Topology of Metric Spaces , Second Edition [Second Edition] 8184870582, 9788184870589

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"From the very beginning the author emphasizes geometric and topological reasoning. Many

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ideas and concepts are illustrated by informative pictures. .... this book is interesting

because it encourages problem solving and topological reasoning. We enjoyed reading this

stimulating book" - An excerpt from a Review in Zentralblatt (Zb1 1106.54001) om

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Topology of Metric Spaces Second Edition

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Topology of

Metric Spaces Second Edition

S. Kumaresan

FOR SALE ONLY IN India, Pakistan, Banglades h, Nepal, Bhutan and Sri Lan ka MARKETING RIGHTS VIO LATION WILL INVITE LEGAL ACTION

we) Narosa Publishing House

New Delhi

Chennai

Mumbai _

Kolkata

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Topology of Metric Spaces, Second Edition 210 pgs.

| 74 figs.

S. Kumaresan

Department of Mathematics and Statistics

University of Hyderabad Hyderabad, India

"Copyright © 2005, 2011, Narosa Publishing House Pvt. Ltd.

First Edition 2005

Second Edition 2011

|

Fourteenth Reprint 2022

NAROSA

PUBLISHING

HOUSE

PVT

LTp.

22 Delhi Medical Association Road, Daryaganj, New Delhi 110 002 35-36 Greams Road, Thousand Lights, Chennai 600 006 306 Shiv Centre, Sector 17, Vashi, Navi Mumbai 400 703

2F-2G Shivam Chambers, 53 Syed Amir Ali Avenue, Kolkata 700 019

www.narosa.com

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electron ic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher. . All export rights for this book vest exclusively with Narosa Publishing House Pvt. Ltd. Unauthorised export is a violation of terms of sale and is subject to legal action.

Printed from the camera-ready copy provided by the Author.

ISBN 978-81-8487-058-9 Published by N.K. Mehra for Narosa Publishing House Pvt. Ltd.,

22 Delhi Medical Association Road, Daryaganj, New Delhi 110 002 Printed in India

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Dedicated to All my Teachers |

especially to N. Ganesan

R. Balakrishnan

V. Ganapathy Iyer M.S. Raghunathan M.S. Narasimhan K. Okamoto R. Parthasarathy

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Preface The aim is to give a very streamlined development of a course in met-

ric space topology emphasizing only the most useful concepts, concrete spaces and geometric ideas. To encourage the geometric thinking, I have chosen large number of examples which allow us to draw pictures and develop our intuition and draw conclusions, generate ideas for proofs. To this end, the book boasts of a lot of pictures. A secondary aim is to treat this as a preparatory ground for 4 general topology course and arm the reader with a repertory of examples. To achieve this, I have adopted the following strategy. Whenever a definition makes sense in an arbitrary topological space or whenever a result is true in an arbitrary topological space, I use the convention ... (metric) space .... This is to tell the reader the sentence makes mathematical sense in any topological space and if the reader wishes, he may assume that the space is

a metric space. See, for example, Def. 4.1.3, Ex. 4.4.12, Def. 5.1.1 and Theorem 5.1.32. On few occasions, I have also shown that if we want

to extend the result from metric spaces to topological spaces, what kind _ of extra conditions need to be imposed on the topological space. These

instances may give the students an idea of why various special types of topological spaces are introduced and studied. A third aim is to use this course aS a surrogate for real analysis to reinforce some of the concepts from basic analysis while dealing with examples such as functions spaces. This also helps the students to gain some perspective of modern analysis. The discerning experts will perceive that I have preferred definitions that use only the primitive concept of an open ball (or open sets) rather than secondary concepts. (See, for instance, the definitions of open sets, bounded sets, dense sets, boundary of a set, nowhere dense sets, etc.)

Teachers who use this book should feel free to use their discretion to

change the order of the topics.

I usually prefer to introduce continu-

ity as early as possible, immediately after open sets and convergent se-

quences are introduced. In fact, I rarely follow the sequence in which the topics appear here. I am of the opinion that open sets, convergence

of sequences, continuity, compactness, connectedness and completeness

should be the focus of attention and nucleus of the course. All other concepts may be developed as and when necessary. Another important suggestion is “Draw pictures”. After all, topology is geometric in nature and most of the proofs can be better explained by drawing pictures. I also suggest that the proofs of the theorem should be explained in detail with appropriate pictures with minimal writing on the board. The students should be asked to write the proofs on their own immediately after this.

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P REF ACR

viii ;

I have included

ercises. Mes

many ‘easy to see’ remarks and observations em may

get a passing mention during a ¢o

88 ey.

a topreciate them at the first instance. If theseUrse,obsebut

r “Aven as exercises, the students may think over them studen's eee t nome fot later use. Also, I have included many pictures in the

a

I have seen students’ classroom notebooks containing no Pictures

y! Many a time, I have shown hoy a ™ a ‘ivearmaylongleadcourse to an inideatopolog of a solution. This, hopefully, wil] eNcourage D eatudents to think geometrically. The book is written With a view

towards self-study by the students. To make them take active participa, tion in the learning of the subject, very often I spell out the strategy of the proof and encourage the reader to complete it on his own. This also shows the students how even long proofs have a simple core idea which

evolves into a more complete and precise proof. This will help them not to miss the wood for the trees. Also, many results are broken into Simple

exercises and they precede the results.

Another noteworthy feature of

this book is that almost all results are followed by typical applications,

Some of the topics/results could be assigned for student seminars.

A few of them are:

Tietze extension

theorem,

Existence

of nowhere

differentiable but everywhere continuous functions, Picard’s existence theorem, topologist’s sine curve, Arzela-Ascoli theorem, Connectednesg and path-connectedness of S” etc. The book may also be used as a supplementary text for courses in general (or point-set) topology so that students will acquire a lot of concrete examples of spaces and maps.

:

.

A prerequisite for the course is an introductory course in real analysis. The book contains approximately 400 exercises of varying difficulty. I have three governing principles when I assign exercises to the students: (1) There should be a reasonable drill by means of straight forward

exercises to test and consolidate the understanding of the definitions and results, (2) standard and typical applications should be given so

that in future students will be able to perceive what results may be used

to solve a problem and (3) mildly challenging to challenging exercise so

that students shouid not think that everything will fall into their lap. They are also meant to build their perseverance. Most of the exercises are given copious hints. Students are advi sed to read the hint only after pondering over the problem for some time.

It is my pleasure to thank the participants of M.T.&.T.S. Programme on whom most of this material was tested. Their enthusiasm and their submission of compilation of the exercises and examples given during my

courses were added to this book.

I than

University who went through a preliminary k A.V. Jayanthan, Goa version of the manuscript

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iy

PREFACE

and pointed out the typos and egregious errors and made suggestions for improvement. Special thanks are to Ajit Kumar who made sense out of my rough sketches, applied his own imagination and drew all the figures in this book. But for his contributions, the book will lack one of the

noteworthy features of the book, namely, profusion of figures.

I also take this opportunity to thank the charged atmosphere that

prevailed in my department! during the month of April 2004.

It was

this which really propelled me into taking up the project of writing this book, since I wanted to work on something that I enjoy and that will help me preserve my sanity. (Yes, it is a case of blatant escapism!) A

Few

words

to the students.

This book is written keeping in

mind the difficulties of the beginners in the subject and with a view towards self-study. I have motivated the concepts geometrically as well as using real-life examples/analogies whenever possible. Most of the common mistakes or misconceptions are pointed out. Lots of easy exercises are given so as to consolidate your understanding. You should try to solve at least half of them. It is my fervent hope that the book will encourage you to think geometrically, put some of the basic tricks, results and examples at your disposal for your future endeavour. Added

for

the

second

edition:

I am

overwhelmed

by the re-

sponses and positive comments on the first edition. Many have sent the list of mistakes and suggestions for improvement. I have incorporated almost all of the corrections/suggestions. I especially thank Mr. Anand Sawant, a participant of M.T.&.T-.S. Programme who went through the book, submitted a list of corrections and made suggestions for improvement Due to a very persistent demand from many of the readers of the first edition, I have included a chapter which contains comments, hints and solutions to almost all the exercises of the book. Needless to say, the reader should try their level best to solve the problems on their own and only if they are stymied with any for a period of time, they should look up its solution. I thank Mr N K Mehra of Narosa Publishers for exerting periodical and gentle pressure on me to include a solutions manual and bring out

a second edition of this title.

I would like to receive suggestions for improvement, corrections and

critical reviews at [email protected]

Hyderabad

1This refers to Department of Mathematics,

S. Kumaresan

University of Mumbai of which I was

a faculty member at the time of preparation of the manuscript for the first edition.

el

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Contents Preface 1

3

4

5

vil

Basic Notions

1.1 1.2 2

.

1

Definition and Examples..........-.-+-.5+. Open Balls and Open Sets...........--2-+-.

Convergence 2.1 Convergent Sequences

1 15

.......---++++2 07s

35 35

2.2

Limit and Cluster Points.................

39

2.3

Cauchy Sequences and Completeness

..........

43

2.4

Bounded Sets

..........-002 2 ee ee eee

48

2.5

Dense Sets...

2.6 2.7

Basis... 0... ee Boundary ofaSet ..............22..

......

Continuity 3.1 Continuous Functions

2.0.0

ee eee

ee ee ees

50 .

53 o-4

..................

56 56

3.2 3.3.

Equivalent Definitions of Continuity .......... Topological Property ............ ep tees

59 72

3.4 3.5

Uniform Continuity...................0. Limit of a Function.................-0.4.

76 82

3.6

Open and closed maps

83

.......

a

ee

Compactness 4.1 Compact Spaces and their Properties. .........

85 85

4.2

Continuous Functions on Compact Spaces

.......

95

4.3 4.4

Characterization of Compact Metric Spaces ...... Arzela-Ascoli Theorem.................4.

100 106

Connectedness

5.1

5.2

Connected,Spaces....

111

2...

0.00.

eee

eee

Path Connected spaces...........0..0005

111

121

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x11

6

CON TENTs

Complete Metric Spaces

123

Spaces . _ | Examples of Complete Metric

6.1

Space... . 0,000 6.2 ‘ Completion of a Metric 00°77 eoem......,...0 6.3 Baire Category Th ple... . |. : Pe. Banach’s Contraction Princi 6.4

7

Answers, Hints and Solutions

| .|. Comments on Exercises in Chapter].. || Comments on Exercises in Chapter2....,00 ..,.0| Comments on Exercises in Chapter3..

71 72 73 74

7.5 7.6

Tt.

r4....0,.| Comments on Exercises in Chapte

19 | 139 1d

159

157 157 165 ne

r5...... || os i Comments on Exercises in Chapte . Chapter6........ : Comments on Exercises in

Bibiography Index

7 ‘or

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Chapter Basic 1.1

1 Notions

Definition and Examples

A metric space is a set in which we can talk of the distance between

any two of its elements. The definition below imposes certain natural conditions on the distance between the points. Definition 1.1.1. Let X be a nonempty set. A function d: X x X ~R is said to be a metric or a distance function on X if d satisfies the following properties:

(i) d(z,y) > 0 for all z,y € X and d(z,y) = 0 iffx = y. (ii) d(z,y) = d(y,x) for all z,y € X.

(iii) d(z,z) < d(x,y) +d(y, z) for all x,y,z € X. Tuis is known as the

triangle inequality. The pair (X,d) is then called a metric space.

Example 1.1.2. The most important example is the set R of real numbers with the metric d(z,y) := |z — y|. Recall the absolute value of a z if2>0

real number: |z|=

¢—-z

ifx |

to+d to—§

to+éd

sede

fr

On 5 dte= ad > 0.

This contradicts our assumption that So f(t) dt =0. Question:

Can you spell out the properties of the (Rie mann) integral

used while deriving the displayed inequality? Remark Riemann conclude if t #1/2

O

1.1.12. Note that if we assume that f is non-negative and integrable on (0, 1] such that h f(t) dt = 0, then we cannot that f = 0 on (0, 1]. For instance, consider the function f(t) = 0 and f(1/2) = 10. Then f is Riemann integrab le on [0,1] and

Io f(t) dt = 0.

Definition 1.1.13. Let V be an inner product space.

Given a vector

z € V, we define the norm or length ||z|| (read as norm of z) as the nonnegative square root of (zx, z), that is, by ||z|] := /(z,z ). The most important examples are the Euclidean space s.

On the

vector space R”, we define the inner product (z,y) := }o_, zyy;. Note

that when n = 2, ||(z,y)|| = /x? + y? is the length of the vector (z, y). The norm function (1) ||z|| > 0 for all x (2) |Jaz|] = lal lz], Furthermore, given

| | : V 4 R has the following properties: € V and ||z|| =0 if and only ifz = 0. 2 € V anda eR. a nonzero vector v € V, the vector defined by

u:= Tel is such that ||u|| = 1 and v = ||v|| u. This u is called the unit vector along v. (See Figure 1.3.) In general, we say that a vector zx € V is of unit norm if |||] = 1.

--°"

.

Figure 1.3: Unit vectors along x and y

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CHAPTER 1. BASIC N

6

OTIONs

an inn er Prog. Theorem 1.1.14 (Cauchy-Schwarz Inequlaity). Let V be uct space. Then we have

\(z,y)| < lei yll for all zy € V. The equality holds iff one of them is a scalar multiple of the other. Proof. If x = 0 or y = 0, then (x,y) = 0 and either (x, z) = 0 or (y,

0. Hence the result. Now consider the case when ||z|| = lly Consider (x — y,x — y). Then 0 0 or (z,y) < 1.

Similarly (rz + y, rz + y) > 0 yields — (x,y) < 1. Hence

(z,y)|

1

]

Be

Tell la

From this we get |(z,y)| < ||x|| !y|.

\2Y

}

2? is an increasing function on [0,00), we deduc e the required inequality. O Definition 1.1.16. Let V be a vector space over R or C. A norm on V is a function | || : V > R satisfying the conditions (i)-(iii) listed in the

last theorem.

The pair (V, | ||) is called a normed linear space, or NLS in short.

Lemma

1.1.17.

Given an NLS (V,||

Then d is a metric on V.

||). we define d(x.y) := |lz—yll.

Proof. We show that d satisfies the triangle inequality. Let us write z—z=(x-—y)+(y-—z) and apply the triangle inequality of the norm:

d(z,z)

=


ICN

8

geometric meaning of the . The next theorem explains the quality in the standar d Bune the equality occurs in the triangle ine quan

cases of many ine metric on R". This is typical of the equality not, have a zen itieg always very special and, more often than

They are

Metric

definition. interpretation. We need the following

real vector space and x,y, 2 ¢ Definition 1.1.20. Let V be any nts x and y iff there exist, ‘ poi the n wee bet lies z nt poi say that the . 0 0, it follows that

d(z,z)+d(z,y)

=

(1-t)|lz-yll+tllz—-4ll

=

d(z,y).

lz — yll

Conversely let us assume that d(x, y) = d(z,z) + d(z,y). We know that lz —y|| < lz — zl +llz — yl]. But we have equality by our assumption.

Therefore, it follows from the equality case of triangle inequality (for a

norm induced by an inner product) that z — z = t(z — y) for some t 20. This shows that (1+ t)z = z+ ty and z = mit ia: Since, t > 0, we see that 0 < a lail [bil < Hall, [|b||,, for all a,b € K”.

(1.5)

Equality holds iff Cy |xx|” = C2 |yx|? for 1 < k < n for some nonzero

constants C, and Cy.

Proof. The strategy is as follows. We take x = Tet and y = Ii, in (1.4) and sum over i. Now, it is a straight forward exercise for the reader.

—

|

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1.1.

DEFINITION AND EXAMPLES

13

If we take z = laf: and y= (pL in (1.4), we get q

1 jal 1 1 1aP

[b,|9

5

pial * oil

ear

2

1.6

[b;|

= Tal Talk:

(1.6)

ee

laP + 12 La pI

“.f

|ai| |b; Tora

2~\ fall, bl,

llalle ” g- —

|:

-.

eee

ail

Simplifying we get 1

whence the inequality. When does equality occur? Goi ng through the proof and recalling when the equality occurs in Young’s inequ ality, we deduce that the equality occurs iff

lzelP

lye?

zl, ~ lvl

=

Ci |r|= ?Co

lel? = Co lyekl’

forl 0 suc

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1.2.

OPEN BALLS AND OPEN SETS

that B(z,r) CU;

c UiesU;.

25

Since zx is arbitrary, it follows that Uje/U;

is open in X. Proof of (iii). Let Uk, 1 < k < n, be a finite collection of open sets in X. We are required to show that their intersection N?_,U, is open.

If Mp=1Uk = 0, then the intersection is open. If not, let z € “=iUk be

arbitrary.

