Set Function T [1 ed.] 9783030650810

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Developments in Mathematics

Sergio Macías

Set Function An Account on F. B. Jones’ Contributions to Topology

Developments in Mathematics Volume 67

Series Editors Krishnaswami Alladi, Department of Mathematics, University of Florida, Gainesville, FL, USA Pham Huu Tiep, Department of Mathematics, Rutgers University, Piscataway, NJ, USA Loring W. Tu, Department of Mathematics, Tufts University, Medford, MA, USA

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Sergio Macías

Set Function T An Account on F. B. Jones’ Contributions to Topology

Sergio Macías Instituto de Matemáticas National Autonomous University of Mexico Ciudad de México, Mexico

ISSN 1389-2177 ISSN 2197-795X (electronic) Developments in Mathematics ISBN 978-3-030-65080-3 ISBN 978-3-030-65081-0 (eBook) https://doi.org/10.1007/978-3-030-65081-0 Mathematics Subject Classification: 54B20, 54C60, 54F15 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Elsa: ¿Cómo poder expresarte Todo lo que hay dentro de mí? Tu forma de ser me indica Que todavía hay gente hermosa. Tu presencia me motiva a seguir Luchando por lo que creo. En este momento me gustaría, además de ser lo que soy, Ser el más grande poeta Y así poder escribirte Los versos más bellos Que nunca nadie En la tierra escribiera Y decirte, de esa manera Muy hermosa, Todo lo que por ti siento. . . Te doy gracias porque eres Mi compañera.

S. M.

In Memoriam

Alfredo Macías María Alvarez James T. Rogers, Jr. Sam B. Nadler, Jr.

Preface

This book is timely. The literature on this subject has become so extensive that it is an overwhelming task for a new researcher to become familiar with what is known. It is nice to see a book organized so elegantly, which will make the subject more readily accessible. There are also a lot of opportunity for further work on the subject, so this book is much more than a complete history of the topic. Such a volume as that could not be written now, as there are too many gaps in our knowledge. I am not including any definitions in this Preface, as they can be found in the text itself. The notion of aposyndesis, introduced by F. B. Jones, dates back more than three quarters of a century, to the early 1940’s. This concept is a useful tool for the study of continua, especially those of intermediate complexity, that is to say, neither locally connected nor indecomposable. The set-valued set function T arises from the analysis of fine detail of when and how a continuum may be, or fail to be, aposyndetic. Professor Macías has provided this book just when it is needed. There is a substantial amount of information available in the literature on the set function T now, and this book provides a coherent and organized presentation of it, along with the most comprehensive bibliography I have seen on the subject. Both current researchers and students will find it to be a helpful reference book. It is also written in such a way that it can be used as a basis for a seminar, or possibly even a single individual, working alone, wishing to learn about the subject from scratch. Through a quirk of history, the set function T , in its earliest form, was called L and was also introduced by Professor Jones. The notation was mostly changed from L to T in the 1950’s and early 1960’s. Jones also introduced a companion set function K; the name of K has not changed. In conversation, at least, T and K have been referred to as dual functions or adjoint functions; I have encountered both terms, but never in print, I think. The function T has been the primary focus of research over the years; K typically has been studied only when its use sheds light on properties of T . The present volume continues that convention; I hope that it may lead to more study of K in its own right. The function T has been both a subject of research in its own right and, increasingly, a powerful tool for investigating problems that, at first glance, seem to have nothing to do with T . vii

viii

Preface

The first chapter provides general background in topology and specifically the theory of continua. The discussion of the function T begins in earnest in the second chapter. I am very glad to have had an opportunity to peruse this volume. I expect that others will also find it interesting and useful. Professor Emeritus of Mathematical Sciences The University of Delaware Newark, Delaware, USA

David P. Bellamy

Introduction

The main purpose of this book is to present Professor F. Burton Jones’ set function T . This topic was treated in [92, Chapters 3 and 5] for metric continua. Here, we present most of those results for Hausdorff continua throughout the book. Whenever we could not find the way to remove the metric hypothesis, the result is presented in its original form. We include new topics not contained in [92, Chapters 3 and 5], as we mention below. The set function T has been used by many authors, for example as a tool to prove results about the semigroup structure of continua [71, 72] and [62], about the existence of a metric continuum that cannot be mapped onto its cone [10] or to characterize spheres [34]. In fact, it seems that [72] is the first paper to use the notation T in print. Even though [71] appeared before [72], the latter was written before than the former. The book has eight chapters. In Chapter 1, we present the basic knowledge needed for the rest of the book on continuous decompositions, topological products and inverse limits (including generalized inverse limits with a single bonding function), uniformities, (Hausdorff) continua, uniformly completely regular maps and hyperspaces. The part on decomposition, continua and hyperspaces is brought from [92, Sections 1.2, 1.7 and 1.8] with the appropriate changes to Hausdorff spaces. Chapter 2 contains the main properties of the set function T . We consider its symmetry, additivity, idempotency and finite powers of it. In Chapter 3, we consider decomposition theorems using T . We present three general decomposition theorems. Also, we give a weak version of Professor F. Burton Jones’ Aposyndetic Decomposition Theorem for homogeneous Hausdorff continua using only the property of Kelley and then present the full version for such continua using the uniform property of Effros. We use the set function T to prove Professor Janusz R. Prajs’ Mutual Aposyndetic Decomposition Theorem for Hausdorff homogeneous continua using the uniform property of Effros and give some relationships between this and Jones’ theorem. We include the statements of most of the results of [92, Section 5.3] to prove Professor James T. Rogers, Jr.’s Terminal Decomposition Theorem for metric homogeneous continua.

ix

x

Introduction

Chapter 4 is about T -closed sets, a topic studied by several authors, for instance [46] and [122]. We present the main properties of this class of sets, a couple of characterizations of T -closed sets: one for Hausdorff continua and the other one for metric continua. A necessary condition to be a T -closed set is given, and we prove that this condition is also sufficient for the class of metric continua with the property of Kelley. We study the class of minimal T -closed sets and the set function T ∞ . We end with the concept of T -growth bound and give an upper bound of the T growth bound for the classes of homogeneous Hausdorff continua with the property of Kelley and type λ continua. We present a relationship of the T -growth bound of continua and monotone and open monotone maps. Chapter 5 deals with the continuity of the set function T . The main difference between the results presented in Section 5.1 and the ones in [92, Section 3.3] is that new results appeared after the publication of [92], and these allow us to remove the hypothesis of point T -symmetry of the continuum in many of the theorems. We present classes of metric continua for which T is continuous. In addition, we include results about the continuity of T on continua. In particular, we show the equivalence of the continuity of T and the continuity of its restriction to continua for the product of two continua. We end this chapter proving that the continuity of the set function T implies the continuity of Professor Jones’ set function K. In Chapter 6, we study the images under the set function T of the hyperspace of closed subsets, 2X , and the hyperspace of singletons, known as the first symmetric product, F1 (X), of a metric continuum X. We show that T (2X ) is an analytic set. We consider finite and countable images of T . We also study connected and compact images of T . Chapter 7 contains applications of the set function T . We study continuously irreducible continua and their hyperspace of subcontinua. Also, we characterize continuously type A metric θ-continua. We present sufficient conditions for continua to be not contractible. We prove that arc-smooth metric continua are strict point T -asymmetric and the reverse implication is true for fans. We consider Rsubcontinua in dendroids. We present two characterizations of local connectedness. We give sufficient conditions for the image under T of a nonempty closed subset of a metric continuum to be a shore set. We consider generalized inverse limits of nonaposyndetic homogeneous metric continua X using the set function T |F1 (X) as a bonding function. We end the chapter giving more relationships between the set functions T and K. Chapter 8 contains the questions from [92, Section 9.2]. We give an update on the advances in answering those questions and include new open ones. I thank Javier Camargo for letting me use some of the pictures he produced. I thank Professor David P. Bellamy for the valuable suggestions he made to improve the book. I really appreciate the time and effort he spent reading the whole book in order to write a very nice Preface. I thank my wife, Elsa, for all her love, help and support during the preparation of this manuscript.

Introduction

xi

I thank the people at Springer, especially Jan Holland, Robinson dos Santos, Elizabeth Loew, Christoph Baumann, Thomas Hempfling, Saveetha Balasundaram, Gomathi Mohanarangan and the Series Editors: Krishnaswami Alladi, Pham Huu Tiep, and Loring W. Tu, for all their help. Ciudad de México, Mexico

Sergio Macías

Contents

1

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Continuous Decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Topological Product and Inverse Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Uniformities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Uniformly Completely Regular Maps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Hyperspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References for Chapter 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 12 19 21 40 42 48

2

The Set Function T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Main Properties of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 T -Symmetry and T -Additivity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Idempotency of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Finite Powers of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References for Chapter 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 49 68 73 84 93

3

Decomposition Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Three Decomposition Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Jones’ Aposyndetic Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Prajs’ Mutual Aposyndetic Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Rogers’ Terminal Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Other Decomposition Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References for Chapter 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

95 95 97 104 110 114 119 122

4

T -Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Main Properties of T -Closed Sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Characterizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Minimal T -closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 The Set Function T ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 T -growth Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References for Chapter 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

123 123 130 134 135 139 143 xiii

xiv

Contents

5

Continuity of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Main Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Examples of Metric Continua for Which T Is Continuous . . . . . . . . . . . 5.3 Continuity of T on Continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Continuity of K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References for Chapter 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

145 145 157 161 164 166

6

Images of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 A Topological Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Finite and Countable Images of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Connected and Compact Images of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References for Chapter 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

167 167 168 178 182

7

Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Continuously Irreducible and Type A Continua . . . . . . . . . . . . . . . . . . . . . . 7.2 Noncontractibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Arc-Smooth Continua. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Local Connectedness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Shore Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Generalized Inverse Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 T and K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References for Chapter 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

183 183 191 193 199 204 205 206 211

8

Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 8.1 Questions About the Set Function T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

Chapter 1

Preliminaries

We gather some of the results of topology of Hausdorff spaces which are useful for the rest of the book. We assume the reader is familiar with the notion of topological space and its elementary properties. We present the proofs of most of the results; we give an appropriate reference otherwise. The topics reviewed in this chapter are: Continuous decompositions, product topology, inverse limits, generalized inverse limits, uniformities, compacta, continua, generalized metric continua and hyperspaces.

1.1 Continuous Decompositions We present a method to construct “new” spaces from “old” ones by “shrinking” certain subsets to points. 1.1.1 Definition. A compactum is a compact Hausdorff space. A map is a continuous function. 1.1.2 Definition. A decomposition of a topological Hausdorff X is a collection of nonempty, pairwise disjoint sets whose union is X. The decomposition is said to be closed if each of its element is a closed subset of X. 1.1.3 Definition. Let G be a decomposition of a Hausdorff topological space X. We define X/G as the set whose elements are the elements of the decomposition G. X/G is called the quotient space. The function q : X → → X/G, which sends each point x of X to the unique element G of G such that x ∈ G, is called the quotient map. 1.1.4 Remark. Given a decomposition of a Hausdorff topological space X, note that q(x) = q(y) if and only if x and y belong to the same element of G. We give a topology to X/G in such a way that the function q is continuous and it is the biggest with this property. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. Macías, Set Function T , Developments in Mathematics 67, https://doi.org/10.1007/978-3-030-65081-0_1

1

2

1 Preliminaries

1.1.5 Definition. Let X be a Hausdorff topological space, let G be a decomposition of X and let q : X → → X/G be the quotient map. Then the topology U = {Γ ⊂ X/G | q −1 (Γ) is open in X} is called the quotient topology for X/G. 1.1.6 Remark. Let G be a decomposition of a Hausdorff topological space X, and let q : X → → X/G be the quotient map. Then a subset Γ of X/G is open (closed, respectively) if and only if q −1 (Γ) is an open (closed, respectively) subset of X. 1.1.7 Definition. Let f : X → → Y be a surjective map between Hausdorff topological spaces. Since f is a function, Gf = {f −1 (y) | y ∈ Y } is a decomposition of X. The function ϕf : X/Gf → → Y given by ϕf (q(x)) = f (x) is of special interest. Note that ϕf is well defined; in fact, it is a bijection, see the diagram on Figure 1.1.

X

f

 q

−→ X/Gf



Y

ϕf

Fig. 1.1 The function ϕf

The next lemma is a special case of the Transgression Theorem [36, 3.2, p. 123]. 1.1.8 Lemma. Let f : X → → Y be a surjective map between Hausdorff topological spaces. If X/Gf has the quotient topology, then the function ϕf is continuous. −1 Proof. If U is an open subset of Y , then ϕ−1 (U ). Since q −1 ϕ−1 f (U ) = qf f (U ) = −1 −1 −1 −1 q qf (U ) = f (U ) and f (U ) is an open subset of X, we have, by the definition of quotient topology, that ϕ−1 f (U ) is an open subset of X/Gf . Therefore, ϕf is continuous.  

1.1.9 Example. Let X = [0, 2π) and let f : X → → S 1 , where S 1 is the unit circle, it be given by f (t) = exp(t) = e . Then f is a continuous bijection. Since Gf is, “essentially,” X, it follows that X/Gf is homeomorphic to X. On the other hand, X is not homeomorphic to S 1 , since X is not compact and S 1 is. Therefore, ϕf is not a homeomorphism. 1.1.10 Definition. A map f : X → Y between Hausdorff topological spaces is said to be open (closed) provided that for each open (closed) subset K of X, f (K) is open (closed) in Y . The following theorem gives sufficient conditions to ensure that ϕf is a homeomorphism:

1.1 Continuous Decompositions

3

1.1.11 Theorem. Let f : X → → Y be a surjective map between Hausdorff topological spaces. If f is open or closed, then the map ϕf : X/Gf → → Y is a homeomorphism. Proof. Suppose f is an open map. Since ϕf is a bijective map, it is enough to show that ϕf is open. Let Γ be an open subset of X/Gf . Since ϕf (Γ) = f q −1 (Γ), ϕf (Γ) is an open subset of Y . Therefore, ϕf is an open map. The proof of the case when f is closed is similar.   Decompositions are also used to construct the cone and suspension over a given space. 1.1.12 Definition. Let X be a Hausdorff topological space and let G = {{(x, t)} | x ∈ X and t ∈ [0, 1)} ∪ {(X × {1})}. Then G is a decomposition of X × [0, 1]. The cone over X, denoted by K(X), is the quotient space (X × [0, 1])/G. The element {X × {1}} of (X × [0, 1])/G is called the vertex of the cone and it is denoted by νX (Figure 1.2).

q

Fig. 1.2 Cone over a space

A proof of the following proposition may be found in [36, 5.2, p. 127]. 1.1.13 Proposition. Let f : X → Y be a map between Hausdorff topological spaces. Then f induces a map K(f ) : K(X) → K(Y ) given by  K(f )(ω) =

νY ,

if ω = νX ∈ K(X);

(f (x), t),

if ω = (x, t) ∈ K(X) \ {νX }.

1.1.14 Definition. Let X be a Hausdorff topological space and let G = {{(x, t)} | x ∈ X and t ∈ (0, 1)} ∪ {(X × {0}), (X × {1})}. Then G is a decomposition of X × [0, 1]. The suspension over X, denoted by Σ(X), is the quotient space (X × [0, 1])/G. The elements {X × {0}} and {X × {1}} of (X × [0, 1])/G are − + called the vertexes of the suspension and are denoted by νX and νX , respectively (Figure 1.3).

4

1 Preliminaries

q

Fig. 1.3 Suspension over a space

1.1.15 Definition. Let X be a Hausdorff topological space and let G be a decomposition of X. We say that G is upper semicontinuous if for each G ∈ G and each open subset U of X such that G ⊂ U , there exists an open subset V of X such that G ⊂ V and such that if G ∈ G and G ∩ V = ∅, then G ⊂ U . We say that G is lower semicontinuous provided that for each G ∈ G any two points x and y of G and each open subset U of X such that x ∈ U , there exists an open subset V of X such that y ∈ V and such that if G ∈ G and G ∩ V = ∅, then G ∩ U = ∅. Finally, we say that G is continuous if G is both upper and lower semicontinuous. 1.1.16 Example. Let X = ([−1, 1] × [0, 1])∪({0} × [0, 2]). For each t ∈ [−1, 1]\ {0}, let Gt = {t} × [0, 1], and for t = 0, let G0 = {0} × [0, 2]. Let G = {Gt | t ∈ [0, 1]}. Then G is an upper semicontinuous decomposition of X (Figure 1.4).

x

Fig. 1.4 Upper semicontinuous decomposition

1.1 Continuous Decompositions

5

1.1.17 Example. Let X = ([0, 1] × [0, 1]) .   × [0, 1], let G1 = {1} × 0, 13 , let G1 = For each   2 t ∈ [0, 1), let Gt =1 {t} 2 {1}× 3 , 1 ,and for  each t ∈  3 , 3 , let G1,t = {(1, t)}. Let G = {Gt | t ∈ [0, 1]}∪ 1 2 {G1,t | t ∈ 3 , 3 } ∪ {G1 , G1 }. Then G is a lower semicontinuous decomposition of X (Figure 1.5).

x

Fig. 1.5 Lower semicontinuous decomposition

The following theorem gives us a way to obtain upper semicontinuous decompositions of compacta. 1.1.18 Theorem. Let f : X → → Y be a surjective map between compacta. If Gf = {f −1 (y) | y ∈ Y }, then Gf is an upper semicontinuous decomposition of X. Proof. Let U be an open subset of X such that f −1 (y) ⊂ U . Note that X \ U is a closed subset of X. Hence, X \ U is compact in X. Then f (X \ U ) is a compact subset of Y . Thus, f (X \ U ) is closed in Y and y ∈ f (X \ U ). Hence, Y \ f (X \ U ) is an open subset of Y containing y. If V = f −1 (Y \ f (X U )), then V is an open subset of X such that f −1 (y) ⊂  \−1 V ⊂ U . Since V = {f (y) | y ∈ Y \ f (X \ U )}, V satisfies the required property of the definition of upper semicontinuous decomposition.   1.1.19 Remark. Let us note that Theorem 1.1.18 is not true without the compactness of X. Let X be the Euclidean plane R2 and let π : R2 → → R be given by π((x, y)) = x. ThenGπ is a decomposition of X which is not upper semicontinuous.

To see this, let U = (x, y) ∈ X x = 0 and y < x1 ∪ {0} × R. Then U is an open set of X such that π −1 (0) ⊂ U , whose boundary is asymptotic to π −1 (0). Hence, for each t ∈ R \ {0}, π −1 (t) ∩ (X \ U ) = ∅. The next theorem gives three other ways to think about upper semicontinuous decompositions.

6

1 Preliminaries

1.1.20 Theorem. If X is a Hausdorff topological space and G is a decomposition of X, then the following conditions are equivalent: (a) G is an upper semicontinuous decomposition; (b) the quotient map q : X → → X/G is closed;  (c) if U is an open subset of X, then WU = {G ∈ G | G ⊂ U } is an open subset of X; WU is a saturated open set;  (d) if D is a closed subset of X, then KD = {G ∈ G | G ∩ D = ∅} is a closed subset of X. Proof. Suppose G is an upper semicontinuous decomposition. Let D be a closed subset of X. By Remark 1.1.6, we have that q(D) is closed in X/G if and only if q −1 q(D) is closed in X. We show that X \ q −1 q(D) is open in X. Let x ∈ X \ q −1 q(D). Then q(x) ∈ X/G \ q(D). This implies that q −1 q(x) ⊂ X \ D. Therefore, since X \ D is open, by Definition 1.1.15, there exists an open set V of X such that q −1 q(x) ⊂ V and for each y ∈ V , q −1 q(y) ⊂ X \ D. Note that x ∈ V and q(V ) ⊂ X/G \ q(D). Thus, V ⊂ X \ q −1 q(D). Therefore, X \ q −1 q(D) is open, since x ∈ V ⊂ X \ q −1 q(D). Now, suppose q is a closed map. Let U be an open subset of X. Since q is a closed map, we have that q −1 (X/G \ q(X \ U )) is an open subset of X such that q −1 (X/G \ q(X \ U )) = WU . (If x ∈ q −1 (X/G \ q(X \ U )), then q(x) ∈ X/G \ q(X \ U ). Hence, q −1 q(x) ⊂ X \ q −1 (q(X \ U )) ⊂ X \ (X \ U )) = U . Thus, x ∈ WU . The other inclusion is clear.) Next, suppose WU is open for each open subset U of X. Let D be a closed subset of X. Then X \ D is open in X. Hence, WX\D is open in X. Since, clearly, KD = X \ WX\D , we have that KD is closed. Finally, suppose KD is closed for each closed subset D of X. To see G is upper semicontinuous, let G ∈ G and let U be an open subset of X such that G ⊂ U . Note that X \ U is a closed subset of X. Hence, KX\U is a closed subset of X. Let V = X \ KX\U . Then V is open, G ⊂ V ⊂ U and if G ∈ G and G ∩ V = ∅, then G ⊂ V . Therefore, G is upper semicontinuous.   1.1.21 Corollary. Let X be a Hausdorff topological space. If G is an upper semicontinuous decomposition of X, then the elements of G are closed. Proof. Let G ∈ G. Let x ∈ G and let q : X → → X/G be the quotient map. Since X is a Hausdorff topological space, {x} is closed in X. By Theorem 1.1.20, q({x}) is closed in X/G. Since q is continuous and q −1 q({x}) = G, G is a closed subset of X.   1.1.22 Theorem. If X is a compactum and G is an upper semicontinuous decomposition of X, then X/G is a Hausdorff space. Proof. To show that X/G is Hausdorff, let q : X → → X/G be the quotient map and let χ1 and χ2 be two distinct points of X/G. Then q −1 (χ1 ) and q −1 (χ2 ) are two disjoint closed subsets of X. Since X is normal, there exist two disjoint open subsets U1 and U2 of X such that q −1 (χ1 ) ⊂ U1 and q −1 (χ2 ) ⊂ U2 . Note that, since G is an upper semicontinuous decomposition, there exist two saturated open

1.1 Continuous Decompositions

7

subsets WU1 and WU2 of X (Theorem 1.1.20) such that q −1 (χ1 ) ⊂ WU1 ⊂ U1 and q −1 (χ2 ) ⊂ WU2 ⊂ U2 . Then q(WU1 ) and q(WU2 ) are two disjoint open subsets of X/G such that χ1 ∈ q(WU1 ) and χ2 ∈ q(WU2 ). Therefore, X/G is a Hausdorff space.   The next theorem gives a characterization of lower semicontinuous decompositions. 1.1.23 Theorem. Let X be a Hausdorff topological space and let G be a decomposition of X. Then G is lower semicontinuous if and only if the quotient map q: X → → X/G is open. Proof. Suppose G is lower semicontinuous. Let U be an open subset of X. We show q(U ) is an open subset of X/G. To this end, by Remark 1.1.6, we only need to prove that q −1 q(U ) is an open subset of X. Let y ∈ q −1 q(U ). Then q(y) ∈ q(U ), and there exists a point x in U such that q(x) = q(y). Since G is a lower semicontinuous decomposition, there exists an open subset V of X containing y such that if G ∈ G and G ∩ V = ∅, then G ∩ U = ∅. Hence, V ⊂ q −1 q(U ). Therefore, q is open. Now, suppose q is open. Let G ∈ G, let x and y be two elements G and let U be an open subset of X such that x ∈ U . Since q is open, V = q −1 q(U ) is an open subset of X such that G ⊂ V . In particular, y ∈ V . Let G ∈ G be such that G ∩ V = ∅. Then G ⊂ V . Thus, q(G ) ∈ q(U ). Hence, there exists a point u ∈ U such that q(u) = q(G ). Since q −1 q(G ) = G , u ∈ G . Thus, G ∩ U = ∅. Therefore, G is lower semicontinuous.   The following corollary is a consequence of Theorems 1.1.20 and 1.1.23: 1.1.24 Corollary. Let X be a Hausdorff topological space and let G be a decomposition of X. Then G is continuous if and only if the quotient map is both open and closed. The next theorem gives us a necessary and sufficient condition on a map f: X → → Y between compacta, to have that Gf = {f −1 (y) | y ∈ Y } is a continuous decomposition. 1.1.25 Theorem. Let X and Y be compacta and let f : X → → Y be a surjective map. Then Gf = {f −1 (y) | y ∈ Y } is continuous if and only if f is open. Proof. If Gf is a continuous decomposition of X, by Theorem 1.1.23, the quotient map q : X → → X/Gf is open. By Theorem 1.1.11, ϕf : X/Gf → → Y is a homeomorphism. Hence, f = ϕf ◦ q is an open map. Now, suppose f is open. By Theorem 1.1.18, Gf is upper semicontinuous. Since q = ϕ−1 f ◦ f and f is open, q is open. By Theorem 1.1.23, Gf is a lower semicontinuous decomposition. Therefore, Gf is continuous.   In the following definition a notion of convergence of sets is introduced. 1.1.26 Notation. The symbol N denotes the set of positive integers.

8

1 Preliminaries

1.1.27 Definition. Let {Xn }∞ n=1 be a sequence of subsets of the Hausdorff topological space X. Then: (1) the limit inferior of the sequence {Xn }∞ n=1 is defined as follows: lim inf Xn = {x ∈ X | for each open subset U of X such that x ∈ U, U ∩ Xn = ∅ for each n ∈ N, save, possibly, finitely many}. (2) the limit superior of the sequence {Xn }∞ n=1 is defined as follows: lim sup Xn = {x ∈ X | for every open subset U of X such that x ∈ U, U ∩ Xn = ∅ for infinitely many indexes n ∈ N}. Clearly, lim inf Xn ⊂ lim sup Xn . If lim inf Xn = lim sup Xn = L, then we say that the sequence {Xn }∞ n=1 is a convergent sequence with limit L = lim Xn n→∞ (Figure 1.6).

Fig. 1.6 lim inf and lim sup

1.1.28 Lemma. Let {Xn }∞ n=1 be a sequence of subsets of the Hausdorff topological space X. Then lim inf Xn and lim sup Xn are both closed subsets of X. Proof. Let x ∈ Cl(lim inf Xn ). Let U be an open subset of X such that x ∈ U . Since x ∈ Cl(lim inf Xn )∩U , we have that lim inf Xn ∩U = ∅. Hence, U ∩Xn = ∅ for each n ∈ N, save, possibly, finitely many. Therefore, x ∈ lim inf Xn . The proof for lim sup is similar.  

1.1 Continuous Decompositions

9

The next theorem tells us that second countable Hausdorff spaces behave like sequentially compact spaces using the notion of convergence just introduced. 1.1.29 Theorem. Each sequence {Xn }∞ n=1 of closed subsets of a second countable Hausdorff space X has a convergent subsequence. 1 ∞ ∞ Proof. Let {Um }∞ m=1 be a countable basis for X. Let {Xn }n=1 = {Xn }n=1 . m ∞ Suppose, inductively, that we have defined the sequence {Xn }n=1 . We define the sequence {Xnm+1 }∞ n=1 as follows: m ∞ m (1) If {Xnm }∞ n=1 has a subsequence {Xnk }k=1 such that lim sup Xnk ∩ Um = ∅, m+1 ∞ m ∞ then let {Xn }n=1 be such subsequence of {Xn }n=1 . m ∞ m (2) If for each subsequence {Xnmk }∞ k=1 of {Xn }n=1 , we have that lim sup Xnk ∩ m+1 ∞ m ∞ Um = ∅, we define {Xn }n=1 as {Xn }n=1 .

Since we have the subsequences {Xnm }∞ n=1 , let us consider the “diagonal subsem ∞ m ∞ quence” {Xm }m=1 . By construction, {Xm }m=1 is a subsequence of {Xn }∞ n=1 . m ∞ We see that {Xm }m=1 converges. m ∞ Let us assume that {Xm }m=1 does not converge. Hence, there exists p ∈ m m lim sup Xm \ lim inf Xm . Let Uk be a basic open set such that p ∈ Uk and m m ∞ m ∞ = ∅ for some subsequence {Xm } of {Xm }m=1 (lim inf Xn is U k ∩ Xm   =1 m ∞ a closed subset of X by Lemma 1.1.28). Clearly, {Xm }=k is a subsequence of k ∞ {Xnk }∞ n=1 . Thus, {Xn }n=1 satisfies condition (1), with k in place of m. Hence, k+1 m ∞ }m=k+1 is a subsequence of {Xnk+1 }∞ lim sup Xn ∩ Uk = ∅. Since {Xm n=1 and m k+1 m lim sup Xm ⊂ lim sup Xn , it follows that lim sup Xm ∩ Uk = ∅. Now, recall m m ∞ that p ∈ lim sup Xm ∩ Uk . Thus, we obtain a contradiction. Therefore, {Xm }m=1 converges.   1.1.30 Theorem. Let X be a sequentially compact Hausdorff space. If {Xn }∞ n=1 is a sequence of connected subsets of X and lim inf Xn = ∅, then lim sup Xn is connected. Proof. Suppose, to the contrary, that lim sup Xn is not connected. Since lim sup Xn is closed, Lemma 1.1.28, we assume, without loss of generality, that there exist two disjoint closed subsets A and B of X such that lim sup Xn = A∪B. Since X is a normal space, there exist two disjoint open subsets U and V of X such that A ⊂ U and B ⊂ V . Thus, there exists N  ∈ N such that if n ≥ N  , then Xn ⊂ U ∪ V . To show this, suppose it is not true. Then for each n ∈ N, there exists mn > n such that Xmn \ (U ∪ V ) = ∅. Let xmn ∈ Xmn \ (U ∪ V ) for each n ∈ N. The set {xmn }∞ n=1 is either finite or has a limit point. Hence, there exists a point x in X such that every open subset of X containing x contains infinitely many elements of {xmn }∞ n=1 . Note that x ∈ X \ (U ∪ V ) and, by construction, x ∈ lim sup Xn , a contradiction. Therefore, there exists N  ∈ N such that if n ≥ N  , then Xn ⊂ U ∪ V . Since lim inf Xn = ∅ and lim inf Xn ⊂ lim sup Xn , we assume, without loss of generality, that lim inf Xn ∩ U = ∅. Then there exists N  ∈ N such that if n ≥ N  , U ∩ Xn = ∅. Let N = max{N  , N  }. Hence, if n ≥ N , then Xn ⊂ U ∪ V and

10

1 Preliminaries

U ∩ Xn = ∅. Since Xn is connected for every n ∈ N, Xn ∩ V = ∅ for each n ≥ N , a contradiction. Therefore, lim sup Xn is connected.   1.1.31 Notation. If (X, d) is a metric space, p is a point of X and ε > 0, then Vεd (p) = {x ∈ X | d(p, x) < ε}. 1.1.32 Lemma. Let X be a compact metric space, with metric d, and let {An }∞ n=1 be a sequence of connected subsets of X such that lim sup An is not connected. Then, for each component L of lim sup An , there exists a subsequence {Ank }∞ k=1 of {An }∞ n=1 such that lim sup Ank = L. Proof. Let L be a component of lim sup An and let P = {pk }∞ k=1 be a countable dense subset of L. For each pk ∈ P , consider V d1 (pk ) and let Ank ∈ {An }∞ n=1 be k such that Ank ∩ V k1 (pk ) = ∅. Observe that P ⊂ lim sup Ank . Since lim sup Ank is closed (Lemma 1.1.28), we have that L ⊂ lim sup Ank . Let q ∈ lim sup Ank and let U be an open subset of X containing q. Thus, ∞ there exists a subsequence {Ankl }∞ l=1 of {Ank }k=1 such that q ∈ lim inf Ankl . It ∞ follows, from the definition of {Ank }k=1 , that there exists a point pkl ∈ P such that Ankl ∩ V d1 (pkl ) = ∅. Since X is a compact metric space, without loss of kl

generality, we may assume that the sequence {pkl }∞ l=1 converges to a point a of X. This implies that a ∈ lim sup Ankl . By Theorem 1.1.30, lim sup Ankl is connected. Since a and q both belong to lim sup Ankl , we have that lim sup Ankl ⊂ L. Thus, q ∈ L. Therefore, lim sup Ankl = L.   The next theorem gives us a characterization of an upper semicontinuous decomposition of a compact metric space in terms of limits inferior and superior. 1.1.33 Theorem. Let X be a compact metric space, with metric d. Then a decomposition G of X is upper semicontinuous if and only if G is a closed decomposition and for each sequence {Xn }∞ n=1 of elements of G and each element Y of G such that lim inf Xn ∩ Y = ∅, then lim sup Xn ⊂ Y . Proof. Suppose G is an upper semicontinuous decomposition of X. By Corollary 1.1.21, G is a closed decomposition. Let {Xn }∞ n=1 be a sequence of elements of G and let Y be an element of G such that lim inf Xn ∩ Y = ∅. Suppose there exists p ∈ lim sup Xn \ Y . Since Y is closed and p is not an element of Y , there exists an open set W of X such that p ∈ W and Cl(W )∩Y = ∅. Let U = X \ Cl(W ). Since G is upper semicontinuous, there exists an open set V of X such that Y ⊂ V and if G ∈ G such that G ∩ V = ∅, G ⊂ U . Let q ∈ lim inf Xn ∩ Y . Then q ∈ lim inf Xn ∩ V . Hence, V ∩ Xn = ∅ for each n ∈ N, save, possibly, finitely many. Thus, W ∩ Xn = ∅ for each n ∈ N, save, possibly, finitely many. This contradicts the fact that p ∈ W ∩ lim sup Xn . Therefore, lim sup Xn ⊂ Y . Now, suppose G is a closed decomposition and let Y be an element of G. Suppose that if {Xn }∞ n=1 is a sequence of elements of G such that lim inf Xn ∩ Y = ∅, then lim sup Xn ⊂ Y .

1.1 Continuous Decompositions

11

To see G is upper semicontinuous, let U be an open subset of X such that Y ⊂ U . For each n ∈ N, let Vn = V d1 (Y ). Suppose that for each n ∈ N, there exists an n element Xn of G such that Xn ∩ Vn = ∅ and Xn ⊂ U . For each n ∈ N, let pn ∈ Xn ∩ Vn . Since X is a compact metric space, {pn }∞ n=1 has a convergent subsequence {pnk }∞ . Let p be the point of convergence of {pnk }∞ k=1 k=1 . Note that p ∈ lim inf Xnk ∩ Y . Hence, lim sup Xnk ⊂ Y . For each k ∈ N, let qk ∈ Xnk \ U . Since X is a compact metric space, the ∞ sequence {qnk }∞ k=1 has a convergent subsequence {qnk }=1 . Let q be the point of convergence of {qnk }∞ . Note that q ∈ Y and q ∈ lim sup Xnk ⊂ lim sup Xnk , =1 a contradiction. Therefore, G is upper semicontinuous.   The following theorem gives us a characterization of a continuous decomposition of a compact metric space in terms of limits inferior and superior. 1.1.34 Theorem. Let X be a compact metric space, with metric d. Then a decomposition G of X is continuous if and only if G is a closed decomposition and for each sequence {Xn }∞ n=1 of elements of G and each element Y of G such that lim inf Xn ∩ Y = ∅, then lim sup Xn = Y . Proof. Suppose G is a continuous decomposition of X. By Corollary 1.1.21, G is a closed decomposition. Let {Xn }∞ n=1 be a sequence of elements of G and let Y be an element of G such that lim inf Xn ∩ Y = ∅. By Theorem 1.1.33, lim sup Xn ⊂ Y . Suppose there exists p ∈ Y \ lim sup Xn . Let U be an open subset of X such that p ∈ U and U ∩ lim sup Xn = ∅ (by Lemma 1.1.28, lim sup Xn is closed). Let q ∈ lim sup Xn ⊂ Y . Since G is a lower semicontinuous decomposition, there exists an open subset V of X such that q ∈ V and if G ∈ G and G ∩ V = ∅, then G ∩ U = ∅. Since q ∈ lim sup Xn and V is an open set containing q, V ∩ Xn = ∅ for infinitely many indexes n ∈ N. Hence, U ∩ Xn = ∅ for infinitely many indexes n ∈ N. Then U ∩ lim sup Xn = ∅, a contradiction. Therefore, lim sup Xn = Y . Now, suppose that G is a closed decomposition of X such that for each sequence {Xn }∞ n=1 of elements of G and each element Y of G such that lim inf Xn ∩ Y = ∅, then lim sup Xn = Y . By Theorem 1.1.33, G is upper semicontinuous. Let Y be an element of G. Let p and q be two points of Y , and let U be an open subset of X such that p ∈ U . For each n ∈ N, let Vn = V d1 (q). Suppose that for n each n ∈ N, there exists an element Xn of G such that Xn ∩Vn = ∅ and Xn ∩U = ∅. Hence, lim sup Xn ∩ U = ∅. For each n ∈ N, let qn ∈ Xn ∩ Vn . Clearly, {qn }∞ n=1 converges to q. Thus, q ∈ lim inf Xn ∩ Y . By hypothesis, Y = lim sup Xn . Hence, Y ∩ U = ∅, a contradiction. Therefore, G is a continuous decomposition.   1.1.35 Theorem. Let X be a compactum and let G be an upper semicontinuous decomposition of X. If D = {D | D is a component of G, for some G ∈ G}, then D is an upper semicontinuous decomposition of X. Proof. Note that D is a decomposition of X, and its elements are closed in X. Let D ∈ D and let G ∈ G be such that D is a component of G. Let U be an open subset of X containing D. Since X is normal, there exists an open subset U  of X such that U  ⊂ U and (Cl(U  ) \ U  ) ∩ G = ∅. Let U  = U  ∪ (X \ Cl(U  )) and let V

12

1 Preliminaries

be an open subset of X such that G ⊂ V and if G ∩ V = ∅, then G ⊂ U  . Let V  = V ∩ U  . If a component D of an element G of G intersects V  , then it lies in U  because G ∩ V = ∅. Since U  and X \ Cl(U  ) are disjoint open subsets and D is contained in their union, D ⊂ U  . Hence, D is upper semicontinuous.  

1.2 Topological Product and Inverse Limits We present basic properties of products of topological spaces, inverse limits of compacta and generalized inverse limits of compacta. 1.2.1 Definition. Let {Xλ }λ∈Λ be a family of nonempty subsets. Then its product is the set: 

 Xλ = f : Λ → Xλ f (λ) ∈ Xλ for all λ ∈ Λ . λ∈Λ

λ∈Λ

If λ0 ∈ Λ, then πλ0 : λth 0 projection map

 λ∈Λ

Xλ → → Xλ0 given by πλ0 (f ) = f (λ0 ) is called the

1.2.2 Remark. Given  a family {Xλ }λ∈Λ of nonempty subsets, the Axiom of Choice guarantees that λ∈Λ Xλ = ∅ if and only if Xλ = ∅ for each λ ∈ Λ. 1.2.3  Notation. If {Xλ }λ∈Λ is a family of nonempty subsets, then an element f ∈ λ∈Λ Xλ is denoted by {xλ }λ∈Λ , where xλ = f (λ). 1.2.4 Definition. Let{Xλ }λ∈Λ be a family of nonempty topological spaces. The product topology for λ∈Λ Xλ has as a base B=

⎧ n ⎨ ⎩

j=1

⎫ ⎬ U πλ−1 (U ) is open in X , j ∈ {1, . . . , n} and n ∈ N . λj λj λj j ⎭

1.2.5 Theorem. Let Z be a topological space and let {X λ }λ∈Λ be a family of nonempty topological spaces. Then a function f : Z → λ∈Λ Xλ is continuous if and only if πλ ◦ f is continuous for every λ ∈ Λ. n Proof. Suppose πλ ◦f is continuous for each λ ∈ Λ. Let j=1 πλ−1 (Uλj ) be a basic j  open subset of λ∈Λ Xλ . Since ⎛ f −1 ⎝

n  j=1

⎞ πλ−1 (Uλj )⎠ = j

n  j=1

f −1 (πλ−1 (Uλj )) j

1.2 Topological Product and Inverse Limits

=

n 

13

(πλj ◦ f )−1 (Uλj ),

j=1

  n −1 we obtain that f −1 π (U is an open subset of Z. Therefore, f is λ j j=1 λj continuous. The other implication is clear.   1.2.6 Theorem. Let {Xλ }λ∈Λ and {Yλ }λ∈Λ be families of nonempty topological spaces. Suppose  that foreach λ ∈ Λ,there exists a mapfλ : Xλ → Yλ . Then the function λ∈Λ fλ : λ∈Λ Xλ → λ∈Λ Yλ given by λ∈Λ fλ ({xλ }λ∈Λ ) = {fλ (xλ )}λ∈Λ is continuous.   Proof. For each λ0 ∈ Λ, let πλ0 : λ∈Λ Xλ → Xλ0 andπλ 0 : λ∈Λ Yλ → Yλ0 be the projection  maps. Let {xλ }λ∈Λ be an element of λ∈Λ Xλ and let λ0 ∈ Λ. Then πλ 0 ◦ λ∈Λ fλ ({xλ }λ∈Λ ) = πλ 0 ({f λ (xλ )}λ∈Λ ) = fλ0 (xλ0 ) = fλ0 ◦ πλ0 ({xλ }λ∈Λ ). Therefore, by Theorem 1.2.5, λ∈Λ fλ is continuous.   The following is Tychonoff’s Theorem, a proof of it may be found in [36, Theorem 1.4, p. 224].  1.2.7 Theorem. Let {Xλ }λ∈Λ be a family of nonempty spaces. Then λ∈Λ Xλ is compact if and only if Xλ is compact for each λ ∈ Λ. 1.2.8 Definition. Let Λ be a nonempty set with an order relation “≤” such that (1) λ ≤ λ for all λ ∈ Λ, (2) If λ1 , λ2 and λ3 belong to Λ and λ1 ≤ λ2 and λ2 ≤ λ3 , then λ1 ≤ λ3 , (3) If λ1 and λ2 belong to Λ, then there exists λ3 ∈ Λ such that λ1 ≤ λ3 and λ2 ≤ λ3 . Then Λ is called a directed set. A subset Λ0 of Λ is a cofinal subset of Λ provided that for each λ ∈ Λ, there exists λ0 ∈ Λ0 such that λ ≤ λ0 . 1.2.9 Definition. Let Λ be a directed set and let {Xλ }λ∈Λ be a family of nonempty topological spaces with Λ as its indexing set. If for each λ and α in Λ, with λ ≤ α, there exists a map fλα : Xα → Xλ such that: (1) fλλ = 1Xλ , where 1Xλ is the identity map on Xλ , and (2) whenever α, β and λ belong to Λ and λ ≤ β ≤ α, we have fλβ ◦ fβα = fλα . Then {Xλ , fλα , Λ} is an inverse system. 1.2.10 Definition. Let {Xλ , fλα , Λ} be an inverse system. The inverse limit of the inverse system {Xλ , fλα , Λ}, denoted by X∞ or by lim{Xλ , fλα , Λ}, is the set ←−

 {xλ }λ∈Λ ∈

 λ∈Λ

α Xλ fλ (xα ) = xλ , whenever λ ≤ α .

14

1 Preliminaries

The elements of the inverse limit are called threads and the maps fλα are called bonding maps. We give X∞ the relative topology from the product topology. 1.2.11 Notation. Let {Xλ , fλα , Λ} be an inverse system whose inverse limit is X∞ . If λ ∈ Λ, then the restriction of the projection map πλ to X∞ is denoted by fλ . 1.2.12 Lemma. Let {Xλ , fλα , Λ} be an inverse system and let λ0 and α be elements of Λ such that λ0 ≤ α. Then fλ0 = fλα0 ◦ fα . Proof. Let {xλ }λ∈Λ be an element of X∞ . Then fλα0 ◦ fα ({xλ }λ∈Λ ) = fλα0 (xα ) = xλ0 = fλ0 ({xλ }λ∈Λ ).   1.2.13 Theorem. Let {Xλ , fλα , Λ} be an inverse system whose inverse limit is X∞ . Then B = {fλ−1 (Uλ ) | λ ∈ Λ and Uλ is open in Xλ } is a base for the topology of X∞ . Proof. Clearly, B is a subbase for X∞ . This implies that the  family of finite n intersections of elements of B forms a base for X∞ . Let U = j=1 fλ−1 (Uλj ) j be a basic open set for X∞ . We show that U ∈ B. Since Λ is a directed set, there exists λ ∈ Λ such that λj ≤ λ for every j ∈ {1, . . . , n}. Let Uλ =

n  

fλλj

−1

(Uλj ).

j=1

Observe that ⎛ fλ−1 (Uλ ) = fλ−1 ⎝

n 

⎞ fλ−1 (Uλj )⎠ = j

j=1 n 

(fλλj ◦ fλ )−1 (Uλj ) =

j=1

n 

fλ−1



fλλj

−1

 (Uλj )

=

j=1 n 

fλ−1 (Uλj ) = U . j

j=1

Therefore, U ∈ B.

  {Xλ , fλα , Λ}

1.2.14 Lemma. Let be an inverse system of compacta whose inverse limit is X∞ . Given two elements λ0 and α of Λ such that λ0 < α, let Sλα0 =

  {xλ }λ∈Λ | xλ0 = fλα0 (xα ) . Then Sλα0 is closed in λ∈Λ Xλ and X∞ =



{Sλα0 | λ0 and α belong to Λ and λ0 < α}.

1.2 Topological Product and Inverse Limits

15

 Proof. Let {xλ }λ∈Λ ∈ λ∈Λ Xλ \ Sλα0 . This implies that xλ0 = fλα0 (xα ). Since Xλ0 is a Hausdorff space, there exist two disjoint open subsets Uλ0 and Wλ0 of Xλ0 such that fλα0 (xα ) ∈ Uλ0 and xλ0 ∈ Wλ0 . Since fλα0 is continuous, there exists an open subset Vα of Xα such that xα ∈ Vα and fλα0 (Vα ) ⊂ Uλ0 . Let U =  (Wλ0 ) ∩ πα−1 (Vα ). Then U is an open πλ−1 0  subset of λ∈Λ Xλ , {xλ }λ∈Λ ∈ U and U ∩ Sλ0 = ∅. Hence, Sλ0 is closed in λ∈Λ Xλ . Clearly, X∞ =



{Sλα0 | λ0 and α belong to Λ and λ0 < α}.  

