The Infinite-Dimensional Topology of Function Spaces

THE INFINITE-DIMENSIONAL TOPOLOGY OF FUNCTION SPACES North-Holland Mathematical Library Board of Honorary Editors: M...

Author: J. van Mill

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THE INFINITE-DIMENSIONAL TOPOLOGY OF FUNCTION SPACES

North-Holland Mathematical Library Board of Honorary Editors: M. Artin, H. Bass, J. Eells, W. Feit, P.J. Freyd, F.W. Gehring, H. Halberstam, L.V. Hormander, J.H.B. Kemperman, W.A.J. Luxemburg, P.P. Peterson, I.M. Singer and A.C. Zaanen

Board of Advisory Editors: A. Bjorner, R.H. Dijkgraaf, A. Dimca, A.S. Dow, JJ. Duistermaat, E. Looijenga, J.P. May, I. Moerdijk, S.M. Mori, J.P. Palis, A. Schrijver, J. Sjostrand, J.H.M. Steenbrink, F. Takens and J. van Mill

VOLUME 64

ELSEVIER Amsterdam - London - New York - Oxford - Paris - Shannon - Tokyo

The Infinite-Dimensional Topology of Function Spaces

Jan van Mill Faculteit der Exacte Wetenschappen, Amsterdam, The Netherlands

2001 ELSEVIER Amsterdam - London - New York - Oxford - Paris - Shannon - Tokyo

ELSEVIER SCIENCE B.V. Sara Burgerhartstraat 25 P.O. Box 211, 1000 AE Amsterdam, The Netherlands © 2001 Elsevier Science B.V. All rights reserved. This work is protected under copyright by Elsevier Science, and the following terms and conditions apply to its use: Photocopying Single photocopies of single chapters may be made for personal use as allowed by national copyright laws. Permission of the Publisher and payment of a fee is required for all other photocopying, including multiple or systematic copying, copying for advertising or promotional purposes, resale, and all forms of document delivery. Special rates are available for educational institutions that wish to make photocopies for non-profit educational classroom use. Permissions may be sought directly from Elsevier Science Global Rights Department, PO Box 800, Oxford OX5 1DX, UK; phone: (+44) 1865 843830, fax: (+44) 1865 853333, e-mail: [email protected]. You may also contact Global Rights directly through Elsevier's home page (http://www.elsevier.nl), by selecting 'Obtaining Permissions'. In the USA, users may clear permissions and make payments through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA; phone: (+1) (978) 7508400, fax: (+1) (978) 7504744, and in the UK through the Copyright Licensing Agency Rapid Clearance Service (CLARCS), 90 Tottenham Court Road, London W1P OLP, UK; phone: (+44) 207 631 5555; fax: (+44) 207 631 5500. Other countries may have a local reprographic rights agency for payments. Derivative Works Tables of contents may be reproduced for internal circulation, but permission of Elsevier Science is required for external resale or distribution of such material. Permission of the Publisher is required for all other derivative works, including compilations and translations. Electronic Storage or Usage Permission of the Publisher is required to store or use electronically any material contained in this work, including any chapter or part of a chapter. Except as outlined above, no part of this work may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the Publisher. Address permissions requests to: Elsevier Science Global Rights Department, at the mail, fax and e-mail addresses noted above. Notice No responsibility is assumed by the Publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Because of rapid advances in the medical sciences, in particular, independent verification of diagnoses and drug dosages should be made. First edition 2001 Library of Congress Cataloging in Publication Data A catalog record from the Library of Congress has been applied for. ISBN: ISSN:

0 444 50557 1 0924-6509

^ The paper used in this publication meets the requirements of ANSI/NISO Z39.48-1992 (Permanence of Paper). Printed in The Netherlands.

Aan de nagedachtenis van Maarten Maurice.

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Contents Introduction

xi

Chapter 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. 1.8. 1.9. 1.10. 1.11.

1. Basic topology Linear spaces Extending continuous functions Function spaces The Borsuk homotopy extension theorem Topological characterization of some familiar spaces The inductive convergence criterion and applications Bing's shrinking criterion Isotopies Homogeneous zero-dimensional spaces Inverse limits Hyperspaces

1 1 21 29 37 41 58 66 70 73 80 95

Chapter 2.1. 2.2. 2.3. 2.4. 2.5.

2. Basic combinatorial topology Affine notions Barycenters and subdivisions The nerve of an open covering Simplices in Rn The Lusternik-Schnirelman-Borsuk theorem

Ill Ill 125 132 138 148

Chapter 3.1. 3.2. 3.3. 3.4. 3.5. 3.6. 3.7. 3.8. 3.9. 3.10. 3.11. 3.12.

3. Basic dimension theory The covering dimension Translation into open covers The imbedding theorem The inductive dimension functions ind and Ind Dimensional properties of compactifications Mappings into spheres Dimension of subsets of Rn and certain generalizations Higher-dimensional hereditarily indecomposable continua Totally disconnected spaces The origins of dimension theory The dimensional kernel of a space Colorings of maps

151 151 157 168 176 183 193 204 210 216 221 227 237

3.13. 3.14.

Various kinds of infinite-dimensionality The Brouwer fixed-point theorem revisited

251 257

Chapter 4.1. 4.2. 4.3.

4. Basic AMR theory Some properties of ANR's A characterization of ANR's and AR's Open subspaces of ANR's

263 263 277 301

Chapter 5.1. 5.2. 5.3. 5.4. 5.5.

5. Basic infinite-dimensional topology Z-sets Extending homeomorphisms in s The estimated homeomorphism extension theorem The compact absorption property Absorbing systems

307 307 311 320 329 343

Chapter 6.1. 6.2. 6.3. 6.4. 6.5. 6.6. 6.7. 6.8. 6.9. 6.10. 6.11. 6.12. 6.13.

6. Function spaces Notation The spaces CP(X): Introductory remarks The Borel complexity of function spaces The Baire property in function spaces Filters and the Baire property in CP(N?) Extenders The topological dual of CP(X) The support function Nonexistence of linear homeomorphisms Bounded functions Nonexistence of homeomorphisms Topological equivalence of certain function spaces Examples

367 368 369 372 377 387 393 399 404 411 416 426 434 445

Appendix A. Preliminaries A.I. Prerequisites and notation A.2. Separable metrizable topological spaces A.3. Limits of continuous functions A.4. Normality type properties A.5. Compactness type properties A.6. Completeness type properties A.7. A covering type property A.8. Extension type properties A.9. Wallman compactifications A.10. Connectivity A.11. The quotient topology A. 12. Homotopies A. 13. Borel and similar sets

457 457 465 468 469 473 479 485 490 494 500 505 510 517

Appendix B.

Answers to selected exercises

527

Appendix C.

Notes and comments

579

Bibliography

597

Special Symbols

613

Author Index

615

Subject Index

619

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Introduction In this book we study function spaces of low Borel complexity. This is a particularly interesting class of spaces; to investigate it one needs a mix of methods and techniques from areas as diverse as general topology, infinite-dimensional topology, functional analysis and descriptive set theory. A striking result is the theorem of Dobrowolski, Marciszewski and Mogilski, which states that all function spaces of low Borel complexity are topologically homeomorphic. A major feature of this book is a complete and self-contained proof of this remarkable fact. In order to understand these details, a solid background in infinitedimensional topology is needed. And for that one needs to know a fair amount of dimension theory as well as ANR theory. The necessary material was partially covered in my previous book 'Infinite-dimensional topology, prerequisites and introduction'. A selection of what was done in that volume can also be found here, but completely revised and in many places expanded with recent results. I chose to take a 'scenic' route towards the Dobrowolski-Marciszewski-Mogilski Theorem, that is linking the results needed for the theorem's proof to interesting recent research developments in dimension theory and infinite-dimensional topology. The first five chapters of this book are intended as a text for graduate courses in topology. For a course in dimension theory, one should cover Chapters 2 and 3 and part of Chapter I . For a course in infinite-dimensional topology, one should cover Chapters 1, 4 and 5. In Chapter 6, which deals with function spaces, I discuss recent research results. It can also be used for a graduate course in topology but its focus is more suited to that of a research monograph than of a textbook; it would, therefore, be more appropriate to use it as a text for a research seminar. This book, consequently, has the character of a textbook as well as a research monograph. In Chapters 1 through 5, unless stated otherwise, all spaces under discussion are separable and metrizable. In Chapter 6 results for more general classes of spaces are presented. In Appendix A we collected for easy reference and for sake of completeness some basic facts that are important in the book. The reader will see that it is not intended as the basis for a course in topology; its purpose is to collect what one should know about general topology, nothing more nothing less. The exercises in the book serve three purposes: to test the reader's understanding of the material, to supply proofs of statements that are used in the text and are not proven therein, and to provide additional information not covered by the text. We included the solutions to selected exercises in

Appendix B. These exercises are important or difficult; they are marked in the text by the symbol >•. If the reader wants to find the meaning of some unfamiliar term in this book, it is best to check Appendix A first, since many basic concepts are defined there. To simplify the search process, in the index all page numbers for terms from Appendix A are italicized. For example, if the reader would like to know the definition of the term 'topologically complete', she or he should look at Page 480. For in the index, the first italicized page number under 'topologically complete' is 480. Finally, I express my indebtedness to Jan Baars, Stoyu Barov, Jan Dijkstra, Tadek Dobrowolski, Klaas Pieter Hart, Michael van Hartskamp, Henryk Michalewski, Witek Marciszewski, Roman Pol, Ruud Salomon and Jan de Vries for their critical reading of parts of the manuscript and their many valuable suggestions for improvements. None of these distinguished colleagues is responsible for the remaining errors, which are mine.

Jan van Mill Bussum, March 29, 2001

CHAPTER 1

Basic topology In this chapter we present some basic facts on the topology of separable metrizable spaces. We discuss linear spaces, inverse limits, hyperspaces, Bing's Shrinking Criterion, etc. Questions about the possibility of extending continuous functions, or creating new continuous functions from old ones, are central in this chapter. Applications include a proof of the topological homogeneity of the Hilbert cube and proofs of topological characterizations of various interesting spaces such as the Cantor set, the unit interval and the spaces of rational and irrational numbers, respectively. Many of the results presented in this chapter are geometrically motivated, although this is not always clear at first sight. For background information, see Appendix A. Our conventions with respect to notation can be found in §A.l. All topological spaces under discussion are separable and metrizable. 1.1. Linear spaces

A linear space is a real vector space L carrying a (separable metrizable) topology with the property that the algebraic operations of addition and scalar multiplication are continuous (warning: a vector space is an algebraic structure which may or may not carry a topology while a linear space is automatically a topological space). Observe that the continuity of the algebraic operations on a linear space show that it is a topological group. A subset A of a linear space L is called convex if for all x, y £ A and a G I we have ax + (1 — a)y G A. Let L be a linear space. If A C L then conv(A) denotes the smallest convex subset of L containing A] this set is called the convex hull of A. A convex combination of elements of A is a vector of the form ^T=i ^ai with a i , . . . , on € A, AI , . . . , An € I and Y%=i ^i = 1 • For each n G N, let convn(A) denote the set of vectors x G L that can be written as a convex combination of at most n vectors from A. Also, put = [J conv n (A)

2

1. BASIC TOPOLOGY

and observe that convoo (A) is the set of all convex combinations of elements of A It is left as an exercise to the reader to present a proof of the following basic lemma (see Exercise 1.1.5). Lemma 1.1.1. Let L be a linear space with subset A. Then (1) conv(A) = (2) if A is finite then conv(A) is compact. A linear space L is called locally convex if the origin of L (which we shall always denote by 0) has arbitrarily small convex neighborhoods. Obviously, the spaces Rn for n G N U {00} with their usual product topologies are locally convex linear spaces under coordinatewise denned addition and scalar multiplication. Let L be a vector space. A norm on L is a function ||-||: L —> [0, oo) having the following properties: (1) (2) (3)

\\x + y\\ < \\x\\ + \\y\\ \\tx\\ = \t\ • \\x\\ \\x\\ = 0

f o r a l l z , y e L, for alU 6 M and x € L, if and only if x = 0.

If || -|| is a norm on L then the function g ( x , y ) = \\x-y\\ defines a metric on L; it is called the metric derived from the norm \\-\\. We call a linear space L normable provided that there exists a norm on it such that the metric derived from this norm is admissible; for obvious reasons such a norm is called admissible. Observe that each normable linear space is locally convex (See Exercise 1.1.6). A normed linear space is a pair (L,\ -||), where L is a vector space and ||-|| is a norm on L. We shall always endow the underlying vector space of a normed linear space with the topology derived from its norm. So we make a formal distinction between normed linear space and normable linear space: a normable linear space may possess many different norms that generate its topology, see Exercise 1.1.8, whereas in a normed linear space the norm is fixed. In topology we make a similar distinction between metric and metrizable spaces. A metrizable space may posses many admissible metrics generating the same topology, whereas in a metric space the metric under consideration is fixed. If (V, || -||) is a normed linear space then

B = {x 6 V : \\x\\ < 1} and

S = x£ B: x = 1

1.1. LINEAR SPACES

3

denote its unit ball and unit sphere, respectively. A subset A of a normed linear space L is called bounded if there exists an e E [0, oo) such that ||a|| < e for every a £ A. A Banach space is a normable linear space for which there exists an admissible norm such that the metric derived from it is complete. So a Banach space is topologically complete by definition, and hence is a Baire space by Theorem A.6.6. Examples of linear spaces. We will now discuss various important examples of linear spaces. Example 1.1.2. The Euclidean spaces W1. The standard norm on W1 is the Euclidean norm which is defined by

As noticed on Page 461, Bn and §n l abbreviate the unit ball and unit sphere in W1. It is clear that W1 is a Banach space since the metric derived from its Euclidean norm is the well-known Euclidean metric which is complete. Example 1.1.3. The space s. The space s is the vector space E°° endowed with the Tychonoff product topology. It is a classical object both in topology and functional analysis and will play a central role in the remaining part of this book. Observe that s is topologically complete by Lemma A.6.2. Its standard complete metric is the following one:

/

CO

I

\ _ V^ 2 _ n n=l

I

\xn-yn\ ~r I n

J/n|

(See Exercise 1.1.1 for the verification.) Since each Mn is normable and since s is in many respects their 'limit', the question naturally arises whether s is normable. We will show below that it is not. Define cr = {x € s : xn = 0 for all but finitely many n G N}. It is clear that cr is a linear subspace of s. Lemma 1.1.4. If L is a linear subspace of s with cr C L then L, endowed with the subspace topology it inherits from s, is not normable. Proof. Assume, to the contrary, that ||-|| is an admissible norm on L. Then U = {xeL: \\x\\ < 1}

1. BASIC TOPOLOGY

is an open neighborhood of the origin of L. By definition of the product topology on s there are an open neighborhood V of 0 in R and an n G N such that

(*)

x

(n^ n ^)n£cc/, n

i=l

oo

i=n+l

where Vi = V for i < n and Rj = IR for i > n. Let y 6 s be defined by yi = 0 if i ^ n + 1 and yn+\ = 1. Since y E a C L and y 7^ 0 it follows that e = ||y|| > 0. By (*), £ y 6 U for every t € E. In particular, ||s/e|| < 1 but also 1 1 Veil — £/e — 1> which is a contradiction. D From the proof of Lemma 1.1.4 it is clear that the interplay between the topology and the linear structure on s prevents it from being normable. (The question naturally arises whether every vector space can be endowed with a norm which is compatible with its linear structure. The answer to this question is in the affirmative, see Exercise 1.1.8.) Consequently, although s seems a natural 'limit' of the spaces IRn, it is notably different from any of its finite dimensional analogs W1 . Example 1.1.5. The spaces C(X) and C*(X). For a nonempty compact space X we let C ( X ) denote the set of all continuous real-valued functions on X. Obviously, C(X) is a vector space; addition of functions and scalar multiplication are defined pointwise. If / e C(X) then define its norm, ||/||, by H/ll =

sup{\f(x)\:xeX}.

(Observe that by compactness of X this supremum is attained. That is: there is an element x e X such that ||/|| = \f(x}\.) It is easily seen that ||-|| : C(X] —> [0, oo) is indeed a norm; it is called the sup-norm on C(X). Consequently, the function (*)

Q(f, 9) = \\f~9\\

defines a metric on C(X) and therefore generates a topology. From now on we shall endow C(X] with this topology. There are other useful and interesting topologies on C(X). In Chapter 6 we shall endow C(X) with the so-called topology of pointwise convergence. It will be clear from our notation which topology on C(X] we are using. For example, C(X) denotes the set C(X] endowed with the above topology, and CP(X) denotes the set C(X) endowed with the topology of pointwise convergence, etc. We claim that C(X) is a Banach space. Let ( f n } n be a £-Cauchy sequence. Then ( f n } n clearly converges pointwise, so / = limn^.^ fn exists and belongs to C(X) by Lemma A. 3.1. It therefore suffices to prove that fn —> f in C(X). But this follows easily from the proof of Lemma A. 3.1.

1.1. LINEAR SPACES

5

The spaces C(X] have the following interesting property that will be used quite frequently in our Chapter 4 on ANR-theory. Lemma 1.1.6. For every compact space X there are a compact space A and an imbedding i: X <—> C(A) such that i[X] is linearly independent. Remark 1.1.7. The linear independence of the set i[X] in the above result is quite interesting, and has proved to be useful in several research papers in infinite-dimensional and related topology. Proof. Let Y be the topological sum of X and a point XQ £ X, and let Q be an admissible metric for Y. Let A be the subspace of C(Y) consisting of all Lipschitz functions /: Y -> R such that f ( x 0 ) = 0. Observe that if / G A then f[Y] C [-diam(y),diam(T)]. This is clear since if y G Y then

l/(i/)l = I/to) - o = l/(y) - /0*o)| < e(y,*o) < diam(r). It is easy to show that A is a closed subspace of C(Y). But even more is true. Claim 1. A is compact. Proof. Let (fm)m be any sequence in A. It suffices to prove that it has a convergent subsequence in A (Theorem A. 5.1). For every n let Un be a finite open cover of [— diam(F), diam(F)] consisting of open sets with diameter at most 7rn' . This cover has a Lebesgue number, say Xn > 0 (Lemma A. 5. 3). By compactness of X there is a finite open cover Vn of X such that mesh(V n ) < i/ 2 A n . Now fix any / G A and V G Vn. Since / is Lipschitz, the diameter of f[V] is at most Xn. Hence f[V] is contained in an element U G U n . So for each / G A there is a function £n(/): V n - > U n

such that for every V € Vn we have Let n = 1. Observe that there are finitely many functions Vi —> Ui only. Hence there is an infinite subset NI C N such that £i(/ m ) = £i(/fc) f°r all integers m, k 6 A7"! . Let m and k be two arbitrary elements of NI , and pick an arbitrary x 6 X. Pick an element F G Vi such that ar G I7. Then both fm(x) and /jfc(x) belong to the same element of Ui, i.e., |/m(^) ~ /fc( x )| < 2"1. Since x was arbitrary, this shows that ||/m — /^|| < 2"1. So now it is clear how to proceed. Let n\ = minA^i and consider the infinite set N\ \ {n\}. There is an infinite subset N% C NI\ {HI} such that for all ra and k in A^ the functions ^(/m) and £2 (/A:) agree. Then by a similar argument as the

6

1. BASIC TOPOLOGY

one above, ||/m — fk\\ < 2~ 2 for all m, k 6 N2. Let n2 = min 7V2. Continuing in this way resursively, we can construct an infinite sequence

ni < n2 < • • • < nm < • • • of natural numbers and a Cauchy sequence (fn )m in C(Y). Since C(Y) is a Banach space, this sequence has a limit and since A is closed, this limit belongs to A. So we conclude that (fnm)m is the desired convergent subsequence Of (fn}n0 Define i : X —> C(A) by the following formula:

t(*0(/) = /(*). Observe that i is well-defined. For i(x) should be an element of C(A), hence should be a function i(x) : A —> M. But an element of A is a function / from Y to R. So the formula tells us that i(x) sends the function / onto its evaluation in the point x £ X C Y. Claim 2. i: X —> i[X] is an isometry. Proof. Let £i,rr 2 G X. Then ||t(xi) - i(x2)\\ = sup{|t(*i)(/) - »

(1) Observe that the last inequality follows from the fact that / is Lipschitz. Define the function g : Y -> M by If y, y' 6 y are arbitrary then

\g(y) - g(y'}\ = \e(y,x2) = \Q(y,x2) -e(y',x2)\ < Q(y,y'}Also, g(xo) — 0. Hence g E A and -i(x2)(g}\ = \Q(xi,x2) - Q(x0, Hence by (1) we get ||Z(:EI) — z(^2)|| = ^(^15^2), as required.

0

It remains to prove that i[X] is a linearly independent subset of C(A). To this end, let #1, . . . , x n +i be distinct elements of ^C. We claim that z(x n +i) is not a linear combination of the i(xi), . . . , i(x n ). Define /i : Y -> E by = mm e ( y , X i ) . 0<^
1.1. LINEAR SPACES

7

Claim 3. h G A. Proof. We shall first prove that h is Lipschitz. To this end, pick arbitrary a, b 6 Y. We may assume without loss of generality that h(b) < h(a). Fix i and j such that h (a) = g(a, Xi) and h(b) = g(b,Xj). Since h (a) = g(a,Xi) we get by the definition of h that g(a,Xi) < g(a,Xj). From this we conclude that

\ h ( a ) - h ( b ) \ =h(a)-h(b) = g(a,Xi} - g(b,Xj) < g(a,Xj) - g(b,Xj) < 0(a,b), as required. Since it is trivial that h(xo) = 0, this completes the proof.

0

Observe that i(xk)(h) = h(xk) = 0 for every k < n, but since the XQ, ..., x n +i are distinct elements of F,

i(xn+i)(h) ^ 0. So i(xn+i} is not a linear combination of the i(x\),..., «(x n ), and this is what we had to prove. D Corollary 1.1.8. Every space is homeomorphic to a linearly independent bounded and closed subspace of a normed linear space. Proof. Let X be a space, and let aX be a compactification of X (Corollary A.4.8). By Lemma 1.1.6 there are a compact space A and an imbedding i: aX —>• C(A) such that i[aX] is linearly independent. By linear independence, if L is the linear hull of i[X] then L n i[aX] = i[X]. Hence i[X] is a closed linearly independent subset of the normed linear space L. Since ||-||: C(A) —> [0, oo) is continuous, and aX is compact, it is clear that i[aX] is a bounded subset of C(A), hence so is i[X]. D If X is not compact then C(X) contains unbounded functions (see Exercise A.5.14 below), and so the formula \\f\\

=sup{\f(x)\:x£X}

does not define a norm on C(X). By considering the subset C*(X) of C(X) consisting of all bounded functions, this problem does not occur; C*(X) endowed with the sup-norm is a Banach space for the same reasons C(Y) is for compact Y. The topology defined here on C*(X) is called the topology of uniform convergence.

8

1. BASIC TOPOLOGY

Example 1.1.9. The space CQ. Put CQ = {x E s : linin-j.oo xn = 0}, and endow it with the norm \\x\\ = sup{|xn| : n E N}. It follows by straightforward calculations that this is indeed a norm compatible with the linear structure on CQ. There is however another way of proving this. Let S be a nontrivial convergent sequence including its limit t, and consider L = {/ E C(S) : /(*) - 0}. Then L is a closed linear subspace of C(S) which clearly can be identified with CQ. So c0 is a closed linear subspace of the function space C(S). The set CQ endowed with a different vector space topology will play a prominent role in our analysis of function spaces later. See Chapter 6 for details. Example 1.1.10. The Hilbert space I2. We saw that the topology on s is very different from the topology on any of its finite dimensional analogs W1. We shall now construct another natural 'limit' of the spaces W1 which behaves better (in this respect). Consider the usual inner product on En given by n (X \ 5I// y/ —

7/ j OC '11' iyi'

If we try to generalize this inner product for E°° then we have to deal with infinite series and it is therefore quite natural to restrict ourselves to the following subset of M°°: oo

I2 — {x E s : /]x2 < oo}. We shall first prove that t2 is a linear subspace of M°°. For every x E t2 we write p(x) = v/X^i X1- ^ x-> V e ^•> tnen Schwarz's inequality applied to En shows that | 5^=i ^^1 — P(X} ' P(y}- From this it follows that

P(x] -p(y] < oo, 1=1 and so oo

since all infinite series considered are convergent. We conclude that for every x, y G I2 we have x + y G i2 . If x e i2 and t G M then trivially tx € I2. Consequently, i2 is indeed a linear subspace of M°° .

1.1. LINEAR SPACES

Since X^i xiVi < °° f°r a^ X->V tion { - , - ) : i2 x £ 2 -^M,

e

^

we

have a well-defined func-

1=1 which is easily seen to be an inner product. Consequently, \\x\\ — p(x] defines a norm on I2 and the metric derived from this norm is: oo

£' We endow I2 with the topology generated by this metric and refer to I2 with this topology as Hilbert space. It is clear that a is a subset of I2 and so Lemma 1.1.4 shows that the topology that i2 inherits from s is different from the topology on i2 which we just defined. We will comment on the precise relation between these topologies later. Lemma 1.1.11. The metric g on I2 defined above is complete. So i2 is also a Banach space. But there is a difference with the spaces C(X] that we discussed before. The norm on i2 is derived from an inner product, while no inner product on C(I) yields its standard norm (Exercise 1.1.9). The topology on s is the topology of 'coordinatewise convergence', see Exercise A.2.2. Topologists usually find such product topologies easier to handle than topologies derived from a norm. However, convergence in £2 can be handled with the same ease, as is stated in the next result. Lemma 1.1.12 (Exercise 1.1.26). Suppose that (x(n)] and x G I2. The following statements are equivalent:

is a sequence in I2,

(1) linin-^oo x(n) = x (in I2), (2) limn_>.00 ||x(n)|| = ||x|| and for every i 6 N, lim^-^oo x(n}i = Xi. From Lemma 1.1.12 it immediately follows that the topology on I2 is finer than the topology that I2 inherits from s. However, more can be concluded. For example, consider the unit sphere 5 = { x e £ 2 : | x | | = l}. Since all points in 5 have the same norm, the topology that 5 inherits from f2 is precisely the same as the topology that S inherits from s, i.e., the topology of 'coordinatewise convergence'. This remark plays an important role in the proof of the Anderson Theorem from ANDERSON [15] that I2 and s are topologically homeomorphic (see also VAN MILL [298, Chapter 6] for a complete proof of this result).

