≥ Ni .

For any x > 0, if as usual we denote by [x] the largest integer that does not exceed x, then x− [x] ∈ [0, 1) and so there exists

i0 ∈ {0, . . . , K − 1} such that x − [x] ∈ [xi0 , xi0 +1 ). Hence 0 ≤ x − ([x] + xi0 ) < δ and  ε    f (x) − f ([x] + xi0 )  < . 2

In particular, for any x ≥ N := max{Ni : i = 0, . . . , K −1}, we have [x] ≥ Ni0 and so

    ε ε |f (x)| ≤ f ([x]+ xi0 ) + f (x)− f ([x]+ xi0 ) < + =  . 2 2 That is, limx→∞ f (x) = 0.

(b) Answer : No, (a) will no longer hold.

Example : Consider the function f : [0, ∞) → R defined by   ⎧ x ∈ n, n + n1 , n ≥ 3 ⎪ ⎨ n(x − n)   f (x) := 2 − n(x − n) x ∈ n + n1 , n + n2 , n ≥ 3 ⎪ ⎩ otherwise. 0

It is readily seen that f is continuous on [0, ∞). For any

x ∈ (0, 1) and any n > x2 , since n+x ∈ (n+ n2 , n+1), we have

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321

f (x + n) = 0. Besides, when x = 0, f (x + n) = f (n) = 0 for all n ∈ N. Hence for any x ∈ [0, 1), f (n + x) → 0 as n → ∞. However, since f (n + n1 ) = 1 for all n ≥ 3, limx→∞ f (x) fails to exist.

26. Suppose that X and Y are metric spaces and f : X → Y is a function. Show that the following two statements are equivalent: (a) f is uniformly continuous. (b) For all sequences {un }n∈N , {xn }n∈N in X, if {dX (un , xn )} → 0 as n → ∞, then {dY (f (un ), f (xn ))} → 0 as n → ∞. Proof . (a)⇒(b): Suppose f is uniformly continuous, then for any ε > 0, there exists δ = δ(ε) > 0 such that dX (u, x) < δ

⇒

dY (f (u), f (x)) < ε .

Let {un }n∈N , {xn }n∈N be two sequences in X such that {dX (un ,

xn )} → 0 as n → ∞. Then there exists N ∈ N such that dX (un , xn ) < δ for all n ≥ N . Hence

dY (f (un ), f (xn )) < ε for all n ≥ N , which implies

{dY (f (un ), f (xn ))} → 0 as n → ∞ . (b)⇒(a): Assume to the contrary that f is not uniformly continuous.

Then there exists ε > 0 such that for any δ > 0, there exists u,

x ∈ X with dX (u, x) < δ but dY (f (u), f (x)) ≥ ε. So for any n ∈ N, we can find un , xn ∈ X such that 1 dX (un , xn ) < and dY (f (un ), f (xn )) > ε . n Notice that that means we have two sequences {un }n∈N , {xn }n∈N in X with {dX (un , xn )} → 0 as n → ∞ but {dY (f (un ), f (xn ))} → 0 as n → ∞, a contradiction.

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4.2 Contraction and Banach’s Fixed Point Theorem A special example of uniformly continuous functions of a metric space X into itself is called a contraction. It is a distance-decreasing function in such a way that the distance between any two distinct points will decrease at least by a factor strictly less than 1. Definition 4.2.1 below will make it precise. It turns out that such functions play an important role in the existence and uniqueness problem of solutions to many differential and integral equations. Definition 4.2.1. A function f : X → X is said to be a contraction of X if there exists a positive number 0 < α < 1 called a contraction constant such that   d f (x), f (y) ≤ α d(x, y) for all x, y ∈ X .

Observe that the contraction constant of any contraction is not unique. In fact, if 0 < α < 1 is a contraction constant of a contraction f , then any number β ∈ [α, 1) is also a contraction constant of f . It is also clear by definition that every contraction of X is uniformly continuous on X. Note that implicitly imposed in Definition 4.2.1, a contraction of X must be a self-map on X, that is, a function which maps X into itself , although it does not have to be surjective. Examples 4.2.2: (i) f : R → R defined by f (x) = x2 is a contraction. Proof . As |f (x) − f (y)| = | x2 − y2 | = 12 |x − y| for any x, y ∈ R, f is a contraction with contraction constant 12 . (ii) f : (2, ∞) → (1, ∞) defined by f (x) = x2 is not a contraction. Proof . Observe that although we do have |f (x) − f (y)| = | x2 − y 1 2 | = 2 |x − y| for any x, y ∈ (2, ∞), since the image of (2, ∞) under f is (1, ∞) ⊂ (2, ∞), according to Definition 4.2.1, f is not a contraction.

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(iii) f : (2, ∞) → (2, ∞) defined by f (x) = 1 + x2 is a contraction.   Proof . Observe first that f (2, ∞) = (2, ∞). Similar to (i), we have   y  1  x    − 1+ |f (x) − f (y)| =  1 +  = x − y  2 2 2 for any x, y ∈ (2, ∞) and so f is a contraction with contraction constant 12 . Remark. Every contraction is distance-decreasing, that is, d(f (x), f (y)) < d(x, y)

for all x = y ∈ X .

(In some books, a distance-decreasing function is known as a contractive mapping). However, the converse is not true. In fact, a function is a contraction means the ratio d(f (x), f (y)) ≤α n .

So the error by using fn to estimate the unique solution f to (*) with respect to d is d(fn , f ) = lim d(fn , fm ) ≤ m→∞

d(f1 , f0 ) n α . 1−α

Now it is easy to compute d(f1 , f0 ) = sup{|f1 (x) − f0 (x)| : x ∈ [0, 1]} = sup{|x| : x ∈ [0, 1]} = 1 ,

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329

and so the error by using fn to estimate the unique solution f to (*) is 1 1 1 1 αn = d(fn , f ) ≤ 1 · n = n−1 . 1−α 2 1− 2 2 So for example, if we want to approximate f within an accuracy of 0.01 at any point in [0, 1], we need 1 2n−1

< 0.01 .

Solving this, we get n ≥ 8. So f8 approximates f with an accuracy of 0.01 at any point in [0, 1]. Remark. For this simple example, it is rather obvious by seeing the first few fn ’s that all fn ’s are scalar multiples of x and it is easy to speculate that this is also the case for its limit f . Hence we may simply put f (x) := kx into (∗) and solve for k. In fact, solving 1 kx = 2 we have k =



1

xy(ky) dy + x ,

0

6 6 and so the unique solution to (∗) is simply y = x. 5 5

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Exercise 4.2 Part A: True or False Questions For each of the following statements, determine if it is true or false. If it is true, prove it. If it is false, give a counterexample or provide proper justification. 1. Every uniformly continuous function f : X → X on a complete metric space X has a fixed point. Answer : False. Example : The function f : R → R given by f (x) := x + 1, x ∈ R, is clearly uniformly continuous on the complete metric space R. But f has no fixed point in R. 2. f : (0, ∞) → R defined by f (x) := ln x is a contraction.

Answer : False. Justification . By Exercise 4.1, Part A, Problem #5, f is not

uniformly continuous. Hence it is not a contraction.

3. f : R → R defined by f (x) := cos(cos x) is a contraction. Answer : True. Proof . By Mean Value Theorem,            cos(cos x) − cos(cos y) = x − y   sin(cos c) sin c

for some c between x and y . As cos c ∈ [−1, 1], we have | sin(cos c)

sin c| ≤ sin 1. Hence          cos(cos x) − cos(cos y) ≤ (sin 1) x − y  for all x, y ∈ R

and so f is a contraction with contraction constant 0 < sin 1 < 1.

4. f : R → R defined by f (x) := ln(1 + ex ) is a contraction. Answer : False.

Uniform Continuity

331

Justification . Since x = ln(1 + ex ) has no solution, f has no fixed point. As R is complete, by Banach’s Fixed Point Theorem, f cannot be a contraction.

5. The function f : [1, ∞) → contraction of [1, ∞).

1

2, ∞



given by f (x) =

x 2

Answer : False. Justification . The function f is not a self map, as f (1) = [1, ∞). Hence by definition, it is not a contraction of [1, ∞). 6. The function f : (−1, ∞) → R defined by f (x) := contraction of [− 12 , 0].

is a

1 2

∈ /

1 is a 1+x

Answer : False. Justification . Note that f is strictly decreasing, f (− 12 ) = 2, and f (0) = 1. Hence f ([− 12 , 0]) = [1, 2] ⊂ [− 12 , 0] and so f is not a contraction of [− 12 , 0]. 7. The function f : (−1, ∞) → R defined by f (x) := contraction of [0, 1].

1 is a 1+x

Answer : False. We have f ([0, 1]) = [ 12 , 1] ⊂ [0, 1]. However, it’s not a contraction. In fact, it is easy to see that

1 |f (x) − f (y)| = |x − y| (1 + x)(1 + y)

for any 0 ≤ x < y ≤ 1 .

By taking x = 0 and letting y approaches 0+ , the right hand side approaches 1 indefinitely and so the contraction constant does not exist.