1.2.15 Theorem. If {Xλ , fλα , Λ} is an inverse system of compacta whose inverse limit is X∞ , then X∞ is a nonempty compactum.  Proof. By Theorem 1.2.7, λ∈Λ Xλ is a compact space. Let S = {Sλα0 | λ0 and α belong to Λ and λ0 < α}. Then S is a collection of closed subsets (Lemma 1.2.14) of the compact space  λ∈Λ Xλ (Theorem 1.2.7). We show that S has the finite intersection property. Let n ∈ N and let Sλα11 , . . . , Sλαnn be elements of S. Since Λ is a directed set, there exists γ ∈ Λ such that αj ≤ γ for every j ∈ {1, . . . , n}. Let xγ ∈ Xγ and for each j ∈ {1, . . . , n}, let xλj = fλγj (xγ ) and xαj = fαγj (xγ ). Consider  the point {zλ }λ∈Λ ∈ λ∈Λ Xλ given by zλj = xλj , zαj = xαj for every j ∈ {1, . . . , n}, zγ = xγ and zλ ∈ Xλ for each λ ∈ Λ \ {λ1 , . . . , λn , α1 , . . . , λn , γ}. α α α Since fλjj (zαj ) = fλjj ◦ fλγj (zγ ) = fλjj ◦ fλγj (xγ ) = fλγj (xγ ) = xλj = n αj zλj , we have that {zλ }λ∈Λ ∈ j=1 Sλj . Thus, S has the finite intersection  α property. Hence, {Sλ0 | λ0 and α belong to Λ and λ0 < α}  = ∅ [36, Theorem 1.3, p. 223]. By Lemma 1.2.14, X is a closed subset of ∞ λ∈Λ Xλ and  α {Sλ0 | λ0 and α belong to Λ and λ0 < α} = X∞ . Therefore, X∞ is a nonempty compact space. The fact that X∞ is a Hausdorff space follows from [36, Theorem 1.3, p. 138]   1.2.16 Theorem. Let {Xλ , fλα , Λ} be an inverse system of compacta whose inverse limit is X∞ . Then the bonding maps are surjective if and only if the projection maps are surjective. Proof. Suppose that for every λ and α in Λ, with λ ≤ α, fλα is surjective. Let μ be an element of Λ.We show that fμ is surjective. Let xμ be a point of Xμ and let K = {xμ } × λ∈Λ\{μ} Xλ . Recall that, by Lemma 1.2.14, X∞ =  α {Sλ0 | λ0 and α belong to Λ and λ0 < α}. By the proof of Theorem 1.2.15, S = {Sλα0 | λ0 and α belong to Λ and λ0 < α} has the finite intersection property. We prove that S ∪ {K} also has the finite intersection property. Let  n ∈ N and n αj = ∅. Since let Sλα11 , . . . , Sλαnn be elements of S. We see that K ∩ S j=1 λj Λ is a directed set, there exists γ ∈ Λ such that μ ≤ γ and αj ≤ γ for all

16

1 Preliminaries

γ γ j ∈ {1, . . . , n}. Since  fμ is surjective, there exists xγ in Xγ such that fμ (xγ ) = xμ . Let {zλ }λ∈Λ in λ∈Λ Xλ be such that zγ = xγ , zμ = xμ and for each j ∈ {1, . . . , n}, zλj = fλγj (zγ ) and zαj = fαγj (zγ ). Clearly, {zλ }λ∈Λ belongs to K. α α Let j ∈ {1, . . . , n}. Note that fλjj (zαj ) = fλjj ◦ fαγj (zγ ) = fλγj (zγ ) = zλj . Hence,    α αj n {zλ }λ∈Λ belongs to Sλjj . Thus, {zλ }λ∈Λ is in K ∩ S λ∈Λ Xλ j=1 λj . Since  is compact (Theorem 1.2.7), {K} ∩ ( S) = ∅. Therefore, by Lemma 1.2.14, there exists {wλ }λ∈Λ in X∞ such that fμ ({wλ }λ∈Λ ) = xμ , and fμ is surjective. The inverse implication is clear by Lemma 1.2.12.  

1.2.17 Theorem. Let {Xλ , fλα , Λ} be an inverse system of compacta with surjective bonding maps, whose inverse limit is X∞ . If A is a nonempty closed subset of X∞ , then {fλ (A), fλα |fα (A) , Λ} is an inverse system with surjective bonding maps and  lim{fλ (A), fλα |fα (A) , Λ} = A = ←−



fλ (A) ∩ X∞ .

(∗)

λ∈Λ

Proof. Given two elements λ and α of Λ such that λ ≤ α, we know that fλα ◦ fα = fλ . Then fλα ◦ fα (A) = fλ (A). Thus, we have that fλα |fα (A) : fα (A) → fλ (A) is surjective. Hence, {fλ (A), fλα |fα (A) , Λ} is an inverse system with surjective bonding maps. Next, we prove the equalities are true. Let {xλ }λ∈Λ ∈ lim{fλ (A), fλα |fα (A) , Λ}. ←−

  λ0 and α Then {xλ }λ∈Λ ∈ λ∈Λ fλ (A) ⊂ λ∈Λ Xλ and for each pair of elements  of Λ, with λ0 ≤ α, fλα0 |fα (A) (xα ) = xλ0 . Therefore, {xλ }λ∈Λ ∈ f λ∈Λ λ (A) ∩ X∞ .    Now, let {xλ }λ∈Λ ∈ λ∈Λ fλ (A) ∩ X∞ . Hence, {xλ }λ∈Λ ∈ λ∈Λ fλ (A) and if λ0 ≤ α, then fλα0 (xα ) = xλ0 . This implies that fλα0 |fα (A) (xα ) = xλ0 . Thus, {xλ }λ∈Λ ∈ lim{fλ (A), fλα |fα (A) , Λ}. ←−   Let {xλ }λ∈Λ ∈ λ∈Λ fλ (A) ∩ X∞ . We prove that {xλ }λ∈Λ ∈ A. For each (xλ0 ). Note that Kλ0 is a nonempty compact subset of λ0 ∈ Λ, let Kλ0 = A ∩ fλ−1 0 A. We see that K = {Kλ }λ∈Λ has the finite intersection property. Let Kλ1 , . . . , Kλn be a finite subfamily of K. Since Λ is a directed set, there exists γ ∈ Λ such that λj ≤ γ for each j ∈ {1, . . . , n}. Let j ∈ {1, . . . , n}. We show that Kγ ⊂ Kλj . Let {yλ }λ∈Λ ∈ Kγ = A ∩ fγ−1 (xγ ). Then fλγj ◦ fγ ({yλ }λ∈Λ ) = fλj ({yλ }λ∈Λ ). n Hence, fλγj (yγ ) = fλγj (xγ ) = xλj and Kγ ⊂ Kλj . Thus, j=1 Kλj = ∅. Therefore,  λ∈Λ Kλ = ∅.  λ ∈ Λ. Hence, Let {zλ }λ∈Λ ∈ λ∈Λ Kλ . Then fλ ({zλ }λ∈Λ ) = xλ for all   zλ = xλ for every λ ∈ Λ. Therefore, {xλ }λ∈Λ ∈ A. Clearly, A ⊂ f (A) ∩ X∞ . λ λ∈Λ   1.2.18 Theorem. Let {Xλ , fλα , Λ} be an inverse system of compacta whose inverse limit is X∞ . Let Λ0 be a cofinal subset of Λ. For each λ ∈ Λ0 , let Aλ be

1.2 Topological Product and Inverse Limits

17

a nonempty closed subset of Xλ and suppose that for every λ and α in Λ0 , with λ ≤ α, fλα (Aα )  ⊂ Aλ . Then {Aλ , fλα |Aα , Λ0 } is and inverse system and α lim{Aλ , fλ |Aα , Λ0 } = λ∈Λ0 fλ−1 (Aλ ). ←−

Proof. Observe that {Aλ , fλα |Aα , Λ0 } is an inverse system. Let λ and α be elements of Λ0 such that λ ≤ α. Since fλα (Aα ) ⊂ Aλ , we have fα−1 (Aα ) ⊂ fλ−1 (Aλ ). Hence, {fλ−1 (Aλ )}λ∈Λ0 is a directed set by inclusion. Let {aλ }λ∈Λ0 ∈ lim{Aλ , fλα |Aα , Λ0 } and let γ ∈ Λ0 . Since ←−

fγ ({aλ }λ∈Λ0 ) = aγ ∈ Aγ ,  −1 {aλ }λ∈Λ0 ∈ fγ−1 (Aγ ). Thus, {aλ }λ∈Λ0 ∈ λ∈Λ0 fλ (Aλ ). Therefore,  −1 lim{Aλ , fλα |Aα , Λ0 } ⊂ λ∈Λ0 fλ (Aλ ). ←−  Let {bλ }λ∈Λ0 ∈ λ∈Λ0 fλ−1 (Aλ ). Then fγ ({bλ }λ∈Λ0 ) = bγ ∈ Aγ , for each γ ∈ Λ0 . Hence, {bλ }λ∈Λ0 ∈ λ∈Λ0 Aλ and given λ and α in Λ0 , fλα (bα ) = bλ .  This implies that {bλ }λ∈Λ0 ∈ lim{Aλ , fλα |Aα , Λ0 }. Therefore, λ∈Λ0 fλ−1 (Aλ ) ⊂ lim{Aλ , fλα |Aα , Λ0 }.

←−

 

←−

As a consequence of the Theorems 1.2.17 and 1.2.18, we obtain: 1.2.19 Corollary. Let {Xλ , fλα , Λ} be an inverse system of compacta whose inverse limit is X∞ . Let Λ0 be a cofinal subset of Λ. If A is a nonempty closed subset X∞ , then  A= fλ−1 (fλ (A)). λ∈Λ0

1.2.20 Theorem. Let {Xλ , fλα , Λ} be an inverse system of connected compacta with surjective projection maps, whose inverse limit is X∞ . Then X∞ is a connected compactum. Proof. By Theorem 1.2.15, X∞ is a nonempty compactum. We show that if X∞ = A ∪ B, where A and B are nonempty closed subsets of X∞ , then A ∩ B = ∅. Since X∞ is compact, both A and B are compact subsets of X∞ . Hence, for every λ ∈ Λ, fλ (A) and fλ (B) are closed subsets of Xλ . Since fλ is surjective, Xλ = fλ (A) ∪ fλ (B). Let λ ∈ Λ and let Zλ = fλ (A) ∩ fλ (B). Since Xλ is connected, Zλ = ∅. If α ∈ Λ is such that λ ≤ α, then define gλα = fλα |Zα . Since fλα (Zα ) = fλα (fα (A)∩fα (B)) ⊂ fλα ◦fα (A)∩fλα ◦fα (B) = fλ (A)∩fλ (B) = Zλ , we have that gλα (Zα ) ⊂ Zλ . Hence, {Zλ , gλ , Λ} is an inverse system of compacta, by Theorem 1.2.15, Z∞ = ∅. Let {zλ }λ∈Λ ∈ Z∞ . Then for each α ∈ Λ, fα ({zλ }λ∈Λ ) ∈ fα (A) ∩ fα (B). By Theorem 1.2.17, {zλ }λ∈Λ ∈ A ∩ B. Therefore, X∞ is connected.   Now, we turn our attention to generalized inverse limits of compacta with a single bonding function.

18

1 Preliminaries

1.2.21 Definition. Let X and Y be compacta. Then a function f : X → 2Y (Definition 1.6.1) is upper semicontinuous if for every point x of X and each open subset U of Y such that f (x) ⊂ U , there exists an open subset V of X such that if x ∈ V , then f (x ) ⊂ U . The image of f , denoted Im(f ), is the set {y ∈ Y | there exists x ∈ X such that y ∈ f (x)}. If Im(f ) = Y , then we say that f is surjective. 1.2.22 Definition. Let X, Y and Z be compacta and let f : X → 2Y and g : Y → 2Z be upper semicontinuous functions. Then the composition of f and g, denoted g ◦ f , is defined by (g ◦ f )(x) = {z ∈ Z | there exists y ∈ Y such that y ∈ f (x) and z ∈ g(y)}. 1.2.23 Notation. Let X be a compactum. Then X ω denotes the product of countably many copies of X with the product topology. The symbol Δω X denotes the diagonal of X ω ; that is, Δω = {(x, x, x, . . .) | x ∈ X}. X 1.2.24 Definition. Let X be a compactum. If f : X → 2X is an upper semicontinuous function, then its generalized inverse limit is ω lim f = {(xn )ω n=1 ∈ X | xn ∈ f (xn+1 ) for each positive integer n}. ←−

The function f is called the bonding function. For each positive integer n, πn : lim f → X is the projection map. ←−

1.2.25 Notation. Let X be a compactum and let G be an upper semicontinuous decomposition of X. Then fG : X → 2X is defined by fG (x) = Gx , where Gx is the unique element of G such that x ∈ Gx . 1.2.26 Lemma. If X is a compactum and G is an upper semicontinuous decomposition of X, then fG is an upper semicontinuous function. Proof. Let x0 ∈ X and let U be an open subset of X such that fG (x0 ) ⊂ U . Since G is upper semicontinuous, there exists an open subset V of X such that fG (x0 ) ⊂ V and if G ∈ G and G ∩ V = ∅, then G ⊂ U . Thus, if x ∈ V , then fG (x) ⊂ U . Therefore, fG is upper semicontinuous.   1.2.27 Lemma. If X is a compactum and G = {G x | x ∈ X} is an upper semicontinuous decomposition of X, then lim fG = {Gω x | x ∈ X}. Moreover, ←− ω ΔX ⊂ lim fG . ←−

Proof. By Lemma 1.2.26, fG is upper semicontinuous. Let (xn )ω n=1 ∈ lim fG . Then ←−

x1 ∈ fG (x2 ) = Gx2 , x2 ∈ fG (x3 ) = Gx3 , etc. Since G is a decomposition, ω Gx1 = Gx2 = · · · . Hence, (xn )ω n=1 ∈ Gx1 . The other inclusion is clear. ω If x ∈ X, then (x, x, x, , . . .) ∈ Gx . Therefore, Δω   X ⊂ lim fG . ←−

1.3 Uniformities

19

1.3 Uniformities We introduce the notion of uniformity of a topological space and give some of its basic properties. Let X be a Hausdorff space. If V and W are subsets of X × X, then −V = {(x , x) ∈ X × X | (x, x ) ∈ V } and V + W = {(x, x ) | there exists x ∈ X such that (x, x ) ∈ V and (x , x ) ∈ W }. We write 1V = V and for each positive integer n, (n + 1)V = nV + 1V . The diagonal of X is the set ΔX = {(x, x) | x ∈ X}. An entourage of ΔX is a subset V of X × X such that ΔX ⊂ V and V = −V . The family of entourages of the diagonal of X is denoted by DX . If V ∈ DX and x ∈ X, then B(x, V ) = {x ∈ X | (x, x ) ∈ V }. By [40, 8.1.3], Int(B(x, V )) is anopen neighbourhood of x. If A is a subset of X and V ∈ DX , then B(A, V ) = {B(a, V ) | a ∈ A}. If V ∈ DX and (x, x ) ∈ V , then we write ρ(x, x ) < V . If (x, x ) ∈ V , then we write ρ(x, x ) ≥ V . We have that if x, x and x are points of X, and V and W belong to DX then the following hold [40, p. 426]: (i) ρ(x, x) < V . (ii) ρ(x, x ) < V if and only if ρ(x , x) < V . (iii) If ρ(x, x ) < V and ρ(x , x ) < W , then ρ(x, x ) < V + W . Let X be a nonempty set. A uniformity on X is a subfamily U of DX such that: (1) (2) (3) (4)

If V ∈ U, W ∈ DX and V ⊂ W , then W ∈ U. If V and W both belong to U, then V ∩ W ∈ U. For  every V ∈ U, there exists W ∈ U such that 2W ⊂ V . {V | V ∈ U} = ΔX .

1.3.1 Definition. A uniform space is a pair (X, U) consisting of a nonempty set X and a uniformity on the set X. 1.3.2 Definition. For any uniformity U on a set X, the family O = {G ⊂ X | for every x ∈ G, there exists V ∈ U such that B(x, V ) ⊂ G} is a topology on the set X [40, 8.1.1]. The topology O is called the topology induced by the uniformity U. 1.3.3 Notation. Let X be a Tychonoff space and let U be a uniformity of X that induces its topology. If V ∈ U, then we define the cover C(V ) = {B(x, V ) | x ∈ X} of X. 1.3.4 Remark. Note that by [40, 8.3.13], for every compactum X, there exists a unique uniformity UX on X that induces the original topology of X.

20

1 Preliminaries

1.3.5 Notation. Let X be a compactum and let UX be the unique uniformity of X that induces its topology. If Y is a subspace of X and U ∈ UX , then UY = U ∩ (Y × Y ). We need the next result [40, 8.3.G]. 1.3.6 Theorem. If X is a compactum, then for every open cover U of X, there exists V ∈ UX such that C(V ) refines U . 1.3.7 Remark. It is known that if X is a compactum and H(X) is the group of homeomorphisms of X, then H(X) is a topological group with the compact-open topology [40, p. 441]. ! 1.3.8 Notation. Let X be a compactum. For every U ∈ UX (Remark 1.3.4), let U be the entourage of the diagonal ΔH(X) ⊂ H(X) × H(X) defined by ! = {(g1 , g2 ) ∈ H(X) × H(X) | ρ(g1 (x), g2 (x)) < U for all x ∈ X}. U ! | U ∈ UX } is a base for a uniformity on It is known that the family U! = {U H(X); the uniformity generated by this family is called the uniformity of uniform convergence induced by UX and is denoted by U" X [40, p. 440]. We also need a special case of [40, 8.2.7]: 1.3.9 Theorem. Let X be a compactum. Then the topology on H(X) induced by the uniformity U" X of uniform convergence coincides with the compact-open topology on H(X), and depends only on the topology induced on X by the uniformity UX (Remark 1.3.4). The following simple lemmas are useful. 1.3.10 Lemma. Let Z be a Tychonoff space, let U be a uniformity of Z that induces its topology, let V ∈ U and let z ∈ Z. Then B(z, nV ) ⊂ IntZ (B(z, (n + 1)V )) for all n ∈ N. Proof. Let n ∈ N and let z  ∈ B(z, nV ) and let z  ∈ B(z  , V ). Since ρ(z  , z  ) < V and ρ(z  , z) < nV , we have that ρ(z  , z) < (n + 1)V . Thus, B(z  , V ) ⊂ B(z, (n + 1)V ). Therefore, by [40, 8.1.2], B(z, nV ) ⊂ IntZ (B(z, (n + 1)V )).   1.3.11 Lemma. Let Z be a compactum and let U be an open cover of Z. If U0 =  {U × U | U ∈ U }, then U0 ∈ UZ .  Proof. Let U0 = {U × U | U ∈ U }. Note that ΔZ ⊂ U0 and −U0 = U0 . Thus, U0 ∈ DZ . Since Z is a compact Hausdorff space, by Theorem 1.3.6, there exists W ∈ UZ such that C(W ) refines U . Since C(W ) refines U , we have that W ⊂ U0 . This implies that U0 ∈ UZ .   The following theorems are [111, Theorem 2.1 and Theroem 2.2], respectively: 1.3.12 Theorem. Let f be a function from a uniform space (X, U) onto a nonempty set Y . Then U(f ) = {V ⊂ Y × Y | (f × f )−1 (V ) ∈ U} is a uniformity for Y .

1.4 Continua

21

1.3.13 Theorem. Let f be a function from a uniform space (X, U) onto the uniform space (Y, U(f )). Then U(f ) = {(f × f )(U ) | U ∈ U}. 1.3.14 Definition. Let (Z, V) and (W, U) be uniform spaces and let g : Z → W be a function. Then g is uniformly continuous if for each U ∈ U, there exists V ∈ V such that for each point z of Z, g(B(z, V )) ⊂ B(g(z), U ). Note the following: 1.3.15 Theorem. Let Z and W be compacta and let g : Z → W be a function. Then g is continuous if and only if g is uniformly continuous Proof. By [129, 35.11], every uniformly continuous function is continuous. By [129, 36.20], each continuous function between compacta is uniformly continuous.  

1.4 Continua We begin with some basic properties of compacta and then we present many properties of continua. We consider the notion of generalized continua and end the section with some results of generalized inverse limits with a single bonding map. 1.4.1 Lemma. Let X be a compactum, and let A be a closed subset of X with a finite number of components. If x ∈ Int(A) and C is the component of A such that x ∈ C, then x ∈ Int(C). Proof. Let C, C1 , . . . , Cn be the components of A. Since x ∈ Int(A), there exists an open subset U of X such that x ∈ U ⊂ A. Since A is closed in X, each of n C, C1 , . . . , Cn is closed in X. Let V = U ∩ X \ j=1 Cj . Then V is an open   n subset of X such that x ∈ V ⊂ A and V ∩ = ∅. Thus, V ⊂ C. j=1 Cj Therefore, x ∈ Int(C).   1.4.2 Definition. Let X be a compactum, and let A and B be two nonempty disjoint subsets of X. A subset C of X is a separator of A and B (or separates X between A and B) if X \ C = U ∪ V , where U and V are separated (i.e., Cl(U ) ∩ V = ∅ and U ∩ Cl(V ) = ∅), A ⊂ U and B ⊂ V . If C = ∅, we say that A and B are separated in X. 1.4.3 Lemma. Let X be a compactum, let a be a point of X and let B be a compact subset of X such that for each b in B, a and b are separated in X. Then a and B are separated in X. Proof. For each b in B, there exist two open and closed subsets Ub and Vb such that X = Ub ∪ Vb , a ∈ Ub and V ∈ Vb . Since B is compact, there exist b1 , . . . , bn in B

22

1 Preliminaries

n n such that B ⊂ j=1 Vbj . Let V = j=1 Vbj . Then V is an open and closed subset of X, B ⊂ V and a ∈ X \ V . Therefore, a and B are separated.   1.4.4 Lemma. Let X be a compactum and let A and B be compact subsets of X. If each point of A and each point of B are separated in X, then A and B are separated. Proof. By Lemma 1.4.3, B and each point of A are separated. Repeating the argument of Lemma 1.4.3, we obtain that A and B are separated.   1.4.5 Lemma. Let X be a compactum, let a and b be two distinct points of X and let {Hλ }λ∈Λ be a set theoretic chain of closed subsets of X ordered by inclusion. If each Hλ contains both a and b but is not the union of two separated sets, one  containing a and the other containing b, then the intersection λ∈Λ Hλ also has this property.  Proof. Let H = λ∈Λ Hλ and suppose that H is the union of two separated sets A and B such that a ∈ A and b ∈ B. Since H is a closed subset of X, we have that A and B are closed subsets of X. Since X is normal, there exist two disjoint open subsets U and V of X such that A ⊂ U and B ⊂ V . For each λ ∈ Λ, Hλ ∩ U = ∅ and Hλ ∩ V = ∅. Let λ ∈ Λ. If Kλ = Hλ ∩ [X \ (U ∪ V )] were empty, then Hλ = (Hλ ∩ U ) ∪ (Hλ ∩ V ) would be a separation of Hλ with a ∈ Hλ ∩ U and b ∈ Hλ ∩ V , a contradiction. Hence, Kλ = ∅. Also, the family {Kλ }λ∈Λ is a set theoretic chain ordered by inclusion; for, given any subset R of X, if Hλ is contained in Hλ , then Hλ ∩ R lies in H λ ∩ R. Thus, {Kλ } λ∈Λ has the finite  intersection property. Since X is compact, λ∈Λ Kλ = ∅. But λ∈Λ Kλ ⊂ λ∈Λ Hλ , which implies that H ∩ [X \ (U ∪ V )] = ∅, a contradiction.   1.4.6 Definition. A continuum is a connected compactum. A subcontinuum is a continuum contained in a space. 1.4.7 Theorem. Let X be compactum and let a and b be two distinct points of X. If X is not the union of two disjoint open sets, one containing a and the other containing b, then X contains a subcontinuum containing both a and b. Proof. Let H = {Hλ }λ∈Λ be the collection of all closed subsets of X, each of which contains {a, b} but in none of which a and b are separated. Since X ∈ H, H = ∅. Partially order H by inclusion. Let L = {Lγ }γ∈Γ be a set  theoretic chain of elements of H, we show that L has a lower bound. Let L = γ∈Γ Lγ . Then L is a nonempty closed subset of X and, by Lemma 1.4.5, L is not the union of two separated sets. Hence, L ∈ H. By the Kuratowsky-Zorn Lemma, H has a minimal element, say H. We show that H is connected. Suppose H is not connected. Then there exist two disjoint closed subsets A and B of X such that H = A ∪ B. Since H ∈ H, H is not the union of two separated sets. Thus, either {a, b} ⊂ A or {a, b} ⊂ B. Either case contradicts the minimality of H. Therefore, H is connected.   The next result is known as The Cut Wire Theorem; it is very useful in continuum theory.

1.4 Continua

23

1.4.8 Theorem. Let X be a compactum and let A and B be nonempty closed subsets of X. If no connected subset of X intersects both A and B, then there exist two disjoint closed subsets X1 and X2 of X such that A ⊂ X1 , B ⊂ X2 and X = X1 ∪ X 2 . Proof. By hypothesis, no component, K, of X intersects both A and B. Note that the components of X are subcontinua of X. Hence, by Theorem 1.4.7, each point of A and each point of B are separated in X. Hence, by Lemma 1.4.4, A and B are separated in X. Therefore, there exist two disjoint closed subsets X1 and X2 of X such that A ⊂ X1 , B ⊂ X2 and X = X1 ∪ X2 .   1.4.9 Example. Let # X = {0} × [−1, 1] ∪

x, sin

   $% 1 2 ∈ R2 x ∈ 0, . x π

Then X is called the topologist sine curve (Figure 1.7).

Fig. 1.7 Topologist sine curve

The following definition is due to Davis and Doyle and they used it in their study of invertible continua [31]: 1.4.10 Definition. Let X be a continuum, and let x ∈ X. We say that X is almost connected im kleinen at x provided that for each open subset U of X containing x, there exists a subcontinuum W of X such that Int(W ) = ∅ and W ⊂ U . We say that X is almost connected im kleinen if it is almost connected im kleinen at each of its points. 1.4.11Example. Let X be the cone over the closure of the harmonic sequence

∞ {0} ∪ n1 n=1 . Then X is called the harmonic fan. Note that X is almost connected im kleinen at each of its points (Figure 1.8).

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1 Preliminaries

Fig. 1.8 Almost connected im kleinen

1.4.12 Definition. Let X be a continuum, and let x ∈ X. We say that X is connected im kleinen at x if for each closed subset F of X such that F ⊂ X \ {x}, there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ F . 1.4.13 Example. Let X be a sequence of harmonic fans converging to a point p; see Figure 1.9. Then X is connected im kleinen at p but X is not locally connected at that point (Definition 1.4.16).

p Fig. 1.9 Connected im kleinen at p

1.4.14 Remark. Note that we do not define connectedness im kleinen globally because, by Theorem 1.4.18, a continuum is connected im kleinen globally if and only if it is locally connected (Definition 1.4.16). The way we define connectedness im kleinen in Definition 1.4.12 is not the usual one. This definition provides the author the answer to the fact that aposyndesis (Definition 1.4.21) is a generalization of connectedness im kleinen. The next theorem presents the usual definition of this concept: 1.4.15 Theorem. If X is a continuum and x ∈ X, then the following are equivalent: (a) X is connected im kleinen at x.

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25

(b) For each open subset U of X such that x ∈ U , there exists an open subset V of X such that x ∈ V ⊂ U and with the property that for each y ∈ V , there exists a connected subset Cy of X such that {x, y} ⊂ Cy ⊂ U . (c) For each open subset U of X such that x ∈ U , there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ U . Proof. Suppose X is connected im kleinen at x. We show (b). Let U be an open subset of X such that x ∈ U . Then X \ U is a closed subset of X not containing x. By hypothesis, there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ (X \ U ) = U . Thus, V = Int(W ) satisfies the required properties. Next, suppose (b). We prove (c). Let U be an open subset of X such that x ∈ U . Let U  be an open subset of X such that x ∈ U  ⊂ Cl(U  ) ⊂ U . By hypothesis, there exists an open subset V of X such that x ∈ V ⊂ U  with the property that for each y ∈ V , there a connected subset Cy of X such that {x, y} ⊂ Cy ⊂ U  .  exists  Let W = Cl y∈V Cy . Then W is a subcontinuum of X such that x ∈ V ⊂ W ⊂ Cl(U  ) ⊂ U . Finally, suppose (c). We show X is connected im kleinen at x. Let F be a closed subset of X such that F ⊂ X \ {x}. Then X \ F is an open subset of X containing x. By hypothesis, there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ F . Therefore, X is connected im kleinen at x.   1.4.16 Definition. Let X be a continuum, and let x ∈ X. We say that X is locally connected at x provided that for each open subset U of X such that x ∈ U , there exists a connected open subset V of X such that x ∈ V ⊂ U . We say X is locally connected if it is locally connected at each of its points. 1.4.17 Lemma. A continuum X is locally connected if and only if the components of the open subsets of X are open. Proof. Suppose X is locally connected. Let U be an open subset of X, and let C be a component of U . For each point x ∈ C, there exists an open connected set Vx such that x ∈ Vx ⊂ U . Then C ∪ Vx is a connected subset of U . Hence, Vx ⊂ C. Therefore, each point of C is an interior point of C. Thus, C is open. Next, suppose the components of open sets are open. Let x ∈ X, and let U be an open subset of X such that x ∈ U . By hypothesis, the component, C, of U containing x is open. Then C is an open connected set such that x ∈ C ⊂ U . Therefore, X is locally connected.   1.4.18 Theorem. A continuum X is connected im kleinen at each of its points if and only if X is locally connected. Proof. Clearly, if X is locally connected, then X is connected im kleinen at each of its points. Suppose X is connected im kleinen at each of its points. Let U be an open subset of X, and let C be a component of U . If x ∈ C, then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ U , Theorem 1.4.15. Since W is a connected subset of U and W ∩ C = ∅, W ⊂ C. Then x is an interior point of C. Thus, C is open. Therefore, by Lemma 1.4.17, X is locally connected.  

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1.4.19 Definition. Let X be a continuum, and let p and q be points of X. We say that X is semi-aposyndetic at p and q provided that there exists a subcontinuum W of X such that {p, q} ∩ Int(W ) = ∅ and {p, q} \ W = ∅. X is semi-aposyndetic if it is semi-aposyndetic at each pair of its points. 1.4.20 Example. Let X be the harmonic fan (Example 1.4.11). Then X is semiaposyndetic at each pair of its points (Figure 1.10).

Fig. 1.10 Semi-aposyndetic continuum

1.4.21 Definition. Let X be a continuum, and let p and q be points of X. We say that X is aposyndetic at p with respect to q provided that there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ {q}. Now, X is aposyndetic at p if X is aposyndetic at p with respect to each point of X \ {p}. We say that X is aposyndetic provided that X is aposyndetic at each of its points (Figure 1.11).

X

W

q

p

Fig. 1.11 Aposyndetic at p with respect to q

1.4.22 Definition. Let X be a continuum and let p and q be points of X. We say that X is mutually aposyndetic at p and q if there exist two disjoint subcontinua K and L of X such that p ∈ Int(K) and q ∈ Int(L). X is mutually aposyndetic if it is mutually aposyndetic at each pair of its points.

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27

1.4.23 Example. Let

∞X be the suspension over the closure of the harmonic sequence {0} ∪ n1 n=1 , with vertexes a and b (Figure 1.12). Then X is mutually aposyndetic.

a

b

Fig. 1.12 Mutually aposyndetic

1.4.24 Remark. The notions of semi-aposyndesis, aposyndesis and mutual aposyndesis resemble those of T0 , T1 and T2 topological spaces, respectively. 1.4.25 Definition. Let X be a continuum, and let p ∈ X. We say that X is semilocally connected at p provided that for each open subset U of X such that p ∈ U , there exists an open subset V of X such that p ∈ V ⊂ U and X \ V only has a finite number of components. We say X is semi-locally connected if X is semi-locally connected at each of its points (Figure 1.13).

Fig. 1.13 Semi-locally connected continuum

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1 Preliminaries

Even though aposyndesis and semi-local connectedness seem to be different concepts, it turns out that globally they are equivalent: 1.4.26 Theorem. A continuum X is aposyndetic if and only if it is semi-locally connected. Proof. Suppose X is aposyndetic. Let x ∈ X. We show that X is semi-locally connected at x. Let U be an open subset of X such that x ∈ U . Since X is aposyndetic, for each y ∈ X \ U , there exists a subcontinuum Wy of X such that y ∈ Int(Wy ) ⊂ Wy ⊂ X \ {x}. Since X \ U is compact, there exist y1 , . . . , ym in m m X \ U such that X \ U ⊂ j=1 Int(Wyj ). Let V = X \ j=1 Wyj . Then V is an open subset of X such that x ∈ V ⊂ U and X \ V only has a finite number of components. Therefore, X is semi-locally connected, since x is an arbitrary point of X. Now, suppose X is semi-locally connected. Let x and y be points of X. We show that X is aposyndetic at x with respect to y. Let U be an open subset of X such that y ∈ U ⊂ X \ {x}. Let U  be an open subset of X such that y ∈ U  ⊂ Cl(U  ) ⊂ U . Since X is semi-locally connected, there exists an open subset V of X such that y ∈ V ⊂ U  and X \ V only has a finite number of components. By Lemma 1.4.1, x is contained in the interior of the component of X \ V containing x. Hence, X is aposyndetic at x with respect to y. Therefore, X is aposyndetic.   Another concept related to aposyndesis is free decomposability. 1.4.27 Definition. A continuum X is freely decomposable if for each pair of distinct points p and q of X, there exist two subcontinua P and Q of X such that X = P ∪ Q, p ∈ P \ Q and q ∈ Q \ P . The next theorem shows that the concepts of free decomposability and aposyndesis coincide. 1.4.28 Theorem. A continuum X is aposyndetic if and only if X is freely decomposable. Proof. Suppose X is freely decomposable. Let p and q be two distinct points of X. Since X is freely decomposable, there exist two subcontinua P and Q of X such that X = P ∪ Q, p ∈ P \ Q and q ∈ Q \ P . Hence, X is an aposyndetic continuum. Now, assume X is an aposyndetic continuum. Let p and q be two distinct points of X. Since X is aposyndetic, there exist two subcontinua A and B such that p ∈ Int(A) ⊂ A ⊂ X \ {q} and q ∈ Int(B) ⊂ B ⊂ X \ {p}. Let U and V be open subsets of X such that U ⊂ A, V ⊂ B, p ∈ U ⊂ X \B and q ∈ V ⊂ X \A. Let Ap be the component of X \V that contains p and let Bq be the component of X \U that contains q. Note that Ap and Bq are subcontinua of X, p ∈ Ap \Bq and q ∈ Bq \Ap . We show that X = Ap ∪ Bq . Suppose this is not true and let L = X \ (Ap ∪ Bq ). Since p ∈ U ⊂ A ⊂ X \ V and q ∈ V ⊂ B ⊂ X \ U , we have that A ⊂ Ap and B ⊂ Bq . Hence, U ∪ V ⊂ Ap ∪ Bq . Thus, L ⊂ X \ (U ∪ V ) = (X \ U ) ∩ (X \ V ). This implies that Cl(L) ⊂ X \ V . Note that Ap ∪ Cl(L) is a closed subset of X \ V that contains Ap properly. Since Ap is a component of X \ V , Ap ∪ Cl(L) is not

1.4 Continua

29

connected. Hence, there exist two disjoint closed subsets R1 and S1 of X such that Ap ∪ Cl(L) = R1 ∪ S1 (Theorem 1.4.8). Without loss of generality, we assume that Ap ⊂ R1 . Thus, S1 ⊂ Cl(L). Since Cl(L) ⊂ X \U , we have that Bq ∪S1 ⊂ X \U . Observe that Bq ∪S1 is a closed subset of X \U containing Bq properly. Since Bq is a component of X \U , Bq ∪S1 is not connected. Thus, there exist two disjoint closed subsets R2 and S2 of X such that Bq ∪ S1 = R2 ∪ S2 (Theorem 1.4.8). Without loss of generality, we assume that Bq ⊂ R2 . Hence, S2 ⊂ S1 . Since R1 ∩ S1 = ∅ and R1 ∩ S2 = ∅, we obtain that (R1 ∪ R2 ) ∩ S2 = ∅. Since X = Ap ∪ Bq ∪ L, we have that X = Ap ∪ Bq ∪ Cl(L). Also, since Ap ∪ Cl(L) = R1 ∪ S1 , X = R1 ∪ S1 ∪ Bq . Moreover, since S1 ∪ Bq = R2 ∪ S2 , we obtain that X = (R1 ∪ R2 ) ∪ S2 . A contradiction to the connectedness of X. Therefore, X = Ap ∪ Bq and X is freely decomposable.   The notion of aposyndesis may be extended as follows: 1.4.29 Definition. A continuum X is aposyndetic at p with respect to the subset K of X if there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ K. 1.4.30 Definition. A continuum X is continuum aposyndetic at p if X is aposyndetic at p with respect to each subcontinuum of X not containing p. X is continuum aposyndetic provided that X is continuum aposyndetic at each of its points. 1.4.31 Definition. A continuum X is freely decomposable with respect to points and continua if for each subcontinuum C of X and each point a ∈ X \ C, there exist two subcontinua A and B of X such that X = A∪B, a ∈ A\B and C ⊂ B\A. 1.4.32 Lemma. Let X be a continuum, and let A be a subcontinuum of X such that X \ A is not connected. If U and V are nonempty disjoint open subsets of X such that X \ A = U ∪ V , then A ∪ U and A ∪ V are subcontinua of X. Proof. Since X \ (A ∪ U ) = V , A ∪ U is closed, hence, compact. Similarly A ∪ V is compact. We show A ∪ U is connected. To see this, suppose A ∪ U is not connected. Then there exist two nonempty disjoint closed subsets K and L of X such that A ∪ U = K ∪ L. Since A is connected, without loss of generality, we assume that A ⊂ K. Note that, in this case, L ⊂ U . Hence, L∩Cl(V ) = ∅. Thus, X = L∪(K ∪Cl(V )), a contradiction, since L and K ∪ Cl(V ) are disjoint closed subsets of X. Therefore, A ∪ U is connected. Similarly, A ∪ V is connected.   1.4.33 Definition. A continuum X is decomposable provided that it can be written as the union of two of its proper subcontinua. We say X is indecomposable if it is not decomposable. We say X is hereditarily decomposable (indecomposable) if each nondegenerate subcontinuum of X is decomposable (indecomposable, respectively). 1.4.34 Lemma. A continuum X is decomposable if and only if X contains a proper subcontinuum with nonempty interior.

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Proof. Suppose X is a decomposable continuum. Then there exist two proper subcontinua, A and B, of X such that X = A ∪ B. Note that X \ B is an open set contained in A. Therefore, Int(A) = ∅. Now, assume A is a proper subcontinuum of X with nonempty interior. If X \ A is connected, then Cl(X\A) is a proper subcontinuum of X and X = A∪Cl(X\A). Hence, X is decomposable. Suppose that X \ A is not connected. Then there exist two nonempty disjoint open subsets U and V of X such that X \ A = U ∪ V . By Lemma 1.4.32, A ∪ U and A ∪ V are subcontinua of X, and X = (A ∪ U ) ∪ (A ∪ V ). Therefore, X is decomposable.   1.4.35 Corollary. A continuum X is indecomposable if and only if each proper subcontinuum of X has empty interior. The next result is known as the Boundary Bumping Theorem. 1.4.36 Theorem. Let X be a continuum and let U be a nonempty, proper, not dense open subset of X. If K is a component of Cl(U ), then K ∩ Bd(U ) = ∅. Proof. Let U be a not dense open subset of X and let K be a component of Cl(U ). Assume that K ∩ Bd(U ) = ∅. Hence, K ⊂ U . Since X is normal, there exists an open subset V of X such that K ⊂ V ⊂ Cl(V ) ⊂ U . Clearly, Cl(V ) is a compactum. Since no connected subset of Cl(V ) intersects both K and Bd(V ) (K is a component of U ), by Theorem 1.4.8, there exists two disjoint closed subsets L and M of X such that Cl(V ) = L ∪ M , K ⊂ L and Bd(V ) ⊂ M . Then X = L ∪ [M ∪ (X \ V )], where L and [M ∪ (X \ V )] are disjoint closed subset of X, a contradiction to the connectedness of X. Therefore, K ∩ Bd(U ) = ∅.   The following corollary is very useful: 1.4.37 Corollary. Let X be a nondegenerate continuum. If A is a proper subcontinuum of X and U is a proper open subset of X such that A ⊂ U , then there exists a subcontinuum B of X such that A  B ⊂ U . Proof. Let A be a proper subcontinuum of X and let U be a proper open subset of X. Since X is normal, there exists an open subset V of X such that A ⊂ V ⊂ Cl(V ) ⊂ U . Let B be the component of Cl(V ) that contains A. By Theorem 1.4.36, B ∩ Bd(V ) = ∅. Therefore, B is a subcontinuum of X and A  B ⊂ U .   1.4.38 Lemma. If X is a continuum such that each of its proper subcontinua is indecomposable, then X is indecomposable. Hence, X is hereditarily indecomposable. Proof. Suppose X is decomposable. Then there exist two proper subcontinua A and B of X such that X = A ∪ B. Note that X \ B is an open subset of X contained in A. On the other hand, there exists a proper subcontinuum, H, of X containing A (Corollary 1.4.37). Hence, H is an indecomposable continuum containing a proper subcontinuum with nonempty interior, which is impossible (Corollary 1.4.35). Therefore, X is indecomposable.  

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1.4.39 Definition. A continuum X is irreducible between two of its points if no proper subcontinuum of X contains both points. A continuum is irreducible if it is irreducible between two of its points. The following results present some of the properties of irreducible continua. 1.4.40 Theorem. Let X be an irreducible continuum between a and b. If C is a subcontinuum of X such that X \ C is not connected, then X \ C is the union of two open and connected sets, one containing a and the other containing b. Moreover, if a ∈ C, then X \ C is connected. Proof. Suppose X \ C is not connected. Then there exist two nonempty disjoint open subsets U and V of X such that X \C = U ∪V . By Lemma 1.4.32, A = C ∪U and B = C ∪ V are subcontinua of X such that X = A ∪ B, A ∩ B = C, A = X and B = X. Since X is irreducible between a and b, {a, b} ∩ C = ∅. If {a, b} ∩ C = ∅, then either A or B is a proper subcontinuum of X containing {a, b}, a contradiction. Therefore, {a, b} ∩ C = ∅. We assume that a ∈ U and b ∈ V . Since A and B are proper subcontinua of X, neither A nor B may contains {a, b}. Since A is a proper subcontinuum of X and a ∈ A, we assert that V = X \ A is connected. To see this, suppose X \ A is not connected. Then there exist two nonempty disjoint open subsets K and L of X such that X \ A = K ∪ L. Since b ∈ X \A, we may assume that b ∈ K. Then, by Lemma 1.4.32, A∪K is subcontinuum of X which is proper and satisfies that {a, b} ⊂ A ∪ K, a contradiction. Therefore, V = X \ A is connected. Similarly, U = X \ B is connected. A similar argument shows that if a ∈ C, then X \ C is connected.   1.4.41 Lemma. Let X be an irreducible decomposable continuum. If Y and Z are two disjoint subcontinua of X, then X \ (Y ∪ Z) has at most three components. Proof. Since X is irreducible, by Theorem 1.4.40, X \ Y has at most two components, U and V , such that they are connected open subsets of X, Z ⊂ U and V may be empty. Similarly, we assume that X \ Z = H ∪ K, where H and K are connected open subsets of X such that Y ⊂ H and K may be empty. Let R = (U \ Z) ∩ (H \ Y ). Note that R is an open subset of X, Cl(R) ∩ Y = ∅ and Cl(R) ∩ Z = ∅. We assert that R is connected. To show this, assume R is not connected. Let C be a component of R. By Theorem 1.4.36, Cl(C) ∩ (Y ∪ Z) = ∅. If Cl(C) ∩ Y = ∅ and Cl(C) ∩ Z = ∅, then V ∪ Y ∪ Cl(C) ∪ Z ∪ K is a subcontinuum of X containing the points of irreducibility of X. Thus, X = V ∪ Y ∪ Cl(C) ∪ Z ∪ K. Since R ∩ (V ∪ Y ∪ Z ∪ K) = ∅, it follows that R ⊂ Cl(C). Hence, R is connected (C ⊂ R ⊂ Cl(C)), a contradiction. Therefore, either Cl(C) ∩ Y = ∅ or Cl(C) ∩ Z = ∅. Let A = {Cl(C) | C is a component of R and Cl(C) ∩ Y = ∅}

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and B = {Cl(C) | C is a component of R and Cl(C) ∩ Z = ∅}. We claim that A and B are both nonempty. Suppose, to the  contrary, that B is empty. Note that V ∪ Y ∪ ( A) is a connected set and that Ais not connected.  Then there exist two separated subsets J and L of V ∪ Y ∪ ( A) such that A = J ∪ L. Note that V ∪ Y ∪ J and V ∪ Y ∪ L are connected sets (Lemma 1.4.32). Hence, either (V ∪ Y ∪ Cl(J)) ∩ Z = ∅ or (V ∪ Y ∪ Cl(L)) ∩ Z = ∅. Suppose (V ∪ Y ∪ Cl(J)) ∩ Z = ∅. Then (V ∪ Y ∪ Cl(J)) ∪ Z ∪ K is a proper subcontinuum of X containing the points of irreducibility of X, a contradiction. Therefore, B = ∅. Similarly, A = ∅.     Since X is a continuum, Cl ( A)∩Cl ( B) = ∅. Let x ∈ Cl ( A)∩Cl ( B).  Since x ∈ Cl ( A), there exists a net {aλ }λ∈Λ of elements of A converging to x [40, 1.6.3]. For each λ ∈ Λ, let Cl(Cλ ) ∈ A be such that aλ ∈ Cl(Cλ ). By Theorem 1.6.7, we assume that the net {Cl(Cλ )}λ∈Λ of subcontinua of X converges (in the Vietoris topology) to a subcontinuum T of X [104, Theorem 4]. Note that T ∩Y = ∅ and x ∈ T . Similarly, there exists a subcontinuum T  of X such that T  ∩ Z = ∅ and x ∈ T  . Consequently, V ∪Y ∪T ∪T  ∪Z∪K is a proper subcontinuum of X containing its points of irreducibility, a contradiction. Therefore, R is connected. Now, observe that X \ [(V ∪ Y ) ∪ (Z ∪ K)] = (X \ V ) ∩ (X \ Y ) ∩ (X \ Z) ∩ (X \ K) = (U ∩ Y ) ∩ (X \ Y ) ∩ (X \ Z) ∩ (H ∩ Z) = U ∩ H = U ∩ (X \ Y ) ∩ (X \ Z) ∩ H = U ∩ (X \ Z) ∩ (X \ Y ) ∩ H = (U \ Z) ∩ (H \ Y ) = R. Since V ∩ (Y ∪ Z ∪ K) = ∅ and K ∩ (Z ∪ Y ∪ V ) = ∅, X \ (Y ∪ Z) = V ∪ R ∪ K.   1.4.42 Definition. A continuum X is weakly irreducible provided that the complement of each finite union of subcontinua of X only has a finite number of components. 1.4.43 Theorem. If X is an irreducible continuum, then X is weakly irreducible. Proof. Let Z1 , . . . , Zn be a finite family of subcontinua of X such that Zj ∩ Zk = ∅ if j = k. For each j ∈ {1, . . . , n}, by Theorem 1.4.40, we assume that X \ Zj = Uj ∪ Vj , where Uj and Vj are open connected subsets of X. Without loss of n n−1 generality, we suppose that j=2 Zj ⊂ V1 and j=1 Zj ⊂ Un . Hence, U1 and Vn j−1 may n be empty. We assume also that, for each j ∈ {2, . . . , n−1}, k=1 Zk ⊂ Uj and k=j+1 Zk ⊂ Vj . For every j ∈ {1, . . . , n−1}, let Rj = (Vj \Zj+1 )∩(Uj+1 \Zj ).