10

1. BASIC TOPOLOGY

Classical theorems. We now present some classical theorems on Banach spaces that will be important later. Open Mapping Theorem 1.1.13. Let T be a continuous linear mapping of a Banach space E onto a Banach space F. Then T is open. Proof. The proof is in three steps. Claim 1. There exists a > 0 such that such that {y € F : \\y\\ < 1} C T[{x G £ : ||x|| < a}].

Proof. For each a > 0 put Ba = {x G -E1 : ||x|| < a}. Since oo

F

=U

n=l

and F is a Baire space (see Page 3), there exists m G N such that T[Bm] has nonempty interior. Since T is linear, it follows easily that T[Bm] is convex, and by the continuity of the algebraic operations on F, so is T[Bm}. In addition, Bm is symmetric, i.e., — Bm = Bm. Again since T is linear, it follows that T[Bm] is symmetric, from which it follows easily that T[Bm] is symmetric as well. Now choose y G F and /3 > 0 such that D(y,/3) C T[Bm]. Letz G F be such that \\z\
T[B This proves that D(Q,/3) C T[Bm] from which it follows easily that

x?(Q,i)cr[B a ], m

where a — /p.

0

In the remaining part of the proof we adopt the notation introduced in Claim 1. Claim 2. {y G F : \\y\ < 1} C T[B2a]. Proof. Let y G F with | y\\ < 1. We shall define recursively a sequence (yn}n in T[Ba] such that for all n, n

" "

" < Tn.

1.1. LINEAR SPACES

11

By Claim 1, there exists y\ 6 T [-£?«] such that \\y — y\\\ < l/zthat yi,..., yn are chosen properly. Then

Suppose

<1 fc=l and therefore, again by Claim 1, there exists yn+i G T[Ba} such that

k=l

It is clear that yn+i is as required. From (*) it easily follows that y = lim

n—>oo

k=l

For every n choose a point xn £ Ba with T(xn) = yn. Since \\xn\\ < a < oo for every n, it is easily seen that the sequence

is Cauchy, and hence that x = lim

n—><x>

exists. Since ||-||: E —>• R is continuous, 00

X

^ f

00

2*

K^/c

k=l

/

' Ot ^— ZiOi.

k=\

By observing that T is continuous and linear we obtain

T(x)= lim

n—>oo

from which we conclude that y 6 T[B^a\.

-0

We are now in a position to prove that T is open. Indeed, let U be a nonempty open subset of E. Let x £ U and choose e > 0 with D(x,e) C [7. Consequently, £?e C [7 — x. It follows by Claim 2 that {?/ e F : Hi/11 < -/2a} C T[Be]. Consequently, since T[5e] C T[U - x] = T[U] - T(x), we have {y£F:\\y-T(x)\\<*/2a}CT[U]. So T[t7] is a neighborhood of T(x) and since x was an arbitrarily chosen point from £7, we are done. D

12

1. BASIC TOPOLOGY

Corollary 1.1.14. A bijective continuous linear function between Banach spaces is a (linear) homeomorphism. Proof. This is clear since such a function is open by Theorem 1.1.13.

D

The following corollary to Theorem 1.1.13 will be of particular importance later. Its converse is trivial (Exercise A. 1.8). Closed Graph Theorem 1.1.15. Let E and F be Banach spaces and let

T: E-+F be linear. If the graph F = {(x,Tx) :xeE} ofT is a closed subset of E x F then T is continuous. Proof. Observe that F is a closed linear subspace of the Banach space Ex F (Exercise A. 1.8) and hence is a Banach space itself. In addition, the function Si: F —> E defined by 5i(x,Tx) = x is clearly continuous (since it is the restriction of a projection). As a consequence, Si is a linear homeomorphism by Corollary 1.1.14. Observe that the function £2: F —>• F defined by 52(x,T(x)) = T(x) is continuous as well. So the function E -> F defined by x H-» (x, Tx] !->• Tx is continuous, being a composition of continuous functions. So T is continuous. D Corollary 1.1.16. Let E and F be Banach spaces additionally endowed with weaker vector space topologies. I f T : E —> F is a linear homeomorphism with respect to the weaker topologies then it is also a linear homeomorphism with respect to the Banach space topologies. Proof. All we need to observe is that by continuity the graph of T is closed in the weaker topology on E x F (Exercise A. 1.8) and hence also in E x F endowed with the product of the Banach space topologies. Simply observe that the product of the weaker topologies is weaker than the product of the Banach space topologies. D The Michael selection theorem. We shall prove that certain setvalued functions admit a continuous selection (for definitions, see below). Let X and Y be sets. A set-valued function F from X to Y is defined to be a function from X to 9(Y) \ {0}, i.e., F: X ->> 7(Y) \ {0}. By the symbol F: X =>Y

we shall mean that F is a set-valued function from X to Y. Let X and Y be topological spaces and let F: X => Y. For every V C Y we put F*=[V] = {x 6 X : F(x] n V / 0}.

1.1. LINEAR SPACES

13

We say that F is lower semi- continuous (abbreviated LSC) provided that for every open subset U of Y, F^[U] is open in X. Observe that if U is a covering of Y then F*=[U] = {F^[U] :U G U } covers X since for every x £ X, F(x) ^ 0. A basic example of an LSC set-valued function is the following one. If X and Y are spaces and /: X —> Y is an open surjection then we define F: Y => X l

by F(y] — f ~ ( y } . The function F is LSC since for every open U C Y we have F*[U] = {y£Y: F(y] n U ^ 0} = {y e Y : f ~ l ( y } n t/ ^ 0} = f[U]. Other examples of LSC set-valued mappings will be presented later. Let X and Y be sets and let F : X =£> Y. A function / : X —> Y is called a selection for F if for all x e X, /(x) G F(x). Since for all £ € JC the set -F(x) is nonempty, by the Axiom of Choice such a selection exists. The question naturally arises whether it is possible to find a continuous selection if X and Y are topological spaces. This question is natural but rather naive. Simple examples show that the answer in general is in the negative. Example 1.1.17. Define F: I => I by (0<x• I the graph of which is contained in the 'graph' of F, which is left as an exercise to the reader as well as to prove that F is LSC. The values of F in the above example are too 'small' for F to admit a continuous selection. If we enlarge these values by for example to require that F(x) = I for all 1/3 < x < 2/3 then F does admit a continuous selection. The following result shows that although not every LSC map has a continuous selection any map with sufficiently many continuous selections must be LSC. Proposition 1.1.18. Let X and Y be spaces and let F: X =>• Y be a setvalued function with the following property: for all x e X and y 6 F(x) there exists a continuous selection f for F such that f ( x ) = y. Then F is LSC.

14

1. BASIC TOPOLOGY

Proof. Let U C Y be open and take x 6 F^[U}. Pick y e F(x)r\U. By assumption there exists a continuous selection /: X —»• Y for F such that f ( x ) — y. Put V^ = f ~ 1 [ U ] , By continuity of /, V is a neighborhood of x. In addition, V C F^[U] since if x' G V then

/(x') € F(z') n tf. We conclude that F^ [U] is open.

D

One of the main results here is the Michael Selection Theorem which states that in normed linear spaces, convex valued LSC set-valued functions admit continuous selections. The following three technical lemmas are needed in the proof of this result. Lemma 1.1.19. Let X and Y be spaces. If F: X => Y is LSC then (1) the function Fc: X =* Y denned by Fc(x) = F(x) is LSC, (2) if f : X -> Y is continuous and Q is an admissible metric for Y and the number r > 0 is such that

then the function G : X =£• Y denned by G(x)=F(x}nB(f(x),r) is LSC.

Proof. For (1), simply observe that for all x G X and open U C Y we have that F(x) H U 7^ 0 if and only if F(x) n U ^ 0. Consequently, for every open subset U C Y the equality F^[U] = F^=[U] holds. For (2) we use (1) to conclude that it suffices to prove that the set-valued mapping G : X =$• Y defined by

(*), r) is LSC. Let V C Y be open and take a point x € G^[V]. There exists y € Y such that Let e = r — g(y: /(x)) and choose 0 < 6 < e such that B(y,S)CB(f(x),r)nV. Since F(x)C\B(y, %) ^ 0 and F is LSC, U0 = F<=[B(y, %)} is a neighborhood of x. In addition, since / is continuous, U\ = f ~ l [ B ( f ( x ) , 5/2)] is also a neighborhood of x. Put U = U0 n U\. We claim that U C G^[V]. To this end, take an arbitrary x' € U. Since x' € f/ 0 , there exists a point y' e F(ar') n B(y, 6/2). In addition, f ( x ' ) G B ( f ( x ) , 5/2) since x' e C/i.

1.1. LINEAR SPACES

15

Consequently,

0(yJ(x))+e(f(x)j(x'))

< %+r - e + % < % + r - 6+% — 7*

and therefore y' e F(x') n B(f(x'),r) the point x' belongs to G^[V].

n V = G(x') n F. We conclude that D

Lemma 1.1.20. Let L be a normed linear space, let X be a space and let F: X =^> L be LSC such that F(x) is convex for every x e X. Then for every r > 0 there exists a continuous function f : X —> L such that for every x € X we have g(f(x),F(x)] < r. Proof. Put By Theorem A. 7. 5 there exists a partition of unity IP on X which is subordinated to F^l'B]. Consequently, for each p G CP there exists bp G L such that Define / : X -> L by

For each a; 6 X there are a neighborhood Ux and a finite subset G(x) of IP such that Ux Hp~l [(0, 1]] ^ 0 if and only if p G G(x). Now fix an arbitrary x £ X. Observe that the restriction of / to Ux is given by

which is a continuous expression in y since G(x) is finite. From this we conclude that / is well-defined and continuous at x. Put 9(x) = {p e G(x] : x € p~l [(0, 1]] }. For each p e 9(«) we have z G p-1 [(0, 1]] CF^[£(& p ,r)] and so there exists yp € F(x] r\B(bp,r}. Observe that

16

1. BASIC TOPOLOGY

This implies that f / \ T [ nf* \

/W

/_. P\x) yp ^

X X

^

/ \ TM T* I

^
• "Jl

X X

^

/ \ II 7 711 O" 1 * 11/I

/ _ . JPW ll°p

II I

ypll 11

= r. x

Since F(x) is convex and Y ^ & ( x P ( } = 1 we have

(Lemma 1.1.1), and consequently, ^>(/(x),F(x)) < r.

D

We need one more technical lemma. Lemma 1.1.21. Let X and Y be spaces and let F: X => Y be LSC. Suppose that A C X is closed and that f : A ->• Y is a continuous selection for the function F\A:A=>Y. Define G: X => Y by r f(~}\ iJ \ x / j

F(x)

(v (^ A\ v x c •"•>•>

(xeX\A).

Then G is LSC. Proof. Let U C Y be open. We first claim that f~l[U] C F*=[U]. Indeed, take an arbitrary x 6 f~1[U]. Then f(x) 6 F(x) H U, hence x € F^[U]. Observe that by continuity of /, the set f~l[U] is open in A. So there is an open V C X such that V n A = f~l[U}. Since f~l[U] C F^[C7] and F^[f7] is open, without loss of generality, V C F^IU]. Consequently, is open.

D

Let (L, || -| |) be a normed linear space. A subset A of L is called complete with respect to \\-\\ if the restriction of the metric g(x,y) = \\x — y\\ to A is complete. Observe that such an A is automatically closed in L and also that every compact subset of L is complete with respect to ||-||. We are now in a position to prove the announced classical result. The Michael Selection Theorem 1.1.22. let X be a space and let be LSC, where (L, ||-||) is a normed linear space. Assume that each F(x) is convex in L and complete with respect to \\-\\. Then for every closed subset A of X and every continuous selection f : A —> L for the function F \A: A=> L,

1.1. LINEAR SPACES

17

there exists a continuous selection g : X —> L for F which extends f . Proof. We shall first prove the theorem in the special case A = 0. By induction on n we shall construct a sequence (fn)n in C(X, L) such that (1) e ( / n , / n + l ) < 2 - ( " - 1 ) , n n

(2) g ( f ( x ) , F ( x ) )

< 2~ for every xtX.

Apply Lemma 1.1.20 with r = 1/2 to find f\: X ->• L such that for every x 6 X,

e(f1(x),F(x))

< %.

Suppose that fn has been defined. Define Fn: X =>• L by Fn(x)=F(x)nB(fn(x),2-n). Then Fn is LSC by Lemma 1.1.19. By another appeal to Lemma 1.1.20 we find h: X —> L such that g ( h ( x ) , F n ( x ) ) < 2~n for every x G X. It is easy to see that fn+i = h is as required. Claim 1. For every x G X, g(x) — limn^.oo fn(x) exists and belongs to F(x}. Proof. Take an arbitrary x G X. By (2) there exists for every n G N a point an G F(x) such that 0(/ n (aO,a n ) < 2~ n . Consequently, by (1) we have Q(an,an+i} < g ( a n , f n ( x ) ) + g(fn(x), fn+i(x)) + g(/ n +i(o;),a n+ i)

We conclude that the sequence (a n ) n is Cauchy and so by completeness of F(x), a = rimn_j.00 an exists. Since 0 ( f n ( x ) , a n ) < 2~n for every n, this implies that limn^.00 fn(x) = g(x) also exists and is equal to a G F(a;). <0 By Lemma A. 3.1 we consequently conclude that the function g is the required continuous selection. Now if A ^ 0, the above special case and Lemma 1.1.21 yield the desired result. D Theorem 1.1.13 and the Michael Selection Theorem imply the following Corollary 1.1.23. Let E and F be Banach spaces and let T: E ^ F be a surjective continuous and linear function. In addition, let KerT denote the kernel ofT. Then there exists a continuous function f : F —> E such that To f = 1F. The function h: E -4 KerT x F defined by h(x) = (x- f ( T x } , T x ] is a homeomorphism.

18

1. BASIC TOPOLOGY

Proof. Define H: F => E by H(y) = T~l(y). Then for each open U C E we have H*=[U] = T[U], so by Theorem 1.1.13, F is LSC. Since the fibers of T are convex and closed by linearity and continuity of T, respectively, the existence of / follows directly from Theorem 1.1.22. The easy proof that h is a homeomorphism is left as an exercise to the reader. D Remarks. We conclude this section by some remarks. We introduced linear spaces, locally convex linear spaces, normable linear spaces and implicitly also inner product spaces. Obviously, the 'underlying' linear space of an inner product space is normable and each normable linear space is locally convex. Since s is locally convex but not normable (Lemma 1.1.4), we see that the class of normable linear spaces is a proper subclass of the class of all locally convex linear spaces. The sup-norm on C(I) violates the parallelogram law and therefore cannot be derived from an inner product (Exercise 1.1.9). As a consequence, the class of normed spaces is strictly larger than the class of inner product spaces. In Exercise 1.1.13 we will present examples of linear spaces that are not locally convex. It is also possible to introduce topological versions of the above concepts. One may ask for example whether every linear space is homeomorphic to a locally convex linear space, whether every locally convex linear space is homeomorphic to a normed linear space and whether every normed linear space is homeomorphic to an inner product space. There is a lot of research in infinite-dimensional topology on these questions, especially when the linear spaces under consideration are (absolutely) Borel. For general (separable metrizable) linear spaces, the answers to these questions are in the negative, see MARCISZEWSKI [269].

Exercises for §1.1. Let L be a linear space. A maximal linearly independent subset B of L (i.e., a subset of L which is maximal with respect to the property of being linearly independent) is called a Hamel basis for L. The Kuratowski-Zorn Lemma easily implies that every linear space has a Hamel basis. It is well-known, and easy to prove, that if B is a Hamel basis for L then each x (E L \ {0} can be written uniquely in the form + Oiix-i + • • • + anxn, with Xi £ B and on € R \ {0} for every i < n (for details, consult any (good) textbook on Linear Algebra). A linear space L is called finite dimensional if it has a finite Hamel basis. Otherwise it is called infinite- dimensional.

1.1. LINEAR SPACES

19

If L is a linear space and x, y £ L then I ( x , y) denotes the straight-line segment from x to y, i.e., I(x, y} = {ax + (1 — a)y : a £ I}.

Let X be a space. For a compact subset K in X and an open U in R. define Topologize C(X] by taking the collection {[K, U] : K C X is compact and U C R. is open} as an open subbase. This topology is called the compact-open topology on C(X). 1. Prove that the formula ,(,.,) - V - - - * ' n=1

*

n

defines an admissible metric on s. 2. Let L be a linear space. Prove that for every x £ L and every convex C C L the set x + C is convex as well. 3. Let L be a linear space. Prove that every translation of L is a homeomorphism. 4. Let L be a linear space and let x £ L. In addition, let W be a local base at the origin of L. Show that x has arbitrarily small neighborhoods of the form x + W , where W £ W. 5. Prove Lemma 1.1.1. 6. Prove that every normable linear space is locally convex. 7. Define a function |||-|||: M" —> E by |||z||| = max{|a;i| : ! < « ' < n}. Prove that || | -||| is a norm on Rn and that it generates the Euclidean topology. 8. Let L be a linear space and let B be a Hamel basis for L. If X — OilXl + 0(2X2 + • • • + anXn,

with Xi G B and on £ R for every i < n, then put ||z|| = |ai| + |a2 + ---- h|a n |. Prove that ||-|| defines a norm on L. 9. Prove that the sup-norm on C(I) cannot be derived from an inner product by showing that it violates the parallelogram law. 0. Prove that for every compact space (X, Q) there exists an isometry i:

X^C(X)

such that (1) for every subset Y C X the image set i[Y] is closed in conv(z[F]), (2) t[X]C{/

20

1. BASIC TOPOLOGY

11. Prove that there does not exist a homeomorphism h: I2 —>• s which is linear, i.e., has the property that h(\x + p,y) — \h(x) + (J,h(y)

for all x, y € I2 and A, /z e It 12. Let A be a bounded subset of a normed linear space L. Prove that if (tn)n is a sequence in R such that tn —> 0 and (an)n is any sequence of elements in A then tnan —>• 0. M3. Let 0 < p < 1 and put lp = {x <E s : E^°=i 1^1" < °°}- Prove that ^ is a linear subspace of s (in the algebraic sense). Define the following metric on lp: oo

e(*,y) = ^2\x^ -yn\p. n=l

Prove that lp with the topology derived from this metric is a linear space which is not locally convex. 14. Let X be a compact space. Prove that the compact-open topology on C(X) coincides with the topology derived from the sup-norm ||-||. 15. Prove that I2 and s are separable. 16. Prove that i2 and I2 x R are linearly homeomorphic. 17. Let X = {x € I2 : (Vn £ N)(|x n | < YJ}. Prove that X and Q are homeomorphic. Let ||-|| be a norm on R". Prove that the topology derived from this norm is the Euclidean topology. In particular, for every x € Rn and e > 0 we have that the set {y (E Rn : \\x — y\\ < e} is compact. 19. Let V be a normed linear space and let W be a finite dimensional linear subspace. Prove that W is closed in V. 20. Let ||-|| be a norm on Rra and let / : Rn -> R be linear but not identically 0. Prove that there exists t > 0 such that

(1) /-1(i)

(2) /- 1 (t) Let V be an infinite-dimensional normed linear space. Prove that there is a sequence (en)n in V such that (1) {en : n e N} is linearly independent, (2) ||eB|| = l ( n € N ) , (3) |je m - en|| > 1 (m, n e N, m ^ n). 22. Let V be an infinite-dimensional normed linear space. Prove that the unit sphere S = {x G V : ||a;|| = 1} is not compact. 23. Let x e W1 and let

Prove that for every a e [0, 1) there exists e > 0 such that B(ax, e) C K. . Prove that every compact convex subset A C Rn with nonempty interior is homeomorphic to Bn and its boundary is homeomorphic to S™" 1 .

1.2. EXTENDING CONTINUOUS FUNCTIONS

21

25. Prove that the metric Q on I2 defined on Page 9 is complete. *>26. Prove Lemma 1.1.12. 27. Let X be a space and let /, g : X —> R with / Isc and g use. Prove that if 9 < /> i- e -j g(x) < f ( x ) f°r every x £ X, then the function F: X =>• R defined by F(x] — [g(x),f(x)] is LSC. Use this to conclude that there exists a continuous h: X —> R such that g < h < f. (See Corollary A.7.6 for a stronger result.) 28. Prove that the function in Corollary 1.1.23 is a homeomorphism. ^•29. Let X be a closed and bounded subspace of a normed linear space L. Prove that A(X) is homeomorphic to the subspace {((1 -t)x,t) :tel,xe X} of L x I.

30. Let E and F be Banach spaces and let (p: E —»• F be linear and continuous. Prove that there exists k G N such that \\ F be a linear homeomorphism. Prove that there exists k E N such that for every x £ E we have

\\x\\/k< M*)|| <*N|. 1.2. Extending continuous functions

Suppose that X, Y and Z are topological spaces with Y a subspace of X and let /: Y —> Z be continuous. In topology it is often of interest to know whether / is the restriction to Y of a continuous function /: X —> Z. Easy examples show that in general this need not be the case. If / is the restriction to F of a continuous function /: X —> Z then we say that / is continuously extendable over X or that / is a continuous extension of /. In this section we shall present examples of spaces Z having the property that if Y is closed in an arbitrary space X, then every continuous function /: Y —>• Z is continuously extendable over X (respectively, over some neighborhood of F). Observe that Urysohn's Lemma can be looked at as a result on extending continuous functions. For let X be a space and let A and B be disjoint closed subsets ofX. If E = A\JF then by Lemma A.4.1 it follows that the continuous function /: E -» I denned by / \ A = 0 and f \ B = I can be extended to a continuous function /: X —> E. The function /: E —>• I is not very interesting and the question naturally arises whether it is also possible to extend more interesting functions. In order to study this question with success, let us first formulate and prove the following technical result.

22

1. BASIC TOPOLOGY

Lemma 1.2.1. Let X be a space, A a closed subset of X, and let A' C A be dense in A. Then there exist a locally finite open cover U of X \A and a sequence of points [au : U E U} in A' such that (1) for all U 6 U and x € U, g(x, au) < 2@(x, A), (2) if Un E U for every n and

lim g(Un,A) = 0

n —>-oo

then lim diam(t/ n ) = 0.

n—>oo

Proof. Let V={B(x,VtQ(x,A)):xeX\A}. Since A is closed, V is an open cover of X \ A. By paracompactness of X \ A (Corollary A. 7. 3) there exists a locally finite open cover U of X \ A that refines V. Since U < V, for each U E U there exists xu E X \ A with UCB(xu,l/4Q(xu,A)). In addition, since A' is dense in A, for each U E U there exists au E -4' with We claim that the C/'s and the Of/'s are as required. Claim 1. For every C7 G U and x G C/ the following inequalities hold:

Proof. The first inequality is easy since

g(x, au) < Q(X,

Also, Q(XU, A] < Q(XU,X) + g(x, A) < %g(xu, A) + g(x, A), from which it follows that %g(xu,A) < g(x,A), as required.

So it remains to verify (2). To this end, assume that Un G U for all n and that lim g(Un,A) = 0 . n—>ao

For each n pick pn G Un such that linin^oo g(pn, A) = 0. By the claim we obtain lim Q(xUn,A) < lim 4/3^Pn, A) = 0. n—too

n—>oo

1.2. EXTENDING CONTINUOUS FUNCTIONS

23

Since Un C B(xun, 1/4^(x[/n, A)) for all n, we get limn^.oo diam(t/n) = 0. So we are done. D An open cover It and a sequence of points {au '• U G U} such as in this lemma is called a Dugundji system for X and A. We now come to the main result in this section. The Dugundji Theorem (Part 1) 1.2.2. Let L be a locally convex linear space and let C C L be convex. Then for every space X with closed subspace A, every continuous function f : A -» C can be extended to a continuous function f : X — > C . Remark 1.2.3. For Part 2 of The Dugundji Theorem, see Page 394. Proof. Let the open cover U of X \ A and the sequence of points (au in A(= A) be a Dugundji system for X and A. In addition, let KU '• X\A for U G U be the Ac-functions with respect to U. Define /: X -» L by (±\

f f(~.\ _ J

f(T\ /W

' ~ i v^

,,-(~\

tt~^\

(rx pfc A\ l ^)i

(x £ X \ A)

We will first prove that / is well-defined and continuous at all points of X\A. To this end, fix an arbitrary x 6 X \ A. Since U is locally finite, there is a neighborhood W of x in X \ A meeting finitely many elements of U only, say C/i, . . . , Un. For U 6 U missing W we clearly have As a consequence, for every y £ W we have

n=l

So J(y] is a convex combination of points in C and therefore belongs to C. By continuity of the ^-functions, it also follows that / |~ W is continuous. It suffices to prove continuity of / at the points of A. Pick an arbitrary element a G A. An arbitrary neighborhood of /(a) is without loss of generality of the form /(a) + W, where W is a convex neighborhood of 0 (Exercise 1.1.4). So let /(a) + W be such a neighborhood. The continuity of / at a implies that there exists 6 > 0 such that B(a, 6) n A C f~l[f(a) + W]. Claim 1. /[B(a, %)} C /(a) + W. Proof. Pick an arbitrary x G -B(a, 5/s)- ~H x E A then there is nothing to prove. So assume without loss of generality that x 0 A. Then Q(x,A) < g(x,a) < %;

24

1. BASIC TOPOLOGY

as a consequence, if x E U E U then

g(a, au) < g(a, x) + g(x, < e(a,x) + 2g(x,A) <S by (1) of Lemma 1.2.1. From this we conclude that if x E U E U then we have au E #(a, 6) (~\ A and so /(at/) E /(a) + W. Consequently, / » - / » = ( J] ^ ( x ) - /(at/)) -/(a)

Since /(at/) - /(a) 6 W for every C7 6 U we see that f ( x ) — /(a) is a convex combination of elements of W. Hence by convexity of W we obviously get that f ( x ) — /(a) E W, as required. 0 We conclude that / is continuous at a.