8. The function f : (−1, ∞) → R defined by f (x) := fixed point in [0, 1]. Answer : True.

1 has a 1+x

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Metric Space Topology: Examples, Exercises and Solutions

Proof√. Solving the equation f (x) = x, one has a unique solution x = 5−1 ∈ [0, 1]. 2 9. Let X be complete. Then every uniformly continuous function f : X → X having a unique fixed point in X must be a contraction of X. Answer : False. Example : The identity function f : R → R, f (x) := x, x ∈ R, is clearly uniformly continuous, with a unique fixed point x = 0, and R is complete. However, f is not a contraction. Part B: Problems 1. Show that the function f : [0, ∞) → R defined by   f (x) := ln 1 + e−x ,

x ∈ [0, ∞) ,

has a unique fixed point. Find explicitly the fixed point. Solution . Note that f is continuous on [0, ∞) and differentiable in −e−x 1 (0, ∞) with f  (x) = . In particular, |f  (x)| ≤ for all 1 + e−x 2 x ∈ (0, ∞). Hence for any x, y ∈ [0, ∞) with x = y , by Mean Value

Theorem,

|f (x) − f (y)| = |f  (c)| |x − y| for some c between x and y . This gives

|f (x) − f (y)| ≤

1 |x − y| 2

and hence f is a contraction, with contraction constant

1 . 2

By

Banach’s Fixed Point Theorem, f has a unique fixed point p. To find p, we can either imitate the proof of Banach’s Fixed Point Theorem, starting with any point x0 ∈ [0, ∞), compute explicitly the

Uniform Continuity

333

sequence {xn }n∈N with xn := f (xn−1 ), n ∈ N, and we will have

p = limn→∞ xn . For example, if we pick x0 = 0, then x1 = f (x0 ) = f (0) = ln 2 , 3 x2 = f (x1 ) = f (ln 2) = ln ,

 2 3 5 = ln , x3 = f (x2 ) = f ln 2 3 

8 5 = ln x4 = f (x3 ) = f ln 3 5 ··· Inductively, it is easy to verify that exn+1 = 1 +

1 for all n ∈ N. exn

As we know that {xn }n∈N converges to p, we have

xn+1

e = lim e

= lim

1+

1



1 . n→∞ n→∞ ep √  1 + √5 1+ 5 p Solving this, we get e = and so p = ln . 2 2 Alternatively , we can√simply solve the equation p = f (p) explicitly 1 + 5 and get p = ln . 2 p

exn

=1+

2. For any k ∈ C([0, 1] × [0, 1]) with |k| ≤ M < 1 on [0, 1] × [0, 1], show that the integral equation  1 2 k(x, y) f (y) dy = ex x ∈ [0, 1] f (x) + 0

has a unique solution f ∈ C[0, 1].

Proof . First observe that if k ≡ 0 on −0, 1], then the given inte2 gral equation has a unique solution f (x) = ex . so we assume that k ≡ 0 on [0, 1] and so M > 0. As usual, we take for granted that (C([0, 1]), d) with metric d defined by d(f, g) := sup{|f (x) − g(x)| : x ∈ [0, 1]}

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Metric Space Topology: Examples, Exercises and Solutions

is complete. Define an operator T : (C[0, 1], d) → C([0, 1], d) by



2

T (f )(x) := ex −

1

k(x, y) f (y) dy 0

for and f ∈ C[0, 1] and x ∈ [0, 1]. Then f is a solution to the given

integral equation if and only if it is a fixed point of T . Now for any f ,

g ∈ C[0, 1], we have

     T (f )(x) − T (g)(x) =  ≤



1

0 1

0

  k(x, y) (f (y) − g(y)) dy 

   k(x, y) f (y) − g(y) dy

≤ M d(f, g) .

So T is a contraction of C[0, 1] with contraction constant 0 < M < 1 and so by Banach’s Fixed Point Theorem, T has a unique fixed point in C[0, 1], which is in turn the unique solution to the given integral equation.

3. Let X := (0, ∞) with the usual Euclidean metric and C 1 (X; X) be the collection of all continuously differentiable positive functions on (0, ∞). If f ∈ C 1 (X, X) satisfies x|f  (x)| ≤ kf (x) for all x ∈ X, where k ∈ (0, 1) is a constant, show that f has a unique fixed point. Proof . For any 0 < x < u, we have    u f  (s)   u |f  (s)|   ds ≤ ds  ln f (u) − ln f (x) =  f (s) x f (s) x  u k ds = k | ln u − ln x| . ≤ x s On X , we consider another metric

    d˜X (u, x) :=  ln u − ln x for all u, x ∈ X .

Uniform Continuity

335

It is not hard to verify that (X, d˜X ) is a complete metric space and from the above estimate, we have

  ˜ x) for all u, x ∈ X d˜ f (u), f (x) ≤ k d(u,

and so f : (X, d˜X ) → (X, d˜X ) is a contraction with contraction constant k ∈ (0, 1). By Banach’s Fixed Point Theorem, f has a

unique fixed point.

4. Let f : X → X be a distance-decreasing function (or a   contractive function) of X, that is, d f (x), f (y) < d(x, y) for any x = y ∈ X. (a) Prove that f has at most one fixed point. (b) If X is complete, is it necessarily true that f must have a fixed point? (c) If X is compact, prove that f has exactly one fixed point. (d) If X is compact, is it necessarily true that f must be a contraction? Proof . (a) Suppose p and q are distinct fixed points of f , then

  d(p, q) = d f (p), f (q) < d(p, q)

which is impossible. So f has at most one fixed point. (b) Answer : No.

Example : Let X := (1, ∞) equipped with the Euclidean √ metric, and f : X → X be defined by f (x) := x, x ∈ X .

Then

√ √ √ √ √ √ |f (x)−f (y)| = | x− y| < | x− y|·| x+ y| = |x−y| . Hence f is distance-decreasing. However, as f (x) < x for all

x ∈ X , f has no fixed point in X .

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Metric Space Topology: Examples, Exercises and Solutions

Alternative Example : Let f : R → R be defined by f (x) := √ x2 + 1, x ∈ R. As shown in Example 4.2.3, f is distancedecreasing. However, it is easily seen that f (x) = x has no solution and so f has no fixed point.   (c) Define g : X → R by g(x) := d x, f (x) for x ∈ X . It is evident that g is continuous on X . If X is compact, g attains its minimum in X at some point, say, a ∈ X . If f (a) = a, since f is distance-decreasing, we have     g(f (a)) = d f (a), f (f (a)) < d a, f (a) = g(a) ,

which contradicts to the assumption that g attains its minimum at a. Therefore, f (a) = a. That is, a is a fixed point of f . Finally, the uniqueness of the fixed point is guaranteed by (a). (d) Answer : No.

Example : Let X := [0, 12 ] equipped with the Euclidean metric, and f (x) := x2 , x ∈ X . Then X is compact and we have

|f (x) − f (y)| = |x2 − y 2 | = |x − y| |x + y| ≤ |x − y| for all x, y ∈ [0,

1 2

]. Note that the equality holds if and only if |x + y| = 1, which forces x = y = 12 . Hence for any x = y ∈ X , we have |f (x) − f (y)| < |x − y| and so f is distance-

decreasing.

But then for x = y ∈ X ,

|f (x) − f (y)| = |x + y| can be |x − y|

arbitrarily close to 1 and so there is no contraction constant.

Hence f is not a contraction of X .

5. Let X be a complete metric space and f : X → X. Suppose there exists n ∈ N such that f (n) := f ◦ f ◦ · · · ◦ f (n iterates) is a contraction of X. (a) Show that f has a unique fixed point in X. (b) Is f necessarily a contraction?

Uniform Continuity

337

Proof . (a) By Banach’s Fixed Point Theorem, f (n) has a unique fixed point, and we denote it as x0 . As f (n) (f (x0 )) = f (n+1) (x0 ) =

f (f (n) (x0 )) = f (x0 ), we see that f (x0 ) is also a fixed point of f (n) . The uniqueness implies that f (x0 ) = x0 , which in turn implies that x0 is a fixed point of f . Finally, it is evident that every fixed point of f must also be a fixed point of f (n) . Hence x0 is the only fixed point of f . (b) Answer : No. Example : Consider the function f : R → R defined by

−x if x ≥ 0 f (x) := −x if x < 0 . 2 Then f 2 (x) =

x 2

is a contraction on the complete metric space

R, but as |f (1) − f (2)| = 1 < |1 − 2|, f is not a contraction. 6. Let X be a complete metric space and f : X → X. For any n ∈ N, let   d f (n) (x1 ), f (n) (x2 ) . an := sup d(x1 , x2 ) x1 =x2 ∞ If n=1 an < ∞, show that f has a unique fixed point in X. ∞ Proof . As an ≥ 0 for all n and n=1 an < ∞, there exists N ∈ N such that aN < 12 . Hence for any x1 = x2 , we have   d f (N ) (x1 ), f (N ) (x2 ) 1 ≤ aN < d(x1 , x2 ) 2 and so f (N ) is a contraction of X with contraction constant

1 . 2

The

assertion now follows from Exercise 4.2, Part B, Problem # 5.

7. Let (X, d) be a complete metric space and T : X → X. If there exists 0 < K < 12 such that       d T (x), T (y) ≤ K[d x, T (x) + d y, T (y) ]

338

Metric Space Topology: Examples, Exercises and Solutions

for all x, y ∈ X, show that T has a unique fixed point in X. Proof . Fix x0 ∈ X . Define a sequence {xn }n∈N in X by xn := T (xn−1 ) for all n ∈ N. By assumption, for each n ∈ N,   d(xn+1 , xn ) = d T (xn ), T (xn−1 )     ≤ K[d xn , T (xn ) + d xn−1 , T (xn−1 ) ] = K[d(xn , xn+1 ) + d(xn−1 , xn )]

and so

K d(xn−1 , xn ) . 1−K

d(xn+1 , xn ) ≤ Inductively, we have

d(xn+1 , xn ) ≤ Since 0 < K
0. (b) Show that f is uniformly continuous on X. (c) Fix a ∈ X. For any n ∈ N, let xn := f (n) (a). Show that {d(xn+1 , xn )} → 0 as n → ∞. (d) Show that {xn }n∈N is a Cauchy sequence. (e) Show that f has a unique fixed point in X. Proof . (a) Fix ε > 0. If ϕ(ε) ≥ ε, since ϕ is increasing, we have ϕ(2) (ε) ≥

ϕ(ε) ≥ ε, and so inductively, ϕ(n) (ε) ≥ ε for all n, which contradicts to the assumption that limn→∞ ϕ(n) (t) = 0 for all t > 0. Hence ϕ(ε) < ε. (b) For any ε > 0, take δ := ε. Then for any x, y ∈ X with d(x, y) < δ = ε, by the monotonicity of ϕ and (a), we have     d f (x), f (y) ≤ ϕ d(x, y) ≤ ϕ(ε) < ε .

Thus f is uniformly continuous on X .

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Metric Space Topology: Examples, Exercises and Solutions

(c) If f (a) = a, then xn = a for all n and there is nothing to prove. So assume that f (a) = a. So d(x1 , a) > 0 and we have

    d(xn+1 , xn ) = d f (xn ), f (xn−1 ) ≤ ϕ d(xn , xn−1 )   ≤ ϕ(2) d(xn−1 , xn−2 )   ≤ · · · ≤ ϕ(n) d(x1 , x)   and so limn→∞ d(xn+1 , xn ) ≤ limn→∞ ϕ(n) d(x1 , x) = 0.