Since U, is open and z € Uj, there exists r, > 0 such that

B(x,rk) C Ug. It is clear what we should do now. Let r := min{r;, : 1 Osuch that B((a,b),€) C B(z,r)x 5 y B(y,s). You need to find 8), kwards. Any € < min{r ~~ d(a,z),s ~ dp y)} See Figure 1.18. Work bac does the jot. y

pS of

e (a,b

(x,y)

y+ =

gar

t

po

pe .:

a al

Emenee

V

et

—t

£

“yg

ata,e)

Figure 1.19:

Figure 1.18: B(z,r) x B(y,s) is

XxY

open in X x Y

U x

5

7%

U xV

is open i

Ex. 1.2.62. Let X,Y be metric spaces. Consider the product set X xy

with the product metric (Ex. 1.1.32.) Let U (respectively, V) be an open set in X (respectively, Y). Show that U x V is open in X x Y. (See

Figure 1.19.) Ex.

1.2.63. Keep the notations as above.

Let W C X x Y

be open

in the product metric. Let px and py denote the projections of X x Y

onto X and Y respectively. Show that px(W) (respectively, py (W)) is open in X (respectively, in Y). (See Fig 1.20.) Y

*

=

-.

py (W)

.

Ww Se,

en ao?

.

px (W)

Figure 1.20: Projections of W on X and Y Ex. 1.2.64. Let (X,d) bea metric space. Define 6 on X as in Ex. 1.1.30

Show that a subset U js d-open iff it is d-open.

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27

1.2.

OPEN BALLS AND OPEN SETS

Ex.

1.2.65. Let X be the set of sequences (z,) such that 0 < z% < 1/k

for all k € N. Define d(z,y) := (SOP, |z% — yel?)!/?. Show that d is a

metric on X.

The metric space (X,d) is called the Hilbert cube and is

usually denoted by I“ or J.

Hint: Note that )-{2,22 < oo for any

xz € I~. Also, recall the triangle inequality in R” for any n.

Ex.

1.2.66. Any open set in the Hilbert cube J© (Ex. 1.2.65) is the

union of open subsets of the form U,

«x

---Un

x

Xnai

X-++&

KXngk

K-o-

’

where X; := [0,1/k] for k € N and where U; are open in Xi, 1 min{vp(a), vp(b)} for a,b€ Q, a £0, b#0anda#

¥b (b) Let d be a metric on a set X. Assy me that it satisfies the ultra-

metric inequality:

d(x, z) < max{d( y),z, d(y, z)}.

(2.1)

Examples of such metrics are the p-adic metrics d, and the metric in Ex. 1.1.28. Prove the following: (i) Equality holds in (2.1) whenever d(z,y) # d(y, z). (ii) Any ball in (x, d) is both closed and open. (iii) Every point in a ball is the cent re of the ball.

(c) Show the sequence (p”) converges in (Q, d,) but not in (Q, d) where

d is the restriction to Q of the absolute valu e metric on R.

(d) What does it mean to say that as equence of integer s converge to

0 in Q with the p-adic metric? (e) Show that the sequence (x,) where Zr) = 3, ro = 33, z3 = 333 and so on is convergent to —1/3 in (Q, ds).

|

2.2.

Limit

and

Definition 2.2.1.

Cluster Points

Let X be a metric space. Let EC

X. A point z € X

is a limit point of E iff for every r > 0, we have B(z,r) NE # @. See

Figure 2.3. How do we extend the notion of a limit point to an arbitrary topo-

logical space? See Ex. 2.2.6 and Definition 2.2.7.

Figure 2.3: Limit Points

Ex. 2.2.2. Let E be a subset of a (metric) space. Show that any r € E is a limit point of E.

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Ex.

2.2.3

CONVE

2.

CHAPTER

40

RGENCp

, it point of A, then it is alg If AC Band x € X isa lim

limit point of B.

in R. nts of (0, 1), (0, 1] and [0,1] poi t limi the d Fin .4. 2.2 Ex. Ex.

limit points of Q in R. 2.2.5. Find the set of all

ning point of E iff every open set contai t limi 8 is z that w Sho 6. 2.2. Ex. with E. z has nonempty intersection

subset of a topological space x A Definition 2.2.7. Let Ec X bea of E if every nonempty open go, point x € X is said to be a limit point E. that contains x contains a point of

Ex.

2.2.8. Let X be a metric space and Ec

X.

A point z isa limit

a. E such that rp, point of E iff there exists a sequence (zn) in fo, This result is false in an arbitrary topological space. It is true

table T; Spaces, a special class of topological spaces known as first coun should analyze the which we do not intend to define! In fact, the reader

logical space : proof and arrive at the conditions to be satisfied by a topo

so that the results continues to be true. The next theorem

and its corollary offer an important

proving that a set E of a metric space is closed.

method of

Theorem 2.2.9. A subset E of a metric space (X,d) is closed iff F contains all its limit points.

Proof. This is an easy exercise. his own.

The reader should attempt a proof on

Assume that EF is closed. Let z be a limit point of E. If z € E, there is nothing to prove. If z ¢ E, then the set U := X \ E is open and z € U. Therefore there exists an r > 0 such that the open ball

B(z,r) C U. Hence, B(z,r) N E = @. This contradicts the assumption

that z is a limit point of E. Therefore we conclude that the statement xz ¢ E is false. In other words, any limit point of E lies in E.

Conversely, if E contains all its limit points, then we want to prove

that E is closed. Consider U := X \ E. We need to prove that U is open.

If U is not open, it follows that there exists z € U such that for every r > 0, the open ball B(z,r) is not contained in U. That is, for every r > 0, there exists a point in the complement of U. Hence x is such that

for every r > 0, the set B(z,r)N E #0. That is, z is limit point of F and it is not in E. This contradicts our hypothesis that E contains atits limit points. Hence we conclude that U is open and hence E is

Cc

.

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2.2.

LIMIT AND CLUSTER POINTS

Al

The following is an immediate corollary of the theorem and Ex. 2.2.8. Corollary 2.2.10.

A subset E of a metric space (X,d) is closed iff the

following holds: If (z,) is a sequence in E converging to x € X, then oO

reek.

Example 2.2.11. Let C := C[0,1] be the set of real valued continuous functions on (0, 1], considered as a subset of B(0, 1] under the norm | ||... Then if f € B[0, 1} is a limit point of C, then f € C. Hence C is closed in B[0,1] by Theorem 2.2.9. Hint: This is a well-known result from real analysis in disguise. See Theorem 6.1.10. Ex. 2.2.12. If A is a nonempty bounded subset of R, then its supremum and infimum are limit points of A. Ex.

2.2.13.

Let E Cc R be a nonempty bounded and closed subset.

Show that sup E, inf EF € E.

- Ex. 2.2.14. Show that the only nonempty subset of R which is both open and closed in R is R. Hint: If A is a nonempty subset of R which is

both open and closed, let r € A. Then (x —¢€,r+€) C A for some e > 0. Let E := sup{b: (x —€,b) C A,b > xz}. Why is E nonempty? Can E be bounded above? If it is bounded above, let 6 := sup E. Conclude that B € A. Use the openness of A to arrive at a contradiction.

Ex. 2.2.15. Let E be a subset of a metric space with the following property: If a sequence (z,) in X converges to a point x € E, then there

exists N € N, such that z, € E for all n > N.

Show that E is open.

Ex. 2.2.16. Show that the diagonal {(x,x) : x € X} is closed in the product metric space X x X. Definition 2.2.17 (Closure of a set). Let (X,d) be a metric space and

AC X.

Let lim(A) denote the set of all limit points of A. Thus

lim(A) := {y € X : For every r > 0, the set B(y,r)

©

A #0}.

Ex. 2.2.18. Show that lim(A) is the smallest closed set containing A. The standard notation for lim(A) is A. It is usually called the closure of A. The last exercise allows us to define the closure of a set in a topological space.

Definition 2.2.19.

Let A be a subset of a topological space. Then the

closure A of A in X is the smallest closed set that contains A. (Why

does this make sense?

How do we know that there exists the smallest

closed set containing A?)

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R2.

CONVE

space B (x,7) Cc B{z,r). Give that in a metric B[z, r]. a proper subset of be n ca r) , B(z example to show that

Ex. 2.2.20. Show

an

to = B[z,r]. Extend the result ,r) B(z t tha w sho R", In Ex. 2.2.21. normed linear spaces.

(a) Q, (5 following subsets of R: the of es sur clo the d Ex. 2.2.22. Fin

(0, 1) U (3. 4]. Z, (c) (-1,0) UN and (d)

the closures of the set n wee bet ion t rela the Ex. 2.2.23. Investigate (metric) space sets A, B in an arbitrary the and B AC B, AN B, AU

point z is a cluster or an accumulation Definition 2.2.24. We say that set B(x,r) 1 E contains a point other of a set E if for each r > 0, t he than x. .

= B(z,r) \ {2}, then BY(z,r) is . called a deleteded

If we let B’(z,r) :

neighbourhood of x.

Thus,

z is a cluster point of E iff every delet

of E. neighbourhood of z contains a point

ks refer to these points An Important Note: Almost all text-boo

distinguish between also as limit points. However in this book, we shall limit points and cluster points. ty, We shall give an analogy of cluster points in real life. If a celebri

in India, say, Amitabh Bachchan or Sachin Tendulkar attends a party

then in any vicinity around him, you will always find some other member we of the party! If you or I were there in a party, it might happen that rs may be sitting in a corner and in our vicinity there are no other membe of the party. Thus, they are cluster points of the set of India people, though they may not be, say, among some tribal people in Africa! Ex. 2.2.25. Show that every point of Z C R is limit point of Z while Z has no cluster point. What are the cluster points of Q in R?

Ex.

2.2.26. Let A := {1/n: n € N} Cc R.

cluster point of A.

Show that 0 is the only

Ex. 2.2.27. Show that every point of a nonempty open set U in R" is

a cluster point of U.

Ex. 2.2.28. Let A be a subset of a metric space X. Then z € X '8 a cluster point of A iff every open set containing x contains infinitely many points of A.

Theorem 2.2.29 (Bolzano Weierstrass Theorem).

Let A be an infinite

bounded subsct of R. Then there is a cluster point of A inR.

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2.3.

CAUCHY SEQUENCES AND COMPLETENESS

43

Proof. Let E := {x € R: x < a for infinitely many a € A}. Let MER be such that - Af 0 there exists a point a € (€—€,@+e)NA other than @ itself.

that such that For,

Since @ — € is not an upper bound for E there is an x € E such &—e« < x. Since z € E there exist infinitely many elements a € A that x < a. Hence there exist infinitely many elements a € A such £—« @+e.

But

then

@+ 0, we can find N EN such that whenever m>N and n> WN, we have d(tm,Zn) < €.

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CHAPTER 2. CONVERGENo,

44

space is Ex. 2.3.2. Show that any convergent sequence in a metric Given z. to ge conver Cauchy. The converse is not true. Hint: Let (zn) What can yoy say € > 0, choose N such that d(an,z) < € for n > N. about d(tm,Zn) if m,n 2 N?

c space? Ex. 2.3.3. What are the Cauchy sequences in a discrete metri Example

2.3.4.

Let X be any nonempty

normed linear space B(X)

set.

Let us consider the

of all bounded real valued functions on y

ig under the sup norm | ||,,. We claim that a sequence (fn) € B(X) . Cauchy iff it is uniformly Cauchy. rmly Cauchy sequence First of all, let us recall the definition of a unifo

of functions on X.

We say that a sequence fn: X > R is uniformly

Cauchy if for a given € > 0, there exists N

€ N with the following

property:

all m,n > N. \fn(x) — fm(x)| < € for all x € X and for Let us assume that (fn) is Cauchy in B (X). Let € > 0 be given. Since (fn) is Cauchy in the NLS, there exists N € Nsuch that || fm — fall, N. Unwinding the definition of the norm,

sup{|fm(x) — fn(z)|: 2 € X} N. In particular, for all z € X, we have

|fm(z) - fn(x)| N. Thus (fn) is uniformly Cauchy on X.

The proof of the converse is left

to the reader.

Ex.

(2.2) O

2.3.5. Show that any Cauchy sequence in a metric space is bounded,

that is, if (z,) is Cauchy in a metric space (X,d), then zp lies in an open ball B(z,r) for all n. Hint: Apply the definition for ¢ = 1 to find N.

Except for finitely many, all x, € B(zn, 1).

The next proposition lists some of the most often used facts about Cauchy sequences. It is in fact a compilation of the facts enunciated in the exercises above.

Proposition 2.3.6. Let (X,d) be a metric space.

1. Any convergent sequence in (X,d) is Cauchy.

The converse is not

true.

2. Any Cauchy sequence (zn) in X is bounded, that is, all rn € B(z,") for some xr € X andr > 0. 3. A Cauchy sequence is convergent iff it has a convergent subsequen©-

-_

|

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9,3.

CAUCHY SEQUENCES AND COMPLETENESS

45

Proof. You must have already proved the results. If not, go ahead and prove them on your own.

Proof of 1. Let (z,) be Cauchy in X. Let € > 0 be given. We need

to find N such that if m,n > N then d(rn,r2m) < €. We are given that Let x, — x. Draw an open ball B(z,¢/2).

(zn) is convergent.

n sufficiently large, xr, € B(a,€/ 2).

Hence for such m,n

For all

we expect that

d(zm,Zn) < 2€ . (See Figure 2.4.) We turn this idea into a proof. N be such that d(z,,z) < €/2. Then for m,n > N, we have

Let

d(In,; 2m) < d(tn,x) + d(z, tm) < €/2+e/2 =e.

Figure 2.4: Convergent sequences are Cauchy

The converse is not true. For, if we take (0,1) with the standard metric, then the sequence (1/n) is Cauchy. Given e > 0, choose N > 1/e. Then, for alln >m>

|1/n

|

WN,

/m|

nm—m

—

nmi

max{2N, no}, we find that 1

TI--

n

>

2--|>

2N | —

: t—

1 aI,

a contradiction. We therefore conclude that the Cauchy sequence (1/n)

is not convergent in (0, 1). Proof of 2. Let us take « = 1. Since (z,) is Cauchy, there exists N

such that d(zn,2m) < 1 for n,m 2 N. In particular, rz, € B(xry,1) for

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CHAPTER 2.

46

CONVERGENCE

alln > N. Let R>

max{d(x,2N),.-. ,d(zn-1,2N); 1}.

Then z, € B(x, R) for alln Ee N. Proof of 3. If the given Cauchy sequence (rp) is convergent, we can take it as the convergent subsequence. Let us prove the converse. Let (z,) be Cauchy in X and assume that (zn,) is a convergent subsequence, converging to z € X. We claim that z, - z.

Let € > 0 be given. Choose N, such that d(am,2n) < €/2 for m,n >

N,. Choose k; such that if k > k;, then d(z,2n,) < €/2. Observe that np > k.

(For, 1 < n; < ng implies ng > 2.

By induction, we see that

np > k.) Let N := max{Nj,,k,}. Then for any n > N, we have d(Zn, 2)


0 be arbitrary. There

exists a positive integer N = N(6) such that for all m > N andn>N, we have |rn — Zm| < 5/2. In particular we have |z,, — rn| < 6/2. Or,

equivalently,

In C (rn — 6/2, ry + 6/2)

for alln > N.

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2.3.

CAUCHY SEQUENCES AND COMPLETENESS

47

From this we make the following observations:

(i) For all n > N, we have z, > ry — 6/2. (ii) If tn > en + 6/2, then n € {1,2,...,N — 1}. Thus the set of n

such that Zn > ry + 6/2 is finite.

We shall apply these two observations below for 6 = 1 and 6 = zr}.

We claim that S is nonempty, bounded above and that sup S is the limit

of the given sequence. From (i), we see that zy — 1 € S. Hence S is nonempty. From (ii) it follows that zy +1 is an upper bound for S. That is, we claim that y < zy + 1 for all y € S. If this were not true, then there exists a y € S such that y > zy + 1 and such that z, > y for infinitely

many n. This implies that z, > ry + 1 for infinitely many n. This contradicts (ii). Hence we conclude that ry + 1 is an upper bound for S. By the LUB axiom, there exists £ € R which is supS. We claim that limz, = ¢. Let e > 0 be given. As @ is an upper bound for S and zn — €/2 € S (by (i)) we infer that ry — €/2 < ¢. Since ¢ is the least upper bound for S and zy + €/2 is an upper bound for S (from (ii)) we see that ¢< ry + €/2. Thus we have ry — €/2< €< an +e/2 or lan — e| < €/2. For n > N we have

lan -—4|


0. Hence we conclude that c? + 1 = 0, which is impossible as c € Q.

Ex. 2.3.19. Prove that (1, | ||,) is complete. Hint: Notice that if (x)

is Cauchy in 1’, then (x?) is Cauchy for each i. Let 2; := lim, 2”. To

show that x := (z;) € I’, use triangle inequality to obtain k

k

k

1

1

1

Dole < Mo lei - 221+ Sle? Why does x" > z? Write a careful proof.