D

Remark 1.2.4. It is a natural problem whether the local convexity assumption in Theorem 1.2.2 can be dropped. This was a fundamental open problem ever since DUGUNDJI'S paper [140] appeared in 1951. It was finally solved in 1994 by CAUTY [87] in the negative. His construction used in an essential way DRANISNIKOV'S result in [139] about the existence of an infinitedimensional compactum with finite cohomological dimension. As a corollary to the Dugundji Theorem we get (cf. Theorem A. 4. 6): The Tietze Theorem 1.2.5. For every space X with closed subspace A, every continuous function from A to E or I can be extended over X. A space X is called an Absolute Retract (abbreviated AR) provided it is a retract of every space Y containing it as a closed subspace. If X is an AR and / : X —> Y is a homeomorphism then Y is an A R as well. Consequently, X is an A R if and only if for every space Y containing a closed subspace Z which is homeomorphic to X, there exists a retraction r: Y —>• Z. Theorem 1.2.7 below implies that a retract of an AR is an AR. A space X is called an Absolute Neighborhood Retract (abbreviated ANR) provided it is a neighborhood retract of every space Y containing it as a closed subspace. The space X = {0, 1} is easily seen to be an ANR but is not an AR; simply observe that by continuity retractions preserve connectivity, and so there does not exist a retraction r:I-*X. Notice that every AR is an ANR. As above, Theorem 1.2.7 below easily implies that X is an A N R if and only if for every space Y containing a closed subspace Z which is homeomorphic to X, Z is a neighborhood retract of Y. Also, every neighborhood retract of an A N R is again an ANR (Proposition 1.2.10).

1.2. EXTENDING CONTINUOUS FUNCTIONS

25

We call a space X an Absolute (Neighborhood) Extensor (abbreviated A(N)E) provided that for every space Y and for every closed subspace A of y, every continuous function /: A —> X can be extended over Y (over a neighborhood (depending on /) of A'mY). We shall prove in Theorem 1.2.7 below that X is an A(N)R if and only if X is an A(N)E. This is of fundamental importance. In the sequel we shall not always conscientiously refer to Theorem 1.2.7 when dealing with A(N)R's. The reader should keep this in mind. We first prove an important fact. Lemma 1.2.6. Every space can be imbedded as a closed subspace of some AE. Proof. This is easy. First observe that every space is homeomorphic to a closed subspace of some normed linear space (Corollary 1.1.8). Now apply Theorem 1.2.2. D We shall now present the announced characterization of A(N)R's. Theorem 1.2.7. Let X be a space. The following statements are equivalent: (1) X is an A ( N ) R , (2) X is an A(N)E. Proof. The implication (2) =>• (1) is trivial. For (1) => (2), let us assume that X is an ANR. The proof for AR's is entirely similar, and shall therefore be omitted. Let Y be a space, A C Y be closed, and /: A —>• X be continuous. By Lemma 1.2.6, we may assume that X is a closed subspace of some AE, say Z. Let G: Y —> Z be a continuous extension of /. Since X is a closed subspace of Z, it follows that X is a neighborhood retract of Z. So let U be a neighborhood of X in Z for which there exists a retraction r: U —>• X. Put V = G~l[U]. Since G[A] C X, V is clearly a neighborhood of A in Y. Let h denote the restriction of r o G to V (observe that h is well-defined since G[V] C U). Then h is continuous and extends / since for every y G A we have h(y) = r(G(y}} = f ( y ) . D Corollary 1.2.8. Every space is homeomorphic to a closed subspace of an AR. Proof. This follows directly from Lemma 1.2.6 and Theorem 1.2.7.

D

Corollary 1.2.9. Let C be a convex subset of a locally convex linear space. Then C is an AR. Proof. This follows from Theorems 1.2.2 and 1.2.7.

D

26

1. BASIC TOPOLOGY

Notice that Cauty's linear space mentioned in Remark 1.2.4 shows that the local convexity assumption in this corollary cannot be dropped. Proposition 1.2.10. A neighborhood retract of an ANR is an ANR. As a consequence, an open subspace of an ANR is an ANR. Proof. Let X be an ANR, A C X closed, U an open neighborhood of A, and r: U —)• A a retraction. In addition, let B be a closed subspace of a space Y and let /: B —> A be continuous. Since X is an ANR there are a neighborhood V of B in Y and a continuous extension g: V —> X of /. Then clearly W = g~l\U] is a neighborhood of A. The function f = ro g: W —> A is continuous and extends / since for every y G B we have

Since an open subspace of a space is clearly a neighborhood retract, the second statement is indeed a consequence of the first. D Proving that a given space is an AR or an ANR is usually a difficult task. We will come back to this in Chapter 4. For the moment we shall prove a few elementary results about AR's and ANR's only. Proposition 1.2.11. A product of countably many nonempty spaces is an AR iff all factors are. Corollary 1.2.12. R n , ln, Q and s are AR's. Proof. Apply Proposition 1.2.11 and Corollary 1.2.9.

D

n

Corollary 1.2.13. For each n > 0, § is an ANR. Proof. Let U = Rn+1 \{(0, 0, . . . , 0)}. Then U is an ANR by Corollary 1.2.12 and Proposition 1.2.10. The function r: U —>• §n defined by x T(T\ — { ' \ \l lXx \l l\

is clearly a retraction. Consequently, S n is a neighborhood retract of an ANR and is therefore an A N R itself (Proposition 1.2.10). D In Chapter 2 we shall prove that no §n is an AR. Proposition 1.2.14. The product of finitely many ANR's is an ANR. Proof. The simple proof is left as an exercise to the reader.

D

A product of a countably infinite number of ANR's need not be an ANR (in contrast to Proposition 1.2.11 on AR's). The following result gives more information on this.

1.2. EXTENDING CONTINUOUS FUNCTIONS

27

Theorem 1.2.15. Let Xn be a nonempty space for every n £ N. For the product X = n^Li Xn, the following statements are equivalent: (1) X is an A N R , (2) each Xn is an ANR and there is an n E N such that Xm is an AR for every m > n.

Proof. Since for arbitrary n, the product X can be factorized as

n^x n x» n

oo

i=l

i=n+l

the implication (2) =$• (1) follows from Propositions 1.2.11 and 1.2.14. For (1) =>• (2), first observe that each Xn can be viewed as a retract of X (cf. the proof of Proposition 1.2.11), hence Xn is an ANR for every n. Now take points xn G Xn, n e N, and let C^ be an AR which contains the space Xn as a closed subset (Corollary 1.2.8). Since X is an ANR and is closed in C = fl^Li Cn (Exercise A. 1.13), there is a neighborhood U of X in C for which there exists a retraction r: C7 -> X. Put x = ( x i , X 2 , . . .)• Since C/ is a neighborhood of x, there are an n G N and neighborhoods V^ of Xi in Cj for every i < n — 1 such that x e W = Vi x • • • x Vn-i x Cn x Cn+i x • • • C [/. Observe that by Proposition 1.2.11, the product ~[\^=nCm is an AR. Now define a function

by An easy check shows that s is a retraction, from which it follows that the product rim=n^™ is an AR. Now apply Proposition 1.2.11. D The following result enables us to create many new A(N)R's from old ones in yet another way. Theorem 1.2.16. Let X = X\ 11X2, where Xi and X% are closed in X, and let X0 =Xi nX2. Then (1) I f X 0 , Xl and X2 are A(N)R's then X is an A ( N ) R . (2) IfX and X0 are A(N)R's then Xi and X2 are A(N)R's. Proof. For (1), assume that XQ, X\ and X% are ANR's. We shall prove that X is an ANR. The proof of (1) for AR's is similar.

28

1. BASIC TOPOLOGY

Assume that X is a closed subset of a space Z. Our task is to construct a neighborhood U of X in Z and a retraction r:U-*X. To this end, define

respectively (cf. the proof of Lemma A. 8.1). It is clear that Z; n X = Xi for i G (0, 1, 2}, that Zi HZ 2 = Z0 and finally that Zi UZ 2 = Z. Also observe that the Zi are closed in Z. Since XQ is closed in ZQ there are a closed neighborhood WQ of XQ in ZQ and a retraction r0 : W0 —>• XQ. Now for i E {1, 2} define Ti : WQ U Xi —>• Xj by rT

i

-

z

- z

Observe that r^ is a retraction. By Theorem 1.2.7 there exists a closed neighborhood Vi of W0 U Xi in Zi such that r; can be extended to a continuous function fi-.Vi^Xi (te{l,2». It is clear that there exists for i 6 {1, 2} a closed neighborhood Ui of Xi in Zi such that Ui C Vi and Ui n ZQ C W0. Then E/i n E/2 C W0

from which it follows that the function f : UiUU2 -^ X defined by fi(z)

(z€Ui),

r2(z)

(z G C7 2 ),

is a well-defined retraction. Since U\ U U2 is a neighborhood of X in the space Z, we are done. For (2), let XQ and X be ANR's. Again, the proof for AR's is similar. Since XQ is an A N R , there are a neighborhood UQ of XQ in X and a retraction r: UQ ->• X0. Define f : Xi U UQ -» Xi by (x 6 A"i),

An easy check shows that r is a retraction. Since X\ U f/o is a neighborhood of Xi and X is an ANR, it now follows that Xi is an A N R as well. The proof for X2 is the same. D Exercises for §1.2. 1. Prove that every ANR is locally contractible. 2. Prove that if A ( X ) is an ANR then so is X.

1.3. FUNCTION SPACES

29

3. Prove Proposition 1.2.11. 4. Let X - {0} U {l/n : n e N}. Prove that A(X) is an example of a contractible space which is not an ANR. 5. Let A C S™" 1 . Observe that Bn \ A is a convex subset of W1 and hence is an AR. Present a direct proof of this. 6. Prove that each AR is path-connected. Prove that every ANR is locally path-connected. Observe that, in particular, every ANR is locally connected. 7. Give an example of a subspace X of R, such that X is an A N R , but XU{p} is not an ANR for some p G R. 8. Let X be the sin (y x )-continuum in the plane. Prove that X is not an ANR. ^•9. Let .4i and A^ be closed subsets of the locally convex linear spaces LI and 1/2, respectively. Prove that for each homeomorphism h: A\ —> AI there exists a homeomorphism H: LI x 1/2 —> LI x Lz such that H(a,0) = (0,/i(a)) for every a G A. 10. Let L be a linear space. In addition, let X be a space, A C X be closed and /: A —> L be continuous. Finally, let It and {au • U £ It} be a Dugundji system for X and A. For each x G X \ A let 8.(x) = {U <E U : x 6 U}.

Define a function F =>• L by

Ffx) = ^ ' ' conv({/(a[/) : U e £ ( x ) } )

(x£X\A).

Prove that F is LSC. 11. Use Exercise 1.2.10 to prove that a normed linear space satisfies the conclusion of the Dugundji Theorem 1.2.2. 12. Let L be a linear space. Assume that for every space X, every LSC function F: X => L such that for every x G X the set F(x) is compact and convex, has a continuous selection. Prove that every linear subspace of L is an AR. 1.3. Function spaces

The space C*(X) from Example 1.1.5 is a special case of a more general construction. Let X and Y be spaces and let Q be an admissible metric on Y. Define Ce(X,Y) = {/ € C(X,Y) : diaine(/[X]) < oo}. It is sometimes convenient to refer to the elements of Ce(X, Y} as bounded functions. We shall endow CB(X, Y) with a useful topology.

30

1. BASIC TOPOLOGY

In §A.2 we observed that @ need not be a metric on C(X, Y). Fortunately, on Ce(X,Y) it is a metric. Lemma 1.3.1. Let X and Y be spaces and let Q be an admissible metric on Y. Then (1) for all /, g 6 Ce(X, Y) we have £(/, g) < oo, (2) the function g: Ce(X,Y) x Ce(X,Y) -> [0, oo) is a metric. Proof. For (1), take an arbitrary point z G X and observe that for all functions /, g G Ce(X, Y) the following holds: Q(f,9) < diam,(/pq) + Q ( f ( z ) , g ( z ) ) + diam, (g(X)) < oo. The proof of (2) is routine and is left as an exercise to the reader.

D

From now on we shall endow CQ(X, Y} with the topology induced by Q. Let us emphasize that the set Ce(X, Y} as well as its topology depend on the choice of the metric Q. It is a natural question to ask when the choice of Q is irrelevant. Lemma 1.3.2. Let X and Y be spaces with X compact. In addition, let Q\ and Q-2 be admissible metrics for Y. Then (1) Cei(X,Y)=Cea(X,Y)=C(X,Y), (2) the topologies on C(X,Y] induced by QI and £2 are the same. Proof. (1) is trivial. For each e > 0, y e Y and i G {1, 2} we put Bi(y,e) = {zeY : Qi(y,z) < e}. For (2), take / 6 Cei(X,Y) and e > 0, arbitrarily. Our aim is to prove that there exists 6 > 0 such that {g € CQ2(X,Y) : g2(f,g)

< 6} C {g e CB1(X,Y) : ^ ( f . g ] < e},

i.e., that the ^2-ball about / with radius 6 is contained in the ^i-ball about / with radius e. To this end, observe that since f[X] is compact, the open cover has a 02-Lebesgue number, say 6 (Lemma A. 5. 3). We claim that this 8 is as required. To see that this is indeed the case, take an arbitrary g 6 CQ2 (X, Y} such that Qi(f,g) < 5. For each x e X we have 02(f(x),g(x)') < 5, so there exists px e X such that { f ( x ) , g ( x } } C J3i(/(p z ), e/2). Consequently, for each element x G X we have Qi(f(x},g(x)) <e, from which it follows by Exercise A. 5. 4 that Qi(f,g) < £•

1.3. FUNCTION SPACES

31

So we conclude that the topology on C(X, Y) induced by £2 is finer than the topology on C(X,Y) induced by 0i. By interchanging the roles of g\ and £2 m the above argument we find that the induced topologies are indeed the same. D Let X and Y be spaces with X compact. From the above lemma we conclude that all the topologies we defined on the set C(X, Y) coincide. So for compact X and any (Y,g) we shall denote the space Ce(X,Y) simply by C(X, Y). The topology on C(X,Y) is called the topology of uniform convergence. Observe that the norm topology on C(X) for compact X defined in Example 1.1.5 coincides with the just defined topology on C(X, R). On Page 19 we defined the so-called compact-open topology on C(X}. This is again a special case of a more general construction. Indeed, let X and Y be spaces and for an arbitrary compact subset K in X and an arbitrary open subset U in Y define [K,U] =

{feC(X,Y):f[K]CU}.

Topologize C(X, Y) by taking the collection {[K, U] : K C X compact and U C Y open} as an open subbase. This topology is called the compact-open topology on the set C(X,Y). We will now show that for compact X and arbitrary Y the compact-open topology on C(X, Y) coincides with the topology of uniform convergence on C(X, Y). This allows us to prove quite easily that C(X, Y) is separable. Proposition 1.3.3. Let X and Y be spaces with X compact. The topology of uniform convergence on C(X, Y) coincides with the compact-open topology on C(X,Y). As a consequence, C(X,Y} is separable. Proof. Let g be an admissible metric on Y. In addition, let K C X be compact and U C Y open. If / G [K, U] then f[K] is a compact subset of U. By Corollary A.5.4 there exists 8 > 0 such that

B(f[K\,S)CU. This clearly implies that if g G C(X, Y} is such that £>(/, g) < 6 then g[K] is contained in U, i.e., {geC(X,Y):Q(f,g)<6}C[K,U}. From this we conclude that [K, U] is open in the topology of uniform convergence on C(X, Y) and hence that the topology of uniform convergence is finer than the compact-open topology.

32

1. BASIC TOPOLOGY

For the converse, let !B and £ be countable open bases for X and Y, respectively, which are both closed under finite unions. For B € *B and E € £ put By the above, each A(B,E) is open in the topology of uniform convergence on C(X,Y). Let A be the (countable) collection of all A(B:E)'s. We claim that the family A* of all finite intersections of elements of A is an open base for C(X, Y) endowed with the topology of uniform convergence. This proves on the one hand that compact-open topology is finer than the topology of uniform convergence and on the other hand that that C(X, Y) has a countable base. Let / G C(X,Y) and e > 0. We shall prove that there exists an element F € A* such that / G F C {g G C(X,Y) : g(f,g) < e}. By compactness of the set f[X], there are finitely many elements of £, say E\ , E-z, • • • •> En, such that (1) (2) for every i < n, diam(Ei) < e.

Let U = {f~l[Ei\ : i < n}. Since "B is a base, there clearly is a a cover V of X consisting of elements of 23 such that V < U. By compactness of X, we may assume that V is finite. For each i < n let Wi be the union of the elements of V the closures of which are contained in f~1[Ei], Since "B is closed under finite unions, W = {Wi : i < n} is a subcollection of !B, "W covers X, and W has the property that the closure of each Wi is contained in f~1[Ei\. (This is so since Wi is a finite union of sets the closures of which are contained in f-^Ei].) Now put

It is clear that / G F. In addition, F is open in the topology of uniform convergence. We claim that F C {g G C(X,Y) : £(/, 5) < e}. To this end, take g G F and x G X. There exists i < n with x G Wi. Since /, g G F, it follows that f[Wi] U g[Wi] C Ei. Consequently, both f(x) and g(x) belong to Ei from which we get by (2) that g(f(x},g(x)] < £. So we are done. D This result can be used to estimate the number of continuous functions. Corollary 1.3.4. Let X and Y be spaces with X compact. Then C(X,Y) has cardinality at most c. Proof. This follows from Proposition 1.3.3 and Exercise A. 2. 16. We now turn to completeness properties of C(X,Y) and CS(X,Y).

D

1.3. FUNCTION SPACES

33

Proposition 1.3.5. Let X and (Y, g) be spaces. Let ( f n } n be a g-Cauchy sequence in C(X,Y] such that for every x G X, lim^^oo fn(x) exists. Then the function f : X -> Y defined by f(x] — limn^oo / n (x) is continuous. In addition, if fn G Ce(X, Y) for every n then f G Cg(X, Y} and f = lim™-^ fn (inCe(X,Y)). Proof. For the first part of the proposition, see Lemma A. 3.1. For the remaining part, suppose that fn G CS(X, Y) for every n. We shall prove that diam(/[A"]) < oo. By Claim 1 in the proof of Lemma A. 3.1 there exists an M G N such that for every x G X, g ( f ( x ) , /M(^)) < 1- Take arbitrary x,z G X. Then g(f(x)J(z})

< g ( f ( x ) , f M ( x ) ) + Q(fM(x], f M ( z ) ) +

g(fM(z}J(z))

< 2 + diam(/ M [A]), so diam(/[A"]) < oo. It remains to prove that / = limTn.00 fn (in C8(X,Y)). again easily from Claim 1 in the proof of Lemma A. 3.1.

But this follows D

Corollary 1.3.6. Let X and (Y, g) be spaces. Then g is complete if and only if g is complete. Proof. Suppose that (Y, g) is complete and let (fn}n be a 0-Cauchy sequence in Ce(X,Y). Fix z G A" arbitrarily and let e > 0. There exists A7" G N such that g(fm fm) < £ for all n,ra > N. Since for all n and m,

we conclude that (fn(z)) is Cauchy in (Y,g). The completeness of (Y, g) and Proposition 1.3.5 now yield that the sequence (/ n )n converges. Now assume that g is complete. Let (yn)n be a ^-Cauchy sequence in Y. For each n let fn : X -> Y be the constant function with value yn. It is easy to see that ( f n ) n is a ^-Cauchy sequence in Ce(X,Y). So / = limn_^oo fn exists and belongs to Ce(X,Y). It is trivial to prove that / is constant and that the sequence (yn)n converges to the unique point in the range of /. D Corollary 1.3.7. Let X be compact and Y topologically complete. the space C(X,Y) is topologically complete.

Then

Groups of homeomorphisms. We will now restrict our attention to spaces of homeomorphisms and will derive some basic properties of them. Let X and Y be spaces and define §(A,F) = {/ G C(X,Y) : f is surjective}. There are spaces X and Y for which §(A", Y) is empty, for example this is the case if X has smaller cardinality than Y.

34

1. BASIC TOPOLOGY

Proposition 1.3.8. Let X and Y be spaces with X compact. Then §>(X, Y} is closed in C(X, Y). Proof. Take an arbitrary / ^ S(X, Y). There exists a point y G Y \ f[X]. So fe[X,Y\{y}]CC(X,Y)\S(X,Y). Since [X, Y \ {y}] is open in C(X,Y) (Proposition 1.3.3), this shows that

is a neighborhood of /. Hence C(X, Y) \ B(X, Y} is open in C(X, Y}.

D

Let X and Y be spaces with X compact, and let e > 0. A continuous function /: X —> Y is called an e-map if for every y G Y, s.

Let e > 0 and put

CE(X, Y) = {/ G C(X, Y) : f is an e-map} and

Se(X,Y)=Ce(X,Y)nS(X,Y), respectively. In addition, put

Lemma 1.3.9. Let X and Y be spaces with X compact. Then C£(X,Y) is open in C(X, Y) for every e > 0. Consequently, S£(X, Y} is a closed subset ofC(X,Y). Proof. Take / G C£(X,Y). Since X is compact, /: X ->• f[X] is a closed map (Exercise A. 5. 5). We claim that for every y G f[X] there exists an open neighborhood Uy (in /[-X"]) such that diam(f-l[Uy})

<e.

This will be achieved in two steps. Take an arbitrary y G f[X]. Then diam(f-l(y))

<e

since / is an £-map. There clearly exists an open neighborhood U of f ~ l ( y ) such that diam(C7) < e. Now by using that / is a closed map, Exercise A. 1.1 5 gives us the required neighborhood Uy. Let 6 > 0 be a Lebesgue number for the open covering

{Uy-.yz f[X}} off[X] (Lemma A. 5. 3). Let g G C(X, Y) be such that g(g, /) < %. We claim that g G C£(X, Y). This clearly suffices. To this end, take an arbitrary y G Y. Since we have 0(f,g) < 5/2 it follows easily that diam (fg~1(y)) < 5. There

1.3. FUNCTION SPACES

35 l

consequently exists a point z 6 f[X] such that fg~~l(y) C Uz. This implies that Since g~1(y) C f~1fg~1(y), we conclude that diam (g~1(y)} < £•, i.e., g is an e-map. That $ £ ( X , Y ) is closed now follows by Proposition 1.3.8. D Let X and Y be spaces. We introduce a few more interesting subsets of C(X, Y). Let 3(X, Y) denote the subset of C(X, Y) consisting of all imbeddings of X into Y , and let 9t(X, Y) denote the set of all homeomorphisms from X onto Y. If X = Y then for H ( X , X ) we shall simply write As usual, ^K(X) is called the autohomeomorphism group of X. Lemma 1.3.10. Let X and Y be spaces with X compact. Then

As a consequence, 3(X,Y) is a G^-subset o f C ( X , Y ) , and 3{,(X,Y) is a G$ subset of both C(X, Y) and S(X, Y). Proof. That 3(X,Y) C f}™=l Cif (X,Y] is a triviality. Pick an arbitrary

Then / is a 1/n-map for every n, hence / is one-to-one. So the compactness of X implies that / is an imbedding (Exercise A. 5. 9). The remaining statements are obvious. D Corollary 1.3.11. Let X and Y be spaces with X compact and Y topologically complete. Then both 3(X, Y} and 'H(X, Y) are topologically complete. Proof. Since C(X,Y) is completely metrizable by Corollary 1.3.6, this follows immediately from Lemma 1.3.10 and Theorem A. 6. 3. D So 'K(X) is complete if X is compact. It will be convenient to explicitly describe a complete metric for "H(X) that generates its topology. Proposition 1.3.12. Let X be a compact space. For /, g 6 'K(X)

define

Then a is a complete metric on "H(X) that generates its topology. Proof. That a is a metric is left as an exercise to the reader. We shall first prove that Q and a generate the same topology on J{,(X}.. Since for

36

1. BASIC TOPOLOGY

all /, g E 3t(X] we have £>(/,#) < cr(/, 0 and every / E Oi(X) there exists S > 0 such that if g E K(X} and g(f,g) < 6 then a(f,g) < E. Choose arbitrary e > 0 and / E ^K(X). By compactness, /-1 is uniformly continuous (Exercise A.5.18) and consequently there exists 7 > 0 such that for all x,y E X with g ( x , y ) < 7 we have g ( f ~ l ( x ) , f ~ l ( y } } < £/2- Let Take g 6 3{(X) such that Q(f,g) < S. Pick an arbitrary x e X and put z = g~l(x). Since £>(/, g) < 6, it follows that Q(f(z),g(z))

= Q(fg-l(x},x}

< 6 < 7.