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By the subset of X. Note that proof of Lemma 1.4.41,  Rj is a connected open n n−1 n X\ = U1 ∪ j=1 Zj j=1 Rj ∪ Vn . Thus, X \ j=1 Zj has, at most, n + 1 components. Therefore, X is weakly irreducible.   1.4.44 Definition. A continuum X is unicoherent provided that for every pair, A and B, of subcontinua of X such that X = A∪B, A∩B is connected. We say that X is hereditarily unicoherent if each subcontinuum of X is unicoherent (Figure 1.14).

Fig. 1.14

Unicoherent

Not Unicoherent

1.4.45 Definition. Let X and Y be continua. A surjective map f : X → → Y is said to be monotone if f −1 (y) is connected for each y ∈ Y . The next lemma gives a very useful characterization of monotone maps. 1.4.46 Lemma. Let X and Y be continua, and let f : X → → Y be a surjective map. Then f is monotone if and only if f −1 (C) is connected for each connected subset C of Y . Proof. Suppose f is monotone and let C be a subset of Y such that f −1 (C) is not connected. Then there exist two nonempty subsets A and B of X such that f −1 (C) = A ∪ B, ClX (A) ∩ B = ∅ and A ∩ ClX (B) = ∅. Observe that if y ∈ C and f −1 (y) ∩ A = ∅, then f −1 (y) ⊂ A (f −1 (y) is connected). Let M = {y ∈ C | f −1 (y) ⊂ A}. Hence, A = f −1 (M ). Similarly, if N = {y ∈ C | f −1 (y) ⊂ B}, then B = f −1 (N ). Note that C = M ∪ N . Suppose there exists y ∈ ClY (M ) ∩ N . Since y ∈ N , f −1 (y) ⊂ B. Since y ∈ ClY (M ), there exists a net {yλ }λ∈Λ of points of M converging to y [40, 1.6.3]. This implies that f −1 (yλ ) ⊂ A for every λ ∈ Λ. Let xλ ∈ f −1 (yλ ). Since X is a compactum, without loss of generality, we assume that {xλ }λ∈Λ converges to a point x [40, 3.1.23 and 1.6.1]. Note that x ∈ ClX (A) and, by the continuity of f , f (x) = y [97, Proposition 3.38]. Thus, x ∈ ClX (A) ∩ f −1 (y) ⊂ ClX (A) ∩ B, a contradiction. Hence, ClY (M ) ∩ N = ∅. Similarly, M ∩ ClY (N ) = ∅. Therefore, C is not connected. The other implication is clear.   1.4.47 Definition. An arc is a continuum with the property that in contains exactly two points whose complement is connected. A metric arc is a metric continuum homeomorphic to [0, 1].

34

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Now, we present the definition of an arc-smooth continuum. This notion is used to study continua which are strictly point T -asymmetric (Definition 7.3.1). 1.4.48 Definition. An arcwise connected continuum X is arc-smooth at p ∈ X if there exists a map α : X → C1 (X) (Definition 1.6.1) such that α(p) = {p}, α(x) is an arc in X joining p and x and if y ∈ α(x), then α(y) ⊂ α(x). The continuum X is arc-smooth if there exists a point at which it is arc-smooth. 1.4.49 Theorem.  If X is an arc-smooth continuum, then for each closed subset H of X, the set x∈H α(x) is a subcontinuum of X.  Proof. Suppose X is arc-smooth at p and let M = x∈H α(x). Note that for each x ∈ H, p ∈ α(x). Hence, M is a connected subset of X. Let z ∈ Cl(M ). Then there exists a net {zλ }λ∈Λ of points of M converging to z [40, 1.6.3]. Thus, for each λ ∈ Λ, there exists xλ ∈ H such that zλ ∈ α(xλ ). Since H is closed in X, without loss of generality, we assume that the net {xλ }λ∈Λ converges to a point x ∈ H [40, 3.1.23 and 1.6.1]. Since X is arc-smooth, α is continuous. Hence, the net of arcs {α(xλ )}λ∈Λ converges to the arc α(x) [97, Proposition 3.38]. Note that, by the properties of α, the net of arcs {α(zλ )}λ∈Λ converges to the arc α(z) and z ∈ α(x). Thus, z ∈ M and M is closed. Therefore, M is a subcontinuum of X.   1.4.50 Definition. Let X be a continuum and let K be a subcontinuum of X. Then K is a terminal subcontinuum of X if for each subcontinuum L of X such that L ∩ K = ∅, then either L ⊂ K or K ⊂ L. 1.4.51 Lemma. Let X be a continuum. If W is a proper terminal subcontinuum of X, then Int(W ) = ∅. Proof. Suppose W is a proper terminal subcontinuum of X and Int(W ) = ∅. Note that Bd(Int(W )) ⊂ W . Let x ∈ X \ W and let C be the component of X \ Int(W ) containing x. By Theorem 1.4.36, C ∩ Bd(Int(W )) = ∅. Hence, C ∩ W = ∅, and C \ W = ∅. Since W is a terminal subcontinuum of X, W ⊂ C, a contradiction to the fact that C ∩ Int(W ) = ∅. Therefore, Int(W ) = ∅.   1.4.52 Lemma. Let X be a continuum. If X is aposyndetic, then X does not contain nondegenerate proper terminal subcontinua. Proof. Suppose Y is a nondegenerate proper terminal subcontinuum of X. Let y ∈ Y and let y  ∈ Y \{y}. Then since X is aposyndetic, there exists a subcontinuum W of X such that y ∈ Int(W ) ⊂ W ⊂ X \ {y  }. Since Y is terminal and Y \ W = ∅, W ⊂ Y . Thus, y is an interior point of Y . Since y is an arbitrary point of Y , all the points of Y are interior points. Hence, Y is a nonempty open and closed proper subset of X. This contradicts the fact that X is connected. Therefore, X does not contain nondegenerate proper terminal subcontinua.   1.4.53 Lemma. Let X and Y be continua. If f : X → → Y is a monotone map and W is a terminal subcontinuum of X, then f (W ) is a terminal subcontinuum of Y . Proof. Let W be a terminal subcontinuum of X. Let K be a subcontinuum of Y such that K ∩ f (W ) = ∅. Since f is a monotone map, f −1 (K) is a subcontinuum

1.4 Continua

35

of X (Lemma 1.4.46) such that f −1 (K) ∩ W = ∅. Hence, since W is a terminal subcontinuum of X, either W ⊂ f −1 (K) or f −1 (K) ⊂ W . This implies that either f (W ) ⊂ K or K ⊂ f (W ) (f is surjective). Therefore, f (W ) is a terminal subcontinuum of Y .   1.4.54 Corollary. Let X and Y be continua. If Z is a terminal subcontinuum of X and h : X → → Y is a homeomorphism, then h(Z) is a terminal subcontinuum of Y . 1.4.55 Definition. A continuum X is homogeneous provided that for every pair of points x1 and x2 of X, there exists a homeomorphism h : X → → X such that h(x1 ) = x2 . 1.4.56 Notation. Let X be a homogeneous continuum, let H(X) be its group of homeomorphisms. If x ∈ X, then define γx : H(X) → X by γx (h) = h(x). 1.4.57 Definition. A homogeneous continuum X is said to be an Effros continuum, if γx is an open map (Notation 1.4.56) for all x ∈ X. 1.4.58 Definition. Let Z be a Tychonoff space and let U be a uniformity that induces the topology of Z. Then Z has the uniform property of Effros with respect to U, provided that for each U ∈ U, there exists V ∈ U such that if z1 and z2 are two points of Z with ρ(z1 , z2 ) < V , there exists a homeomorphism h : Z → → Z such that h(z1 ) = z2 and ρ(z, h(z)) < U for all z ∈ Z. The entourage V is called an Effros entourage for U . A homeomorphism h : Z → → Z satisfying ρ(z, h(z)) < U , for all z ∈ Z, is called a U -homeomorphism. 1.4.59 Theorem. Let Z be a connected Tychonoff space and let U be a uniformity that induces the topology of Z. If Z has the uniform property of Effros with respect to U, then Z is a homogeneous space. Proof. Let z1 and z2 be two points of Z and let U ∈ U be such that ρ(z1 , z2 ) ≥ U . Since Z has the uniform property of Effros with respect to U, there exists an Effros entourage V for U . Let V  ∈ U be such that 2V  ⊂ V . Note that {Int(B(z, V  )) | z ∈ Z} is an open cover of Z. Since Z is connected, by [24, (2F2)], there exist w1 , . . . , wn in Z such that z1 ∈ Int(B(w1 , V  )), z2 ∈ Int(B(wn , V  )) and Int(B(wj , V  )) ∩ Int(B(wk , V  )) = ∅ if and only if |j − k| ≤ 1. Let x0 = z1 and let xn+1 = z2 . For each j ∈ {1, . . . , n − 1}, let xj ∈ Int(B(wj , V  )) ∩ Int(B(wj+1 , V  )). Let j ∈ {0, . . . , n}. Since ρ(xj , xj+1 ) < V , there exists a U homeomorphism hj : Z → → Z such that hj (xj ) = xj+1 . Let h = hn ◦ · · · ◦ h0 . Then h : Z → → Z is a homeomorphism such that h(z1 ) = z2 . Therefore, Z is a homogeneous space.   1.4.60 Theorem. If X is an Effros continuum, then X has the uniform property of Effros. Proof. By Theorem 1.3.9, the topology induced by U" X on H(X) coincides with the compact-open topology. Also, U! is a base for the uniformity U" X . Let x0 be a point of X and let U ∈ UX (Remark 1.3.4). Let U  ∈ UX be such that 2U  ⊂ U . Since X " ))) is an Effros continuum, the map γx0 is an open map. Then γx0 (Int(B(1X , U

36

1 Preliminaries

is an open neighbourhood of x0 in X. Then there exists Vx0 ∈ UX such that " ))). Thus, if x ∈ B(x0 , Vx ), then there exists B(x0 , Vx0 ) ⊂ γx0 (Int(B(1X , U 0 " )) such that h(x0 ) = x and ρ(z, h(z)) < U  for every z ∈ X. h ∈ Int(B(1X , U Observe that {Int(B(x, Vx )) | x ∈ X} is an open cover of X. Since X is compact, by Theorem 1.3.6, there exists V ∈ UX such that C(V ) refines {Int(B(x, Vx )) | x ∈ X}. Let x1 and x2 be points of X such that ρ(x1 , x2 ) < V . Then there exists x3 in X such that x1 and x2 both belong to Int(B(x3 , Vx3 )). Let h1 and h2 be elements " )) such that h1 (x3 ) = x1 and h2 (x3 ) = x2 . Let h = h2 ◦ h−1 . of Int(B(1X , U 1 Then h is a homeomorphism of X and h(x1 ) = x2 . Let z be a point of X. Since −1 −1 −1  ρ(z, h(z)) = ρ(z, h2 ◦ h−1 1 (z)), ρ(z, h1 (z)) = ρ(h1 ◦ h1 (z), h1 (z)) < U , −1 −1   ρ(h1 (z), h2 ◦ h1 (z)) < U and 2U ⊂ U , we obtain that ρ(z, h(z)) < U . Therefore, X has the uniform property of Effros.   As a first consequence of the uniform property of Effros, we have: 1.4.61 Theorem. Let X be a continuum with the uniform property of Effros. Then the following are equivalent: (1) (2) (3) (4) (5)

X X X X X

is locally connected; is locally connected at some point; is connected im kleinen at some point; is almost connected im kleinen at every point of X; is almost connected im kleinen at some point of X.

Proof. By Theorem 1.4.59, X is homogeneous. We only prove that (5) implies (1), the other implications are clear. Suppose X is almost connected im kleinen at x0 . Let A be an open subset of X such that x0 ∈ A. Let U ∈ UX be such that B(x0 , U ) ⊂ A. Let U  ∈ UX be such that 2U  ⊂ U . Since X has the uniform property of Effros, there exists an Effros entourage V for U  . Without loss of generality, we assume that V ⊂ U  . Since X is almost connected im kleinen at x0 , there exists a subcontinuum W of X such that W ⊂ Int(B(x0 , V ) and Int(W ) = ∅. Let w0 ∈ Int(W ). Then ρ(x0 , w0 ) < V . Hence, there exists a U  -homeomorphism h : X → → X such that h(w0 ) = x0 . Thus, x0 ∈ Int(h(W )). Since for each w ∈ W , ρ(x0 , w) < V , ρ(w, h(w)) < U  and V ⊂ U  , we obtain that for every w ∈ W , ρ(x0 , h(w)) < 2U  . Hence, since 2U  ⊂ U , we have that h(W ) ⊂ B(x0 , U ) ⊂ A. Thus, X is connected im kleinen at x0 . Since X is homogeneous, X is connected im kleinen at each of its points. Therefore, X is locally connected, Theorem 1.4.18.   1.4.62 Definition. A metric continuum X is a graph provided that it can be written as the union of finitely many metric arcs any two of which are either disjoint or intersect only in one or both of their end points. 1.4.63 Definition. A metric continuum X is of type λ provided that X is irreducible and each indecomposable subcontinuum of X has empty interior. 1.4.64 Remark. By [119, Theorem 10, p. 15], a metric continuum X is of type λ if and only if it admits a monotone upper semicontinuous decomposition G such that

1.4 Continua

37

each element of G is nowhere dense and X/G is a metric arc. Each element of G is called a layer of X. 1.4.65 Definition. A metric continuum X of type λ for which G (Remark 1.4.64) is continuous is a continuously irreducible continuum. 1.4.66 Definition. A continuum X is a θ-continuum (θn -continuum for some n ∈ N) provided that for each subcontinuum K of X, X \ K only has finitely many components (X \ K has at most n components). 1.4.67 Definition. Let X be a metric θ-continuum (metric θn -continuum). We say that X is of type A provided that it admits a monotone upper semicontinuous decomposition D whose quotient space is a graph. X is of type A if, in addition, the elements of the decomposition have empty interior. 1.4.68 Definition. A metric θ-continuum (metric θn -continuum) of type A for which the decomposition D is continuous is a continuously type A θ-continuum (θn -continuum). 1.4.69 Lemma. Let X be a continuously type A θ-continuum and let q : X → → D be the quotient map, where D is a graph. If K is a proper subcontinuum of X, then q(K) = D. Proof. Let K be a proper subcontinuum of X. If IntX (K) = ∅, then, since q is an open map and D is a finite graph, q(K) is a degenerate subcontinuum of D. Thus, q(K) = D. Assume that IntX (K) = ∅. Since X is θ-continuum, X \ K only has a finite number of components, say C1 , . . . , C . Note that each Cj is an open subset of X, Lemma 1.4.1. Suppose q(K) = D. Since C1 is open, q(C1 ) is an open connected subset of D. Let χ0 ∈ IntD (q(C1 )) and let {χj }∞ j=1 be a sequence of distinct elements of IntD (q(C1 )) converging to χ0 . Since D is a graph, D \ {χj }∞ j=0 has   ∞ −1 (χk ) . Then, since q(K) = infinitely many components. Let L = K ∪ j=0 q D, L is a subcontinuum of X. Also, since K ∩ C1 = ∅ and, for each j ∈ N ∪ {0}, C1 ∩ q −1 (χj ) = ∅, we obtain that X \ L has infinitely many components, a contradiction to the fact that X is a θ-continuum. Therefore, q(K) = D.   A proof of the next result may be found in [45, Theorem 3.4]. 1.4.70 Theorem. A continuum X is weakly irreducible if and only if X is a θcontinuum. Next, we consider the notion of generalized metric continuum. 1.4.71 Definition. A generalized metric continuum is a locally compact, connected, metric space 1.4.72 Definition. A generalized metric continuum Z is an exhausted metric σcontinuum provided that there exists a sequence {Kn }∞ n=1 of subcontinua of Z such

38

that Kn ⊂ Int(Kn+1 ), for all n ∈ N, and Z = is an exhaustive sequence of subcontinua of Z.

1 Preliminaries

∞ n=1

Kn . The sequence {Kn }∞ n=1

1.4.73 Example. Let X be the topologist sine curve, Example 1.4.9, and let Z = X \ {(0, 0)}. Then Z is a generalized metric continuum that is not an exhaustive σ-continuum. 1.4.74 Definition. A Hausdorff space Z is continuumwise connected provided that for each pair of points z1 and z2 of Z, there exists a subcontinuum W of Z such that {z1 , z2 } ⊂ W . 1.4.75 Theorem. Let Z be a generalized metric continuum. Then the following are equivalent: (1) Z is an exhausted metric σ-continuum; (2) For every pair of points p and q of Z, there exists a subcontinuum K of Z such that {p, q} ⊂ Int(K); (3) Z is continuumwise connected and for each point p in Z, there exists a subcontinuum K of Z such that p ∈ Int(K); (4) For every point p of Z, there exists a subcontinuum K of Z such that p ∈ Int(K); (5) For each compact subset L of Z, there exists a subcontinuum K of Z such that L ⊂ Int(K); (6) For every compact subset L of Z, there exists a subcontinuum K of Z such that L ⊂ K; (7) The hyperspace of all compact subsets of Z is arcwise connected; (8) The hyperspace of all compact subsets of Z is continuumwise connected. Proof. Note that if Z is an exhausted metric σ-continuum, then, by definition, for every pair of points p and q of Z, there exists a subcontinuum K of Z such that {p, q} ⊂ Int(K). Now, suppose (2) holds. Then Z is continuumwise connected and for each point p of Z, there exists a subcontinuum K of Z such that p ∈ Int(K). Assume (3) is true. Then for every point p of Z, there exists a subcontinuum K of Z such that p ∈ Int(K). Now suppose (4) holds, we show (3) is true. Let p be a point of Z and let κ(p) be the union of all subcontinua of Z containing p. Note that (4) implies that κ(p) is an open subset of Z. Since the family {κ(z) | z ∈ Z} forms a decomposition of Z, κ(p) is a closed subset of Z. Since Z is connected, κ(p) = Z. Thus, Z is continuumwise connected. Next, we assume (4) and prove (5) holds. Let L be a compact subset of Z. For each point l ∈ L, let Kl be a subcontinuum of Z  such that l ∈ Int(Kl ). Since L is n compact, there exist l1 , . . . , ln in L such that L ⊂ j=1 Int(Klj ). Let z be a point of Z. Since (3) is true, Z is continuumwise connected. Hence, for each jn ∈ {1, . . . , n}, there exists a subcontinuum Mj such that {z, lj } ⊂ Mj . Let K = j=1 (Klj ∪Mj ). Then K is a subcontinuum of Z and L ⊂ Int(K). Clearly, if (5) is true, then (6) holds.

1.4 Continua

39

Assume (6), we show (1). We use mathematical induction. Let {Ln }∞ n=1 be a sequence of compact subsets of Z such that L ⊂ Int(L ) for all n ∈ N and Z = n n+1 ∞ L . By (6), there exists a subcontinuum K of Z such that L ⊂ Int(K n 1 1 1 ). n=1 Suppose K1 , . . . , Kn are subcontinua of Z such that Kj ⊂ Int(Kj+1 ) for each j ∈ {1, . . . , n−1}. By (6), there exists a subcontinuum Kn+1 of Z such that Ln+1 ∪ Kn ⊂ Int(Kn+1 ). Hence, {Kn }∞ n=1 is an exhaustive sequence of subcontinua of Z. Therefore, Z is an exhausted metric σ-continuum. Next, suppose (6) and we prove (7). Let L1 and L2 be compact subsets of Z. By (6), there exists a subcontinuum K of Z such that L1 ∪ L2 ⊂ K. By [105, (1.8)], there exist two arcs α, β : [0, 1] → C1 (K) such that α(0) = L1 , α(1) = K, β(0) = L2 and β(1) = K. Thus, α([0, 1]) ∪ β([0, 1]) contains an arc joining L1 and L2 . Clearly, (7) implies (8). To finish, we show that if (8) is true, then (6) holds. Let L be a compact subset of Z and let z be a point of Z. Since the hyperspace of compact subsets of Z is continuumwise connected, there exists a subcontinuum K of this space containing  L and {z}. Let K = K. By [105, (1.49)], K is a subcontinuum of Z containing L.   We end this section with some results of generalized inverse limits of metric continua with a single bonding function. 1.4.76 Theorem. If X is a metric continuum and G is a monotone upper semicontinuous decomposition of X, then lim fG is an infinite-dimensional decomposable ←− metric continuum. Proof. By [65, Theorem 3.2] and Lemma 1.2.27, lim fG is compact and connected. ←− By [92, Lemma 1.1.7 and 1.1.8], lim fG is a metric space. Since, by Lemma 1.2.27, ←− if G ∈ G, then Gω ⊂ lim fG , we have that lim fG is infinite-dimensional. ←−

←−

ω To see that lim fG is decomposable let (xn )ω n=1 and (yn )n=1 be two points of ←−

ω lim fG . Then there exist two points z1 and z2 of X such that (xn )ω n=1 ∈ Gz1 ←−

ω ω ω ω and (yn )ω n=1 ∈ Gz2 . Then Gz1 ∪ ΔX ∪ Gz2 is a proper subcontinuum of lim fG ω containing (xn )ω n=1 and (yn )n=1 . Therefore, lim fG is decomposable. ←−

←−

 

1.4.77 Remark. Let X be a metric continuum and let G be an upper semicontinuous decomposition of X. If G  = {K | K is a component of G, for some G ∈ G}, then G  is an upper semicontinuous decomposition of X (Theorem 1.1.35) and lim fG  is a metric subcontinuum of lim fG . ←−

←−

1.4.78 Theorem. If X is a metric continuum and G = {Gx | x ∈ X} is a continuous decomposition of X, then G ∗ = {Gω x | x ∈ X} is a continuous decomposition of lim fG . Moreover, lim fG /G ∗ is homeomorphic to X/G. ←−

←−

40

1 Preliminaries

Proof. Let q : X → → X/G be the quotient map, and let π1 : lim fG → X be the ←−

projection map to the first factor space. Note that G ∗ = {(q ◦ π1 )−1 (χ) | χ ∈ X/G}. Hence, G ∗ is an upper semicontinuous decomposition of lim fG (Theorem 1.1.18). ←−

ω points of Gω Let (xn )ω x , and let U be an open subset of n=1 and (yn )n=1 be two  m ω lim fG such that (xn )n=1 ∈ U . Let j=1 πj−1 (Uj ) be a basic open set such that ←− m −1 (xn )ω n=1 ∈ j=1 πj (Uj ) ⊂ U , where πj : lim fG → X is the projection map ←−

to the jth factor space. Note that for each j ∈ {1, . . . , m}, xj ∈ Gx ∩ Uj . Since G is a continuous decomposition, for each j ∈ {1, . . . , m}, there exists an open subset Vj such that yj∈ Vj and if G ∈ G is such that G ∩ Vj = ∅, we have that m G ∩ Uj = ∅. Let V = j=1 πj−1 (Vj ). Let Gω ∈ G ∗ be such that Gω ∩ V = ∅. Then G ∩ Vj = ∅ for every j ∈ {1, . . . , m}. By the construction of the Vj s, we have that G ∩ Uj = ∅ for all j ∈ {1, . . . , m}. For each j ∈ {1, . . . , m}, let zj ∈ G ∩ Uj , and m −1 ω for each j ≥ m + 1, let zj ∈ G. Then (zn )ω n=1 ∈ G ∩ j=1 πj (Uj ) . Hence Gω ∩ U = ∅. Therefore, G ∗ is continuous. Let q ∗ : lim fG → lim fG /G ∗ be the quotient map. Note that q ◦ π1 is constant ←− ←− on each of the fibres of q ∗ and q ∗ is constant on each of the fibres of q ◦ π1 . Then, by [36, 3.2, p. 123], there exist two maps h : lim fG /G ∗ → X/G and h : X/G → ←−

lim fG /G ∗ such that h ◦ q ∗ = q ◦ π1 and h ◦ q ◦ π1 = q ∗ . Note that h and h are ←−

homeomorphisms. Therefore, lim fG /G ∗ is homeomorphic to X/G. ←−

 

1.5 Uniformly Completely Regular Maps We extend the notion of completely regular maps [37] to uniform spaces and present the properties needed to prove the F. B. Jones’ Aposyndetic Decomposition Theorem (Theorem 3.3.8). 1.5.1 Definition. Let X and Y be Tychonoff spaces, let U and U be uniformities inducing the topologies of X and Y , respectively. Let g : X → → Y be a uniformly continuous and surjective function. Then g is uniformly completely regular with respect to U and U if for each U ∈ U, there exists V ∈ U such that if y and y  are two points of Y such that ρY (y, y  ) < V , there exists a homeomorphism h : g −1 (y) → → g −1 (y  ) such that ρX (z, h(z)) < U for all z ∈ g −1 (y). 1.5.2 Theorem. Let X and Y be compacta. If g : X completely regular map, then g is open.

→ →

Y is a uniformly

Proof. Let W be an open subset of X and let y ∈ g(W ). Then there exists x ∈ W such that g(x) = y. Also, there exists U ∈ UX such that BX (x, U ) ⊂ W . Since g is uniformly completely regular, there exists V ∈ UY such that if y  is a point of Y and ρY (y, y  ) < V , there exists a homeomorphism h : g −1 (y) → → g −1 (y  ) such that −1 ρX (z, h(z)) < U for every z ∈ g (y). In particular, ρX (x, h(x)) < U . Hence,

1.5 Uniformly Completely Regular Maps

41

h(x) ∈ W and g(h(x)) = y  . Thus, BY (y, V ) ⊂ g(W ). Since y is an arbitrary point of g(W ), g(W ) is an open subset of Y . Therefore, g is an open map.   1.5.3 Definition. Let D be the class of metric spaces or the class of normal spaces. By an absolute neighbourhood retract for the class D we mean a space Y in D such that for each Z ∈ D for which there exists an embedding i : Y → Z such that i(Y ) is a closed subset of Z, there exist an open subset U of Z containing i(Y ) and a map r : U → → i(Y ) such that r(r(u)) = r(u) for all u ∈ U . 1.5.4 Remark. Note that an absolute neighbourhood retract Y for the class of metric spaces is an absolute retract for the class of normal spaces if and only if Y is separable and topologically complete [61, Theorem 4.1, p. 86]. 1.5.5 Theorem. Let X be a continuum. If A is a terminal subcontinuum of X, if B is a nonempty subset of X disjoint from A, and if f : A → Y is a map from A into a compact metric absolute neighbourhood retract Y , then there exists a map F : X → Y such that F |A = f and F |B is homotopic to a constant map. Proof. Since Y is a metric absolute neighbourhood retract, there exist an open subset U of X containing A and a map g : U → Y such that g|A = f [99, 1.5.2]. Without loss of generality, we assume that U ∩ B = ∅. By [61, Theorem 7.1, p. 96], Y is locally contractible. Hence, there exists an open neighbourhood V of a point a of A such that V ⊂ U and g|V is homotopic to a constant map. Note that A\V and X \U are two closed subsets of X \V such that no connected subset of X \ V intersects both A \ V and X \ U . (If K is a connected subset of X \ V such that K ∩ (A \ V ) = ∅ and K ∩ (X \ U ) = ∅, then Cl(K) is a continuum in X \ V intersecting A and X \ A. Since A is terminal, A ⊂ Cl(K). This implies that V ∩ Cl(K) = ∅, a contradiction.) By Theorem 1.4.8, there exist two disjoint closed subsets X1 and X2 of X such that X \ V = X1 ∪ X2 , A \ V ⊂ X1 and X \ U ⊂ X2 . Note that X1 ∪A and X2 are two disjoint closed subsets of X. Then, by Urysohn’s Lemma, there exists a map  h : X → → [0, 1] such that  h(X 1 ∪ A) = {0} and h(X2 ) = {1}. Let M = h−1 0, 12 and let N = h−1 12 , 1 . Then X = M ∪N , A ⊂ M , X \ U ⊂ N and M ∩ N ⊂ V \ A. Since g|M ∩N is homotopic to a constant map (because M ∩ N ⊂ V and g|V is homotopic to a constant map), by [24, (15.A.1)], there exists a map k : N → Y such that k is homotopic to a constant map and k|M ∩N = g|M ∩N . Let F : X → Y be given by F (x) =

 g(x),

if x ∈ M,

k(x),

if x ∈ N.

Then F is well defined and continuous. Note that F |A = g|A = f and F |B = k|B . Since k|B is homotopic to a constant map (because B ⊂ X \ U ⊂ N and k is homotopic to a constant map), F is the desired extension of f .  

42

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1.5.6 Definition. A continuum X is cell-like if each map of X into a compact metric absolute neighbourhood retract is homotopic to a constant map. 1.5.7 Definition. Let X and Y be continua and let f : X → → Y be a surjective map. Then f a cell-like map if for each y ∈ Y , f −1 (y) is a cell-like subcontinuum of X. 1.5.8 Theorem. Let X and Z be nondegenerate continua. Suppose g : X → → Z is a monotone uniform completely regular map. If z1 is a point of Z such that g −1 (z1 ) is a terminal subcontinuum of X, then g is a cell-like map. Proof. Let Y be a compact metric absolute neighbourhood retract and let f : g −1 (z1 ) → Y be a map. Let z2 ∈ Z \ {z1 }. By Theorem 1.5.5, there exists an extension F : X → Y of f such that F |g−1 (z2 ) is homotopic to a constant map. Since Y is a compact metric absolute neighbourhood retract, by [61, Theorem 1.1, p. 111], there exists ε > 0 such that for any space A and any two maps k, k  : A → Y such that d(k(a), k  (a)) < ε, for all a ∈ A, where d is a metric compatible with Y , k and k  are homotopic. Since F is uniformly continuous (Theorem 1.3.15), there exists V ∈ UX such that if ρX (x, x ) < V , then d(F (x), F (x )) < ε. Let Z  = {z ∈ Z | F |g−1 (z) is homotopic to a constant map}. Note that, since z2 ∈ Z  , we have that Z  = ∅. We show that Z  is closed in Z. Let z ∈ Cl(Z  ). Since g is uniformly completely regular, there exists W ∈ UZ such that if z  is a point of Z such that ρZ (z, z  ) < W , then there exists a homeomorphism h : g −1 (z) → g −1 (z  ) such that ρX (x, h(x)) < V for each x ∈ g −1 (z). Observe that there exists a point z  ∈ Z  such that ρZ (z, z  ) < W . Thus, there exists a homeomorphism h as described above. Let x ∈ g −1 (z). Then ρX (x, h(x)) < V . Hence, d(F (x), F (h(x))) = d(F |g−1 (z) (x), (F |g−1 (z ) ) ◦ h(x)) < ε. Thus, F |g−1 (z) and (F |g−1 (z ) ) ◦ h are homotopic. Since z  ∈ Z  , (F |g−1 (z ) ) ◦ h is homotopic to a constant map. As a consequence of this, F |g−1 (z) is homotopic to a constant map, and z ∈ Z  . Therefore, Z  is closed in Z. A similar argument proves that Z  is open in Z. Since Z  is a nonempty, closed and open subset of Z and Z is connected, Z  = Z. This implies that F |g−1 (z1 ) = f is homotopic to a constant map. Since Y and f are arbitrary, g −1 (z1 ) is a cell-like continuum. Since the fibres of uniformly completely regular maps defined on continua are homeomorphic, g is a cell-like map.  

1.6 Hyperspaces We give the definition of the main hyperspaces associated with a continuum. We present some of their elementary properties. We introduce the property of Kelley for continua and we relate it to the uniform property of Effros. 1.6.1 Definition. Given a compactum X, we define its hyperspaces as the following sets:

1.6 Hyperspaces

43

2X = {A ⊂ X | A is closed and nonempty}, and for each n ∈ N Cn (X) = {A ∈ 2X | A has at most n components}, Fn (X) = {A ∈ 2X | A has at most n points}. Fn (X) is called n-fold symmetric product of X and Cn (X) is called n-fold hyperspace of X (Figure 1.15).

Fig. 1.15 Hyperspaces

1.6.2 Remark. Given a continuum X, let us observe that for each n ∈ N, Fn (X) ⊂ Cn (X),

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Cn (X) ⊂ Cn+1 (X), and that Fn (X) ⊂ Fn+1 (X). The n-fold symmetric products were defined by Borsuk and Ulam [15]. These hyperspaces have been studied by many people. Some recent results and references about n-fold symmetric products may be found in [79, 80] and [48]. It is not clear who defined or where the n-fold hyperspaces were defined. A study of the n-fold hyperspaces for metric continua is presented in [92, Chapter 6], see also [81]. 1.6.3 Notation. Let X be a compactum. Given a finite collection of nonempty subsets of X, U1 , . . . , Um , we define U1 , . . . , Um  =  A ∈ 2X

m Uk and A ∩ Uk = ∅ for each k ∈ {1, . . . , m} A⊂ k=1

(Figure 1.16).

Fig. 1.16 The set U1 , . . . , Um 

1.6.4 Notation. Let X be a compactum and let n ∈ N. If U1 , . . . , Um are nonempty subsets of X, then U1 , . . . , Um n = U1 , . . . , Um  ∩ Cn (X). A proof of the next theorem may be found in [105, (0.11)]: 1.6.5 Theorem. Let X be a compactum. If B = {U1 , . . . , Un  | U1 , . . . , Un are open subsets of X and n ∈ N}, then B is a basis for a topology of 2X .

1.6 Hyperspaces

45

1.6.6 Definition. The topology for 2X given by Theorem 1.6.5 is called the Vietoris topology. By [98, 4.9.6 and 4.13.5] we have: 1.6.7 Theorem. If X is a compactum, then 2X and C1 (X) are compacta. 1.6.8 Lemma. Let X bea continuum. If K is a subcontinuum of 2X and K ∈ K, then each component of K intersects K.  Proof. First, note that K ∈ 2X [98, 2.5.2]. Suppose that there exists a component C of K such that K ∩ C =∅. Hence, byTheorem 1.4.8, there exist two disjoint closed subsets K1 and K2 of K such that K = K1 ∪K2 , K ⊂ K1 and C ⊂ K2 . Let K1 = {L ∈ K | L ⊂ K1 } and K2 = {L ∈ K | L ∩ K2 = ∅}. Then K1 and K2 are disjoint closed subsets of K. Also, K = K1  ∪ K2 , a contradiction to the fact that K is connected. Therefore, each component of K intersects K.   ∞ 1.6.9 Lemma. If X is a compactum, then F(X) = n=1 Fn (X) is dense in 2X . Proof. Let U be an open subset of 2X and let A ∈ U . By Theorem 1.6.5, there exist open subsets U1 , . . . , Um of X such that A ∈ U1 , . . . , Um  ⊂ U . For each j ∈ {1, . . . , m}, let xj ∈ Uj . Then {x1 , . . . , xm } ∈ U1 , . . . , Um  ∩ F(X). Hence, {x1 , . . . , xm } ∈ U ∩ F(X). Therefore, F(X) is dense in 2X .   1.6.10 Lemma. Let X be a continuum and let n ∈ N. Then X is locally connected if and only if Fn (X) is locally connected. Proof. Suppose X is locally connected. Then X n is locally connected. By [98, → Fn (X) is continuous and surjective. Proposition 2.4.3], the function fn : X n → Hence, Fn (X) is locally connected. Suppose Fn (X) is locally connected. Since F1 (X) is homeomorphic to X, we may assume that n ≥ 2. Let x0 be a point of X and let U be an open subset of X such that x0 ∈ U . Since Fn (X) is locally connected, there exists a connected open subset W of Fn (X) such that {x0 } ∈ W ⊂ Cl(W) ⊂ U  ∩ Fn (X). By Theorem 1.6.5, there exist open subsets V1 , . . . , Vm of X such X that {x0 } ∈ V1 , . . . , Vm  ∩ Fn (X) ⊂ W.  Since Cl(W) is a subcontinuum of 2 and Cl(W) ∩ C1 (X), by Lemma 1.6.8, Cl(W) is  a subcontinuum of X.  mSince Cl(W) ⊂ U , we have that Cl(W) ⊂ U and x0 ∈ Cl(W). Let V = j=1 Vj , and let x  ∈ V . Then {x} ∈  V1 , . . . , Vm  ∩ Fn (X) ⊂ W. Hence, x ∈ Cl(W), and V ⊂ Cl(W). Thus, Cl(W) is a connected neighbourhood of x0 contained in U . Thus, X is connected im kleinen at each of its points. Therefore, X is locally connected (Theorem 1.4.18).   1.6.11 Definition. Let X be a continuum, we define a uniformity on 2X as follows: If U ∈ UX (Remark 1.3.4), then let 2U = {(A, A ) ∈ 2X × 2X | A ⊂ B(A , U ) and A ⊂ B(A, U )}. Let BX = {2U | U ∈ UX }. Then BX is a base for a uniformity, denoted by 2UX [40, 8.5.16]. Observe that the topology generated by 2UX coincides with the Vietoris topology [98, 3.3]. Note that 2X is compact and Hausdorff, Theorem 1.6.7. Thus, 2UX is unique (Remark 1.3.4), and

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2UX = {W ⊂ 2X × 2X | there exists U ∈ UX such that 2U ⊂ W}. Given n ∈ N, we consider the restriction of 2UX to Cn (X) and Fn (X) and denoted by Cn (UX ) and Fn (UX ), respectively. 1.6.12 Remark. Let X be a continuum. We need to consider the hyperspace 22 U with the corresponding uniformity given by Definition 1.6.11, 22 X .

X

1.6.13 Lemma. Let X be a continuum and let A and B be elements of 22 . If   U U ∈ UX and ρ22X (A, B) < 22 , then ρ2X ( A, B) < 2U . X

Proof. Let U ∈ UX and let A and B be elements of 22 such that ρ22X (A, B) <    U 22 . Observe that, by [98, 2.5.2], A and B both belong to 2X . Let a ∈ A. U Then there exists A ∈ A such that a ∈ A. Since ρ22X (A, B) < 22 , A ⊂ U B(B, 22 ). Hence, there exists B ∈ B such that ρ2X (A, B) < 2U . This implies  that A⊂ B(B, U ). Thus, there exists b ∈ B such that ρ (a, b) < U . Hence, A⊂ X      B ( B, U ). Similarly, B ⊂ B ( A, U ). Therefore, ρ2X ( A, B) < 2U .   X

1.6.14 Definition. Let f : X → Y be a map between compacta. Then 2f : 2X → 2Y given by 2f (A) = f (A) is called the induced map between the hyperspaces of closed subsets of X and Y . For each n ∈ N, the functions Cn (f ) : Cn (X) → Cn (Y ) and Fn (f ) : Fn (X) → Fn (Y ) given by Cn (f ) = 2f |Cn (X) and Fn (f ) = 2f |Fn (X) are called the induced map between the n-fold hyperspaces of X and Y and the induced map between the n-fold symmetric products of X and Y , respectively. A proof of the following two theorems may be found in [98, 5.10]. 1.6.15 Theorem. Let f : X → Y be a map between compacta. Then the functions 2f , Cn (f ) and Fn (f ) (n ∈ N) are continuous. 1.6.16 Theorem. Let f : X → Y be a map between compacta. If f is open, then the function (f ) : 2Y → 2X given by (f )(B) = f −1 (B) is continuous. 1.6.17 Definition. A continuum X has the property of Kelley provided that for each U ∈ UX , there exists V ∈ UX such that for any two points p and q of X with ρX (p, q) < V , and each P ∈ C1 (X) with p ∈ P , there exists Q ∈ C1 (X) such that q ∈ Q and ρC1 (X) (P, Q) < C1 (U ). The entourage V is called a Kelley entourage for C1 (U ). W. J. Charatonik [23] and W. Makuchowski [95] defined the pointwise version of the property of Kelley as follows: 1.6.18 Definition. A continuum X has the property of Kelley at a point x0 if for every subcontinuum L of X containing x0 and each open subset A of C1 (X) containing L, there exists an open subset K(x0 , L, A) of X such that x0 ∈ K(x0 , L, A) and if x ∈ K(x0 , L, A), then there exists a subcontinuum M of X

1.6 Hyperspaces

47

such that x ∈ M and M ∈ A. The open set K(x0 , L, A) is called a Kelley set for A, L and x0 . 1.6.19 Theorem. A continuum X has the property of Kelley if and only if X has the property of Kelley at each of its points. Proof. Suppose X has the property of Kelley. Let p be an element of X, let P ∈ C1 (X) be such that p ∈ P and let A be an open subset of C1 (X) such that P ∈ A. Note that there exists U ∈ UX such that B(P, C1 (U )) ⊂ A. Let V be a Kelley entourage for C1 (U ). Hence, IntX (B(p, V )) is an open subset of X. Let q ∈ IntX (B(p, V )). Thus, ρX (p, q) < V . Then there exists Q ∈ C1 (X) such that q ∈ Q and ρC1 (X) (P, Q) < C1 (U ). This implies that Q ∈ A. Since p is an arbitrary point of X, X has the property of Kelley at each of its points. Now, assume that X has the property of Kelley at each of its points and let U ∈ UX . Since X has the property of Kelley at each of its points, for every point z of X and each Z ∈ C1 (X), there exists a Kelley set K(z, Z, IntC1 (X) (B(Z, C1 (U )))) for IntC1 (X) (B(Z, C1 (U ))), Z and z. Note that {K(z, Z, IntC1 (X) (B(Z, C1 (U )))) | z ∈ X and Z ∈ C1 (X)} is an open cover of X. By Lemma 1.3.11, V = {K(z, Z, IntC1 (X) (B(Z, C1 (U )))) × K(z, Z, IntC1 (X) (B(Z, C1 (U )))) | z ∈ X and Z ∈ C1 (X)} belongs to UX . Let p and q be points of X such that ρX (p, q) < V and let P ∈ C1 (X) be such that p ∈ P . Then, by the definition of V , we have that q ∈ K(p, P, IntC1 (X) (B(P, C1 (U )))). Thus, there exists Q ∈ C1 (X) such that Q ∈ IntC1 (X) (B(P, C1 (U ))) and ρC1 (X) (P, Q) < C1 (U ). Therefore, X has the property of Kelley.   For the following results we use the pointwise version of the property of Kelley. 1.6.20 Theorem. Let X be a continuum with the property of Kelley. Let A be a nonempty closed subset of X and let W be a subcontinuum of X such that Int(W ) = ∅ and W ∩ A = ∅. If w ∈ W , then there exists a subcontinuum K of X such that w ∈ Int(K) and K ∩ A = ∅. Proof. Let w ∈ W . Since X is normal, there exists an open subset U of X such that W ⊂ U and Cl(U ) ∩ A = ∅. Let U = U, Int(W )1 . Note that W ∈ U . Also observe that if L ∈ U , then L ⊂ U and L ∩ W = ∅. Since X has the property of Kelley at w, there exists a Kelley set V for U and w. Let v ∈ V \ {w}. Then there exists a subcontinuum Lv of X such that v ∈ Lv and Lv ∈ U . Let Lw = W , and  let K = Cl ( {Lv | v ∈ V }). Then K is a subcontinuum of X, w ∈ V ⊂ K and K ⊂ Cl(U ) ⊂ X \ A.  

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1.6.21 Corollary. Let X be a continuum with the property of Kelley, let W be a subcontinuum of X, with nonempty interior, and let A be a nonempty closed subset of X such that W ∩ A = ∅. Then there exists a subcontinuum M of X such that W ⊂ Int(M ) and M ∩ A = ∅. Proof. Let w ∈ W . By Theorem 1.6.20, there exists a subcontinuum Kw such that w ∈ Int(K) and K ∩ A = ∅. Note that {Int(Kw ) | w ∈ W } forms  an open cover of n W , by compactness, there exist w1 , . . . , wn in W such that W ⊂ j=1 Int(Kwj ). n Let M = j=1 Kwj . Then M is a subcontinuum of X, W ⊂ Int(M ) and M ∩ A = ∅.   Next, we show that continua with the uniform property of Effros have the property of Kelley. 1.6.22 Theorem. If X is a continuum with the uniform property of Effros, then X has the property of Kelley. Proof. By Theorem 1.6.19, it is enough to show that X has the property of Kelley at each of its points. Let x0 be a point of X, let K be a subcontinuum of X with x0 ∈ K and let U be an open subset of C1 (X) containing K. Then there exists U ∈ UX (Remark 1.3.4) such that B(K, C1 (U )) ⊂ U . Since X has the uniform property of Effros, there exists an Effros entourage V ∈ UX for U . Let x ∈ Int(B(x0 , V )). Then there exists a U -homeomorphism h : X → → X such that h(x0 ) = x. Then h(K) is a subcontinuum of X, x ∈ h(K) and h(K) ∈ B(K, C1 (U )) ⊂ U . Therefore, X has the property of Kelley.   As a consequence of Theorems 1.4.60 and 1.6.22, we obtain. 1.6.23 Corollary. If X is an Effros continuum, then X has the property of Kelley.

References for Chapter 1 Section 1.1: [24, 36, 60, 92]. Section 1.2: [24, 36, 48, 65]. Section 1.3: [40, 48, 49, 111, 129] Section 1.4: [1, 31, 36, 40, 45, 52, 55, 65–67, 75, 90–92, 97, 104, 105, 119, 127]. Section 1.5: [24, 37, 38, 61, 92, 99]. Section 1.6: [15, 23, 24, 40, 48, 79–81, 90–92, 95, 98, 105].