Remark 2.3.20. The concept of Cauchy sequences does not exist in an arbitrary topological space.

2.4

Bounded

Sets

Definition 2.4.1. We define A to be bounded in (X, d) if there exists

to € X and FR > 0 such that AC B(zo, R). See Figure 2.5.

Ex.

2.4.2. Show that a subset A of a metric space (X,d) is bounded

iff for every x € X there exists r > Q such that Ac Figure 2.6.)

B(z,r).

(See

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2.4.

BOUNDED SETS

Figure 2.5: Bounded set

49

Figure 2.6: Illustration for Exer-

cise 2.4.2

Ex.

2.4.3. Show that the union of a finite number of bounded sets in a metric space is bounded.

Definition 2.4.4. Let A bea nonempty subset of a metri c space (X, d). The diameter diam (A) of A is defined by

diam (A) := sup{d(z, y) : z,y € A}, in the extended real number system.

Ex. 2.4.5. Show that (a) the diameter diam (B(z,r)) < 2r and that (b) the strict inequality can occur. Ex. 2.4.6. Show that a subset A is bounded iff either it is empty or its diameter diam (A) is finite. (This is the standard definition of a bounde d set in a metric space.) Ex. 2.4.7. Let (z,) be a convergent sequence in a metric space (X,d). Show that the set {z,} is bounded. More generally, if (an) is Cauchy in

X, show that the set {z,} is bounded in (X,d).

Ex. 2.4.8. Show that diam (B(z,r)) = 2r for any ball in an NLS. Hint: You may work with R” and with the ball B(0,r). Note that z € B(0,r) iff —z € B(0,r).

Ex.

2.4.9. Let (X,|| ||) be an NLS. Show that A C X is bounded iff

there exists M > 0 such that ||z|| < M for all x € A.

Ex. 2.4.10. Which vector subspaces of an NLS are bounded subsets?

Ex. 2.4.11. Show that the set O(n) of all orthogonal matrices (that is,

the set of matrices satisfying AA’ = J = A'A) is a bounded subset of M(n.R). Here M(n, R) is considered as an NLS as in Ex. 1.1.33.

Ex. 2.4.12. Show that the set SL(n,R) of all n x n real matrices with

determinant 1 is not bounded in A/(n,R) for n > 2. (The metric is as

in Ex. 1.1.33.)

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CHAPTER 2. CONVERGENCE

50

the Ex. 2.4.18. Let G be a subgroup of the multiplicative group C* of

non-zero complex numbers. Assume that as a subset of C it is bounded.

Show that |g| = 1 for all g € G. 2.4.14.

Ex.

Consider R with the standard

metric d and the metric

6(x,y) := min{d(z, y),1}. Then (R,6) is bounded while (R,d) is not.

Remark

2.4.15. The moral of the last example is that ‘boundedness’

is metric specific. R with the standard metric is unbounded while with

respect to 6 := min{1,d}, it is bounded. However the topologies induced

by d and 6 are the same.

Remark 2.4.16. There is no concept of bounded sets in an arbitrary topological space.

2.5

Dense Sets

Definition 2.5.1. We say that a subset D c X of a metric space is dense in X if for any given x € X and r > 0, we have B(z,r)N DF 9. In other words, any non-empty open set in X must contain a point of D. A subset D C X of a topological space is dense in X if for every: nonempty open set U c X, we have DNU # 9, that is U intersects D non-trivially. A real life analogy will be useful. Imagine a lake with an abundant

supply of fish. Then when you throw a net, however small, into the lake, the net will trap a fish. Then we say that the fish is dense in the lake.

Figure 2.7: Dense set

Figure 2.8: Not dense

Another example would be the set of people with tonsured head in Tirupati Balaji temple. In any ‘small neighbourhood’ of devotees in the

temple, you will always find one with a tonsured head. (Do not stretch the real life examples too far!) Ex. 2.5.2. Show that Q is dense in R. Is R \ Q dense in R?

Can you think of a countable dense subset in R?? in R"?

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2.5. DENSE SETS

'

Example 2.5.3. A rational numbe r of the form m/2" is called a dyadic rational. We claim the set D of dyadic rationals is dense in R. It is enought to show that given a < bt here exists r € D such that a < r < b. is such that a 1. In the interval (2"a, 2”b), we can then find an integer m so that a f(z).

We say that f is continuous on a subset AC X each a € A.

if f is continuous at

:

Figure 3.1: Continuity of f at x

Ex.

3.1.2. Show that any constant map from a metric space to another

is continuous.

| /

Ex.

3.1.3.

Show that the identity map z »

metric space (X, d) to itself.

Ex. 3.1.4. Show that the map z + 2? from R

x is continuous from a

to itself is continuous.

Example 3.1.5. We show that the maps R? > R given by a: (z,y)7 xr+y and p: (x,y) + ry are continuous.

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3.1. CONTINUOUS FUNCTIONS

57

° Let (z,y) € R* 2 be arbitrary. We show that a is continuous at (x,y).

Let (2 Yn) : (z,y). By Ex. 2.1.7, 2, + x and Yn + y. By the stand ard result on algebra of limits ye

of sequences in R (see also Ex. 2.1.9), it follows 7 t+y. That is, (zy, Yn) > a(z,y). Hence a is continuous

A similar argument proves the contin uity of » on R?2.

Ex. 3.1.6. Show that the projection map s p;: R" — R given by p;(z) = aj (1 0 such that f(z’) # 0 for all x’ € B(z,r) and the function g(z’) := 1/f(z') from B(z,r) to R is continuous at z. Analogous results hold if we replace R by C.

Proof. Let rn — x. Then (f + 9)(tn) = f(tn) + 9(rtn). We are given

that f(zn) >

f(z) and g(zn) — g(x).

From the algebra of limits of

sequences, it follows that f(zn) + g(tn) > f(z) + 9(z) = (f + g)(z). This proves the continuity of f +g at x. The other statements are proved similarly.

oO

Remark 3.1.10. Do you appreciate the ease with which you proved these results thanks to our definition of continuity? The core of the argument falls upon established facts on convergent sequences in R or in C!

Ex. 3.1.11. Use Ex. 3.1.6 and Theorem 3.1.9 to conclude that any poly-

Nomial function p(z1,...,Zn) in the variables x},..., 2, is continuous on

R". (Give examples of such functions!)

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CHAPTER 3. CONTINUITy

ss

). Show that f is q Ex. 3.1.12. Consider M(2,R). Let f(A) = det(A 2.1.18. function. Hint: Ex. 3.1.11 and Ex.

continuous Can you think of a generalization?

function fz defined by f2(y) := Ex. 3.1.13. Fix z © X. Show that the d(x,y) is continuous. Ex.

3.1.14. Find a continuous function f: C* > S):= {z EC:

1} such that f(z) =z for z € 5’.

|2|=

continuous on C. Ex. 3.1.15. Show that the conjugation map z+ 2 is Proposition 3.1.16. Composite of continuous functions 1s continuous: € X Let X,Y,Z be metric spaces. Let f: X + Y be continuous atx and g: Y + Z be continuous at y = f(x). Then the composite map x € X go f: X > Z is continuous at ! Proof. This is easy. at f of / Let ty, > x. Then y, := f(2n) > y = f(z) by the continuity “ 2. Since g is continuous at y, it follows that.9(yn) — g(y) or what is the 3.1.17. Ex. metrics.

In this exercise, .

U

/

same, 9(f(In)) + 9(f(z)).

the product sets are given the product

Show that the following maps are continuous: r+ y. (1) the vector addition map R” x R" — R® given by(z,y) +> az. (a,x) by (2) the scalar multiplication map R x R” — R®” given

(3) the inner product map R” x R" — R

given by (z,y) > (x,y).

Ex. 3.1.18. Let f,g: X — R be continuous functions on a metric space

X.

Show that the map y: X —> R? given by y(z) = (f(z), 9(z)) is

Ex.

3.1.19.

continuous.

map.

Let X,Yi,Y2 be metric spaces.

Let f: X — Y, x Yo bea

Write f(z) = (f1(z), fo(z)) in an obvious notation. Show that f

is continuous iff each of f;, fo is continuous.

Ex. 3.1.20. Let f: R? — R be continuous. Show that the map g: R? R given by g(z,y) := f(x + y,z — y) is continuous.

Ex. 3.1.21. The map A> A‘ from M(n,R) to M(n,R) is continuous.

Here A‘ denotes the transpose of the matrix A such that the (i, j)-th entry of A’ is the (j,i)-th entry of A.

Ex. 3.1.22. Show that the squaring map A +> A? on M(n,R) is com

tinuous. Can you think of generalizations?

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3.2.

EQUIVALENT DEFINITION S

OF CONTINUITY

59

induced

metric. Show that the vector spac e of continuous functions on X is linearly isomorphic to the vector space of all convergent sequences

in R. =

In

Ex.

Hint: Given f € C(X ), consider the real sequence (z,) where f(1/n).

3.1.24.

Keep the notation as in the last

exercise. NU {oo}. Let f: Y + X be the bijection defined by

Consider

Y

=

f(n) =1/n forneN and f(oo) = 0.

Using this bijection, we transfer the metric on X to Y. (See Ex. 1.2.7 5.) Let (A,d) be a metric space and (an) be a sequence in A. Let a € A. Then a, — ain A iff the functi on ~: Y — A defined by setting y(n) = a,

for n € N and (oo)

- 3.2

= a is continuous at oo.

Equivalent Definitions of Continuit y

In this section we shall give the standard e-d definition of continuity as well another one which involves open sets. The latter one generalizes

to topological spaces. We shall also show how to use the latter one to give an ‘easy’ way of finding open or closed sets and allows us to decide

whether a set is open or closed. The c¢-é definition leads us naturally to the definition of uniform continuity. Theorem 3.2.1 (Equivalent Characterizations of Conti nuity). Let X,Y be metric spaces. Let f: X + Y bea function. Then the following are equivalent:

(a) f is continuous at x € X. (b) Given any € > 0, there exists a 5 > 0 such that if d(x, 2’) < 6,

|

then d( f(x), f(z'))

(b): We shall prove this by contradiction.

Assume that for

some € > 0 there exists no 6 > 0 with the required property.

Thus, if

we take 6 = 1/n, then there exists an x’ such that d(z, x! ) < 1/n but

d( f(x), f(z')) > e. Let us call this x’ as z, to emphasize its dependence on 6 =1/n. As d(tn,x) < 1/n, the sequence r, — x. (See Figure 3.2.)

By our assumption, f(z,) must converge to f(x). Hence we deduce that d(f(zn), f(x)) + 0. This does not happen, since d(f (zn), f (z)) > € for : all n. This is the desired contradiction.

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CHAPTER 3. CONTINUrpy

60

f(z1) f Z)

ae -

Pa T

z3

:

2

e EE

ee r

.

.

en

sorte,

UN ‘

De

iid

ee

see

‘

Tey

Ge

:

;

Be

a

tree? T |-

Me, oy

B(z,1/3)

> B(r,1/2)

L____» B(z,1)

Figure 3.2: a =>

(b) =»

f(zn)

z2

f(z3)

Hea)

f(z4)

b of Theorem 3.2.1

(c): Let V be given as in (c). Since V is open and f(z) € V,

there exists an ¢ > 0 such that B(f(zx),€) C V. Since we assume (b), for

this ¢, there exists 6 > 0 with the property stated in (b). That is, if we let U := B(z,6), then U is an open set that contains x and is such that f(U) CV. (See Figure 3.3.)

f-(V) Figure 3.3: b =>

(c) =>

c of Theorem 3.2.1

(a): Let z, + x. We need to show that f(z,) + f(z). Lete >0

be given. Consider V := B( f(x),¢). This is an open set containing f(z).

So, by (c), there exists an open set U containing x such that f(U) c V. Since U is open and z € U, there exists r > 0 such that B(z,r) Cc U.

Since z, — x, there exists N € N such that if n > N, we have sn €

B(z,r). It follows that f(a) € f(B(z,r)) c f(U) Cc B(f(z),€) for all

n> WN. That is, f(z,) + f(x). Therefore, f is continuous at x.

Ex. 3.2.2. To gain practice, show all possible two way implications o the last theorem: (a)

(b), (b) 0. Then there exists 6 > 0 such that f(x) > f(a)/2 for all x € B(a,d). Hint: We needed this when we wish to prove that || ||, satisfies || f ||, = 0 iff f =0

in Example 1.1.10. (See Figure 3.4.)

|

How to extend the notion of continuity to functions between topolog-

ical spaces? Part (c) of Theorem 3.2.1 suggests the following definition.

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CHAPTER 3. CONTINUrpy

62 q—

t

o

{

f(a)-

c= {)

f(a)te=2Oe)

Figure 3.4: Illustration for Exercise 3.2.4 Definition 3.2.5. Let X,Y be topological spaces and f: X > Y bea map.

We say that f is continuous at r if given an open set V 5 f(z),

Ex.

3.2.6. Show that Ex. 3.1.2, Ex. 3.1.3, Ex. 3.2.4, Theorem 3.1.9

we can find an open set U > x such that f(U) C V.

and Proposition 3.1.16 remain valid (after suitable modifications in the statements!) for arbitrary topological spaces. (How will you modify Ex. 3.2.47) Ex.

3.2.7. Let X,Y be (metric) spaces. Show that a map f: xXxyY

is continuous iff for every open set V C Y, its inverse image f “1(V) is open in X. Ex. 3.2.8. Let the notation be as in the last exercise. Let By be basis of open sets for the topology on Y. (See Definition 2.6.1.) Show that f

is continuous iff f~!(B) is open in X for all Be By.

Ex. 3.2.9. Let X,Y be (metric) spaces. Show that a map f: X + Y is continuous iff for every closed set V C Y, its inverse image f —1(V) is closed in X.

Ex. 3.2.10. One can use Ex. 3.2.7 and Ex. 3.2.9 to show that certain sets are open or closed.

For instance the set of points (x,y) € R? such that cos(x?) + 2° -

47y > e* — y? is an open subset of R?. For, the function y(z,y) = cos(x?) + x3 — 47y — (e* — y”) is continuous on R? and the set under : consideration is f~!(—co, 0).

As a concrete example, let us show that a rectangle (a,b) x (c,d)

is open in R?. (This is Ex. 1.2.35.) Where do we look for continuous

functions? We look at the defining properties of the set. A point (z,y)

lies in the rectangle iff x € (a,b) and y € (c,d). Now (a,b) and (c,d) are

open and the projections p(z,y) = 2 and po(z,y) = y are continuous. Thus, the set Uj := p; (a,b) is open and U2 := po Me, d) is open. The rectangle is the intersection of these open sets and:hence is open.

Ex. 3.2.11. Redo Ex. 1.2.16, Ex. 1.2.29, Ex. 1.2.30, Ex. 1.2.32, Ex. 1.2.35, Ex. 1.2.58, Ex. 1.2.59, Ex. 1.2.102, Ex. 1.2.97 and Ex. 1.2.97. Ex.

3.2.12. Show that the set of all invertible matrices in M (2,R) is

open. (Hint: Ex. 3.1.12.)

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EQUIVALENT DEFINITION S OF CONTINUITY

63

Ex. 3.2.13. Show that the set SL( n,R) of matrices in M(n,R) with determinant one is a closed subset of (n,R). Ex. 3.2.14. Show that the set

O(n,R ) of orthogonal matrices in M(n, R) one is a closed subset of 4 (n,R) How about the set SO(n, R) of orthogonal matrices in M (n,R) with determinant 1? Ex.

3.2.15.

Show that the set of all nilpoten t matrices in M (n,R) is is said to be nilpotent if At = 0 for some k € N.) Hint: Ex. 3.1.22. An infinite union of closed sets need not be closed. So, you need something from linea r algebra also!

closed. (Recall a matrix A ¢ YW (n,R)

Ex. 3.2.16. Let (X,d) bea metric space. Cons ider the distance function

d: X x X +R. We equip X x X with the product metric (Ex. 1.1.32). Show that d is continuous. Ex. 3.2.17. ~Y be continuous. Show that theLetset X,Y EF := be{xmetric € X : spaces. f(r) # Let g(z)}f,9:X is

This exercise has an extension to arbitrary topol ogical spaces pro-

vided that Y is Hausdorff.

Ex. 3.2.18. Let X,Y be metric spaces and D Cc X be dense. Let f,g: X — Y be continuous functions such that f(x) = g(x) for all z € D. Show that f = g on X. This has an extension to topological spaces with an extra assumption on Y. Let X,Y be topological spaces. Assume that Y is Hausdorff. Let Dc

X be dense in X.