As a consequence, We conclude that 6(g~1,f~i) < Eli by Exercise A. 5.4. Therefore,

Now let (fn)n be a cr-Cauchy sequence in "H(X}. Then (fn)n is a QCauchy sequence in C(X, X) and therefore the limit / = limn_>oo fn exists and belongs to C(X, X] (Corollary 1.3.6). Similarly, by the definition of cr, the limit g = limn^.oo f~l exists and belongs to C(X,X). It is easily seen that fog = lx=gof from which it follows that / 6 'H(X). D Lemma 1.3.10 and Proposition 1.3.12 both imply by Theorem A. 6. 6 the following Corollary 1.3.13. If X is compact then 3i(X) is a Baire space. Exercises for §1.3. 1. Let N denote the discrete space of natural numbers. Prove that C(N, K; |-|) is not separable. 2. Let X, Y and Z be compact spaces. For / € C(Z, X) and g,h£ C(X, Y) prove that Q(g° f,ho f ) < g(g,h). In addition, show that if / is surjective then

3. Let X be compact, let /, g G C(X, X) such that g is a homeomorphism. Prove that

1.4. THE BORSUK HOMOTOPY EXTENSION THEOREM

37

4. Let X be a compact space. Prove that the function f : M(X) x :K(X)-» K(X) defined by ^(/,S) = /o 5 " 1 is continuous (i.e., 'K(X) is a topological group). 5. Prove that the function /: I —>• I defined by (0 < x < V 4 ),

x

belongs to the closure of IK(I) in C(I,I). (Hence IK (A") is even for compact A not necessarily a closed subspace of C(X, A).) 6. Prove that IK (I) has exactly two components. 7. Give an example of two nontrivial continua A and Y such that S(A", Y) is empty. 1.4. The Borsuk homotopy extension theorem The aim of the present section is to prove that a continuous function / from a closed subspace A of a space X into an A N R Z is extendable over X if and only if / is homotopic to an extendable function g: A —> Z. We shall need the following simple lemma: Lemma 1.4.1. Let A be a closed subset of a space X. Then for every neighborhood V of B — (X x {0}) U (A x I) in X x I there is a continuous function a: X x I —>• V which is the identity on B. I U

V 0

x

A

Figure 1. Proof. Let TT: X x I ->• X be the projection. Then by compactness of I, TT is closed by Exercise A.5.8. We may assume without loss of generality that V is open. So F = (X x I) \ U is a closed set which misses A x I. Hence

U = X\ 7t[F}

38

1. BASIC TOPOLOGY

is an open neighborhood of A in X such that U = U x I C V . Now let ft: X ^l be a Urysohn function (Lemma A. 4.1) such that 0\A=1,

0\X\U = Q.

Define a: X x I -»- X x l b y a(x,t) = (x,{3(x) • t). Then a. is clearly continuous. It is easily seen that it restricts to the identity on B. We shall prove that a[X x I] C V, thereby showing that a is as required. To this end, take (x, t) e X x I arbitrarily. If x £ U then j3(x) = 0 and consequently, a(x,t) = (x,0) 6 B C V. On the other hand, if x £ U then a(z, f) = (x, /3(:r) • t) € U x I C V . D We now come to the main result in this section. The Borsuk Homotopy Extension Theorem 1.4.2. Let A be a closed subspace of a space X, let Z be an A N R and let H : A x I —> Z be a homotopy such that HQ is extendable to a continuous function f : X —> Z. Then there is a homotopy F : X x I -> Z such that (1) ^o = /, (2) for every tel,Ft \A = Ht.

Proof. Using the notation as in Lemma 1.4.1, define a function £: B —>• Z

by

^,t) = lf(x) \H(x,t)

(»^,t = 0), (x e A,tei).

Since X x {0} and Axl are closed in B, it follows easily that £ is continuous. As Z is an ANR and as B is closed in X x I, we can find a neighborhood V of B in .X x I such that £ can be extended to a continuous function £' : F —> Z. Let a : X x I -> V be as in Lemma 1.4.1. Define F : X x I ->• Z by the formula

It is easily seen that F is as required.

D

Theorem 1.4.2 has some interesting corollaries. Corollary 1.4.3. Let A be a closed subset of a space X . Let Z be an A N R and let f : A —>• Z be continuous. The following statements are equivalent: (1) / can be extended over X, (2) / is homotopic to an extendable function g: A —>• Z.

1.4. THE BORSUK HOMOTOPY EXTENSION THEOREM

39

Proof. That (1) => (2) is clear since / ~ / by Lemma A. 12.1(1). For (2) =>• (1), let H : A x I -> Z be a homotopy with H0 = g and HI = f . Since g is extendable, / is extendable as well (Theorem 1.4.2). D Our next goal is to investigate contractibility within the class of ANR's. Theorem 1.4.4. Let X be an AR. Then X is contractible. Proof. By Corollary 1.1.8, X can be thought of as a closed subset of a normed linear space L. Fix an arbitrary point p £ L. Since the function F : L x I —>• L defined by the formula ( x , t ) \->t-p+(l-t)-x is a contraction, it follows that X is contractible, being a retract of L (Theorem A. 12.4). D These results yield the following interesting characterization of AR's. Corollary 1.4.5. Let X be a space. The following statements are equivalent:

(1) X is an AR, (2) X is a contractible A N R . Proof. The implication (1) =$• (2) is trivial by Theorem 1.4.4.

For (2) =» (1), let H: X xl->X be a homotopy such that HQ = 1 and HI is constant, say with constant value c. In addition, let Y be a space, A C Y be closed and /: A —)• X be continuous. Define F: A x I —>• X by Then F is clearly a homotopy with F0 — f and FI a constant function. Trivially, FI can be extended to a continuous function from Y to X. As X is an A N R , by Corollary 1.4.3 it follows that FQ = f can be extended over Y. We conclude that X is an AR. D For a nontrivial generalization of Corollary 1.4.5, see Theorem 4.2.20. Cones. Let X be a space and consider its cone A(JC). We observed in Exercise 1.2.2 that if A(X) is an ANR, then so is X. It is an interesting question whether the converse holds. Theorem 1.4.6. For a space X the following statements are equivalent:

(1) X is an ANR, (2) A(X) is an A N R ,

40

1. BASIC TOPOLOGY

(3) A (AT) is an AR.

Proof. Observe that (3) =>• (2) is trivial, and (2) =>• (1) follows from the above observation. So we only need to prove the implication (1) =>• (3). To this end, first observe that we may think of X as a closed and bounded subspace of a normed linear space L (Corollary 1.1.8). The cone A (A") is by Exercise 1.1.29 homeomorphic to the subspace

K = {(t,tx)

:tel,xeX}

of I x L (for notational simplicity, we switched the coordinates of L and II , and applied the homeomorphism t \-> 1 — £ of I). Since X is an A N R , there are an open neighborhood U of X in L and a retraction r : U —> X. Let V be an open neighborhood of L \ U such that V n X = 0 (Corollary A. 4. 3). By Lemma A. 4.1 there is a continuous function a: L —>• I such that a \ X = 1 and a \ V = 0. Define a function /3 : I x L —>• K by

t),ta(Vt)r(*/t))

(otherwise).

We claim that (3 is a retraction from I x L onto /f . This suffices for the proof, since clearly II x L is an AR (Corollary 1.2.9 and Proposition 1.2.11) and a retract of an AR is an AR. Notice that /2(0,Q) = (0,0). In addition, if x G X and t G II \ {0} are arbitrary then tx/t = x E U and so fi(t,tx) = (ta(x),ta(x)r(x))

= (t,tx).

This proves that j3 \ K is the identity. So all there remains to prove is that (3 is continuous, and this is left to the reader in Exercise 1.4.2. D Notice that the implication (2) =$> (3) also follows from Corollary 1.4.5. Simply observe that A(AT) is contractible (Exercise A. 12. 5).

Exercises for

§1.4.

1. Let A be a closed subspace of a space X and let T be the subspace (X x {0}) U (A x I)

of X x I. Prove that for every space Y, if a continuous function /: T —> Y can be extended over (X x {0}) U t/, where U is a neighborhood of A x I in X x I, then / can be extended over X x I. ^•2. Prove that the function /3 in the proof of Theorem 1.4.6 is continuous.

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

41

1.5. Topological characterization of some familiar spaces The aim of this section is to present topological characterizations of two familiar spaces, namely, the Cantor middle-third set C and the closed unit interval I. (See §1.9 for topological characterizations of Q and P.) Topological characterization of the Cantor set. A space X is called zero- dimensional if it is nonempty and has a base consisting of clopen sets, i.e., if for every point x E X and for every neighborhood U of x there exists a clopen subset C C X such that x G C C U. It is clear that a nonempty subspace of a zero-dimensional space is again zero-dimensional and that products of zero-dimensional spaces are zero-dimensional. Observe that by Exercise A. 2. 12 it follows that a space is zero-dimensional if and only if it has a countable base consisting of clopen sets. It is clear that no nontrivial connected space is zero-dimensional. As a consequence, if a space contains a nontrivial connected subspace then it is not zero-dimensional. Proposition 1.5.1. Let X be zero- dimensional. Then for every open cover U of X there exists a refinement V of U consisting of pairwise disjoint clopen sets. Proof. Let U be an open cover of X. Since X is zero-dimensional, it can be refined by a clopen cover £. We may assume without loss of generality that £ is countable, say £ = {En : n 6 N} (Corollary A. 2. 3). Now put Vi = EI and put Vn = En\[j Ei i
for every n > 2. Then V = {Vn : n 6 N} is as required.

D

Corollary 1.5.2. Let X be a space. Then X is zero-dimensional if and only if for all disjoint closed sets A and B there exists a clopen subset U C X such that A C U and B n U = 0. Proof. First assume that X is zero-dimensional. For every x £ X let Ux be a neighborhood of x such that Ux D A = 0 or Ux fl B = 0 (here we use that A and B are disjoint closed sets). By Proposition 1.5.1 there is a refinement V of {Ux : x € X} consisting of pairwise disjoint clopen sets. Now put = \J{ V G V:

It is easy to see that U is as required. If for all disjoint closed subsets A and B of X there exists a clopen set C with A C C C X \ B, then the clopen subsets of X clearly form a base. So then X is zero-dimensional. D

42

1. BASIC TOPOLOGY

We shall now characterize the zero-dimensional subspaces of the real line R. Proposition 1.5.3. A nonempty subspace X of R is zero-dimensional if and only if it does not contain any (nondegenerate) interval. Proof. Assume that X C R is zero-dimensional. If X contains a nontrivial interval E then X contains a nontrivial connected subspace and is therefore not zero-dimensional. Now assume that X C R is such that it contains no nontrivial intervals. Then the set D = R \ X is dense in R. The collection {(di,d 2 ) r\X : di < d2 and c?i,d 2 € D} is easily seen to be a base for X consisting of clopen subsets of X. Consequently, X is zero-dimensional. D This proposition gives us a rich supply of zero-dimensional spaces. For example, the rational numbers Q, the irrational numbers P, the product QxP, etc., are all zero-dimensional. The following example is of particular interest: Example 1.5.4. The Cantor middle-third set C. From I = [0,1] remove the interval (Y 3 , 2/3), i.e., the 'middle-third' interval. From the remaining two intervals, again remove their 'middle-thirds', and continue in this way. At stage i of the construction we have a disjoint family 3^ of 2l closed subintervals of I, each of length 3~ l . The union of 3~; is denoted by Hi. Observe that Hi is a closed subset of I and hence is compact. The intersection of the H^s is called the Cantor middle-third set, C. Clearly, C is closed in I and hence compact. Also, C is zero-dimensional by Proposition 1.5.3. For if (a, b) is a nondegenerate subinterval of R then there exists i £ N such that 3~z < b — a which implies that (a, b) cannot be contained in C since C can be covered by a disjoint family of closed intervals each of length 3~ z .

Figure 2.

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

43

Observe that if F 6 ^ then F n C / 0. For at stage i + 1 of the construction there exists an interval AI £ 3^+1 such that A\ C F. In addition, at stage i + 2 of the construction, there exists an interval AI e 3^+2 such that AI C AI. Proceeding in this way inductively, it is easy to construct a decreasing sequence (An}n of elements of 3*i+n such that An C F for every n. By compactness of I, 0 / fl^Li An C C, as desired. It is easy to see that C is the subspace of I consisting of all points that have a triadic expansion in which the digit 1 does not occur, i.e., the set oo

x=

\

51 ^T : Xi

G

{°' 2} for every *'

We claim that the set C has no isolated points. If x belongs to C and a < x < b then choose i so large that 3~z < min{;r — a, b — x}. Take F G 3^ with x & F. Then F C (a, 6). There are precisely two disjoint elements of 3~i+i that are contained in F. By the above, both these elements contain at least one point of C. We conclude that (a, b) D C contains at least two points. The Cantor set C is a very interesting space. There exists a simple topological characterization of it which we shall now present. Theorem 1.5.5. C is (topologically) the unique nonempty zero-dimensional compact space without isolated points. Proof. Since in Example 1.5.4 we showed that C is a zero-dimensional compact space without isolated points, it suffices to prove that if X and Y are such spaces then they are homeomorphic. To this end, let X and Y be compact zero-dimensional spaces without isolated points. By induction on n, we shall construct elements m(n) G N and finite disjoint clopen covers Un = {Un^ : i < m(n)},

V = {Vn^ : i < m(n}}

of X and Y, respectively, consisting of nonempty sets, such that and for n > 2,

(1) mesh(U n ) < Vn, meshCV n ) < Yn, (2) U n 1. Pick an arbitrary i < m(ri). By Proposition 1.5.1 there exist covers £ and 3r of C/n)i and Vn>i, respectively, consisting of pairwise disjoint clopen sets, such that (4) £ <

(5) y-

44

1. BASIC TOPOLOGY

By compactness of Un,i and Vn,i, the covers £ and 3~ are finite. We assume that each element of £ as well as 3" is nonempty. We claim that without loss of generality we may assume that £ and 3" have the same cardinality. To see this, pick an arbitrary E £ £. Since X has no isolated points, there exist distinct x,y E E. In addition, since X is zero-dimensional, its clopen sets form a base and so there is a clopen subset C C E such that x E C and y $. C. So E can be split into two nonempty clopen sets, namely, C and E\C. So we can replace £ by a cover having |£| + 1 elements. By repeating the same procedure if necessary, we can therefore ensure that £ and 3~ have the same cardinality. We conclude that for i < m(n) there are covers £; and 3^ of Un^ and Vn,i, respectively, such that (6) |£t| = |^|,

(7) 8,i and 3^ are pairwise disjoint and consist of nonempty clopen sets, each of diameter less than l/n. Put Un+i — LJl^i ^"i and Vn_|_i = U™LI 3V By construction, we can enumerate lin+i and Vn+i in such a way that (3) is satisfied for n + l. This completes the inductive construction. Observe that IJ^Li ^ *s a base for X and, similarly, that U^Li "^n ^s a base for Y (Exercise A.2.15). We will now use these covers for the construction of the required homeomorphism. If x E X then for each n E N there is a unique i(n,x) < m(n) such that x E C/ n ,i(n,x)- By (3) it follows that the collection {Vn^n^}n is decreasing and since Y is compact, we get H^Li Ki,i(n,z) 7^ 0- So by (1), the intersection H^Li ^n,i(n,z) consists of precisely one point. By interchanging the roles of X and F, these remarks imply that the function /: X —> Y defined by f ( x ) = y <£=^ for every n e N and i < m(n)

(x 6 C7n)i iff y G V^^),

is a bijection. Also / is continuous since for n e N and i < m(n), f~l[Vn>i] is equal to C/^^. Similarly, f~l is continuous. So we conclude that / is a homeomorphism. D A space homeomorphic to C is called a Cantor set from now on. The characterization of C allows us to recognize many spaces as Cantor sets. Corollary 1.5.6. Let X be any compact nonempty zero-dimensional space. Then X x C is a Cantor set. Proof. Since both X and C are zero-dimensional it easily follows that X x C is compact and zero-dimensional. Since C has no isolated points it follows that .Y x C has no isolated points. An appeal to Theorem 1.5.5 now yields the desired result. D

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

45

Since {0,1}°° is compact and zero-dimensional, and has clearly no isolated points, we conclude that, in particular, C « {0,1}°°. This consequence can also be verified directly: the function /: {0,1}°° —>• C defined by 00

£1 \

\~^

C)

X

i

/(*) = L 6 t=l -^ is a homeomorphism (see Page 43). Corollary 1.5.7. Every zero-dimensional space can be imbedded in C. Proof. Let X be zero-dimensional and let U be a countable base for X consisting entirely of clopen sets. Let {t/n : n 6 N} enumerate U and for every n define fn: X -> (0,1} by fn(x) = 0 <^ x e Un.

Now define an imbedding e: X —> {0,1}°° by putting e(x)n = fn(x]

(n G N)

(this is in fact the same imbedding as in the proof of Corollary A.4.4). Since by the above C « {0,1}°°, this completes the proof. D So every zero-dimensional space can be imbedded in R. This fact admits a nontrivial generalization that will be proved in §3.3. By Cantor's standard diagonal argument, it follows that {0,1}°° is uncountable. So C is uncountable as well. (To be precise, its cardinality is c.) Since by Corollary 1.5.6, C is homeomorphic to C x C, this remark implies that there is a family A consisting of uncountably many pairwise disjoint Cantor sets in R. (Let < ^ : C x C —> C b e a homeomorphism and put A — {^[{c} x C] : c € C}.) Since there are only countably many rational numbers in R, one of the elements of A misses Q, i.e., there exists a Cantor set K in R consisting entirely of irrational numbers. Let P denote the set of irrational numbers in R. Since as was observed in Corollary 1.5.7, every zero-dimensional space can be imbedded in C, we consequently get: Corollary 1.5.8. Every zero-dimensional space can be imbedded in P. We now aim at proving that for every compact space X there exists a continuous surjection from C onto X. Lemma 1.5.9. The function f : C x C —> J denned by /(x,y) = x — y is a continuous surjection. Proof. Since C C E, / is well-defined and clearly continuous. Consider the set FI — [0, 1/3] U [2/3,1], i.e., the first approximation to C. It is a triviality to verify that FI — FI, i.e., the set {x — y : x, y G FI}, is equal to J. By induction

46

1. BASIC TOPOLOGY

one can prove that all approximations Fn, n G N, to C that were used to define C, have the same property (the induction step can be performed by noting that Fn — (l/3-Fn-i)\j(l/3- Fn-i + 2/3), a union of two shrunken copies of Fn-i.}. Since the Fn decrease and C = H^Li Fn, an easy compactness-type argument shows that C-C-JJ, as required.

D

We now come to the announced Theorem 1.5.10. Every compact space is a continuous image of C. Proof. By Lemma 1.5.9 and Exercise A.1.13 it follows that there exists a continuous surjection /: C°° —>• Q. Now let X be any compact space. According to Corollary A.4.4, we may assume that X C Q. Put Y — f ~ l [ X } . The restriction g = f\Y:Y->X is a continuous surjection. Observe that Y is zero-dimensional. By Corollary 1.5.6, C x Y and C are homeomorphic. The required continuous surjection a: C —>• X can therefore be obtained as the composition

a = (3 o TT o g, where TT: C x Y —> Y is the projection and j3: C —>• C x Y is an arbitrary homeomorphism. D Remark 1.5.11. Theorem 1.5.10 can be used to construct so-called 'space filling curves', i.e., continuous surjections from I onto E2. For let g: C —> I 2 be a continuous surjection (Theorem 1.5.10). Then, by Corollary 1.2.12, g can be extended to a continuous surjection g: I —»• E 2 . The Cantor set can be imbedded in 'many' spaces. Since it is uncountable, it cannot be a subspace of any countable space. There are also examples of uncountable spaces of which it is not a subspace, but it is not so easy to construct such a space. That such spaces exist follows however from Corollary 1.5.14 below. Before we present the next theorem, we first fix some notation. By S we mean the set consisting of all finite sequences consisting of O's and 1's. (This includes the empty sequence.) If s G £ then \s\ denotes its length, i.e., if s = 0 then \s = 0 and if s = i\ii... in, where ij G {0,1} for each j < n, then \s\ — n. If s — i\i^ . • -in G S and k G (0,1} then s^k G S denotes the sequence i\i<± • • • ink. If s G S U {0,1}°° then t G S is an initial segment of s if there is an n G N such that t is equal to the sequence consisting of the first n elements of s. Finally, if s G Eu{0,l}°° then s extends t G S if t is an initial segment of s.

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

47

Theorem 1.5.12. Let X be topologically complete and let f : X —>• Y be a continuous surjection. If Y is uncountable then X contains a Cantor set K such that f [ K is one-to-one (and hence an imbedding). Proof. Let Q be an admissible complete metric on X which is bounded by y2 (Exercise A. 1.6). Since Y is uncountable, X contains an uncountable subset on which / is one-to-one. This set contains an uncountable dense-in-itself subspace, say E (Exercise A.2.13). For s G £ we shall define a nonempty open subset Us of X having the following properties:

(1) t / s n £ ^ 0 ,

_

(2) if t 6 E and t extends s then U± C f/ s ,

(3) /[^ 0 ]n/[t7 s ^] = 0, (4) diamt/ s < 2 M . We put £70 = X. Now suppose that we constructed Us for all s 6 S such that |s| < n. Fix an arbitrary s G E such that |s| = n. By (1) and the fact that E is dense-in-itself, we may pick two distinct points x,y e Usr\E. There exists e > 0 such that D(x,e) U D(y,e) C Us. Since f(x] ^ f(y) and / is continuous, we may additionally assume that f[D(x,e)] r\f[D(y,e}] — 0 and that e < 2' s l. Define Us~0 — B(x,e) and Us~1 — B(y,e), respectively. It is easily seen that our choices are as required. Now for s G {0,1}°° put As = \\{Ut '• s extends t}. By (1), (2) and (4) and the fact that Q is complete it follows easily that As consists of a single point of X, say xs. Define a function £: {0,1}°° —> X by

f(s) =xs. We claim that £ is an imbedding. That £ is one-to-one is clear since (3) implies that / o £ is one-to-one. By Exercise A.5.9 it therefore suffices to prove that £ is continuous. Since by Exercise A.2.15 the collection [Us : s 6 S} is a local base at every point of the range of £, it suffices to prove that £-1[t/t] is open for every t G S. But this is clear since ^[Ut] = {s e {0,1}°° : s extends t} is a basic clopen subset of (0,1}°°.

D

Since the cardinality of the Cantor set is c, we immediately obtain from this that 'for topologically complete spaces the Continuum Hypothesis holds'. By this we mean that a topologically complete space cannot be of cardinality strictly between uj and c. (Observe that by Exercise A.2.16 it also cannot be of cardinality larger than c.)

48

1. BASIC TOPOLOGY

Corollary 1.5.13. Let X be an uncountable analytic space. Then X contains a Cantor set. As a consequence, if X is analytic then either X\ < uo or \X\ = c. Proof. Let Y be topologically complete, and let /: Y -» X be a continuous surjection. If \X\ < uj then there is nothing to prove. So assume that X is uncountable. Then Y is uncountable, and hence by Theorem 1.5.12 there exists a Cantor set K C Y such that / \K is one-to-one, and hence a topological imbedding (Exercise A.5.9). From this we get \X\ > c. Finally, \X\ < c by Exercise A.2.16. D This result enables us to prove the following interesting result. Corollary 1.5.14. Let X be a space. Then we can decompose X into two sets neither of which contains an uncountable topologically complete subspace (in particular, neither contains an uncountable compact subspace). Proof. If \X\ < c then by Corollary 1.5.13 there is nothing to prove. So assume that \X\ = c. First observe that the number of closed subsets of X is at most c (Exercise A.2.16). So there are also at most c closed subsets of X of cardinality c and we can list them as {Aa : a < A} (no repetitions permitted) for some cardinal A < c. By transfinite induction on a < A we will pick distinct points xa,ya E Aa such that {x<*,ya} n {xp,yp : /3 < a} = 0. This is easily done since each \Aa\ = c and the number of points that has to be avoided at stage a is smaller than c. Now put A = {xa : a < A} and B = X \ A, respectively. We claim that A and B are as required. To this end, let Y C A be a topologically complete subspace. If Y were uncountable, then it would contain a Cantor set by Theorem 1.5.12, and hence it would contain some Aa. But this would contradict ya G Aa \ A. The proof for B is similar. D We now present an interesting application of Corollary 1.3.11, which is a partial generalization of Theorem 1.5.12. Corollary 1.5.15. Let X be a topologically complete space, let K be compact, and assume that X contains an uncountable family ofpairwise disjoint homeomorphs of K. Then X contains a copy of the product C x K. Remark 1.5.16. Observe that the uncountably many pairwise disjoint homeomorphs of K may be irregularly imbedded. This result shows that they may be replaced by a Cantor set of 'regularly' imbedded copies of K.

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

49

Proof. By Corollary 1.3.11 it follows that 3(K, X] is topologically complete, so we can pick an arbitrary complete admissible metric p for it and use the same technique as in the proof of Theorem 1.5.12 to create a Cantor set subspace of it. By assumption, 3(K, X] contains an uncountable subspace A consisting of functions having pairwise disjoint images. We perform our construction with that subspace. Since A is uncountable it contains an uncountable dense-in-itself subspace, say £ (Exercise A. 2. 13). We proceed as in the proof of Theorem 1.5.12, with one slight adaptation only. Recall that in each step of the construction we picked in some nonempty open subset U C 3(K, X) meeting £ two distinct elements /o,/i 6 £, together with small enough neighborhoods VQ,VI C U of /o and /i , respectively. Here we make the neighborhoods even smaller (see below). An inspection of the proof of Theorem 1.5.12 shows that this does not create problems. We do not only want to get a Cantor set C C 3(K, X) at the end of the process, but we additionally want the collection {h[K] : h 6 6}

(*)

to be pairwise disjoint. But this can easily be achieved. For recall that /o[/r]n/i[tf]=0. So by compactness of K there exists e > 0 such that

5e( (Corollary A. 5.4). Put

, Y) : g(fi, h) <e}

(t = 0, 1).