(d) For any ε > 0, by (a), ε − ϕ(ε) > 0 and so by (c), there exists

N ∈ N such that

d(xn+1 , xn )
0 is an upper bound for |K| in [a, b ] × [a, b ]. Next,

we have

    2 F (f )(x) − F 2 (g)(x)  x       = |λ|  K(x, y) F (f )(y) − F (g)(y) dy  a  x     |K(x, y)| F (f )(y) − F (g)(y) dy ≤ |λ| a  x 2 2 (y − a) dy ≤ |λ| M d(f, g) a

(x − a)2 d(f, g) = |λ|2 M 2 2 and so by induction,

|F n (f )(x) − F n (g)(x)| ≤ |λ|n M n

(x − a)n d(f, g) . n!

Hence

d(F n (f ), F n (g)) ≤ |λ|n M n

(b − a)n d(f, g) . n!

In particular, whenever n is large enough, or to be precise, whenever n

|λ|n M n (b−a) n!

< 1, F n is a contraction. By Exercise 4.2, Part B, Problem # 5, F has a unique fixed point f ∈ C[a, b ], which is in turn the unique solution to the Volterra Equation.

10. Estimate the solution in C[0, 1] to the integral equation  1 1 2 (x + y 2 )f (y) dy + x f (x) = 3 0 within an error of 0.01 everywhere in [0, 1]. Solution . We refer the readers to Example 4.2.7, and consider the operator F : C[0, 1] → [0, 1] defined by  1 1 2 F (f )(x) := (x + y 2 )f (y) dy + x , x ∈ [0, 1] . 3 0

Uniform Continuity

343

Using the same notations as those in Example 4.2.7, we have λ =

1 , 3

M = sup{x2 + y 2 : (x, y) ∈ [0, 1] × [0, 1]} = 2, b = 1, a = 0. So λM (b − a) = 23 < 1 and so F is a contraction, with contraction constant α = λM (b − a) = 23 . By Banach’s Fixed Point Theorem, F has a unique fixed point f which is also the unique solution to the given integral equation. To approximate the unique solution, we pick as the starting point the simple function

f0 := 0 . Then we have

f1 = F (f0 ) = x ,  1 1 2 1 1 , (x + y 2 )ydy + x = x2 + x + f2 = F (f1 ) = 3 0 6 12 f3 = F (f2 ) = · · · . Note from the proof of Banach’s Fixed Point Theorem, we have

d(fn , fm ) ≤

d(f1 , f0 ) n α for all m > n . 1−α

So the error by using fn to estimate the unique solution f is

d(fn , f ) = lim d(fn , fm ) ≤ m→∞

Now it is easy to compute

d(f1 , f0 ) n α . 1−α

d(f1 , f0 ) = sup{|f1 (x) − f0 (x)| : x ∈ [0, 1]} = 1 , and so the error by using fn to estimate the unique solution f is

1 1 αn = d(fn , f ) ≤ 1−α 1−

2 3

n 2n 2 = n−1 . 3 3

In particular, to approximate f within an error of 0.01 in d, we solve

2n 3n−1

< 0.01

and get n ≥ 15, hence we could use f15 as an approximation.

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Metric Space Topology: Examples, Exercises and Solutions

11. Let F : D ⊂ R2 → R be a continuous function satisfying a Lipschitz condition in y, that is, there exists L > 0 such that   F (x, y1 ) − F (x, y2 ) ≤ L|y1 − y2 |

for all (x, y1 ), (x, y2 ) ∈ D. Show that for any (x0 , y0 ) ∈ D, the initial value problem ⎧ ⎨ dy (x) = F x, y(x) dx ⎩ y(x0 ) = y0

for all x ∈ I

(*)

has a unique solution y = y(x) on the interval I = [x0 −δ, x0 +δ] for suitably chosen δ > 0. Proof . Observe first that y = y(x) is a solution to the given initial value problem (*) if and only if

y(x) = y0 +



x

x0



  F t, y(t) dt ,

x∈I .

(**)



Consider the metric space C(I), d with

   d(f, g) := sup f (x) − g(x) : x ∈ I .

We again take it for granted that (C(I), d) is complete. Define

T : C(I) → C(I) h → T (h)

by

(T h)(x) := y0 +



x

x0

  F t, h(t) dt ,

x∈I .

Then y = y(x) solves (**) if and only if it is a fixed point of T . Thus, by Banach’s Fixed Point Theorem, the problem reduces to showing

Uniform Continuity

345

that T is a contraction on I = [x0 − δ, x0 + δ] with a suitably chosen

δ, which is by now rather straight forward. In fact, by

   d(T f, T g) = sup (T f )(x) − (T g)(x) : x ∈ I  

 x     |F (t, f (t)) − F (t, g(t))|dt : x ∈ I ≤ sup  x0 

  x       f (t) − g(t) dt : x ∈ I ≤ sup L x0    ≤ sup L · d(f, g) · (x − x0 ) : x ∈ I ≤ δL d(f, g) ,

we see that T is a contraction with contraction constant δL whenever

δ
0 and every x ∈ S, there exists N ∈ N such that |fn (x) − f (x)| < ε for every n ≥ N . Note that naturally, the natural number N is dependent on the given positive number ε and also on the point x ∈ S under consideration. So in general, when ε > 0 or x ∈ S varies, the number N will also vary accordingly. Therefore, in general, if we want to make this observation clearer, we would say that for any ε > 0 and any x ∈ S, there exists N = N (ε, x) ∈ N such that |fn (x) − f (x)| < ε for every n ≥ N (ε, x) . Now a natural question is: what common properties of the fn ’s can be passed on to their pointwise limit f ? For instance, we pose the following questions: Question 1 : If fn is continuous for each n ∈ N and {fn } → f pointwisely, must f be continuous? Observation: Let c ∈ S be any point in S and we are to determine whether f is continuous at c. By definition, we are to check whether it is true that limx→c f (x) equals f (c). Now by assumption, {fn } → f pointwisely on S and so limn→∞ fn (x) = f (x) for every x ∈ S. Hence in particular,   lim f (x) = lim lim fn (x) . x→c

x→c

n→∞

Uniform Convergence

351

On the other hand, for each n ∈ N, fn is continuous at c. So we have limx→c fn (x) = fn (c). Moreover, we have limn→∞ fn (c) = f (c) and so   lim lim fn (x) = lim fn (c) = f (c) . n→∞

x→c

n→∞

Combining, the question reduces to determining whether we have the equality     ? lim lim fn (x) = lim lim fn (x) . x→c

n→∞

n→∞

x→c

In other words, Question 1 is to determine whether the two limits limx→c and limn→∞ commute.

Question 2 : If {fn } → f pointwisely on an interval [a, b ] ⊂ R, fn is integrable over [a, b ] for every n ∈ N, and f is integrable over [a, b ], is it necessarily true that lim

n→∞



b

?

fn (x)dx =

a



b

f (x)dx . a

Observation: As limn→∞ fn (x) = f (x) for every x ∈ S, the problem reduces to determining whether it is true that lim

n→∞



b

?

fn (x)dx = a



b

lim f (x)dx .

a n→∞

That is, the problem is to determine whether limn→∞ can be taken out from the integral sign, or equivalently, whether lim n→∞ commutes with integration. Question 3 : If {fn } → f pointwisely on an interval (a, b) ⊂ R and fn is differentiable in (a, b) for every n ∈ N, is it necessarily true that f is differentiable in (a, b) with f  = limn→∞ fn ? Observation: As limn→∞ fn (x) = f (x) for every x ∈ S, the problem reduces to determining whether it is true that limn→∞ fn is

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Metric Space Topology: Examples, Exercises and Solutions

differentiable, and, if limn→∞ fn exists, whether it is true that ?

lim fn (x) =

n→∞



 lim fn (x) ,

n→∞

x ∈ (a, b) .

That is, the problem is to determine whether limn→∞ commutes with differentiation. The following simple Examples show that in general, the answers to all these three questions are negative. Example 5.1.1. For any n ∈ N, let fn : [0, 1] → R be defined by fn (x) := xn , x ∈ [0, 1]. Then fn (x) → f (x) =



0 if x ∈ [0, 1) 1 if x = 1

pointwisely on [0, 1]. Note that all fn ’s are continuous on [0,1] but the pointwise limit f is not. Example 5.1.2. For any n ∈ N, let fn : [0, 1] → R be defined by fn (x) := n2 (1 − x)xn , x ∈ [0, 1]. Then fn (x) → f (x) :≡ 0 pointwisely on [0, 1]. Observe that 

0

1

fn (x)dx =

n2 (n + 1)(n + 2)

for all n ∈ N

and so lim

n→∞



0

1

n2 = 1 = 0 = fn (x)dx = lim n→∞ (n + 1)(n + 2)



1

0

Hence in general, limit and integration do not commute.

f (x)dx .

Uniform Convergence

Example 5.1.3. defined by

353

(i). For any n ∈ N, let fn : (−1, 1) → R be 

1 , x ∈ (−1, 1) . n It is clear that each fn is differentiable in (−1, 1) and fn (x) :=

x2 +

fn (x) → f (x) := |x| pointwisely in (−1, 1) but f is not differentiable at 0. So the pointwise limit of a sequence of differentiable functions may not be differentiable. (ii). Let {fn }n∈N be the sequence of differentiable functions on R defined by 1 fn (x) := √ sin nx , x ∈ R , n ∈ N . n Observe that for any x ∈ R, 1 1 |fn (x)| = √ sin nx ≤ √ → 0 as n → ∞ , n n

thus {fn } → f :≡ 0 pointwisely on R. Note that f is differentiable in R with f  ≡ 0. However, √ fn (x) = n cos nx , x ∈ R

does not converge at any x ∈ R. Hence in particular, limit and differentiation do not commute. Although these examples are rather disappointing in the sense that they gave negative answers to the three questions posed at the beginning of the section, a careful analysis shows that they share a common feature that while the sequences of functions {fn }n∈N are all pointwisely convergent, the speeds of the convergence at various points are very different. In other words, for the same ε > 0, at different points x ∈ S, we need very different corresponding integer N ∈ N. That is, there is no single N ∈ N that is good for all points x ∈ S. This important observation leads to the following definition.

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Metric Space Topology: Examples, Exercises and Solutions

Definition 5.1.4. A sequence of functions {fn }n∈N on S is said to converge uniformly on S to a function f , denoted by {fn } → f uniformly on S, if for any ε > 0, there exists N ∈ N such that fn (x) − f (x) < ε

for all n ≥ N and all x ∈ S .