Let f,g: X — Y be continuous functions such

that f(x) = g(x) for all x € D. Then show that f = g on X. Hint: Last

exercise. Ex.

3.2.19. Let f: R > R be a continuous additive group homomor-

Ex.

3.2.20. Let D be dense in a (metric) space X. Let Y be another

phism. Show that f(z) = Az for z € R where \ = f(1).

(metric) space. Assume that f: X — Y

that D’ = f(D) is dense in Y.

is continuous and onto. Show

Ex. 3.2.21. Let X,Y be metric spaces. Let f,g: X — Y be continuous.

Then the set {x € X : f(x) = 9(zx)} is closed in X. | | In particular, if f: X — X is continuous, the fixed point set {x €

X : f(x) = x} is closed.

Remark: The results are false for general topological spaces. If we assume that Y is Hausdorff then the results are true. Compare Ex. 3.2.17.

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CHAPTER 3. CONTINYrpy

Ex. 3.2.22. Define the map f: R > R by setting f(x) = |r ~ 1. Show that f is continuous. More generally, let J = [—1, 1] and let

f(x) := inf{|z — t|: t € [-1, 1]}.

Show that f is continuous. Find an explicit expression/formula for f. Draw its graph. Can you interpret f(r) as a geometric relation between x and the interval J? Example 3.2.23. Let A bea nonempty subset of a metric space (X, d). Define

da(z) := inf{d(z,a): a € A},

re X.

Then dy is continuous. (Geometrically, we think of d4 (xz) as the distance of x to A.) We give a proof even though it is easy, because of the importance of this result.

Let z,y € X and a € A be arbitrary.

We have, from the

triangle inequality d(a,x) < d(a,y) + d(y,z), d(a, y)

2

d(a, x)

d(a, y)

2

da(z)

- d(y, x)

— d(y, z),

(3.1)

since d(a,x) > inf{d(a’,x) : a’ € A} := da(z).

The inequality (3.1)

says that da(x) — d(y,x) is a lower bound for the set {d(a,y) : a € A}. Hence the greatest lower bound of this set, namely, inf{d(a,y) : a € A} is greater than or equal to this lower bound, that is,

da{y) > da(x) — d(y, 2). Therefore, d4(y) — d(x) > —d(y,xz) or what is the same,

da(x) — da(y) < d(y,z). If we interchange z and y in this inequality, we see that +(d,(z) da(y)) < d(z,y), that is, |da(r) — da(y)| < d(z,y). Continuity of dy, follows. One may also arrive at this as follows:

d(a,x) < d(a,y) + d(y,z).

(3.2)

Now we are in the following situation. We have two families A := {ai: i € I} and B := {b; : i € I} of real numbers (bounded below) indexed by the same set J and with the property that a; < b; for eachi € J. Let

a := inf{a;} and b := inf{b,}. Then we claim that a < b. For, a < a {or

each i and hence a < }; for each i. Thus, a is a lower bound for the s#

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3.2.

EQUIVALENT DEFINITIONS OF CONTINUITY

65

B. Since 6 is the greatest lower bound for the set B, we conclude that

a < b, as claimed. Applying this claim to (3.2), we obtain

da(x) < da(y) + d(y,z). The rest of the argument is as above.

Remark 3.2.24. In view of Ex 3.1.13 and Example 3.2.23, we see that

there exist non-constant real valued continuous functions on any metric

space X. In fact, they are ‘abundant’ in the sense that given any two distinct points z,y € X, there exists f: X + R such that f(z) =0 and f(y) = 1. Thus, any pair of distinct points are “separated” by means of a continuous function! In fact, we can say more. Any two disjoint closed subsets of a metric space can be separated by means of a contin uous function. See Theorem 3.2.35. In an arbitrary topological space, one cannot assert the existe nce of

non-constant real valued continuous functions, leave alone being able to separate disjoint closed sets! We need to impose extra conditions to ensure such a possibility. This leads one to the definition of completely regular and normal spaces. notions.

As usual, we refrain from introducing these

Ex. 3.2.25. Let A = (0,1) C R. Draw the graph of the function da.

Ex. 3.2.26. Let c € R. What is dg(zx)?

Ex. 3.2.27. Let p := (a,b) € R? and A be the z-axis. What is da(p)? Ex.

3.2.28. What is d4(p) where A := {(z,y) € R? : 2? + y? = 1}?

Find an explicit expression.

Ex. 3.2.29. This exercise assumes knowledge of inner product spaces. Let V be a real (finite dimensional) inner product space and W a vector

subspace. What is dw(z) for x € V?

Ex. 3.2.30. Show that z is a limit point E iff de(x) = 0. Conclude that x € A iff d4(xr) = 0. In particular, if A is closed, then d(x) = 0 iff rE A.

Definition 3.2.31 (Distance between two subsets). Let A and B be two nonempty subsets of a metric space X. We define the distance d(A, B) between them by setting d(A, B) := inf{d(a,b):a€ A,be B}. Ex.

3.2.32.

d(F,,F2)=0.

Find

two closed sets F, and

F> which

Hint: N and {n+ 4 >n€N,n > 2}.

are disjoint but

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CHAPTER 3.

66

CONTINurry

B) between A and B where Ex. 3.2.33. Find the distance d(A,

(a) A=Qand B=R\Q.nonempty

| subset of R. (b) A= Q and B is any of axes a zy = 1 and B is the union ol rb pe hy ar gul tan rec the is (c) A

ry = 0.

metric space (X ,d). Show Ex. 3.2.34. Let A be a nonempty subset in a and

any € > 0 (i) directly that the set {x € X;da(z) < €} is open for (ii) by using the continuity of da.

be two disjoint closed Theorem 3.2.35 (Urysohn’s Lemma). Let A,B us function f{: XR subsets of a metric space. There exists a continuo

on B.

=1 < f Az is continuous.

In fact, we can prove more. The joint map from M(n,R) x R" to R"

given by (A,z) ++ Az is continuous.

.

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EQUIVALENT DEFINITIONS OF CONTINUITY

67

Ex. 3.2.38. Let (X, | ||) be an NLS. Show that | | : X = R is continyous. Ex. 3.2.39. (a) Let X,Y be normed linear spaces. Let T: X + Y bea

linear map (a) Assume that there exists C > 0 such that |Tz|| < C |x| for all z € X. Show that T is continuous on X. (b) Prove that T is

continuous iff it is continuous at 0 € X.

Use this to show that T is

continuous iff there exists a constant C > 0 such that ||Tz|| (C{0,1], | lloo) given by Df = f’, the derivative of f. Show that D is not continuous.

Ex. 3.2.44. Find a continuous function f: (a,b) + R which is bijective

and such that f~! is also continuous.

Ex. 3.2.45. Let f: R — R be such that f—!(a,00) and f~!(—0o,b) are

open for any a,b € R. Show that f is continuous.

Ex. 3.2.46. Let A be an open subset of a metric space (X,d). Let (Y, d)

be another metric space. A function f: X — Y is said to be continuous on A iff f is continuous at each a € A. Show that f is continuous on A

iff f: (A,d) — (Y,d) is continuous.

Is the result true if A is arbitrary?

Ex. 3.2.47. Let f: (X,d) — (Y,d) be continuous. Let A c X be open.

Show that the restriction f |4 of f to A is a continuous function from

the metric space (A,d) to (Y,d).

metric.)

(Here the metric on A is the induced

Lemma 3.2.48 (Gluing Lemma). Let X and Y be (metric) spaces. (1) Let {U; : i € I} be a family of open sets such that UU; = X.

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CHAPTER 3. CONTINUrTy

68

fi . vi that there exists a continuous function

Assume

each

7 2 ie

i € I with the property that f;(z) = fj(z) for all x E ” i \ fa € I], = fi(z) fre, Then the function f: X + Y defined by setting f(z) is well-defined and continuous on X.

U;A; = (2) Let {Aj :i € F} be a finite family of closed sets such that

each X. Assume that there exists a continuous function fi: A; > Y for

A;

i € F with the property that fi(x) = f;(x) for all xe

Aj and

i,j € F. Then the function f: X + Y defined by setting f(x) := fi(z) if x € A; is well-defined and continuous on X.

Proof. The proofs are straight forward and the reader should attempt

them on his own. The only hint needed is that if U C X is open (re spectively closed) and A C U is open (respectively closed) in U, then A is open (respectively closed) in X.

Let us prove (1). Let V Cc Y be open. Then we have to show that

f~*(V) For, c fi(z) € the set

is open in X. Now, € f-1(V) NU; iff € V, that is iff ee U;N f\(V) ™ U; is open in

we observe f~!(V) U; and f(x) € V, fz\(V). Since f;: U; and hence open

NU; = f;'(V) Nn U;. that is, iff € U; and U; + Y is continuous, in X. Thus, it follows

that f~1(V) is the union of open sets and hence is open:

f(V) = fFUV) NX = fol) n (UUs) =U; (F(V) NU).

This proves-the continuity of f on X. The proof of (2) is very similar, except that we use the characterization of continuity by means of inverse images of closed sets. Let C C Y

be a closed set. We shall show that f~!(C) is closed in X.

As in the

earlier case, we find that f-'(C)M.A; = f-1(C)N A; and that it is closed

in A; and hence in X. We then express f —1(V) as a finite union of closed sets of this form.

fC) =f (C)NX

=F-E)N (UiAi) = Ui (F71(C)

This completes the proof.

Ai) .

An immediate, though trivial, application is Ex.

3.2.49.

tinuous.

Show that the absolute-value functi on | |: R > R is con-

A most useful application is the foll owing Ex.

3.2

.50. 9(0). Define

Let f,9 : [0,1] +

h(t) = fe

X be continuous.

g(2t-1)

A ssume that f(1)1) =

0 x’. We claim that there exists ng such that if n > n3, then’d(an, bn) < 6. Since an > Z z', for 6/3, there exist n1,n2 € N such that and by,

n>n, => d(dn,z) < 6/3 and n > nq => d(bn,2’) < 5/3. (3.17) ,

It follows that if d(z, x’) < 6/3 then for n > ng := max{nj, ni}

d(an, bn) < d(an,x) + d(x, 2’) + d(x’, bn) < 3(5/3) = 6.

(3.18)

In view of (3.16) and (3.18), we get

d(z,x') < 6/3 andn>n3z

=>

Now since f(an) — g(x) and f(b,) k,,kg € N such that

d(f(an), f(bn)) ki, d(f(bn), 9(2’)) < e/3 for n > ke.

(3.20)

a

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34.

UNIFORM CONTINUITY

81

It follows from (3.20) » (3.19) and (3.20) that for r,z’ € X with d(z,2z’) < 6/3

d(9(z),9(2'))


0 (called a Lipschitz constant of f) such that fo, all x1, 22 € X, we have

d( f(x1), f(x2)) < Ld(x1, 22). Show that any Lipschitz continuous function is uniformly continuous. Ex.

3.4.16. Let T: X — Y be a linear map between two NLS. Show

that T is continuous iff it is Lipschitz.

Ex. 3.4.17. Let f: R > R be differentiable with | f'(z)| 0, there exists a 6 > 0 such

that for all x € BY(a,6) MD, we have d(f(z),y) < «. (See Figure 3.8.) If

such a y exists, it is called the limit of f as x > a and is denoted by limz_,q f(z) = y. Note that we have used the definite artic le ‘the’ and it is justified in the next exercise. Ex.

3.5.2. Keep the notation

of the definition. If there exist yi, yoeY such that lim,,. f(z) = y) and limz+a f(z) = yo, then yi = yo. Hint: You learnt to prove the uniqueness of the limi t of a convergent sequence.

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3.6.

OPEN AND CLOSED MAPS

Fi

igure 3.88: limli f(z)

83

=

y

Ex. 3.5.3. How do you assign a meaning to lim,,, f(x) = y if f: DC X -Y is a function between two arbitrary topological spaces? Do you think uniqueness of limit still holds true? What additional condition on

X (or on Y) will be needed to ensure uniqueness?

Ex. 3.5.4. Can you reformulate the notion of continuity at a point a in terms of the limit of a function as x > a?

3.6

Open and closed maps

These are very useful concepts in topology. Definition 3.6.1. Let X,Y be (metric) spaces. A map f: X — Y is said to be open if the image f(U) is open for every open set U c X. It is said to be closed if f(K) is closed for every closed subset K Cc X. Ex. 3.6.2. Show that the projection map 7;: R" — R given by 7;(z) =

z; is open.

It is not closed.

Hint: Consider {(x,y) : « > 0 and zy =

1} C R? and the projection onto the z-axis.

Ex.

3.6.3. Let X := (IR,d) be the usual real line and Y := (R,6) be

the set R with discrete metric 6. Show that the identity map 1: X

>~ Y

is not continuous but it is open as well as closed. (See Ex. 3.6.2.) On the other hand, 1: Y — X is continuous which is neither open

nor closed.

Ex. 3.6.4. A bijection f: X — Y is open iff it is closed. Ex. 3.6.5. Let A be subset of a (metric) space given the induced topol-

ogy. When is the inclusion map a + a of A into X open? Closed?

When is it

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84

CHAPTER 3. CONTINUrTy

Ex. 3.6.6. Prove the following:

a map g: Y Zig (a) If f: X + Y is a continuous surjection, then

open if go f is open.

(b) If g: Y — Z is a continuous injection, then a map f: X 4 y ;,

open if go f is open.

(c) Do (a) and (b) remain true if we replace ‘open’ by ‘closed’?

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Chapter 4

Compactness 4.1

Compact

Spaces and their Properties

The basic intuitive idea of a compact subset of a (metric) space is that it is a generalization of a finite set.

Most professional mathematicians

may subscribe to this view but no author of a book would like to put it in print. But you have it here! Look for instances where this crude intuition helps us conjecture results.

We set. A covers Let family

first introduce the concept thorough knowledge of this etc. X be any nonempty set. of subsets of X indexed by

of an indexed family of subsets of a will be needed to understand open Let J be another nonempty set. A the indez set I is a map I — P(X)

of I into the power set of X, that is, the set of all subsets of X. We denote the image of 7 € I by A; or some such notation. Then the family

is denoted by {A, : i € J}. (Compare this with the case of sequences. A

sequence in X is a function from N to X and in practice, we denote it by (rn) etc.) Let us look at some examples.

Example 4.1.1. Let X = R? and J = (0,00). For any r € J, we let C, = {(z,y) € R? : 2? + y? = r7} be the circle of radius r with centre at the origin. Here the family is {C, : r € [0, 00)}.

Example 4.1.2. Let X = R and J = Q‘, the set of positive rational numbers. For any r € Q*, we let J, denote the interval (—r,r). Here the family is {J, : r € Qr}.

Let a family {A; : i € J} of subsets of X indexed by I be given. Let ACI be asubset. Then the family {A; : i € A} is called a subfamily of

the given family {A; : i € I}.

The subset Ujc,.A; of X defined by

:

WienAy = {x € X : x € A; for some iin A} 85

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CHAPTER 4. COMPACTNEg¢

is called the union of the subfamily.

Let us again look at some examples,

2], then the union of he subfam. In Example 4.1.1, if we take A = (1, S 4) on {(z,y) € R° : 1 so ily {C, : r € [1, 2]} is the annular regi B(0, 1) in R or the closed ball If we take A = (0, 1], it is the closed disk

the union of the subfamily In Example 4.1.2, if we take A = N, then

family Observe further that it is the union of the

{Jn :n € N} is R. {J,:7r € QT}!

Definition 4.1.3. Let X be a (metric) space and A C Xx.

A family

of subsets {U; : i € I} is called an open cover of A if each U; is open

(Note that U; need not be contained in A and they are

and A C U,U;.

required to be open in X.)

If J C ] is a subset of J such that A C Ue Uj, we then say {U;: i € J} is a subcover of the given open cover of A. If there exists a finite subset J C I such that the A C UjeyUj, then

we say that the given cover admits a finite subcover for A. Example

4.1.4.

Consider the family {(1/n,1)

: 1 € N,n

> 2}.

We

show that this is an open cover of (0,1). We need to show that (0,1) C

US24(1/n,1). Let x € (0,1). We want to show that z € (1/n, 1) for some n > 2. If there exits such an n, then 1/n < x < 1. Hence n > 1/z. By Archimedean property of R, N is not bounded above in R. Hence 1/z is not an upper bound for N. It follows that there exists N € N such that

N > 1/zx or 1/N < x < 1. Thus the given family is an open cover of

(0, 1).

We claim that the given cover admits no finite subcover for (0, 1). Suppose that it does. Then (0,1) c Ufe1 (1/n;,1) for some nj, n2,..., Nk. Note that (1/m,1) Cc (1/n,1) for m < n. If we set N := max{n,;:1< j < k}, then UF_1(1/n;, 1) = (1/N,1). Hence we deduce that (0,1) c

(1/N,1).