If hi G Jj for i = 0, 1 then hQ[K] n hi[K] = 0. So now we replace V; by the set Vi n 2F; for i = 0, 1. These neighborhoods are clearly as required. So at the end of our process we indeed get a Cantor set C C 3(K, X } having the property that the collection in (*) is pairwise disjoint. Now define the function £ : G x K —> X by

Then £ is clearly continuous. It is moreover one-to-one by the fact that the collection in (*) is pairwise disjoint. So it is an imbedding by an application of Exercise A. 5. 9. D Remark 1.5.17. An uncountable analytic space contains a Cantor set by Corollary 1.5.13. So if the compact space K contains one point only, then for K Corollary 1.5.15 not only holds for topologically complete spaces but also for the broader class of analytic spaces. The question therefore naturally arises whether Corollary 1.5.15 is also true for analytic spaces. This question was considered in BECKER, VAN ENGELEN and VAN MILL [45]; it was shown there that it is undecidable.

50

1. BASIC TOPOLOGY

We say that a space X is totally disconnected if for all distinct points x, y in X there exists a clopen set C in X such that x E C but y (£ C. It is clear that every zero- dimensional space is totally disconnected. Even more is true. Every space X that admits a one-to-one continuous function into a zero-dimensional space is totally disconnected. The question naturally arises whether every totally disconnected space is zero-dimensional. If this were true then checking whether a given space is zero-dimensional would be simpler. However, the answer to this question is in the negative, as the following example shows. Example 1.5.18. Erdos' space. Put E — {x G t2 : Xi G Q for every i}. We claim that E is totally disconnected but not zero-dimensional. On Page 9 it was observed that the topology on I2 is finer than the topology that i2 inherits from R°° . We conclude that E admits a one-to-one continuous function into the product Q°° , which is zero-dimensional (see Exercise 1.5.2). This implies that E is totally disconnected. We shall now prove that E is not zero-dimensional by showing that if U is open in i2 and 0 6 U C {x e i2 : \\x\\ < 1} then U D E is not clopen in E. To this end, let U be such a neighborhood of 0. We shall inductively define a sequence g^ e Q, i & N, such that for every i,

Let qi = 0 and assume that the QJ are defined for j 2. For 0 < ra < i put x(i,m) = ( g i , . . . , f t _ i , m / i? 0,0, . . . ) . Observe that x(i,0) — Xi-\ e U and that x ( i , i ] — (
g(xi,(? \U) < Q(x(i,m0),x(i,m0 + 1))

This completes the inductive construction. By (*) we have i Y^ n2 < 1

N

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

51

and hence

We conclude that the point q = (q\,q-2, • • • ) belongs to £ 2 , and hence to E. Since clearly, lim Xi — q (in £ 2 ) i—>oo

(Lemma 1.1.12), it follows by (*) that on the one hand q belongs to the closure of U fl E in E, while on the other hand Q(q,P\U)=]im

i—too

Q(Xi,P\U)=Q,

so q $ U. We conclude that U fl E is not closed in £", hence not clopen. Topological characterization of the closed unit interval. We shall now present a topological characterization of the closed unit interval I . The characterization below in Theorem 1.5.19 is not very appealing and its proof is rather involved. But it turns out to be very useful as will be demonstrated in the proof of the Mazurkiewicz Theorem 1.5.22 about the existence of topological copies of I in certain spaces. Theorem 1.5.19. Let Y be a continuum containing two distinct points x and y. Assume that for every z G Y\{x,y} there exist closed sets Lz and Uz ofY such that (1) LZ(JUZ=Y, (2) xeLz,y£ Uz, (3) Lz n Uz = {z}. Then Y is homeomorphic to I. Remark 1.5.20. In Exercise A. 10. 6 we proved that every continuum has at least two noncut points. From the conditions in Theorem 1.5.19 it obviously follows that Y \ {x,y} consists entirely of cut points. So the points x and y are the only noncut points of Y. Proof. We first establish the following Claim 1. For every z 6 Y \ {x,y}, Lz and Uz are continua. Proof. Suppose that for certain z £ Y \ {x,y}, Lz is not a continuum. Then Lz can be written as the union of two disjoint nonempty relatively open sets, say E and F. Without loss of generality, z £ E. Then F is contained in A = Y \ Uz and since A is open in F, this easily implies that F is open in Y. However, since F is closed in Lz and Lz is closed in Y, it also follows that F is closed in Y. Consequently, F is a nonempty clopen subset of F,

52

1. BASIC TOPOLOGY

which contradicts the fact that Y is connected. (See also Exercise A.10.5.) The proof for Uz is similar. 0 Define an order ^ on Y by putting:

Claim 2. Let d, e e Y \ {x,y}. The following statements are equivalent: (4) d * e,

(5) d e L e , (6) Ld C L e , (7) e E C/d,

(8) Ue C t/d-

Proof. The implication (4) =>» (5) is trivial. For (5) =J> (6) and (5) => (8), assume that d & Le. If d = e then there is nothing to prove, so assume that d ^ e. This implies d <£ Ue, for otherwise d G Le n Ue - {e}.

Observe that Y \ {d} is the disjoint union of the open sets Ld\{d},

Ud\{d}.

Since y e f/e n (t/d \ {d}), it therefore follows by connectivity of Ue (Claim 1) that Ue C (t/d \ {c?}) (hence (8) follows) which also implies that

Ld C X \ Ue C Le (hence (6) follows). Observe that (8) => (7) is trivial. The implication (7) =^ (6) can be proved in precisely the same way as (5) => (8). Since (6) =$• (4) follows by the definition of =^, we are done. 0 Claim 3. =<; is a linear order on Y. Proof. This follows easily from Claims 1 and 2. Let D be the set of all rational numbers between 0 and 1, i.e., D = Qn(o,l). Claim 4. There is a subset E C D and a one-to-one function f:E-+Y\{x,y} such that (9) for all d,e£E, f(d) =$ /(e) iff d < e, (10) E is dense in I, (11) f[E] is dense in Y.

<0

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

53

Proof. Let 23 = {Bn : n £ N, n even} be a base for F consisting of nonempty open sets. In addition, let 7 — {Fn : n £ N, n odd} be a base for D, also consisting of nonempty open sets. Observe that by connectivity, every Bn is infinite. In addition, since Q has no isolated points, the same holds for every Fn. By induction on n we shall pick a point en £ D and a point f ( e n ) £ Y such that the following conditions are satisfied: (12) if n is odd then en £ Fn, (13) if n is even then /(e n ) £ f? n , (14) / r { e i , . . . , e n } is one-to-one and order preserving (in the sense of (9)). For n = 1 pick an arbitrary point e\ in F\ and an arbitrary point f ( e \ ] in y \ {x,?/}. Observe that by connectivity of Y such a choice is possible. Now suppose that we completed the construction of the points

There are two cases to consider of course. Suppose first that n + I is odd. Pick an arbitrary point e n +i £ Fn+\ \ {ei, . . . , en} and define G = {m < n : em < e n +i},

H — {m < n : em > e n +i},

respectively. Let a = max{/(em) : m £ G} and b = min{/(em) : ra £ H}. Since / is one-to-one, a ^ b and so La n t/& = 0. In addition, La and E/6 are closed so that by connectivity of Y there exists a point /(e n +i) in F strictly between a and 6 (with respect to the order =^). It is easy to see that en+i and /(e n +i) are as required. For n + 1 even we can proceed by a similar argument. 0 Now fix t £ I for a moment and put At = {Lf(d) : d £ E and t < d} U {£//(<*) : d £ E and d < t}. Claim 5. At has the finite intersection property and ^\At consists of precisely one point. Proof. That At has the finite intersection property follows easily from the definition of =<(. Consequently, the compactness of Y implies that f ] A t ^ 0. Now suppose that there exist two distinct points a and b £ f\At- Without loss of generality, a =<; b. Since both La and Ub are closed in Y and disjoint, by connectivity and by the fact that f[E] is dense, there exists d £ E \ {t} such that f(d) lies strictly between a and b. Now if t < d then £/(d) € At which implies that b £ f ] A t C I//(d), which is a contradiction. However, if d < t then t//(d) £ At which implies that a £ f\At C t^/(d), which is also a contradiction.

54

1. BASIC TOPOLOGY

Define a function g : I —> Y by

Claim 6. g is a homeomorphism. Proof. First observe that g is order-preserving. Therefore, since g \ E = /, / is one-to-one and £ C I is dense, it follows that g is injective. Also observe that the range of g is dense in Y since it contains f[E]. By compactness of I it therefore suffices to prove that g is continuous (Exercise A. 5. 9). To this end, let U C Y be open, and take an arbitrary point t e g ~ l [ U ] . By compactness of Y, Claim 5 implies that there exist dQ,di € E such that do < t < d\ and £/(«M n U f ( d o ) C U. A straightforward verification shows that ( d o , d i ) C ^~ 1 [C7], i.e., g~1[U] is a neighborhood of t. <> So we are done with the proof of the theorem.

D

As an application we shall now prove the Mazurkiewicz Theorem about the existence of topological copies of E in certain spaces. Let X be a space and let a, b 6 X. A simple chain connecting a and b is a collection U\ , . . . , Un of open subsets of X such that

(1) « e (2) & e (3) ^ n C 7 j ^ 0 iff H - j | < 1. Lemma 1.5.21. Let X be a connected space and let U be an open cover o f X . For any two points x and y in X there exists a simple chain connecting x and y consisting of elements of U. Proof. Let V be the set of all points in X which are connected to x by a simple chain of elements of U. Then V is clearly open and x E V. We shall prove that V is closed as well. The connectivity of X will then imply that V = X. Take an arbitrary v e V. There exists U 6 U containing v, pick a point t € UnV. Then x and t can be connected by a simple chain C/i, . . . , Un of elements of U. Observe that U n Un ^ 0 and define k = mm{i
1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

55

The Mazurkiewicz Theorem 1.5.22. Let X be topologically complete, connected and locally connected. If x and y are distinct points of X then there exists an imbedding $: I —>• X such that $(0) = x and <&(!) = y.

Figure 3.

Proof. Let Q be an admissible complete metric on X. For each n 6 N we shall construct a simple chain t/n,i> • • • •> UnjTn(n) from x to y having the following properties (1) for each i < m(n), Un>i is connected and diam(t/ ni ;) < l/n, (2) for all ii < i% < m(n + 1) there exist j\ < J2 < m(ri) such that CU and The existence of t/i,i, . . . , t/i, m (i) follows from Lemma 1.5.21. Assume that the sets t/n,i> . . . , £/ n ,m(n) have been constructed for certain n. For j smaller than m(n) pick a point PJ £ f/ n ,j n t/nj+i ; put po — ^ and Pm(n) — VBy another application of Lemma 1.5.21 for each 1 < j; < m(n) there exists a simple chain Vj from PJ~I to p^ in t/^^- consisting of connected sets such that (3) if V € Vj then V C UnJ, (4) for every V e Vj, diam(y) < V n+ i. Unfortunately, we cannot simply join these chains together, because of doubling back (see Figure 3). We can obtain the desired simple chain by the following procedure. For every 1 < j < m(n) let Vj = {V^i, . . . , Vj>n(j)}. Put = min{/c < n(l) : (3A <

n

0)}-

Let A = max{/i < n(2) : Vi ;7r n V^ 7^ 0}- Now replace the collection Vi by {Vi,i,...,Vi ) 7 r }, and, similarly, V2 by |y 2 ,A, • • • , ^2,n(2)}- Repeat this with the 'new' Vi U V2 and the 'old' V 3 , etc. At the end of the process, the union of the 'new' Vj's is clearly the required simple chain. This completes the inductive construction.

56

1. BASIC TOPOLOGY

For every n G N put m(n) /

*-- n —

I I

Unjii

i=l

and let

We claim that Y is homeomorphic to the closed unit interval I. Observe that y is a closed subset of X and that x,y G Y. Claim 1. y is a continuum. Proof. That Y is compact follows from the construction and Corollary A. 5. 2. Now apply Proposition A. 10. 7. <0> We claim that Y satisfies the conditions mentioned in Theorem 1.5.19. Let z G Y \ {x,y}. For each n, at least one and at most two of the £/ nj i's contain z. Let An be the union of all the t/n/s preceding these and Bn be the union of all the Un/s following these. Put

n=l

n— 1

Observe that both A and B are open subsets of Y. Now if d &Y \{z} then (1) easily implies that for certain n, d G An U Bn. So A U B = Y \ {z}. We claim that A n 5 = 0. To the contrary, assume that for some n,p € N there exists d 6 yln fi Sp. Without loss of generality, n < p. Let i\ < m(p) be such that z G f/p,^. Since d £ Bp there exists i such that i\ j D An = 0, which is impossible since d G An n [7niJ- . Therefore z 0 C7n> j . Now since z G f/p,H C Unjl ,

ji < j,

we obtain J7n)J- C Bn. But this is clearly a contradiction since d G Unj n A n and Anr\Bn = 0. Put Lz = A U {z} and l^ = B U {2;}, respectively. By Theorem 1.5.19 we conclude that Y w I.

D

Corollary 1.5.23. Let X be topologically complete and locally connected. IfUCX is open and connected and x and y are distinct points of U then there exists an imbedding $: I -> U such that $(0) = x and $(1) —y. Proof. By Theorem A. 6. 3, U is topologically complete. So it satisfies the conditions in Theorem 1.5.22. D

1.5. TOPOLOGICAL CHARACTERIZATION OF SOME FAMILIAR SPACES

57

Corollary 1.5.24. Let X be topologically complete, connected and locally connected. Then X is path-connected and locally path-connected. Remark 1.5.25. A disconnected space cannot be path-connected. So connectivity is essential in the Mazurkiewicz Theorem. The same holds for both completeness and local connectivity. Examples: • A connected and locally connected space which is not path-connected. For every i £ N let i—l

oo

Wi = U [-1 + Vi, 1 - Viln * {l}i X JJ n=l

[-1 + l/i} 1 - l/,]n.

n=i+l

It will be shown in Exercise 1.5.17 that the subspace oo

W=\JWi i—l

of Q is as required. • A connected, topologically complete space which is not locally connected. The sin(1/x)-continuum in the plane is clearly as required. Exercises for §1.5. A spaxe X is called nowhere locally compact if no point has a neighborhood with compact closure. A metric g on a set X is called non-Archimedean if e(x, y) < max{£(;r, z), Q(Z, y ) } for all x, y, z G X.

1. Prove that every space of cardinality less than the cardinality of R is zerodimensional. 2. Let Xn be zero-dimensional for every n G N. Prove that H^Li Xn is zero-dimensional. 3. Let X be a space. Suppose that for every n G N there is a discrete collection Jn of closed subsets of X such that (i) \jyn = x, (2) mesh(Jn) < y n . Prove that X is zero-dimensional. 4. Let X be a compact zero-dimensional space. Prove that the collection of clopen subsets of X is countable. (Observe that N has uncountably many clopen subsets, so the compactness assumption on X is essential.) 5. Let X be a zero-dimensional space and let /: X -> Y be an open and closed continuous surjection. Prove that Y is zero-dimensional. 6. Let X be zero-dimensional. Prove that X admits a zero-dimensional compactification.

58

1. BASIC TOPOLOGY

7. Let X be a zero-dimensional cr-compact space. Prove that there is a nullsequence (An)n of compact subsets of X such that An n Am = 0 if n ^ m and Ur=i An = X. 8. Let X be a locally connected and locally compact space. In addition, let U C X be connected and open. Prove that if K C U is compact then there is a continuum C C [7 which contains .K". 9. Let X be a Peano continuum and let U and V be connected open subsets of X such that [7 U V = X. Prove that for all compact subsets A and B of X with A C. U and B C V there are subcontinua /C and L m X such that A C K C [7 and B C L CV while moreover K\J L = X. 10. Prove that for every space X there exist a zero-dimensional space Y and a continuous surjection /: Y —*• X. 11. Let X be a space. Prove that the following statements are equivalent: (1) X is zero-dimensional, (2) X has an admissible metric which is non-Archimedean. 12. Let X be locally compact and totally disconnected. Prove that X is zerodimensional. 13. Let X be totally disconnected. Prove that if x € X and K C X \ {x} is compact then there is a clopen C C X such that x £ C C X \ K. 14. Let £7 be Erdos' space. Prove that E is homeomorphic to E x E. 15. Prove that both Q are P are nowhere locally compact. 16. Prove that if infinitely many factors in a product H^L} Xn are not compact then it is nowhere locally compact. 17. Prove that the space W described on Page 57 is connected and locally connected, but not path-connected.

1.6. The inductive convergence criterion and applications In this section we present a method enabling us to construct new homeomorphisms from old ones, namely, the Inductive Convergence Criterion. The method is illustrated with applications: we deduce the topological homogeneity of Q and we present results on countable dense homogeneity. Lemma 1.6.1. Let (X,g) be a complete metric space and let (An}n be a sequence of subsets of X. Suppose that (xn)n is a Cauchy sequence in X such that for every n, (*)

e(zn+i, xn) < 3~n • min{g(xl, A*) : 1 < i < n}.

Then lin^^oo xn does not belong to (J^Li -^-n-

1.6. THE INDUCTIVE CONVERGENCE CRITERION AND APPLICATIONS

59

Proof. Take an arbitrary n G N and let x = linin-^oo xn. We shall prove that x $_ An. To this end, observe that condition (*) implies that for arbitrary m we have Q(xm+n,x(m_i]+n)

< 3-« m ~ 1 )+ n ) •e(xn, An) < 3~m • g(xn,An).

From this it follows that TO

' ^\~^ni

Q(Xm+ni Xn) ^ Z—~i i=l

n

>'>

and as

we obtain oo £}{ Jb j Jt>ri)

Hill

77T.—$-00

tr ^^TTl-f-Tl 5 JL/TI )

_

/

* ^

^

y\t^rn ; -*~*~7l)

/2 c: V Tl 5 -**-7lJ ;

so x $. An.

D

Observe that in the above lemma the distance between x n +i and xn 'depends' only on the points £1,... ,xn. Hence if one wishes to choose the points (xn)n inductively so that (*) is satisfied, at stage n + I the choice of xn+i is subject to n - hence finitely many - conditions. Let X be a compact space and let (hn)n be a sequence in ^K(X). It is clear that for each n e N the function fn = hn o • • • o hi

belongs to 'H(X). If / = lim.^-^ fn exists then it will be denoted by lim hn o • • • o hi

n—>oo

and is called the infinite left product of the sequence (hn}n. We want to find conditions on the sequence (hn)n which ensure that linin^oo hno- • -ohi exists and belongs to "H(X). The Inductive Convergence Criterion 1.6.2. Let (hn)n be a sequence in "H(X) such that X is compact. In addition, assume that for all n G N, (1) Q(hn+ (2) e(hn+l,lx} Then h — limn_^00 hn o • • • o hi exists and belongs to 'H(X). Proof. For every n, put fn — hn o • • • o hi. Condition (1) and Exercise 1.3.2 imply that for all n, (3)

£(/n+i, /n) = e(hn+i o • • • o hi, hn o • • - o hi) = e(hn+i,lx) < 2~ n .

60

1. BASIC TOPOLOGY

We conclude that the sequence (fn)n

ls

£-Cauchy and consequently,

/ = lim fn n—>-oo

exists (Corollary 1.3.6). Since fn E S ( X , X ) for every n, Proposition 1.3.8 implies that / 6 S ( X , X ) . Now by (2), (3), Lemmas 1.6.1 and 1.3.10 we obtain n=l

as required.

D

The above result tells us that if the sequence (@(hn, 1)) converges rapidly to 0, then the infinite left product limn^.00 hn o • • • o h\ is a homeomorphism. It turns out that we are never interested in the precise speed at which (@(hn,l)} converges. We are always in the pleasant situation that while inductively defining the sequence (hn)n, we are able to choose the next homeomorphism 'sufficiently close' to the identity. This simplifies life considerably. Application 1: Topological homogeneity of the Hilbert cube.

Every nontrivial topological group has fixed-point free homeomorphisms. Simply observe that if (G, °) is a topological group then every translation of the form x i->- g o x, where g is not the identity element of G, is a fixed-point free homeomorphism. If Q were a topological group then the proof of its homogeneity would be trivial by the above remarks. But Q is not a topological group since it has the fixed-point property, as we will see later in Corollary 2.4.6. It is geometrically obvious that no I n , n G N, is homogeneous. For n = I this is a triviality. Simply observe that E\{0} is connected while I\{ l/2} is not. For larger n the proof is more complicated. See the proof of Theorem 2.4.14 for more details. As we announced, we wrill prove that Q is homogeneous, thereby demonstrating a striking difference between the 'finite dimensional' and the 'infinitedimensional' situation. We put 5= n-1

respectively. We call s the pseudo-interior and B(Q) the pseudo-boundary ofQ.

1.6. THE INDUCTIVE CONVERGENCE CRITERION AND APPLICATIONS

61

This notation is slightly confusing since s was already our abbreviation for R°°. But M°° and 0^=1 (~~1» l)n are homeomorphic and so this abuse of notation is not as bad as it seems at first glance. Lemma 1.6.3. Suppose that x,y e s C Q. Then there is a homeomorphism h 6 3t(Q) with h(x) = y. Proof. Since for each i, Xi, yi 6 (— 1, l) t , there is an element hi G ^K(JJ;) such that hi(xi) = yl, we may take h — h\ x /i2 x • • • (Exercise A. 1.13). D We see that all points in the pseudo-interior are topologically equivalent. Consequently, if we show that every point in Q can be 'homeomorphed' into s, then we have shown that Q is homogeneous. We need a preliminary lemma. Lemma 1.6.4. Suppose that x G Q, that m e N, and that e > 0. Then there is an element h 6 'H(Q) such that (1) e(h,!Q) <e,

(2) /i(z) m e(-i,i) m , (3) h does not affect

the first m — 1 coordinates of any point, i.e.,

h(y)i = yi

for all i < m — 1 and y e Q.

«UI m

1 l

— z,2~( n ~ 1 )

Figure 4. Proof. If |x m | 7^ 1, let /i = IQ. Therefore, without loss of generality assume that xm = 1. Let n > m be such that 2~( n ~ 2 ) < e. It is geometrically obvious that there is a homeomorphism / : J I m x J I n ^ j r m x J I T l such that (4) / ( p , g ) - ( p , g ) i f p < l - 2 - ( - 1 ) , (5) /[{I} x JJn] is contained in JJm x {1} and its projection on J m is contained in the open interval (1 — 2~( r i ~ 1 ), 1).

62

1. BASIC TOPOLOGY

For a moment we think of Q as (JJm x JJ n ) x R. where R denotes the product of all remaining factors. We define h = f x IR. So h only affects the n-th and m-th coordinate of any point. To verify that h is as required, first observe that (2) and (3) are clearly satisfied. For (1), notice that for every x £ Q, Q(h(x),x) = 2~m • \h(x)m -xm\+2-n- \h(x)n-xn\ < i.2-("- 1 ) +2~n - 2

< £.

From this it follows that g(h, IQ) < e (Exercise A. 5. 4).

D

Lemma 1.6.5. Let x G Q. There is an h G M(Q) with h(x) e s. Proof. The homeomorphism h shall be of the form lim hn o • • • o hi ,

n—>-

where (hn)n is an inductively constructed sequence of homeomorphisms of Q. To ensure convergence in the limit, at each stage of the construction the next homeomorphism is constructed in accordance with the smallness conditions in the Inductive Convergence Criterion 1.6.2. Specifically, the sequence (hn}n shall satisfy the following conditions: (l)n hn does not affect the first n — I coordinates of any point, (2) n (hn o • • • o hi(x))i G (-1, 1) for every i < n, (3)n Q(hn, IQ) is so small that the conditions of the Inductive Convergence Criterion 1.6.2 are satisfied. Apply Lemma 1.6.4 to find a homeomorphism / 6 "H(Q} such that /Or)! 6 (-1,1) and define hi = f . Then hi is as required since (l)i and (3)i are empty conditions. Now suppose that hi has been defined for every i < n. The Inductive Convergence Criterion 1.6.2 gives us a magic e > 0 and tells us that we must choose the next homeomorphism e-close to the identity. We are not interested at all in the precise value of e. The only thing we need is that such e > 0 exists. By Lemma 1.6.4 there is an element /in+i G 3t(Q) such that g(hn+i, IQ) < £, hn+i does not affect the first n coordinates of any point, and (hn+i(hno.-.ohl(x)))n+1 It is evident that hn+i is as required.

G (-1,1).

1.6. THE INDUCTIVE CONVERGENCE CRITERION AND APPLICATIONS

63

Now put h = linin-^oo hn o • • • o hi. By construction, h G ^K(Q). continuity of the projections 7rm : Q —>• J m we find that for every ra,

By

h(x)m = ( lim /in o • • • o /ii(x)) m n—>oo

= lim (hn o • • • o hi(x)) n>m

m

i.e., /i(x) 6s.

D

We now come to the announced result: Theorem 1.6.6. The Hilbert cube Q is homogeneous. Proof. Take x,y G Q. By Lemma 1.6.5 we find /ii,/&2 € ^(Q) such that both hi(x) and /i2(y) belong to s. In addition, by Lemma 1.6.3, there is an element / G ^K(Q) with f ( h i ( x ) ) = h2(y). Then takes x onto ?/.