In this case, we write f = limn→∞ fn uniformly on S and call f the uniform limit of {fn }n∈N on S. Note that in Definition 5.1.4, the integer N ∈ N depends only on ε but not on x ∈ S. That is, the same N ∈ N is good for all points x ∈ S, or equivalently, we have a uniform N for all x ∈ S. Remark. It is obvious by definition that uniform convergence implies pointwise convergence. Theorem 5.1.5. Let {fn }n∈N be a sequence of functions on S. The following statements are all equivalent: (a) {fn } → f uniformly on S. (b) For any ε > 0, there exists N ∈ N such that 

sup fn (x) − f (x) : x ∈ S < ε

(c) sup{|fn (x) − f (x)| : x ∈ S} → 0

for all n ≥ N .

as n → ∞.

Proof. (a) ⇒ (b): Suppose {fn } → f uniformly on S. For any ε > 0, there exists N ∈ N such that

Hence

fn (x) − f (x) < ε 2

for all n ≥ N and all x ∈ S .

 ε

sup |fn (x) − f (x) : x ∈ S ≤ < ε for all n ≥ N . 2

(b) ⇒ (a): It follows from definition of uniform convergence on S.

Uniform Convergence

355





 (b) ⇔ (c): Observe that sup |fn (x) − f (x) : x ∈ S n∈N is a sequence of real numbers. The equivalence of (b) and (c) now follows immediately from definition of limit of a sequence of real numbers.  Back to the three Questions posed at the beginning of this Section. With the stronger condition that the convergence of the sequence of functions {fn }n∈N is uniform, we have an affirmative answer to Question 1: Theorem 5.1.6. The uniform limit of a sequence of continuous functions is continuous. More precisely, let {fn }n∈N be a sequence of functions on S such that {fn } → f uniformly on S. If fn is continuous at c ∈ S for every n ∈ N, then so is f . In particular, lim lim fn (x) = lim f (x) = f (c) = lim fn (c) = lim lim fn (x) .

x→c n→∞

x→c

n→∞

n→∞ x→c

Proof. For any ε > 0, since {fn } → f uniformly on S, there exists N ∈ N such that fN (x) − f (x) < ε 3

for all x ∈ S .

Since fN is continuous at c, there exists δ > 0 such that fN (x) − fN (c) < ε 3

for all x ∈ S with |x − c| < δ .

Hence for any x ∈ S with |x − c| < δ, f (x) − f (c) ≤ f (x) − fN (x) + fN (x) − fN (c) + fN (c) − f (c) ε ε ε  < + + =ε. 3 3 3

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Corollary 5.1.7. The uniform limit of a sequence of uniformly continuous functions is uniformly continuous. Proof. The proof is analogous to that of Theorem 5.1.6, noting that here c can be an arbitrary point in S. In fact, for any ε > 0, since {fn } → f uniformly on S, there exists N ∈ N such that fN (x) − f (x) < ε 3

for all x ∈ S .

Since fN is uniformly continuous in S, there exists δ > 0 such that fN (x) − fN (y) < ε 3

for all x, y ∈ S with |x − y| < δ .

Hence for any x, y ∈ S with |x − y| < δ, f (x) − f (y) ≤ f (x) − fN (x) + fN (x) − fN (y) + fN (y) − f (y) ε ε ε  < + + =ε. 3 3 3

Under the condition of uniform convergence, we also have an affirmative answer to Question 2: Theorem 5.1.8. Let {fn }n∈N be a sequence of continuous functions on [a, b ] which converges uniformly to f on [a, b ]. Then 

b

lim fn =

a n→∞



a

b

f = lim

n→∞



b

fn .

a

Proof. By Theorem 5.1.6, being the uniform limit of the sequence of continuous functions {fn }n∈N , f is also continuous. Hence all fn ’s and f are integrable over [a, b ]. Now as {fn } → f uniformly on [a, b ], for any ε > 0, there exists N ∈ N such that fn (x) − f (x)
0, there exists N ∈ N s.t. 

d(fn , f ) = sup fn (x) − f (x) < ε for all n ≥ N x∈[a,b ]

 ⇐⇒ d(fn , f ) → 0 as n → ∞.



Recall that in a general metric space, a sequence is convergent implies it is Cauchy but the converse is in general not true unless the metric space under consideration is complete. However, for a sequence of functions on S, uniform convergence is equivalent to “uniform Cauchyness”. Definition 5.1.10. A sequence of functions {fn }n∈N on S is said to be uniformly Cauchy in S if for any ε > 0, there exists N ∈ N such that fn (x) − fm (x) < ε for all n, m ≥ N and all x ∈ S .

Uniform Convergence

359

Note that in Definition 5.1.10, similar to the definition of uniform convergence, the integer N ∈ N depends only on ε but not on x ∈ S. That is, the same N ∈ N is good for all points x ∈ S, or equivalently, we have a uniform N for all x ∈ S. Theorem 5.1.11 (Cauchy’s criterion for uniform convergence of sequences). Let {fn }n∈N be a sequence of real-valued functions on S. Then {fn } → f uniformly on S for some function f if and only if {fn }n∈N is uniformly Cauchy in S. Proof. (⇒) Obvious by triangle inequality. (⇐) Since {fn }n∈N is uniformly Cauchy in S, for any x ∈ S, 

fn (x) n∈N is a Cauchy sequence of real numbers. Since R is complete, this Cauchy sequence is convergent in R. Naturally the limit of this sequence depends on the point x and so we can call it f (x). Do this for all x ∈ S and we obtain a well-defined function f : S → R given by f (x) := limn→∞ fn (x) for every x ∈ S. Now as {fn }n∈N is uniformly Cauchy in S, for any ε > 0, there exists N ∈ N such that fn (x) − fm (x) < ε for all n, m ≥ N and any x ∈ S .

Hence for any n ≥ N and any x ∈ S, we have

fn (x) − f (x) = lim fn (x) − fm (x) ≤ ε . m→∞

Therefore,

 d(fn , f ) = sup |fn (x) − f (x)| : x ∈ S ≤ ε

for all n ≥ N . Hence {fn }n∈N is uniformly convergent on S.



Remark. At first sight, Theorem 5.1.11 is rather astonishing, as for a general metric space, Cauchy sequences may not converge unless the space is complete. Here for Theorem 5.1.11, it seems that no

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completeness is required. But then the careful readers would be able to tell that certain completeness condition has already been implicitly imposed, namely, the completeness of the target space R. In fact, it is easy to see that the completeness of R is critically used in the proof and so the result is not really peculiar at all. Remark. It’s time to do a little recap here. For any sequence of real-valued functions {fn }n∈N on [a, b ] ∈ R, the following conditions are all equivalent: (i) {fn } → f uniformly on [a, b ]. (ii) For any ε > 0, there exists N ∈ N such that |fn (x) − f (x)| < ε for all n ≥ N and all x ∈ [a, b ] . (iii) For any ε > 0, there exists N ∈ N such that 

sup |fn (x) − f (x) : x ∈ [a, b ] < ε for all n ≥ N .



(iv) sup |fn (x) − f (x) : x ∈ [a, b ] → 0 as n → ∞. (v) {d(fn , f )} → 0 as n → ∞, where d is the “sup” metric on C[a, b ] defined by 

d(f, g) := sup |fn (x) − f (x) : x ∈ [a, b ] for f, g ∈ C[a, b ] .

(vi) {fn }n∈N is uniformly Cauchy.

So to prove that {fn } → f uniformly on [a, b ], it suffices to prove any of the preceding equivalent statements. Specifically, if the fn ’s are given with explicit formula and the pointwise limit of {fn }n∈N exists and is known or easily computed, in general, it would be easier to check the uniform convergence by checking conditions (ii), (iii), (iv), or (v). On the other hand, if the explicit formulae of fn ’s are not given or are very complicated, and the pointwise limit of {fn }n∈N is not known, then we may have to resort to condition (vi).

Uniform Convergence

361

To the contrary, in order to prove that {fn } → f uniformly on [a, b ], it suffices to show any of the following equivalent statements which are precisely the negation of the preceding equivalent statements: (i) {fn } → f uniformly on [a, b ]. (ii) There exists ε > 0 such that for any N ∈ N, |fn (x) − f (x)| ≥ ε for some n ≥ N and some x ∈ [a, b ] . (iii) There exists ε > 0 such that for any N ∈ N, 

sup |fn (x) − f (x) : x ∈ [a, b ] ≥ ε

for some n ≥ N .



 (iv) sup |fn (x) − f (x) : x ∈ [a, b ] → 0 as n → ∞. (v) {d(fn , f )} → 0 as n → ∞, where d is the “sup” metric on C[a, b ] defined by 

d(f, g) := sup |fn (x) − f (x) : x ∈ [a, b ] for f, g ∈ C[a, b ] .

(vi) {fn }n∈N is not uniformly Cauchy, that is, there exists ε > 0 such that for any N ∈ N, we can find n, m ≥ N and x ∈ S such that |fn (x) − fm (x)| ≥ ε . Example 5.1.12. We investigate the convergence of the sequence {fn }n∈N on [0, 1] defined by fn (x) := xn , x ∈ [0, 1]. As shown in Example 5.1.1, fn (x) → f (x) =



0 if x ∈ [0, 1) 1 if x = 1

(*)

pointwisely on [0, 1] as n → ∞. Since all fn ’s are continuous on [0, 1] but the pointwise limit f is not, by Theorem 5.1.6, the convergence is not uniform on [0, 1].

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Instead of proving it indirectly as above, it is more illuminating and helps us better understand the behavior of the fn ’s if we prove it by first principle: Observe that for any fixed n ∈ N, no matter how large or small it is, by choosing t ∈ [0, 1) close enough to 1, the value of fn (t) = tn can be made as close to 1 as we wish. Hence we pick ε := 12 , then for every n ∈ N, we can find t ∈ [0, 1) such that |fn (t) − f (t)| = |fn (t)| > 12 = ε. This shows statement (ii) in the preceding Remark and so the convergence {fn } → f is not uniform on [0, 1]. Alternatively, by the same observation, we have

 1 sup fn (x) − f (x) : x ∈ [0, 1] > for any n ∈ N 2 which is statement (iii) in the preceding Remark and so the convergence {fn } → f is only pointwise but not uniform on [0, 1].