This is absurd, since 1/2N € (0,1) but (1/2N) ¢ (1/N,1).

Hence the claim follows. Suppose we add two open intervals, say, Jo, J; with the property that

0 € Jo and 1 € J to the above family. Then the collection {(1/n,1) : n > 2} U {Jo, Ji} is an open cover of [0,1]. We claim that this cover admits a finite subcover of [0, 1]! ff

\

“$0

c

cf

a

RG

\

7

4

f

\

I-e

“|

\

1

ite

a

7

Figure 4.1: Illustration for Example 4.1.4 Since 0 € Jo, there exists 6 > 0 such that (—5,5)

C Jo.

There

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4.

COMPACT SPACES AND THEIR PROPERTIES

87

exists N € N such that 1/N < 6, by Archim edean property of R. Then (0, 1] Cc Jo U Ji U

(1/N,

1).

Let x € [0,1]. If

(Verify. )

=O or 1, there x € JoU J. If 0 a; >a.

Also, ¢ is

the least upper bound for the set {a, : k € N} and 6 is its upper bound. Hence a0,!} as k — 00.

Since c € [a,b] and {U; : i € I} is an open cover of [a, b], there exists

a € I such that c € U,.

Since U, is open (this is the only place where

we use the fact that the sets in {U; : i € I} are open!), there exists < > 0 such that (c — €,c++€) C Ug. Since a, > c and by — c, there exist positive integers k; and k2 such that a, € (c—e,c+e) fork >

k,; and & € (c—eé,c+e) fork

> ko.

Hence if k > ko := max{ky, kp}, then ax, by € (c—e€,c+e). Therefore, [ax, by] C (c-—€,c +e) C Ug. This implies that {U; : i € I} admits a finite subcover for J, (if k > ko).

This contradicts our assumption on J. This contradiction shows that out original assumption that {U; : i € I} does not admit a finite subcover

for J is wrong. Hence the theorem follows.

‘Steen

"We used the fact that 2~* — 0. It follows from the observation alk en, This is proved by induction. Consequently, 2~* < 1 /k.

O

that k < 2* for

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CHAPTER 4. COMPACTNp¢. !

00

Bx. 4.1.16. Find all compact subsets of a discrete metric space. X.

4.1.10.

mple of an uncountabe compact While Theorem 4.1.15 gave an exa pact set which is count. set, the next exercise gives an example of a com ably infinite.

pact in R. Can yoy Ex. 4.1.17. Show that {1/n:n € N} U {0} is com

generalize this? (in a metric space?)

ed and Theorem 4.1.18. Any compact subset of a metric space 1s clos

bounded.

Proof. Let K Cc X be compact. Fix any zo € X. Consider the open cover {B(zo,n) : n € N}. This admits a finite subcover for K, say,

{B(xo,nj):1< 7 m, it follows that K C B(xo, N) for N := max{n;:1 0 by continuity of f at z. Since X is compact, the

Open cover {B(xr,5,) : xz € X} admits a finite subcover, say, {B(zi, 4) lsi< n} where 6; = 6;,. One may be tempted to believe that if we

set 6 := min{d; : 1 < i < n}, it might work. See where the problem is. Once you arrive at a complete proof, you may refer to the proof below.

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98

pg, CHAPTER 4. COMPACTN proof. Given 0. The following result gives three characterizations of compact metric spaces. Each of these is quite useful. One use is, of course, to decide

whether a given metric space is compact or not.

The other is to ap-

ply them to derive some results about compact metric spaces.

illustrate their uses below.

Most often for theoretical purpose, charac-

terization (2) of the theorem seems to be efficient.

and Ex. 4.4.10.

We shall

See Theorem 4.4.8

Condition (4) is quite useful when dealing with prob-

lems, where we have some idea of the convergence in the given metric

space. See Ex. 4.3.19-Ex. 4.3.24.

Only experience will teach us which

characterization is useful or most expedient in a given context.

Spaces). For a Theorem 4.3.15 (Characterization of Compact Metric

metric space (X,d), the following are equivalent: subcover. (1) X is compact: every open cover has a finite

(2) X is complete and totally bounded.

|

cluster point. (3) Every infinite subset of X has a rgent subsequence. (4) Every sequence in X has a conve

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APTER 4. COMPA CTNEss CH

102

: the fami Sisnj

Bk “3 mpact. co be ) ,d (X t Le Proof. (1) = (2): X. Let {B(zi: of r ve co en op an {B(z,€) | z € X} is is totally bounded.

e . Hence X for every k EN ther en be a finite subcover Th X. in ce en qu uchy se 5 Now let (zn) be @ Ca %- We choose nx’s < 1/k for all n > d(tn,tn,) exits n, such that

ng sequence. Let that (n,) is an increasi

1 | d(x,2n,) > Ef

Uy = {2 Ee xX

Now tn ¢ Ux for n > Nk.

Then U; is open. 3

cover of U,’s cover X:

n> max{n},...,M%m}-

For,

if they

did, say,

Hence no finite subUm,U;,

=

X

we

h 1 < k < m. Then tn ¢ Ux for any k wit

take

This

Thus there exists x € X \ Up Up. implies that {U;,} cannot cover X. — 7 Since (rn) is Cauchy we see But then d(z,2n,) < ¢- Hence tn, X is complete. We have thus also converges to lim, Zn,- Thus that zp, shown (1) implies (2). et of X. (2) => (3): Let E be an i nfinite subs subset of X such that X = UzerF,

Let F,, be a finite

B(x,+). Then for n = 1 there exists

te. As a second step, replace E z, € F, such that EN B(zx,1) is infini 1) is contained by Fo. The infinite set EN B(2a),

by EN B(zi,1) and F, nce there exists 22 € F2 such that in the finite union Urer, B(z, 1/ 2). He

EN B(a1,1)/N

B(z2, 1/2) is infinite.

B(zk, ;)) is infinite. Inductively choose zn € Fp such that BM (N2_, we see that

1) Since there is a € EN B(zm, +) N B(rn,

1

1

2

d(tm,In) < d(Zm,a) + d(a,Zn) < - +o7 0. Question: Draw pictures. Are

truth?

ct. Hint: Ex. 4.4.20. Show that the Hilbert cube (Ex. 1.2.65) is compa you may prove seShow that it is complete and totally bounded. Or, ally prefer the quential compactness by a diagonal argument. I person first, as it is more instructive.

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Chapter

5

Connectedness We say that a (metric) space is connected if it is a ‘single piece.’ This is

a very difficult notion to be formulated precisely.

If we look at R \ {0},

we would think of it consisting of two pieces, namely, one of positive

numbers and the other of negative numbers.

Similarly, if we consider an

ellipse or a parabola, it is in a single space while a hyperbola has two distinct pieces. If we remove a

single point from a circle, it still remains

as a single piece. We start with a definition which is unintuitive but with

examples and further exploration see that it captures our intuitive ideas.

5.1

Connected

Spaces

Definition 5.1.1. A (metric) space X is said to be connected if the only sets which are both open and closed in X are @ and the full space X.

A subset A of a (metric) space X is said to be connected if A is a connected space when considered as a (metric) space with the induced (or subspace) topology. In the case of metric spaces, this amounts to

saying that (A, 4) is connected, where 6 is the restriction of the metric d on X to A.

Ex. 5.1.2. R is connected (Ex. 2.2.14) while R*, the subset of nonzero real numbers is not connected. Ex. 5.1.3. Show that a (metric) space X is not connected iff there exist

two disjoint proper non-empty subsets A and B such that A and B are both Open and closed in X and X = AUB. In such a case, we say that the pair (A, B) is a disconnection of X.

Ex. 5.1.4. Let X be a set such that |X| > 2 with discrete metric. Show X is not connected. Ex, 5.1.5. When is a finite subset of a metric space connected? 111

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112

CHAP

TER 5.

CONNEC TEDNp..

of ant characterization rt po im st mo the is The following result

nected spaces.

uo, ted iff every contin ec nn co is X e ac sp pological | Theorem 5.1.6. A to nction. fu nt ta ns co a is 1} {+ 15 connecteg function f: X > the subspace topology, h wit d we do en X, of A subset A on {+1} wa constant functi A> f: on cti fun s uou iff every contin a continuons space and f: X > {+1} ted nec con a be X Let of. Pro is non. a constant function. If f

that f is function. We want to show Then A = f73(1) and B = f71(-1). constant, then it is onto. Let are B sets of X such that A and sub pty -em non nt joi dis are A and B . This is a of X and X = AU B.(Why?) s set sub sed clo and n ope h bot nt. contradiction. Therefore f is consta not connected. Therefore there Conversely, let us assume that X is ets A an exist two disjoint proper non-empty subs and X = AUB. Now we define a and B are both open and closed in X map f: X > {+1} as 1

(=) = tt

ifreA

if x € B.

stant function. (Why?) This Then f: X — {+1} is a continuous n on-con

completes the proof.

' Proposition 5.1.7.

The interval [a,b] C R is connected.

e theorem from real analysis, Proof. If we assu me the intermediate valu J is an interval and if f: J > we can give a short and elegant proof. If that

e exist x,y € J such {+1} is an onto continuous function, then ther there = 1. By the intermediate value theorem,

f(z) = —1 and f(y) 0, a contradiction to our exists z between x and y such that f (z) =

Hence no such f exists and assumption that f takes only the values +1. hence J is connected.

the intermediate Instead, we may adapt the argument used to prove is connected and value theorem to show directly that any interval [a, b] deduce the general result form this.

that Assume that {a, 6] is not connected. Then there is U C (a, b| such

0 4U # [a,b] and U is both open and closed in [a,b]. Let V = [a,b] \U:

that Without loss of generality assume that a € U. We intend to show

U = [a,b] so that V = 0. Consider E := {x € [a,6] : (a,z] C U}. Since a € U and U is op

there exists an ¢ > 0 such that [a,a+e¢) C U. Hence [a,a+ ¢/2} (Uo

a+e/2€ Eso that E #9. E is clearly bounded above by 0. ‘Thus by

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:

5.4:

CONNECTED SPACES 113 1B axiom there exists a real numb

ar

ot © € R which is sup £. Note

cs}.that c € E. ;" weNe claim For eachn N,cis not oor E. We can therefore find z, € E such that 1/n c-1/n o

een

min = & Since Zn € EF, rz, € U. Since U is closed jin [a,b] (with to the subspace topolo gy) and c € [a,b], we see that ¢ = limz res : ectU

Now (a,c) = Ula, a,c)

7 I/n] Ula, zn]. As each of [a, rn] C U we see that C U. This along with the fact that ¢ ¢ Y allows us to conclud that (a,c] C U and hence c€ E.

cee

We now show that c = b. This wil] com plete the proof. dU is open there exists an Since cE U (relatively)

open

subset a If c < 6, then there exists an N € N such that (c This mean

in

ing i

s that the set [a,c + 1/(2N)] C [aye] U( e—1/Nee + LIN) € E. This contradicts the

y. Thusc+1/(2N)

Ne + Uy ore

fact that c = supE Therefore our assumption that c < b is wrong. Th us c = b, “This completes the proof of connecte dness of [a, b. An alternate argument runs as follows.

We claim that c € U.

Let, if possible, ¢c € V. Since V is open in

[a, b], there exists 6 > O such that c € (c—6,c+6)N[a, b] CV. Since

c—6 < c= sup E, there exists x € E such that ¢—6 < + 0 such that c € (c — 6,c + 6)N [a,b] C U. In partic ular, (c—46,c] C U. Since c-6 < c, there exists x € E such that

c-—5 0. Consider the continuous function f: [0,00) — [0,00) given by f (t) = t". By Archimedean property of R, there exists N € N such that N > a. We have 0 = f(0) < a and f(N) = N" > N >a. So, applying the

intermediate value theorem to the restriction of f to the interval [0, NJ, we see that there exists x € (0, N] such that x” = a. Uniqueness is easy and left to the reader.

O

Theorem 5.1.24. Any polynomial with real coefficients and of odd degree has a real root. That is, if p(x) = anz™ + an_1z"-1+4+---+a,;z2+ 49, a; €R for0 0} be the upper hemisphere. Let D,, := {u€ R": ||ull < 1}. Note

that

D, is convex and hence connected. We claim that D, is homeomorphic to S". Consider the map f+: Dn + S® given by fy(u) = (t1,...,Un, yi- \|u||?). Clearly, f+ is bijective and continuous. Since f, is a bijective continuous map of a compact space to a metric space, it is a homeomorphism.

In any case, sn

being the

continuous image of the connected set D,,, is connected. Similarly, we

show that the lower hemisphere S™ := {x € S" : z_4; < 0} is the image

of f-: Dn + S® given by f_(u) = (u1,...,un,—1/1 — |[ull?). Clearly, the intersection StS"

= {xz € R"+! : 2,4, = 0} is nonempty. Hence

we conclude that S” is connected by (1) of Theorem 5.1.8. We have already seen that S” is the union of two sets homeomorphic to R® (see Example 3.3.5) with nonempty intersection. The connectedness of S” follows from this.

_

Ex. 5.1.28. Let A be the union of the following subsets of R?: S L,;

:= :=

Lo Lz

:= :=

{(z,y):22+y? =1} {(z,y):2>1 and y=0} {(z,y):2< —land y=0} {(z,y):y>1landz=0}

La = {(z,y):y 1. Let A be ; | claim that R” \ A is connected. R” \ A lie in By Ex. 5.1.9, it suffices to show that any two points of a connected set. Recall that

(p,q) :={c ER": 2c =(1-t)pttg,0 (z,y)

(z0,yo)

(z,yo)

X x {yo} —» X

Figure 5.2: Connectedness of the product Now the point (2, yo) lies in both sets X x {yo} and {z}xY.

The re-

srictions of f to either of these sets are continuous and henc e constants. We see that f(z0, Yo)

= f(z, yo)) for all x € X and simi larly, f(z,y) =

{(x,yo) for all y € Y. In particular, f(z,y) = f(z, yo) = f(z0, yo). (See

Figure 5.2 on page 119.)

Q

Ex. 5.1.33. Show that the annulus {x € R?: 1 < l|z|| < 2} is connected. Hint: Continuous image of a connected set is connected. Ex. 5.1.34. Show that S”, the unit sphere in R"+! is connected. Hint: Can you think of a continuous map from R"*! \ {0} to onto S$"? The next exercise is very typical of the way connectedness hypothesis

is used in the proof.

Ex. 5.1.35. We say that f: X — Y is a locally constant function if for each z € X, there exists an open set U; containing z with the property that f is a constant on U;.

If X is connected, then any locally constant function is constant on

X. Is the converse true?

Ex. 5.1.36. Let U be an open connected subset of R" and f: U +R

be a differentiable function such that Df(p) = 0 for all p € U. Then f i$ 4 constant function.

Let f: X — R be a nonconstant continuous function on a hence X connected (metric) space. Show that f(X) is uncountable and Ex. 5.1.37.

8s uncountable.

Ex. 5.1.38. Let (X,d) be a connected metric space.

Assume that X

cardinality of X has at least two elements. Then |X| > |R|, that is, the continuous sat least that of R. Hint: Can you think of a nonconstant’ function?

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12h

CHAPTER §. CONNECTEDNp«,

c space. Ex. 5.1.39. Let (X,d) be an unbounded connected metri thay 2 ¢ X andr > 0 be arbitrary. Show that there exists y € X such

d(x,y) =r.

Ex. 5.1.40. Which of the following sets are connected subsets of R2? (a) {(x,y) € R?: 2? +y? = 1}.

(b) {(z,y) € R?: y= 27}.

(c) {(2,y) €R?: zy = 1).

(d) {(x,y) € R? : zy = @ for some fixed a € R}.

(e) {(x,y) € R? : (x?/a”) + (y?/b*) = 1} for some a > b > 0.

Ex.

5.1.41. Let X be the union of open disk in R? along with the

tangent line z = 1. Show that X is connected.

Theorem 5.1.11 shows that connectedness is a topological property. We can use'this to show that certain spaces are not homeomorphic.

Ex. 5.1.42. Show that a circle or a line or a parabola in R? is not homeomorphic to a hyperbola.

Ex.

5.1.43. Show that R is not homeomorphic to R?. Hint: Observe

that if f: X — Y is a homeomorphism and if f(A) = B for a subset AC X, the the restriction of f to X \ A is a homeomorphism of X \ A to

Y \ B. Recall that Ex. 5.1.30 says that R? minus a point is connected.

Ex. 5.1.44. Let X be the union of axes given by zy = 0 in R?. Is it homeomorphic to a line, a circle, a parabola or the rectangular hyperbola zy=1?