Chapter 2

The Set Function T

We prove basic results about the set function T defined by F. Burton Jones [68] to study the properties of metric continua. We define this function on compacta, and then we concentrate on continua. In particular, we present some of the well known properties (such as connectedness im kleinen, local connectedness, semilocal connectedness, etc.) using the set function T . The notion of aposyndesis was the main motivation of Jones to define this function. We present some properties of a continuum when it is T -symmetric and T -additive. We give properties of continuum on which T is idempotent, idempotent on closed sets and idempotent on contina. We also present results about the set functions T n , when n ∈ N.

2.1 Main Properties of T 2.1.1 Definition. Given a topological space X, the power set of X, denoted by P(X), is P(X) = {A | A ⊂ X}. 2.1.2 Remark. Let X be a topological space. Let us note that if A is a subset of X, then its closure satisfies Cl(A) = {x ∈ X | for each open subset U of X such that x ∈ U, we have that U ∩ A = ∅}. As we see below (Definition 2.1.3), this property is similar to the definition of the function T . 2.1.3 Definition. Let X be a compactum. Define T : P(X) → P(X) © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. Macías, Set Function T , Developments in Mathematics 67, https://doi.org/10.1007/978-3-030-65081-0_2

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50

by T (A) = {x ∈ X | for each subcontinuum W of X such that x ∈ Int(W ), we have that W ∩ A = ∅}, for every subset A of X. The function T is called Jones’ set function T . 2.1.4 Remark. In general, when working with the set function T , we usually work with complements. Hence, for any compactum X and any subset A of X, we have that T (A) = X \ {x ∈ X | there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ A}. 2.1.5 Remark. Let X be a compactum. If A is a subset of X, then A ⊂ T (A), and T (A) is closed in X. Hence, the range of T is 2X ∪ {∅} (Definition 1.6.1), i.e., T : P(X) → 2X ∪ {∅}. The next proposition gives a relation between aposyndesis and the set function T . 2.1.6 Proposition. Let X be a compactum, and let A be a subset of X. If x ∈ X \ T (A), then X is aposyndetic at x with respect to each point of A. Also, X is aposyndetic at x with respect to A. Proof. Let x ∈ X \ T (A), and let a ∈ A. Since x ∈ X \ T (A), by Definition 2.1.3, there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ A. In particular, W ⊂ X \ {a}. Therefore, X is aposyndetic at x with respect to a and with respect to A.   2.1.7 Proposition. Let X be a compactum. If A and B are subsets of X and A ⊂ B, then T (A) ⊂ T (B). Proof. Let x ∈ X \ T (B). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ B. Since X \ B ⊂ X \ A, x ∈ Int(W ) ⊂ W ⊂ X \ A. Therefore, x ∈ X \ T (A).   2.1.8 Corollary. Let X be a compactum. If A and B are subsets of X, then T (A)∪ T (B) ⊂ T (A ∪ B). 2.1.9 Remark. Let us note that the reverse inclusion of Corollary 2.1.8 is, in general, not true (Example 2.1.18). It is an open question to characterize continua X such that T (A ∪ B) = T (A) ∪ T (B) for each pair of closed subsets A and B of X. Let us see a couple of examples. 2.1.10 Example. Let X be the Cantor set. Then T (∅) = X. To see this, suppose there exists a point x ∈ X \ T (∅). Then there exists a subcontinuum W of X such

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51

that x ∈ Int(W ) ⊂ W ⊂ X \ ∅ = X. Since X is the Cantor set, X is totally disconnected and perfect. Hence, no subcontinuum of X has interior. Therefore, W cannot exist. Hence, T (∅) = X. Note that, by Proposition 2.1.7, T (A) = X for each subset A of X.  ∞ 2.1.11 Example. Let X = {0} ∪ n1 n=1 . Then T (∅) = {0}. This follows from the fact that the only subcontinuum of X with empty interior is {0}. Hence, by Remark 2.1.5 and Proposition 2.1.7, T (A) = {0} ∪ A for each subset A of X. 2.1.12 Remark. Note that if X is as in Example 2.1.10 or 2.1.11, then T |2X : 2X → 2X is continuous, where 2X has the topology given by the Hausdorff metric [92, Theorem 1.8.3] or the Vietoris topology. The following theorem gives us a characterization of compacta X for which T (∅) = ∅. 2.1.13 Theorem. Let X be a compactum. Then T (∅) = ∅ if and only if X only has finitely many components. Proof. Suppose X only has finitely many components. Let x ∈ X, and let C be the component of X such that x ∈ C. Hence, C is a subcontinuum of X and, by Lemma 1.4.1, x ∈ Int(C). Thus, each point of X is contained in the interior of a proper subcontinuum of X. Therefore, T (∅) = ∅. Now, suppose T (∅) = ∅. Then for each point x ∈ X, there exists a subcontinuum Wx of X such that x ∈ Int(Wx ) ⊂ Wx ⊂ X \ ∅. Hence, {Int(Wx ) | x ∈ X} is an open nX is compact, there exist x1 , . . . , xn in X such that ncover of X. Since X = j=1 Int(Wxj ) ⊂ j=1 Wxj ⊂ X. Thus, X is the union of finitely many continua. Therefore, X only has finitely many components.   2.1.14 Corollary. If X is a continuum, then T (∅) = ∅. Let us see more examples. 2.1.15 Example. Let X be the cone over the Cantor set, with vertex νX and base B (Figure 2.1). Since the Cantor set is totally disconnected and perfect, it follows

Fig. 2.1 Cantor fan

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that the subcontinua of X having nonempty interior must contain νX . Hence, T ({νX }) = X. If r ∈ X \ {νX }, then T ({r}) is the line segment from r to the base B. Also, T (B) = B. 2.1.16 Example. Let Y = X ∪ X  , where X is the cone over the Cantor set, with vertex νX and base B, and X  is another copy of the cone over the Cantor set, with vertex νX  ∈ B (Figure 2.2). In this case, T ({νX }) = X, T ({νX  }) = X  and T T ({νX }) = T 2 ({νX }) = Y . Note that this implies, in general, that the function T is not idempotent (Definition 2.3.1).

Fig. 2.2 Double Cantor fan

2.1.17 Example. Let X be'the topologist sine curve, Example 1.4.9. Let J = X \ &  1  x, sin x 0 < x ≤ π2 . Since X is not locally connected at any point of J, if p ∈ J, then T ({p}) = J. Also, T ({q}) = {q} for each q ∈ X \ J. The following example shows that, in general, T (A ∪ B) = T (A) ∪ T (B) does not hold. 2.1.18 Example. Let X be the suspension over the closure of the harmonic sequence (Example 1.4.23). Note that for each x ∈ X, T ({x}) = {x}. In particular,

Fig. 2.3 Suspension over the closure of the harmonic sequence

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T ({a}) = {a} and T ({b}) = {b}; however, T ({a, b}) consists of the limit segment from a to b (Figure 2.3). 2.1.19 Theorem. Let X be a continuum. If A is a subset of X, then T (A) intersects each closed separator of A and some point of T (A). Proof. Suppose the theorem is not true. Then there exist p ∈ T (A) and a closed subset B of X such that X \ B = H ∪ K, A ⊂ H, p ∈ K, where H and K are disjoint open subsets of X and B ∩ T (A) = ∅. Hence, for each b ∈ B, there exists a subcontinuum Wb of X such that b ∈ Int(W nb ) ⊂ Wb ⊂ X \ A. Since B is compact, there exist b1 , . . . , bn in B such that B ⊂ j=1 Int(Wbj ). n Since Cl(K) \ K ⊂ B, by Theorem 1.4.36, Y = K ∪ j=1 Wbj only has a finite number of components. Note that Y is closed in X, since it is the union of a finite number of continua and Cl(K) \ K ⊂ Y . Since p ∈ K and K ⊂ Y , p ∈ Y . Let C be the component of Y such that p ∈ C. By Lemma 1.4.1, p ∈ Int(C). This implies that p ∈ X \ T (A), a contradiction. Therefore, B ∩ T (A) = ∅.   Now, we present several consequences of Theorem 2.1.19. 2.1.20 Corollary. Let X be a continuum. If A is a subset of X, then each component of T (A) intersects Cl(A). Proof. Suppose the corollary is not true. Then there exists a component C of T (A) such that C ∩ Cl(A) = ∅. Hence, by Theorem 1.4.8, there exist two disjoint closed subsets H and K of T (A) such that T (A) = H ∪ K, Cl(A) ⊂ H and C ⊂ K. Since X is a normal space, there exist two disjoint open subsets U and V of X such that H ⊂ U and K ⊂ V . Let B = X \ (U ∪ V ). Then B is a closed separator of A and C and B ∩ T (A) = ∅, a contradiction to Theorem 2.1.19.   2.1.21 Corollary. Let X be a continuum. If A is a subset of X, then the cardinality of the components of T (A) is at most equal to the cardinality of the components of Cl(A). 2.1.22 Corollary. Let X be a continuum, and let A and B be two closed subsets of X. If K is a component of T (A ∪ B) such that K \ (T (A) ∪ T (B)) = ∅, then K ∩ A = ∅ and K ∩ B = ∅. Proof. Since X is a continuum, by Corollary 2.1.14, T (∅) = ∅. By Corollary 2.1.20, K ∩ (A ∪ B) = ∅. Now, by Theorem 2.1.47, T ((A ∪ B) ∩ K) = K. Since K \ (T (A) ∪ T (B)) = ∅, (A ∪ B) ∩ K meets both A and B. Thus, K ∩ A = ∅ and K ∩ B = ∅.   2.1.23 Corollary. Let X be a continuum, and let A and B be closed subsets of X. If T (A ∪ B) = T (A) ∪ T (B), then there exists a subcontinuum K of X such that K ⊂ T (A ∪ B), K ∩ A = ∅ and K ∩ B = ∅. Proof. Let K be a component of T (A ∪ B) such that K \ (T (A) ∪ T (B)) = ∅ and apply Corollary 2.1.22.   The next result is due to H. S. Davis [120, p. 18].

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2.1.24 Theorem. Let X be a compactum, and let A and B be nonempty closed subsets of X. Then the following are equivalent: (1) T (A) ∩ B = ∅. (2) There exist two closed subsets M and N of X such that A ⊂ Int(M ), B ⊂ Int(N ) and T (M ) ∩ N = ∅. Proof. Assume (1), we show (2). For each b ∈ B, there exists a subcontinuum Wb of X such that b ∈ Int(Wb )  ⊂ Wb ⊂ X \ A.Since B is compact, there exist n n b1 , . . . , bn in B such that B ⊂ j=1 Int(Wbj ) ⊂ j=1 Wbj ⊂ X \ A. Since X is a normal space, there V of X such that A ⊂ U ⊂ Cl(U ) ⊂  exist two open sets U and  n n X \ j=1 Wbj and B ⊂ V ⊂ Cl(V ) ⊂ j=1 Int(Wbj ). Let M = Cl(U ) and let N = Cl(V ). Then M and N are closed subsets of X such that A ⊂ Int(M ), B ⊂ Int(N ) and T (M ) ∩ N = ∅. The other implication follows easily from Proposition 2.1.7.   The next corollary is a consequence of proof of Theorem 2.1.24. 2.1.25 Corollary. Let X be a continuum, and let A be a closed subset of X. If x ∈ X \ T (A), then there exists an open subset U of X such that A ⊂ U and x ∈ X \ T (Cl(U )). 2.1.26 Theorem. Let X be a continuum, and let A be a nonempty subset of X. Then T (A) is disconnected if and only if there exist a nonempty proper subset B of A and a closed subset W of X such that W ∩ A = ∅, W only has finitely many components and Int(W ) separates B from A \ B in X. Proof. Suppose T (A) is disconnected. Then T (A) = K1 ∪ K2 , where K1 and K2 are nonempty disjoint closed subsets of X. Let B = A ∩ K1 . Then A \ B = A ∩ K2 . By Theorem 2.1.19, B is a nonempty proper subset of A. Let U be an open subset of X such that K1 ⊂ U and Cl(U ) ∩ K2 = ∅. For each z ∈ Bd(U ), there exists a subcontinuum Wz of X such that z ∈ Int(Wz ) ⊂ Wz ⊂ X \ A. mSince Bd(U ) is compact, there exist z1 , . . . , zm in Bd(U ) such that Bd(U ) ⊂ j=1 Int(Wzj ). m Let W = j=1 Wzj . Then W only has finitely many components and Bd(U ) ⊂ Int(W ). Hence, X \Int(W ) = [U ∩(X \Int(W ))]∪[(X \Cl(U ))∩(X \Int(W ))]. Note that B ⊂ U ∩ (X \ Int(W )) and A \ B ⊂ (X \ Cl(U )) ∩ (X \ Int(W )). Now, assume that there exist a nonempty proper subset B of A and a closed subset W of X such that W ∩ A = ∅, W only has finitely many components and Int(W ) separates B from A \ B in X. Let x ∈ Int(W ), and let W1 be the component of W containing x. Then x ∈ Int(W1 ) ⊂ W1 ⊂ X \ A (Lemma 1.4.1). Thus, x ∈ X \ T (A). Hence, T (A) ⊂ X \ Int(W ) = L1 ∪ L2 , where L1 and L2 are nonempty disjoint closed subsets of X, with B ⊂ L1 ∩ T (A) and A \ B ⊂ L2 ∩ T (A). Therefore, T (A) is not connected.   2.1.27 Theorem. Let X be a continuum. If W is a subcontinuum of X, then T (W ) is also a subcontinuum of X.

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Proof. By Remark 2.1.5, T (W ) is a closed subset of X. Suppose T (W ) is not connected. Then there exist two disjoint closed subsets A and B of X such that T (W ) = A ∪ B. Since W is connected, without loss of generality, we assume that W ⊂ A. Since X is a normal space, there exists an open subset U of X such that A ⊂ U and Cl(U ) ∩ B = ∅. Note that Bd(U ) ∩ T (W ) = ∅. Hence, for each z ∈ Bd(U ), there exists a subcontinuum Kz of X such that z ∈ Int(Kz ) ⊂ Kz ⊂ X \  W . Since Bd(U ) n is compact, there exist z1 , . . . , zn in Bd(U ) such that Bd(U ) ⊂ j=1 Int(Kzj ) ⊂   n n j=1 Kzj (Figure 2.4). Let V = U \ j=1 Kzj . Let Y = X \ V = (X \    n U) j=1 Kzj . By Theorem 1.4.36, Y only has a finite number of components. Observe that B ⊂ X \ Cl(U ) ⊂ X \ U ⊂ Y . Hence, B ⊂ Int(Y ). Let b ∈ B, and let C be the component of Y such that b ∈ C. By Lemma 1.4.1, b ∈ Int(C). By construction, C ∩ W = ∅. Thus, b ∈ X \ T (W ), a contradiction. Therefore, T (W ) is connected.  

Fig. 2.4 Covering the boundary

The set function T may be used to define some of the properties of continua like connectedness im kleinen or local connectedness. In the following theorems we show the equivalence between both types of definitions. The next theorem gives us a local definition of almost connectedness im kleinen using T . 2.1.28 Theorem. Let X be a continuum. If p ∈ X, then X is almost connected im kleinen at p if and only if for each subset A of X such that p ∈ Int(T (A)), p ∈ Cl(A). Proof. Suppose X is almost connected im kleinen at p. Let A be a subset of X such that p ∈ Int(T (A)). Let N = {Uλ }λ∈Λ be an open local base about p. Without loss of generality, we assume that Λ is a directed set. Then there exists a cofinal subset  Λ of Λ such that for every λ ∈ Λ , p ∈ Uλ ⊂ Cl(Uλ ) ⊂ T (A) and λ∈Λ Uλ = {p}. Since X is almost connected im kleinen at p, for each λ ∈ Λ , there exists a subcontinuum Wλ of X such that Int(Wλ ) = ∅ and Wλ ⊂ Uλ ⊂ Cl(Uλ ) ⊂ T (A). Hence, Wλ ∩ A = ∅ for each λ ∈ Λ . Let xλ ∈ Wλ ∩ A, for every λ ∈ Λ . Note that,

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by construction, the net {xλ }λ∈Λ converges to p and {xλ }λ∈Λ ⊂ A. Therefore, p ∈ Cl(A). Now suppose that p ∈ X satisfies that for each subset A of X such that p ∈ Int(T (A)), p ∈ Cl(A). Let U be an open subset of X such that p ∈ U . Let V be an open subset of X such that p ∈ V ⊂ Cl(V ) ⊂ U . If some component of Cl(V ) has nonempty interior, then X is almost connected im kleinen at p. Suppose, then, that all the components of Cl(V ) have empty interior. Let A = Bd(V ). Then A is a closed subset of X and p ∈ X \ A. We show that V ⊂ T (A). To this end, suppose there exists x ∈ V \ T (A). Thus, there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \A. Since all the components of Cl(V ) have empty interior, and W is a subcontinuum with nonempty interior, we have that W ∩ (X \ V ) = ∅ and, of course, W ∩ V = ∅. Since W is connected, W ∩ Bd(V ) = W ∩ A = ∅, a contradiction. Hence, V ⊂ T (A). Since p ∈ V ⊂ T (A), by hypothesis, p ∈ Cl(A) = A, a contradiction. Therefore, Cl(V ) has a component with nonempty interior, and X is almost connected im kleinen at p.   The next theorem gives us a global definition of almost connectedness im kleinen using T . 2.1.29 Theorem. A continuum X is almost connected im kleinen if and only if for each closed subset, F , of X, Int(F ) = Int(T (F )). Proof. Suppose X is almost connected im kleinen. Let F be a closed subset of X. Since F ⊂ T (F ) (Remark 2.1.5), Int(F ) ⊂ Int(T (F )). Let x ∈ Int(T (F )). Then, by Theorem 2.1.28, x ∈ Cl(F ) = F . Hence, Int(T (F )) ⊂ F . Therefore, Int(T (F )) ⊂ Int(F ) and Int(T (F )) = Int(F ). Now, suppose that Int(F ) = Int(T (F )) for each closed subset F of X. Let x ∈ X, and let U be an open subset of X such that x ∈ U . Since Int(X \U )∩U = ∅, by hypothesis, Int(T (X \ U )) ∩ U = ∅. Hence, there exists y ∈ U such that y ∈ X \ T (X \ U ). Thus, there exists a subcontinuum K of X such that y ∈ Int(K) ⊂ K ⊂ X \ (X \ U ) = U . Therefore, X is almost connected im kleinen at x.   2.1.30 Theorem. Let X be continuum and let p be a point of X. Then X is connected im kleinen at p if and only if for each subset A of X such that p ∈ T (A), p ∈ Cl(A). Proof. Let p ∈ X. Assume X is connected im kleinen at p. Let A be a subset of X and suppose p ∈ X \ Cl(A). Hence, Cl(A) is a closed subset of X \ {p}. Since X is connected im kleinen at p, there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ A. Therefore, p ∈ X \ T (A). Now, suppose that p ∈ X satisfies that for each subset A of X such that p ∈ T (A), p ∈ Cl(A). Let D be a closed subset of X such that D ⊂ X \ {p}. Since p ∈ X \ D, by hypothesis, p ∈ X \ T (D). Then there exists a subcontinuum K of X such that p ∈ Int(K) ⊂ K ⊂ X \ D. Therefore, X is connected im kleinen at p.  

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2.1.31 Corollary. Let X be continuum and let p be a point of X. Then X is connected im kleinen at p if and only if for each closed subset A of X such that p ∈ T (A), p ∈ A. The following theorem follows from Definition 1.4.19. 2.1.32 Theorem. Let X be a continuum, and let p and q be points of X. Then X is semi-aposyndetic at p and q if and only if either p ∈ X \T ({q}) or q ∈ X \T ({p}). The next theorem follows from the definition of aposyndesis. 2.1.33 Theorem. Let X be a continuum, and let p and q be points of X. Then X is aposyndetic at p with respect to q if and only if p ∈ X \ T ({q}). The following theorem is a global version of Theorem 2.1.33. 2.1.34 Theorem. A continuum X is aposyndetic if and only if T ({p}) = {p}, for each p ∈ X. Proof. Suppose X is aposyndetic. Let p ∈ X. Then for each q ∈ X \ {p}, X is aposyndetic at q with respect to p. Hence, by Remark 2.1.5 and Theorem 2.1.33, q ∈ X \ T ({p}). Therefore, T ({p}) = {p}. Now, suppose T ({x}) = {x} for each x ∈ X. Let p and q be points of X such that p = q. Since T ({q}) = {q}, p ∈ X \ T ({q}). Thus, by Theorem 2.1.33, X is aposyndetic at p with respect to q. Since p and q are arbitrary points of X, X is aposyndetic.   2.1.35 Theorem. Let X be a continuum, and let p be a point of X. Then X is semi-locally connected at p if and only if T ({p}) = {p}. Proof. Suppose X is semi-locally connected at p. By Remark 2.1.5, {p} ⊂ T ({p}). Let q ∈ X \ {p}, and let U be an open subset of X such that p ∈ U and q ∈ X \ Cl(U ). Since X is semi-locally connected at p, there exists an open subset V of X such that p ∈ V ⊂ U and X \ V only has finitely many components. By Lemma 1.4.1, q belongs to the interior of the component of X \ V containing q. Hence, q ∈ X \ T ({p}). Therefore, T ({p}) = {p}. Now, suppose T ({p}) = {p}. Let U be an open subset of X such that p ∈ U . Since T ({p}) = {p}, for each q ∈ X \ U , there exists a subcontinuum Wq of X such that q ∈ Int(Wq ) ⊂ Wq ⊂ X \ {p}. Note that {Int(Wq ) | q ∈ X \ U } is an open cover of X \ U . Since this set is compact, there existq1 , . . . , qn in X \ U n n such that X \ U ⊂ j=1 Int(Wqj ). Let V = X \ j=1 Wqj . Then V is an open subset of X such that p ∈ V ⊂ U and X \ V only has finitely many components. Therefore, X is semi-locally connected at p.   As a consequence of Theorems 2.1.34 and 2.1.35, we have the following. 2.1.36 Corollary. A continuum X is aposyndetic if and only if it is semi-locally connected. The next theorem gives us a characterization of local connectedness in terms of the set function T .

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2.1.37 Theorem. A continuum X is locally connected if and only if for each closed subset A of X, T (A) = A. Proof. Suppose X is locally connected. Let A be a closed subset of X. Then X \ A is an open subset of X. Let p ∈ X \A. Then there exists an open subset U of X such that p ∈ U ⊂ Cl(U ) ⊂ X \ A. Since X is locally connected, there exists an open connected subset V of X such that p ∈ V ⊂ U . Hence, Cl(V ) is a subcontinuum of X such that p ∈ V ⊂ Cl(V ) ⊂ Cl(U ) ⊂ X \ A. Consequently, p ∈ X \ T (A). Thus, T (A) ⊂ A. By Remark 2.1.5, A ⊂ T (A). Therefore, T (A) = A. Now, suppose T (A) = A for each closed subset A of X. Let p ∈ X, and let U be an open subset of X such that p ∈ U . Then X \ U is a closed subset of X such that p ∈ X \ U . Since X \ U = T (X \ U ), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ (X \ U ) = U . Thus, by Theorem 1.4.15, X is connected im kleinen at p. Since p is an arbitrary point of X, by Theorem 1.4.18, X is locally connected.   Theorem 2.1.37 may be strengthened as follows. 2.1.38 Theorem. Let X be a continuum. Then X is locally connected if and only if T (Y ) = Y for each subcontinuum Y of X. Proof. If X is locally connected, by Theorem 2.1.37, T (Y ) = Y for each subcontinuum Y of X. Now suppose that for every subcontinuum Y of X, T (Y ) = Y . We show that X is locally connected. To this end, by Lemma 1.4.17, it is enough to see the components of open subsets of X are open. Let V be an open subset of X. Let p ∈ V . Since T ({p}) = {p} and X \V is compact, there exist finitely many subcontinua contained in X \ {p} whose interiors cover X \ V . Let W = {W 1 , . . . , W m } be  m such a collection of smallest possible cardinality. Let U = X \ j=1 Wj , and let P be the component of Cl(U ) such that p ∈ P . We show that p ∈ Int(P ) ⊂ m P ⊂ V . By Theorem 1.4.36, each component of Cl(U ) intersects j=1 Wj . By the minimality of m, we assert that no component of Cl(U ), except P , can intersect more than one of the Wj ’s. To see this, let P  be a component of Cl(U ), different from P , and suppose that P  intersects both Wj and Wk , where j and k are in {1, . . . , m} and j = k. Note that, in this case, P  ∪ Wj ∪ Wj is a continuum. Let W  = {W1 , . .. , Wm , P ∪ Wj ∪ Wk } \ {Wj , Wk }. Then W  has m − 1 elements  ∪ (P  ∪ Wj ∪ Wk ), a contradiction to the minimality of and X \ V ⊂ l=j Wl l=k m. For every j ∈ {1, . . . , m}, let Aj be the union of all the components of Cl(U ), which intersect Wj . Then Aj ∩ Ak ⊂ P , for each j and k in {1, . . . , m} such that j = k. Now, since p ∈ T (Wj ) = Wj , for each j ∈ {1, . . . , m}, there exists a subcontinuum Kj of X such that p ∈ Int(Kj ) ⊂ Kj ⊂ X \ Wj . Then Kj ∩ (Aj \ P ) = ∅. To see this, suppose there exists x ∈ Kj ∩ (Aj \ P ). Let L be the component of Kj ∩ Cl(U ) such that x ∈ L. By Theorem 1.4.36, L intersects the boundary of Kj ∩ Cl(U ) in Kj . This boundary is contained m in =1 W . (If y ∈ BdKj (Kj ∩ Cl(U )), then y ∈ Cl(U ). If y ∈ U , then

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y ∈ Kj ∩U ⊂ Kj ∩Cl(U ) and(Kj ∩U )∩(Kj \(Kj ∩Cl(U ))) = ∅, a contradiction. m Therefore, y ∈ Cl(U ) \ U ⊂ =1 W .) Hence, there exists k ∈ {1, . . . , m} \ {j} such that L ∩ Wk = ∅. But L ⊂ M , for some component M of Cl(U ), M = P and M ⊂ Aj ,so that M ∩ Wk = ∅, a contradiction.  m m m Thus, K (A \ P ) = ∅, and it follows that j=1 Kj ⊂ P . ∩ j j j=1 j=1   m Since p ∈ Int j=1 Kj , p ∈ Int(P ). Therefore, P is a connected neighborhood of p in X contained in V . Since p is an arbitrary point of V , every component of V is open.   Our next three results are the consequence of Theorem 2.1.38. 2.1.39 Theorem. Let X be an aposyndetic continuum. If for each subcontinuum W of X, X \ W only has finitely many components, then X is locally connected. Proof. To show that X is locally connected, by Theorem 2.1.38, it suffices to prove that T (W ) = W for every subcontinuum W of X. Let W be a subcontinuum of X, we know that W ⊂ T (W ) (Remark 2.1.5). Let x ∈ X \ W . Since X is aposyndetic, for each w ∈ W , there exists a subcontinuum Kw of X such that w ∈ Int(Kw ) ⊂Kw ⊂ X \ {x}. Since , . . . , wn in W such n W is compact, there exist w1 n n that W ⊂ j=1 Int(Kwj ) ⊂ j=1 Kwj ⊂ X \ {x}. Let K = j=1 Kwj . Then K is a subcontinuum of X such that W ⊂ Int(K) ⊂ K ⊂ X \ {x}. By hypothesis, X \ K only has finitely many components. Let C be the component of X \ K containing x. Then, since X \ K is open and only has finitely many components, C is open in X. Thus, Cl(C) is a subcontinuum of X such that x ∈ Int (Cl(C)) ⊂ X \ Int(K) ⊂ X \ W . Hence, x ∈ X \ T (W ). Therefore, T (W ) = W , and X is locally connected.   2.1.40 Corollary. Let X be an aposyndetic metric continuum such that dim(C1 (X)) < ∞. Then X is locally connected. Proof. Since dim(C1 (X)) < ∞, it follows that there exists n ∈ N such that C1 (X) does not contain an n-cell. Hence, using a similar proof to the one given for [92, Theorem 6.1.11], we have that X does not contain n-ods. Note that this implies that the complement of each subcontinuum of X has at most n − 1 components. The corollary now follows from Theorem 2.1.39.   2.1.41 Theorem. Let X be an aposyndetic continuum. If T (W ) = W for each subcontinuum W of X with nonempty interior, then X is locally connected. Proof. Let A be a subcontinuum of X, and let x ∈ X \ A. Since X is aposyndetic, for each a ∈ A, there exists a subcontinuum Wa of X such that a ∈ Int(Wa ) ⊂ there exist a1 , . . . , an in A such that A ⊂ W na ⊂ X \ {x}. Since A iscompact, n j=1 Int(Waj ). Let W = j=1 Waj . Note that A ⊂ W , Int(W ) = ∅ and x ∈ X \ W . Hence, by hypothesis, T (W ) = W . Since A ⊂ W , by Proposition 2.1.7, T (A) ⊂ T (W ) = W . Thus, x ∈ X \ T (A). Therefore, T (A) = A. Now the theorem follows from Theorem 2.1.38.   2.1.42 Theorem. A continuum X is continuum aposyndetic if and only if X is locally connected.

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Proof. Suppose X is continuum aposyndetic. It follows from the definition that if W is a subcontinuum of X, then T (W ) = W . Hence, by Theorem 2.1.38, X is locally connected. The reverse implication is clear.   As a consequence of Theorem 2.1.42, we obtain the following. 2.1.43 Theorem. A continuum X is freely decomposable with respect to points and continua if and only if X is locally connected. Proof. Suppose X is freely decomposable with respect to points and continua. Then X is continuum aposyndetic. Thus, by Theorem 2.1.42, X is locally connected. Now, assume X is locally connected. Let C be a subcontinuum of X, and let a ∈ X \ C. Consider the quotient space X/C. Then X/C is a locally connected continuum. Since locally connected continua are aposyndetic, by Theorem 1.4.28, X/C is freely decomposable. Let q : X → → X/C be the quotient map, and let χ be the point of X/C corresponding to q(C). Note that q is a monotone map. Since X/C is freely decomposable, there exist two subcontinua Γ1 and Γ2 of X/C such that X/C = Γ1 ∪ Γ2 , q(a) ∈ Γ1 \ Γ2 and χ ∈ Γ2 \ Γ1 . Hence, X = q −1 (Γ1 ) ∪ q −1 (Γ2 ), a ∈ q −1 (Γ1 ) \ q −1 (Γ2 ) and C = q −1 (χ) ⊂ q −1 (Γ2 ) \ q −1 (Γ1 ). Therefore, X is freely decomposable with respect to points and continua.   2.1.44 Theorem. Let X be a continuum. Then X is indecomposable if and only if T ({p}) = X for each p ∈ X. Hence, T (A) = X for each nonempty closed subset A of X. Proof. Suppose X is indecomposable. By Corollary 1.4.35, each proper subcontinuum of X has empty interior. Therefore, T ({p}) = X for each p ∈ X. Now, suppose X is decomposable. Then there exist two proper subcontinua A and B of X such that X = A ∪ B. Let a ∈ A \ B, and let b ∈ B \ A. Then b ∈ Int(B) ⊂ B ⊂ X \ {a}. Hence, b ∈ X \ T ({a}). Therefore, T ({a}) = X.   2.1.45 Lemma. Let X be a compactum, and let A be a nonempty closed subset of X. Then p ∈ X \ T (A) if and only if there exist a subcontinuum W and an open subset Q of X such that p ∈ Int(W ) ∩ Q, Bd(Q) ∩ T (A) = ∅ and W ∩ A ∩ Q = ∅. Proof. Let p ∈ X \ T (A). Then there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ A. Since X is a normal space, there exists an open subset Q of X such that p ∈ Q ⊂ Cl(Q) ⊂ Int(W ). Hence, Bd(Q) ∩ T (A) = ∅ and W ∩ A ∩ Q = ∅. Now, suppose there exist a continuum W and an open subset Q of X such that p ∈ Int(W ) ∩ Q, Bd(Q) ∩ T (A) = ∅ and W ∩ A ∩ Q = ∅. Since Bd(Q) is a compact set and Bd(Q) ∩ T (A) = ∅, there W1 , . . . , Wn , nexist a finite collection, m of subcontinua of X such that Bd(Q) ⊂ j=1 Int(Wj ) ⊂ j=1 Wj ⊂ X \ A. If W ⊂ Q, then, clearly, p ∈ X \ T (A). Assume W \ Q = ∅. By Theorem 1.4.36, the closure of each mcomponent of W ∩ Q must intersect at least one of the Wj ’s, since Bd(Q) ⊂ j=1 Wj . Hence,   m H = (W ∩ Q) ∪ j=1 Wj only has a finite number of components. Let K be

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the component of H such that p ∈ K. Since p ∈ Int(W ) ∩ Q, by Lemma 1.4.1, p ∈ Int(K) and, of course, K ∩ A ⊂ H ∩ A = ∅. Thus, p ∈ X \ T (A).   2.1.46 Lemma. Let X be a compactum, and let A be a subset of X. If T (A) = M ∪ N , where M and N are disjoint closed sets, then T (A ∩ M ) = M ∪ T (∅). Proof. Suppose p ∈ T (A ∩ M ) \ (M ∪ T (∅)). Since p ∈ X \ T (∅), there exists a subcontinuum W of X such that p ∈ Int(W ). Since X is a normal space, there exists an open subset Q of X such that N ⊂ Q and Cl(Q) ∩ M = ∅. Note that p ∈ Int(W ) ∩ Q and Bd(Q) ∩ T (A ∩ M ) ⊂ Bd(Q) ∩ T (A) = ∅, and W ∩ (A ∩ M ) ∩ Q ⊂ Q ∩ M = ∅. Then, by Lemma 2.1.45, p ∈ X \ T (A ∩ M ), thus contradicting the assumption. Now, suppose p ∈ (M ∪ T (∅)) \ T (A ∩ M ). Since p ∈ X \ T (A ∩ M ) and ∅ ⊂ A∩M , p ∈ X \T (∅). Hence, p ∈ M . Since X is a normal space, there exists an open subset Q of X such that M ⊂ Q and Cl(Q)∩N = ∅. Since p ∈ X \T (A∩M ), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ (A ∩ M ). Observe that p ∈ Int(W ) ∩ Q and Bd(Q) ∩ T (A) = ∅. Since Q ∩ N = ∅, W ∩ A ∩ Q = W ∩ (A ∩ M ) = ∅. Hence, by Lemma 2.1.45, p ∈ X \ T (A ∩ M ) ⊂ X \ M , thus contradicting our assumption.   2.1.47 Theorem. Let X be a compactum, and let A be a closed subset of X. If K is a component of T (A), then T (A ∩ K) = K ∪ T (∅). Proof. Let K be a component of T (A). Let Γ(K) = {L ⊂ T (A) | K ⊂ L and L is open and closed in T (A)}. Note that the collection of {A ∩ L | L ∈ Γ(K)} fails to be a filterbase if for some L ∈ Γ(K), A ∩ L = ∅. In this case, the conclusion of Lemma 2.2.9 holds (an argument similar to the one given in the proof of Theorem 2.2.10 shows this). Note that Lemma 2.1.46 remains true even if A ∩ M = ∅. Hence, by Lemma 2.1.46, for each L ∈ Γ(K), T (A ∩ L) = L ∪ T (∅). Therefore, by Lemma 2.2.9, the following sequence of equalities establishes the theorem:    T (A ∩ K) = T ( {A ∩ L| L ∈ Γ(K)}) = {T (A ∩ L) | L ∈ Γ(K)} = {L∪ T (∅) | L ∈ Γ(K)} = {L | L ∈ Γ(K)} ∪ T (∅) = K ∪ T (∅). (The fact that {L | L ∈ Γ(K)} = K follows from Theorem 1.4.7.)   The following corollary says that for each nonempty closed subset A of a continuum X, the components of T (A) are also in the image of T . 2.1.48 Corollary. Let X be a continuum, and let A be a closed subset of X. If K is a component of T (A), then K = T (A ∩ K). Proof. Let A be a closed subset of X, and let K be a component of T (A). By Theorem 2.1.47, T (A ∩ K) = K ∪ T (∅). Since T (∅) = ∅, Corollary 2.1.14, K = T (A ∩ K).   2.1.49 Corollary. Let X be a continuum, and let W1 and W2 be subcontinua of X. If T (W1 ∪ W2 ) = T (W1 ) ∪ T (W2 ), then T (W1 ∪ W2 ) is a continuum.

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Proof. Suppose T (W1 ∪ W2 ) is not connected. Then there exist two disjoint closed subsets A and B of X such that T (W1 ∪ W2 ) = A ∪ B. By Lemma 2.1.46 and Corollary 2.1.14, T ((W1 ∪ W2 ) ∩ A) = A and T ((W1 ∪ W2 ) ∩ B) = B. Suppose W1 ⊂ A. If W2 ⊂ A, then A = T ((W1 ∪W2 )∩A) = T (W1 ∪W2 ), a contradiction. Thus, W2 ⊂ B. Hence, T (W1 ) = A and T (W2 ) = B, which implies that T (W1 ∪ W2 ) = T (W1 ) ∪ T (W2 ), a contradiction. Therefore, T (W1 ∪ W2 ) is connected.   2.1.50 Corollary. Let X be a continuum. If p and q are in X such that T ({p, q}) = T ({p}) ∪ T ({q}), then T ({p, q}) is a continuum. The next theorem gives relationships between the images of maps and the images of T . We subscript T to differentiate the continua on which T is defined. 2.1.51 Theorem. Let X and Y be continua, and let f : X → → Y be a surjective map. If A is a subset of X and B is a subset of Y , then the following hold. (a) (b) (c) (d) (e)

TY (B) ⊂ f TX f −1 (B). If f is monotone, then f TX (A) ⊂ TY f (A) and TX f −1 (B) ⊂ f −1 TY (B). If f is monotone, then TY (B) = f TX f −1 (B). If f is open, then f −1 TY (B) ⊂ TX f −1 (B). If f is monotone and open, then f −1 TY (B) = TX f −1 (B).

Proof. We show (a). Let y ∈ Y \ f TX f −1 (B). Then f −1 (y) ∩ TX f −1 (B) = ∅. Thus, for each x ∈ f −1 (y), there exists a subcontinuum Wx of X such that x ∈ Int(Wx ) ⊂ Wx ⊂ X \ f −1 (B). Since f −1 (y) is compact,  there exist  x1 , . . . , xn in n n −1 −1 −1 f (y) such that f (y) ⊂ j=1 Int(Wxj ). Note that (B) = ∅ j=1 Wxj ∩f   n and, for each j ∈ {1, . . . , n}, f −1 (y) ∩ Wxj = ∅. Hence, f j=1 Wxj ∩ B = ∅,   n and f is a continuum (y ∈ f (Wxj ) ∩ f (Wxk ) for every j and k in j=1 Wxj {1, . . . , n}).   n Observe that Y \ f X \ j=1 Int(Wxj ) is an open subset of Y , and it   n n . To see this, note that, since j=1 Int(Wxj ) ⊂ is contained in f W x j j=1   n n n n ⊂ j=1 Wxj , X \ j=1 Wxj ⊂ X \ j=1 Int(Wxj ). Then f X \ j=1 Wxj     n n f X \ j=1 Int(Wxj ) and, consequently, Y \ f X \ j=1 Int(Wxj ) ⊂     n n Y \ f X \ j=1 Wxj . Since f is onto, Y \ f X \ j=1 Int(Wxj ) ⊂      n n = f f X \ X \ j=1 Wxj j=1 Wxj . Therefore, we obtain Y \     n n f X \ j=1 Int(Wxj ) ⊂ f j=1 Wxj .  n n Also, since f −1 (y) ⊂ Int(W ), we have that {y}∩ f X \ j=1 xj j=1    n Int(Wxj ) = ∅. Hence, y ∈ Y \ f X \ j=1 Int(Wxj ) . Therefore, y ∈ Y \ TY (B). We prove (b). First, we see f TX (A) ⊂ TY f (A). Let y ∈ Y \ TY f (A). Then there exists a subcontinuum W of Y such that y ∈ Int(W ) ⊂ W ⊂ Y \ f (A). It

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follows that f −1 (y) ⊂ f −1 (Int(W )) ⊂ f −1 (W ) ⊂ f −1 (Y ) \ f −1 f (A) ⊂ X \ A. Hence, f −1 (y) ⊂ X \ A. Since f is monotone, f −1 (W ) is a subcontinuum of X (Lemma 1.4.46). Thus, since f −1 (y) ⊂ f −1 (Int(W )) ⊂ f −1 (W ) ⊂ X \ A, f −1 (y) ∩ TX (A) = ∅. Therefore, y ∈ Y \ f TX (A). Now, we show TX f −1 (B) ⊂ f −1 TY (B). Let x ∈ X \ f −1 TY (B). Then f (x) ∈ Y \TY (B). Hence, there exists a subcontinuum W of Y such that f (x) ∈ Int(W ) ⊂ W ⊂ Y \ B. This implies that x ∈ f −1 (f (x)) ⊂ f −1 (Int(W )) ⊂ f −1 (W ) ⊂ X \ f −1 (B). Since f is monotone, f −1 (W ) is a subcontinuum of X. Therefore, x ∈ X \ TX f −1 (B). We prove (c). By (a), TY (B) ⊂ f TX f −1 (B). Since TX f −1 (B) ⊂ f −1 TY (B), by (b), f TX f −1 (B) ⊂ f f −1 TY (B) = TY (B). Therefore, f TX f −1 (B) = TY (B). We show (d). Let x ∈ X \ TX f −1 (B). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ f −1 (B). Hence, f (x) ∈ f (Int(W )) ⊂ f (W ) ⊂ Y \ B. Since f is open, f (Int(W )) is an open subset of Y . Therefore, f (x) ∈ Y \ TY (B). Note that (e) follows directly from (b) and (d).   The following two theorems are applications of Theorem 2.1.51. 2.1.52 Theorem. Let X and Y be continua, and let f : X → → Y be a surjective map. If X is locally connected, then Y is locally connected. Proof. By Theorem 2.1.37, it suffices to show that TY (B) = B for each closed subset B of Y . Let B be a closed subset of Y . By Remark 2.1.5, B ⊂ TY (B). By Theorem 2.1.51 (a), TY (B) ⊆ f TX f −1 (B). Since X is locally connected, TX f −1 (B) = f −1 (B). Hence, TY (B) ⊆ f f −1 (B) = B. Therefore, TY (B) = B, and Y is locally connected.   2.1.53 Theorem. The monotone image of an indecomposable continuum is an indecomposable continuum. Proof. Let f : X → → Y be a monotone map, where X is an indecomposable continuum. Let y ∈ Y . Then, by Theorem 2.1.51 (b), TX f −1 ({y}) ⊂ f −1 TY ({y}). Since X is indecomposable, by Theorem 2.1.44, TX f −1 ({y}) = X. Hence, Y = f (X) ⊂ f f −1 TY ({y}) ⊂ Y . Thus, TY ({y}) = Y . Therefore, by Theorem 2.1.44, Y is indecomposable.   2.1.54 Definition. Let X and Z be continua, and let f : X → → Z be a map. Then f is an atomic map if for each subcontinuum K of X such that f (K) is nondegenerate, then K = f −1 (f (K)). 2.1.55 Lemma. Let X and Y be continua, and let f : X → → Y be an atomic map. Then for every x ∈ X, f −1 (f (x)) ⊂ TX ({x}). Proof. Let x ∈ X, and let z ∈ X \ TX ({x}). Then there exists a subcontinuum W of X such that z ∈ IntX (W ) ⊂ W ⊂ X \ {x}. Since f −1 (f (x)) is a terminal subcontinuum of X [92, Theorem 8.1.25], by Lemma 1.4.51, we have that f −1 (f (x)) ∩ W = ∅. In particular, z ∈ X \ f −1 (f (x)). Therefore, f −1 (f (x)) ⊂ TX ({x}).  

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2.1.56 Lemma. Let X and Y be continua, where Y is aposyndetic. If f : X → → Y is a monotone map, then TX ({x}) ⊂ f −1 f (x) for each x ∈ X. Proof. Let x ∈ X, and let x ∈ X \ f −1 (f (x)). This implies that f (x) = f (x ). Hence, since Y is aposyndetic, there exists a subcontinuum W of Y such that f (x ) ∈ IntY (W ) ⊂ W ⊂ Y \ {f (x)}. Thus, x ∈ f −1 (f (x )) ⊂ IntX (f −1 (W )) ⊂ f −1 (W ) ⊂ X \ f −1 (f (x)) ⊂ X \ {x}. Since f is monotone, f −1 (W ) is a subcontinuum of X, Lemma 1.4.46. It follows that x ∈ X \ TX ({x}). Therefore, TX ({x}) ⊂ f −1 (f (x)).   Since atomic maps are monotone [92, Theorem 8.1.24], as a consequence of Lemmas 2.1.55 and 2.1.56, we have the following. 2.1.57 Corollary. Let X and Y be continua, where Y is aposyndetic, and let f: X → → Y be an atomic map. Then for every x ∈ X, TX ({x}) = f −1 (f (x)). 2.1.58 Definition. Let X and Y be continua. A surjective map f : X → → Y is quasi-monotone, provided that for every subcontinuum Q of Y , with nonempty interior, f −1 (Q) only has finitely many components, and if C is a component of f −1 (Q), then f (C) = Q. 2.1.59 Theorem. Let X and Y be continua, and let f : X → → Y be a quasimonotone map. If A is a subset of X and B is a subset of Y , then (1) f TX (A) ⊂ TY f (A) and TX f −1 (B) ⊂ f −1 TY (B); (2) TY (B) = f TX f −1 (B); (3) if f is also open, then f −1 TY (B) = TX f −1 (B). Proof. First, we show that f TX (A) ⊂ TY f (A). To this end, let y ∈ Y \ TY f (A). Then there exists a subcontinuum W of Y such that y ∈ Int(W ) ⊂ W ⊂ Y \ TY f (A). Let x ∈ f −1 (y) and let Cx be the component of f −1 (W ) containing x. Note that x ∈ Int(f −1 (W )). Since f is quasi-monotone, f −1 (W ) only has a finite number of components. Hence, by Lemma 1.4.1, x ∈ Int(Cx ). Also, Cx ⊂ X \ f −1 f (A) ⊂ X \ A. Thus, x ∈ X \ TX (A). Since x is an arbitrary point of f −1 (y), f −1 (y) ⊂ X \ TX (A). Therefore, y ∈ Y \ f TX (A). Next, we prove TX f −1 (B) ⊂ f −1 TY (B). Let x ∈ X \f −1 TY (B). Then f (x) ∈ Y \TY (B). Hence, there exists a subcontinuum W of Y such that f (x) ∈ Int(W ) ⊂ W ⊂ Y \ B. Let C be the component of f −1 (W ) that contains x. Observe that x ∈ Int(f −1 (W )). Since f is quasi-monotone, f −1 (W ) only has a finite number of components. Thus, by Lemma 1.4.1, x ∈ Int(C). Also, C ⊂ f −1 (Y \ B) ⊂ X \ f −1 (B). As a consequence of this, x ∈ X \ TX f −1 (B). Now, we see that TY (B) = f TX f −1 (B). By Theorem 2.1.51 part (a), we have that TY (B) ⊂ f TX f −1 (B). By part (1) of this theorem, TX f −1 (B) ⊂ f −1 TY (B). Hence, f TX f −1 (B) ⊂ f f −1 TY (B). Since f is surjective, f f −1 TY (B) = TY (B). Therefore, TY (B) = f TX f −1 (B). Suppose f is also open. By Theorem 2.1.51 part (d), f −1 TY (B) ⊂ TX f −1 (B). Also, by part (1) of this theorem, TX f −1 (B) ⊂ f −1 TY (B). Therefore, f −1 TY (B) = TX f −1 (B).  