D

Application 2: Countable dense homogeneity. A space X is called countable dense homogeneous provided that for all countable and dense subsets D,E C X there exists an element / G W(X) with f[D] = E. The topological sum of S1 and §2 is an example of a countable dense homogeneous space which is not homogeneous. But for connected spaces it will turn out that countable dense homogeneity implies homogeneity. Let X be a space and fix an arbitrary point x G X. Let r(x) denote the type of x, i.e., Observe that r(x) is invariant under "H(X}, that is, for all y G T(X) and all h G 'H(X) the image h(y) also belong to T(X). Theorem 1.6.7. If X is countable dense homogeneous and x G X then T(X) is clopen in X . Proof. We will show that r(x) is clopen. Claim 1. r(x) is closed. Proof. To the contrary, assume that there exists p G T(X) \ T(X). Let E be a countable dense subset of r(x), and let F be a countable dense subset ofU = X\ T(X). Then both

A = E U F,

B = A U {p}

64

1. BASIC TOPOLOGY

are countable dense subsets of X and hence there exists by assumption an element h e 'K(X) with h[A] = B. Since p 0 h[E] because h[r(x}] C T(X), there exists b E -F with /i(6) = p. But then /&[£/] is a neighborhood of p and hence intersects r(x). But this implies that for some q 6 C/ we have that /i(g) G r(x), i.e., q 6 T(#). This is a contradiction. 0 Claim 2. T(X) is open. Proof. It is easy to see that r(x) is open if and only if Intr(ar) 7^ 0. Striving for a contradiction, suppose therefore that Intr(x) = 0. By Claim 1 it follows that r(x) is a closed subset of X with empty interior, hence is nowhere dense. Let E C U = X \ T(X] be countable and dense in X. There is an element h e 3t(X) with h[E] = EU{x}. Pick e £ E such that /i(e) = x. But then e G r(a;), which is a contradiction. So we are done.

D

Corollary 1.6.8. A connected countable dense homogeneous space is homogeneous. So for connected spaces, countable dense homogeneity is a strong form of homogeneity. There is an example of a connected homogeneous space which is not countable dense homogeneous, see VAN MILL [294] for details. A space X is called strongly locally homogeneous provided that it has an open base U such that for all U e U and points x,y G U there exists an element h E 3i(X) such that h(x) ~ y and h(z) = z for z 0 U. Such a homeomorphism is said to be supported on U. It is geometrically obvious that the Euclidean spaces W1 , n G N, are strongly locally homogeneous. Before we come to the main result in this Application, we make a few remarks. Let X be a locally compact space. Observe that if / G 'H(X) then the function af : aX —>• aX defined by

af(x) = ^ J v ' \ oo

,

x

(a; = oo),

belongs to CK(aX). It will sometimes be convenient to think of a locally compact space as subspace of its Alexandroff compactification. Theorem 1.6.9. Let X be locally compact and strongly locally homogeneous. Then X is countable dense homogeneous. Proof. Let A — {01,02,...} and B = { & i , & 2 ) - - - } be faithfully indexed dense subsets of X. The hypothesis of strong local homogeneity implies that for each neighborhood U of a point x G X, and for any dense G C X, there exists a homeomorphism of aX which is supported on U and takes x into G

1.6. THE INDUCTIVE CONVERGENCE CRITERION AND APPLICATIONS

65

(use that G n U ^ 0). Using the Inductive Convergence Criterion 1.6.2, we construct a sequence (hn}n of homeomorphisms of aX such that its infinite left product h is a homeomorphism and such that the following conditions (which ensure h[A] = B and h \ X G IK(JQ) are satisfied: (1) hn o • • • o hi (di) = h-2i ° • • • o h\_ (di) G B for each i and n > 2i, (2) (hn° • • • °hi)~l(bi) = (hii+i o • • -ohi)~l(bi) G A for each i and each n > li + l, (3) for all n, hn(oo) = oo. Assume hi, . . . , h• X is limited by U provided that for every x G X there exists U G 1C such that {x,h(x)} C U. ^•1. Let (X, £>) be a complete space, and let (U ra ) n be a sequence of open covers and (hn)n a sequence of homeomorphisms of X satisfying the following conditions:

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1. BASIC TOPOLOGY

(1) (2) (3) (4) Prove

lln is a star-refinement of lt n -i, mesh(Un) < 2"71, (hn-i o • • • o hi)[Un] has mesh less that 2~n, hn is limited by Un. that h = limn^.oo hn ° • • • ° h\ is a homeomorphism of X.

2. Use Exercise 1.6.1 to show that a topologically complete space which is strongly locally homogeneous is countable dense homogeneous. 3. Prove that both s and B(Q) are dense in Q. Prove that B(Q) is cr-compact. Are s and B(Q) homeomorphic? 4. Prove that an open subspace of a strongly locally homogeneous space is strongly locally homogeneous. 5. Let X be a compact space with closed subspace A. Prove that there is a continuous function /: X —> Q such that (2) / \ X \ A is one-to-one. 6. Let X be a homogeneous space. Prove that the following statements are equivalent: (1) X is a Baire space. (2) X is not meager in itself. 7. Let X be a space and x £ X. Prove that for every h e !K(A") we have h[r(x}] = r(x). 1.7.

Bing's shrinking criterion

In this section we provide a second method for constructing new homeomorphisms from old ones, namely Bing's Shrinking Criterion, which is one of the most powerful tools in geometric topology. As an application we will prove that Q is homeomorphic to its own cone. Let X and Y be compact spaces. A continuous surjection /: X —>• Y is called a near homeomorphism provided that for every e > 0 there is a homeomorphism h: X —> Y such that g(h, f) < e. Clearly, / e C(X,Y) is a near homeomorphism if and only if / belongs to the closure of 3-C(X, Y) in C(X, Y}. Simple examples show that a near homeomorphism need not be a homeomorphism (Exercise 1.3.5). Near homeomorphisms are very important in geometric topology. Let X and Y be compact spaces. A continuous surjection /: X —> Y is called shrinkable provided that for every e > 0 there is a homeomorphism h: X —> X satisfying (1) diam (hf~1(y}}

< e for every y € Y (i.e., / o h~l is an e-map),

Shrinkability means that the point-inverses of / can be uniformly shrunk to small sets by a homeomorphism of X that, from the standpoint of the

1.7. BING'S SHRINKING CRITERION

67

space F, does not change / very much. The reader should pause to get used to this definition. Ding's Shrinking Criterion says that shrinkablity is the same as being a near homeomorphism. If the domain of the function under consideration has a rich supply of homeomorphisms, e.g., if it is a manifold, then verifying shrinkability is usually easier than verifying directly that the function in question is a near homeomorphism. Lemma 1.7.1. Let X and Y be compact spaces and let f : X —> Y be shrinkable. Then for each e > 0 and for each (p 6 'K(X) there exists an element ip £ ^K(X) such that (1) f oip is an e-map, (2) Q ( f o < f > , f o i j , ) < £ . Proof. Let 7 > 0 be such that for A C X, if diam(A) < 7 then (*)

diam (p"1 [A]) < e

(use the fact that (p~l is uniformly continuous (Exercise A. 5. 18)). Since / is shrinkable, there is a homeomorphism h G 3t(X) such that (3) g(f,foh)<£, (4) V y e r jdiam^-Hy)) < 7Since £(/, / ° h) < e, it follows that g(fh~l(p,fhh~1(p} so that

< £ (Exercise 1.3.2),

We claim that f h ~ l ( p is an e-map. If so, then ijj = h~l(p is as required. To this end, take an arbitrary y E Y and consider (fh~l(p)~l(y} = ( p ~ l h f ~ l ( y } . Since diam ( h f ~ l ( y ) } < 7 we see by (*) that diam ((p~1hf~1(y)} < e, and this is what we had to prove. D Bing's Shrinking Criterion 1.7.2. Let X and Y be compact spaces and let f : X —>• Y be a continuous surjection. Then f is a near homeomorphism if and only if f is shrinkable. Proof. First assume that / is a near homeomorphism. Let e > 0. There is a homeomorphism g: X —>• Y such that g(g, /) < £/2. Find 7 > 0 such that if A C Y has diameter less than 7 then 5-1[A] has diameter less than e (use the fact that g~l is uniformly continuous (Exercise A. 5. 18)). Let p: X —>• Y be a homeomorphism such that g(p, /) < min{Y2? e/2J and put h = g~l ° p. We claim that h is a shrinking homeomorphism. Since £>(/,p) < £/2 and

= Q(g,f)

68

1. BASIC TOPOLOGY

(we used Exercise 1.3.2 twice), it follows that g(fjh]

< g(f,p) + g(pjh) < */2 + */2 = e.

Now take an arbitrary y G Y. If z G p(f~l(y)) then since g(p, /) < 0/2 we get g(y,z) < T/2- So, diam (p(f~1(y})) < 7 from which it follows that diam (g~lp(f~l(y}}} < e. We conclude that diam (h(f~1(y))} < £. Now assume that / is shrinkable. Let e > 0 and let ho be the identity homeomorphism on X. By applying Lemma 1.7.1 inductively, we find that there exists a sequence (hn}n>\ in 'K(X) such that if we put pn = f ° hn for each n > 1, then the following conditions are satisfied: (1) pn is a (2) Q(pn+i,pn) <3~n-e and (3) g(pn+i,pn) < 3 ~ n

By (2), the sequence (pn)n is Cauchy. Since g is complete, the function p = lim pn n—too

is well-defined and is continuous by Lemma A. 3.1. Observe that by Proposition 1.3.8, p G S(X,y). Again by (2), it follows that

i-l

Now apply Lemma 1.6.1 to the sequence (pn)n to obtain

n=l

i.e., p G ^(X^Y) (Lemma 1.3.10). We conclude that / is a near homeomorphism. D Corollary 1.7.3. Let X and Y be compact spaces and let /: X —>• Y be shrinkable. Then X and Y are homeomorphic. Application: The cone over the Hilbert cube is the Hilbert cube. Recall that for a compact space the cone over X is the space we get from X x I by identifying X x {1} to a single point oo (see Page 515). It is geometrically clear that for each n the cone over In is homeomorphic to I n+1 and so naively one would expect, by taking the 'limit' as n goes to infinity, that A(Q) « Q. That this is indeed true will be demonstrated here as an application of Bing's Shrinking Criterion. Lemma 1.7.4. For every n G N, t G [— 1, 1) and e > 0 there exists a homeomorphism h: J J n x J I ^ - J I n x J I such that

1.7. BING'S SHRINKING CRITERION

69

(1) h \ J n x [—!,£] is the identity function, (2) the diameter of h[$n x {!}] is less than e. Proof. The lemma is geometrically obvious. Alternatively, observe that the Lemma is clear for n — 1 since then h can be found such as in the proof of Lemma 1.6.4, and proceed inductively. D Lemma 1.7.4 allows us to conclude that: Theorem 1.7.5. A(Q) is homeomorphic to Q. Proof. Let p: Q x E —>• A(Q) be the natural quotient map. We will show that p is shrinkable. From Corollary 1.7.3 it wall then follow that

n ~n v TT ~ /\(n\ \cS —t* /-^ \cJ —f xx Jl r*~* L i \V\os T/ /11 Let e > 0 and let U be a neighborhood of oo G A(Q) having diameter less than e. There exists t G (0,1) such that (Q x [t, 1)) U {00} C U. Find n G N such that oo

]T 2~ m < e/2 m=n

and applying Lemma 1.7.4 find a homeomorphism h: J n x I —>• J n x I such that (1) /i C JJn x [0,t] = 1, (2) diam(/i[JJn x {!}]) < e/2.

Define // G tt(g x E) by (we make a few obvious identifications here). By the special choice of n it follows easily that dia,m(H[Q x {!}]) < e. We claim that g(p,po H} < e. Take a point (x, s) G Q x I. If s t then (p(x,s),pH(x,s)} C U. Therefore, since diam(L r ) < e, we find that g(p(x, s),pH(x, s)) < £. We conclude that p is shrinkable. D Consider the point oo G A(Q). Basic closed neighborhoods of this point have the form Ut = ( Q x [ t , l ) ) u { c x D }

(te (0,1)).

Observe that the boundary of Ut is equal to Q x {t}. Consequently, Theorem 1.7.5 implies that oo has arbitrarily small closed neighborhoods U such that (1) U is homeomorphic to Q, (2) Fr([7) is homeomorphic to Q.

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1. BASIC TOPOLOGY

By the homogeneity of Q (Theorem 1.6.6), each point of Q has arbitrarily small closed neighborhoods U satisfying (1) and (2). In particular, each point of Q has arbitrarily small closed neighborhoods with contractible boundaries. This demonstrates a striking difference with the finite dimensional situation.

Exercises for §1.7.

1. Let X be a compact space. Let Z be the space we obtain from A(X) x I by identifying A(X) x {1} to a single point, and let TT: A(X) x I —>• Z be the decomposition map. Prove that TT is shrinkable and conclude from this that A(ApQ) « A(X) x I. 2. Let X be a compact space and let Y = A(X) x Q. Prove that Y is homeomorphic to its own cone 3. Let X, Y and Z be compact spaces and let / : X —>• Y and g : Y —> Z be near homeomorphisms. Prove that gof : X —>• Z is a near homeomorphism. 4. Let p: Q —>• 0^2^* be the projection. Prove that p is a near homeomorphism.

1.8. Isotopies

We will now present the third and last method for constructing new homeomorphisms from old ones by means of so-called isotopies. In comparison to the Inductive Convergence Criterion and Bing's Shrinking Criterion, at first glance this method does not seem to be as important as the other two since direct appealing applications are hard to find. However, it turns out that this method is as fundamental as the preceding two. Let X and Y be spaces. Recall that a homotopy from X to Y is a continuous function H: X x I —>• Y. We call a continuous function

H: X x K ^Y, where K is any compact space, a K-homotopy. If t G K then the function Ht: X —> Y defined by Ht(x) — H(x,i) is called the t-th level of H. A K-isotopy from X to Y is a K-homotopy H: X x K —> Y each level of which is a homeomorphism from X onto Y. If K C M is a compact interval then a K-homotopy (resp. K-isotopy) is simply called a homotopy (resp. isotopy). The proof of the following proposition is a triviality and is left as an exercise to the reader.

1.8. ISOTOPIES

71

Proposition 1.8.1. For each n G N let Hn : Xn x Kn ->• Yn be a Kn-isotopy. Then the function OO

H: l[Xnx n=l

OO

OO

*[[Kn-> JJr n , n=l

n=l

defined by H(x,t)n = Hn(xn,tn) is a

(n e N),

n^Li Kn-isotopy.

We shall now present our last method for constructing new homeomorphisms from old ones. Theorem 1.8.2. Let K be a compact space. If H: X x K —>• X is a Khomotopy and a: Y —>• K is continuous then so is the function /: X xY -*X xY

defined by (1) f ( x , y ) = (H(x,a(y»,y)Moreover, if X and Y are compact and H is a K-isotopy then f is a homeomorphism. Remark 1.8.3. Observe that the function / defined in (1) is 'level preserving', i.e., / does not change the second coordinate of any point. Proof. That / is continuous is trivial. Assume next that X and Y are compact and that H is a K-isotopy. Since / is 'level preserving' and since each 'level' of / is a homeomorphism from X onto Jf, it follows that / is surjective. By compactness we therefore only need to show that / is one-toone (Exercise A.5.9). To this end, take distinct (a, &),(«', 6') eXxY. If b / b' then clearly /(a, b) / /(a',6')- If b = b' then a(b) = a(fc') and a / a'. Consequently, /f (a,a(6)) ^ H(a',a(b')) since Ha^ is one-to-one. D Application: Extending certain homeomorphisms. Isotopies are very useful in situations where one wants to extend a given homeomorphism. This will be demonstrated in a simple situation here. The same ideas will be used to derive much more complicated results in §5.2. Let E, F C K. be compact subsets and let /: E —> F be a homeomorphism. It is easy to see that it is not always possible to extend / to a homeomorphism /: 1R —>• E. For example, let E = F = {0,1,2} and define / by/(0) = l,/(l) = O a n d / ( 2 ) = 2. Identify R and the £-axis in R 2 . Although in general a map / between compact subsets of E cannot be extended over R, an easy application of

72

1. BASIC TOPOLOGY

Exercise 1.2.9 yields that if / is a homeomorphism then it can be extended to a homeomorphism /: R2 —> R 2 . We shall now present a different proof of this which much more than the solution of Exercise 1.2.9 illustrates the technique of extending homeomorphisms that we are going to use in §5.2. Theorem 1.8.4. Let E,F C R be compact and let f : E —> F be a homeomorphism. Then f can be extended to a homeomorphism /: R2 -> R 2 . Proof. We shall prove the theorem for the interval (—1,1) instead of R. Since ( — 1, 1) and R are homeomorphic, we are allowed to do this. So let E,F C (—1,1) be compact and let /: E —>• F be a homeomorphism. Let K C (—1,1) be a compact interval containing both E and F. For each t G K let Ht '• J —> J be the unique homeomorphism taking [—!,£] linearly onto [—1,0] and [t, 1] linearly onto [0,1]. It is easily seen that the function H : JJ x K —> J defined by H(x,t) — Ht(x) is an isotopy (Exercise 1.8.1). Let F C ( — 1, 1)2 be the graph of the function /, i.e.,

Our aim is to find a homeomorphism h: J 2 —> J 2 that takes F onto {0} x F. W7e achieve this by applying Theorem 1.8.2. Define the function (p: F —>• K as follows: K. Now define h : J 2 —> J 2 by h(x,y) = ( H ( x , a ( y ) ) , y ) . By Theorem 1.8.2 it follows that h is a homeomorphism. If (x,/(x)) G F then (1)

h ( x , f ( x ) ) = (H(x,a(f(x»)JW)

= (H(x,x)J(x))

= (0, /(*)).

We conclude that h is as required. By precisely the same argumentation we can find a homeomorphism g: JJ2 —>• JI2 such that for every (x, /(x)) G F, (2)

g(x,f(x))=(x,Q).

Now define ^ : J 2 -> JJ2 by £(x, y) = (y, x) and put / = £ o h o g~l . Then / is a homeomorphism of J 2 and it is easily seen that /[(—I, I) 2 ] = ( — 1, 1)2. We claim that / extends / and is therefore as required. Take an arbitrary x G -E1. Then

(apply (1) and (2)).

D

1.9. HOMOGENEOUS ZERO-DIMENSIONAL SPACES

73

Exercises for §1.8.

1. Let -1 < a J be the unique homeomorphism taking [—!,£] linearly onto [—1,0] and [£, 1] linearly onto [0,1]. Prove that the function H: Jx K-» J

denned by H ( x , t ) = Ht(x) is an isotopy. ^•2. Prove that Q is strongly locally homogeneous. 3. Let D C Q be countable and dense. Prove that Q \ D is homogeneous. 1.9. Homogeneous zero-dimensional spaces Homeomorphism theory in the class of zero-dimensional spaces is particularly elegant. It turns out that various classical zero-dimensional spaces such as Q and F can be topologically characterized as nicely as the Cantor set. Homogeneous zero-dimensional spaces. Our first aim is to characterize the class of all homogeneous zero-dimensional spaces. Lemma 1.9.1. Let X be a zero-dimensional space and let x,y 6 X. If for each e > 0 there are clopen neighborhoods U and V of x and y, respectively, such that (1) diam(C7) < e, diam(V) < e, (2) U is homeomorphic to V, then there is a homeomorphism h: X —> X with h(x] — y. Proof. For each n > 0 we will construct a clopen neighborhood Un of y and a clopen neighborhood Vn of x and for each n > 1 an element hn € "K(X} such that (3) (4) (5) (6)

diam(C/n) < 2~ n , diam(K) < 2~ n , UnUhn^o-.-oh^Vn] C C7n_! and h^1 o • • • o h'^ [Un] U Vn CVn-i, hn°...ohi[Vn] = Un, hn is supported on Un-\.

Put UQ = VQ — X and suppose that for some n > 0, Vi and U% have been constructed for all 0 < i < n — 1 and hi for all 1 < i < n — 1. If hn-i o ••• ohi(x) = y then let V be any clopen neighborhood of x of diameter less than 2~ n such that both V and U = hn-\ o • • • o hi[V] have diameter less than 2~n. Put

74

1. BASIC TOPOLOGY

Then our choices are obviously as required. Suppose therefore that

hn-i o ••• ohi(x) -£ y. Let V C Vn-i be a clopen neighborhood of x and U C Un-\ be a clopen neighborhood of y such that (7) diam(V) < 2~ n , diam(t7) < 2~ n , (8) V is homeomorphic to [7, (9) hn_i o • • • o hi[V] CC7 n _!,

(10) fcn_i o... 0 /i 1 [v r ]nt/ = 0.

Let /: /in-i ° • • • ° hi [V] —> C7 be a homeomorphism. Define Un = U,Vn = V and /i n : A" —>• X by

{

x f(x)

(x ^C/ n U/i n _! o - - - o h i [ V n ] ) , (x£hn-io...ohi[[Vn]),

f-l(x]

(xeun).

It is clear that Un, Vn and hn satisfy our inductive requirements. If p £ X and g(p, x) > 2~n then p $ Vn and so hn ° • • • ° hi (p) £ hn o • • • o hi [Vn] = Un.

By (6) this implies that (11)

hk°'--°hi(p) = hno---ohi(p)

for all k > n. Consequently, if we define h: X —>• X by h = lim hn ° • • • o hi n—»oo

then /i is well-defined. Observe that by (4) and (5), h(x) = y. We claim that h is a homeomorphism. If g(p,x} > 2~n then, as remarked above, hn-i o • • • o / i 1 ( p ) £ C7n which implies by (11) that /i(p) ^ C/n. Since /i(x) = y, this implies that h is one-to-one. In addition, if g(p,y) > 2~n then p £ Un and therefore by (4) if

q = h~l o - - - o h ~ l ( p ) then h(q) = p. We conclude that h is surjective. By (3) and (6), h is continuous (Lemma A.3.1). It is clear that h is open at all points but x. We therefore only check openness of h at x. This will give us that h is a homeomorphism. To this end, let V be any neighborhood of x. Let n > 0 be so large that Vn C V. Then clearly y = h(x) e Un = h[Vn] C h[V]. We therefore conclude that h[V] is a neighborhood of y, which implies that h[V] is open. D Lemma 1.9.1 will be used among other things in Exercises 1.9.1 and 1.9.2 for the construction of a partition of E into two homeomorphic homogeneous parts.

1.9. HOMOGENEOUS ZERO-DIMENSIONAL SPACES

75

Corollary 1.9.2. Let X be zero-dimensional. The following statements are equivalent: (1) X is homogeneous. (2) if x, y E X then x and y have arbitrarily small homeomorphic clopen neighborhoods. Strongly homogeneous zero-dimensional spaces. Let X be zerodimensional. We say that X is strongly homogeneous provided that all nonempty clopen subsets of X are homeomorphic (and hence are homeomorphic to X). By Corollary 1.9.2 it follows that a strongly homogeneous (zero-dimensional) space is homogeneous. Corollary 1.9.3. A strongly homogeneous (zero-dimensional) space is homogeneous. A homogeneous zero-dimensional space need not be strongly homogeneous (VAN DOUWEN [128]), but at least the following weaker statement holds. Lemma 1.9.4. A homogeneous zero-dimensional space is strongly locally homogeneous. Proof. Let x £ X and let U be an arbitrary clopen neighborhood of x. In addition, pick an arbitrary y E U \ {x}. We claim that there is a homeomorphism h: X —>• X sending x onto y and which moreover restricts to the identity on the complement of U. Since X is homogeneous, x and y have arbitrary small clopen neighborhoods which are homeomorphic by homeomorphisms sending x onto y. So we can find disjoint clopen neighborhoods E and F of x and y, respectively, and a homeomorphism f : E —> F such that (1) E(JF CU, (2) f ( x ) = y.