Ex. 5.1.45. Let A Cc X. What does it mean to say that the characteristic function x, is continuous? Can you say something more about A if X is connected?

4 Ex. 5.1.46. Give an example of a sequence ¢A,) of connected subsets of R? such that A,41 C A, for n € N, but 9,,A,, is not connected. Ex. 5.1.47. Show that no nonempty open subset of R is homeomorphic to an open subset of R?.

Ex.

5.1.48. The notation is as in Ex. 2.1.19.

connected subsets of (Q, d,) are singletons.

Show that the maximal

What are the maximal connected subsets of Q with the standard metric?

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g, PATH CONNECTED SPACES 5.2

°

Path Connected spaces nition 5.2.1. Let X be a (metric) space. A path in X

isa contin-

ip be 8 path joining the points x and y or simply a path from r to 1 cee Figure 5.3. We

also say that zx is path connected to y.

oT

X

Figure 5.3: Path between x and y

Ex. 5.2.2. If z is path connected to y, then y is path connected to z. Hint: Define o(t) := y(1 — t) for t € [0,1]. Then show that o co nnects y

_ tox. (The path o is called the reverse path of +.)

Ex. 5.2.3. Give at least two paths in R? that connect (—1,0) and (1.0) and pass through (0,1). > Ex. 5.2.4. Show that if x is path-connected to y and y is path connected to z in X, then z is path-connected to z.

More precisely, prove the following. Let 7;: [0,1] > X, i = 1,2, be two paths such that +(1) = 72(0). Then show that there exists a path

93: [0,1] + X such that 43(0) = 71(0), y3(1/2) = 11(1) = 72(0) and

7(1) = y2(1). Hint: If you think of (0, 1] as the time interval, your train 1 should cover the distance between y;(0) and 7(1) along the railway

track +,(0,1] within half the time and +2 should take over from there. To wit, consider

13(1) =

fut) vo(2t—1)

€ [01/2] ift ift € (1/2, 1).

Definition 5.2.5. A (metric) space X is said to be path connected

if

path y: [0,1] > X for any pair of points x and y in X, there exists a

such that +(0) = x and 7(1) = y-

Ex. 5.2.6. Show that any interval in R is path connected.

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CHAPTER 5.

99

CONNECTEDNp.,

\

4

. ~~ two points cay ted nec con h pat is R" ce Example 5.2.7. The spa y(t) = 2+ t(y — 2), forO 0. Show that the e-ne

ighbourhood of A defined by U,(A) := {rE R": da(z) < ¢} is path-connected.

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Chapter 6

Complete Metric Spaces give two most We recall the definition of a complete metric space and and one eximportant examples of complete metric spaces of functions ample of an incomplete metric space of functions.

6.1

Examples of Complete Metric Spaces

Recall that a metric space (X,d) is said to be complete if every Cauchy

is complete sequence in X is convergent in X. We have shown that R” for n EN. We start with an example next only in importance to R.

Proposition 6.1.1.

The normed linear space (B(X), || lo) 1s complete.

Proof. Let a Cauchy sequence (f,) be given. By Example 2.3.4, the sequence of functions (f,) is uniformly Cauchy on X. Fix z € X. Consider the sequence of scalars (f,(x)). By (2.2), it is a Cauchy sequence in R (or in C, if we are dealing with complex valued

functions).

| |

| | |

| |

For definiteness sake, we shall assume that we are working

with real valued functions. (The proof for complex valued functions is exactly the same.) Since R is complete, the sequence converges to a real number r € R, which we denote by f(x), to show the dependence of a = limpsoo fn(7)

on xz. Thus, for each x € X, we get a real number f(z) such that limn—oo fn(z) = f(x). (What we have shown so far is that the sequence fn converges to some function f: X — R pointwise.) To show that

fn — f in the metric space, we first of all need to show that the function z+» f(z) lies in B(X), that is, f is bounded on X. Next we need to show that f, — f in the metric space, that is, in view of Example 2.1.11,

we need to prove that f, — f uniformly on X. If we accomplish thes

two tasks, the proof of the theorem is complete. 128

4

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pl

pXA MPLES

OF COMPLETE

METRIC

SPACES

129

Sj ince (fn) is. Ca Therefore. there nin, ue fn < M for n EN. In particular, | fna( > ; z)|< Mf or all z € ON Fix z € X. Since f, (zr) > f(z), gi ven € = 1) th fn *

frst Show that f isis bounded.

js bounded by Ex. 2.3.5.

* hich MAY depend on zx and so we may denote t by

, (s)- f(z)| < 1 for n > N(x). It follows that fn

Y

N

.

metric 0 such X and ere exi:sts N

N(z)) such that

I= = MFC) — frveay(z)] + Frvyay(x)} SIF) ~ frey(2)| + [Fgey(x)} < 1+ Af.

have therefore shown that |f(x)| < M +1 for all x =

if

f uniformly on X.

and see the remark after the proof.)

(This proof is delicate

Let ¢ > 0 be given.

Since (f, )

is Cauchy in the metric, there exists np € N such that l fn — fl ce /2 for m,n > 1. We claim that | f(z) — f,(xr)| < « for all n > no and for all r¢X. Since fn(z) + f(z), for the given € > 0, there exists N (x) EN such that |fm(z) — f(z)| < €/2 for m > N(z). We have, for all n > no, f(z) - fa(z)|

=

f(t) — fm(z) + fm(zx) — fn(x)|

(for any m € N)

If (x) — fm(x) + fm(x) — fn(z)|

(in particular for any m € N with m > N (x))

lf(z) — fm(xz) + fm(z) — fr(z)|


N(z) and m > No)

¢/2+6¢/2=e.

That is, f, + f uniformly on X.

O

Remark 6.1.2. The proof in the last paragraph is not the one found in any textbook.

The standard proof for the boundedness of the limit

function f and the convergence of f, to f in the metric is by passing to

the limit as n —> oo keeping m fixed in the inequalities of the kind: \fn(x) — f(z)| < € for m,n > N.

we are While this proof is correct, the beginner needs to be told that

using the continuity of the distance function (in this case, the absolute as ‘curry leaves value). We refer to the trick we have employed above

r and trick’. In south Indian cooking, one uses curry leaves to add flavou throw the flavouring agent.

In our proof, the curry leaves are 72'S.

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>

CHAPTER 6. COMPLETE METRIC sp, Ons

130

Proof of Note also that we have employed this trick even earlier. See the (3) of Proposition 2.3.6. Theorem 6.1.3 (Weierstrass M-Test). Let X be any nonempty set. 7 6

(fn) be a sequence in B(X,R) such that there exists My > 0 such tha, I fnlloc < Mn and such that )°7~) Mn < 00. Then the series >™° | h

is uniformly convergent, that is, the sequence (sn) of partial sums of th,

series is uniformly convergent on X. (Recall that s,(x) := Dewi Se (2), ) Proof. We need only show that the sequence (s,) is Cauchy in the the

metric space (B(X,R), | ||,,). Let € > 0 be given. Since Yona Atn < 00,

the partial sums of the series }>°°, M, is convergent and in partic. ular Cauchy

hems

For the given

in R.

¢, there exists

N

€

N such

Mk < € for n > m > N. Hence we have, for n 2 m 2 N,

lA

[sn —Sm|l.. =

sup

xrEX

S—

k=mn+1

sup >

lA

TEX pim+1 n =m+1


R

be differentiable at x € R.

Let (s,) and (tn) be

sequences such that sn < x < ty with lim ty — sp = 0, that is, sn 7

xz and t, + x. Show that the sequence ((y(tn) — (Sn)|/(tn — $n)

is convergent.

2. A rational number of the form m/2" is called a dyadic rational. Show that the set D of dyadic rationals is dense in R. (See Exam-

ple 2.5.3.) Hence conclude that given z € R and n€N, there exist

dyadic rationals s,,t, such that s, = j/2" and t, = (7 + 1)/2" and 8, < z < tn.

|

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| EXAMPLES OF COMPLETE METRIC spacpe

6"

‘

pet f: R-

R be defined by f om ait : (x) := min{q

3 where [z] stands for the greatest integer

ae

Dea

131 [7] + 1)},

praw the) =graph of )f. Show that it igs continuous on R (be that f(x a(x, Z). Note that 0 S f(z) < 1/2 for ie serve all >.

_ Show that f is periodic of period 1, that is,

ey eR.

M(t) =_ f(r +1) for

5, Consider the half intervals [0,1 /2\ and [1/2 1).

Note that if say, $ < ¢, lie in the same half interval, 8, t, then the d iffere nce quotient , [s() — f(s)|/(t — s) is the slope of the chord joi ning (s, f(s)) and (t, f(t) and it is either 1 or —}.

g. Let fo = f and fn(z) := 27" f(2"x) fi, fg on the ‘ ‘same graph sheet’.

forn

>] = = Draw graphs of ap

Observe that 0 S fn(z) < fords 7, Let 9(z) := dopco Se(z). Show that g is a continuous function on R.

8, Let s = m2 be a dyadic rational. Then for any k > n, we have f(s) = 0. Hence conclude that g(s) = ar, Sx(s). 7

9. Let z € R be fixed.

Choose dyadic rationals Sn,tn as in Item 2.

Let Qn := 9(tn) — 9(8n)/(tn — Sn). In view of Item 3, to show that

g is not differentiable at z, it is enough to show that the sequence ([9(tn) — 9(Sn)]/(tn — $n)) does not converge. Hint: Write sp = 3/2” so that th = (j + 1)/2". For 0 N and for all « € X.

€ N such that

In particular this

exists inequality is true for n = N. By the continuity of fy at zo, there

§ > 9 (or an open set U 3 zp in X) such that ifz € B(zo,4) (or ifx € U), then, |fw(xz) — fn(zo)| < €/3. Thus, we have for any x as above,

f(z) - F(zo)|


0, there exists N @(2n41,2n) m> WN, we

oo. We need to show that (r,) isi convergent. Since eenol=1UIn41, AEned 2 n) < suffices to show that (z,) is complete, it is Cauchy. Note that Lk =m 42k, R41) + 0 since )n=1 4(Tn+1, Zn) is con vergent. Hence Given

have

n-—1

fore)

d(im,In) < >= (rk, 2k+1) < > O( rp, 244 1) € as n> 00.

(6.4)

For, if € > 0 is given, by the Cauchy nature of (r,), there exists N €N

such that d(tm,2n) < €/2 for m,n > N. We have, for n > N,

d(in,€):= lim d(tn,zm) 0 be given. Let (an) be . ° th such N € N exists there . since (an) is Cauchy, ’

‘

co all m,n

2 N.

So,

consider

if we

Qn

E

Xx,

then

a

at

represe

.

Ntative of

(Qn, am)


N We then have for all m,n > N a d(im,Zin)


max{1/e,N}.

yields

Then for all m,n > no, the above inequality d(Zn,,Zn) < 3e.

Thus (z,) is Cauchy in X. Let its equivalence class be £ € X. We now show

that a, — € as n —> 00.

d(an,€) < d(an, Zn) + d(Zn,€) < 1/n + d(Zn,€) > 0, 88 d(Zn,£) + 0 (as n — 00) by (6.4). ix o d),

;

completes the proof of the fact that (X,d) is a completion of 0

Remark 6.2.13.

“ompleteness of R.

struction?)

Note that Cantor’s construction also depends on the

(Do you see where we needed it in the above con-

space tions of a given metric ple com all t tha s say a The next lemm This is known as the ™ the same in the sense that they are isometric.

ueness of completions.

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CHAPTER 6.

144 .

Lemma 6.2.14 (Uniqueness

COMPLETE METRIC SPACE,

of Completion ). If (Y, d) and (Z,d) are tu

e (X,d), completions of the same metric spac of Y onto Z. Proof. We shall only sketch a proof.

then there is an isomety,

f(X) a f: X + Y be the map such that f is isometry and in Y. Let g be the analogous map from X to Z.

.

is denge

that Now given any y € Y, there exists a sequence (tn) in X such

it follows that f(tn) 7 y. Since go f~!: f(X) > g(X) is an isometry, to a point, (g(zn)) will be a Cauchy sequence in Z and hence converges p(y) does not say, z € Z. We define y(y) = z. (It is easy to show that

is the depend on the choice of the sequence (r,) converging to y.) This O ; | required isometry. Remark 6.2.15. We can give an alternate proof using Theorem 3.4.12.

The map y := gof~!: f(X) > Z is an isometry and hence is a uniformly continuous map from the dense subset f(X) to the complete metric space

Z. Hence it extends to a unique continuous map, say, Y: Y — Z. It is

easy to see that w is an onto isometry.

Ex. 6.2.16. Let (X,d) be a metric space and D C X be a dense subset. Assume that every Cauchy sequence in D is convergent in X, that is, if (t,) is a Cauchy sequence in D then there exists z € X such that In — x. Show that (X,d) is complete. Ex. 6.2.17. Assume that (X,d) is not complete. Prove that there exists a uniformly continuous function f: X — (0,00) such that the infrex f(z) = 0.

Ex. 6.2.18. Show that a metric space is compact iff every real valued continuous function on X attains a maximum. Note that this exercise is very much weaker than Ex. 4.2.4 on Page 96.

6.3

Baire Category Theorem

Definition 6.3.1. A subset A C X of a (metric/topological) space is

said to be nowhere dense in X, if given any nonempty open set U, we

can find a nonempty open subset V C U such that ANV =9. A prototypical example is a line in R?. See the next exercise. Se

also Figure 6.4 Take sometime to understand the definition. In the past, dense sets

were known as everywhere dense sets. Nowhere dense sets are diamet-

rically opposite to dense sets. If we want to Say a set is not everywhere

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Wy

“SS

galRE CATEGORY THEOREM

‘3

145

Figure 6.4: Nowhere dense sets get, how shall we formulate it?

here but may be dense ‘somewhere’.

gn (0, 1).

The set m ay not be dense ey-

For instance, look at th

It is not dense everywhere but dense in any open interval

Jc (0,1). Again, if we want to

say a set is nowhere dense in X, how do ‘e formulate it? If you think over these questions, you will gain insight

sto our definition.

.

fx. 6.3.2. Show that the set Z is nowhere sense in R. QcR?

How about

Show that the x-axis given by y = 0 is nowhere dense in R2. Ex. 6.3.3. Let V be any proper vector subspace of R". Show that V is nowhere dense in R”.

More generally, let X be a normed linear space. Let V be any proper closed vector subspace of X. Then V is nowhere dense in X. (Compare this with Ex. 1.2.51.)

Ex. 6.3.4. Show that the set {1/n : n € N} U {0} is nowhere dense. Ex.

6.3.5. Show

that A

C X

is nowhere

dense in X iff the interior

of the closure of A is empty, that is, (A)° = 0. (This is the standard definition.)

Ex. 6.3.6. Use the definition of nowhere dense sets in Ex. 6.3.5 to solve Ex 6.3.2-Ex, 6.3.4.

Theorem 6.3.7 (Baire Category Theorem). Let (X,d) be a complete

metric space.

. (1) Let (Un)nen be a sequence of open dense subsets of X.

¥ nonempty,

Then NnUn

such (2) Let (Fr)nen be a sequence of nonempty closed subsets of X In = UnF,. Then at least one of Fn’s has nonempty interior.

union of ther words, a complete metric space cannot be a countable

dense closed subsets.

A

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IC sp4c, TR ME TE LE MP CO 6. R TE CHAP

146

are equivalent. p the st atements -

observe e tha b ‘rstt observ X \G is closed and no. Proof. We firs Where th "s © f the theorem follows 8 f from iff i*8of CP se one and denany ent ope.n Hen tem is ense sta the . te the Proving Hence d : we confine ourselves to So, complements. ther by taking other f . first. (1). We shal of on versi ger stron @ In fact, we shall we shall prove . X. show that N,Up is dense in ey t U is dense in X. Let > tha ve pro to e hav We . Let U = 1Un Since y, .

7) : and r > 0 be given. We need to show that B(z,us 7 exists z € B(z, dense and B(z,r) is open there is open, there exists r, such that 0< 17 < 1/2 and

a 0.

ince B(x, rnu, Blz,71] c B(x, r) 7

open set B(xz1,71) and the dense U,. We repeat this argument for the

set Up to get zp € B(zi,71)

U2. Again, we can find rz such tha

0 < rp < 2-2 and Blze,ra] C B(z1,71) U2. See Figure 6.5. Proceeding

this way, for each n € N, we get zn € X and an r, with the properties B[zn,Tn]

C

Un and

NM

B(tn-1:Tn-1)

0


x €0 X is an isolated point in the metric space such that B(z, r) A(X

Let (X.d) be compact.

infinite?

\ {z}) =0.