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Now, we present some relations between T and inverse limits due to H. S. Davis [30]. 2.1.60 Lemma. Let {Xλ , fλα , Λ} be an inverse system of continua whose inverse limit is X . Let {pλ }λ∈Λ be a point of X , and let A be a nonempty closed subset of X . If there exist λ0 ∈ Λ, two open subsets Uλ0 and Vλ0 of Xλ0 and an inverse system {Wλ , fλα |Wα , Λ} of subcontinua such that pλ0 ∈ Uλ0 , fλ0 (A) ⊂ Vλ0 and for each α ≥ λ0 , (fλα0 )−1 (Uλ0 ) ⊂ Wα ⊂ Xα \ (fλα0 )−1 (Vλ0 ), then {pλ }λ∈Λ ∈ X \ T (A). Proof. Let W = lim{Wλ , fλα |Wα , Λ}. Then W is a subcontinuum of X , ←−

(Uλ0 ) ⊂ W since, for each α ≥ λ0 , (fλα0 )−1 (Uλ0 ) ⊂ Theorem 1.2.20. Note that fλ−1 0  α −1 −1 α −1 Cl((fλ0 ) (Uλ0 )) ⊂ Cl(fα (fλ0 ) (Uλ0 )) ⊂ Wα and W = α≥λ0 fα−1 (Wα ) (Uλ0 ), {pλ }λ∈Λ ∈ Int(W ). (Corollary 1.2.19). Since {pλ }λ∈Λ ∈ fλ−1 0 (Vλ0 ) = Since, for every α ≥ λ0 , fα−1 (Wα ) ⊂ X \ fα−1 (fλα0 )−1 (Vλ0 ) = X \ fλ−1 0  −1 −1 fλ0 (Xλ0 \Vλ0 ) ⊂ X \A, α≥λ0 fα (Wα ) ⊂ X \A. Thus, W ∩A = ∅. Therefore, {pλ }λ∈Λ ∈ X \ T (A).   2.1.61 Lemma. Let {Xλ , fλα , Λ} be an inverse system of continua whose inverse limit is X . Let {pλ }λ∈Λ be a point of X , and let A be a nonempty closed subset of X . If {pλ }λ∈Λ ∈ X \ T (A), then there exist λ0 ∈ Λ, two open subsets Uλ0 and Vλ0 of Xλ0 and an inverse system {Wλ , fλα |Wα , Λ} of subcontinua such that pλ0 ∈ Uλ0 fλ0 (A) ⊂ Vλ0 and, for every α ≥ λ0 , (fλα0 )−1 (Uλ0 ) ∩ fα (X ) ⊂ Wα ⊂ fα (X ) \ (fλα0 )−1 (Vλ0 ). Proof. Since {pλ }λ∈Λ ∈ X \ T (A), there exists a subcontinuum W of X such that {pλ }λ∈Λ ∈ Int(W) ⊂ W ⊂ X \ A. By Proposition 1.2.13, there exist λ(p) ∈ Λ −1 and an open subset Uλ(p) of Xλ(p) such that {pλ }λ∈Λ ∈ fλ(p) (Uλ(p) ) ⊂ Int(W). Similarly, for each {aλ }λ∈Λ ∈ A, there exist λ(a) ∈ Λ and an open subset Uλ(a) −1 of Xλ(a) such that {aλ }λ∈Λ ∈ fλ(a) (Uλ(a) ) ⊂ X \ W. Since A is compact, m −1 1 m there exist {aλ }λ∈Λ , . . . , {aλ }λ∈Λ in A such that A ⊂ j=1 fλ(aj ) (Uλ(aj ) ).

λ0 −1 Let λ0 = max{λ(p), λ(a1 ), . . . , λ(am )}. Define Uλ0 = (fλ(p) ) (Uλ(p) ) and m λ0 −1 Vλ0 = j=1 (fλ(aj ) ) (Uλ(aj ) ). For each λ ∈ Λ, let Wλ = fλ (W). Let α ≥ λ0 , and let zα ∈ (fλα0 )−1 (Uλ0 ) ∩ fα (X ). Since zα ∈ fα (X ), there exists {zλ }λ∈Λ ∈ X such that fα ({zλ }λ∈Λ ) = zα . α Since fλ(p) (zα ) ∈ Uλ(p) , fλ(p) ({zλ }λ∈Λ ) ∈ Uλ(p) . Therefore, −1 {zλ }λ∈Λ ∈ fλ(p) (Uλ(p) ) ⊂ Int(W) ⊂ W.

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Thus, zα = fα ({zλ }λ∈Λ ) ∈ fα (W) = Wα . Hence, (fλα0 )−1 (Uλ0 ) ∩ fα (X ) ⊂ Wα . Now, let wα ∈ Wα = fα (W). Then there exists {wλ }λ∈Λ ∈ W such that fα ({wλ }λ∈Λ ) = wα . Clearly, wα ∈ fα (X ). Suppose wα ∈ (fλα0 )−1 (Vλ0 ). α Then there exists j ∈ {1, . . . , m} such that fλ(a ∈ Uλ(aj ) . Hence, j ) (wα ) fλ(aj ) ({wλ }λ∈Λ ) ∈ Uλ(aj ) . Therefore, {wλ }λ∈Λ ∈ X \ W. This contradiction establishes that Wα ⊂ fα (X ) \ (fλα0 )−1 (Vλ0 ).   2.1.62 Theorem. Let {Xλ , fλα , Λ} be an inverse system of continua whose inverse limit is X . Let {pλ }λ∈Λ be a point of X , and let A be a nonempty closed subset of X . Then the following are equivalent: (a) {pλ }λ∈Λ ∈ X \ T (A); (b) there exist λ0 ∈ Λ, two open subsets Uλ0 and Vλ0 of Xλ0 and an inverse system {Wλ , fλα |Wα , Λ} of subcontinua such that pλ0 ∈ Uλ0 , fλ0 (A) ⊂ Vλ0 and, for every α ≥ λ0 , (fλα0 )−1 (Uλ0 ) ∩ fα (X ) ⊂ Wα ⊂ fα (X ) \ (fλα0 )−1 (Vλ0 ). Proof. Suppose (a). Then Lemma 2.1.61 shows (b). Now, suppose (b). Note that if W = lim{Wλ , fλα |Wα , Λ}, then for each λ ∈ Λ, ←−

fλ (W) = fλ (X )∩Wλ , fλ (W) is a continuum and W = lim{fλ (W), fλα |fα (W) , Λ} ←− (Proposition 1.2.17). The relations in Lemma 2.1.60 imply that (fλα0 )−1 (Uλ0 ) ∩ fα (X ) ⊂ Wα ⊂ fα (X ) \ (fλα0 )−1 (Vλ0 ).   2.1.63 Theorem. Let {Xλ , fλα , Λ} be an inverse system of continua, with surjective bonding maps, whose inverse limit is X . Let {pλ }λ∈Λ be a point of X , and let A be a nonempty closed subset of X . Then the following are equivalent: (a) {pλ }λ∈Λ ∈ X \ T (A); (b) there exist λ0 ∈ Λ, two open subsets Uλ0 and Vλ0 of Xλ0 and an inverse system {Wλ , fλα |Wα , Λ} of subcontinua such that pλ0 ∈ Uλ0 , fλ0 (A) ⊂ Vλ0 and, for every α ≥ λ0 , (fλα0 )−1 (Uλ0 ) ⊂ Wα ⊂ Xα \ (fλα0 )−1 (Vλ0 ). Proof. Note that, since the bonding maps are surjective, the projection maps are surjective also (Theorem 1.2.16). Thus, the theorem now follows directly from Theorem 2.1.62.   2.1.64 Corollary. Let {Xλ , fλα , Λ} be an inverse system of continua, with surjective bonding maps, whose inverse limit is X . Let A be a nonempty closed subset of X , and let Λ0 be a cofinal subset of Λ. Then,  λ∈Λ0

fλ−1 TXλ fλ (A) ⊂ TX (A).

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Proof. Let {pλ }λ∈Λ ∈ X \ TX (A). By Theorem 2.1.63, there exist λ0 ∈ Λ, two open subsets Uλ0 and Vλ0 of Xλ0 and an inverse system {Wλ , fλα |Wα , Λ} of subcontinua such that pλ0 ∈ Uλ0 , fλ0 (A) ⊂ Vλ0 and, for every α ≥ λ0 , (fλα0 )−1 (Uλ0 ) ⊂ Wα ⊂ Xα \ (fλα0 )−1 (Vλ0 ). Since Λ0 is a cofinal subset of Λ, there exists α ∈ Λ0 such that λ0 ≤ α. Then pα ∈ (fλα0 )−1 (Uλ0 ) ⊂ Wα ⊂ Xα \ (fλα0 )−1 (Vλ0 ) ⊂ Xα \ fα (A). Thus, pα ∈ Xα \TXα fα (A). Hence, {pλ }λ∈Λ ∈ X \ fα−1 TXα fα (A). Therefore, {pλ }λ∈Λ ∈ X \ α∈Λ0 fα−1 TXα fα (A).   2.1.65 Corollary. Let {Xλ , fλα , Λ} be an inverse system of continua, with surjective bonding maps, whose inverse limit is X . Let A be a nonempty closed subset of X , and let Λ0 be a cofinal subset of Λ. If the bonding maps are monotone, then 

fλ−1 TXλ fλ (A) = TX (A).

λ∈Λ0

 Proof. Let {pλ }λ∈Λ ∈ X \ λ∈Λ0 fλ−1 TXλ fλ (A). Then there exist λ0 ∈ Λ0 such that pλ0 ∈ Xλ0 \ TXλ0 fλ0 (A). Thus, there exist a subcontinuum Wλ0 of Xλ0 and two open subsets Uλ0 and Vλ0 of Xλ0 such that pλ0 ∈ Uλ0 ⊂ Wλ0 ⊂ Xλ0 \ Vλ0 ⊂ Xλ0 \ fλ0 (A). Let α ∈ Λ0 be such that λ0 ≤ α, and let Wα = (fλα0 )−1 (Wλ0 ). Since the bonding maps are monotone, Wα is a subcontinuum of Xα (Lemma 1.4.46). Clearly, {Wλ , fλγ |Wγ , λ, α ∈ Λ0 and λ0 ≤ λ ≤ γ} is an inverse system of subcontinua and (fλα0 )−1 (Uλ0 ) ⊂ Wα ⊂ Xα \(fλα0 )−1 (Vλ0 ). Hence, by Theorem 2.1.63, {pλ }λ∈Λ ∈ X \ TX (A). The other inclusion is given by Corollary 2.1.64.   The next corollary is a consequence of Davis’ work. 2.1.66 Corollary. Let {Xλ , fλα , Λ} be an inverse system of aposyndetic continua, with surjective bonding maps, whose inverse limit is X . If the bonding maps are monotone, then X is aposyndetic. Proof. Let Λ0 be a cofinal subset  of Λ. Let {pλ }λ∈Λ ∈ X . By Corollary 2.1.65, we have that TX ({{pλ }λ∈Λ }) = λ∈Λ0 fλ−1 TXλ ({pλ }). Since each factor space is aposyndetic, by Theorem 2.1.34,  λ∈Λ0

fλ−1 TXλ ({pλ }) =



fλ−1 ({pλ }) = {{pλ }λ∈Λ }

λ∈Λ0

(Corollary 1.2.19). Therefore, TX ({{pλ }λ∈Λ }) = {{pλ }λ∈Λ }. Since {pλ }λ∈Λ is arbitrary, by Theorem 2.1.34, X is aposyndetic.   2.1.67 Theorem. Let X be a continuum with the property of Kelley, and let W1 and W2 be disjoint subcontinua with nonempty interior. Then T (W1 ) ∩ T (W2 ) = ∅.

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Proof. By Corollary 1.6.21, there exists a subcontinuum M1 such that W1 ⊂ Int(M1 ) and M1 ∩W2 = ∅. This implies that W1 ∩T (W2 ) = ∅. By Corollary 1.6.21, there exists a subcontinuum M2 such that T (W2 ) ⊂ Int(M2 ) and M2 ∩ W1 = ∅.   Hence, T (W2 ) ∩ T (W1 ) = ∅.

2.2 T -Symmetry and T -Additivity We study the properties of continua for which the set function T is symmetric and additive. 2.2.1 Definition. A continuum X is T -symmetric if for each pair, A and B, of closed subsets of X, A ∩ T (B) = ∅ if and only if B ∩ T (A) = ∅. We say that X is point T -symmetric if for each pair, p and q, of points of X, p ∈ T ({q}) if and only if q ∈ T ({p}). 2.2.2 Theorem. If X is a weakly irreducible continuum, then X is T -symmetric. Proof. Let A and B be two closed subsets of X. Suppose A ∩ T (B) = ∅. Hence, for each a ∈ A, there exists a subcontinuum Wa of X such that a ∈ Int(Wa ) ⊂ Wa ⊂ X \ B. Since A is compact, there exist a1 , . . . , an in A such that A ⊂ n n n j=1 Int(Waj ) ⊂ j=1 Waj ⊂ X \ B. This implies that B ⊂ X \ j=1 Waj ⊂   n X\ j=1 Int(Waj ) ⊂ X \ A.   n Let b ∈ B. Since X is weakly irreducible, X \ j=1 Waj only has a finite number of components that are open subsets of X. Let C be the X\   component of  n n j=1 Waj such that b ∈ C. Then b ∈ C ⊂ Cl(C) ⊂ X \ j=1 Int(Waj ) ⊂ X \ A. Thus, b ∈ X \ T (A). Therefore, B ∩ T (A) = ∅. Similarly, if B ∩ T (A) = ∅, then A ∩ T (B) = ∅. Therefore, X is T -symmetric.   2.2.3 Corollary. If X is an irreducible continuum, then X is T -symmetric. Proof. Let X be an irreducible continuum. By Theorem 1.4.43, X is weakly irreducible. Hence, by Theorem 2.2.2, X is T -symmetric.   As a consequence of Theorem 1.4.70 and Theorem 2.2.2, we have the following. 2.2.4 Corollary. If X is a θ-continuum, then X is T -symmetric. 2.2.5 Theorem. Let X be a T -symmetric continuum, and let p be a point of X. Then X is connected im kleinen at p if and only if X is semi-locally connected at p. Proof. Suppose X is connected im kleinen at p. Let q ∈ T ({p}). Since X is T symmetric, p ∈ T ({q}). Thus, p ∈ {q} (Corollary 2.1.31). Hence, p = q and T ({p}) = {p}. Therefore, by Theorem 2.1.35, X is semi-locally connected at p.

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Now, suppose X is semi-locally connected at p. Then T ({p}) = {p} (Theorem 2.1.35). Let A be a closed subset of X such that p ∈ T (A). Hence, since X is T -symmetric, A ∩ T ({p}) = ∅. Thus, p ∈ A. Therefore, by Corollary 2.1.31, X is connected im kleinen at p.   2.2.6 Definition. A continuum X is T -additive if for each pair, A and B, of closed subsets of X, T (A ∪ B) = T (A) ∪ T (B). The next theorem gives us a sufficient condition for a continuum X to be T additive. 2.2.7 Theorem. Let X be a continuum. If for each point x ∈ X and any two subcontinua W1 and W2 of X such that x ∈ Int(Wj ), j ∈ {1, 2}, there exists a subcontinuum W3 of X such that x ∈ Int(W3 ) and W3 ⊂ W1 ∩ W2 , then X is T -additive. Proof. Let A1 and A2 be closed subsets of X, and let x ∈ X \ T (A1 ) ∪ T (A2 ). Hence, there exist two subcontinua W1 and W2 of X such that x ∈ Int(Wj ) ⊂ Wj ⊂ X \ Aj , j ∈ {1, 2}. By hypothesis, there exists a subcontinuum W3 of X such that x ∈ Int(W3 ) and W3 ⊂ W1 ∩ W2 . Thus, x ∈ Int(W3 ) ⊂ W3 ⊂ X \ (A1 ∪ A2 ). Hence, x ∈ X \ T (A1 ∪ A2 ). Therefore, by Corollary 2.1.8, T (A1 ∪ A2 ) = T (A1 ) ∪ T (A2 ).   We need the following definition to give a characterization of T -additive continua. 2.2.8 Definition. Let X be a compactum. A filterbase Γ in X is a family Γ = {Aω }ω∈Ω of subsets of X having two properties: (a) for each ω ∈ Ω, Aω = ∅ and (b) for each ω1 , ω2 ∈ Ω, there exists ω3 ∈ Ω such that Aω3 ⊂ Aω1 ∩ Aω2 . 2.2.9 Lemma. Let X be a  compactum. If Γ is a filterbase of closed subsets of X,  then T ( {G | G ∈ Γ}) = {T (G) | G ∈ Γ}.   Proof. By Proposition  2.1.7, T ( {G | G ∈ Γ}) ⊂ {T (G) | G ∈ Γ}. Let p ∈ X \ T ( {G | G ∈ Γ}).  Then there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \( {G | G∈ Γ}). Since W is compact, there exist n G1 , . . . , Gn in Γ such that W ∩ j=1 Gj = ∅. Since Γ is a filterbase, there exists n G ∈ Γ such  that G ⊂ j=1 Gj . Note that W∩ G = ∅. Thus, p ∈X \ T (G). Hence, p ∈ X \ {T (G) | G ∈ Γ}. Therefore, T ( {G | G ∈ Γ}) = {T (G) | G ∈ Γ}.   The next theorem gives us a characterization of T -additivity on continua. 2.2.10 Theorem. Let X be a continuum. Then X is T -additiveif and only if for each  family Λ of closed subsets of X whose union is closed, T ( {L | L ∈ Λ}) = {T (L) | L ∈ Λ}.

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Proof. Suppose X is T -additive. By Proposition 2.1.7,   {T (L) | L ∈ Λ} ⊂ T {L | L ∈ Λ} .  Now, suppose x ∈ X \ {T (L) | L ∈ Λ}. Then for each L ∈Λ, let F (L) = {A ⊂ X | A is closed andL ⊂ Int(A)}. If L = ∅, then T (L) = {T (A) | A ∈ F (L)}. (Clearly T (L) ⊂ {T (A) | A ∈ F (L)}. Let z ∈ X \ T (L). Then there exists a subcontinuum W of X such that z ∈ Int(W ). Let A be anyproper closed subset of X \W . Then A ∈ F (L) and z ∈ X \T (A). Hence, z ∈ X \ {T(A) | A ∈ F (L)}.) If L = ∅, thenF (L) is a filterbase of closed subsets of X. Since {A | A ∈ F (L)} = L, T (L) = {T (A) | A∈ F (L)}, by Lemma 2.2.9. Hence, for each L ∈ Λ, x ∈ X \ {T (A) | A ∈ F (L)}; and thus, there exists, for each L ∈ Λ, AL ∈F (L) such that x ∈ X \ T (AL ). Note that {Int(AL ) | L ∈ Λ} is an open cover of {L | L ∈ Λ}. Since this set is compact, there exist L1 , . . . , Lm in Λ such that m Int(ALj ). {L | L ∈ Λ} ⊂ j=1

   m m Since, by hypothesis and mathematical induction, T j=1 ALj = j=1 T (ALj ),  m T ( {L | L ∈ Λ}) ⊂ since for every j ∈ {1, . . . , m}, j=1 T (ALj ). Now,  x ∈X \ T (ALj ), it follows that x ∈ X \ T ( {L | L ∈ Λ}). Thus, we have that  T ( {L | L ∈ Λ}) ⊂ {T (L) | L ∈ Λ}.   2.2.11 Theorem. Each T -symmetric continuum is T -additive. Proof. Let X be a T -symmetric continuum, and let A and B be two closed subsets of X. By Corollary 2.1.8, T (A) ∪ T (B) ⊂ T (A ∪ B). Let x ∈ T (A ∪ B). Then {x} ∩ T (A ∪ B) = ∅. Since X is T -symmetric, T ({x}) ∩ (A ∪ B) = ∅. Hence, either T ({x}) ∩ A = ∅ or T ({x}) ∩ B = ∅. Thus, since X is T -symmetric, {x} ∩ T (A) = ∅ or {x} ∩ T (B) = ∅; that is, x ∈ T (A) or x ∈ T (B). Then x ∈ T (A) ∪ T (B). Therefore, X is T -additive.   2.2.12 Theorem. If X is a hereditarily unicoherent continuum, then X is T additive. Proof. Let X be a hereditarily unicoherent continuum, and let A and B be two closed subsets of X. By Corollary 2.1.8, T (A) ∪ T (B) ⊂ T (A ∪ B). Let x ∈ X \ (T (A) ∪ T (B)). Then x ∈ T (A) and x ∈ T (B). Hence, there exist subcontinua WA and WB of X such that x ∈ Int(WA ) ∩ Int(WB ) ⊂ WA ∩ WB ⊂ X \ (A ∪ B). Since X is hereditarily unicoherent, WA ∩ WB is a subcontinuum of X. Therefore, x ∈ X \ T (A ∪ B).   The following corollary is a consequence of Theorem 2.2.10; however, we present a different proof based on Corollary 2.1.25.

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2.2.13 Corollary.  If X is a T -additive continuum and if A is a closed subset of X, then T (A) = a∈A T ({a}). Proof. Let A be a closed subset of the T -additive continuum X.  If a ∈ A, then {a} ⊂ A and T ({a}) ⊂ T (A), by Proposition 2.1.7. Hence, a∈A T ({a}) ⊆ T (A). Now, let x ∈ X be such that x ∈ X \ T ({a}) for each a ∈ A. Then, by Corollary 2.1.25, for every a ∈ A, there exists an open subset Ua of X such that a ∈ Ua and x ∈ X \ T (Cl(Ua )). Since A is compact  and {Ua }a∈A is an open n cover of A, there exist a1 , . . . , an in A such that A ⊂ j=1 Uaj . Since for each n j ∈ {1, . . . , n}, x ∈ X \ T (Cl(Uaj )), x ∈ X \ j=1 T (Cl(Uaj )). Thus, x ∈   n X \T Cl(U ) (since X is T -additive). Hence, x ∈ X \ T (A).   a j j=1 2.2.14 Theorem. A continuum X is locally connected if and only if X is aposyndetic and T -additive. Proof. Suppose X is a locally connected continuum. Then, by Theorem 2.1.37, T (A) = A for each closed subset A of X. Hence, T ({p}) = {p} for each point p ∈ X and T (A ∪ B) = A ∪ B = T (A) ∪ T (B) for each pair of closed subsets A and B of X. Therefore, X is aposyndetic (Theorem 2.1.34) and T -additive. Now, suppose X is aposyndetic and T -additive. Let A be  a closed subset of X. Since X is T -additive, by Corollary 2.2.13, T (A) = a∈A T ({a}). Since X is aposyndetic, by Theorem   2.1.34, T ({a}) = {a} for each a ∈ A. Thus, T (A) = a∈A T ({a}) = a∈A {a} = A. Therefore, by Theorem 2.1.37, X is locally connected.   As a consequence of Theorems 2.2.2, 2.2.11 and 2.2.14, we have the following. 2.2.15 Corollary. If X is an aposyndetic weakly irreducible continuum, then X is locally connected. 2.2.16 Corollary. If X is an aposyndetic hereditarily unicoherent continuum, then X is locally connected. Proof. Let X be an aposyndetic hereditarily unicoherent continuum. Since X is hereditarily unicoherent, by Theorem 2.2.12, X is T -additive. Then, since X is an aposyndetic T -additive continuum, by Theorem 2.2.14, X is locally connected.   2.2.17 Theorem. A continuum X is T -symmetric if and only if X is point T symmetric and T -additive. Proof. Suppose X is T -symmetric. By Theorem 2.2.11, X is T -additive. Clearly, X is point T -symmetric. Now, suppose X is point T -symmetric and T -additive. Let A and B be two closed subsets of X. Suppose that A ∩ T (B) = ∅ and B ∩ T (A) = ∅. Let x ∈ B ∩ T (A). Then x ∈ B and x ∈ T (A). Note that x ∈ X \ A (if x  ∈ A, then x ∈ A∩B ⊂ A∩T (B), a contradiction). Since X is T -additive, T (A) = a∈A T ({a}), Corollary 2.2.13. Thus, there exists a ∈ A such that x ∈ T ({a}). This implies

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that a ∈ T ({x}) (since X is point T -symmetric). Consequently, since x ∈ B, a ∈ T ({x}) ⊂ T (B) (Proposition 2.1.7). Hence, a ∈ A ∩ T (B), a contradiction to our assumption. Therefore, B ∩ T (A) = ∅, and X is T -symmetric.   2.2.18 Theorem. Let X be a semi-aposyndetic irreducible metric continuum. Then X is homeomorphic to [0, 1]. Proof. Since X is an irreducible metric continuum, by Corollary 2.2.3, X is T -symmetric. Let x and y be points of X. Since X is semi-aposyndetic, by Theorem 2.1.32, either x ∈ X \ T ({y}) or y ∈ X \ T ({x}). This implies that x ∈ X \ T ({y}) if and only if y ∈ X \ T ({x}), by the T -symmetry of X. Hence, T ({x}) = {x} for each x ∈ X. Therefore, X is aposyndetic (Theorem 2.1.34). Since X is T -symmetric, X is T -additive (Theorem 2.2.11). Since X is an aposyndetic T -additive continuum, X is locally connected (Theorem 2.2.14). Thus, X is an irreducible locally connected metric continuum. Consequently, X is an irreducible arcwise connected metric continuum [60, Theorem 3-15]. Therefore, X is homeomorphic to [0, 1].   2.2.19 Theorem. The monotone image of a T -symmetric continuum is T symmetric. Proof. Let X be a T -symmetric continuum, and let f : X → → Y be a monotone map. Let A and B be two closed subsets of Y such that A ∩ TY (B) = ∅. Then, by Theorem 2.1.51 (c), A ∩ f TX f −1 (B) = ∅. Thus, f −1 (A) ∩ TX f −1 (B) = ∅. Since X is T -symmetric, TX f −1 (A) ∩ f −1 (B) = ∅. Hence, f (TX f −1 (A) ∩ f −1 (B)) = f TX f −1 (A) ∩ B = TY (A) ∩ B = ∅. Therefore, Y is T -symmetric.   2.2.20 Theorem. The monotone image of a T -additive continuum is T -additive. Proof. Let X be a T -additive continuum, and let f : X → → Y be a monotone map. Let A and B be two closed subsets of Y . Then, by Theorem 2.1.51 (c), TY (A∪B) = f TX f −1 (A ∪ B) = f TX (f −1 (A) ∪ f −1 (B)) = f (TX f −1 (A) ∪ TX f −1 (B)) = f TX f −1 (A) ∪ f TX f −1 (B) = TY (A) ∪ TY (B).   As a consequence of Theorem 2.1.59, we have the next results. 2.2.21 Theorem. Let X and Y be continua, and let f : X → Y be a quasimonotone map. If X is TX -symmetric, then Y is TY -symmetric. Proof. Let B1 and B2 be two closed subsets of Y such that B1 ∩ TY (B2 ) = ∅. By Theorem 2.1.59 part (2), TY (B2 ) = f TX f −1 (B2 ). Hence, B1 ∩f TX f −1 (B2 ) = ∅. Since f −1 (B1 ) ∩ TX f −1 (B2 ) ⊂ f −1 (B1 ) ∩ f −1 f TX f −1 (B2 ) = ∅, we have that f −1 (B1 ) ∩ TX f −1 (B2 ) = ∅. Thus, by TX -symmetry of X, TX f −1 (B1 ) ∩ f −1 (B2 ) = ∅. Since f (TX f −1 (B1 ) ∩ f −1 (B2 )) = f TX f −1 (B1 ) ∩ B2 and, by Theorem 2.1.59 part (2), TY (B1 ) = f TX f −1 (B1 ), we obtain that TY (B1 ) ∩ B2 = ∅. Therefore, Y is TY -symmetric.  

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2.2.22 Theorem. Let X and Y be continua, and let f : X → Y be a quasimonotone map. If X is TX -additive, then Y is TY -additive. Proof. Let B1 and B2 be closed subsets of Y . By Theorem 2.1.59 part (2), we have TY (B1 ∪ B2 ) = f TX f −1 (B1 ∪ B2 ) = f (TX (f −1 (B1 ) ∪ f −1 (B2 ))). Since X is TX -additive, TX (f −1 (B1 ) ∪ f −1 (B2 )) = TX f −1 (B1 ) ∪ TX f −1 (B2 ). Thus, f (TX (f −1 (B1 ) ∪ f −1 (B2 ))) = f (TX f −1 (B1 ) ∪ TX (f −1 (B2 ))) = f TX f −1 (B1 ) ∪ f TX f −1 (B2 ) = TY (B1 ) ∪ TY (B2 ) (Theorem 2.1.59 part (2)). Therefore, TY (B1 ∪ B2 ) = TY (B1 ) ∪ TY (B2 ), and Y is TY -additive.  

2.3 Idempotency of T We present a characterization of idempotency of T and consequences of this. We also give a sufficient condition for the idempotency of T on closed sets and give an example that shows this condition is not necessary. In 1980, David P. Bellamy asked: If X and Y are indecomposable continua, is T idempotent on X × Y ? Even for only the closed sets of X × Y ?. We prove that the set function T is not idempotent on the product of two indecomposable continua. This gives a negative answer to the first question, and the second one remains open (Question 8.1.3). We also prove that T is never idempotent on the cone and suspension over an indecomposable continuum. We give an example of a decomposable metric continuum Z such that T is not idempotent on the family of closed sets of either Z × [0, 1] or K(Z) or Σ(Z). We consider continua for which T is idempotent on continua. We present a couple of classes of continua on which idempotency of T on continua implies idempotency of T on closed sets. 2.3.1 Definition. Let X be a continuum. We say that T is idempotent on X provided that T 2 (A) = T (A) for each subset A of X. We say that T is idempotent on closed sets (on continua) if T 2 (A) = T (A) for each closed subset (subcontinuum) A of X. 2.3.2 Proposition. Let X be a continuum, and let Z be a nonempty closed subset of X such that T 2 (Z) = T (Z). If T (Z) = A ∪ B, where A and B are nonempty closed subsets of X, then T (A ∪ B) = T (A) ∪ T (B) = A ∪ B. Proof. Since T 2 (Z) = T (Z), by Remark 2.1.5 and Proposition 2.1.7, we have T (Z) = A ∪ B ⊂ T (A) ∪ T (B) ⊂ T (A ∪ B) = T 2 (Z) = T (Z).

 

2.3.3 Proposition. Let X be a continuum. If Z is a nonempty closed subset of X such that T 2 (Z) = T (Z), then T (Z) =



{T ({w}) | w ∈ T (Z)}.

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 Proof. Let x ∈ {T ({w}) | w ∈ T (Z)}. Then there exists w ∈ T (Z) such 2 2 that x ∈ T ({w}).  Since T (Z) = T (Z), T ({w}) ⊂ T (Z) = T (Z). Hence, x ∈ T (Z), and {T ({w}) | w ∈ T (Z)} ⊂ T (Z). The other inclusion is clear.   2.3.4 Theorem. Let X be a continuum. Then T is idempotent on X if and only if for each subcontinuum W of X and each point x ∈ Int(W ), there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ M ⊂ Int(W ). Proof. Suppose T is idempotent on X. Let W be a subcontinuum of X, and let x ∈ Int(W ). Hence, x ∈ X\T (X\W ). Since T is idempotent, x ∈ X\T 2 (X\W ). Thus, there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ M ⊂ X \ T (X \ W ) ⊂ X \ (X \ W ) = W . Therefore, x ∈ Int(M ) ⊂ M ⊂ Int(W ). Now, suppose the condition stated holds. We show T is idempotent. By Remark 2.1.5, for each subset A of X, T (A) ⊂ T 2 (A). Let B be a subset of X, and let x ∈ X \ T (B). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ B. By hypothesis, there exists a subcontinuum M such that x ∈ Int(M ) ⊂ M ⊂ Int(W ). Since Int(W ) ⊂ X \ T (B), x ∈ Int(M ) ⊂ M ⊂ X \ T (B). Hence, x ∈ X \ T 2 (B). Therefore, T is idempotent.   2.3.5 Corollary. Let X be a continuum. If T is idempotent on X, W is a subcontinuum of X and K is a component of Int(W ), then K is open. Proof. Let W be a subcontinuum of X, and let K be a component of Int(W ). If x ∈ K, then, by Theorem 2.3.4, there exists a subcontinuum M of X such that x ∈ Int(M ) ⊂ M ⊂ Int(W ). Hence, M ⊂ K and x is an interior point of K. Therefore, K is open.   2.3.6 Definition. Let X be a continuum. A subcontinuum M of X is a continuum domain if M = Cl(Int(M )). A continuum domain M is a strong continuum domain provided that Int(M ) is connected. 2.3.7 Corollary. Let X be a continuum. If T is idempotent on X, x ∈ X and W is a subcontinuum of X such that x ∈ Int(W ), then there exists a strong continuum domain M of X such that x ∈ Int(M ) ⊂ M ⊂ W . Proof. Let W be a subcontinuum of X, and let x ∈ Int(W ). By Corollary 2.3.5, the component, K, of W containing x is open. Let M = Cl(K). Then M is a strong continuum domain and x ∈ Int(M ) ⊂ M ⊂ W .   2.3.8 Theorem. Let X be a continuum, and let A be a subset of X. If T 2 (A) = T (A), then the components of X \ T (A) are open and continuumwise connected. Proof. If A = ∅ or T (A) = X, then the result is clear. Let A be a nonempty subset of X such that T (A) = X. Let L be a component of X \ T (A), and let x ∈ L. Since T 2 (A) = T (A), x ∈ X \ T 2 (A). Hence, there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ T (A). Since L is a component of X \ T (A) and L ∩ W = ∅, W ⊂ L. Thus, x is an interior point of L. Therefore, L is open.

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We show that L is continuumwise connected. Let  and  be two points of L. For each  ∈ L, there exists a subcontinuum W of X such that  ∈ Int(W ) ⊂ W ⊂ X \ T (A). Observe that W ⊂ L for all  ∈ L. Also note that {Int(W ) |  ∈ L} forms an open cover of L, since L is connected, there exist 1 , . . . , n in L such that {Int(W1 ), . . . , Int(Wn )} formsa chain such that  ∈ Int(W1 ) and n  ∈ Int(Wn ) [24, (2.F.2)]. Hence, W = j=1 Wj is a subcontinuum of X such     that { ,  } ⊂ W ⊂ L. Therefore, L is continuumwise connected. A similar result to Corollary 2.3.7 is true when T is idempotent on closed sets. 2.3.9 Theorem. Let X be a continuum such that T is idempotent on closed sets. If A is a nonempty closed subset of X and x ∈ X \ T (A), then there exists a strong continuum domain W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ T (A) ⊂ X \ A. Proof. Let A be a nonempty closed subset of X, and let x ∈ X \ T (A). Since T is idempotent on closed sets, we have that x ∈ X \ T 2 (A). By Corollary 2.1.25, there exists an open subset U of X such that T (A) ⊂ U and x ∈ X \ T (Cl(U )). Let L be the component of X \ T (Cl(U )) containing x. Since T is idempotent on closed sets, by Theorem 2.3.8, L is open. Let W = Cl(L). Then W is a strong continuum domain of X such that x ∈ Int(W ) ⊂ W ⊂ X \ U ⊂ X \ T (A).   2.3.10 Theorem. Let X be an aposyndetic continuum such that T is idempotent on closed sets. If T (W ) = W for each strong continuum domain W of X, then X is locally connected. Proof. Note that, by Theorem 2.1.38, it is enough to show that T (A) = A for each subcontinuum A of X. Clearly T (X) = X. Let A be a proper subcontinuum of X. By Remark 2.1.5, A ⊂ T (A). Let x ∈ X \ A. Since X is aposyndetic, by Theorem 2.3.9, for each a ∈ A, there exists a strong continuum domain Wa of X such that a ∈ Int(Wa ) ⊂  Wa ⊂ X \ {x}. Since n A is compact, there exist n a1 , . . . , an in A such that A ⊂ j=1 Int(Waj ) ⊂ j=1 Waj ⊂ X \ {x}. Observe   n n that j=1 Int(Waj ) is a connected open set. Let W = Cl j=1 Int(Waj ) . Then W is a strong continuum domain such that A ⊂ W ⊂ X \ {x}. By hypothesis, T (W ) = W . Hence, by Proposition 2.1.7, T (A) ⊂ T (W ) = W ⊂ X \ {x}. Thus, x ∈ X \ T (A). Therefore, T (A) = A, and X is locally connected.   The next theorem gives a sufficient condition to have T idempotent on closed sets. 2.3.11 Theorem. Let X be a continuum. If for each nonempty subset A of X and every subcontinuum K of X such that Int(K) = ∅ and A ∩ K = ∅, there exists a subcontinuum W of X such that K ⊂ Int(W ) ⊂ W ⊂ X \A, then T is idempotent on closed sets. Proof. Suppose that for each nonempty subset A of X and every subcontinuum K of X such that Int(K) = ∅ and K ∩ A = ∅, there exists a subcontinuum W of X such that K ⊂ Int(W ) ⊂ W ⊂ X \ A. Let A be a nonempty closed subset of X, and let x ∈ X \ T (A). Then there exists a subcontinuum K of X such that

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x ∈ Int(K) ⊂ K ⊂ X \ A. By hypothesis, there exists a subcontinuum W of X such that K ⊂ Int(W ) ⊂ W ⊂ X \ A. This implies that x ∈ X \ T 2 (A). Hence, T 2 (A) ⊂ T (A). By Remark 2.1.5 and Proposition 2.1.7, T (A) ⊂ T 2 (A). Therefore, T is idempotent on closed sets.   The next example shows that the converse of Theorem 2.3.11 is not true even for metric continua. 2.3.12 Example. Let # X = ({0} × [−1, 2]) ∪

x, sin

   π (% 1 , | x ∈ 0, x 2

    let K = {0} × 0, 32 and let A = {0} × − 31 , − 12 . Then T is idempotent on closed sets, also, Int(K) = {0} × 1, 32 , K ∩ A = ∅, and if W is a subcontinuum of X such that K ⊂ Int(W ), then A ⊂ W . 2.3.13 Theorem. Let X be a continuum. If T is idempotent on X and T ({p, q}) is a continuum for all p and q in X, then X is indecomposable. Proof. Suppose X is decomposable. Then, by Corollary 2.3.7, there exists a strong continuum domain W of X. Let p0 and q0 be any two points in X \ Int(W ). Note that, by Theorem 2.3.4, T ({p0 , q0 }) ∩ Int(W ) = ∅. Since T ({p0 , q0 }) is connected and T ({p0 , q0 })∩Int(W ) = ∅, p0 and q0 belong to the same component of X \ Int(W ). Thus, X \ Int(W ) is a continuum. By Theorem 2.3.4, there exists a subcontinuum M , with nonempty interior, such that M ⊂ Int(X \ Int(W )) ⊂ X \ W . Then let K = X \ (Int(W ) ∪ Int(M )), and let p1 and q1 be any two points of K. By Theorem 2.3.4, T ({p1 , q1 }) ⊂ K. Hence, K is a continuum also. Now, let p ∈ Int(M ), and let q ∈ Int(W ). Observe that, by Theorem 2.3.4, T ({p, q}) ∩ Int(K) = ∅. Then T ({p, q}) ⊂ X \ Int(K) ⊂ W ∪ M . Hence, T ({p, q}) is not a continuum, a contradiction.   2.3.14 Theorem. Let X be a point T -symmetric continuum for which T is idempotent on singletons (Definition 3.1.1). Suppose there exists a point p ∈ X such that T ({p}) = X. Then X is indecomposable. Proof. Let p be a point of X such that T ({p}) = X, and let q ∈ X \ {p}. Then either p ∈ X \ T ({q}) or p ∈ T ({q}). If p ∈ X \ T ({q}), then, by point T symmetry, q ∈ X \ T ({p}) = ∅, a contradiction. Hence, p ∈ T ({q}). Thus, T ({p}) ⊂ T 2 ({q}) = T ({q}) (T is idempotent of singletons). Then T ({q}) = X. Therefore, T ({x}) = X for each point x of X, and X is indecomposable, Theorem 2.1.44.   Now, we give a characterization of locally connected continua. 2.3.15 Theorem. Let X be a continuum such that T is idempotent on continua. Then X is locally connected if and only if T |C1 (X) : C1 (X) → → C1 (X) is surjective. Proof. First note that, by Remark 2.1.5, A ⊂ T (A) for all subsets A of X. Thus, T ({x}) = {x} for each x ∈ X. Hence, X is aposyndetic, Theorem 2.1.34.

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Let A ∈ C1 (X). Since T |C1 (X) is surjective, there exists A ∈ C1 (X) such that T (A ) = A. Thus, since T is idempotent on continua, T (A) = T 2 (A ) = T (A ) = A. Therefore, by Theorem 2.1.38, X is locally connected. The converse implication is clear since for locally connected continua X, T is the identity map on 2X , Theorem 2.1.37. In particular, T |C1 (X) is the identity map on C1 (X).   2.3.16 Corollary. Let X be a continuum such that T is idempotent on closed sets. Then X is locally connected if and only if T : 2X → → 2X is surjective. 2.3.17 Theorem. Let X and Y be continua such that TX is idempotent on continua, and let f : X → → Y be a quasi-monotone map. If for each subcontinuum Z of Y , there exists a subcontinuum W of X such that TX (W ) = f −1 (Z), then Y is locally connected. Proof. Let Z be a subcontinuum of Y . Since f is quasi-monotone, by Theorem 2.1.59 (2), we have that TY (Z) = f TX f −1 (Z). By hypothesis, there exists a subcontinuum W of X such that TX (W ) = f −1 (Z). Hence, since TX2 (W ) = TX (W ), we have that f TX f −1 (Z) = f TX TX (W ) = f TX (W ) = f f −1 (Z) = Z. Thus, TY (Z) = Z. Therefore, by Theorem 2.1.38, Y is locally connected.   The next theorem is an application of Theorem 2.3.11. 2.3.18 Theorem. If X is a continuum with the property of Kelley, then T is idempotent on closed sets. Proof. Let A be a nonempty closed set of X. We know that T (A) ⊂ T 2 (A). Let x ∈ X \ T (A). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ A. By Corollary 1.6.21, there exists a subcontinuum M of X such that W ⊂ Int(M ) and M ∩ A = ∅. By Theorem 2.3.11, T is idempotent on closed sets.   2.3.19 Remark. Theorem 2.3.18 cannot be improved to obtain that the set function T is idempotent, even for metric continua. Let X be the product of a solenoid and a simple closed curve. Then X is a homogeneous metric continuum. By [92, Theorem 4.2.32], TX is idempotent on closed sets and, by Theorem 2.3.22, TX is not idempotent. Next, we prove that the set function T is not idempotent on the product of two continua one of which is indecomposable. In particular, T is not idempotent on the product of two indecomposable continua. We need the following. 2.3.20 Lemma. Let X be an indecomposable continuum, let x0 ∈ X and let Y be a continuum. If W is a subcontinuum of X × Y such that IntX×Y (W ) = ∅, then W ∩ ({x0 } × Y ) = ∅. Proof. Let πX : X × Y → → X be the projection map. Note that πX (W ) is a subcontinuum of X such that IntX (πX (W )) = ∅. Thus, since X is indecomposable, πX (W ) = X, Corollary 1.4.35. Let w ∈ W be such that πX (w) = x0 . Then w ∈ W ∩ ({x0 } × Y ). Therefore, W ∩ ({x0 } × Y ) = ∅.  

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2.3.21 Corollary. If X is an indecomposable continuum, x0 ∈ X and Y is a continuum, then TX×Y ({x0 } × Y ) = X × Y . 2.3.22 Theorem. If X is an indecomposable continuum and Y is a continuum, then TX×Y is not idempotent on X × Y . Proof. Let (x0 , y0 ) ∈ X × Y , let A = ({x0 } × Y ) \ {(x0 , y0 )} and let (x, y) ∈ (X × Y ) \ A, where y = y0 . Let U be an open subset of X such that x ∈ U ⊂ ClX (U ) ⊂ X \ {x0 }. Let W = (ClX (U ) × Y ) ∪ (X × {y0 }). Then W is a subcontinuum of X × Y such that (x, y) ∈ IntX×Y (W ) ⊂ W ⊂ (X × Y ) \ A. Hence, TX×Y (A) = X × Y . Since TX×Y (A) is closed, ClX×Y (A) ⊂ TX×Y (A). Since A is dense in {x0 } × Y , ClX×Y (A) = {x0 } × Y . Hence, by 2 Corollary 2.3.21, X × Y ⊂ TX×Y ({x0 } × Y ) ⊂ TX×Y (A). Therefore, TX×Y is not idempotent on X × Y (Figure 2.5).  