The interested readers can easily prove that the convergence is not uniform on [0, 1] by establishing (iv) , (v) , or (vi) in the preceding remark. Although the convergence of {fn } → f on [0, 1] is only pointwise but not uniform, it is interesting to observe that it becomes uniform when restricted to a slightly smaller domain [0, a] for any 0 ≤ a < 1. In fact, for any ε > 0, let N ∈ N be large enough such that aN < ε. Then |fn (x) − f (x)| = xn ≤ an ≤ aN < ε for all n ≥ N and all x ∈ [0, a] . Hence {fn } → f uniformly on [0, a].

Alternatively, for any ε > 0, let N ∈ N be large enough such that 2aN < ε. Then |fn (x) − fm (x)| = |xn − xm | = xm |xn−m − 1| ≤ xN (|x|n−m + 1) ≤ 2aN < ε

for all n > m ≥ N . Thus {fn }n∈N is uniformly Cauchy in [0, a] and so it is uniformly convergent there. Since the pointwise limit of {fn }n∈N is f , we conclude that {fn } → f uniformly on [0, a].

Uniform Convergence

363

The interested readers are encouraged to check whether the convergence is uniform in [0, 1). With Cauchy’s Criterion for uniform convergence at hand, we are now in a position to prove the following long overdue result. Corollary 5.1.13.

For any f, g ∈ C[a, b ], define

 d(f, g) := sup |f (x) − g(x)| : x ∈ [a, b ] .

  Then C[a, b ], d is a complete metric space.

Proof. The fact that d is a well-defined metric on C[a, b ] has been shown before. So it remains to show completeness. Let {fn }n∈N be a Cauchy sequence in C[a, b ]. Then for any ε > 0, there exists N ∈ N such that d(fn , fm ) < ε for all m, n ≥ N . Hence fn (x)−fm (x) ≤ d(fn , fm ) < ε for all m, n ≥ N and all x ∈ [a, b ] .

That is, {fn }n∈N is uniformly Cauchy in [a, b ] and so by Theorem 5.1.11, it is uniformly convergent on [a, b ] to some function f ∈ C[a, b ]. By Theorem 5.1.6, we have f ∈ C[a, b ]. Finally, by Theorem 5.1.9, we have {d(fn , f )} → 0 as n → ∞, that is, {fn } → f in (C[a, b ], d) and so (C[a, b ], d) is complete. 

Finally, we shall address Question 3. Theorems 5.1.6 and 5.1.8 naturally lead us speculate that Question 3 would also have an affirmative answer in case the convergence of the sequence of functions {fn }n∈N is uniform. However, the following Example shows that unfortunately, it is not the case.

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Metric Space Topology: Examples, Exercises and Solutions

Example 5.1.14.

Let us revisit Example 5.1.3 again.

(i) Observe that the sequence of differentiable functions  1 fn (x) := x2 + , x ∈ (−1, 1), n ∈ N n actually converges uniformly to f (x) := |x| in (−1, 1). In fact, for all x ∈ (−1, 1) we have  1 fn (x) − f (x) = x2 + − |x| n  2 1 x + n − |x|2 =  x2 + 1 + |x| n 1

≤ n

1 n

1 = √ →0 n

as n → ∞ .

Hence {fn } → f uniformly in (−1, 1). However, it is clear that f is not differentiable at 0. Hence the uniform limit of a sequence of differentiable functions may not be differentiable. (ii) Observe that the sequence of differentiable functions 1 fn (x) := √ sin nx , n

x∈R,

n∈N

actually converges uniformly to f ≡ 0 on R. In fact, sup{|fn (x) − 0| : x ∈ R}   1 1 = sup √ sin nx : x ∈ R = √ → 0 as n → ∞ . n n

Hence {fn } → f uniformly on R. Clearly, f  ≡ 0 on R but fn (x) =

√ n cos nx ,

x∈R,

does not converge at any point in R. Hence the derivative of the uniform limit of a sequence of differentiable functions may not be

Uniform Convergence

365

equal to the limit of the sequence of derivatives (the latter may not even exist.) Example 5.1.14 shows that in general, the uniform limit of a sequence of differentiable functions may not necessarily be differentiable, and even if it is differentiable, its derivative may not be equal to the limit of the sequence of derivatives. While that is a bit disappointing, we do have an interesting result which says that if a sequence of functions is convergent at one point and if the sequence of derivatives is uniformly convergent, then the sequence of functions is uniformly convergent to a differentiable function, with derivative equals to the uniform limit of the sequence of derivatives. More precisely, we have Theorem 5.1.15. Let {fn }n∈N be a sequence of differentiable functions on a bounded open set S := (a, b) ⊂ R. Suppose (i) {fn } → some function g uniformly on S, and

 (ii) there exists x0 ∈ S such that fn (x0 ) n∈N converges,

then

(a) {fn } → some function f uniformly on S, and



(b) f is differentiable, with f  = g = (limn→∞ fn ) on S. Proof. For every n ∈ N, by our experience in elementary calculus, we consider the difference quotient of the differentiable function fn as follows. For any fixed c ∈ S, we define ⎧ ⎨ fn (x) − fn (c) if x = c gn (x) := x−c ⎩  if x = c . fn (c)

Then by construction, gn is continuous at c. On the other hand, by Mean Value Theorem, for any x = c,     fn (x) − fm (x) − fn (c) − fm (c) gn (x) − gm (x) = x−c   (t) = fn (t) − fm

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Metric Space Topology: Examples, Exercises and Solutions

for some t ∈ S between x and c. Since {fn }n∈N converges uniformly on S, it is uniformly Cauchy in S and so the same is true for {gn }n∈N . (a) In order to show that {fn }n∈N is uniformly convergent, in view of Theorem 5.1.11, it suffices to show that it is uniformly Cauchy. By the construction above, we have fn (x) − fn (c) = gn (x)(x − c) for all x ∈ S and so   fn (x) − fm (x) = fn (c) − fm (c) + (x − c) gn (x) − gm (x)

for all x ∈ S. Note that this is true for any fixed c ∈ S. In

 particular, we take c := x0 . As fn (x0 ) n∈N converges, there exists N1 ∈ N such that fn (x0 ) − fm (x0 ) < ε 2

for all n, m ≥ N1 .

On the other hand, since {gn }n∈N is uniformly Cauchy in S, there exists N2 ∈ N such that gn (x) − gm (x)
0, there exists N ∈ N such that |fn (x) − f (x)| < 2ε and |gn (x) − g(x)| < 2ε for all x ∈ S and all n ≥ N . Hence     fn +gn (x)− f (x)+g(x) ≤ fn (x)−f (x) + gn (x)−g(x) < ε for all x ∈ S and all n ≥ N . That is, {fn + gn } → f + g uniformly

on S .

2. The sequence of functions {fn }n∈N on [0, ∞) defined by fn (x) := 1 , x ∈ [0, ∞), is uniformly convergent on [0, ∞). 1+nx Answer : False.



1 if x = 0 0 if x = 0 pointwisely on [0, ∞). Since f is not continuous on [0, ∞), the convergence is not uniform on [0, ∞). Justification . It is evident that {fn (x)} → f (x) :=

3. The sequence of functions {fn }n∈N on [0, ∞) defined by fn (x) := 1 , x ∈ [0, ∞), is uniformly convergent on (0, ∞). 1+nx Answer : False. Justification . It is evident that {fn (x)} → f (x) :≡ 0 pointwisely on (0, ∞). Although f is continuous, the convergence is still not uniform. In fact, take ε = 12 . Then for every n ∈ N, the point xn := n1 would give

|fn (x) − f (x)| =

1 ≥ε. 2

Uniform Convergence

369

4. The sequence of functions {fn }n∈N on [0, 1] defined by fn (x) := n i i=0 x (1 − x), x ∈ [0, 1], is uniformly convergent on [0, 1]. Answer : False.

Justification . It is evident that {fn (x)} → f (x) :=



1 if x ∈ [0, 1) 0 if x = 1

pointwisely on [0, 1]. Since f is not continuous on [0, 1], the convergence is not uniform on [0, 1].

5. The sequence of functions {fn }n∈N on [0, ∞) defined by fn (x) := x 1+nx , x ∈ [0, ∞), is uniformly convergent on [0, ∞). Answer : True. Proof . It is clear that fn (0) = 0 for all n ∈ N and so {fn (0)} → 0 as n → ∞. On the other hand, for x > 0, we have 0 < fn (x) ≤ x 1 nx = n → 0 as n → ∞. Hence {fn } → f :≡ 0 pointwisely on [0, ∞). It is not hard to see that the convergence is actually uniform. 1 In fact, for any ε > 0, let N ∈ N be large enough such that N < ε. 1 1 Then |fn (x) − 0| ≤ n ≤ N < ε for all n ≥ N and all x ∈ (0, ∞). Thus {fn } → 0 uniformly on (0, ∞). Combining, we have {fn } → 0 uniformly on [0, ∞). 6. The sequence of functions {fn }n∈N on [0, 1] defined by fn (x) := x(1 − x)n , x ∈ [0, 1], is uniformly convergent on [0, 1]. Answer : True. Proof . It is clear that fn (0) = 0, fn (1) = 0 for all n ∈ N and so {fn (0)} → 0 and {fn (1)} → 0 as n → ∞. On the other hand, for any 0 < x < 1, we have 0 < fn (x) ≤ (1 − x)n → 0 as n → ∞. Hence {fn } → f :≡ 0 pointwisely on [0, 1]. To show that the convergence is actually uniform, observe first that by elementary calculus, the maximum of the function fn (x) on [0, 1] is attained at

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Metric Space Topology: Examples, Exercises and Solutions

the point

1 , n+1

and so for any x ∈ [0, 1], we have

 n 1 1 nn 1− |fn (x)−f (x)| = x(1−x) ≤ = . n+1 n+1 (n + 1)n+1 n

This suggests that for any ε > 0, we take N ∈ N large enough such

that

Then

NN (N +1)N +1

< ε. This is possible since limN →∞

|fn (x) − f (x)| ≤

NN (N +1)N +1

= 0.

nn NN ≤ 0

pointwisely on [0, ∞). Since f is not continuous on [0, ∞), the convergence is not uniform.