Now define h: X -> X by

{ Then h is clearly as required.

p

(p 0 E U F),

f(p)

(P € £),

/-'(P) (peF). D

Lemma 1.9.5. The Cantor set C is strongly homogeneous and strongly locally homogeneous. Proof. This is clear from the topological characterization of C (see Theorem 1.5.5) since every nonempty clopen subspace of C is compact, zerodimensional and has no isolated points. D

76

1. BASIC TOPOLOGY

Topological characterization of Q and F. We now apply the results obtained so far to get interesting topological characterizations of various classical zero-dimensional spaces. Theorem 1.9.6. The space of all rational numbers Q is topologically the unique nonempty countable space without isolated points. Proof. It is clear that Q is countable, nonempty, and has no isolated points. It therefore suffices to prove that all nonempty countable spaces without isolated points are homeomorphic. To this end, let X and Y be such spaces. By Exercise 1.5.1, both X and Y are zero-dimensional. They consequently have zero-dimensional compactifications aX and aY by Exercise 1.5.6. Both aX and aY clearly have no isolated points because X and Y both have no isolated points. So aX w aY & C by Theorem 1.5.5. Since C is strongly locally homogeneous (Lemma 1.9.5), it is countable dense homogeneous by Theorem 1.6.9. So there is a homeomorphism /: aX —>• aY with f[X] = Y. In particular, X w Y. D Remark 1.9.7. Our next result is a topological characterization of the space of irrational numbers P. There is a proof of this characterization along the same lines as the proof of Theorem 1.9.6. But it is simpler by using the completeness of P to follow the method in the proof of Theorem 1.5.5. Theorem 1.9.8. The space of all irrational numbers P is topologically the unique nonempty, topologically complete, nowhere locally compact and zerodimensional space. Proof. That P is topologically complete, zero-dimensional and nowhere locally compact, follows easily from Remark A.6.5, Proposition 1.5.3 and Exercise 1.5.15. Claim 1. Every nonempty open subset of a nowhere locally compact zerodimensional space can be decomposed into infinitely many pairwise disjoint nonempty clopen sets. Proof. Let X be nowhere locally compact and zero-dimensional. Let U C X be nonempty and open. The proof of Proposition 1.5.1 shows that we can decompose U into nonempty clopen subsets of X (possibly, a finite number). So without loss of generality we may assume that U is clopen. Since X is nowhere locally compact (Exercise 1.5.15), U is not compact. It consequently has an open cover U without finite subcover. The proof of Proposition 1.5.1 shows that this cover can be refined by a clopen partition (here clopen means clopen inX). This partition is necessarily infinite since if it were finite then U would have a finite subcover. Observe that in the proof of Theorem 1.5.5 we proved that if X and Y are zero-dimensional compact spaces without isolated points then they have

1.9. HOMOGENEOUS ZERO-DIMENSIONAL SPACES

77

isomorphic trees of clopen sets which form bases of X and Y, respectively. A path P through these trees has a unique point as its interesting because of compactness and the fact that the elements of P have smaller and smaller diameters. The trees are finitely branching, i.e., each node has finitely many successors only. With topologically complete, zero-dimensional nowhere locally compact spaces similar trees of clopen sets can be constructed (their construction is easier than in the case of Cantor sets). By applying Claim 1 we can ensure that all levels of the trees are infinite, that they have countable height and that they are infinitely branching, i.e., each node has infinitely many (hence ui many) successors. So the trees are obviously isomorphic. By using complete metrics, one can again ensure that for each path P the intersection of its elements is a single point. So the proof can then be completed such as in the proof of Theorem 1.5.5. D Corollary 1.9.9. P and N°° are homeomorphic. Proof. Use Exercise 1.5.16 to conclude that N°° is nowhere locally compact. Zero-dimensionality and completeness follow from Proposition 1.5.3, Exercise 1.5.2, Remark A.6.5 and Lemma A.6.2. So we are done by Theorem 1.9.8. D Corollary 1.9.10. If X is topologically complete then it is a continuous image of P. Proof. By Theorem A.6.3, X is homeomorphic to a closed subspace of E°°. Claim 1. IR°° is a continuous image of P. Proof. Since P°° w (N00)00 w N°° % P (Corollary 1.9.9), it suffices to prove that M is a continuous image of P. Consider the zero-dimensional product space Y — NX C. Since {n} x C maps onto [—n, n] by Theorem 1.5.10, Y maps onto E. But P x Y satisfies the conditions mentioned in Theorem 1.9.8 (cf. the proofs of Theorem 1.5.10 and Corollary 1.9.9) and hence is homeomorphic to P. So we conclude that P maps onto Y and so in turn maps onto R. 0" So we are done.

D

This yields the following characterization of analytic spaces. Corollary 1.9.11. Let X be a space. The following statements are equivalent: (1) X is analytic, (2) X is a continuous image of P.

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1. BASIC TOPOLOGY

A Hurewicz-type theorem. The space Q is clearly not a Baire space. It is in fact a 'test space' for the Baire property. Theorem 1.9.12. Let X be a space. If X is not a Baire space then X contains a closed subspace homeomorphic to Q. Proof. Let 9 be a countable collection of dense open subsets of X such that ("") S is not dense. There exists a nonempty open subset V C X such that

y n f | S = 0. Then V contains no isolated points. For if x is an isolated point of V then clearly x E V and so, V being open in X, x is an isolated point of X. But then x belongs to every dense subset of X, i.e.,

So by Theorem 1.9.6 it suffices to prove the following: Claim 1. Let Y be a space having no isolated points. If 9 is a countable collection of dense open sets in Y then Y has a countable subspace K having no isolated points such that K \ K C f] 9. Proof. For y € Y, if A is a collection of subsets of Y we say that A converges to y if every neighborhood of y contains all but finitely many A 6 A. It is easy to see that if y e Y and U is a neighborhood of y then there is a pairwise disjoint infinite collection V of nonempty open subsets of U \ {y} such that V converges to y and U V C U (Exercise 1.9.4). Enumerate 9 as {Gn '• n E N}. We will construct a sequence

{Un : n e N} of pairwise disjoint collections of nonempty open sets in Y and a sequence {Kn : n e N} of countable subsets of F, as follows. Let Ui = {Y}. At stage n + 1, where n > 0, for each U e Un, pick an arbitrary element kn(U) e U Pi Gn and choose an infinite pairwise disjoint collection Vn(U) of nonempty open subsets of Y \ {kn(U}} that converges to kn(U), and satisfies (1) \JVn(U) CUnGn.

Let Un+l = \J{Vn(U) :UeUn}

and Kn = {kn(U) : U G Un}. Note that

1.9. HOMOGENEOUS ZERO-DIMENSIONAL SPACES

79

(2) Kn C U Un, U Un+i C |J Un \ Kn and |J U n+ i C Gn. This completes the construction of the desired sequences. Put K = U^°=i Kn- Then K is clearly countable. It remains to show that K has no isolated points and has the property that K \ K C P) 9For each n and U £ Un the subset {kn+i(V) : V G V n+ i([/)} of Kn+i converges to kn(U) since Vn(U) converges to kn(U), and it does not contain kn(U) since kn(U) $ Kn+i. So kn(U] is not an isolated point of K. We will next show that 1C \ K C f| 9. It follows from (2) that (3) For all j G N and U G Uf.

Ur\K C {kj(U)} \J\JV j ( U ) . To see this, consider any j, s G N and C7 € Uj. If s > j then

and so Ks n C7 C U(T/ G U J+i : F C U} = \J Vj(U). Moreover, if s = j then kj(U] eKsnU = {k0(V) :VnUJ:Q} = {kj(U)}. Finally, if s < j then Ks n U C Ks n |J Uj+i C Ks n |J U s+ i = 0. The reason why the proof works is that by (1) and (3) we get: (4) For every J E N and U 6 Uj we have K n U C U. Now consider any x G K \K. We claim that for every j there is an element Uj e Uj which contains x. Since \J Uj C Gj this implies that

which is as required. Let U\ = Y. Next, consider any j ' G N and assume that we already found Uj. Since x £ K but x 7^ kj(Uj) and Vj(Uj) converges to kj(Uj) we conclude by (3) that there is an element UJ+i G Vj(Uj) with

x G c/j+i n /c. Then x G t/j+i because of (4). This completes the construction of the sequence {Un : n G N}. <> So we are done.

D

Corollary 1.9.13. Let X be a space. Then every closed subspace of X is a Baire space if and only if X has no closed subspace homeomorphic to Q. Proof. Necessity is clear. To prove the sufficiency, it is enough to prove that if X is not a Baire space then it contains a closed homeomorph of Q. But this is the statement of Theorem 1.9.12. D

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Exercises for §1.9. 1. Let A be a subset of R such that A+Q = A. Prove that A is homogeneous. 2. Let

A = { A C R : Q C A , A + Q = A, An

(J

A + mr = 0}.

n£Z\{0}

(1) Prove that A / 0. (2) Let 3C be a chain (under inclusion) of elements of A. Prove that (3) Apply the Kuratowski-Zorn Lemma to conclude that A contains a maximal element, say AQ. (4) Prove that if x £ K. \ AO then there exist n G Z \ {0} and a G AO such that x = a + mr. (5) For each n <E Z \ {0} let An = {x G R \ AO : (3 a <E A 0 )(or = a + mr)}. Prove that the collection {An : n G Z} partitions R. (6) Prove that for every n € Z we have An = An + Q. (7) Prove that R can be decomposed into two homeomorphic homogeneous sets. 3.

(1) Prove that there is a Borel subset E C R such that EnaR\E. (2) Prove that if E is a homogeneous subset of R such that E ~ R \ E then E is not Borel.

4. Let Y be a space without isolated points. Prove that if y £ y and £7 is a neighborhood of y then there is a pairwise disjoint infinite collection V of nonempty open subsets of U\{y} such that V converges to y and (J V C [7. 1.10. Inverse limits Sometimes it is possible to 'approximate' a complicated space 'arbitrarily closely' by less complicated objects. In this section we shall formally define what we mean by this and we shall derive an interesting 'approximation theorem'. On Page 137 we will discuss Freudenthal's Approximation Theorem about approximations of spaces by polyhedra. An inverse sequence is a sequence of pairs (Xn,fn)n of spaces Xn and continuous functions fn: Xn+i —> Xn. The spaces Xn are called coordinate spaces and the mappings fn are called bonding maps. The inverse limit of the inverse sequence (Xn,fn)n, denoted by is defined to be the following subspace of the product of the Xn: oo

G JJ Xn : (Vn e N)(/ n (z n + i) = z n ) . n=l

1.10. INVERSE LIMITS

81

It will sometimes be convenient to let Xoo denote the inverse limit of the sequence (Xn, fn)n. For every n 6 N, the restriction to hjn(JC n ,/ n ) n of the projection onto the n-th factor of the product O^Li Xn shall be denoted by /£°. By definition it follows that for every n, fn o f^_l = f^°. The functions /£° are called the projections of the inverse sequence. In addition, for ra > n the composition fm ° • • • ° fn • Xm+i -> Xn +1

shall be denoted by /™ . Observe that /£+1 = /n. Finally, /™ is defined to be the identity function from Xm to Xm. Lemma 1.10.1. Let (Xn,fn)

be an inverse sequence. Then the collection

is a base for the topology of]nn(Xn,fn)nIn addition, this base is closed under finite unions and finite intersections. Proof. Let U be a relatively open subset of X^. In addition, let x £ U. By the definition of the product topology, there exist N € N and for every i < N an open oen subset Vi C Xi such that N

oo

n

i=l

i=N+l

Since x\ G Vi, x% G V-2 and f\(x-2] = a?i, it follows that Continuing in this way, we obtain that AT

An easy verification shows that which is as required. The second part of the lemma follows easily from the fact that m

foo _

foo

/ n Jm Jn for all m > n. Simply observe that a basic open subset of .Y^ that 'depends' on the n-ih element of the inverse sequence, i.e., is of the form

for certain open U C X, also 'depends' on the m-th element of the sequence for every m > n. This is true since for every m > n and U C Xn we have

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1. BASIC TOPOLOGY

So for a finite collection of basic open sets U in Xoo it is possible to identify an index TV € N such that every U £ U 'depends' on XN. D We will now present a few illuminating examples of inverse sequences and their limits. Example 1.10.2. Consider the inverse sequence (A r n ,/ n ) n , where Xn = X and fn is the identity on X for every n. This is a trivial inverse sequence. A moments reflection shows that its inverse limit is (homeomorphic to) X. The next example shows that inverse limits are generalizations of products (see the exercises for details). Example 1.10.3. For every n let Xn be a space. The sequence n

I I -<*-rm

ra=l

where rn : Om^i Xm —> 0™=! Xm is the projection, is an inverse sequence with the infinite product

71=1

as its inverse limit. Example 1.10.4. Let X be a space and for every n let Xn be a subspace of X such that Xn+i C Xn. Form an inverse sequence (Xn,in)n by letting in be the inclusion Xn+i °->- Xn. Then \Ym.(Xn,in}n is homeomorphic tofT=l*n.

From this example we see that an inverse limit of nonempty spaces can be empty. Since the intersection of a decreasing sequence of nonempty compact spaces is nonempty, for compact spaces one cannot get such an example (see Lemma 1.10.10(2) below). Example 1.10.5. Let X be a compact space. By Corollary A. 4. 4 we may assume that X is a subspace of Q. For every n let Xn = rn[X], n

where rn : Q —> J is the projection. In addition, let fn : Xn+i —>• Xn be defined by fn(Xi,...,Xn+i]

A trivial verification shows that

= (xi,...,Xn). A Hm(A n , fn)n = X .

There are several simple but useful observations that we have to make about inverse limits. First, let (Xn, fn)n be an inverse system, and let A C N be infinite. Write A as {mi, . . . , m n , . . . } in such a way that

mi < m-2 < • • • < mn < • • • .

1.10. INVERSE LIMITS

83

The sequence (Xmi, fm*+l)i is also an inverse sequence. By abuse of notation it will be denoted by (A A n ,/ n ) n e ^. It is called an inverse subsequence Of

(Xn,fn)n.

Lemma 1.10.6. If (Xn, fn}n^A is an inverse subsequence o f ( X n , f n ) n the limits \im(Xn, fn}n^A and ]im(Xn, fn)n are homeomorphic. Proof. The function (p: \im(Xn,fn)n

then

—>• Hm(A"n, fn}n^A defined by

is easily seen to be a homeomorphism.

D

So passing to a subsequence makes no difference in the limit (as with ordinary limits). Lemma 1.10.7. Let (Xn,fn)n and (Yn,gn}n be inverse sequences. Suppose that for every n G N there is a continuous function rn : Xn —>• Yn such that the diagram Xn 4- X

V JLn

g " V *•'J-n+l

commutes. Then the function (p: lim(Xn, fn)n -> lim(Yn,gn)n defined by <{>((zn)n) = (rn(xn)}n is continuous. Proof. This is clear since (p is the restriction to hm(J^ n ,/ n ) n of a product map from f]£Li Xn ->> H^Li Yn (Exercise A. 1.13). D These simple results will sometimes be used in the following situation. Suppose that we are inductively defining an inverse sequence (Xn, fn)n and that we would like to construct it in such a way that Hm(X n ,/ n ) n admits a selfmap (p with certain special properties. This is sometimes achieved by constructing for every n a function (pn : Xn+i -> Xn such that the diagram

commutes. By Lemma 1.10.6 it follows that the inverse sequences (Xn,fn)n and (Xn,fn}n>2 have the same limit, and Lemma 1.10.7 shows that the functions (pn induce a canonical selfmap of that limit.

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1. BASIC TOPOLOGY

Lemma 1.10.8. Let (Xn: fn)n be an inverse sequence with surjective bonding maps. Then the projections of (Xn, fn)n are also surjective. Proof. Fix n G N and take an arbitrary x 6 Xn. Since fn is surjective, there exists xn+i G Xn+i with fn(xn+i) = x. Similarly, there exists x n+2 6 Xn+2 with / n+1 (x n+2 ) = xn+i, etc. Define

etc. The sequence (xn)n 6 O^Li Xn belongs to hm(.Y n ,/ n ) n and since f°°((r } } — ^n r — ^i r Jn \\^n)n)

we are done.

D

Remark 1.10.9. Let (Xn,fn)n be an inverse sequence. For every i define the subspace X(i) of fl^Li Xn by oo

X(i) = {x 6 J| Xn : (Vn < i)(fn(xn+l)

= x n )}.

n=l

It is clear that X(i) is closed in H^Li Xn and nonempty and that X(i + l) CX(i)

for every i, and, finally, that

1=1

So ^oo is a closed subspace of H^Li Xn. This shows that if IP is a topological property which is 'closed hereditary and (countably) productive' then X^ has O5 if all Xn have O3. This simple observation allows us to prove many properties of inverse limits. Compactness and completeness properties of inverse limits. It is natural to ask when an inverse limit is compact or topologically complete. Lemma 1.10.10. (1) The inverse limit of an inverse sequence of topologically complete spaces is topologically complete. (2) The inverse limit of an inverse sequence of nonempty compact spaces is nonempty and compact. Proof. Observe that compactness and completeness are closed hereditary and productive (Lemma A.6.2). So we are done by Remark 1.10.9. D

1.10. INVERSE LIMITS

85

Connectivity properties of inverse limits. Detecting connectivity in topological spaces is important. For products this is easy: a product is connected if and only if all factors are. But inverse limits do not share this property with products, as the following example shows. Example 1.10.11. A disconnected inverse limit of connected spaces. Let {qn : n 6 N} be an enumeration of the rational numbers in I. For every n G N define 00

Xn = ( I x { 0 , l } ) U (J ({
Then X\ D X% D • • • D Xn ! > • • • , every Xn is connected and

n-l

is disconnected. So we are done by Example 1.10.4. Observe that the inverse limit in this example is compact. So compactness of the inverse limit is not enough to guarantee connectivity. If all spaces involved are continua, then there are no problems. Lemma 1.10.12. The inverse limit of an inverse sequence of continua is a continuum. Proof. Let (Xn, fn)n be an inverse sequence of continua. Let the spaces X ( i ) for every i be defined as in Remark 1.10.9. We claim that every X ( i ) is a continuum. Indeed, fix i and define a function (p: f l l i ^-j ~^ -^"(0 by Then ip is evidently a continuous surjection. It therefore follows that X ( i ) is a continuum since Hyl; ^-j ^ s It now suffices to apply Proposition A. 10. 7. D The dyadic solenoid. As an example of the power of inverse limits, we construct a very peculiar space here. Consider the following inverse system gi < f gi , f ... , f gi , f where / : S1 —> S1 is the function f(z} = z2 (complex multiplication). The inverse limit of this system is denoted by £2 and is called the dyadic solenoid. We list a few properties of £2 that we can derive easily from previous results. First, it is a continuum by Lemma 1.10.12. In addition, since S1 is a topological group under complex multiplication, and / is a continuous homomorphism, it follows easily that £2 is a closed subgroup of the product group (S1)00. So £2 is a topological group and hence a homogeneous space.

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1. BASIC TOPOLOGY

A continuum X is called indecomposable if it is not the union of two proper subcontinua. So I is not indecomposable (= decomposable) since it is the union of [0, 1/2] and [l/2,1]. Lemma 1.10.13. Let X be a continuum. The following statements are equivalent: (1) X is indecomposable, (2) every proper subcontinuum of X is nowhere dense. Proof. That (2) =£> (1) is clear since no space is the union of two nowhere dense subsets. For (1) => (2), suppose that X contains a proper subcontinuum C with nonempty interior U. Let £ be the collection of components ofY = X\U. Then every E 6 £ meets U \ U by Corollary A.10.5. So every E e £ meets C but misses C7. If |£| = 1 then clearly X is the union of two proper subcontinua. Suppose therefore that £ has more than one element. Then Y contains a proper nonempty relatively clopen subspace W by Proposition A.10.3(1). Every member from £ either is contained in W or is disjoint from it. So W is the union of a family of subcontinua of Y which all meet C. So C U W is a proper subcontinuum of X. Similarly, C\J(Y\ W) is a proper subcontinuum of X. Since they cover X, we arrive at a contradiction. D Corollary 1.10.14. A nondegenerate locally connected continuum is decomposable. Proof. Let x and y be distinct points in X and let U be a connected neighborhood of x whose closure misses y. Then L7" is a proper subcontinuum of X which is not nowhere dense. So X is decomposable by Lemma 1.10.13. D Theorem 1.10.15. The dyadic solenoid is indecomposable. Proof. Let C C S2 be a proper subcontinuum. For every n put Cn — f%°[C]. Since C is proper there exists n E N such that Cn is a proper subset of S1. Observe that Cn is a subinterval of S1. Now f ~ l [ C n ] is the union of two pairwise disjoint nonempty subintervals of S1 of arc-length exactly one half the arc-length of Cn. In addition, it contains C n +i. By connectivity, Cn+i is contained in one of them. We conclude that the arc-length of Cn+i is less than or equal to half of the arc-length of Cn. So for large N the arc-length of CN is arbitrarily small. Now assume that C has nonempty interior. Then there are i E N and a nonempty open interval U C S1 such that (fi°)~l[U] C C. But the arclength of f~l[U] is the same as the arc-length of U, etc. So the arc-length of CN does not tend to 0 for large N. This is a contradiction.

1.10. INVERSE LIMITS

87

So we conclude that C has empty interior. Since C was arbitrary, we conclude that £ 2 is indecomposable by Lemma 1.10.13. D Remark 1.10.16. It is easy to show that the dyadic solenoid contains an arc. For every n put In = {ei7Vt :0
Then gn = f \ I n +i —»• I n is a homeomorphisrn. The inverse limit of the inverse sequence (I n?
C=\J C(B) B£-B is equal to the composant E of x. It is clear that C is a subset of E. Now if y 6 E then there is a proper subcontinuum K of X such that both x and y belong to K. Since K is proper, there is an element B e 23 which misses K. But then K C C(B) and hence y G C(B] C C. So E is the union of countably many proper subcontinua of X. An application of Lemma 1.10.13 therefore yields the desired result. D A continuum Z with Z2 w Z but Z°° $ Z. As a second example of the power of inverse limits, we construct another very peculiar space here. Most of the known examples of compact spaces X such that X2 w X (which implies that Xn w X for every n] have the property that X00 w X. Think of the Cantor set and the Hilbert cube as examples of such spaces. We will use inverse limits to demonstrate that this is not always the case. Theorem 1.10.18. Let Y be a locally compact space homeomorphic to its own square. Then Y has a compactification homeomorphic to its own square.

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1. BASIC TOPOLOGY

Moreover, if the one-point compactification of Y is a continuum then this compactification can be chosen to be a continuum as well. Proof. The compactification X of Y is the inverse limit of a suitable inverse sequence (anY, fn}n of compactifications anY of Y, where for each n, fn is a continuous function from an+\Y —> anY which restricts to the identity on Y. This implies that f[an+iY \Y] = anY \ Y by Exercise A.4.2. In order to ensure that X2 w X we will construct the inverse sequence in such a way that for each n E N there is a homeomorphism hn: an+iY —>• anY x anY so that the diagram h

a2Y

<-^—

a3Y

<

•••

a2Y x a2Y i •• • commutes. Let a\Y be the one-point compactification of Y and let h be any homeomorphism from Y onto Y2. There is a compactification a2Y of y such that h extends to a homeomorphism hi: a2Y —> aiY x a\Y. Since a\Y is the smallest compactification of Y, the identity mapping on Y extends to a mapping /i: a2Y -» oiF. Similarly, there is a compactification a^Y of y such that h extends to a homeomorphism h2: a3Y —> a2Y x a2Y. Define /2 : a3Y ->• a 2 y

by

f2 = hl1 o(fi x fi)oh2. It is easily seen that /2 f y is the identity on Y. In the same way define the compactification a^Y', and the functions /ts and /s, etc. Since the hn's are homeomorphisms, it follows from Lemma 1.10.7 applied twice and Lemma 1.10.10(2) that X = Lirr^a^y, / n ) n is a compactification of Y homeomorphic to its own square (cf. Exercise 1.10.9). Observe that if aiY is a continuuum then so are all the other a n y's. As a consequence, X is a continuum as well by Lemma 1.10.12. D Now consider the Cantor set C, and let B be the topological sum of NX C and a one-point space. Clearly, B2 = ((N x C) © {pt}) x ((N x C) 0 {pt}) w (N2 x C 2 ) © (N x C) 0 (N x C) © {pt} « (N x C) 0 {pt} w JB. Put Y = B x [0,1) x Q.

Lemma 1.10.19. Y2 w y.

1.10. INVERSE LIMITS

89

Proof. Simply observe that F2 = (B x [0,1) x Q) x (B x [0,1) x Q)

&(BxB)x ([0,1) x [ 0 , l ) ) x ( Q x Q ) « £ x ([0,1) x I ) x Q w 5 x [0,1) x Q

= y. So we are done.

D

Let Z be the compactification of Y we get from Lemma 1.10.19. Observe that the one-point compactification of Y is connected since it is the cone over B x Q. From this it follows that Z is a continuum as well. Lemma 1.10.20. Z ^ Z°°. Proof. Striving for a contradiction, suppose that h is a homeomorphism from Z onto Z°°. Since Y, being locally compact, is open in Z, h[Y] is open in Z°°. In addition, since B has an isolated point, Y contains an open copy of [0.1) x Q. Therefore, Z°° contains an open copy of [0,1) x Q, say F. Since F is open and nonempty, it contains a basic open subset. So there are an integer n and for each i < n a nonempty open set Ui C Z so that Ui x U-2 x • • • x Un x Z x Z x • • • C F.

Hence Z is (homeomorphic to) a retract of some open subset of F, or, equivalently, Z is homeomorphic to a retract of some open subset of [0,1) x Q. Since [0,1) x Q is locally connected, this implies that Z is locally connected (Exercise A.12.4). Its open subspace Y is therefore locally connected as well. But this implies that C is locally connected, which is a contradiction. D Remark 1.10.21. In view of the existence of a compact space homeomorphic to its own square but not to its countable infinite product, the question naturally arises whether there is a compact space X which can be mapped onto X2 but not onto X00. Interestingly, the answer to this question is in the negative (see Exercise 1.10.8). Brown's Approximation Theorem. Before we are in a position to formulate and prove an interesting 'approximation theorem' for inverse sequences, we have to fix some notation. Let (Xn,fn)n be an inverse sequence with inverse limit X^. It is sometimes useful to have an explicit admissible metric for XQQ . For every n £ N let dn be an admissible metric for Xn which is bounded by 1 (Exercise A. 1.6). The formula oo

g(x,y) = ^1"ndn(xn,yn) n=l

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1. BASIC TOPOLOGY

defines an admissible metric for the product of the Xn (Exercise A. 1.10), and hence also for XOQ. Observe that g is also bounded by 1. Lemma 1.10.22. Let (Xn,fn)n be an inverse sequence. In addition, for every n let dn be an admissible metric for Xn which is bounded by I. Then with respect to the metric Q for XOQ — hjn^n,

fn)n

defined above, we have that every /£° : Xoo —> Xn is a 2~(n~^ -mapping. Proof. This is easy. Fix n £ N and let p, q e ^oo be such that

fn(p) = fn(o)Observe that pi — qi for every i < n. Consequently,

We conclude that /£° is a 2~( n ~ 1 )-mapping.