Can the set of isolated points be countably

Ex. 6.3.12. Show that any countable complete metric space has an isolated point. Ex. 6.3.13. Let X be an infinite dimensional complete normed linear a Show that X cannot be countable dimensional.

Ex. 6.3.14. Use Baire’s theorem to show that R isi uncountablele. ' Ex, 6.3.15 (Uniform Boun lete dedness Principle). Toei fmetiona ee X.

Soo en ns & & temrily of combines a ee tho t (fall C,

Assume that for all x € X, there exi sts C, > 0 suc set U C X anda lor all f © F. Then there

exists “nstant C such that lf(z)| < C a OT the faced ve U. for all f €

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N

CHAPTER 6. COMPLETE METRic SPAcp

148

application of Baire’s theorem The next result. is a beautiful . ty,Uch rspective, recall tj ePoin cont;then.. not be that uses both the versions! To put it In persP ne d Whi

.

Ntinuo

Wh wild to, be ot cann t the uniform limit is. However, the pointwise limi

ions limit of a sequence of continuous functSe

eedaes

Theorem 6.3.16. Let X be a complete metric space. Let fy: x _,) ’ a sequence of continuous functions. Assume that there ezists g Functog

f:X + R such that fa(z) converges to f(x) for each x € x, The

there exists a dense subset D of X such that each point of D is , Poin of continuity of f.

Proof. Fix ¢ > 0. Define, for eachk € N,

Ey(e) := {2 € X : |fn(z) — Sm(z)| S$ &, for all m,n > ky}. Then we claim that E;(e) is closed for each k. Reason:

Fix m,n

> k.

Then the set Ey/"(e)

:= {re

x.

\fn(z) — fm(x)| < €} is a closed subset of X, since |f, — fra| is continuous. Now, since Ex(€) = Omn>eE,’”(€), the claim follows.

It is easy to show that X = UL, E,(e). Reason:

Let zp € X. Since fn(z0) + f(Zo), the sequence (fn(z0))

is Cauchy. Hence for the given < > 0, there exists ko such that for m,n > ko, we have |fm(x0) — fn(zo)| < ¢. Hence we conclude that

Zo € Ex, (€).

Since X is a complete metric space, at least one of F;,(€) should have

nonempty interior.

open subset of X.

Let U, := U,Int (Ex(e)).

Then U, is a nonempty

Let U, := Uj/,. We claim that each Up is dense in X. Reason: It is enough if we show that every closed ball B := B[z,r] has non-empty intersection with U,. (Why?) Observe that the closed set (and hence a complete metric space) B is the union of a

countable family of closed sets: B = Un(BNM E,(1/n)). By Baire

category theorem, at least one of them has nonempty interior, say, Int (BN E,(1/n)) 4 0. Since Int (BN Ey (1/n)) Cc BN Int Ex(1/n),

it follows that B[x,r] NU, # 0 and hence the claim is proved.

Let D :=,Un. By Baire category theorem, D is dense in X. We claim that every z € D is a point of continuity of f.

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‘3 BAAl

pe CATEGORY THEOREM

n: Fix p € D. Let € > 0 be given Choose JN Osuch that B(z,r) = {zx}.

Ex. 1.2.38: Q is not open in R by the observation made in the solution of Ex. 1.2.36. A better argument runs as follows: If Q is open and if x € Q, then there exists r > 0 such that (x — r,x +r) C Q. But by the density of irrationals, there exists an irrational number y such that r—-r 0, then Bif,r) C E.

Ex. 1.2.59: If f € B,(0,1), let ¢:=1-— Jo \f(t)| dt. Then B.(f,e) c B,(01)., Ex. 1.2.60: Draw pictures for the kind of functions mentioned in the hint. You will be able to write explicit formulas for such functions. It is most important to understand the hint than writing explicit formulas

for functions and do a routine verification. A little more explicitly, how about a function whose graph is a tri-

ingle with a base (1/2 —1/N,1/2+1/N) and a peak M > Oat x = 1/2 tnd zero outside the interval (1/2 — 1/N,1/2 +1/N)? One such function is:

{=

0

fO [et - 22| i=]

i=1

k

= Vles— atl + lle” — 2a. 1=1

The argument uses curry leaves trick.

Ex. 2.4.2: If EC B(a,r) and z € X, for any z € E, we have d(z,z)
0, by the definition of density, we

of a set. Use the last exercise and the definition of closure

2.5.6:

Ex. 2.5.8: Observe that {mV/2: m € Z} is a closed set. Ex. 2.5.9: Recall that finite sets are closed in metric spaces.

the Ex. 2.5.10: Any subset of a discrete metric space is closed. Hence only dense subset of such a space is the whole space. Ex. 2.5.11: Let p € X. Then p is not a limit point of A = X \ {p}Why and what does this mean? Ex. 2.5.12:

AUB.

(a) If any nonempty open set meets A, certainly it will meet

(b) Consider A = Q and B its complement in R.

Ex. 2.5.13: Let U be any nonempty open set. Observe UN (AN B) = (UN A)

B. Use the density of B.

Ex. 2.5.14: How about R minus a (nonempty) finite set? Or, R \ Z?

Ex. 2.5.15: Weierstrass approximation theorem says that the subset P

of all polynomial functions is dense in (C[0, ’ 1],|| ’

||.) oo

/*

Al

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-3. COMMENTS ON EXERCISES IN CHAPTER 3

171

Ex. 2.6.3: Given D such that DN B # @ for each Bc B, let V bea nonempty open set. If x € V, there exists B. € B such thatre Bc

V.

(Why?) Hence @4B,NDCVND

Ex. 2.6.4: Observe that B((z,y),¢) = B(z,e) x Bly,¢).

Ex. 2.7.3: Ifx ¢ {0,1}, then z € U := (—cc,0) U(0,1) U (1,00). Since

U is open,

there exists € > 0 such that

Ex.

OA=A.

(x — ze,r + €) C

U.

Hence

x

lies in an open set which either has no intersection with A or with the complement of A. Thus, z cannot be a boundary point of A. 2.7.4:

Ex. 2.7.5: Given u with |u| = 1 note that any B(u,r) will contain tu and (1 —t)u for all sufficiently small t > 0. If x is not of unit norm, then z € B(0,1) U {y: |ly|l > 1}, union of two open sets. This shows (how?)

that x cannot be a boundary point of B(0, 1).

For the last part, consider a discrete metric space.

Ex. 2.7.6: If A C B(0,1) is a finite subset, then 0(B(0,1) \ A) = {x : ||z|| = 1}UA. Draw pictures and see how you can turn that into a proof. Ex. 2.7.7: Did you draw a picture? The boundary is {z : |z| = 1} U {t : 0 Ti

for each i. Ex. 3.1.7: Recall Ex. 2.1.10 on Page 37. Ex. 3.1.8: Same hint. as for the last exercise. Ex. 3.1.11: All that is needed is already spelled out in the exercise! Ex. 3.1.13: Refer to Ex. 1.1.38.

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172

CHAPTER

7.

ANSWERS,

HINTS AND SOLUTIONg

Ex. 3.1.14: Consider f(z) = z/|z|. Why is f continuous? Ex. 3.1.17: Ex. 2.1.9 on Page 37 may be of use.

Ex. 3.1.18: If z, + z, we need to show that y(tn) — (7). Recay Ex. 2.1.7 (page 37) and use the continuity of f and g.

Ex. 3.1.19:

You need Ex. 2.1.10 on Page 37

Ex. 3.1.20: Composition of continuous functions.

Ex. 3.1.21: When n = 2, the map is something like (a,b, c,d) (a, c.b,d) from R¥ to itself. If Ay = (af;) + A = (aij, then Al = (4,.) where bf; = ak.. Recall Ex. 2.1.15 on Page 38. Ex. 3.1.22:

If A, — A, we need to show that A? — A?.

In viey of

Ex. 2.1.15, this happens iff the ij-the entry of A? converges to the ij-th entry of A. But the 1j-th entry of A? is a polynomial in the entries of Ax. Perhaps, you should work out the case when n = 2 explicitly ty understand what is going on.

Ex. 3.1.23: Given a continuous function f: X — R, define a sequence

(zn) by setting rz, := f(1/n) and f(0) = x. By continuity of f at 0. it follows that f(1/n) :=z, > x = f(0). Conversely, given a convergent sequence (z,,) converging to z, define

f: X — R as above. Ifa # 0 in X, then any sequen ce (a;,) in X converging to a is eventually a constant. (Why?) Hence f is continuous

at any a # 0.

If = 0a and a, = 1/n, — a, then, given¢ > 0, choose

k, such that for k > k,, we have |a,| < €. Similarly, there exists ke such that |zn, — z| < € for k > ko. Let ko = max{k,,k 2}. Then for k > ko,

we have

ltn, — 2| < €, ie.

[f(1/ng) — £(0)| < e,ice.

f(a.) — f(a)| x such that f(U) CV. Thus, we see that

9(f(U)) CW. That is, (go f)(U) c W. Ex. 3.2.4 is modified as follow

8: Iff: X > R is continuous at a € X with f(a) > 0, there exists an open set U 5 a such that f(x) > f(a)/? forz Ee U.

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73. COMMENTS ON EXERCISES IN CH APTER 4 Ex. 8.2.7: Let ref

j: X >

Y be continuous.

Let Vc

173 Y be open.

Let

and t is, open set (sUz ). 3 Sinr cesuchf isthacon (xr) U,EVc fy f(Uuou t tin ,) s c atV. x Tha there. existe ss anan

f-1(V) is the union of U, as x varies over f-1(y) Let f satisfy the condition '

open set with f(r) € V.

and r € X be

Since U :=

f-1

f(U) c V. (If ; you are looking for a

(V)

i.ven.

Bive enn. and Let

the basis By.

f-(V) is open.

VcJ

Hence

Y be

pen and & € U, clearly in the context

iso

spaces, there exists r > 0 such that B(x,Proof C U. y € fv) says that f(y) E V, that is (x,r)

Ex. 3.2.8: Let Vv CY

.

For an

» f(B(z,r)) c V.)

yy

of metric E Br(r,r),

be open. Then Vv = U;B; is a unio n of sets from

Since f-!(V) = f-1 (UiBi) = Uif~"B;, it follows that

Ex. 3.2.9: If B C Y is closed, then V :-= y \ B is open in Y. Hence f-1(V) is open. But, f-(Y\B) = X \ f~1(B). Hence f~'(B) is closed. The other way implication is also proved similarly.

Ex. 3.2.11:

Let us work out some of the old exercises.

Ex. 1.2.29: Consider the continuous function f: R? > R defined by f(x,y) = zy. Then the set {ry 4 0} is the inver se image under f of the open set R \ {0}. Ex. 1.2.58:

Consider the function y: C[0,1] > R given by y(f) =

f(0). Then y is continuous and the set E = py *(R\ {0}).

Ex. 1.2.102: Consider the continuous function f: R® = R given by f(z,y.z) = ax + by+cz—d. Then the plane is f~*(0) and hence is

closed.

Ex. 3.2.12:

An n x n matrix is invertible iff its determinant is nonzero.

Consider f: A1(n,R) — R given by f (A) = det A. Then the set of invertible matrices if f—!(IR\ {0}) and hence is open. (Why is f continuous?)

Ex. 3.2.13: The last hint should give you ideas.

Ex. 3.2.14:

Consider the maps (X,Y)

+

(X,Y‘) from M(n,R) x

M(n,R) + M(n,R) x M(n,R) and (X,Y) 4 XY from M(n,R) x M(n,R) > AJ(n,R). Their composition ¢ is (X,Y) +> XY‘. O(n,R) of orthogonal matrices is y~ (J).

The set

SO(n,R) is closed in M(n,R) since it is closed i O(n,R) (Why? For, | | det is continuous.) Ex. 3.2.15:

A well-known fact:

A matrix A € M(n, R) is nilpotent 7

A" = 0. Now the map f: M(n,R) > M(n,R) given by f(X) = X" is |

Continuous.

Ex. 3.2.16: Let (21,1) and (z2,y2) be given. We shall use the samme ation. symbol d for metrics on the three spaces under consider

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174

CHAPTER

7.

ANSWERS,

HINTS AND SOLUTIONg

context you should know which d is being referred to. It is part of the

exercise! We estimate |d(2y :y1)

~

d(x2, y2)|

< <
0 is given, we may take 0 < ¢/2. independent of the point under consideration.

Note that 6 jg

Ex. 3.2.17: Draw pictures. Let x € E. Choose open sets V and W in Y such that f(x) € V and g(z) € W and VOW = @. Then

reU:=f(v)nf-!(w). Ex.

3.2.18:

If f(p) # g(p), using the Hausdorff property,

there exist

disjoint open sets V and W containing f(p) and g(p) respectively. Now,

U, := f~"(V) and U2 := g~!(W) are open sets with p € U := U; N Up.

Since D is dense there exists q € DU.

This implies that f(9) = 9(q)

and hence f(g) = 9(q) € VN W, a contradiction.

Ex. 3.2.19: Show by induction that f(n) = nf (1) for any n € Z and then that f(m/n) = (m/n)f(1) for m/n € Q. Define g(x) = f(1)z. To show that f = g, you need the last exercise.

Ex. 3.2.20: Let V C be nonempty open. Then U = f —!(V) is nonempty (why?) and open. There exists x € DNU. Then f(x) € f(D)N F(U) =

DNV.

Ex. 3.2.21:

Is this related to Ex. 3.2.17?

Ex. 3.2.22: More generally, if we let f(z) := d(x,a), a € X fixed, then f is continuous:

f(z) — F(y)| = |d(x,a) - d(y,a)| < d(z,y). The expression for f is f(x) =

l-z

ifr1 The geometric meaning of f (x) is that it is the “distance of z to the

set [—1, 1].” Ex. 3.2.25: Can you arrive at the followin g expression of d4?

da(z)=

—z

ifz]

40

if x € (0,1).

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73, COMMENTS ON EXERCISES IN CHAPTER 3

175

Now it must be easy to draw the graph of d,! Ex. 3.2.26: dg(z) = 0 for z ER. Ex. 3.2.27: da((a,b)) = |b]. Ex.

3.2.28:

Draw

pictures.

If p =

(a,b)

#

0, the point on the cir-

cle nearest to p is p/||p||. Hence da(p) = |1— Va? ae

. The picture gives a geometric proof using Pythagoras theorem. Or, you may also use the Lagrange’s method of constrained extremum.

Since d((a,

b), (z,y)) = /o? + b? + 1 — 2(azx + by), you may wish to find the maximum of f(x,y) = ax + by

subject to the constraint g(x,y) = 2? + y2 — 1.

Ex. 3.2.29: For inspirations, look at Ex. 3.2.27. Let V = W @ W' be the orthogonal decomposition. wi + w2 in a unique way. If w € W is arbitrary, then

d(z,w)? = |lz-wl?

by Pythagoras theorem. d(z, W)

=

| w2 |.

=

=

Then any z =

((w, —w) + uw», (wv, — w) + wo)

|lw — wll? + [wel],

Hence d(zx, w) is a minimum iff w — w,. Hence ;

For inspirations, look at Ex. 3.2.27. Ex. 3.2.30: If dz(x) = 0, then for each n € N, there exists z, € E such that d(z,zp) N, we have z, € A. Thus;

(tn)n>Nn is a sequence in A converging to a. Since f: (A,d) 7 (% d)

is continuous, it follows that f(zn) +

f(a).

That is, f: X

7 Y 8

continuous at a.

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COMMENTS ON EXERCISES IN CHAPTER 3 Consider

f(z)

a

1

if

I

€

Q

and

f(x)

=

0

if

T

¢

Q.

_ The

sequence

z ¥2/n > 0, but f(r) does not conv erge to f(0). But f:Q3R us tant and hence is continuous.

js 8 con®

Bx 3.2.47: Similar to the last one. EX.

3.2.49:

Corisider

fi(zx)

=

—£

on

A,

=

(—>0, 0]

and

fo(x)

= zon Ay = (0,00). Clearly, A; is closed, f; is continuous on Aj,i=1,2. Also,

ANB.. The new function f is continuous on R. But {0} =function his the modulus j is agree

Ex. 3.2.50: If you have done the last exercise, this is similar. Ex, 3.2.51: This is similar to the last two exercises.

Ex. 3.3.4: Consider f(x) = c+ ££(2—a) from [a,b] to [c, d]. Can

you arrive at f geometrically? (The length b—a of [a, b] is ‘stretched /magnified’ (or ‘contracted’) to d — c of [c,d]. Thus the magnif ication factor is (d-c)/(o— a). If x € [a,b], the distance of x to a is x — a. When z is mapped to a point say y € [c,d] we wish that the distance y —cis magnified by the same factor and so on.) Ex. 3.3.8: Consider f(x) = 1/z. Ex. 3.3.9: Observe that (i) translations of an N LS are homeomorphisms,

and (ii) the ‘magnification maps’ z +> Az for 0 # X €R are homeomorphisms. Recall Ex. 3.3.8 and B(z,r) =x+rB(01).