Fig. 2.5 T not idempotent on products

2.3.23 Corollary. If X and Y are indecomposable continua, then TX×Y is not idempotent on X × Y .

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2.3.24 Theorem. Let X and Y be continua. If Z is a closed subset of X × Y such that πX (Z) = X and πY (Z) = Y , where πX and πY are the projection maps, then TX×Y (Z) ⊂ πX (Z) × πY (Z). Proof. Let (x, y) ∈ (X × Y ) \ (πX (Z) × πY (Z)). Without loss of generality, we assume that x ∈ X \ πX (Z). Since πX (Z) is closed in X, there exists an open set U of X such that x ∈ U ⊂ ClX (U ) ⊂ X \ πX (Z). Let y  ∈ Y \ πY (Z). Let W = (ClX (U ) × Y ) ∪ (X × {y  }). Then W is a subcontinuum of X × Y such that (x, y) ∈ IntX×Y (W ) ⊂ W ⊂ (X × Y ) \ πX (Z) × πY (Z). Therefore, TX×Y (Z) ⊂ πX (Z) × πY (Z).   2.3.25 Corollary. Let X and Y be continua. If Z is a closed subset of X × Y such that TX×Y (Z) = X × Y , then either πX (Z) = X or πY (Z) = Y . 2.3.26 Corollary. Let X and Y be continua. If A and B are nonempty proper closed subsets of X and Y , respectively, then TX×Y (A × B) = A × B. Regarding chainable metric continua [92, Definition 2.4.3], we have the next result. 2.3.27 Theorem. Let X and Y be indecomposable chainable metric continua, and let Z be a subcontinuum of X × Y . Then TX×Y (Z) = X × Y if and only if either πX (Z) = X or πY (Z) = Y . Proof. If TX×Y (Z) = X × Y , then, by Corollary 2.3.25, either πX (Z) = X or πY (Z) = Y . Now, suppose πX (Z) = X. Let W be a subcontinuum of X × Y such that IntX×Y (W ) = ∅. We show that W ∩ Z = ∅. Note that, since X is indecomposable, πX (W ) = X, Corollary 1.4.35. Suppose that W ∩ Z = ∅. Then there exists ε > 0 such that ε < d(W, Z). Note that πY (Z) is a chainable continuum, by [92, Theorem 2.4.10]. Let f: X → → [0, 1] and g : πY (Z) → → [0, 1] be ε-maps [92, Theorem 2.4.22]. Then f × g : X × πY (Z) → → [0, 1] × [0, 1] is an ε-map (we use the “max” metric). Hence, (f × g)(Z) ∩ (f × g)(W ) = ∅, otherwise f × g would not be an ε-map. Since πX ((f × g)(W )) = f (πX (W )) = [0, 1] and πY ((f × g)(Z)) = g(πY ((Z)) = [0, 1], we have that (f × g)(Z) ∩ (f × g)(W ) = ∅ [102, Theorem 130, p. 158], a contradiction. Therefore, W ∩ Z = ∅, and TX×Y (Z) = X × Y .   As a consequence of Theorems 2.3.24 and 2.3.27, we have the following. 2.3.28 Corollary. Let X and Y be indecomposable chainable metric continua. If Z is a subcontinuum of X × Y , then either TX×Y (Z) = X × Y or TX×Y (Z) ⊂ πX (Z) × πY (Z). Now, we turn our attention to cones, q : X × [0, 1] → → K(X) denotes the quotient map. 2.3.29 Lemma. Let X be an indecomposable continuum and let x0 ∈ X. If W is a subcontinuum of K(X) such that IntK(X) (W ) = ∅, then W ∩ q({x0 } × [0, 1]) = ∅.

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Proof. If νX ∈ W , then it is clear that W ∩q({x0 }×Y ) = ∅. Suppose that νX ∈ W . Then, since q is monotone, q −1 (W ) is a subcontinuum of X ×[0, 1] (Lemma 1.4.46) such that IntX×[0,1] (q −1 (W )) = ∅. By Lemma 2.3.20, q −1 (W ) ∩ ({x0 } × [0, 1]) = ∅. Therefore, W ∩ q({x0 } × [0, 1]) = ∅.   2.3.30 Corollary. If X is an indecomposable continuum and x0 ∈ X, then TK(X) q({x0 } × [0, 1]) = K(X). 2.3.31 Theorem. If X is an indecomposable continuum, then TK(X) is not idempotent on K(X). Proof. Let (x0 , t0 ) ∈ X × (0, 1), let A = (X × [0, 1]) \ {(x0 , t0 )} and let (x, t) ∈ (X × (0, 1)) \ A, where t = t0 . Construct W as in the proof of Theorem 2.3.22 in such a way that W ∩ (X × {1}) = ∅. Then q((x, t)) ∈ IntK(X) (q(W )) ⊂ q(W ) ⊂ K(X) \ q(A). Thus, TK(X) q(A) = K(X). Since q(A) is dense in q({x0 }×Y ), ClK(X) (q(A)) = q({x0 }×Y ). Hence, by Corollary 2.3.30, 2 q(A). Therefore, TK(X) is not idempotent K(X) ⊂ TK(X) q({x0 } × Y ) ⊂ TK(X) on K(X) (Figure 2.6).  

Fig. 2.6 T not idempotent on cones

Next, we consider suspensions. The results about suspensions are similar to the ones about cones. We state them and present the necessary changes. Here, q : X × [0, 1] → → Σ(X) denotes the quotient map. 2.3.32 Lemma. Let X be an indecomposable continuum and let x0 ∈ X. If W is a subcontinuum of Σ(X) such that IntΣ(X) (W ) = ∅, then W ∩ q({x0 } × [0, 1]) = ∅.

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Proof. If either ν + ∈ W or ν − ∈ W , then it is clear that W ∩ q({x0 } × [0, 1]) = ∅. Suppose that {v + , v − } ∩ W = ∅. Then, since q is monotone, q −1 (W ) is a subcontinuum of X × [0, 1] (Lemma 1.4.46) such that IntX×[0,1] (q −1 (W )) = ∅. By Lemma 2.3.20, q −1 (W )∩ ({x0 }× [0, 1]) = ∅. Therefore, W ∩ q({x0 }× [0, 1]) = ∅.   2.3.33 Corollary. If X is an indecomposable continuum and x0 ∈ X, then TΣ(X) q({x0 } × [0, 1]) = Σ(X). 2.3.34 Theorem. If X is an indecomposable continuum, then TΣ(X) is not idempotent on Σ(X). Proof. Let (x0 , t0 ) ∈ X × (0, 1), let A = (X × [0, 1]) \ {(x0 , t0 )}, and let (x, t) ∈ (X × (0, 1)) \ A, where t = t0 . Construct W as in the proof of Theorem 2.3.22 in such a way that W ∩ (X × {1}∪ X × {0}) = ∅. Then q((x, t)) ∈ IntΣ(X) (q(W )) ⊂ q(W ) ⊂ Σ(X) \ q(A). Thus, TΣ(X) q(A) = Σ(X). Since q(A) is dense in q({x0 } × [0, 1]), ClΣ(X) (q(A)) = q({x0 } × [0, 1]). Hence, by Corollary 2.3.33, Σ(X) ⊂ 2 q(A). Therefore, TΣ(X) is not idempotent on Σ(X). TΣ(X) q({x0 } × Y ) ⊂ TΣ(X)   Now, we give a decomposable metric continuum Z such that T is idempotent on the family of closed sets of neither Z × [0, 1] nor K(Z) nor Σ(Z). In particular, T is not idempotent on either Z × [0, 1] or K(Z) or Σ(Z). 2.3.35 Example. Let us consider the Knaster continuum X, [92, Example 2.4.7], and let X  be the reflection of X in R2 with respect to the origin p = (0, 0). Let Z = X ∪ X  , [92, Example 2.4.9] (Figure 2.7). It is well known that X is an indecomposable continuum, [92, Remark 2.4.8]. To see that TZ×[0,1] is not idempotent on the family of closed subsets of Z × [0, 1], let us note that, using Corollary 2.3.21, we have that TZ×[0,1] ({p} × [0, 1]) = Z × [0, 1]. Let z ∈ X \ {p}. Then, using again Corollary 2.3.21, we obtain that TZ×[0,1] ({z} × [0, 1]) = X × [0, 1]. In particular, {p} × [0, 1] ⊂ TZ×[0,1] ({z} × [0, 1]). Hence, TZ×[0,1] TZ×[0,1] ({z} × [0, 1]) = TZ×[0,1] (X × [0, 1]) = Z × [0, 1]. Therefore, TZ×[0,1] is not idempotent on the family of closed subsets of Z × [0, 1]. To show that TK(Z) is not idempotent on the family of closed subsets of K(Z), let us observe that, using Corollary 2.3.30, we obtain that TK(Z) (q({p} × [0, 1])) = K(Z). Let z ∈ X \ {p}. Then, using Corollary 2.3.30, we have that TK(Z) (q({z} × [0, 1])) = K(X). Observe that q({p} × [0, 1]) ⊂ TK(Z) (q({z} × [0, 1]). Thus, TK(Z) TK(Z) (q({z} × [0, 1])) = TK(Z) (K(X)) = K(Z). Therefore, TK(Z) is not idempotent on the family of closed subsets of K(Z).

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To see that TΣ(Z) is not idempotent on the family of closed subsets of Σ(Z), let us note that, using Corollary 2.3.33, we obtain that TΣ(Z) (q({p} × [0, 1])) = Σ(Z). Let z ∈ X \ {p}. Then, by Corollary 2.3.33, we have that TΣ(Z) (q({z} × [0, 1])) = Σ(X). Hence, q({p} × [0, 1]) ⊂ TΣ(Z) (q({z} × [0, 1]). Thus, TΣ(Z) TΣ(Z) (q({z} × [0, 1])) = TΣ(Z) (Σ(X)) = Σ(Z). Therefore, TΣ(Z) is not idempotent on the family of closed subsets of Σ(Z). Next, we consider the idempotency of T on continua. Note that the idempotency of T on closed sets implies the idempotency of T on continua. The next example shows that the converse is not true for metric continua.

Fig. 2.7 Double Knaster

2.3.36 Example. There exists a metric continuum X such that T is idempotent on continua, and T is not idempotent on closed sets. Let X be the continuum defined by the closure of the union of a sequence {Xn }∞ n=1 of harmonic suspensions such that of the size of Xn+1 is 12 the size of Xn . Also, one of the vertexes of X2 is the mid point of the limit segment of X1 , and one of the vertexes of X3 is the mid point of the limit segment of X2 , etc., as shown in Figure 2.8. Note that T is idempotent on continua. The limit segment in Xn is denoted by x2n−1 x2n , for each n ∈ N. Also, observe that given n ∈ N: • T ({x1 , x2 , x4 , x6 , . . . , x2n }) = x1 x2 ∪ {x4 , . . . , x2n }; • T 2 ({x1 , x2 , x4 , x6 , . . . , x2n }) = x1 x2 ∪ x3 x4 ∪ {x6 , . . . , x2n }; .. . • T n ({x1 , x2 , x4 , x6 , . . . , x2n }) = x1 x2 ∪ x3 x4 ∪ . . . ∪ x2n−1 x2n ; and • T n+1 ({x1 , x2 , x4 , x6 , . . . , x2n }) = T n ({x1 , x2 , x4 , x6 , . . . , x2n }). Therefore, T is not idempotent on closed sets.

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Fig. 2.8 Sequence of harmonic suspensions

2.3.37 Theorem. Let X be a T -additive continuum. If T is idempotent on continua, then T is idempotent on closed sets. Proof. Let A be a nonempty  closed subset of X. Since X is T -additive, by Corollary 2.2.13, T (A) = {T ({a}) | a ∈ A}. Note thatT ({x}) is a continuum, Theorem 2.1.27, for all x ∈ X. Hence, T 2 (A) = T ( {T ({a}) | a ∈ A}) =  {T 2 ({a}) | a ∈ A} = {T ({a}) | a ∈ A} = T (A). The second equality is by T -additivity (Corollary 2.2.13), and the third equality is by the idempotency of T on continua.   2.3.38 Corollary. If X is either a weakly irreducible or a hereditarily unicoherent continuum, then the idempotency on continua of T implies the idempotency on closed sets of T . Proof. Suppose X is weakly irreducible. Then X is T -symmetric, Theorem 2.2.2 Hence, X is T -additive, Theorem 2.2.11. If X is hereditarily unicoherent, then X is T -additive, Theorem 2.2.12. In both cases, the corollary now follows from Theorem 2.3.37.   As a consequence of Theorem 1.4.43 and Corollary 2.3.38, we obtain that the following. 2.3.39 Corollary. If X is an irreducible continuum, then the idempotency on continua of T implies the idempotency of closed sets of T . 2.3.40 Proposition. Let X be a continuum and let n ∈ N. If A = A1 ∪ · · · ∪ An ∈ Cn (X), where A1 , . . . , An are the components of A, is such that T (A) ∈ Cn (X) \ Cn−1 (X), then T (A) = T (A1 ) ∪ · · · ∪ T (An ).

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Proof. Suppose that T (A) = D1 ∪· · ·∪Dn , where D1 , . . . , Dn are the components of T (A). Since A ⊆ D1 ∪ · · · ∪ Dn and Dj ∩ A = ∅ for each j ∈ {1, . . . , n}, Corollary 2.1.20, we may assume that Aj ⊆ Dj for every j ∈ {1, . . . , n}. Observe that T (Aj ) = T (A∩Dj ) = Dj , Corollary 2.1.48, for all j ∈ {1, . . . , n}. Therefore, T (A) = T (A1 ) ∪ · · · ∪ T (An ).   2.3.41 Theorem. Let X be a continuum, and let n ∈ N. If T is idempotent on continua, then T 2n (A) = T 2n−1 (A) for each A ∈ Cn (X). Proof. We do the proof by induction over n. Since T is idempotent on continua, T 2 (A) = T (A) for each A ∈ C1 (X). Suppose that T 2k−2 (A) = T 2k−3 (A) for every A ∈ Ck−1 (A). Let B ∈ Ck (X). We show that T 2k (B) = T 2k−1 (B). Note that T (B) ∈ Ck (X) (Corollary 2.1.21). If T (B) ∈ Ck−1 (X), then T 2k−2 T (B) = T 2k−3 T (B), and T 2k−1 (B) = T 2k−2 (B). Hence, suppose that T (B) ∈ Ck (X) \ Ck−1 (X). By Proposition 2.3.40, T (B) = T (B1 ) ∪ · · · ∪ T (Bk ), where B1 , . . . , Bk are the components of B. We consider two cases. Case 1. T 2 (B) ∈ Ck−1 (X). Then T 2k−2 T 2 (B) = T 2k−3 T 2 (B). Hence, T 2k (B) = T 2k−1 (B). Case 2. T 2 (B) ∈ Ck (X) \ Ck−1 (X). By Proposition 2.3.40, T 2 (B) = T (T (B1 ) ∪ · · · ∪ T (Bk )) = T 2 (B1 ) ∪ · · · ∪ T 2 (Bk ). Since T is idempotent on continua, T 2 (B) = T (B). Therefore, T 2k (B) = T 2k−1 (B) for every B ∈ Ck (X).   2.3.42 Remark. Observe that Example 2.3.36 shows that Theorem 2.3.41 cannot be improved to obtain a global bound for all the elements of C∞ (X) =  ∞ ∞ n=1 Cn (X). Also in the same example, note that if A = {x1 , p} ∪ {x2n }n=1 , where limn→∞ x2n = p, then T k (A) = T k+1 (A) for any k ∈ N.

2.4 Finite Powers of T We study the properties of continua using the set functions T n , n ∈ N. 2.4.1 Definition. Let X be a compactum. If A is a subset of X, then let T 0 (A) = A. If A is a subset of X and n ∈ N, then T n (A) = T T n−1 (A). 2.4.2 Remark. Let X be a compactum. Observe that, by Remark 2.1.5, for each n ∈ N and every subset A of X, T n (A) is a closed subset of X. Hence, T n (A) is compact, for all n ∈ N. Also, by Proposition 2.1.7, if B is a subset of X and A ⊂ B, then T n (A) ⊂ T n (B) for each n ∈ N. In addition, by Theorem 2.1.27, if A is a subcontinuum of X, then T n (A) is a subcontinuum of X for all n ∈ N. 2.4.3 Definition. Let X be a compactum, and let n ∈ N. Then X is point T -nsymmetric provided that for each pair of points p and q of X, p ∈ T n ({q}) if and only if q ∈ T n ({p}).

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2.4.4 Lemma. Let X be a θ-continuum. Let x1 and x2 be two points of X, and let j and k be in N ∪ {0}. If j < k and T j ({x1 }) ∩ T k−j ({x2 }) = ∅, then T j+1 ({x1 })) ∩ T k−j−1 ({x2 }) = ∅. Proof. Let z ∈ T j ({x1 }). Then z ∈ X \ T k−j ({x2 }). Hence, there exists a subcontinum Wz of X such that z ∈ Int(Wz ) ⊂ Wz ⊂ X \ T k−j−1 ({x2 }). Note that {Int(Wz ) | z ∈ T j ({x1 })} is an open cover of T j ({x1 }). Since T j ({x1 }) is compact there exist z1 , . . . , zm in T j ({x1 }) such m(Remark 2.4.2),  m j that T ({x1 }) ⊂ l=1 Int(Wzl ) ⊂ l=1 Wz ⊂ X \ T k−j−1 ({x2 }). Thus, l m j since T ({x1 }) is a continuum (Remark 2.4.2), Hence, m l=1 Wzl is a continuum. m if z  ∈ T k−j−1 ({x2 }), then z  ∈ X \ l=1 W ⊂ X \ Int(W zl ) ⊂ l=1 m zl X \ Tj ({x1 }). Since X is a θ-continuum and l=1 Wzl is a subcontinuum of X, m X \ l=1 Wzl only has finitely (Theorem 1.4.70). Let A m many open components  be the component of X \ l=1 Wzl that contains z . Then A is an open subset m of X. Hence, z  ∈ A ⊂ Cl(A) ⊂ l=1 Int(Wzl ) ⊂ X \ T j ({x1 }). Thus, z  ∈ X \ T j+1 ({x1 }). Therefore, T j+1 ({x1 })) ∩ T k−j−1 ({x2 }) = ∅.   As a consequence of Theorems 1.4.43 and 1.4.70 and Lemma 2.4.4, we have the following. 2.4.5 Corollary. Let X be an irreducible continuum between a and b. Let x1 and x2 be two points of X, and let j and k in N ∪ {0}. If j < k and T j ({x1 }) ∩ T k−j ({x2 }) = ∅, then T j+1 ({x1 })) ∩ T k−j−1 ({x2 }) = ∅. 2.4.6 Theorem. If X is a θ-continuum and n ∈ N, then X is point T -n-symmetric. Proof. We apply Lemma 2.4.4 several times for k = n. Let x1 and x2 be two points of X and suppose j = 0. Then {x1 } ∩ T n ({x2 }) = T 0 ({x1 }) ∩ T n ({x2 }) = ∅ implies that T ({x1 })∩T n−1 ({x2 }) = ∅. The equality T ({x1 })∩T n−1 ({x2 }) = ∅ implies that T 2 ({x1 }) ∩ T n−2 ({x2 }) = ∅. Continuing with this process, we obtain that ∅ = T n ({x1 }) ∩ T 0 ({x2 }) = T n ({x1 }) ∩ {x2 }. Therefore, X is point T -nsymmetric.   Also, from Theorems 1.4.43, 1.4.70 and 2.4.6, we obtain the following. 2.4.7 Corollary. If X is an irreducible continuum between a and b, then X is point T -n-symmetric. 2.4.8 Theorem. Let X be a θ-continuum. If there exists an n ∈ N such that T n+1 ({x}) = T n ({x}) for each x in X, then Gn = {T n ({x}) | x ∈ X} is a decomposition of X. Proof. Let p and q be two points of X, and suppose that there exists x ∈ T n ({p}) ∩ T n ({q}). By Remark 2.4.2, we have that T n ({x}) ⊂ T n (T n ({p})) = T n ({p}). By Theorem 2.4.6, p ∈ T n ({x}) and, by Remark 2.4.2, T n ({p}) ⊂ T n (T n ({x})) = T n ({x}). Hence, T n ({p}) = T n ({x}). Similarly, T n ({q}) = T n ({x}). Therefore, T n ({p}) = T n ({q}), and Gn is a decomposition.  

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2.4.9 Definition. A metric continuum X is said to be n-indecomposable provided that X is the union of n continua such that no one of them is a subset of the union of the others and X is not the union of n + 1 such continua. In the next result, we present characterizations of n-indecomposable metric continua given by C. E. Burgess [16, Theorem 1], [17, Theorems 1, 4 and 5] and [18, Theorem 7]. 2.4.10 Theorem. Let X be a metric continuum. Then the following are equivalent: (1) X is n-indecomposable, for some n ≥ 2; (2) X is the union of n indecomposable continua such that no one of them is a subset of the union of the others and it is irreducible about some n points; (3) X is the union of n indecomposable continua such that no one of them is a subset of the union of the others and it is irreducible about some finite set; (4) X is the union of n indecomposable continua M1 , . . . , Mn such that for each k ∈ {1, . . . , n}, there exists a composant Ck of Mk , where Ck ∩ Mj = ∅, for each j ∈ {1, . . . , n}, j = k; (5) T (F1 (X)) is finite and |T (F1 (X))| ≥ 2. 2.4.11 Theorem. Let X be an irreducible metric continuum between a and b, and suppose that T n ({a}) = X and T m ({a}) = X for any m < n. If T k ({a}) is a k-indecomposable continuum and if T k+1 ({a}) = T k ({a}), then T k+1 ({a}) is a (k + 1)-indecomposable continuum. Proof. Since X is an irreducible continuum between a and b, by Theorem 1.4.40, T k+1 ({a})\T k ({a}) is connected for k+1 < n. Hence, Cl(T k+1 ({a})\T k ({a})) is an indecomposable continuum, otherwise, T m ({a}) = X for some m < n.   2.4.12 Theorem. A necessary and sufficient condition for a metric irreducible continuum X, between a and b, to be n-indecomposable is that T ({a}) is indecomposable and T n ({a}) = X and T m ({a}) = X for any m < n. Proof. We show the condition is necessary. Note that for each k ∈ {1, . . . , n − 1}, T k+1 ({a}) = T k ({a}). Thus, if T ({a}) is indecomposable, applying Theorem 2.4.11 several times, we obtain that T n ({a}) = X is an n-indecomposable continuum. By Theorem 2.4.10, if X is n-indecomposable, then X is the union of nindecomposable continua such that no one of them is a subset of the union of the others. Hence, the condition is sufficient.   2.4.13 Theorem. Let X be a weakly irreducible continuum. If Λ is a family of subsets of X whose union is closed, then T n ( {L | L ∈ Λ}) =  closed n {T (L) | L ∈ Λ}, for each n ∈ N. Proof. For n = 1, by Theorem 2.2.2, X is T -symmetric and, by Theorem 2.2.11, X is T -additive. Hence, by Theorem 2.2.10, we have that if Λ is a family of closed

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  subsets of X whose union is closed, then T ( {L | L ∈ Λ}) = {T (L) | L ∈ Λ}. Now, the theorem follows from mathematical induction.   As a consequence of Theorems 2.4.13, 1.4.43 and 1.4.70, we obtain the following. 2.4.14 Theorem. Let X be an irreducible continuum. If Λ is a  family of closed  subsets of X whose union is closed, then T n ( {L | L ∈ Λ}) = {T n (L) | L ∈ Λ}, for every n ∈ N. 2.4.15 Theorem. Let X be a hereditarily unicoherent continuum.  If Λ is a family n of closed subsets of X whose union is closed, then T ( {L | L ∈ Λ}) =  n {T (L) | L ∈ Λ}, for all n ∈ N. Proof. By Theorem 2.2.12, X is T -additive. Thus, by Theorem 2.2.10, we have that if Λ is a family  of closed subsets of X whose union is closed, then T ( {L | L ∈ Λ}) = {T (L) | L ∈ Λ}. Now, the theorem follows from mathematical induction.   2.4.16 Theorem. If X is a hereditarily unicoherent continuum and K and L are subcontinua of X such that K ∩ L = ∅, then T n (K) ∩ T n (L) = T n (K ∩ L), for each n ∈ N. Proof. For n = 1, by Proposition 2.1.7, T (K ∩ L) ⊂ T (K) ∩ T (L). Suppose there exists x ∈ T (K) ∩ T (L) \ T (K ∩ L). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ (K ∩ L). Since X is hereditarily unicoherent, W ∩ K and W ∩ L are connected. Note that W ∩ K = ∅, otherwise x ∈ Int(W ) ⊂ W ⊂ X \ K and x ∈ X \ T (K). Similarly, W ∩ L = ∅. Since W ∩ (K ∩ L) = ∅, W ∩ K and W ∩ L are disjoint subcontinua. Since K ∩ L = ∅, K ∪ L is a subcontinuum of X. Since X is hereditarily unicoherent, W ∩ (K ∪ L) is a subcontinuum of X and W ∩ (K ∪ L) = (W ∩ K) ∪ (W ∩ L), a contradiction. Therefore, T (K) ∩ T (L) ⊂ T (K ∩ L) and the theorem is true for n = 1. Let n ∈ N, by Remark 2.4.2, T n (K) and T n (L) are subcontinua of X. By Remark 2.1.5 and Proposition 2.1.7, K ⊂ T (K) ⊂ T n (K) and L ⊂ T (L) ⊂ T n (L). Since K ∩ L = ∅, T n (K) ∩ T n (L) = ∅. Therefore, the theorem now follows by mathematical induction.   2.4.17 Theorem. Let X be a continuum, let K1 , . . . , Kh be subcontinua of X and h let K = j=1 Kj . Then, for each n ∈ N ∪ {0}, either T n (K) is a subcontinuum of X or T n (K) = M1 ∪ M2 , where M1 and M2 are two disjoint closed subsets of X, and for Lj = Mj ∩K (j ∈ {1, 2}), K = L1 ∪L2 and T n (K) = T n (L1 )∪T n (L2 ). Proof. The theorem is true for n = 0 because T 0 (L1 ∪ L2 ) = L1 ∪ L2 . Suppose the result is true for n = k − 1, but it is false for n = k. Note that, if T k−1 (K) is a continuum, by Theorem 2.1.27, T k (K) is a continuum. Hence, if T k (K) is disconnected, then T k−1 (K) is disconnected. Since the theorem is not true for n = k, T k (K) = M1 ∪ M2 , where M1 and M2 are disjoint closed subsets of X. Let Mj = T k−1 (K) ∩ Mj (j ∈ {1, 2}). Then M1 and M2 are disjoint closed subsets of

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X and T k−1 (K) = M1 ∪ M2 . By Corollary 2.1.20, we obtain that both M1 and M2 are nonempty. Hence, Lj = Mj ∩ K = Mj ∩ K (j ∈ {1, 2}). Since the theorem is true for n = k − 1, T k−1 (K) = T k−1 (L1 ) ∪ T k−1 (L2 ). Since X is normal, there exist open subsets U1 and U2 of X such that Mj ⊂ Uj (j ∈ {1, 2}) and Cl(U1 ) ∩ Cl(U2 ) = ∅. Suppose there exists x ∈ M1 \ T k (L1 ). Then there exists a subcontinuum Wx of X such that x ∈ Int(Wx ) ⊂ Wx ⊂ X \ T k−1 (L1 ). For each w ∈ Wx ∩ Bd(U1 ), we have that w ∈ X \ T k (K). Thus, there exists a subcontinuum Ww of X such that w ∈ Int(Ww ) ⊂ Ww ⊂ X \ T k−1 (K). Since Wx ∩ Bd(U1 ) is compact, there exist w1 , . . . , wl in Wx ∩ l l Bd(U1 ) such that Wx ∩ Bd(U1 ) ⊂ j=1 Int(Wwj ). Observe that j=1 Wwj ∪ (U1 ∩ Wx ) only has a finite number of components (Theorem 1.4.36). Let S be the l component of j=1 Wwj ∪ (U1 ∩ Wx ) that contains x. Since Wx ∩ T k−1 (L1 ) = ∅ and T k−1 (L2 ) ⊂ M2 ⊂ U2 , we obtain that Cl(U1 ) ∩ T k−1 (L2 ) = ∅. Thus, (Wx ∩ Cl(U1 )) ∩ T k−1 (K) = ∅. Since T k−1 (K) = T k−1 (L1 ) ∪ T k−1 (L2 ) and Ww ∩ T k−1 (K) = ∅, for all w ∈ Wx ∩ Bd(U1 ), we have that S ∩ T k−1 (K) = ∅. Hence, x ∈ Int(Wx ) ∩ U1 ⊂ S ⊂ X \ T k−1 (K). This implies that x ∈ X \ T k (K), a contradiction to the fact that x ∈ M1 ⊂ T k (L1 ). Thus, M1 ⊂ T k (L1 ). Similarly, M2 ⊂ T k (L2 ). As a consequence of the above, T k (K) = M1 ∪ M2 ⊂ T k (L1 ) ∪ T k (L2 ) ⊂ T k (K). Hence, T k (K) = T k (L1 ) ∪ T k (L2 ). Therefore, the theorem is true by mathematical induction.   2.4.18 Corollary. h Let X be a continuum, let K1 , . . . , Kh be subcontinua of X and let K = j=1 Kj . Then, for each n ∈ N, either T n (K) is a continuum or l T n (K) = j=1 T n (Lj ), where l ≤ h and T n (L1 ), . . . , T n (Ll ) are pairwise disjoint continua. The next result extends Corollary 2.1.49. 2.4.19 Corollary. Let X be a continuum and let K1 and K2 be subcontinua of X. Then, for each n ∈ N, either T n (K1 ∪ K2 ) is a continuum or T n (K1 ∪ K2 ) = T n (K1 ) ∪ T n (K2 ). 2.4.20 Theorem. Let X be a continuum, let p1 , . . . , pk be points of X and let n ∈ k N. If X = j=1 T n ({pj }), then X = T n ({p1 , . . . , pk }). Proof. By Remark 2.4.2, we have that for each j ∈ {1, . . . , k}, T n ({pj }) ⊂ T n ({p1 , . . . , pk }). Thus, X=

k

T n ({pj }) ⊂ T n ({p1 , . . . , pk }).

j=1

Therefore, X = T n ({p1 , . . . , pk }).

 

2.4.21 Definition. Let X be a continuum, let A be a nonempty subset of X and let n ∈ N. Then A is a T n -set if and only if

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(1) if a ∈ A, then T n ({a}) ⊂ A; and (2) if p ∈ X \ A, then T n ({p}) ∩ A = ∅. A T n -set A is total if and only if for each T n -set B contained in A, we have that B = A. 2.4.22 Proposition. If X is a T -symmetric continuum, then each T -closed subset of X (Definition 4.1.1) is a T n -set for each n ∈ N. Proof. Let A be a T -closed set and let a ∈ A. Then, by Proposition 2.1.7, T ({a}) ⊂ T (A) = A. Hence, for each n ∈ N, T n ({a}) ⊂ A. Now, let p ∈ X \A. We show, by mathematical induction, that T n ({p})∩A = ∅. For n = 1, suppose that T ({p}) ∩ A = ∅. Let a ∈ T ({p}) ∩ A. Since X is point T -symmetric, p ∈ T ({a}) ⊂ A (by the previous paragraph), a contradiction. Thus, T ({p}) ∩ A = ∅. Suppose T n ({p}) ∩ A = ∅ and assume that T n+1 ({p}) ∩ A = ∅. Let a ∈ T n+1 ({p}) ∩ A. Since X is T -symmetric, we have that ∅ = T ({a}) ∩ T n ({p}) ⊂ A ∩ T n ({p}), a contradiction. Hence, T n+1 ({p}) ∩ A = ∅. Therefore, A is a T n -set for all n ∈ N.   2.4.23 Theorem. Let X be a continuum, and let n ∈ N. If A is T n -set and A1 is a component of A, then A1 is a T n -set. Proof. Let A be a T n -set, and let A1 be a component of A. Let a ∈ A1 . Then, since A is a T n -set, T n ({a}) ⊂ A. Since T n ({a}) is a continuum (Remark 2.4.2), T ({a}) ∩ A1 = ∅ and A1 is a component of A, we have that T n ({a}) ⊂ A1 . Next, let p ∈ X \ A1 and assume that there exists q ∈ T n ({p}) ∩ A1 . If p ∈ X \ A, then, since A is a T n -set, T n ({p}) ∩ A = ∅. Thus, T n ({p}) ∩ A1 = ∅, a contradiction. Thus, p ∈ A. This implies that T n ({p}) ⊂ A. Since T n ({p}) is a continuum (Remark 2.4.2), T ({p}) ∩ A1 = ∅ and A1 is a component of A, we have that T n ({p}) ⊂ A1 . In particular, p ∈ A1 , a contradiction. Hence, T n ({p}) ∩ A1 = ∅. Therefore, A1 is a T n -set.   2.4.24 Corollary. Let X be a continuum, and let n ∈ N. If A is a total T n -set, then A is connected. Proof. Let A be a total T n -set, and let A1 be a component of A. By Theorem 2.4.23, A1 is a T n -set and A1 ⊂ A. Hence, by Definition 2.4.21, we have that A1 = A. Therefore, A is connected.   2.4.25 Theorem. Let X be a continuum, let {Am }∞ m=1 be a sequence of nonempty subsets of X and let n ∈ N. If for each m ∈ N, Am is a T n -set, then both ∞ ∞ n m=1 Am and m=1 Am are T -sets. ∞ Proof. Let a ∈ m=1 Am . Then, for every m ∈ N, a ∈ Am . Let m ∈ N. Since is a T n -set, T n ({a}) ⊂ Am . Hence, since m ∈ N is arbitrary, T n ({a}) ⊂ A m ∞ ∞ m=1 Am . Let p ∈ X \ m=1 Am . Then there exists m0 ∈ N such that p ∈ X \

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∞ Am0 . SinceAm0 is T n -set, T n ({p})∩Am0 = ∅. Hence, T n ({p})∩ m=1 Am = ∅. ∞ Therefore, m=1 Am is T n -set. ∞ Let a ∈ m=1 Am . Then there ∞exists m0 ∈ N such that ∞a ∈ Am0 . Since Am0 is n n a T -set, T ({a}) ⊂ Am0 ⊂ m=1 Am . Let p ∈ X \ m=1 Am . Then, for each n n ({p}) ∩ Am = ∅ for all m ∈ N, p ∈ X \ Am . Since ∞every Am is a T -set, T∞ n m ∈ N. Hence, T ({p}) ∩ m=1 Am = ∅. Therefore, m=1 Am is a T n -set.   2.4.26 Theorem. Let X be an irreducible continuum between q and s, and let A be a subcontinuum of X. Then A is a T -set if and only if for each n ∈ N, A is a T n -set. Proof. Suppose A is T -set. Let a0 ∈ A. Then, by hypothesis, T ({a}) ⊂ A. Hence, by Proposition 2.1.7, T 2 ({a0 }) ⊂ T (A). Thus, by mathematical induction, T n ({a0 }) ⊂ T n−1 (A). We prove that A is a T -closed set. Let x ∈ X \ A. Since A is a T -set, T ({x}) ∩ A = ∅. Thus, for each a ∈ A, there exists a subcontinuum Wa of X such that a ∈ Int(Wa ) ⊂ Wa ⊂ X \ {x}. Since A is compact, there k k exist a1 , . . . , ak in A such that A ⊂ j=1 Int(Waj ) ⊂ j=1 Waj ⊂ X \ {x}. k Since A a continuum, j=1 Waj is a subcontinuum of X. Since X is irreducible, k by Theorem 1.4.40, X \ j=1 Waj has at most two open components. Let K be the k component of X \ j=1 Waj that contains x. Hence, Cl(K) is a subcontinuum of X such that x ∈ K ⊂ Cl(K) ⊂ X \ A. Thus, x ∈ X \ T (A), and T (A) = A. It now follows that T n−1 (A) ⊂ A. Therefore, T n ({a0 }) ⊂ T n−1 (A) ⊂ A. Next, we show that if p ∈ X \ A, then T n ({p}) ∩ A = ∅. Suppose there exists x ∈ T n ({p}) ∩ A. Since X is point T -n-symmetric (Corollary 2.4.7), we have that p ∈ T n ({x}) and, by the previous paragraph, T n ({x}) ⊂ A. Thus, p ∈ A, a contradiction. Hence, T n ({p}) ∩ A = ∅. Therefore, A is a T n -set for all n ∈ N.   2.4.27 Definition. Let X be a continuum, let p ∈ X and let n ∈ N. A subset Vp of X is a T n -chain composant from p of X if and only if Vp = {x | there exist T n ({x0 }), . . . , T n ({xm }) such that x0 = p, xm = x and T n ({xj }) ∩ T n ({xk }) = ∅ if and only if |j − k| ≤ 1}. Note that {T n ({x0 }), . . . , T n ({xm })} forms a chain contained in Vp . This composant V is an essential T n -chain composant from p if and only if each link in these finite chain has an essential part in the chain union; that is, each link is not contained in the union of the other links. 2.4.28 Theorem. Let X be a continuum, and let n ∈ N. If Vp is a T n -chain composant from p of X, then Vp is a connected T n -set. Proof. Observe that,  for each x ∈ Vp , if {T n ({x0 }), . . . , T n ({xm })} is the chain m joining p and x, then j=0 T n ({xj }) is a connected set (Remark 2.4.2). Hence, Vp is connected.

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To see that Vp is a T n -set, let x ∈ Vp . Then, by Definition 2.4.27, T n ({x}) ⊂ Vp . Now, let x ∈ X \ Vp and assume there exists x ∈ T n ({x}) ∩ Vp . Then there exists a chain {T n ({x0 }), . . . , T n ({xm })} such that x0 = p and xm = x . Then {T n ({x0 }), . . . , T n ({xm }), T n ({x})} contains a chain from p to x. Hence, x ∈ Vp , a contradiction. Thus, T n ({x}) ∩ Vp = ∅. Therefore, Vp is a T n -set.   2.4.29 Theorem. Let X be a continuum, and let n ∈ N. Then V = {Vx | x ∈ X}, the family of T n -chain composants of X is a monotone decomposition of X. Proof. By Theorem 2.4.28, the elements of V are connected. Let p and q be elements of X, and assume there exists x ∈ Vp ∩ Vq . Since x ∈ Vp , there exists a chain {T n ({x0 }), . . . , T n ({xm })} such that x0 = p and xm = x. Since x ∈ Vq , there exists a chain {T n ({z0 }), . . . , T n ({zl })} such that z0 = q and zl = x. Then {T n ({x0 }), . . . , T n ({xm }), T n ({z0 }), . . . , T n ({zl })} contains a chain from p to q. Therefore, Vp = Vq and V is a decomposition of X.   2.4.30 Theorem. Let X be a continuum, let A be a nonempty subset of X and let n ∈ N. Then A is a total T n -set if and only if A is a T n -chain composant. Proof. Suppose A is a total T n -set. By Theorem 2.4.29, V = {Vx | x ∈ X} is a monotone decomposition of X. Let a ∈ A. Then Va consists of the union of all finite chains of T n ({x})’s from a. By condition (2) of Definition 2.4.21, we have that A contains the union of all these finite chains from a. Hence, Va ⊂ A. By Theorem 2.4.28, Va is a T n -set and, since A is a total T n -set, we obtain that Va = A. Now, assume that A is a T n -chain composant and let a ∈ A. Then, by Theorem 2.4.29, A = Va . By Theorem 2.4.28, A is a T n -set. By condition (2) of Definition 2.4.21, Va cannot contain a proper T n -set. Therefore, A = Va is a   total T n -set. 2.4.31 Definition. Let X be a continuum, and let L be a proper subset of X. Then L is a C-set if for each subcontinuum K of X such that K ∩ L = ∅, either K ⊂ L or L ⊂ K. 2.4.32 Remark. Let X be a continuum. Then C-sets are connected with empty interior. Note that each proper terminal subcontinuum X is a C-set. If X is an indecomposable continuum with more than one composant, then the composants of X are C-sets. 2.4.33 Theorem. Let X be a continuum, and let L be a C-set in X. If K is a subset of X such that K ⊂ Cl(L), then Cl(L) ⊂ T (K). Proof. Suppose there exists x ∈ Cl(L) \ T (K). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ K. Note that L ∩ Int(W ) = ∅. If W ∩ (X \ L) = ∅, then, since L is a C-set, L ⊂ W . Hence, K ⊂ Cl(L) ⊂ W , a contradiction. Thus, W ∩ (X \ L) = ∅. Then W ⊂ L, a contradiction to the fact that C-sets have empty interior (Remark 2.4.32). Therefore, Cl(L) ⊂ T (K).  

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2.4.34 Theorem. Let X be a continuum, let n ∈ N and let P be a subset of X such that T n (P ) is subcontinuum of X that is a C-set. If K is a subset of X such that K ⊂ T n (P ), then T n (P ) = T n (K). Proof. Let K be a subset of X such that K ⊂ T n (P ). By Theorem 2.4.33, T n (P ) ⊂ T (K) ⊂ T n (K), Remark 2.4.2. By hypothesis, K = T 0 (K) ⊂ T n (P ). Suppose that T n (K) ∩ (X \ T n (P )) = ∅. Hence, there exists h ∈ N such that T h (K) ⊂ T n (P ) and T h+1 (K) ∩ (X \ T n (P )) = ∅. Let x ∈ T h+1 (K) \ T n (P ). Then there exists a subcontinuum W of X such that x ∈ Int(W ) ⊂ W ⊂ X \ T n−1 (P ). Since x ∈ T h+1 (K), we have that W ∩ T h (K) = ∅. But T h (K) ⊂ T n (P ), and there exists a point w ∈ W ∩ T h (K) ⊂ T n (P ). Also, x ∈ W \ T n (P ). Since T n (P ) is a C-set, by Definition 2.4.31, T n−1 (P ) ⊂ T n (P ) ⊂ W , a contradiction to the choice of W . Therefore, T n (K) ⊂ T n (P ) and T n (K) = T n (P ).   2.4.35 Theorem. Let X be a continuum, let n ∈ N and let P be a subcontinuum of X such that Int(T n (P )) = ∅. Assume also that P has the property: if p ∈ T n (P ), then T n (P ) ⊂ T n ({p}). If n = 1, then T (P ) is an indecomposable continuum. If n ≥ 2, then T n (P ) may contain uncountably many indecomposable continua. Proof. By Theorem 2.1.27, T (P ) is a subcontinuum of X. Suppose that T (P ) is the union of two proper subcontinua A and B. Since Int(T (P )) = ∅, there exists a point x ∈ Int(T (P )). Suppose there exists an open subset R of X such that x ∈ R ⊂ A. Let b ∈ T (P ) \ A. Then x ∈ R ⊂ A ⊂ X \ {b}. Thus, x ∈ X \ T ({b}). By hypothesis, T (P ) ⊂ T ({b}), a contradiction. Hence, R ⊂ A. Let a ∈ T (P ) ⊂ B and let z ∈ R \ A, which is open. Then z ∈ R \ A ⊂ B ⊂ X \ {a}. Thus, z ∈ X \ T ({a}). By hypothesis, T (P ) ⊂ T ({a}), a contradiction. Therefore, T (P ) is an indecomposable continuum. For n ≥ 2, consider the following example in R3 . Let K be the Knaster continuum, [92, Example 2.4.7], in the (x, z)-plane tangent to the z-axis at p = (0, 0, 0) and lying to the right of the z-axis. Let X  be the set obtained by revolving K about the z-axis, and let C be a quarter of a circle in the (x, y)-plane traced by a point q ∈ K∩(x-axis). Let C  be a copy of the Cantor ternary set onC, and let Kc be the indecomposable continuum in X  through c ∈ C  . Let X = {Kc | c ∈ C  }. Then X is a continuum. Let P = {p}. Note that T 2 ({p}) = X and, for each x ∈ T 2 ({p}), T 2 ({x}) = T 2 ({p}). Thus, the hypothesis is satisfied and X is the union of uncountably many metric indecomposable subcontinua.   2.4.36 Theorem. Let n ∈ N, and let X be a point T -n-symmetric continuum. Then the following are equivalent: (1) if x ∈ T n ({p}), then T n ({x}) ⊂ T n ({p}); (2) if x ∈ X \ T n ({p}), then T n ({x}) ∩ T n ({p}) = ∅. Proof. Suppose (1) is true and (2) is false. Let x ∈ X \ T n ({p}), and let z ∈ T n ({p}) ∩ T n ({x}). By (1), we have that T n ({z}) ⊂ T n ({p}). Since z ∈ T n ({x}) and X is point T -n-symmetric, x ∈ T n ({z}) ⊂ T n ({p}), a contradiction. Therefore, (1) implies (2).

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Now, assume (2) is true and (1) is false. Let z ∈ T n ({p}) be such that there exists w ∈ T n ({z}) \ T n ({p}). Since X is point T -n-symmetric, we obtain that z ∈ T n ({w}). Hence, z ∈ T n ({w}) ∩ T n ({p}). By (2), T n ({w}) ∩ T n ({p}) = ∅, a contradiction. Therefore, (2) implies (1).   As a consequence of Definition 2.4.21 and Theorem 2.4.36, we have the following. 2.4.37 Corollary. Let n ∈ N, and let X be a point T -n-symmetric continuum. If for each p ∈ X, x ∈ T n ({p}) implies that T n ({x}) ⊂ T n ({p}), then T n ({p}) is a T n -set.

References for Chapter 2 Section 2.1: [3–5, 29–31, 60, 91, 92, 96, 101, 120]. Section 2.2: [3–5, 8, 96]. Section 2.3: [3, 19, 24, 53, 86, 92, 94, 102]. Section 2.4: [16–18, 20, 32, 33, 118, 125].

Chapter 3

Decomposition Theorems

The set function T has also been used to prove several decomposition theorems. Here, first, we present three general decomposition theorems. Then we prove a Hausdorff version, for continua with the uniform property of Effros, of Jones’ Aposyndetic Decomposition Theorem and Prajs’ Mutual Decomposition Theorem. We present the way to show Rogers’ Terminal Decomposition Theorem for homogeneous metric continua, and we end with other decomposition theorems.

3.1 Preliminaries We present the necessary results to show the first three decomposition theorems of next section. 3.1.1 Definition. Let X be a continuum. We say that T is idempotent on singletons if T T ({x}) = T ({x}) for all x ∈ X. The next definition is based on a property obtained by Bellamy and Lum in [11, Lemma 5]. 3.1.2 Definition. Let X be a continuum, and let z ∈ X. We say that T ({z}) has property BL provided that T ({z}) ⊂ T ({x}) for each x ∈ T ({z}). 3.1.3 Lemma. Let X be a decomposable continuum for which T is idempotent on singletons, and let z ∈ X. If T ({z}) has property BL, then T ({w}) = T ({z}) for every w ∈ T ({z}). Proof. Let z ∈ X be such that T ({z}) has property BL, and let w ∈ T ({z}). Since w ∈ T ({z}) and T is idempotent on singletons, T ({w}) ⊂ T T ({z}) = T ({z}). Hence, T ({w}) ⊂ T ({z}). Since T ({z}) has property BL, T ({z}) ⊂ T ({w}). Therefore, T ({w}) = T ({z}).  