8. The sequence of functions {fn }n∈N on [0, ∞) defined by fn (x) := tan−1 (nx), x ≥ 0, is uniformly convergent on [a, ∞) for any a > 0. Answer : True. Proof . It is clear that {fn } → f :≡

π 2

pointwisely on [a, ∞). To

show that the convergence is actually uniform, we first observe that

π |fn (x) − f (x)| = tan−1 (nx) − = cot−1 (nx) . 2

Hence we need to show that for n ∈ N large enough, cot−1 (nx) will be smaller than a prescribed positive number ε > 0 for all x ≥ a.

Uniform Convergence

371

Now by the fact that cot−1 is monotonically decreasing on [0, ∞), we have

cot−1 (nx) < ε ⇐⇒

nx

⇐⇒

n

for all x ≥ a

> cot ε for all x ≥ a cot ε for all x ≥ a > x

and so this suggests that we pick N ∈ N such that N >

cot ε a .

Then

for any n ≥ N and any x ≥ a, we have

n≥N >

cot ε cot ε ≥ a x

and so

|fn (x) − f (x)| = cot−1 (nx) < ε . Thus the convergence is uniform.

9. The sequence of functions {fn }n∈N on [0, ∞) defined by fn (x) := tan−1 (nx), x ≥ 0, is uniformly convergent on (0, ∞). Answer : False. Justification . Using the analysis of Exercise 5.1, Part A, Problem #8, we see that the convergence is uniform on (0, ∞) if and only if for any given ε > 0, there exists N ∈ N large enough such that |fn (x)−f (x)| = cot−1 (nx) < ε for all x ∈ (0, ∞) and all n ≥ N , or

n>

cot ε x

This requires N >

for all x ∈ (0, ∞) and all n ≥ N .

cot ε for all x ∈ (0, ∞), which is clearly x

impossible. Thus the convergence is not uniform.

10. Let {fn }n∈N be a sequence of functions uniformly convergent to f on R. If all fn ’s are discontinuous at 0, then f is discontinuous at 0.

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Metric Space Topology: Examples, Exercises and Solutions

Answer : False. Example : For any n ∈ N, let fn : R → R be defined by fn (x) :=



1 n

0

if x ∈ Q

otherwise .

Then none of the fn ’s is continuous at 0 but {fn } → f :≡ 0 uniformly on R and f is continuous everywhere.

11. Let {fn }n∈N be a sequence of functions on S ⊂ R. If {fn }n∈N is uniformly convergent on every subset of S which is closed in R, then {fn } is uniformly convergent on S.

Answer : False. Example : Consider S := (0, 1) ⊂ R and for every n ∈ N, fn (x) := xn , x ∈ S . By Example 5.1.12, {fn }n∈N is uniformly convergent to f :≡ 0 on every subset [a, b ] ⊂ S . Now let C ⊂ S be a subset of S which is closed in R. It is evident that C is contained in some [a, b ] ⊂ S . Hence {fn }n∈N is uniformly convergent to f ≡ 0 on C . But then by Example 5.1.12 again, {fn }n∈N is not uniformly convergent to f ≡ 0 on S .

12. The pointwise limit of a sequence of bounded functions is bounded. Answer : False. Example : Consider the sequence of functions {fn }n∈N on (0, 1) given n by fn (x) := nx+1 , x ∈ (0, 1). Clearly, {fn (x)} → f (x) := x1 pointwisely on (0, 1). Furthermore, for each n ∈ N, n ≤ fn ≤ n on (0, 1) , n+1 hence each fn is bounded on (0, 1) but the pointwise limit f (x) = is not.

1 x

Uniform Convergence

373

13. The uniform limit of a sequence of bounded functions is bounded. Answer : True. Proof . Let {fn }n∈N be a sequence of functions on S ⊂ R such that each fn is bounded on S and {fn } → f uniformly on S . So there exists N ∈ N such that |fn − f | < 1 on S for all n ≥ N . Since fN is bounded on S , there exists MN > 0 such that |fN | ≤ MN on S . Hence

|f | ≤ |fN | + |fN − f | < MN + 1 on S . That is, the uniform limit f is bounded on S .

14. The uniform limit of a sequence of contractions {fn }n∈N is also a contraction. Answer : False. Example : For any n ∈ N, let fn : [0, 1] → [0, 1] be defined by fn (x) := (1 − n1 ) x, x ∈ [0, 1]. Then for every n ∈ N, fn is a contraction on [0, 1] with contraction constant 0 < 1 − n1 < 1. Furthermore, it is not hard to see that {fn }n∈N converges uniformly to the function f (x) := x on [0, 1]. But f is not a contraction. Details are left to the readers.

15. If a sequence of functions {fn }n∈N converges pointwisely to a continuous function f , then the convergence is uniform. Answer : False. Example : For any n ∈ N, let fn : R → R be defined by fn (x) := 1 . Then {fn } → f :≡ 0 pointwisely on R. However, 3 n (x − n1 )2 + 1 for any n ∈ N, we have       fn 1 − f 1 = fn 1 − 0 = 1 , n n n thus the convergence is not uniform.

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Metric Space Topology: Examples, Exercises and Solutions

16. If a sequence of continuous functions {fn }n∈N converges uniformly to f and a sequence of uniformly continuous functions {gn }n∈N converges uniformly to g, then {fn ◦ gn }n∈N converges uniformly to f ◦ g.

Answer : False. Example : For any n ∈ N, let fn , gn : R → R be defined by fn (x) := x2 + n1 and gn (x) := x + n1 , respectively. Then each fn is continuous on R and each gn is uniformly continuous on R. Furthermore, {fn }n∈N converges uniformly in R to the function f (x) := x2 and {gn }n∈N converges uniformly in R to the function g(x) := x. Observe also that fn ◦ gn (x) = (x + n1 )2 + n1 and f ◦ g(x) = x2 , x ∈ R. Now for all n ∈ N, we have   2  1 1 − n2 + fn ◦ gn (n) − f ◦ g(n) = n + n n 1 1 + ≥2, 2 n n does not converge uniformly to f ◦ g . =2+

thus {fn ◦ gn }n∈N

17. If {fn }n∈N converges uniformly to f in R, then {g ◦ fn }n∈N converges uniformly to g ◦ f in R for any continuous function g on R. Answer : False. Example : For any n ∈ N, let fn : R → R be defined by fn (x) := x + n1 , x ∈ R. It is easy to check that {fn }n∈N converges uniformly to the function f (x) := x on R. Let g : R → R be defined by g(x) := x2 , x ∈ R. Then g is continuous on R. Since 2  1 1 2x + 2 g ◦ fn (x) = x + = x2 + n n n for all x ∈ R and all n ∈ N, we have {g ◦ fn (x)} → g(x) = x2 pointwisely on R. However, as for every n ∈ N we have

  1 1 2 2n 2 + 2 − n = 2+ 2 ≥ 2 , g ◦ fn (n) − g ◦ f (n) = n + n n n the convergence is not uniform.

Uniform Convergence

375

18. If {fn }n∈N converges uniformly to f in R, then for any function g on R, {fn ◦ g}n∈N converges uniformly to f ◦ g in R.

Answer : True. Proof . By the uniform convergence {fn } → f in R, for any ε > 0, there exists N ∈ N such that fn (y) − f (y) < ε for any n ≥ N and any y ∈ R .

In particular,

    fn g(x) − f g(x) < ε for any n ≥ N and any x ∈ R .

Thus {fn ◦ g} → f ◦ g uniformly in R.

19. If {fn }n∈N is a sequence of continuous functions on [0, 1] which is convergent uniformly to f on [0, 1], then for any continuous function g on R, {g ◦ fn }n∈N converges uniformly to g ◦ f on [0, 1]. Answer : True. Proof . Since continuous functions on compact sets are bounded, each fn is bounded. Since {fn } → f uniformly on [0, 1], there exists N ∈ N such that fn (x) − fm (x) < 1 for any n, m ≥ N and any x ∈ [0, 1] .

In particular, if |fN | ≤ M on [0, 1], then |fn | ≤ M + 1 on [0, 1] for

all n ≥ N and so without loss of generality, we simply assume that

|fn | ≤ M + 1 for all n ∈ N. Since g is continuous, it is uniformly continuous on the compact set S := [−M − 1, M + 1], which is a superset of the range of all fn ’s. So for any ε > 0, there exists δ > 0 such that

g(y1 ) − g(y2 ) < ε for any y1 , y2 ∈ S with |y1 − y2 | < δ .

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Metric Space Topology: Examples, Exercises and Solutions

Since {fn } → f uniformly on [0, 1], there exists K ∈ N such that

fn (x) − f (x) < δ for all x ∈ [0, 1] and all n ≥ K .

Therefore,

    (x) − g f (x) g f < ε for all x ∈ [0, 1] and all n ≥ K . n Thus {g ◦ fn } → g ◦ f uniformly on [0, 1].

20. If a sequence of uniformly continuous functions {fn }n∈N converges uniformly to f and a sequence of functions {gn }n∈N converges uniformly to g, then {fn ◦ gn } converges uniformly to f ◦ g. Answer : True. Proof . Let {gn }n∈N be a sequence of functions on S ⊂ R such that {gn } → g uniformly on S , and {fn }n∈N be a sequence of uniformly continuous functions on T ⊂ R such that {fn } → f uniformly on  T . In order to make sense, we assume that T ⊃ n∈N gn (S). By Corollary 5.1.7, being the uniform limit of a sequence of uniformly

continuous functions, f is uniformly continuous on T . Hence for any

ε > 0, there exists δ > 0 such that f (y1 ) − f (y2 ) < ε for any y1 , y2 ∈ T with y1 − y2 < δ . 2 Since {gn } → g uniformly on S , there exists N1 ∈ N such that

Hence

gn (x) − g(x) < δ for any n ≥ N1 and any x ∈ S .

    ε f gn (x) − f g(x) < 2

for any n ≥ N1 and any x ∈ S .

On the other hand, as {fn } → f uniformly on S , there exists N2 ∈ N such that

fn (y) − f (y) < ε 2

for any n ≥ N2 and all y ∈ T .

Uniform Convergence

377

In particular,

    ε fn gn (x) − f gn (x) < 2

for any n ≥ N2 and all x ∈ S .

Therefore,

    fn gn (x) − f g(x)         ≤ fn gn (x) − f gn (x) + f gn (x) − f g(x) ε ε < + =ε 2 2

for any n ≥ max{N1 , N2 } and any x ∈ S , that is, {fn ◦gn } → f ◦g uniformly on S .