Ex. 3.3.10: By Ex. 3.2.37, both the maps z++ Az and 2+> A7!z are

continuous.

Ex. 3.3.11:

Assume that the circle is given by xz? + y? = 1 and the

ellipse is given by (z/a)? + (y/b)? = 1. Think of an invertible 2 x 2 diagonal matrix A. Use the last exercise and Ex. 3.3.7. Or, use the standard parametrizations of these and set up an obvious map between them. If you are lucky, that is a homeomorphism. Ex. 3.3.12: How will you parametrize the parabola? Ex, 3.3.14:

Any subset of N is d-open as well as 6-open.

identity map is a homeomorphism. Ex, 3.3.15:

Are (—1,1) and R homeomorphic?

Ex. 3.3.12 will help us.

Hence the

Then Ex. 3.3.3 and

Ex. 3.3.16: Let D be a countable dense subset of a topological space X. Let f: X + Y be a homeomorphism. it is dense in Y thanks to Ex. 3.2.20.

Then f(D) is countable and

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CHAPTER 7. ANSWERS, HINTS AND SOLUTIONS

Ex. 3.4.4: (a) Look at the estimate

nix —y|.

|

S

|


0, choose 0 < 6 < 1.

Ex. 3.4.8: False. Consider N with the discrete metric and f:NOR defined by f(k) =k.

Ex. 3.4.9: Let A C R™ be bounded. If f(A) is not bounded, then there exists a sequence (a,) in A such that |f(a,)| > k for k € N. Since (ax) is bounded, by Bolzano-Weierstrass

subsequence, say, a;, + a € R™. hence (f (ax,)) is bounded.

theorem,

there exists a convergent

By continuity,

f (ax,) —

f(a) and

Ex. 3.4.10: Given a Cauchy sequence (z,,), we need to show that (f(tn)) is Cauchy. If € > 0 is given, let 56 > 0 be chosen by uniform continuity of f. There exists N such that for m,n > N, we have d(tm,In) < 6. Hence d(f(zm), f(tn)) < € for msn > N. The converse is not true. Consider f: R — R be defined by f(z) =

x*. Let (tn) be Cauchy. Then z, € [-M, M] for some M > 0. f is uniformly continuous on [—Af.M] by Example 3.4.3. (We can also see this as follows.

On this interval, f is differentiable with lf'(x)| < 2M.

By (c) of Ex. 3.4.4, f is uniformly continuous on this interval.) Hence it

carries the Cauchy sequence (z,,) to a Cauchy sequence by the first part. Ex. 3.4.11: Let « > 0 given. Let 6 correspond to the unifor m continuity

of f. Then there exists N € N such that d(z,sn) < 6 forn > N. Hence d(f (zn), f(sn)) N. To see the converse, we argue by contradiction.

Assume

that for a

given € > 0, no 6 can be found. For each 6 = 1/n, we can find In, Sn € X

such that d(rp,8n) < 1/n but d(f(7n), f(sn)) > €. Now go ahead and

complete the proof.

Ex. 3.4.15-Ex. 3.4.17: Straight forward.

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4. COMMENTS ON

EXERCISES IN CHAPTER 4

179

Ex. 3.5.3: This is an exploratory exercise for your personal development

and so I refrain from giving you a hint!

Ex. 3.5.4: f is continuous at a iff limzg f(x) = f(a).

7.4

Comments on Exercises in Chapter 4

Ex. 4.1.5: Understand the arguments of Example 4.1.4. Both the problems can be dealt with in a similar way.

Note that (5 — 5n,3 + dn) = (1/n,1 — (1/n)). Use Archimedean property to find k,m such that 1/k < z and 1/m < 1 ~ z for r € (0,1). If n is the maximum of k,m, then z € (1/n,1 — (1/n)).

Hence the sets

form an open cover of (0,1). If this admits a finite subcover, this means

that (0,1) Cc (1/N,1 —1/N) for N, which is absurd.

Ex. 4.1.7: (a) Fix r > 0. {B(z.r): 2 € X}. (b) Fix re X. {B(z,n): n€ N}. (c) {X \{z}: re X}. Ex. 4.1.8: Given € > 0, let 6, correspond to ¢ and the continuity at zé€ X. Then {B(z,6,) : x € X} is an open cover. Ex. 4.1.9: If 0 € U of an open cover, then there exists (—e,¢) C U. Hence all but finitely many 1/n € (—e,¢€). The remaining finitely many can be covered by a finite number of open sets in the given cover.

Ex. 4.1.10: How about {{n} : n € Z}? Or, look at {Un : n € N}, where Un, := {k EZ: |k| 1. Then {z": n € N} is an unbounded subset

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180

1}.

CHAPTER

7.

ANSWERS,

HINTS AND SOLUTIONS

Argue as above replacing x with |z| to conclude H c {z € C*: |z| =

Ex. 4.1.23:— Ex. 4.1.25: do not require any more hints. Ex. 4.1.26: By Ex. 1.2.112, the set E is closed in Q. It is also bounded

in Q. (Why?)

The open cover (—/2 — 1, /2+ 4) :n€N}

is an open

cover of E. This cover does not admit a finite subcover. To see this use Archimedean property to see that EF Cc (-/2 — v: V2+ W) for some N. Now use the density of rationals to arrive at a contradiction.

Ex. 4.1.33: in R".

(a) Compact, since the set is closed (why?)

(b) is not compact, since it is not bounded.

compact,

since it is not bounded.

bounded (why?)

(Why?)

in R? and hence is compact.

and bounded

(Why?)

(d) is closed

(c) is not

(why?)

and

(e) A conic section in R2

is bounded and closed iff it is cither empty, a circle or an ellipse. These

are the only compact conic sections. (f) is compact.

(g) a nonzero vector

subspace is not bounded (why? have we seen this earlier?) and hence is not compact.

©

Ex. 4.1.34: Not bounded and hence not compact.

Ex. 4.1.35: O(n.R) is earlier shown to be closed and bounded in M(n,R) = R”. (Ex. 2.4.11 and Ex. 3.2.14). Hence O(n,R) is com-

pact.

Ex. 4.1.36: None of GL(n,R), SL(n,R) and the set of nilpot ent matrices are bounded and hence not compact. Ex.

4.2.3:

If K C R” is not compact, either it is not bounded or it is

not closed. Assume that it is not bounded. Fix p € K.. This means that for any r > 0, K is not a subset of B(p,r). Hence the continuous function f(x) := d(x,p) is unbounded. If K is not closed, then a limit

point, say, a € R” does not lie in K. The function g(x) := 1/d(z, a) is not bounded on K.

Ex. 4.2.4: Please see my article “Characterization of Compact Metriz-

able spaces via metrics and continuous functi ons” available at http: //mtts.org. in/expository-arti cles/ Ex. 4.2.7: f attains its infimum at some point a € X. Hence f(z) > f(a) for all z € X.

Ex. 4.2.8: If {V; : i © J} is an open cover of Y, then {U;} is an open cover of X where U; = f~'(V;). If {Ui,:l 0, there exists ko € N such that d(x,, 21)

0° =

(3:

where x,

eS

(z

»Le2:°°"?

> ko.

(7.2)

=

eZ

1/2 IZkn

= an!

< E,

for

k,l

n=]

This shows S f lor any fixed n, the sequence (Zkn)Z2, is Cauchy and hencé Oe an In € R. Since Tren € [0,1/k] for all k, the limit zn © ,1/k]. Hence x = (Zn) lies in the Hilbert cube.

a

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185

75, COMMENTS ON EXERCISES IN CHAPTER 5 We claim that x, — z in X.

The proof is similar to the one in the

To prove total boundedness:

Let « > 0 be given. Let N be such that

hint for Ex. 2.3.19.

9. ko? < €/2. Consider K = {x € X : tp = 0,n > N}.

Since K

is homeomorphic to the compact set TI, (0, 1/k] c R¥, K is compact. Hence there exist an €/2-net {z1,... ,lm} for K. It is now easy to see

|

that X = Uf. B(a;,€).

7.5

Comments on Exercises in Chapter 5

Ex. 5.1.4: Fix z € X. Then {x} is a proper nonempty open set. Ex. 5.1.5: A finite subset A of a metric space is connected iff it is a singleton. IfA = {a),...,an} with n > 2, let r := min{d(a;,a;) : t > 2}. Then B(a,,r)M A = {a;}. Thus {a} is a nonempty proper open subset of A. Ex. 5.1.9: Let f: X — {+1} be continuous. Fix p € X. We may assume f(p) = 1. If x € X, there exists a connected subset A 35 z,p. Hence f is a constant on A and hence f(z) = 1. Since z is arbitrary, we conclude that f is a constant. _ Ex. 5.1.10: If J is any interval, and if x,y € J, then [z,y] C J. know that [z,y] is connected. Apply the last exercise.

Ex. 5.1.13:

We

Consider the continuous map f: (0,27] — R? given by

f(t) = (cost,sint). The image is the unit circle.

Ex. 5.1.14: (a) Consider f(x,y) = z/ |x|. Then f is continuous and its image is +1.

(b) Same as in (a). But why does f make sense? (c) Consider f(z,y) = y/|y|. (Why does f make sense?) Then f

continuous and its image is +1.

is

Ex. 5.1.15: The continuous map f(A) := det A/(|det A]) has both +1 as its values. (Prove this.)

Ex. 5.1.16: If A is orthogonal, AA‘ = J so that det(AA') = 1. Since

det(AA‘') = det(A)?, we see that detA € {+1}.

Produce orthogonal

matrices A and B such that det A = 1 and detB = -1.

Ex. 5.1.17: Any element of SO(2) is of the form (ort .

sint

a

cost

‘)

You

can think of an obvious continuous from the unit circle in R? onto SO(2). Ex. 5.1.18: The map y: X — X x Y given by (zr) = (z, f(z)) is continuous. (Why?) What is its image?

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CHAPTER

186

/7.

ANSWERS,

HINTS

AND

SOLUTIONS

subsets of R are inter. ted nec con , For on. let sing y) Cc 4 Ex. 5.1.21: It must be a say, 7, Y € A, then (z, al, ion rat one n tha e vals. If A contains mor ationals. due to the density of irr But (z, y) will have irrationals image wij] ction f existed, then its fun a h suc if For No. : .22 Ex. 5.1 one point. ge wl ll have more than ima The R. of set sub ted nec con a be uncount able. (Why?) But the image (Why?) Hence the image must be is countable. (Why?) imum, say, Af and minEx. 5.1.25: We know that f assumes the max ,6)) c imum, say, m on [a,b]. Then f({a,6]) = [m, M}. Clearly, f({a

orem, [m, M]. The other way inclusion is by the intermediate value the

since M = f(x) and m = f(x2) for some 71,22 € (a, db}.

Ex. 5.1.26: Consider h = g — f. Apply intermediate value theorem. Ex. 5.1.28: one.

Solve the next exercise.

See how

it applies to the present

Ex. 5.1.29: You need to apply Theorem 5.1.8. Ex. 5.1.30:

We offer two hints, both based on geometric thinking.

(i)

Given any two points v,w € X := R*\{(0,0)}, consider the line segment

[v,w] :-= {(1 —t)v+tw:0 0, there exists ko such that | 2* —2||.. < €/3. particular, for a fixed n, we have zn~p

—

In.

We

want

In

to estimate

Ita — Zml:

IZn — Lm| S |2n — Lkn| + |Lkn — Lem + |Lem — Lm -

The middle term on the right is less than ¢/3 for all k > k;. (Why?) If

we choose k > max{ko, ki}, then the result follows. (Note that k is then

curry leaf here.)

Ex. 6.1.8: If f is an isometry and (z,) is Cauchy, then (f(z,)) is Cauchy. If X is complete, then z, — x for some x. Hence f(z,) > f(z). Ex. 6.1.9: The map f: ((0,1],6) — ({1, 00), d) is an isometry. Ex. 6.1.11: This is just an observation.

Nothing to prove.

Ex. 6.1.12: If (y,) is Cauchy in Y, let «, € X be the unique pre-image

of y. Then d(tm,2n) < d(ym; Yn) Shows that (zp) is Cauchy in X. Ex. 6.1.14:

We repeat the hint.

of Theorem 6.1.13.

Go through the proof of (b)

=

(c)

Ex. 6.1.19: Assume that (X,d) is a metric space in which every Cauchy Sequence is eventually constant. Then any subset A of X is closed. Hence

Sn’s of the exercise are closed.

What is the diameter of S,? What is the hypothesis on the diameter

of F,,’s in the Cantor’s intersection theorem? If you understand this hint, you can think of familiar examples which

illustrate the point of the exercise. For instance N with discrete metric

or N with the metric induced by the inclusion N c R.

Ex. 6.1.20: Routine.

If (an, yn) is Cauchy in the product metric, then

Tn) and (y,) are Cauchy in their respective spaces.

Ex, 6.1.21: If (z,) is Cauchy with distinct terms and which is not “Onvergent, consider A := {x2n41 : k € N}. What is your choice of B?

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CHAPTER

190

ANSWERS,

7.

Show that A and B are closed. d(A,

B)

—_

Using the fact (tn) is Cauchy, show that

0.

Ex. 6.1.23:

Yes.

Ex. 6.1.24:

No.

The proof is similar to the proof of completeness of

the Hilbert cube and other sequence spaces @.

Lkn

=

SOLUTIONS

HINTS AND

= 1/k for n < k and

Consider x* € coo defined by Ten

Oifn>k.

Ex. 6.1.25: It is similar to the way we induced a metric 6 on (0, 1] using the bijection f(z) = 1/z from (0, 1] onto [1, 00). Show that the sequence (n) is Cauchy but not convergent. Ex. 6.1.26: Convergence in (IRN, d) is the same as coordinatewise con-

vergence as in the finite dimensional R”.

Ex. 6.1.27: If (xn) is Cauchy, let A := {mn : nN € N}. Then A is a closed

and bounded set.

Ex. 6.1.28: Note that (—1,1) is homeomorphic, say, via f, to R. Using the bijection, we may transfer the standard (absolute value) metric d on R to get a new metric 6 on (—1,1). The metric 6 and the usual metric d (the restriction of the metric on R) are equivalent. Since f is an isometry of ((—1, 1), 6) to (R,d), ((—1,1),6) is complete. But ((—1,1),d) is not. Ex. 6.1.29: Since C; ||z — y|l, < llc —yllp < Ca llz — yl), we see that (an) is Cauchy in || ||, iff it is Cauchy in || ||. If (zn) is convergent to x

in | ||,, the second part of the inequality shows the convergence in | ||9.

Ex. 6.2.2: For example, the translation r++ z + a for a fixed vector a. ‘Ex.

6.2.3:

Observe

that if 2,y

€ R” are column

vectors,

then the

Euclidean inner product is given by (z,y) = y'z, where we identify the

1 x 1-matrix with its entry. Hence (Az, Ay) = y' A‘ Ar.

Ex.

6.2.4:

Compute

(Ru(x)

— Ru(y), Ru(z)

— Ru(y)).

Ex. 6.2.6: Density of Q and R \ Q in R shows that the completion in each case is R.

Ex. 6.2.7: If f: X — Y

of X onto Y.

is the ‘completion’ ma

Pp;

then

;

f is an isometry

Ex. 6.2.10: No, for, the proof already uses the fact that R is complete.

Ex. 6.2.16:

Go through the argument of completeness of (X,d) in the

,

Cantor’s construction. Ex. 6.2.17:

If X is not complete, let f: X +

We shall identify X with f. Letae

Y \ X.

The,

aw

.

continuous function on X which takes only positive valen.

ones

(z,a) is &

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76. COMMENTS ON EXERCISES IN CHAPTER 6

191

Ex. 6.2.18: To show that X is compact, we need to show that X is

complete and totally bounded. Assume that X is not complete. Let the

notation be as in the last hint. Consider h = 1 /g. Then h is unbounded

and continuous on X.

If X is complete but not compact, then X is not totally bounded. Hence there is an € > O for which no finite e-net exists. Let r1€xX

be arbitrary. By induction construct zn) ¢ Ui, B(zi,e). Note that {B(zn,€/2)} is a pairwise disjoint collection. Draw pictures. Define —

’

2E B(a2n,€/ ’ 2) otherwise

‘6

fu(z) fn’s are continuous.

2nd(z,¢n ’ ), E

Consider f = }> fn. Note that for each x € X , at

most one summand f,(zx) (on the right) is nonzero.

Ex. 6.3.2: Given a nonempty open interval J, if it contains an integer

k, then (kK—e€,k+