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. Macías, Set Function T , Developments in Mathematics 67, https://doi.org/10.1007/978-3-030-65081-0_3

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3.1.4 Corollary. Let X be a decomposable continuum for which T is idempotent on singletons. If z1 and z2 are two points of X such that both T ({z1 }) and T ({z2 }) have property BL, then either T ({z1 }) = T ({z2 }) or T ({z1 }) ∩ T ({z2 }) = ∅. Proof. Let z1 and z2 be points of X such that both T ({z1 }) and T ({z2 }) have property BL. Suppose T ({z1 }) ∩ T ({z2 }) = ∅. Let z3 ∈ T ({z1 }) ∩ T ({z2 }). Then, by Lemma 3.1.3, T ({z1 }) = T ({z3 }) = T ({z2 }).   The proof following lemma is very similar to the proof of Proposition 2.3.3. 3.1.5 Lemma. Let X be a decomposable continuum for which T is idempotent on singletons. If z ∈ X, then T ({z}) =



{T ({w}) | w ∈ T ({z})}.

Proof. Let z ∈ X, and let w0 ∈ T ({z}).Since T is idempotent on singletons, T ({w0 }) ⊂ T T ({z}) = T ({z}). Hence, {T ({w}) | w ∈ T ({z})} ⊂ T ({z}). The other inclusion is clear.   The proof of the next theorem is based on a technique of Bellamy and Lum [11, Lemma 5]. 3.1.6 Theorem. If X is a decomposable continuum for which T is idempotent on singletons, then, for each x ∈ X, there exists z ∈ T ({x}) such that T ({z}) has property BL. Proof. Let x ∈ X. Then by Lemma 3.1.5, we have that T ({x}) =



{T ({w}) | w ∈ T ({x})}.

Let Gx = {T ({w}) | w ∈ T ({x})}. Partially order Gx by inclusion. Let H = {T ({wλ })}λ∈Λ be a (set theoretic) chain of elements of Gx . We show that H has a lower bound in Gx .  Since H isa (set theoretic) chain of continua (Theorem 2.1.27), λ∈Λ T ({wλ }) = ∅. Let w0 ∈ λ∈Λ T ({wλ }). Then, ) T ({w0 }) ⊂ T

 λ∈Λ



* T ({wλ })

⊂



T T ({wλ }) =

λ∈Λ

T ({wλ }) ⊂ T ({x}).

λ∈Λ

Hence, by Kuratowski–Zorn Lemma, there exists z ∈ T ({x}) such that T ({z}) is a minimal element; that is, each w ∈ T ({z}) satisfies that T ({z}) ⊂ T ({w}). Therefore, T ({z}) has property BL.  

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3.1.7 Lemma. Let X be a continuum, let A and L be two nonempty closed subsets of X and let {Aλ }λ∈Λ be a net of nonempty closed subsets of X converging to A. If the net {T (Aλ )}λ∈Λ converges to L, then L ⊂ T (A) ⊂ T (L). Proof. We prove that L ⊂ T (A). Let l ∈ L, and let W be a subcontinuum of X such that l ∈ Int(W ). Since {T (Aλ )}λ∈Λ converges to L, there exists λ0 ∈ Λ such that T (Aλ ) ∩ Int(W ) = ∅ for all λ ≥ λ0 . This implies that Aλ ∩ W = ∅ for every λ ≥ λ0 . By [40, 3.1.23 and 1.6.1], there exist Λ ⊂ Λ and zλ ∈ Aλ ∩ W , for each λ ∈ Λ , and a point z in X such that {zλ }λ∈Λ converges to z. Note that z ∈ A ∩ W . Since W is an arbitrary subcontinuum of X containing l in its interior, l ∈ T (A). Hence, L ⊂ T (A). Since Aλ ⊂ T (Aλ ), {Aλ }λ∈Λ converges to A and {T (Aλ )}λ∈Λ converges to L, we have that A ⊂ L. Thus, T (A) ⊂ T (L).   3.1.8 Definition. Let X be a continuum. Define ξ : X → → F1 (X) by ξ(x) = {x}. 3.1.9 Remark. Let X be a continuum. Since X is a Hausdorff space, the function ξ given in Definition 3.1.8 is a homeomorphism [98, p. 153]. 3.1.10 Theorem. Let X be a continuum with the property of Kelley. If G = {T ({x}) | x ∈ X} is a decomposition of X, then T ({x}) is a terminal subcontinuum of X for all x ∈ X. Proof. Suppose there exists a point x0 in X such that T ({x0 }) is not a terminal subcontinuum of X (by Theorem 2.1.27, T ({x0 }) is a continuum). Then there exists a subcontinuum L of X such that L∩T ({x0 }) = ∅, L\T ({x0 }) = ∅ and T ({x0 })\ L = ∅. Let p ∈ L \ T ({x0 }) and  ∈ L ∩ T ({x0 }). Since G is a decomposition, without loss of generality, we assume that x0 ∈ T ({x0 })\L. Since p ∈ L\T ({x0 }), there exists a subcontinuum W of X such that p ∈ Int(W ) ⊂ W ⊂ X \ {x0 }. Let K = L ∪ W . Then K is a subcontinuum of X, p ∈ Int(K) and K ⊂ X \ {x0 }. By Theorem 1.6.20, there exists a subcontinuum M of X such that  ∈ Int(M ) and M ⊂ X \ {x0 }, a contradiction to the choice of . Therefore, T ({x}) is a terminal subcontinuum of X for all x ∈ X.  

3.2 Three Decomposition Theorems We present three decomposition theorems: one for the class of point T -symmetric continua X for which T is idempotent on singletons; the second one is for the class of continua X for which T |F1 (X) is continuous and T is idempotent on singletons; and the third for the class of continua for which the set function T is continuous. We restrict ourselves to decomposable nonlocally connected continua; since it is known that T is a constant map on indecomposable continua, Theorem 2.1.44, and T is the identity map on locally connected continua, Theorem 2.1.37.

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The proof of the following theorem is similar to the one given for Theorem 5.1.1. 3.2.1 Theorem. Let X be a continuum. If G = {T ({x}) | x ∈ X} is a decomposition of X, then G is upper semicontinuous. Hence, X/G is a continuum. Proof. Let x0 ∈ X, and let U be an open subset of X such that T ({x0 }) ⊂ U . Let K = {x ∈ X | T ({x}) ∩ (X \ U ) = ∅}. We show that K is closed in X. Let x ∈ Cl(K). Then there exists a net {xλ }λ∈Λ of points of K converging to x [40, 1.6.3]. Since each xλ ∈ K, there exists zλ ∈ T ({xλ }) \ U for every element λ ∈ Λ. Without loss of generality, we assume that the net {zλ }λ∈Λ converges to a point z [40, 3.1.23 and 1.6.1]. Note that z ∈ X \ U . We prove that z ∈ T ({x}). To this end, suppose that z ∈ X \ T ({x}). Then there exists a subcontinuum H of X such that z ∈ Int(H) ⊂ H ⊂ X \ {x}. Since {zλ }λ∈Λ converges to z and {xλ }λ∈Λ converges to x, there exists λ0 ∈ Λ such that zλ ∈ Int(H) and xλ ∈ X \ H for all λ ≥ λ0 . This implies that zλ ∈ X \ T ({xλ }) for each λ ≥ λ0 , a contradiction. Thus, z ∈ T ({x}) \ U , and K is closed. Let V = X \ K. Then V = {x ∈ X | T ({x}) ⊂ U }, T ({x0 }) ⊂ V and V is open in X. Therefore, G is upper semicontinuous. By Theorem 1.1.22, X/G is a Hausdorff space. Therefore, X/G is a continuum.   3.2.2 Corollary. Let X be a continuum such that TX is idempotent on singletons. If G = {TX ({x}) | x ∈ X} is a decomposition of X, then X/G is an aposyndetic continuum. Proof. By Theorem 3.2.1, X/G is a continuum. Let q : X → → X/G be the quotient map. By Theorem 2.1.27, q is a monotone map. Let χ ∈ X/G, and let x ∈ X be such that TX ({x}) = q −1 (χ). Since q is monotone, by Theorem 2.1.51 (c), TX/G ({χ}) = qTX q −1 (χ) = qTX TX ({x}) = qTX ({x}) = qq −1 (χ) = {χ}, the third equality is true because TX is idempotent on singletons. Therefore, X/G is an aposyndetic continuum, Theorem 2.1.34.   3.2.3 Lemma. Let X be a continuum, and let z ∈ X. If G = {T ({x}) | x ∈ X} is a decomposition and W is a subcontinuum of X such that T ({z}) ∩ Int(W ) = ∅, then T ({z}) ⊂ W . Proof. Let W be a subcontinuum of X, and let z be a point of X such that T ({z}) ∩ Int(W ) = ∅. Let x ∈ T ({z}) ∩ Int(W ), and suppose there exists y ∈ T ({z}) \ W . Thus, x ∈ Int(W ) ⊂ W ⊂ X \ {y}; that is, x ∈ T ({y}). Since G is a decomposition, T ({x}) = T ({z}) = T ({y}), a contradiction. Therefore, T ({z}) ⊂ W .  

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Now we are ready to prove our first decomposition theorem. 3.2.4 Theorem. Let X be a decomposable point TX -symmetric continuum for which TX is idempotent on singletons. Then, G = {TX ({x}) | x ∈ X} is an upper semicontinuous decomposition of X such that the quotient space X/G is an aposyndetic continuum. Proof. Let x ∈ X. Then by Theorem 3.1.6, there exists z ∈ TX ({x}) such that TX ({z}) has property BL. Since X is point TX -symmetric and z ∈ TX ({x}), we have that x ∈ TX ({z}). Thus, since TX ({z}) has property BL, TX ({x}) = TX ({z}), Lemma 3.1.3. Hence, TX ({x}) has property BL. Therefore, by Corollary 3.1.4, G = {TX ({x}) | x ∈ X} is a decomposition of X. Thus, by Theorem 3.2.1, G is upper semicontinuous and, by Corollary 3.2.2, X/G is an aposyndetic continuum.   As an application of Theorem 3.2.4, we show the next result due to Javier Camargo. 3.2.5 Theorem. Let X be an irreducible metric continuum. If T is idempotent on singletons, then T is idempotent on closed sets. Proof. If X is indecomposable, by Theorem 2.1.44, T is idempotent on closed sets. Hence, suppose X is decomposable. By Corollary 2.2.3, X is T -symmetric; in particular, X is point T -symmetric. By Theorem 3.2.4, G = {T ({x}) | x ∈ X} is an upper semicontinuous decomposition of X such that X/G is an aposyndetic continuum. Let q : X → → X/G be the quotient map. Since for each x ∈ X, q −1 q(x) = T ({x}) and T ({x}) is a continuum (Theorem 2.1.27), we have that q is a monotone map. Thus, X/G is an irreducible metric continuum [75, Theorem 3, p. 192]. In fact, by Theorem 2.2.18, X/G is an arc. Let A be a closed subset of X. Thus, T ({a}) = q −1 q(a) ⊂ T (A), for each a ∈ A. Hence, q −1 q(A) ⊂ T (A). Let z ∈ X \ q −1 q(A). Then q(z) ∈ X/G \ q(A). Since X/G is an arc, there exists a subcontinuum W of X/G such that q(z) ∈ IntX/G (W) ⊂ W ⊂ X/G \ q(A). Thus, q −1 (W) is a subcontinuum of X (Lemma 1.4.46), z ∈ IntX (q −1 (W)) and q −1 (W) ∩ A = ∅. This implies that z ∈ X \ T (A). Hence, q −1 q(A) = T (A) and, by Theorem 4.1.25, T (A) is a T -closed set. Therefore, T 2 (A) = T (A) for each closed subset A of X, and T is idempotent on closed sets.   A proof of the following result may be found in [73, 2.1]. The theorem was originally proved by E. Dyer. 3.2.6 Theorem. Let X and Y be nondegenerate metric continua. If f : X → → Y is a surjective, monotone and open map, then there exists a dense Gδ subset W of Y having the following property: for every y ∈ W , for each subcontinuum B of

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f −1 (y), for all x ∈ Intf −1 (y) (B) and for every neighbourhood U of B in X, there exist a subcontinuum Z of X containing B and a neighbourhood V of y in Y such −1 that x ∈ IntX (Z), (f |Z ) (V ) ⊂ U and f |Z : Z → → Y is a monotone surjective map. Now, we are ready to prove our second decomposition theorem. 3.2.7 Theorem. Let X be a decomposable continuum for which TX |F1 (X) is continuous and for which TX is idempotent on singletons. Then, G = {TX ({x}) | x ∈ X} is a continuous decomposition of X such that the quotient space X/G is an aposyndetic continuum. Moreover, all the elements of G are nowhere dense in X; and if X is metric, then there exists a dense Gδ subset W of X/G such that if q(z) ∈ W, then TX ({z}) is an indecomposable metric continuum, where q: X → → X/G is the quotient map. Proof. We show that G is a decomposition of X. To this end, let M = {x ∈ X | TX ({x}) has property BL} (by Lemma 3.1.3, if x ∈ M , then TX ({z}) = TX ({x}) for all z ∈ TX ({x})). Note that for every m ∈ M , TX ({m}) ⊂ M . By Theorem 3.1.6, M = ∅, and M ∩ TX ({x}) = ∅ for all x ∈ X. We show that M = X. First, we prove that M is closed in X. Let x ∈ Cl(M ). Then there exists a net {mλ }λ∈Λ of elements of M converging to x [40, 1.6.3]. Let z ∈ TX ({x}). Since TX is continuous on singletons, the net {TX ({mλ })}λ∈Λ converges to TX ({x}) [97, Proposition 3.38]. Hence, for each λ ∈ Λ, there exists zλ ∈ TX ({mλ }) such that {zλ }λ∈Λ converges to z [104, Theorems 2 and 4]. Since TX is continuous on singletons, the net {TX ({zλ })}λ∈Λ converges to TX ({z}). Since TX ({zλ }) = TX ({mλ }) for all λ ∈ Λ, we have that TX ({z}) = TX ({x}). Therefore, M is closed. Now, we see that M = X. Suppose this is not true. Assume first that M is not connected. Then there exist two nonempty closed subsets M1 and M2 of X such that M = M1 ∪ M2 . Since X is normal, there exist two disjoint open subsets U1 and U2 of X such that M1 ⊂ U1 and M2 ⊂ U2 . Let j ∈ {1, 2}. Observe that for each mj ∈ Mj , TX ({mj }) ⊂ Mj ⊂ Uj . Since TX is upper semicontinuous (Theorem 5.1.1), for every mj ∈ Mj , there exists an open subset  Vmj of X such that mj ∈ Vmj and TX ({z}) ⊂ Uj for each z ∈ Vmj . Let Vj = {Vmj | mj ∈ Mj }. Clearly, Vj is an open subset of X, Vj ⊂ Uj , Mj ⊂ Vj and TX ({z}) ⊂ Uj for every z ∈ Vj . Let Rj = {x ∈ X | TX ({x}) ⊂ X \ Mj } = ξ −1 ((TX |F1 (X) )−1 (X \ Mj 1 )),

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101

and let R = {x ∈ X | TX ({x}) ∩ M1 = ∅ and TX ({x}) ∩ M2 = ∅} = ξ −1 ((TX |F1 (X) )−1 (X, M1 1 ∩ X, M2 1 )). Note that Rj is open in X and R is closed. Also, Vk ⊂ Rj , j and k in {1, 2} and k = j, and (V1 ∪ V2 ) ∩ R = ∅. Thus, Rj = ∅, j ∈ {1, 2}. By Theorem 3.1.6, if x ∈ X, then TX ({x}) ∩ M = ∅. Hence, X ⊂ R ∪ R1 ∪ R2 , and X = R ∪ R1 ∪ R2 . Observe that if x ∈ R1 ∩ R2 , then TX ({x}) ⊂ X \ (M1 ∪ M2 ), a contradiction to Theorem 3.1.6. This implies that R1 ∩ R2 = ∅. As a consequence of this, since X is connected, R = ∅. Let x ∈ R ∩ (Cl(R1 ) ∪ Cl(R2 )). Without loss of generality, we assume that x ∈ Cl(R2 ). Then, by [40, 1.6.3], there exists a net {rλ }λ∈Λ of elements of R2 converging to x. Since x ∈ R, TX ({x}) ∩ M2 = ∅. Also, since M2 ⊂ V2 , we have that TX ({x}) ∈ X, V2 1 . Since TX is continuous on singletons, there exists λ0 ∈ Λ such that TX ({rλ0 }) ∈ X, V2 1 . Let z ∈ TX ({rλ0 }) ∩ V2 . By the construction of V2 , TX ({z}) ⊂ U2 . By Theorem 3.1.6, TX ({z}) ∩ M = ∅. Since M1 ∩ U2 = ∅, we have that TX ({z}) ∩ M2 = ∅. Since TX is idempotent on singletons, TX ({z}) ⊂ TX2 ({rλ0 }) = TX ({rλ0 }), and TX ({rλ0 }) ∩ M2 = ∅, a contradiction to the fact that rλ0 ∈ R2 . Thus, M cannot be disconnected. Hence, M is connected. Let x ∈ X \ M . By Theorem 3.1.6, there exists m0 ∈ M such that TX ({m0 }) ⊂ TX ({x}). Since X is normal, there exist two disjoint open subsets U and V of X such that M ⊂ U and x ∈ V . Since TX is upper semicontinuous (Theorem 5.1.1), there exists an open subset W of X such that  m0 ∈ W and TX ({z}) ⊂ U , for all z ∈ W . Let K = Cl ( {TX ({z}) | z ∈ W }) ∪ M . Since M is a subcontinuum of X and TX ({z}) ∩ M = ∅ for each z ∈ W , we have that K is a subcontinuum of X. Note that K ⊂ Cl(U ) ⊂ X \ V and m0 ∈ W ⊂ K. Thus, K ∩ {x} = ∅ and m0 ∈ X \ TX ({x}), a contradiction to the fact that TX ({m0 }) ⊂ TX ({x}). Therefore, M = X, and G = {TX ({x}) | x ∈ X} is a decomposition of X. Hence, X is point TX -symmetric and, by Theorem 3.2.4, X/G is an aposyndetic continuum. Since TX |F1 (X) is continuous, in fact, G is a continuous decomposition of X. Note that, by Theorem 2.1.27, q is a monotone map. Also, since G is a continuous decomposition, by Corollary 1.1.24, q is an open map. Since q is an open map, all the elements of G are nowhere dense in X. Suppose X is a metric continuum. Since the quotient map q is surjective, monotone and open, let W be the dense Gδ subset of X/G given by Theorem 3.2.6. Let ω ∈ W, and let z ∈ X be such that q(z) = ω. Suppose TX ({z}) is decomposable. Then there exist two subcontinua H and K of TX ({z}) such that TX ({z}) = H ∪ K. Let x ∈ H \ K, and let U be an open subset of X such that H ⊂ U and K \ U = ∅. By Theorem 3.2.6, there exist a subcontinuum Z of X containing H and a neighbourhood V of ω in X/G such that x ∈ IntX (Z) and (q|Z )−1 (V) ⊂ U . Since x ∈ TX ({z}) ∩ IntX (Z), by Lemma 3.2.3, TX ({z}) ⊂ Z. Observe that this implies that TX ({z}) ⊂ (q|Z )−1 (V) ⊂ U , a contradiction.   Therefore, TX ({z}) is an indecomposable metric continuum.

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As a consequence of Theorems 2.3.14 and 3.2.7, we obtain the following. 3.2.8 Corollary. Let X be a continuum for which T |F1 (X) is continuous and for which T is idempotent on singletons. If there exists a point p of X such that T ({p}) = X, then X is indecomposable. Next, we prove our third decomposition theorem. 3.2.9 Theorem. Let X be a continuum for which TX is continuous. Then, G = {TX ({x}) | x ∈ X} is acontinuous decomposition of X, X/G is a locally connected continuum   and  TX 2X is homeomorphic to 2X/G . (In particular, if X is metric, then TX 2X is homeomorphic to the Hilbert cube.) Moreover, all the elements of G are nowhere dense in X; and if X is metric, then there exists a dense Gδ subset W of X/G such that if q(z) ∈ W, then TX ({z}) is an indecomposable metric continuum, where q: X → → X/G is the quotient map. Proof. Since T is continuous, by Theorem 5.1.6, T is idempotent. Hence, by Theorem 3.2.7, G is a continuous decomposition such that X/G is an aposyndetic continuum. Let q : X → → X/G be the quotient map. By Theorem 2.1.27, q is a monotone. By Theorem 1.1.24, q is an open and surjective map. Note that q −1 TX/G (Γ) = TX q −1 (Γ) for each subset Γ of X/G (Theorem 2.1.51 (e)). Hence, q is a TXX/G continuous surjective open map. Since TX is continuous, TX/G is also continuous (Theorem 5.1.4). Since aposyndetic continua for which T is continuous are locally connected (Corollary we have that X/G is locally connected.  5.1.17),  To see that TX 2X is homeomorphic to 2X/G , let (q) : 2X/G → 2X be given by (q)(Γ) = q −1 (Γ). By Theorem 1.6.16, (q) is continuous. Note that 2q ◦  X/G X/G is a homeomorphism. (q) = 12X/G . In particular, (q) : 2 → (q) 2     Thus, it is enough to show that TX 2X = (q) 2X/G . Let Γ ∈ 2X/G . Then TX (q)(Γ) = TX q −1 (Γ) = q −1 TX/G (Γ) = q −1 (Γ) = (q)(Γ); the second equality is true by Theorem third  2.1.51  (e), and the  equality  X is X X/G 2 , and (q) 2 ⊂ T . valid by Theorem 2.1.37.Thus, (q)(Γ) ∈ T X X 2  Now, let K ∈ TX 2X . Then there exists A ∈ 2X such that T (A) =  −1 X −1 K. We prove that K = (q)q(A). Note that q q(A) = {q q(a) | a ∈  A} = {TX ({a}) | a ∈ A}. Since G is a decomposition, X is point TX symmetric. Hence, X is TX -additive (Theorem 5.1.10). Since X is TX -additive,  {TX ({a}) | a ∈ A}  = TX (A)  (Corollary  2.2.13).  Thus, (q)q(A) = TX (A) = K, K ∈ (q) 2X/G and TX 2X ⊂ (q) 2X/G .       Therefore, TX 2X = (q) 2X/G . Since (q) 2X/G is homeomorphic to   2X/G , we see that TX 2X is homeomorphic to 2X/G . Suppose X is metric. Then X/G is homeomorphic to X/G is a metric continuum [92, Theorem   1.3.7]. Since 2 the Hilbert cube [105, (1.97)], TX 2X is homeomorphic to the Hilbert cube. Now the theorem follows from Theorem 3.2.7.  

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3.2.10 Corollary. Let X be a hereditarily decomposable T -additive metric continuum. Then T is continuous if and only if X is locally connected. Proof. If X is locally connected, then T is the identity map (Theorem 2.1.37). Hence, T is continuous. Suppose X is a hereditarily decomposable T -additive metric continuum. Note that if X is also aposyndetic, then, by Theorem 2.2.14, X is locally connected. Assume that X is not aposyndetic. Since T is continuous, by Theorem 3.2.7, G = {T ({x}) | x ∈ X} is a continuous decomposition of X. Since X is not aposyndetic, by Theorem 2.1.34, there exists x0 ∈ X such that T ({x0 }) is nondegenerate. Then, by the continuity of T , there exists an open subset U of X such that x0 ∈ U , and for each x ∈ U , T ({x}) is nondegenerate. Thus, by Theorem 3.2.9, there exists x1 ∈ U such that T ({x1 }) is indecomposable, a contradiction. Hence, X is aposyndetic. Therefore, X is locally connected.   3.2.11 Corollary. Let X be a hereditarily decomposable and hereditarily unicoherent metric continuum. Then T is continuous if and only if X is locally connected. Proof. Since X is hereditarily unicoherent, X is T -additive, Theorem 2.2.12. Now the corollary follows form Corollary 3.2.10.   3.2.12 Theorem. Let X be a decomposable continuum with the property of Kelley. If TX is continuous on singletons and G = {TX ({x}) | x ∈ X}, then TX (Z) = q −1 TX/G q(Z) for each nonempty closed subset Z of X, where q: X → → X/G is the quotient map. Proof. Without loss of generality, we assume that X is not aposyndetic. Note that by Theorem 2.3.18 and Theorem 3.2.7, G is a decomposition of X. Let Z be a nonempty closed subset of X. We divide the proof in six steps. Step 1. q −1 q(Z) ⊂ TX (Z). Let x ∈ q −1 q(Z). Then q(x) ∈ q(Z). Thus, there exists z ∈ Z such that q(z) = q(x). This implies that TX ({z}) = TX ({x}). Hence, since TX is idempotent on closed sets (Theorem 2.3.18), TX ({x}) = TX ({z}) ⊂ TX (Z). Therefore, x ∈ TX (Z), and q −1 q(Z) ⊂ TX (Z). Step 2. TX (Z) = q −1 qTX (Z). Clearly, TX (Z) ⊂ q −1 qTX (Z). Let x ∈ q −1 qTX (Z). Then q(x) ∈ qTX (Z). Hence, there exists y ∈ TX (Z) such that q(y) = q(x). This implies that TX ({y}) = TX ({x}). Thus, since TX is idempotent on closed sets (Theorem 2.3.18), we have that TX ({y}) ⊂ TX (Z). Hence, x ∈ TX (Z), and q −1 qTX (Z) ⊂ TX (Z). Therefore, TX (Z) = q −1 qTX (Z). Step 3. If Z is connected and IntX (Z) = ∅, then Z = q −1 q(Z). Clearly, Z ⊂ q −1 q(Z). Let x ∈ q −1 q(Z). Then q(x) ∈ q(Z). Thus, there exists z ∈ Z such that q(z) = q(x). This implies that TX ({z}) = TX ({x}). Since TX ({x}) is a nowhere dense terminal subcontinuum of X

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(Lemma 1.4.51 and Theorems 3.1.10 and 2.1.27), TX ({x}) ∩ Z = ∅ and IntX (Z) = ∅, we have that TX ({x}) ⊂ Z. In particular, x ∈ Z. Therefore, Z = q −1 q(Z). Step 4. qTX (Z) ⊂ TX/G q(Z). Let χ ∈ X/G \ TX/G q(Z). Then there exists a subcontinuum W of X/G such that χ ∈ IntX/G (W) ⊂ W ⊂ X/G \ q(Z). From these inclusions, we obtain that q −1 (χ) ⊂ IntX (q −1 (W)) ⊂ q −1 (W) ⊂ X \q −1 q(Z) ⊂ X \Z. Hence, since q is monotone (Theorem 2.1.27), q −1 (χ) ∩ TX (Z) = ∅. Thus, qq −1 (χ) ∩ qTX (Z) = ∅. Therefore, χ ∈ X/G \ qTX (Z), and qTX (Z) ⊂ TX/G q(Z). Step 5. TX/G q(Z) ⊂ qTX (Z). Let χ ∈ X/G \ qTX (Z). Then {χ} ∩ qTX (Z) = ∅. This implies that q −1 (χ) ∩ q −1 qTX (Z) = ∅. Hence, by Step 2, q −1 (χ) ∩ TX (Z) = ∅. Since TX is idempotent on closed sets (Theorem 2.3.18), q −1 (χ) ∩ TX2 (Z) = ∅. Thus, there exists a subcontinuum W of X such that q −1 (χ) ⊂ IntX (W ) ⊂ W ⊂ X \ TX (Z) ⊂ X \ Z. From these inclusions, since q is an open map (G is a continuous decomposition), we obtain that {χ} = qq −1 (χ) ⊂ IntX/G (q(W )) ⊂ q(W ) ⊂ q(X \Z). To finish, we need to show that q(W ) ∩ q(Z) = ∅. Suppose there exists χ ∈ q(W ) ∩ q(Z). Then, by Steps 3 and 1, q −1 (χ ) ⊂ q −1 q(W )∩q −1 q(Z) = W ∩q −1 q(Z) ⊂ W ∩TX (Z), a contradiction to the election of W . Hence, q(W )∩q(Z) = ∅, and χ ∈ X/G \ TX/G q(Z). Therefore, TX/G q(Z) ⊂ qTX (Z). Step 6. qTX (Z) = TX/G q(Z). The equality follows from Steps 4 and 5. From Steps 2 and 6, we have that TX (Z) = q −1 qTX (Z) = q −1 TX/G q(Z). Therefore, TX (Z) = q −1 TX/G q(Z).

 

3.3 Jones’ Aposyndetic Decomposition We prove a weak version of F. Burton Jones’ Aposyndetic Decomposition Theorem only using the property of Kelley. Note that we obtain that the decomposition is only upper semicontinuous. Then we give a full version of the theorem using the uniform property of Effros. We begin showing that the set function T commutes with homeomorphisms. 3.3.1 Lemma. Let X and Y be continua, and let h : X → → Y be a homeomorphism. If A is a subset of X, then h(TX (A)) = TY (h(A)).

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Proof. Let A be a subset of X. Let y ∈ Y \ h(TX (A)). Then h−1 (y) ∈ X \ TX (A). Hence, there exists a subcontinuum W of X such that h−1 (y) ∈ Int(W ) ⊂ W ⊂ X \ A. This implies that y ∈ Int(h(W )) ⊂ h(W ) ⊂ Y \ h(A). Thus, y ∈ Y \ TY (h(A)). Now, let y ∈ X \ TY (h(A)). Then there exists a subcontinuum K of Y such that y ∈ Int(K) ⊂ K ⊂ Y \ h(A). This implies that h−1 (y) ∈ Int(h−1 (K)) ⊂ h−1 (K) ⊂ X \ A. Thus, h−1 (y) ∈ X \ TX (A). Hence, y ∈ Y \ h(TX (A)). Therefore, h(TX (A)) = TY (h(A)).   The next theorem is the weak version of F. Burton Jones’ Aposyndetic Decomposition Theorem. 3.3.2 Theorem. Let X be a homogeneous continuum with the property of Kelley. If G = {T ({x}) | x ∈ X}, then (1) G is an upper semicontinuous decomposition of X; (2) the elements of G are homogeneous, terminal subcontinua of X which are mutually homeomorphic; (3) the quotient space, X/G, is an aposyndetic homogeneous continuum, and it does not contain nondegenerate proper terminal subcontinua; (4) the quotient map q : X → → X/G is an atomic map. Proof. Let x be a point of X. We show that if z ∈ T ({x}), then T ({z}) = T ({x}). Since X has the property of Kelley, by Theorem 2.3.18, T is idempotent on closed sets. Hence, T ({z}) ⊂ T ({x}). Also, by Theorem 3.1.6, there exists x0 ∈ T ({x}) such that T ({x0 }) has property BL. Since X is homogeneous, there exists a homeomorphism h : X → → X such that h(x0 ) = x. Since z ∈ T ({x}), h−1 (z) ∈ T ({x0 }). Hence, by Lemma 3.1.3, T ({h−1 (z)}) = T ({x0 }). Thus, by Lemma 3.3.1, T ({x}) = h(T ({x0 })) = h(T ({h−1 (z)})) = T ({z}). Therefore, G is a decomposition of X. By Theorem 3.2.1, G is upper semicontinuous. The fact that each element of G is a terminal subcontinuum of X follows from Theorem 3.1.10. Also, by Corollary 3.2.2, X/G is an aposyndetic continuum. By Lemma 1.4.52, X/G does not contain nondegenerate proper terminal subcontinua. Also, by Lemma 3.3.1, we have that the elements of G are homogeneous and mutually homeomorphic. We show that X/G is a homogeneous continuum. Since G is an upper semicontinuous decomposition of X, by Theorem 1.1.22, X/G is a Hausdorff space. Therefore, X/G is a continuum. Let q : X → → X/G be the quotient map. Let χ1 and χ2 be two elements of X/G. We define ζ : X/G → → X/G as follows: let x1 and x2 be elements of X such that q(x1 ) = χ1 and q(x2 ) = χ2 . Since X is homogeneous, there exists a homeomorphism h : X → → X such that h(x1 ) = x2 . Note that, by Lemma 3.3.1, h(T ({x1 })) = T ({x2 }); that is, h(q −1 (χ1 )) = q −1 (χ2 ). Define

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ζ(χ) = q ◦ h(q −1 (χ)). Then ζ is well defined, and ζ(χ1 ) = χ2 . To see that ζ is continuous, let U be an open subset X/G. Then q −1 (U ) is a saturated open subset of X. Since h is a homeomorphism, h−1 (q −1 (U )) is a saturated open subset of X (Lemma 3.3.1). Hence, q(h−1 (q −1 (U ))) = ζ −1 (U ) is an open subset of X/G. Thus, ζ is continuous. Similarly, ζ −1 is defined and is continuous. Therefore, X/G is homogeneous. To see that the quotient map is atomic, let K be a subcontinuum of X such that q(K) is nondegenerate. Clearly, K ⊂ q −1 q(K). Let x ∈ q −1 q(K). Then there exists k ∈ K such that q(k) = q(x). Thus, q −1 q(x) ∩ K = ∅. Since q −1 q(x) is a terminal subcontinuum of X, Theorem 3.1.10, and q(K) is nondegenerate, we have that x ∈ q −1 q(x) ⊂ K. Hence, q −1 q(K) ⊂ K. Therefore, q is an atomic map.   3.3.3 Corollary. Let X be a decomposable homogeneous continuum with the property of Kelley. If x ∈ X, then T ({x}) is the maximal terminal proper subcontinuum of X containing x. Proof. If X is aposyndetic, then T ({x}) = {x} (Theorem 2.1.34). Since aposyndetic continua do not contain nondegenerate proper terminal subcontinua (Lemma 1.4.52), T ({x}) is the maximal terminal proper subcontinuum of X containing x. Suppose X is not aposyndetic, and let x ∈ X. Then, by Theorem 3.3.2, T ({x}) is a terminal subcontinuum of X. Suppose K is a terminal proper subcontinuum of X such that T ({x})  K. Let q : X → → X/G be the quotient map. By Lemma 1.4.53, q(K) is a terminal subcontinuum of X/G. Since X/G does not contain proper nondegenerate terminal subcontinua (Theorem 3.3.2), q(K) = {q(x)}. Hence, K = q −1 q(x) = T ({x}), a contradiction. Therefore, T ({x}) is the maximal terminal proper subcontinuum of X containing x.   As a consequence of Theorem 3.3.2, we have the following: 3.3.4 Corollary. If X is an arcwise connected homogeneous continuum with the property of Kelley, then X is aposyndetic. 3.3.5 Definition. Let G be a decomposition of a compactum X, and let H be a family of homeomorphisms of X. We say that H respects G provided that for each pair, G and G , of elements of G and every h ∈ H, either h(G) = G or h(G) ∩ G = ∅. The following result is useful to prove Theorem 3.3.8. 3.3.6 Theorem. Let X be a continuum with the uniform property of Effros, and let G be a decomposition of X such that the elements of G are nondegenerate closed sets of X. If the homeomorphism group, H(X), of X respects G, then the following hold: (1) G is a continuous decomposition of X. (2) The elements of G are homogeneous and mutually homeomorphic.

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(3) The quotient space X/G is homogeneous. (4) If q : X → → X/G is the quotient map, then q is uniformly completely regular. Proof. By Theorem 1.4.59, X is a homogeneous continuum. First we show that G is upper semicontinuous. Let G be an element of G, and let W be an open subset of X containing G. For each g ∈ G, there exists Ug ∈ UX such that exist g1 , . . . , gn in G such that B(g, 2U ng ) ⊂ W . Since G is compact,there n G ⊂ j=1 IntX (B(gj , Ugj )). Let U = U j=1 gj . Then U ∈ UX . Let V be an  Effros entourage for U . Let W  = {IntX (B(g, V )) | g ∈ G}, and let G ∈ G be such that G ∩ W  = ∅. Let x ∈ G ∩ W  . Then there exists g0 ∈ G such that x ∈ IntX (B(g0 , V )). This implies that ρX (x, g0 ) < V . Thus, there exists a U -homeomorphism h : X → → X such that h(g0 ) = x. Since H(X) respects G and h(G) ∩ G = ∅, we have that h(G) = G . Let g ∈ G. Then there exists j ∈ {1, . . . , n} such that g ∈ IntX (B(gj , Ugj )). Since U ⊂ Ugj , ρX (h(g), g) < Ugj . Also, ρX (g, gj ) < Ugj . Hence, ρX (h(g), gj ) < 2Ugj . This implies that h(g) ∈ W . Therefore, G ⊂ W , and G is an upper semicontinuous decomposition. Next, we see that G is lower semicontinuous. Let G ∈ G, let g1 and g2 be two elements of G and let W be an open subset of X containing g1 . Then there exists U ∈ UX such that B(g1 , U ) ⊂ W . Let V be an Effros entourage for U . Let G ∈ G be such that G ∩ IntX (B(g2 , V )) = ∅, and let g  ∈ G ∩ IntX (B(g2 , V )). Then ρX (g  , g2 ) < V . Thus, there exists a U -homeomorphism h : X → → X such that h(g2 ) = g  . Since H(X) respects G and h(G) ∩ G = ∅, we obtain that h(G) = G . Then h(g1 ) ∈ G and ρX (g1 , h(g1 )) < U . This implies that h(g1 ) ∈ G ∩ B(g1 , U ) ⊂ G ∩ W . Hence, G is lower semicontinuous. Therefore, G is a continuous decomposition of X. Observe that, since X is homogeneous and H(X) respects G, all the elements of G are homeomorphic and homogeneous. Now, we prove that X/G is a homogeneous continuum. Since G is an upper semicontinuous decomposition of X, by Theorem 1.1.22, X/G is a Hausdorff space. Therefore, X/G is a continuum. Let q : X → → X/G be the quotient map. Let χ1 and χ2 be two elements of X/G. Let x1 and x2 be points of X such that q(x1 ) = χ1 and q(x2 ) = χ2 . Since X is homogeneous, there exists a homeomorphism h : X → → X such that h(x1 ) = x2 . Since H(X) respects G, we have that h(q −1 (χ1 )) = q −1 (χ2 ). Define ζ : X/G → → X/G by ζ(χ) = q ◦ h(q −1 (χ)). Then ζ is well defined (H(X) respects G), and ζ(χ1 ) = χ2 . Since G is a continuous decomposition, q is an open map. Hence, since q is open and h is continuous, ζ is continuous. Note that ζ −1 : X/G → → X/G

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is given by ζ −1 (χ) = q ◦ h−1 (q −1 (χ)), and it is continuous. Thus, ζ is a homeomorphism and, therefore, X/G is a homogeneous continuum. Next, we demonstrate that the quotient map q : X → → X/G is uniformly completely regular. Let U ∈ UX , and let V ∈ UX be an Effros entourage for U . Then, by Theorem 1.3.13, Ω = (q × q)(V ) ∈ UX/G . Let χ and χ be two elements of X/G such that ρX/G (χ, χ ) < Ω. Then there exist x and x in X such that ρX (x, x ) < V , q(x) = χ and q(x ) = χ . Hence, there exists a U -homeomorphism h: X → → X such that h(x) = x . Since H(X) respects G and h(q −1 (χ)) ∩ −1  q (χ ) = ∅, we have that h(q −1 (χ)) = q −1 (χ ). Thus, h|q−1 (χ) : q −1 (χ) → → q −1 (χ ) is a homeomorphism such that ρX (z, h|q−1 (χ) (z)) < U for all z ∈ q −1 (χ). Therefore, q is a uniformly completely regular map.   ˇ 3.3.7 Definition. A continuum X is said to be acyclic provided that the first Cech 1 ˇ cohomology group with integer coefficients is trivial; that is, H (X, Z) = {0}. We are ready to state and prove F. Burton Jones’ Aposyndetic Decomposition Theorem. 3.3.8 Theorem. Let X be a decomposable continuum with the uniform property of Effros. If G = {TX ({x}) | x ∈ X}, then the following hold: (1) G is a continuous, monotone and terminal decomposition of X. (2) The elements of G are cell-like, acyclic, homogeneous and mutually homeomorphic continua. (3) The quotient map q : X → → X/G is uniformly completely regular and atomic. (4) The quotient space, X/G, is an aposyndetic homogeneous continuum, and it does not contain nondegenerate terminal subcontinua. (5) If X is metric, then the elements of G are indecomposable continua of the same dimension as X, and the quotient space is a one-dimensional metric continuum. Proof. By Theorem 1.4.59, X is a homogeneous continuum. By Theorem 2.1.27, the elements of G are continua. Since H(X) respects G (Lemma 3.3.1), by Theorem 3.3.6, G is a continuous decomposition of X, the elements of G are homogeneous and mutually homeomorphic and the quotient map is uniformly completely regular. By Theorem 3.3.2, the elements of G are terminal subcontinua, the quotient space is aposyndetic, it does not contain nondegenerate proper terminal subcontinua and the quotient map is atomic. By Theorem 1.5.8, q is a cell-like map. Hence, the elements of G are cell-like continua. Since the elements of G are celllike continua, we have, in particular, that every map from an element of G into a simple closed curve is homotopic to a constant map. Hence, by [35, Theorem 8.1], the elements of G are acyclic. Part (5) follows from [92, Theorem 5.1.18].   3.3.9 Remark. Since every Effros homogeneous continuum has the uniform property of Effros, Theorem 1.4.60, Theorem 3.3.8 is true for Effros homogeneous continua. Now we present some consequences of Theorem 3.3.8. We need the following:

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3.3.10 Definition. A continuum X is semi-indecomposable provided that for every pair of subcontinua W1 and W2 of X such that IntX (W1 ) = ∅ and IntX (W2 ) = ∅, we have that W1 ∩ W2 = ∅. The next result shows that the semi-indecomposability of a decomposable continuum with the uniform property of Effros is equivalent to the semi-indecomposability of its quotient space with respect to Jones’ decomposition. 3.3.11 Theorem. Let X be a decomposable continuum with the uniform property of Effros, and let G = {TX ({x}) | x ∈ X}. Then X is semi-indecomposable if and only if X/G is semi-indecomposable. Proof. By Theorem 1.4.59, X is a homogeneous continuum. By Theorem 3.3.8, G is a continuous monotone decomposition, and the elements of G are terminal subcontinua of X. Let q : X → → X/G be the quotient map. Then q is monotone and open. If X/G is not semi-indecomposable, then there exist two disjoint subcontinua W1 and W2 of X/G such that IntX/G (W1 ) = ∅ and IntX (W2 ) = ∅. Hence, q −1 (W1 ) and q −1 (W2 ) are two disjoint subcontinua of X (Lemma 1.4.46) such that IntX (q −1 (W1 )) = ∅ and IntX (q −1 (W1 )) = ∅. Therefore, X is not semiindecomposable. Suppose X is not semi-indecomposable. Then there exist two disjoint subcontinua W1 and W2 of X such that IntX (W1 ) = ∅ and IntX (W2 ) = ∅. Since the elements of G are terminal subcontinua of X, by Lemma 1.4.51, the elements of G are nowhere dense. Hence, since IntX (W1 ) = ∅, if w ∈ W1 , then TX ({w}) ⊂ W1 . Similarly, the same is true for the elements of W2 . Let j ∈ {1, 2}. Thus,  Wj = {TX ({w}) | w ∈ Wj }. This implies that q(W1 ) and q(W2 ) are disjoint subcontinua of X/G such that IntX (q(W1 )) = ∅ and IntX (q(W2 )) = ∅. Therefore, X/G is not semi-indecomposable.   3.3.12 Theorem. Let X be a continuum with the uniform property of Effros, and let G = {TX ({x}) | x ∈ X}. Then X/G is locally connected if and only if TX is continuous. Proof. By Theorem 1.4.59, X is homogeneous. Suppose X/G is locally connected. If X is an indecomposable continuum, then, by Theorem 2.1.44, TX is a constant map, and X/G is a point. We assume that X is decomposable. By Theorem 3.3.8, G is a terminal continuous decomposition of X, and the quotient map is monotone and open. Let B be a proper subcontinuum of X, and let x ∈ X \ B. Since TX ({x}) is a terminal subcontinuum of X, either TX ({x}) ∩ B = ∅ or B ⊂ TX ({x}). In any case, q(B) is a proper subcontinuum of X/G. Therefore, by Theorem 5.1.2, TX is continuous. The converse implication follows directly from Theorem 3.2.9.   3.3.13 Theorem. Let X be a continuum with the uniform property of Effros. Let G = {TX ({x}) | x ∈ X}, and let q : X → → X/G be the quotient map. If (q) : 2X/G → 2X is given by (q)(Γ) = q −1 (Γ), then TX (2X ) ⊂ (q)(2X/G ). Moreover, TX (2X ) = (q)(2X/G ) if and only if TX is continuous.

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Proof. By Theorem 1.4.59, X is a homogeneous continuum. By Theorem 3.3.8, G is a monotone continuous decomposition of X. Hence, q is a monotone and open map. Thus, (q) is continuous, Theorem 1.6.16. To show that TX (2X ) ⊂ (q)(2X/G ), let K ∈ TX (2X ). Then there exists Z ∈ 2X such that TX (Z) = K. By Theorems 1.6.22 and 3.2.12, TX (Z) = q −1 TX/G q(Z) = (q)TX/G q(Z). Since K = TX (Z), we have that K = (q)TX/G q(Z). Therefore, TX (2X ) ⊂ (q)(2X/G ). Now, suppose that TX (2X ) = (q)(2X/G ). Let Γ ∈ 2X/G . Then, by Theorem 2.1.51 (c), TX/G (Γ) = qTX q −1 (Γ) = 2q ◦ TX ◦ (q)(Γ). Since (q)(Γ) ∈ TX (2X ), there exists Z ∈ 2X such that TX (Z) = (q)(Γ). Hence, TX ◦ (q)(Γ) = TX TX (Z) = TX (Z) = (q)(Γ) (the second equality is true by Theorems 1.6.22 and 2.3.18). Thus, TX/G (Γ) = 2q ◦ (q)(Γ) = Γ. This implies that X/G is locally connected, Theorem 2.1.37. Hence, by Theorem 3.3.12, TX is continuous. If TX is continuous for X, by the proof of Theorem 3.2.9, TX (2X ) = (q)(2X/G ).  