Part B: Problems 1.

(a) Find the pointwise limit of the sequence of the functions {e−nx }n∈N on [0, 1]. (b) Find the limit of the sequence {e−nx }n∈N in (C[0, 1], d), where 1/p  1 |f (x) − g(x)|p dx d(f, g) := 0

with p ≥ 1. (c) What is your observation? Solution . (a) It is elementary to see that −nx

{e

} → f (x) :=



0 if x ∈ (0, 1] 1 if x = 0 .

(b) It is easy to see that −nx

d(e

, 0) =



0

1

−pnx

e

dx

1/p

=



1 − e−pn pn

1/p

→0

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Metric Space Topology: Examples, Exercises and Solutions

as n → ∞, thus {e−nx } → 0 in (C[0, 1], d).

(c) Convergence in different contexts can be very different.

2. Consider the sequence of functions {fn }n∈N on [0, 1] defined by fn (x) := sin[xn (1 − xn )], x ∈ [0, 1]. (a) Does {fn (x)}n∈N converge pointwisely on [0, 1]? (b) Does {fn (x)}n∈N converge uniformly on [0, 1]? 1 (c) Does limn→∞ 0 fn (x) dx exist?

Solution . (a) Answer : Yes. Proof . Clearly fn (0) = fn (1) = 0 for all n ∈ N. For any x ∈ (0, 1), since sin y ≤ y for all y ∈ (0, 1), we have |fn (x)| ≤ xn (1 − xn ) ≤ xn → 0 as n → ∞. Therefore, {fn } → 0 pointwisely on [0, 1]. (b) Answer : No.   Justification . Since fn

n

1 2

= sin 14 , we have

 1 sup |fn (x) − 0| : x ∈ [0, 1] ≥ 4

for all n ∈ N

and so the convergence cannot be uniform. (c) Answer : Yes, and the limit is 0. Hence in particular,

lim

n→∞



1

fn (x) dx = 0 = 0



1

lim fn (x) dx .

0 n→∞

Justification . Similar to (a), we have 

0

1

 fn (x) dx ≤

1

xn (1−xn ) dx = 0

n →0 (n + 1)(2n + 1)

as n → ∞.

3. Consider the sequence of functions {fn }n∈N on R defined by 1 2 2 fn (x) := e−n x , x ∈ R. n

Uniform Convergence

379

(a) Show that {fn (x)}n∈N converges pointwisely on R to some function f . Find f . (b) Determine whether {fn }n∈N converges uniformly on R. (c) Show that limn→∞ fn = f  on R (that is, {fn } → f  pointwisely on R). (d) Show that the convergence {fn } → f  cannot be uniform on any non-degenerate interval containing 0. Solution . (a) Since |fn (x)| ≤ on R.

1 n

for all x ∈ R, {fn } → f :≡ 0 pointwisely

(b) For the same reason as that in (a), we have

 1 sup |fn (x) − f (x)| : x ∈ R ≤ → 0 as n → ∞ . n

Hence {fn } → f ≡ 0 uniformly on R.

(c) We have fn (x) = −2nxe−n

2 2

x

for all x ∈ R, n ∈ R. Clearly,

= 0 for all n ∈ N. On the other hand, for any x = 0, 2 2 by elementary analysis, we have {ne−n x } → 0 as n → ∞. Hence fn → 0 ≡ f  pointwisely on R. (d) Let I ⊂ R be a non-degenerate interval with 0 ∈ I . For any N ∈ N, there exists n ≥ N such that n1 ∈ I or − n1 ∈ I . 1 Without loss of generality, assume that n ∈ I . Then fn (0)

  1  2  = , sup |fn (x) − 0| : x ∈ I ≥ fn n e

thus the convergence {fn } → f  cannot be uniform on I .

4. For each n, define gn : R → R by gn (x) :=



0 b+

1 n

if x = 0 or x ∈ /Q, a if x = b , a, b ∈ Z, a = 0, b > 0, (a, b) = 1 .

Show that {gn }n∈N is uniformly convergent on R.

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Metric Space Topology: Examples, Exercises and Solutions

Proof . Note that for any n, m ∈ N,  0 if x = 0 or x ∈ /Q, gn (x)−gm (x) = 1 1 a if x = b , a, b ∈ Z, b > 0, (a, b) = 1 . n − m Hence

1 1 |gn (x) − gm (x)| ≤ − n m

for all x ∈ R .

As { n1 }n∈N is a Cauchy sequence, {gn }n∈N is uniformly Cauchy. By

Cauchy’s criterion, {gn }n∈N is uniformly convergent on R.

5. Let {fn }n∈N be a sequence of uniformly continuous functions on S ⊂ R, and {xn }n∈N a sequence in S with {xn } → x ∈ S as n → ∞. (a) If {fn }n∈N converges uniformly to some function f on S, show that {fn (xn )} → f (x) as n → ∞. (b) If the convergence {fn } → f is only pointwise, would the assertion in (a) still be true? Proof . (a) As {fn } → f uniformly on S , for any ε > 0, there exists

N1 ∈ N such that

|fn (x) − f (x)|
0 such that ε |f (p) − f (q)| < for any p, q ∈ S with |p − q| < δ . 2 Since {xn } → x as n → ∞, there exists N2 ∈ N such that

|xn − x| < δ for all n ≥ N2 . Combining, we have

|fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| < ε for any n ≥ max{N1 , N2 }. The assertion follows.

Uniform Convergence

381

(b) In case the convergence {fn } → f is only pointwise, the assertion in (a) fails to hold.

Example : Consider the sequence of functions {fn }n∈N on [0, 1] defined by fn (x) := xn , x ∈ [0, 1]. We have seen that  0 if x ∈ [0, 1) {fn (x)} → f (x) := if x = 1 1 pointwisely but not uniformly on [0, 1]. Let xn := 1 −

1 n

∈

[0, 1] for all n ∈ N. We have {xn } → 1 as n → ∞. But then n  1 1 fn (xn ) = 1 − → = 1 = f (1) as n → ∞ . n e

6. Show that every uniformly convergent sequence of bounded functions is uniformly bounded. More precisely, let {fn }n∈N be a sequence of functions on S ⊂ R which is uniformly convergent. If each fn is bounded on S, i.e., supx∈S |fn (x)| < ∞ for each n ∈ N, then {fn }n∈N is uniformly bounded, that is, sup {|fn (x)| : x ∈ S, n ∈ N} < ∞. Proof . As {fn }n∈N is uniformly convergent, it must be uniformly Cauchy. So there exists N ∈ N such that |fn (x) − fN (x)| < 1 for all n ≥ N and all x ∈ S . It follows that

|fn (x)| ≤ |fN (x)| + 1 for all n ≥ N and all x ∈ S . For any n ∈ N, write Mn := supx∈S |fn (x)|. If we set

M := max {M1 , M2 , . . . , MN −1 , 1 + MN } > 0 , it is easy to see that

sup |fn (x)| ≤ M

x∈S

for all n ∈ N .

That is, {fn }n∈N is uniformly bounded.

382

Metric Space Topology: Examples, Exercises and Solutions

7. (Dini’s Theorem) Let {fn }n∈N be a sequence of functions on S ⊂ R. Suppose (i) S is compact,

(ii) fn is continuous for every n ∈ N, and

(iii) for each x ∈ S, {fn (x)}n∈N ↓ 0 as n → ∞,

show that {fn } → 0 uniformly on S.

Proof . Let ε > 0 be given. For any n ∈ N, since fn is continuous, the   subset On := fn−1 (−∞, ε) = {x ∈ S : fn (x) < ε} ⊂ S is open. Furthermore, for any x ∈ O(n), since {fn (x)}n∈N is decreasing,

we have

fn+1 (x) ≤ fn (x) < ε and so x ∈ On+1 . Hence {On }n∈N is increasing. On the other hand,

for each x ∈ S , since {fn (x)} ↓ 0 as n → ∞, there exists nx ∈ N

such that fnx (x) < ε and so x ∈ Onx . Thus {On : n ∈ N} is an open cover of S . By the compactness of S , there exists N ∈ N such that S =

N

n=1 On .

Since {On }n∈N is increasing, we have S = ON .

In particular, for any n ≥ N and any x ∈ S = ON ,

0 ≤ fn (x) ≤ fN (x) < ε . This shows the {fn } → 0 uniformly on S .

8. In Exercise 5.1, Part A, Problem #7, is condition (i), i.e., the compactness of S, necessary? Answer : Yes, compactness is necessary. Example : For any n ∈ N, consider fn : R → R given by fn (x) :=

x . n

Clearly, fn is continuous for every n ∈ N and fn (x) ↓ 0 as n → ∞ for every x ∈ R, but R is not compact. Note that since for every

Uniform Convergence

383

n ∈ N,

|x| 

 sup fn (x) − f (x) : x ∈ R = sup :x∈R =∞, n

we have



 sup fn (x) − f (x) : x ∈ R → ∞ as n → ∞

and so the convergence {fn } → 0 is not uniform on R.

9. In Exercise 5.1, Part A, Problem #7, is condition (iii), i.e., the monotonicity of {fn }n∈N , necessary? Answer : Yes, monotonicity is necessary. Example : Consider the sequence of functions {fn }n∈N on [0, 1] given by

⎧ 2 ⎪ ⎨n x fn (x) := −n2 x + 2n ⎪ ⎩ 0

1 n ≤ n2

0≤x≤ 1 n 2 n

≤x

≤x≤1.

Clearly fn is continuous on X for each n, [0, 1] is compact, and fn →

f ≡ 0 pointwisely on X . However, the sequence {fn (x)}n∈N is not monotone. Observe that



 sup fn (x) − f (x) : x ∈ X = n → ∞ as n → ∞ ,

hence the convergence is not uniform.

10. In Exercise 5.1, Part A, Problem #7, is condition (ii), i.e., fn is continuous for all n necessary? Answer : Yes, save for finitely many n’s, the continuity of fn ’s is necessary.

Example : Consider the sequence of functions {fn }n∈N on [0, 1] given by

⎧ ⎪ ⎨ 0 if x = 0 fn (x) := 1 if 0 < x < 1/n ⎪ ⎩ 0 if n1 ≤ x ≤ 1 .