3.4 Prajs’ Mutual Aposyndetic Decomposition We begin with the definition of the relation that provides the base for Prajs’ theorem. 3.4.1 Definition. Let X be a continuum, and let x1 and x2 be two points of X. We say that x1 and x2 are related, x1 ∼ x2 , provided that X is not mutually aposyndetic at them; that is, if Wx1 and Wx2 are two subcontinua of X such that x1 ∈ IntX (Wx1 ) and x2 ∈ IntX (Wx2 ), then we have that Wx1 ∩ Wx2 = ∅. 3.4.2 Theorem. Let X be a continuum with the property of Kelley. Then the relation “∼” is an equivalence relation. Proof. It is clear that the relation is reflexive and symmetric. We show that it is transitive. Suppose that there exist three points x1 , x2 and x3 in X such that x1 ∼ x2 , x2 ∼ x3 , but x1 ∼ x3 . Since x1 ∼ x3 , there exist two disjoint subcontinua W1 and W3 such that x1 ∈ IntX (W1 ) and x3 ∈ IntX (W3 ). By Theorem 2.1.67, T (W1 ) ∩ T (W3 ) = ∅. Since x1 ∼ x2 , for each subcontinuum W of X such that x2 ∈ IntX (W ), we obtain that W ∩ W1 = ∅. This implies that x2 ∈ T (W1 ). Similarly, x2 ∈ T (W3 ). Thus, T (W1 ) ∩ T (W3 ) = ∅, a contradiction. Therefore, x1 ∼ x3 , and “∼” is an equivalence relation.   3.4.3 Notation. We denote the equivalence class of x with respect to “∼” by Qx . 3.4.4 Lemma. Let X be a continuum with the property of Kelley. If x is a point of X, then Qx is a closed subset of X. Proof. Let x be a point of X, and let z ∈ X \ Qx . Then there exist two disjoint subcontinua Wx and Wz such that x ∈ Int(Wx ) and z ∈ Int(Wz ). Hence, Int(Wz ) ⊂ X \ Qx . Therefore, Qx is a closed subset of X.  

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3.4.5 Lemma. Let X be a continuum with the property of Kelley, and let x be a point of X. If h : X → → X is a homeomorphism, then h(Qx ) = Qh(x) . Proof. Let z ∈ X \h(Qx ). Since h is a homeomorphism, h−1 (z) ∈ X \Qx . Hence, there exist two disjoint subcontinua Wx and Wz of X such that h−1 (z) ∈ IntX (Wz ) and x ∈ IntX (Wx ). Thus, h(Wz ) and h(Wx ) are disjoint subcontinua of X such that z ∈ IntX (h(Wz )) and h(x) ∈ IntX (h(Wx )). This implies that z ∈ X \ Qh(x) . Now, let z ∈ X \ Qh(x) . Then there exist two disjoint subcontinua Wz and Wx such that z ∈ IntX (Wz ) and h(x) ∈ IntX (Wx ). Hence, h−1 (Wz ) and h−1 (Wx ) are two disjoint subcontinua of X such that h−1 (z) ∈ IntX (h−1 (Wz )) and x ∈ IntX (h−1 (Wx )). Thus, h−1 (z) ∈ X \ Qx . This implies that z ∈ X \ h(Qx ). Therefore, h(Qx ) = Qh(x) .   3.4.6 Corollary. If X is a continuum with the property of Kelley and Q = {Qx | x ∈ X}, then the group of homeomorphisms of X, H(X), respects Q. 3.4.7 Lemma. If X is a continuum with the property of Kelley and x is a point of X, then  Qx = {T (W ) | W is a subcontinuum of X and x ∈ Int(W )}. Proof. Let z ∈ X \ Qx . Then there exist two disjoint subcontinua Wx and Wz of X such that x ∈ Int(Wx ) and z ∈ Int(Wz ). Hence, z ∈ X \ T (Wx ). Thus, z∈X\



{T (W ) | W is a subcontinuum of X and x ∈ Int(W )}.

 Let z ∈ X \ {T (W ) | W is a subcontinuum of X and x ∈ Int(W )}. Then there exists a subcontinuum Wx of X such that x ∈ Int(Wx ) and z ∈ X \ T (Wx ). Thus, there exists a subcontinuum Wz of X such that z ∈ Int(Wz ) ⊂ Wz ⊂ X \ Wx . This implies that z ∈ X \ Qx .   3.4.8 Lemma. Let X be a continuum with the property of Kelley. If W is a subcontinuum of X with nonempty interior, then T (W ) = q −1 qT (W ), where Q = {Qx | x ∈ X}, and q : X → → X/Q is the quotient map. Proof. Let W be a subcontinuum of X with nonempty interior. Clearly, T (W ) ⊂ q −1 qT (W ). Let x ∈ X \ T (W ). Since T is idempotent on closed sets (Theorem 2.3.18), x ∈ X \ T 2 (W ). Thus, there exists a subcontinuum Wx of X such that x ∈ Int(Wx ) ⊂ Wx ⊂ X \T (W ). By Corollary 1.6.21, there exists a subcontinuum KW of X such that T (W ) ⊂ Int(K) and Wx ∩ KW = ∅. This implies that Qx ∩ T (W ) = ∅. Hence, q(x) ∩ qT (W ) = ∅. Thus, q −1 q(x) ∩ q −1 qT (W ) = ∅, and x ∈ X \ q −1 qT (W ). Therefore, T (W ) = q −1 qT (W ).   We are ready to state and prove Janusz R. Prajs’ Mutual Aposyndesis Decomposition Theorem. 3.4.9 Theorem. Let X be a decomposable continuum with the uniform property of Effros. If Q = {Qx | x ∈ X}, then the following hold:

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(1) Q is a continuous decomposition of X. (2) The elements of Q are homogeneous and mutually homeomorphic closed subsets of X. (3) The quotient map q : X → → X/Q is uniformly completely regular. (4) The quotient space, X/Q, is a mutually aposyndetic homogeneous continuum. Proof. By Theorem 1.4.59, X is a homogeneous continuum. By Theorem 1.6.22, X has the property of Kelley. By Theorem 3.4.2, Q is a decomposition of X. By Corollary 3.4.6, the group of homeomorphisms of X respects Q. Hence, by Theorem 3.3.6, Q is a continuous decomposition of X, the elements of Q are homogeneous and mutually homeomorphic closed subsets of X, the quotient map is uniformly completely regular and the quotient space is a homogeneous continuum. We need to prove that X/Q is mutually aposyndetic. To this end, let χ1 and χ2 be two distinct elements of X/Q. Let x1 and x2 be points of X such that q(x1 ) = χ1 and q(x2 ) = χ2 . Then, X is mutually aposyndetic at x1 and x2 . Hence, there exist two disjoint subcontinua Wx1 and Wx2 of X such that x1 ∈ IntX (Wx1 ) and x2 ∈ IntX (Wx2 ). By Theorem 2.1.67, we have that TX (Wx1 ) ∩ TX (Wx2 ) = ∅. Since Q is a continuous decomposition, q is open. Thus, by Lemma 3.4.8, qTX (Wx1 ) ∩ qTX (Wx2 ) = ∅, χ1 ∈ IntX/Q (qTX (Wx1 )) and χ2 ∈ IntX/Q (qTX (Wx2 )). Therefore, X/Q is mutually aposyndetic.   3.4.10 Remark. Since every Effros homogeneous continuum has the uniform property of Effros, Theorem 1.4.60, Theorem 3.4.9 is true for Effros homogeneous continua. The next theorem gives us a sufficient condition to have that Jones’ decomposition coincides with Prajs’ decomposition. 3.4.11 Theorem. Let X be a continuum with the uniform property of Effros. If G = {TX ({x}) | x ∈ X} is Jones’ decomposition, Q = {Qx | x ∈ X} is Prajs’ decomposition and X/G is mutually aposyndetic, then TX ({x}) = Qx for each x ∈ X. In particular, G = Q. Proof. By Theorem 1.4.59, X is a homogeneous continuum. By Theorem 3.3.8, G is a continuous decomposition of X. Let x1 be a point of X. Let x2 ∈ X \ TX ({x1 }). Then TX ({x1 }) ∩ TX ({x2 }) = ∅. Let q : X → → X/G be the quotient map. Note that q(x1 ) = q(x2 ). Since X/G is mutually aposyndetic, there exist two disjoint subcontinua W1 and W2 of X/G such that q(x1 ) ∈ IntX/G (W1 ) and q(x2 ) ∈ IntX/G (W2 ). Then, since q is monotone, q −1 (W1 ) and q −1 (W2 ) are two disjoint subcontinua of X (Lemma 1.4.46) such that x1 ∈ IntX (q −1 (W1 )) and x2 ∈ IntX (q −1 (W2 )). Hence, X is mutually aposyndetic at x1 and x2 . Thus, x1 ∈ X \ Qx2 and x2 ∈ X \ Qx1 . This implies that Qx1 ⊂ TX ({x1 }). Clearly, TX ({x1 }) ⊂ Qx1 . Therefore, TX ({x1 }) = Qx1 .   As a consequence of Theorems 5.1.29 and 3.4.11, we have the following.

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3.4.12 Corollary. Let X be a nonaposyndetic continuum with the uniform property of Effros, let G = {TX ({x}) | x ∈ X} be Jones’ decomposition and let Q = {Qx | x ∈ X} be Prajs’ decomposition. If TX is continuous for X, then TX ({x}) = Qx for each x ∈ X. In particular, G = Q. 3.4.13 Theorem. Let X be a decomposable continuum with the uniform property of Effros, and let Q = {Qx | x ∈ X} be Prajs’ decomposition. If the elements of Q are connected, then X is not semi-indecomposable. Proof. By Theorem 1.4.59, X is a homogeneous continuum. By Theorem 3.4.9, Q is a continuous decomposition of X. Suppose that the elements of Q are connected. Since X/Q is mutually aposyndetic, X/Q is not semi-indecomposable. Note that the quotient map q is monotone. Now the rest of the proof is equal to the first part of the proof of Theorem 3.3.11.   For the next results, we adopt the following notation. 3.4.14 Notation. Given a continuum X with the uniform property of Effros, let G = {TX ({x}) | x ∈ X} be Jones’ decomposition, let XJ = X/G and let qJ : X → → XJ be the quotient map. Let Q = {Qx | x ∈ X} be Prajs’ → XP be the quotient map. decomposition, let XP = X/Q and let qP : X → For XJ , let QJ = {Qζ | ζ ∈ XJ } be Prajs’ decomposition, let XJP = XJ /QJ and let qJP : XJ → → XJP be the quotient map. 3.4.15 Theorem. Let X be a continuum with the uniform property of Effros. With the notation in Notation 3.4.14, qJ (Qx ) = QqJ (x) for all x ∈ X. Proof. By Theorem 1.4.59, X is a homogeneous continuum. Let x be a point of X, and let ζ ∈ XJ \ QqJ (x) . Then there exist two disjoint subcontinua Wζ and WqJ (x) such that ζ ∈ IntXJ (Wζ ) and qJ (x) ∈ IntXJ (WqJ (x) ). Since qJ is monotone (Theorem 3.3.8), we have that qJ−1 (Wζ ) and qJ−1 (WqJ (x) ) are disjoint subcontinua of X (Lemma 1.4.46) such that qJ−1 (ζ) ⊂ IntX (qJ−1 (Wζ )) and x ∈ IntX (qJ−1 (WqJ (x) )). Thus, qJ−1 (ζ) ∩ Qx = ∅. This implies that ζ ∈ XJ \ qJ (Qx ). Now, let ζ ∈ XJ \ qJ (Qx ). Then qJ−1 (ζ) ∩ qJ−1 qJ (Qx ) = ∅. In particular, −1 qJ (ζ) ∩ Qx = ∅. Let z ∈ X be such that qJ (z) = ζ. Hence, there exist two disjoint subcontinua Wz and Wx of X such that z ∈ IntX (Wz ) and x ∈ IntX (Wx ). Since qJ is atomic and open (Theorem 3.3.8), qJ (Wz ) and qJ (Wx ) are two disjoint subcontinua of XJ such that ζ = qJ (z) ∈ IntXJ (qJ (Wz )) and qJ (x) ∈ IntXJ (qJ (Wx )). Thus, ζ ∈ XJ \ QqJ (x) . Therefore, qJ (Qx ) = QqJ (x) .   3.4.16 Corollary. Let X be a continuum with the uniform property of Effros. With the notation in Notation 3.4.14, XJP is homeomorphic to XP . Proof. By Theorem 1.4.59, X is a homogeneous continuum. Note that for each x ∈ X, qJ−1 qJ (x) = TX ({x}) ⊂ Qx = qP−1 qP (x). Hence, by the Transgression → XP such that g ◦ qJ = qP . Theorem [36, 3.2, p. 123], there exists a map g : XJ → Let χ ∈ XP , and let x ∈ X be such that qP (x) = χ. Then g −1 (χ) = qJ (Qx ) =

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QqJ (x) (the last equality is true by Theorem 3.4.15). Now, let α ∈ XJP , and let −1 (α) = Qζ . Since qJ is surjective, there ζ ∈ XJ be such that qJP (ζ) = α. Then qJP exists z ∈ X such that qJ (z) = ζ. Hence, by Theorem 3.4.15, Qζ = QqJ (z) . Thus, −1 (α), and vice versa. for each χ ∈ XP , there exists α ∈ XJP such that g −1 (χ) = qJP Hence, by the Transgression Theorem, there exist two maps h : XP → → XJP and k : XJP → → XP such that h ◦ g = qJP and k ◦ qJP = g. Then h and k are inverse of each other. Therefore, XJP is homeomorphic to XP .  

3.5 Rogers’ Terminal Decomposition We state most of the results needed to prove James T. Rogers’, Jr., Terminal Decomposition Theorem for metric homogeneous continua. The proofs of these results may be found in [92, Section 5.3]. We begin with a discussion of Poincaré model of the hyperbolic plane H ([2, 41] and [117]). Let H be the interior of the closed unit disk D in R2 , and let S 1 be its boundary. 3.5.1 Definition. A geodesic in H is the intersection of H and a circle C in R2 that intersects S 1 orthogonally (straight lines through the origin are considered circles centred at ∞). 3.5.2 Definition. A reflection in a geodesic C ∩ H is a Euclidean inversion in the circle. A hyperbolic isometry of H is a composition of reflections in geodesics. 3.5.3 Definition. The set H with the family of isometries is the Poincaré model of the hyperbolic plane. The boundary S 1 of H, which is not in H, is called the circle at ∞. Let F be a double torus. The universal covering space of F may be chosen to be H ([2] and [117]) and the group of covering homeomorphisms to be a subgroup of the orientation-preserving isometries of H. Each covering homeomorphism (except the identity map 1H ) is a hyperbolic isometry. The pertinent property for us is that a hyperbolic isometry, when extended to the circle at ∞, has exactly two fixed points in S 1 and none in H ([2, Theorem 9-3, p. 132] and [117, p. 410]). The universal covering map σ : H → → F may be chosen to be a local isometry. 3.5.4 Definition. A geodesic in F is the image under σ of a geodesic in H. A geodesic is simple if it has no transverse intersection. 3.5.5 Definition. A simple closed geodesic is a geodesic in F that is a simple closed curve. A simple closed curve in F is essential if it does not bound a disk. Assume that the universal covering space (H, σ) of F is constructed with the following properties. The geodesic H ∩ x-axis maps to a simple closed geodesic C1 under σ. The geodesic H ∩ y-axis maps to a simple closed geodesic C2 under σ. Furthermore, C1 ∪ C2 is a figure-eight W , and C1 ∩ C2 = {v}.

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3.5.6 Theorem. Let σ : H → → F be the universal covering map of F. Let M be +. If L is + = σ −1 (M ). Suppose L is a component of M a subspace of F, and let M compact, then σ|L is one-to-one. 3.5.7 Theorem. Let σ : H → → F be the universal covering map of F. Let M be + = σ −1 (M ). Suppose L is a component of M +. If M is a subspace of F, and let M arcwise connected, then σ(L) = M . 3.5.8 Definition. Let Q be the Hilbert cube, and let σ × 1Q : H × Q → → F×Q be the universal covering map of F × Q. Let X be a metric continuum essentially embedded in W × Q (this means that the embedding is not homotopic to a constant map; note that this eliminates some continua from consideration). Let f : X → → W −1 ˜ , , + be be the projection map. Let X = (σ × 1Q ) (X), and let f : X → → W , the projection map. If K is a component of X, then the set E(K) = {z ∈ S 1 | z is a (Euclidean) limit point of f˜(K)} is called the set of ends of K. + = σ −1 (W ). Then W + is the universal covering space of W , 3.5.9 Notation. Let W + ) ∩ S 1 is a Cantor set; call it Z the “infinite snowflake” in Figure 3.1. The set Cl(W [113, p. 341].

Fig. 3.1 Infinite snowflake

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3.5.10 Lemma. Let X be a metric continuum. If  : X → W is a map that is not homotopic to a constant map, then X can be essentially embedded in W × Q. 3.5.11 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a metric continuum essentially embedded in W × Q. If , = (σ × 1Q )−1 (X), then no component of X , is compact. X 3.5.12 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a homogeneous metric continuum essentially embedded in W × Q, , = (σ × 1Q )−1 (X). Then there exists δ > 0 such that if x and let X ˜ ∈ K, x ˜ ∈ K   ˜ , and d(˜ x, x ˜ ) < δ (where K and K are components of X), then E(K) = E(K ). Compactify H × Q with S 1 × Q. Shrink each set of the form {z} × Q, where z ∈ S , to a point to obtain another compactification of H × Q. This time the remainder is S 1 . Let π : (H × Q) ∪ S 1 → → H ∪ S 1 be the map of this latter compactification onto the disk D obtained by naturally extending the projection map. Note that the , is just f˜. (If x , then σ ◦ f˜(˜ restriction of π to X ˜ ∈ X, x) = f ◦ (σ × 1Q )(˜ x) ∈ W . ˜ + Hence, f (˜ x) ∈ W .) 1

3.5.13 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a homogeneous metric continuum essentially embedded in W × Q. , = (σ × 1Q )−1 (X) and K is a component of X, , then E(K) is either a twoIf X point set or a Cantor set. Furthermore, E(K) contains a dense subset, each point of which is a fixed point of a hyperbolic isometry ϕ such that ϕ × 1Q is a covering homeomorphism. 3.5.14 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a nonaposyndetic homogeneous metric continuum essentially , = (σ × 1Q )−1 (X), and let K be a component of X. , embedded in W × Q. Let X Let Y = Cl(H×Q)∪S 1 (K). Then Y \ K = E(K), and Y is connected im kleinen at each point of E(K). 3.5.15 Corollary. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a nonaposyndetic homogeneous metric continuum essentially , = (σ × 1Q )−1 (X), and let K be a component of X. , embedded in W × Q. Let X Let Y = Cl(H×Q)∪S 1 (K). Then Y is aposyndetic at each point of E(K). 3.5.16 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a nonaposyndetic homogeneous metric continuum essentially , = (σ × 1Q )−1 (X), and let K be a component of X. , Let embedded in W × Q. Let X Y = Cl(H×Q)∪S 1 (K). Then the collection G = {TY ({˜ x}) | x ˜ ∈ Y} is a continuous, terminal decomposition of Y with the following properties: (1) The quotient space Y/G is aposyndetic. (2) TY ({˜ x}) is degenerate if x ˜ ∈ Y \ K. (3) The nondegenerate elements of G are mutually homeomorphic, cell-like, indecomposable, homogeneous metric continua of the same dimension as Y. (4) K/G  is homogeneous, where G  = {TY ({˜ x}) | x ˜ ∈ K}.

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3.5.17 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a nonaposyndetic homogeneous metric continuum essentially , = (σ × 1Q )−1 (X), and let K be a component of X. , embedded in W × Q. Let X ˜ ∈ K, then (σ × 1Q )|TY ({˜x}) is one-to-one. Let Y = Cl(H×Q)∪S 1 (K). If x 3.5.18 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a nonaposyndetic homogeneous metric continuum essentially , = (σ × 1Q )−1 (X), and let K1 and K2 be components embedded in W × Q. Let X , Let Yj = Cl(H×Q)∪S 1 (Kj ), j ∈ {1, 2}. If x of X. ˜1 ∈ K1 and x ˜2 ∈ K2 are such that (σ ×1Q )(˜ x1 ) = (σ ×1Q )(˜ x2 ), then TY1 ({˜ x1 }) is homeomorphic to TY2 ({˜ x2 }), and (σ × 1Q )(TY1 ({˜ x1 })) = (σ × 1Q )(TY2 ({˜ x2 })). 3.5.19 Lemma. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a nonaposyndetic homogeneous metric continuum essentially , = (σ × 1Q )−1 (X). If Z, is a subcontinuum of X , such embedded in W × Q. Let X , of H × Q such that (σ × 1Q )|Z is one-to-one, then there exists an open subset U , , that Z ⊂ U and (σ × 1Q )|U is one-to-one. 3.5.20 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a nonaposyndetic homogeneous metric continuum essentially , = (σ × 1Q )−1 (X), and let K be a component of X. , embedded in W × Q. Let X Let Y = Cl(H×Q)∪S 1 (K). If x ˜ ∈ K, then (σ × 1Q )(TY ({˜ x})) is a maximal terminal proper, cell-like subcontinuum of X. We are ready to state and prove James T. Rogers’, Jr., Terminal Decomposition Theorem. 3.5.21 Theorem. Let σ × 1Q : H × Q → → F × Q be the universal covering map of F × Q. Let X be a homogeneous metric continuum that admits a map into the figure-eight W that is not homotopic to a constant map. Hence, we , = (σ × 1Q )−1 (X). If may consider X essentially embedded in F × Q. Let X , and Y = x})) | x ˜ ∈ K, where K is a component of X, G = {(σ × 1Q )(TY ({˜ Cl(H×Q)∪S 1 (K)}, then G is a continuous decomposition such that the following hold: (1) G is a monotone and terminal decomposition of X. (2) The elements of G are mutually homeomorphic, indecomposable, cell-like terminal, homogeneous metric continua. (3) The quotient space, X/G, is a homogeneous metric continuum. (4) X/G does not contain any proper, nondegenerate terminal subcontinuum. (5) If X is decomposable, then X/G is an aposyndetic metric continuum; in fact, the decomposition G is F. Burton Jones’ decomposition (Theorem 3.3.8). (6) If the elements of G are nondegenerate, then they have the same dimension as X and X/G is one-dimensional. Proof. Observe that, by [92, Theorem 4.2.31], X has the property of Effros. Also, note that, by Theorem 3.5.20, G is a collection of maximal proper terminal cell-like subcontinua of X.

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We show that G is a decomposition of X. Let x ˜1 and x ˜2 be two elements of , such that (σ × 1Q )(TY ({˜ X x1 })) ∩ (σ × 1Q )(TY2 ({˜ x2 })) = ∅ (here, Kj is the 1 , such that x component of X ˜j ∈ Kj and Yj = Cl(H×Q)∪S 1 (Kj ), j ∈ {1, 2}). Let z ∈ (σ×1Q )(TY1 ({˜ x1 }))∩(σ×1Q )(TY2 ({˜ x2 })). Then there exists z˜j ∈ TYj ({˜ xj }) such that (σ × 1Q )(˜ zj ) = z, j ∈ {1, 2}. Hence, by Theorem 3.5.18, TY1 ({˜ x1 }) is x2 }) and (σ × 1Q )(TY1 ({˜ x1 })) = (σ × 1Q )(TY2 ({˜ x2 })). homeomorphic to TY2 ({˜ Since σ × 1Q is a covering map, X=



x})) | x ˜ ∈ K, where K is a component of {(σ × 1Q )(TY ({˜ , and Y = Cl(H×Q)∪S 1 (K)}. X,

Therefore, G is a decomposition of X. Now, we prove that the homeomorphism group of X, H(X), respects G and then apply Theorem 3.3.6 to conclude that G is continuous. , such that Let h ∈ H(X), and let x ˜1 and x ˜2 be two points of X x1 }))) ∩ (σ × 1Q )(TY2 ({˜ x2 })) = ∅. h((σ × 1Q )(TY1 ({˜ x1 })) is terminal, h((σ × 1Q )(TY1 ({˜ x1 }))) is terminal Since (σ × 1Q )(TY1 ({˜ (Corollary 1.4.54). Thus, since (σ × 1Q )(TY2 ({˜ x2 })) is a maximal terminal subcontinuum, x1 }))) ⊂ (σ × 1Q )(TY2 ({˜ x2 })). h((σ × 1Q )(TY1 ({˜ x1 })) ⊂ h−1 ((σ × 1Q )(TY2 ({˜ x2 }))). Hence, This implies that (σ × 1Q )(TY1 ({˜ −1 (σ × 1Q )(TY1 ({˜ x1 })) = h ((σ × 1Q )(TY2 ({˜ x2 }))). Consequently, h((σ × 1Q )(TY1 ({˜ x1 }))) = (σ × 1Q )(TY2 ({˜ x2 })). Therefore, H(X) respects G. Since H(X) respects G, by Theorem 3.3.6, the elements of G are mutually homeomorphic homogeneous continua, and the quotient space X/G is homogeneous. The fact that the nondegenerate elements of G have the same dimension as X follows from [112, Theorem 8]. The proof of the indecomposability of the elements of G is similar to the one given in Theorem 3.2.7. The fact that X/G is one-dimensional may be found in [115, Theorem 6]. Next, we show that X/G does not contain nondegenerate proper terminal subcontinua. To this end, let Γ be a nondegenerate terminal proper subcontinuum of X/G. Let q : X → → X/G be the quotient map. Then q is a monotone map. Thus, q −1 (Γ) is a subcontinuum of X (Lemma 1.4.46). Note that q −1 (Γ) =



{(σ × 1Q )(TY ({˜ x})), | x ˜ ∈ K, where K is a component

, Y = Cl(H×Q)∪S 1 (K) and q((σ × 1Q )(TY ({˜ x}))) ∈ Γ}. of X,

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Since Γ is a nondegenerate proper subcontinuum of X/G, q −1 (Γ) is a proper subcontinuum of X, and if q((σ × 1Q )(TY ({˜ x}))) ∈ Γ, (σ × 1Q )(TY ({˜ x}))  q −1 (Γ). Hence, q −1 (Γ) is not a terminal subcontinuum of X (each element (σ × 1Q )(TY ({˜ x})) of G is a maximal terminal proper subcontinuum of X). Thus, there exists a subcontinuum Z of X such that Z ∩ q −1 (Γ) = ∅, Z \ q −1 (Γ) = ∅ and q −1 (Γ) \ Z = ∅. Since q(Z ∩ q −1 (Γ)) = q(Z) ∩ Γ, q(Z) is a subcontinuum of X/G such that q(Z) ∩ Γ = ∅. Since Γ is a terminal subcontinuum of X/G, either q(Z) ⊂ Γ or Γ ⊂ q(Z). If q(Z) ⊂ Γ, then q −1 (q(Z)) ⊂ q −1 (Γ). Hence, Z ⊂ q −1 (Γ), a contradiction. Suppose, then, that Γ ⊂ q(Z). This implies that q −1 (Γ) ⊂ q −1 q(Z). Thus, for each x ∈ q −1 (Γ), there exists z ∈ Z such that q(z) = q(x). Hence, q −1 q(x) ∩ Z = ∅. Since q −1 q(x) is a terminal subcontinuum of X, either Z ⊂ q −1 q(x) or q −1 q(x) ⊂ Z. If Z ⊂ q −1 q(x), then Z ⊂ q −1 (Γ), a contradiction. Thus, q −1 q(x) ⊂ Z. Since x is an arbitrary point of q −1 (Γ), q −1 (Γ) ⊂ Z, a contradiction. Therefore, X/G does not contain nondegenerate terminal proper subcontinua. Next, suppose X is decomposable. Let G  = {TX ({x}) | x ∈ X} be Jones’ decomposition (Theorem 3.3.8). We show that G = G  . Thus, by Theorem 3.3.8, X/G is aposyndetic. Let (σ × 1Q )(TY ({˜ x})) ∈ G. Let x ∈ (σ × 1Q )(TY ({˜ x})).Then (σ × 1Q )(TY ({˜ x})) ∩ TX ({x}) = ∅. Since both of these continua are maximal terminal subcontinua of X (Corollary 3.3.3), (σ × 1Q )(TY ({˜ x})) = TX ({x}). Thus, x})) ∈ G  . Hence, G ⊂ G  . (σ × 1Q )(TY ({˜ , such Let TX ({x}) ∈ G  . Since σ × 1Q is a covering map, there exists x ˜ ∈ X , x) = x. Let K be the component of X such that x ˜ ∈ K. Then that (σ × 1Q )(˜ (σ × 1Q )(TY ({˜ x})) ∩ TX ({x}) = ∅. Since both of these continua are maximal x})) = TX ({x}). Thus, TX ({x}) ∈ G. terminal subcontinua of X, (σ × 1Q )(TY ({˜ Hence, G  ⊂ G. Therefore, G = G  .  

3.6 Other Decomposition Theorems We start with the next technical lemma. 3.6.1 Lemma. Let X be a continuum of type λ. Let G be the finest monotone upper semicontinuous decomposition of X such that each element of G is nowhere dense and X/G is an arc, and let q : X → → [0, 1] be the quotient map. If x ∈ X and A ⊂ q −1 (q(x)), then T (A) ⊂ q −1 (q(x)). Proof. Let y ∈ X \ q −1 q(x). Then q(y) = q(x). Hence, there exists a closed subinterval [r, t] of [0, 1] such that q(y) ∈ Int[0,1] ([r, t]) and q(x) ∈ [0, 1] \ [r, t]. This implies that q −1 q(x) ∩ q −1 ([r, t]) = ∅. Since q is monotone, Lemma 1.4.46, q −1 ([r, t]) is a subcontinuum of X. By construction, y ∈ IntX (q −1 ([r, t]). Therefore, y ∈ X \ T (A), and T (A) ⊂ q −1 q(x).   Our first decomposition theorem is as follows.

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3.6.2 Theorem. If X is a continuum of type λ, then {T 2 ({x}) | x ∈ X} is the finest monotone upper semicontinuous decomposition G of X such that each element of G is nowhere dense and X/G is an arc. Proof. Let G be the finest monotone upper semicontinuous decomposition of X such that each element of G is nowhere dense and X/G is an arc, and let q : X → → [0, 1] be the quotient map. Observe that G = {q −1 q(x) | x ∈ X}. Let x ∈ X. Note that T ({x}) ⊂ q −1 q(x), Lemma 3.6.1. By [119, Theorem 18, p. 26], there exists z ∈ q −1 q(x) such that T ({z}) = q −1 q(x). Since X is T symmetric, Corollary 2.2.3, and x ∈ T ({z}), we have that z ∈ T ({x}). Hence, by Lemma 3.6.1, q −1 q(x) = T ({z}) ⊂ T 2 ({x}) ⊂ q −1 q(x). Thus, T 2 ({x}) = q −1 q(x). Since x is an arbitrary point of X, G = {T 2 ({x}) | x ∈ X}.   3.6.3 Notation. Let Z be a set, and let G be a decomposition. For each point z of Z, let G(z) be the unique element of G that contains z. If H is another decomposition of Z, we say that G refines H, written G < H, if and only if G(z) ⊂ H(z) for all z ∈ Z. If {Gγ }γ∈Γ is a collection of decompositions of Z, let the common intersection  element be G(z) = γ∈Γ {Gγ (z) | Gγ (z) ∈ Gγ }. Note that GΓ = {G(z) | z ∈ Z} is a decomposition of Z. Also, GΓ is precisely the greatest lower bound of {Gγ }γ∈Γ under the partial order < on the family of all decompositions of Z. 3.6.4 Definition. Let Z be a set, and let G = {Gγ }γ∈Γ be the family of all decompositions of Z that have a certain property of P . We say that G is the core decomposition of Z with respect to P if and only if G = GΓ and G ∈ G. Our second decomposition theorem is a special case of [46, Theorem 2.5]. 3.6.5 Theorem. If X is a continuum, then there exists a core decomposition G of X with respect to the property: “G is upper semicontinuous with T -closed elements” (Definition 4.1.1). 3.6.6 Theorem. If X is a continuum, then there exists a unique decomposition G of X such that (1) G is the core decomposition of X with respect to being upper semicontinuous with TX -closed elements (Definition 4.1.1); (2) G is monotone and X/G is aposyndetic; (3) G is the core decomposition of X with respect to being monotone with X/G aposyndetic. Proof. Let G be the core decomposition of X with respect to being upper semicontinuous with TX -closed elements, given by Theorem 3.6.5. Hence, G satisfies (1). Let q : X → → X/G be the quotient map, and let χ ∈ X/G. Since q is monotone, by Theorem 2.1.51 (c), TX/G ({χ}) = qTX q −1 ({χ}) = qq −1 ({χ}) = {χ} (the elements of G are TX -closed sets). Thus, by Theorem 2.1.34, X/G is aposyndetic. Let D = {D | D is a component of an element of G}. Then, by Theorem 1.1.35, D is an upper semicontinuous decomposition of X, and, by Theorem 4.1.15, the elements of D are TX -closed sets. By the definitions of D and G, both inequalities

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D < G and G < D hold. Hence, G = D, and G is monotone. Thus, G satisfies (2). If H is any monotone decomposition of X such that X/H is aposyndetic, then, by Theorem 3.6.5, the elements of H are TX -closed. Hence, G < H. Thus, G satisfies (3).   A special case for metric θ-continua of Theorem 3.6.6 is the following [45, Theorem 6.1]: 3.6.7 Theorem. If X is a metric θ-continuum, then there exists a unique decomposition G such that G has the following three characterizations: (1) G is the core decomposition of X with respect to being upper semicontinuous with T -closed elements (Definition 4.1.1); (2) G is the core decomposition of X with respect to being monotone, and X/G is aposyndetic; (3) G is the core decomposition of X with respect to being monotone, and X/G is a graph. Proof. By Theorem 3.6.6, there exists a unique core decomposition G of X such that G satisfies (1) and (2). We show (3). By [92, Theorem 1.7.3] and [45, Theorem 3.1], X/G is a metric θ-continuum, and, by [45, Theorem 3.11], X/G is locally connected. In fact, by [45, Theorem 4.7], X/G is a graph. Thus, G is monotone, and X/G is a graph. Let H be any monotone decomposition of X such that X/H is a graph. Then, in particular, X/H is aposyndetic. Hence, by (2), G < H. Therefore, G satisfies (3).   The next result is [46, Lemma 3.6]. 3.6.8 Theorem. Let X be a continuum, and let n ∈ N. If K = {T n ({x}) | x ∈ X} is a decomposition of X and T ∞ ({x}) = T n ({x}) (Definition 4.4.1) for all x ∈ X, then K is the core decomposition G of Theorem 3.6.6. 3.6.9 Theorem. Let X be a metric θ-continuum. If n ∈ N is such that T n+1 ({x}) = T n ({x}) for every point x in X, then G = {T n ({x}) | x ∈ X} is the core decomposition of Theorem 3.6.6. Proof. By Theorem 2.4.8, G = {T n ({x}) | x ∈ X} is a decomposition of X. Hence, by Theorem 3.6.8, G is the core decomposition of Theorem 3.6.6.   A proof of the following theorem may be found in [54, Theorem 2]: 3.6.10 Theorem. Let X be a metric θn -continuum. Then X admits a monotone, upper semicontinuous decomposition G such that the elements of G have void interior and the quotient space X/G is a graph if and only if Int(T (H)) = ∅ for every subcontinuum H of X with void interior. Furthermore, G = {T 2n ({x}) | x ∈ X}.

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References for Chapter 3 Section 3.1: [11, 84, 86, 90, 98]. Section 3.2: [43, 53, 73, 83, 84, 86, 90, 91, 97]. Section 3.3: [35, 90–92, 110]. Section 3.4: [90, 91, 110]. Section 3.5: [2, 41, 92, 112–115, 117]. Section 3.6: [43, 45, 46, 54].

3 Decomposition Theorems

Chapter 4

T -Closed Sets

The family of T -closed sets have been considered by several authors, for example [46] and [122]. We introduce the family of T -closed sets of a continuum X and present its main properties. We also give a characterization of T -closed sets. We consider minimal T -closed sets and the set function T ∞ . We introduce the T growth bound of a continuum.

4.1 Main Properties of T -Closed Sets We start with a couple of definitions and give some results that provide examples of families of T -closed sets. 4.1.1 Definition. Given a continuum X, a subset A of X is a T -closed set provided that T (A) = A. We denote the family of T -closed sets of a continuum X by T(X). Note that X ∈ T(X). 4.1.2 Definition. A continuum X is m-aposyndetic provided that for each subset K of X with m points, K ∈ T(X). We say X is countable closed aposyndetic if for each countable closed subset K of X, K ∈ T(X). 4.1.3 Theorem. If X is a continuum and A is a closed subset of X such that T (A) is totally disconnected, then A ∈ T(X). Proof. Let A be a closed subset of X such that T (A) is a totally disconnected subset of X. By Remark 2.1.5, A ⊂ T (A). By Corollary 2.1.20, each component of T (A) intersects Cl(A) = A. Since all the components of T (A) are singletons, T (A) ⊂ A.   4.1.4 Corollary. Let X be a continuum such that T (2X ) = 2X . If K is a nonempty closed and totally disconnected subset of X, then K ∈ T(X).

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 S. Macías, Set Function T , Developments in Mathematics 67, https://doi.org/10.1007/978-3-030-65081-0_4

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Proof. Let K be a nonempty closed and totally disconnected subset of X. Since T (2X ) = 2X , there exists L ∈ 2X such that T (L) = K. Since K is totally disconnected, by Theorem 4.1.3, T (L) = L. Thus, K = T (L) = L, and K ∈ T(X).   The next theorem tells us that T of a product of continua behaves like the identity map at the product of two proper closed subsets. This result was already proved in Corollary 2.3.26, but we include another proof here. 4.1.5 Theorem. Let X and Y be continua. If A and B are proper closed subsets of X and Y , respectively, then A × B ∈ T(X × Y ). Proof. Let A and B be two proper closed subsets of X and Y , respectively. By Remark 2.1.5, A × B ⊂ T (A × B). Let (x, y) ∈ (X × Y ) \ (A × B); without loss of generality, we assume that x ∈ X \ A. Then there exists an open subset U of X such that x ∈ U ⊂ Cl(U ) ⊂ X \ A. Hence, (Cl(U ) × Y ) ∩ (A × B) = ∅. Let z ∈ Y \ B. Then (X × {z}) ∩ (A × B) = ∅. Thus, (x, y) ∈ (Cl(U ) × Y ) ∪ (X × {z}) ⊂ (X × Y ) \ (A × B). Since (Cl(U ) × Y ) ∪ (X × {z}) is a continuum containing (x, y) in its interior, (x, y) ∈ (X ×Y )\T (A×B). Therefore, A × B ∈ T(X × Y ).   As a consequence of Theorem 2.1.34 and Theorem 4.1.5, we have the following corollary: 4.1.6 Corollary. If X and Y are continua, then X×Y is an aposyndetic continuum. 4.1.7 Theorem. Let X and Y be continua. If A and B are two closed totally disconnected subsets of X and Y , respectively, then for each closed subset K of A × B, K ∈ T(X). Proof. Let K be a closed subset of A × B. By Theorem 4.1.5, T (A × B) = A × B. Since K is a closed subset of A × B, by Proposition 2.1.7, T (K) ⊂ T (A × B) = A × B. Hence, T (K) is totally disconnected. By Theorem 4.1.3, K ∈ T(X).   4.1.8 Theorem. Let X and Y be continua. If K is a countable closed subset of X × Y , then K ∈ T(X × Y ). In particular, X × Y is m-aposyndetic for each m ∈ N. Proof. Let K be a countable closed subset of X × Y . Let πX : X × Y → → X and πY : X × Y → → Y be the projection maps. Since K is countable and closed, πX (K) and πY (K) are countable closed subsets of X and Y , respectively. Then   K ⊂ πX (K) × πY (K). Thus, by Theorem 4.1.7, K ∈ T(X × Y ). 4.1.9 Theorem. If X is a metric continuum and K is a countable closed subset of K(X), then K ∈ T(K(X)). Proof. Let K be a countable closed subset of K(X). We assume first that νX ∈ K. Then q −1 (K) is a countable closed subset of X × [0, 1]. Hence, by Theorem 4.1.8, TX×[0,1] q −1 (K) = q −1 (K). Since q is monotone, by Theorem 2.1.51 (c), we have that TK(X) (K) = qTX×[0,1] q −1 (K). Therefore, TK(X) (K) = qq −1 (K) = K.

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If νX ∈ K and νX is an isolated point of K, then K \ {νX } is a closed subset of K(X). Then, by the previous paragraph, TK(X) (K \ {νX }) = K \ {νX }. Hence, we have that K ∈ T(K(X)). Next, we assume that νX ∈ K. Then note that q −1 (K) = (X × {1}) ∪ ∞ ∞ {(xn , tn )}∞ n=1 , also suppose that ClX×[0,1] ({(xn , tn )}n=1 )\ {(xn , tn )}n=1 ⊂ X × −1 {1}. Let (x, t) ∈ (X × [0, 1]) \ q (K). We assume that t ∈ (0, 1), the case t = 0 is similar. Since q −1 (K) is closed and X × [0, 1] is metric, there exist an open set U of X such that x ∈ U and an ε > 0 such that 0 < t − ε < t + ε < 1 and ClX (U ) × [t − ε, t + ε] ⊂ (X × [0, 1]) \ q −1 (K). Let s ∈ [t − ε, t + ε] \ {tn }∞ n=1 , and let W = (X × {s}) ∪ (ClX (U ) × [t − ε, t + ε]). Then W is a subcontinuum of X × [0, 1] such that (x, t) ∈ IntX×[0,1] (W ) ⊂ W ⊂ (X × [0, 1]) \ q −1 (K). Hence, (x, t) ∈ (X × [0, 1]) \ TX×[0,1] q −1 (K). Thus, TX×[0,1] q −1 (K) = q −1 (K). Since q is monotone, by Theorem 2.1.51 part (c), we have that TK(X) (K) =   qTX×[0,1] q −1 (K). Therefore, TK(X) (K) = qq −1 (K) = K. 4.1.10 Theorem. If X is a metric continuum and K is a countable closed subset of Σ(X), then K ∈ T(Σ(X)). Proof. Let K be a countable closed subset of Σ(X). Let ν + and ν − be the vertexes of Σ(X). We assume first that {ν + , ν − } ∩ K = ∅. Then q −1 (K) is a countable closed subset of X × [0, 1]. Hence, by Theorem 4.1.8, TX×[0,1] q −1 (K) = q −1 (K). Since q is monotone, by Theorem 2.1.51 (c), TΣ(X) (K) = qTX×[0,1] q −1 (K). Therefore, TΣ(X) (K) = qq −1 (K) = K. Next, we assume that ν + ∈ K and ν − ∈ Σ(X) \ K. If ν + is an isolated point of K, a similar argument to the one given in Theorem 4.1.9 shows that TΣ(X) (K) = K. Suppose that q −1 (K) = (X × {1}) ∪ {(xn , tn )}∞ n=1 , also ∞ assume that ClX×[0,1] ({(xn , tn )}∞ ) \ {(x , t )} ⊂ X × {1}. Let (x, t) ∈ n n n=1 n=1 (X × [0, 1]) \ q −1 (K). Without loss of generality, we assume that t = 0. Since q −1 (K) is closed and X × [0, 1] is metric, there exist an open set U of X such that x ∈ U and an ε > 0 such that 0 < t − ε < t + ε < 1 and ClX (U ) × [t − ε, t + ε] ⊂ (X × [0, 1]) \ q −1 (K). Let s ∈ [t − ε, t + ε] \ {tn }∞ n=1 , and let W = (X × {s}) ∪ (ClX (U ) × [t − ε, t + ε]). Then W is a subcontinuum of X × [0, 1] such that (x, t) ∈ IntX×[0,1] (W ) ⊂ W ⊂ (X × [0, 1]) \ q −1 (K). Hence, (x, t) ∈ (X × [0, 1]) \ TX×[0,1] q −1 (K). Thus, TX×[0,1] q −1 (K) = q −1 (K). Since q is monotone, by Theorem 2.1.51 (c), we have that TΣ(X) (K) = qTX×[0,1] q −1 (K). Therefore, TΣ(X) (K) = qq −1 (K) = K. Now, suppose that {ν + , ν − } ⊂ K. If either ν + or ν − or both are isolated points of K, a similar argument to the one given in Theorem 4.1.9 shows that TΣ(X) (K) =   ∞ K. Assume that q −1 (K) = (X ×{1})∪(X ×{0})∪{(xn , tn )}∞ n=1 ∪{(xn , tn )}n=1 . ∞ ∞ Also suppose that ClX×[0,1] ({(xn , tn )}n=1 ) \ {(xn , tn )}n=1 ⊂ X × {1} and   ∞ ClX×[0,1] ({(xn , tn )}∞ n=1 ) \ {(xn , tn )}n=1 ⊂ X × {0}. Let (x, t) ∈ (X × (0, 1)) \ q −1 (K). Since X × [0, 1] is metric, there exist an open set U of X and an ε > 0 such that 0 < t − ε < t + ε < 1 and ClX (U ) × [t − ε, t + ε] ⊂  ∞ (X × [0, 1]) \ q −1 (K). Let s ∈ [t − ε, t + ε] \ ({tn }∞ n=1 ∪ {tn }n=1 ) and let

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W = (X × {s}) ∪ (ClX (U ) × [t − ε, t + ε]). Then W is a subcontinuum of X × [0, 1] such that (x, t) ∈ IntX×[0,1] (W ) ⊂ W ⊂ (X × [0, 1]) \ q −1 (K). Hence, (x, t) ∈ (X × [0, 1]) \ TX×[0,1] q −1 (K). Thus, TX×[0,1] q −1 (K) = q −1 (K). Since q is monotone, by Theorem 2.1.51 (c), we have that TΣ(X) (K) = qTX×[0,1] q −1 (K).   Therefore, TΣ(X) (K) = qq −1 (K) = K. 4.1.11 Theorem. If X is a metric continuum, then T(X) is a Gδ subset of 2X . Proof. Given ε > 0, define Θε = {A ∈ 2X | T (A) ⊂ Vε (A)}. Note that if A ∈ Θε , then H(A, T (A)) < ε. If A ∈ Θε , then define Mε (A) =



 H Vsε (A) ∩ {B ∈ 2X | T (B) ⊂ V(1−s)ε (A)} .

0