384

Metric Space Topology: Examples, Exercises and Solutions

Clearly, [0, 1] is compact and {fn } ↓ 0 pointwisely on [0, 1], but none





1 of the fn ’s is continuous on [0, 1]. As fn ( 2n ) − 0 = 1 for all n ∈ N,

the convergence is not uniform.

11. Let {fn }n∈N be a sequence of functions defined on S ⊂ R. Suppose (i) S is compact, (ii) fn is continuous for every n ∈ N, and (iii) there is a continuous function f : S → R such that {fn (x)}n∈N ↓ f (x) as n → ∞ for every x ∈ S, show that {fn } → f uniformly on S. What if the pointwise limit f is not continuous on S? Solution . For any n ∈ N, define gn := fn − f . As {fn } ↓ f in S , {gn } ↓ 0 in S . As all fn ’s and f are continuous, all gn ’s are continuous. Hence all conditions in Exercise 5.1, Part A, Problem #7 are satisfied and so {gn } → 0 uniformly on S . Hence {fn } → f

uniformly on S .

However, in case the pointwise limit f is not continuous, we may not have uniform convergence.

Example : Consider the sequence of functions {fn }n∈N on [0, 1] given 1 by fn (x) := 1+nx , x ∈ [0, 1]. Clearly, [0, 1] is compact, fn is con 1 if x = 0 . tinuous for all n ∈ N, and {fn (x)} ↓ f (x) := 0 if x = 0 Note that f is not continuous. Observe that as shown in Exercise 5.1, Part A, Problem #2, the convergence {fn } → f is not uniform. 12. Let {fn }n∈N be a sequence of continuous real-valued functions defined and uniformly bounded (cf. Exercise 5.1, Part B, Problem #6) on S ⊂ R. Suppose for each x ∈ S, the sequence of real numbers {fn (x)}n∈N is monotonically decreasing. Determine whether {fn } is uniformly convergent on S. Answer : No.

Uniform Convergence

385

Example: Let S := [0, 1] and for all n ∈ N,

fn (x) :=



−nx

−1

0≤x≤ 1 n

1 n

0 such that I ⊂ [−M, M ], that is, |x| ≤ M for all x ∈ I . For any ε > 0, let N ∈ N be large enough such that N > M ε . Then |fn (x) − f (x)| =

M M |x| ≤ ≤ 0, since { n }n∈N is Cauchy, there is N ∈ N such

1 that | n −

1 m|


0, there exist N1 , N2 ∈ N such that

ε 2M ε |gn (x) − g(x)| < 2M

|fn (x) − f (x)|
0 such that |f (x1 ) − f (x2 )| < ε for any x1 , x2 ∈ R with |x1 − x2 | < δ . Let {an }n∈R be any sequence with {an } → 0 as n → ∞. Then

there exists N ∈ N such that

|an | < δ for all n ≥ N . Hence for any x ∈ R and any n ≥ N , we have |(x − an ) − x| =

|an | < δ and so

|fan (x) − f (x)| = |f (x − an ) − f (x)| < ε . Thus {fan } → f uniformly on R.

(⇐): It is more convenient to use contrapositive arguments. So suppose {fan } → f uniformly whenever {an } → 0 as n → ∞, but f

is not uniformly continuous on R. Then there exists ε > 0 such that for each n ∈ N, we can find xn , yn ∈ R with |xn − yn | < 1/n but

|f (xn ) − f (yn )| ≥ ε. For any n ∈ N, write an := xn − yn . By construction, {an } → 0 as n → ∞. So by assumption, {fan } → f uniformly on R. Hence there exists N ∈ N such that |fan (x) − f (x)| < ε for all n ≥ N and all x ∈ R . In particular, we have

|f (yN ) − f (xN )| = |f (xN − aN ) − f (xN )|

= |faN (xN ) − f (xN )| < ε ,

which is a contradiction. Therefore, f is uniformly continuous.

Uniform Convergence

389

5.2 Series of Functions From our experience with elementary calculus, associated to an infinite sequence of real numbers there is a corresponding formal infinite series of real numbers with the same convergence as the sequence. Conversely, associated to any formal infinite series of real numbers, there is a corresponding infinite sequence of real numbers with the same convergence as the series. To be more precise, starting with an infinite sequence of real numbers {an }n∈N , we define 

b1 bk+1

:= a1 := ak+1 − ak ,

k∈N,

∞ and form the formal infinite series k=1 bk . Observe that for every n ∈ N, we have n  bk = an . k=1

Furthermore, the infinite sequence {an }n∈N is convergent if and only ∞ if the formal infinite series k=1 bk is convergent, and in this case we write n ∞   bk = bk . lim an = lim n→∞

n→∞

k=1

k=1

Conversely, starting with a formal infinite series for any n ∈ N, the n-th partial sum sn :=

n 

∞

k=1 ak ,

we define,

ak ,

k=1

and form the infinite sequence of partial sums {sn }n∈N . Then the ∞ formal infinite series k=1 ak is convergent if and only if the infinite sequence of partial sums {sn }n∈N is convergent, and in this case lim sn = lim

n→∞

n→∞

n 

k=1

ak =

∞ 

k=1

ak .

390

Metric Space Topology: Examples, Exercises and Solutions

So there is a natural 1 − 1 correspondence between infinite sequences of real numbers and formal infinite series of real numbers. In certain situations, it is easier to deal with sequences, while for other situations, since we have the mechanism of addition, it could be easier to deal with infinite series. Analogously, there is a natural 1 − 1 correspondence between infinite sequences of functions and formal infinite series of functions. Namely, starting with an infinite sequence of functions {fn }n∈N on S ⊂ R, we define 

g1 gk+1

:= f1 := fk+1 − fk ,

k∈N,

and form the formal infinite series of functions every n ∈ N, we have n  gk = fn .

∞

k=1 gk .

Then for

k=1

Furthermore, the infinite sequence of functions {fn }n∈N is convergent pointwisely/uniformly on S if and only if the formal infinite series of ∞ functions k=1 gk is convergent pointwisely/uniformly on S, and in this case n ∞   gk = gk . lim fn = lim n→∞

n→∞

k=1

k=1

Conversely, starting with a formal infinite series of functions on S, we define, for any n ∈ N, the n-th partial sum sn :=

n 

∞

k=1 fk

fk ,

k=1

and form the infinite sequence of partial sums {sn }n∈N . Then the ∞ formal infinite series k=1 fk is convergent pointwisely/uniformly

Uniform Convergence

391

on S if and only if the infinite sequence of partial sums {sn }n∈N is convergent pointwisely/uniformly on S, and in this case lim sn = lim

n→∞

n→∞

n 

fk =

k=1

∞ 

fk .

k=1

Again, in certain situations, it would be easier to deal with series of functions instead of sequences of functions. Hence it will be handy to display the series version of the results in the preceding section. ∞ Definition 5.2.1. Let k=1 fk be a formal series of functions on S. For any n ∈ N, The n-th partial sum of the series is defined as sn (x) :=

n 

fk (x) ,

k=1

x∈S .

∞ The series k=1 fk is said to converge uniformly on S if the sequence of partial sums {sn }n∈N is uniformly convergent on S, that is, there is a function f : S → R such that {sn } → f uniformly on S. In this case f is said to be the uniform limit of the series and we write ∞ 

fk (x) = f (x)

uniformly on S .

k=1

Theorem 5.2.2 (Cauchy’s criterion for uniform convergence ∞ of infinite series). n=1 fn converges uniformly on S if and only if for every ε > 0, there exists N ∈ N such that n+p  fk (x) < ε k=n+1

for all n ≥ N , p ∈ N, and all x ∈ S .

(*)

 Proof. Consider the sequence of partial sums sn := nk=1 fk . In view ∞ of Theorem 5.1.11, n=1 fn is uniformly convergent on S if and only

392

Metric Space Topology: Examples, Exercises and Solutions

if {sn }n∈N is uniformly convergent on S if and only if {sn }n∈N is uniformly Cauchy on S if and only if for any ε > 0, there exists N ∈ N such that n+p  fk (x) = sn+p (x) − sn (x) < ε k=n+1



for all n ≥ N , p ∈ N and all x ∈ S.

Corollary 5.2.3. If fn : S → R is continuous at c ∈ S for each n ∈  N and ∞ n=1 fn (x) = f (x) uniformly on S, then f is also continuous at c. n Proof. Consider the sequence of partial sums sn := k=1 fk . Then by assumption, {sn } → f uniformly on S. Since each fk is continuous on S, the same is true for each sn . By Theorem 5.1.6, being the  uniform limit of {sn }n∈N , f is continuous on S. A very useful tool to check uniform convergence of series of functions is the following ∞ Theorem 5.2.4 (Weierstrass M-Test). Let n=1 fn be a formal series of functions on S satisfying (i) for each n ∈ N, fn ≤ an on S for some an ∈ R, and ∞ (ii) n=1 an converges, ∞ then n=1 fn converges uniformly on S.  Proof. Since ∞ n=1 an is convergent, for any ε > 0, there exists N ∈ ∞ N such that k=N +1 ak < ε. Thus for all n ≥ N , p ∈ N, and all x ∈ S, n+p n+p n+p ∞     fk (x) ≤ ak ≤ ak < ε . fk (x) ≤ k=n+1

k=n+1

k=n+1

The result now follows from Theorem 5.2.2.

k=N +1



Uniform Convergence

393

Example 5.2.5. We have seen in Example 5.1.12 that for every 0 ≤ a < 1, the sequence of functions {fn }n∈N on [0, a] defined by fn (x) := xn is uniformly convergent on [0, a]. It would be constructive to see how this can be shown by first converting the sequence of functions to its corresponding formal series of functions and then proving the uniform convergence of the latter by using Weierstrass M-Test, as follows. Following the prescription at the beginning of this Section, we define 

g1

:= f1

gn

:= fn − fn−1 ,

n≥2.

Then the sequence of functions {fn }n∈N converges uniformly on [0, a]  if and only if the series of functions ∞ n=1 gn converges uniformly on [0, a], and in this case they have the same limit. So the problem boils ∞ down to showing the uniform convergence of n=1 gn on [0, a]. Now for any x ∈ [0, a], we have |g1 (x)| = |f1 (x)| = x ≤ a ,

|gn (x)| ≤ |fn (x)| + |fn−1 (x)| = xn + xn−1 ≤ an + an−1 ,

n≥2.

Since 0 ≤ a < 1, we have a+

∞   n  a2 + a