D

Brown's Approximation Theorem 1.10.23. Let(Xn,fn)n be an inverse sequence of compact spaces with inverse limit X^,. If each fn is a near homeomorphism, then so is each f£°. In particular, if each fn is a near homeomorphism then Xoo is homeomorphic to X\ (and hence to every Xn). Proof. We shall prove that /f0 : Xoo —>• X\ is a near homeomorphism. For each n, let dn be an admissible metric for Xn which is bounded by 1 and let Q be the metric for X^ introduced above. Let E > 0. Inductively, we shall construct for every n G N a homeomorphism hn : Xn+i —>• Xn such that the functions gn : X^ —> X\ defined by gn = hi o • • • o hn o f^_l have the following properties: (1) di(5i,/r)< e / 2 , (2) dl(gn,gn+l) <3~n-e, (3) di(gn,gn+1)<3-n-inm{di(gi,Si/.(X00,Xl))

:l
Since j\ is a near homeomorphism and /f° = j\ o /^°, it is clear that h\ can be constructed so that condition (1) is met. Suppose that the homeomorphisms hi are defined for 1 0. Since gn = hlo---ohno f™+1 =h1o---ohno fn+1

1.10. INVERSE LIMITS

91

and fn+i is a near homeomorphism, we can find a homeomorphism hn+i '• Xn+2 —> Xn+i such that g(hi o---ohno fn+l o f™+2,hi o---ohno hn+i o /£j.2) < 8 (use the fact that hi o • • • o hn is uniformly continuous (Exercise A. 5. 18)). It is clear that hn+i is as required, which completes the inductive construction of the hn. By (2) it follows that the sequence (gn)n is -oo

exists. Since gn £ ^ ( X o c ^ X i ) for every n, Proposition 1.3.8 implies that Observe that (1) and (2) imply that di(/i°,0)<e. Also, (3), Lemmas 1.6.1 and 1.3.10 show that

n=l

which is as required.

D

Dynamical systems. A dynamical system is a pair (X, f) where X is a space and /: X —>• X is continuous. In topological dynamics one studies the behaviour of the iterates of /. Inverse limits play a role in certain considerations there. We will not go into this, but we remark that the following considerations that will play a role at one point in our chapter about dimension theory were motivated by well-known ideas in topological dynamics. Let (X, /) be a dynamical system, and consider the following inverse sequence

X < f X < f ••• < f X < f with inverse limit X^. So the same map is repeated infinitely often in the inverse system under consideration. Define the so-called shift a: X00 -» X00 as follows: cr(xi,x2,...) = (/ (zi), £1,2:2,... )• It is clear that a is well-defined. The compositions / o TTi 7Tn_i

(n — 1), ) > ^ 1), i\ (n

are all continuous from which it follows that <j is continuous. It is in general not true that a is a homeomorphism.

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Define another shift function r: X00 —> X00 by T(>i,Z2,...) = (ar2,X3,...)It is clear that r is continuous as well. A moments reflection shows that <7[Xoo] C Xoo,

r[Xoo] C Xo

Moreover, if x £ A^ then cr(r(x)) = (7(2:2, £ Similarly, r(cr(;r)) = We conclude that a \ X^ is a homeomorphism, with inverse r \ X^. By abuse of notation, we will denote a \ X^ also by a. Lemma 1.10.24. Let (AT, /) be a dynamical system, and let a: X^ —>• Xc denote the shift homeomorphism. Then for each n the diagram Y

A.

' n Y \- .A no

,00

Y <'" Y AOO A

commutes. As a consequence, if f is fixed-point free then so is a. Proof. This is clear since if x e X^ then / o f£°(x) = f ( x n ) and which is equal to f ( x \ ) if n = I and x n _i — f ( x n ) if n > 1. This implies that if x is a fixed point of a then f£°(x) is a fixed point of /, which establishes the second statement. D So in some vague sense, every continuous function /: X —>• X can be 'approximated' by a homeomorphism (cf. Lemma 1.10.22). Exercises for §1.10. A space X is called isotopically homogeneous provided that for every x, y £ X there exists an isotopy H : X x I —>• X such that HO = lx and HI(X) = y. 1. Let A" be the inverse limit of an inverse sequence of copies of I. Prove that X has the fixed-point property. 2. Prove that the circle S1 is not homeomorphic to the inverse limit of an inverse sequence of copies of I. 3. Prove that the statement in Example 1.10.3 is correct.

1.10. INVERSE LIMITS

93

4. Prove that the statement in Example 1.10.4 is correct. 5. Let (X n , fn) be an inverse system with open surjective (continuous) functions. Prove that for every n € N the function is open. 6. Let (Xn,fn)n be an inverse sequence such that every fn is a homeomorphism. Prove that hm(X n ,/ n ) n is homeomorphic to X\ (and hence to every Xn). 7. Let (Xn, fn)n be an inverse sequence. In addition, for every n let An C Xn be such that fn[An+l] C An; put gn — fn f A n +i : An+i —> An . Prove that the inverse limit and the subspace (x £ hm(Xn,/n)n : (Vn G N)(x n G A n )} of Um(Xn, fn)n are homeomorphic. ^•8. Assume that X is a compact space. Prove that X maps onto X2 if and only if it maps onto X00 . 9. Let (anX,fn)n be an inverse sequence consisting of compactifications of the space X. Assume moreover that for every n the function fn restricts to the identity on X. Hence aiX < a2X <••• X —> anX restricts to the identity on X. (Hence a^X > anX for every n.) 10. Let (Xm fn) be an inverse sequence and let Y be a space such that for each n there is a continuous function gn : Y —> Xn such that fnogn+i — gn. Prove that the function g : Y —>• ]\m(Xn, fn)n defined by

g(y) = (g^(y)}n is well-defined and continuous. 11. Let X be a space and let

Ji C ^2 C • • • C yn C • • • be an increasing sequence of Wallman bases of X. (1) Prove that for every n there is a continuous function which extends the identity on X. (2) Prove that 3 = (Jn°=1 3n is a Wallman base of X. (3) Prove that the inverse limit of the sequence (a; (X,3~n ),/n)n and u; (X", 9") are equivalent compactifications of X.

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12. Let (Xn, fn)n be an inverse sequence with compact spaces. Prove that if A and B are disjoint closed subsets of X^ then there exists an N E N such that for each n > N the sets /~[A] and f%°[B] are disjoint. 13. Let X and Y be compact spaces and let /: X —>• Y and g: Y —> X be continuous functions such that both g o f : X —> X and f o g : Y —>• Y are near homeomorphisms. Prove that X and Y are homeomorphic. 14. Prove that the statement in Example 1.10.5 is correct. 15. Prove that in the statement in Remark 1.10.5 the compactness assumption is essential. 16. Prove that an indecomposable continuum is not path-connected. 17. Give an example of a homogeneous continuum which is not isotopically homogeneous. 18. Let X be a compact space and let /: X —> X be continuous. Prove that

K=f\f[X\ i=l

is nonempty and also that f[K] = K. 19. Let X be an indecomposable continuum and let E be one of its composants. Prove that if E is not dense then E is closed. 20. Let X be an indecomposable continuum and let E and F be distinct composants of X. Prove that E and F are disjoint. 21. Prove that for a continuum X the following statements are equivalent: (1) X is hereditarily indecomposable. (2) If C, D are arbitrary subcontinua of X such that C n D ^ 0 then C C D or D C C. 22. Let X be a hereditarily indecomposable continuum, and let /: X —> Y be monotone. Prove that Y is a hereditarily indecomposable continuum. 23. Let X be an indecomposable continuum. Prove that the family of composants of X is uncountable. It can be shown that the dyadic solenoid £2 cannot be imbedded in the plane. It is an interesting question whether there is a planar indecomposable continuum. This question will be considered in the remaining part of this section. Let o, b and c be three distinct points in the plane. Inductively, construct finite simple chains Ci, 62, • • • of open balls having among other things the following properties: (1) mesh(e ri ) < 2 ~ n , (2) Cn < C n _ i . But these are not all conditions. We require the C n 's to follow a certain pattern. The collection Ci is a simple chain from a to c through b, 63 is a simple chain from a to b through c, 63 is a simple chain from b to c through a (see Figure 5). Then repeat this process at infinitum. For every n let Cn = |J Cn and let P = H^Li Cn-

1.11. HYPERSPACES

95

24. Prove that P is a continuum. 25. Prove that F is indecomposable.

Figure 5. 1.11. Hyperspaces In this section we shall introduce the hyperspace of a topological space. Hyperspaces are used later to construct various interesting examples. Let X be a space. The collection of all non-empty compact subsets of X shall be denoted by 1X . We will endow 1X with a suitable topology. For E,F e 2X define the Hausdorff distance Qn(E,F} between E and F by

QH(E,F] = infje > 0 : E C B(F,e) and F C B(E,e)}. The compactness of E and F implies that gn(E,F) e [0, oo) (Exercise 1.11.1). Lemma 1.11.1. g# : 2X x 2 X —> [0, oo) is a metric. Proof. We shall only present a proof of the triangle inequality, the other verifications being trivial. Take E, F, G e 2 X , let e > 0 and put 6 = £/2. By definition we have (*)

E C B(F,6 + QH(E,F}) and F C B(G, 6 + QH(F,G}}.

Take an arbitrary point x G E. By (*) there exists y E F such that

Analogously there exists z G G such that e(y,z

We conclude that Q(X,Z)

<e

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Therefore, since x was an arbitrary point of E, E C B(G,e + QH(E,F] + QH(F,G)). Similarly, G C B(E,e + QH(E,F] + QH( Since e was an arbitrary positive number, we obtain as required.

D

The metric QH is called the Hausdorff metric on 2X . We endow 2X with the topology derived from this metric and call it the hyperspace of X. Define e: X —»• 2X by e(x] — {x}. It is easily seen that e is an isometry, hence an imbedding, and that e[X] is a closed subset of 2X (Exercise 1.11.3). It will sometimes be convenient to identify X and e[X], Compactness and completeness properties of hyperspaces. We

will show that compactness and completeness are 'preserved' by the hyperspace operator. We need a simple lemma first. Lemma 1.11.2. Let (X, Q) be a space and (An)n a Cauchy sequence in 2X . LetA= (J~=i^n. Then (1) The metric p = Q \ (A x A) is totally bounded. (2) If AOO — linin-^oo An exists in 2X then it is equal to (f) {x eX : ( V n G N)(3x n e A n )(lim n _, 00 a; n = x)}, and also to (+)

\\n-l U m > n ^ T i -

(3) If AOO — linin^oo An exists in 2X then A is compact and oo

A = Avo U |J An. n=l

(4) If Q is complete then limn-^ An exists in 2X (and hence has to be equal to the sets (f) and (J) in (2)J. Proof. For (1), let e > 0. There exists A" e N such that QH(An,Am) < l/±e for all n,m > N. Consider AN and let -F C AN be a finite 1/4e-net (by Theorem A. 6.1, g\(Apf x AN) is totally bounded, whence the desired net exists by Exercise A. 6.2). Fix n > N. Since QH(AN,An) < l/±e, An C B(AN, 1/4£:)So for every x e An there exists y £ AN such that Q(X, y) < V4e. For such a y there exists z e F such that Q(z,y] < l/4e. We conclude that F is a y2e-net for An U AN- Since n was arbitrary, F is even a l/2e-net for \Jn>N An, and hence an e-net for \Jn>N An- Since \Jn=i An is compact, it has a finite e-net

1.11. HYPERSPACES

97

by the same reason AN has one. The union of these two finite e-nets is a finite e-net for A. So p is totally bounded by Exercise A.6.2. For (2), let B = [x G X : (Vn G N)(3xn G ^(lim^^ xn = x)} and C = n^Li Um>n Am, respectively. We will prove that AOO CB C C C AOO. Pick an arbitrary x G AQO. For each n G N there exists a point xn G An such that ^(x,x n ) < 2g(x,An). We claim that the sequence (xn)n converges to x. To this end, let e > 0. Since (An)n converges to A^, there exists TV G N such that for each m > N, A^ C B(Am, l/<2^}- From this it follows that for m > N we have ^(x, A m ) < l/2£, i.e., 0(:r,:rm) < e. Hence AOQ C 5. That B C C is trivial. Striving for a contradiction, assume that there exists a point x G C\Aoo. Let e = g(x,AOQ) and observe that e > 0. Since A n —>• AOO in 2 X , there exists N G N such that for every n > N we have An C 5(Aoo, V2£)- As a consequence, (^ 4 XV G (J A n

(

which is a contradiction. For (3), let U be a cover of A* = AOO U (J^li ^™ by open subsets of X. Since AOO is compact, there is a finite y C U such that AOO C y 3. By Corollary A.5.4 there exists e > 0 such that B(A00^e) C y J. Since (A n ) n converges to AQQ, there exists TV G N such that An C 5(Aoo,£) C y ^ for n > TV, and so n>N

Since 9" is finite and Un_T1 An is compact, we conclude that A* can be covered by finitely many elements of U. So A* is compact, and since ^4^ C A by (2), it follows that A* C A. But since (J^Li An is dense in A and is contained in A*, it also follows that A C A*. For (4), we first observe that by (1) the metric p — Q \ (A x A] is totally bounded. Since A is closed, p is also complete. So A is compact by Theorem A.6.1. Put K = H^Li U m > n ^ ™ - Then K is the intersection of a decreasing sequence of nonempty closed subsets of the compact space A, and hence is nonempty and compact itself. Hence K G 2 X . To prove that An ->• K in 2 X , let £ > 0. By the compactness of K, there exists mi G N such that (5)

AnCB(K,e).

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Since (An)n is Cauchy, there exists ra2 E N such that QH(An,Am) < £/2 for n, ra > 7712- As a consequence, for n > rri2,

(6)

^ C |J ATO TT1>TO2

From (5) and (6) it now follows that @H(K,An) < e for n > max(rai,m2), which is as desired. D Theorem 1.11.3. Let X be a space with admissible metric Q. Then (1) If Q is complete then so is QH(2) If Q is totally bounded then so is QHProof. Observe that (1) follows from Lemma 1.11.2(4). For (2), assume that Q is totally bounded. Let E > 0 and let y be a finite open cover of X with nonempty sets of diameter less than e. For every F e 3 pick XF G F. For A e 2X, put BA = {xF : F n A ^ 0 } . Since J covers X, A C \J{F e y : F n A 7^ 0}, from which it easily follows that QH(A,BA) < £• Since 3* is finite, this shows that 2X can be covered by finitely many er-balls. D Corollary 1.11.4. If X is topologically complete then so is 2X . If X is compact then so is 2X . Proof. Both statements follow from Theorems 1.11.3 and A. 6.1.

D

Hyperspace maps. Let X and Y be spaces and let /: X —> Y be continuous. By Exercise 1.11.7 we may regard X and Y to be (canonical) subspaces of 2X and 2 y , respectively. There exists a canonical continuous extension 2^ of / over 2X , which is called the hyperspace map of /. Indeed, define 2-^ : 2X —>• 2Y in the obvious way by We shall prove below that hyperspace maps are continuous. They are of crucial importance in the process of analyzing hyperspaces. Lemma 1.11.5. Let X and Y be spaces and let /: X —>• Y be continuous. Then the hyperspace map 2? : 2X —> 2Y is continuous. Proof. Suppose that (An)n is a sequence in 2X converging to A 6 2X . Then obviously (An)n is Cauchy in 2X , and hence A — (J^Li An is compact by Lemma 1.11.2(3). As a consequence / \ A is uniformly continuous (Exercise A. 5. 18). Now let e > 0 and pick S > 0 such that if x, y £ A are arbitrary points with g(x,y) < 6 then 0(/(x),/(y)) < e. If D,E C A are nonempty

1.11. HYPERSPACES

and compact and 0H(D,E) < 6 then gH(f[D],f[E]) in 2Y.

99

< e. So, f[An] ->• f[A] D

The union operator. We shall now consider a more complicated 'hyperspacemap'. Let X be a space. Consider the space 22 , i.e., the hyperspace of the hyperspace of X. Each point A G 22 is a compact subset of 2X and therefore can be regarded as a 'compact' family of compact subsets of X. We shall prove that the union of such a family is compact. From this it follows that the 'union-operator' defined by \J(A)=\jA, is well-defined. We shall also prove that this operator is continuous. Lemma 1.11.6. Let X be a space and let A belong to 22 . Then \JA is a compact subset of X. Proof. Take a sequence (xn)n in \JA. We shall prove that it has a convergent subsequence in \JA. For each n there exists An G A such that xn e An. By compactness of the subspace A of 2X , we may assume without loss of generality that the sequence (An)n converges to a point A 6 A. From Lemma 1.11.2(3) it follows that B — A U U^=1 An is compact. Hence (xn)n has a convergent subsequence in B (Theorem A. 5.1). Since B C (J.A, it follows that this convergent subsequence is convergent in \JA. Now apply Theorem A. 5.1. D So from Lemma 1.11.6 we conclude that for every Jf , the union-operator

is well-defined. We shall now prove that it is continuous. Proposition 1.11.7. Let X be a space. Then the union-operator |J:2 2 "^2 X is Lipschitz and hence is continuous. Proof. Let QHH denote the Hausdorff metric on 22 . Take arbitrary points £ and y in 22 and let QHH (£)3~) — £• Fix £ > 0 and pick an arbitrary element x G U £. There exists E G £ such that x G E. Since E G £ there exists F G 3 such that QH(E, F) < 6 + e. This implies that E C B(F, 5 + e) and so there exists y G F C \J y such that Q(X, y) < 6 + e. We conclude that

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By interchanging the roles of £ and 3 we therefore conclude that

Since e was chosen arbitrarily, we get £>#(U £, (J IF) < $, as required.

D

Corollary 1.11.8. Let X andY be spaces and let g : X —> 2Y be continuous. Then the function g:2x^2Y defined by

g(A) = |J g(x) x£A

is continuous. Proof. The hyperspace map 29 : 2X —> 22 is continuous by Lemma 1.11.5. In addition, by Proposition 1.11.7, the union-operator (J : 22 —> 2Y is also continuous. Since g — [J o 25, we are done. D Let X be a space and for each n € N define yn(X] = {A e 2X : \A\ < n}.

Observe that ^(X) C ?2(X) C • • • C 3n(X) C • • • . The following simple consequence of Exercise 1.11.6, is basic for hyperspaces. Theorem 1.11.9. Let X be a space containing more than one point and let n € N. Then the function fn : Xn -»• 3n(X) denned by

is continuous. Corollary 1.11.10. Let X be a space and let AI, . . . ,An be nonempty subsets of2x. Then the function f : HlLi ^ ~> 2^ defined by f(A1,...,An)=A1\J---UAn is continuous. Proof. Again, let \J: 22 —> 2X denote the union-operator. By Theorem 1.11.9 we get that the function g: HlLi ^ ~* ^ defined by

is continuous. Since / = \J og, we are done.

D

1.11. HYPERSPACES

101

A canonical base for 1X . We used the Hausdorff distance to endow 2X with a natural topology. There is a different way to do this in terms of a canonical base for 2X yielding the so-called Vietoris topology on 2X . We will show that both topologies on 2X are the same. The alternative description of the topology on 2X in terms of a canonical base is sometimes very convenient. Let X be a space. For every finite collection A of subsets of X put (A) = {B e2x :B C\jA and for every A G A, A n B ^ 0}. Lemma 1.11.11. Let X be a space. Then (1) if A is a finite family of open subsets of X then (A} is open in 2X , (2) the collection 'B(X) = {(3) : 3 is a finite family of open subsets of X} is an open base for 2X which is closed under finite intersections. Proof. For (1), let .A be a finite family of open subsets of X and let D G (A}. For each A G A pick an arbitrary point XA £ An D. Since A is finite and consists of open sets, and D is compact, there exists by Corollary A. 5. 4 an e > 0 such that (3) for every A e A, B(xA,e) C A, (4) Now take E G 2X such that QH(D,E] < e. By (4), E C \JA. Fix an arbitrary element A G A. Since D C B(E,e), there exists an x G E such that Q(XA,X) < e. By (3), x belongs to A, i.e., E intersects A. This proves that the ball (in 2X) about D of radius e is contained in (.A). For the first part of (2), take arbitrary D G 2X and e > 0. By compactness of D, there is a finite family A of open subsets of X such that (5) D C\JACB(D,e}, (6) for every A&A,AnD^$, (7) for every A G A, diam(A) < e. So D G (A). Observe that (A) is open by (1). We claim that (.A) is contained in the e-ball in 2X about D. To this end, take an arbitrary E G (.A). By (5), we need only show that D C B(E, e). But this is trivial. For take an arbitrary element x G D. By (5) there exists A G A containing x. Since E G (.A) we have EnA 7^ 0, which implies by (7) that there exists x' G E with g(x, x') < e. We conclude that x G B(E,e). Finally, let fo = {Ui, . . . , Un} and 3\ = {Vi, . . . , Vm} be arbitrary collections open subsets of X. Put U = (J^o and V — U ^ i * respectively. An easy verification shows that

W n (9-1) - ({u n vlt . . . , u n vm, Ui n v, . . . , un n V}),

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as required.

D

Corollary 1.11.12. Let X be a space. Then

= |J n—l X

is dense in 2 . Connectivity in hyperspaces. We now turn to connectivity properties of 2*. Proposition 1.11.13. Let X be a space. Then (1) X is connected iff2x is connected, (2) X is locally connected iff 2X is locally connected. Proof. We prove (1). First suppose that X is connected. Since y^X) is dense in 2X (Corollary 1.11.12), it suffices to prove that it is connected, and for this it suffices to prove that every Jn(X) is connected. However, this follows directly by Theorem 1.11.9 since for each n, yn(X) is the image of Xn under the map fn : Xn -»• 2X denned by Conversely, assume that 2X is connected. We shall prove that X is connected. Striving for a contradiction, assume that X is not connected. So it is possible to write X as UuV, where U and V are disjoint nonempty open sets. Clearly,

In addition, ({U}},({V}}, and ({U, V}) are pair wise disjoint, nonempty and open by Lemma 1.11.11(1). This contradicts the connectivity of 2X . We prove (2).

Now assume that X is locally connected. Take A e 2X and let U be a neighborhood of A in 2X . By Lemma 1.11.11(2) there exists a finite family V of open subsets of X such that A £ (V) C U. Since X is locally connected, every component of every V e V is open (Exercise A. 2. 8). By compactness of A we may therefore assume without loss of generality that V consists of connected open sets. We claim that (V) is connected. First observe that for every V G V we may identify 2V and the subspace {A e 2X : A C V} of 2X (Exercise 1.11.7). Let V = {Vi, . . . , Vn}. Now define the function

i-l

1.11. HYPERSPACES

by

103

n

g(A1:...,An)= \jAi. 1=1 Then g is continuous by Corollary 1.11.10. By the first part of the Theorem, 2Vl is connected for every i < n. So (V) is connected, being the range of g. Conversely, assume that 2X is locally connected. Let x 6 X and let U be an open neighborhood of x in X. Since ({U}) is an open neighborhood of {x} in 2X (Lemma 1.11.11(1)), by local connectivity of 2X there is a connected neighborhood C of {x} in 2X such that {x} e C C {{£/}). Then V = C n X is a neighborhood of x in X and clearly

V CC = |J C C [7. We claim that C is connected. Suppose that this is not true. Let E and F be disjoint nonempty relatively open nonempty subsets of C that cover C. There are disjoint open subsets E',F' C X such that E1 n C = E and F' n C = F (Lemma A.8.1). Now observe that ( { E ' } ) , ({Ff}) and ({E' , F'}} are disjoint nonempty open subsets of 2X (Lemma 1.11.11(1)), that Gn({E'}} and en({F'}) are nonempty and that G C ({E'}) U({F'}) U ({E',F'}). This contradicts the connectivity of C. D Hyperspaces of subcontinua. Let X be a space and put G(X) = {A € 2X : A is connected}. It is called the hyperspace of subcontinua of X and is of particular interest in the case that X is a continuum. Lemma 1.11.14. Let X be a space. Then G(X) is a closed subspace of2x. So if X is compact then so is Proof. Let A be an element of 2X which does not belong to Q(X). Write A as the union of two disjoint relatively clopen nonempty subsets E and F. By Corollary A.4.3 there are disjoint open subsets E' and F' in X with E C E' and F C F' . Then ({E',F'}) is an open neighborhood of A which clearly misses C(X). The second part of the lemma follows from the first part and Corollary 1.11.4. D Whitney maps. A fundamental tool in hyperspace theory is the concept of a Whitney map which we will now discuss. Let X be a compact space. A Whitney map for 2X is a continuous function g: 2X —> [0, oo) having the following properties: (1) if A, B G 2X and A is a proper subset of B then g(A) < g ( B ) ,

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(2) g ( { x } ) = 0 for every x e X. Observe that (1) implies that if A G 2X and A\ > 1 then g(A] > 0. There are several ways to construct such functions. We will discuss only one of them. Consider the function space C(X,I). It is separable by Proposition 1.3.3. So let us fix a sequence (fi)i of functions in C(X, I) which form a dense subset of C(X, I). Define oj: 1X ->• I by

i—l

It is clear that uj is well-defined. Proposition 1.11.15. Let X be a compact space. The function u;: 2X —>• E denned in (*) is a Whitney map for 2 X . Proof. Let us first prove that for a fixed i the function 2X -> I defined by A H-> diam/; [A] is continuous. This is clear since it is the composition of the functions

here 2^ is the hyperspace map of /; (which is continuous by Lemma 1.11.5) and diam: 21 -> I is the function that sends every element A G 21 onto its diameter (which is continuous by Exercise 1.11.12). So u; is continuous by Exercise A. 3.1. Let A, .B G 2X be such that A is a proper subset of B. Fix an arbitrary element x e £?\ A There is by Lemma A. 3.1 a continuous function /: X -> 1 such that /[A] = {0} and /(x) = 1. Let n > 2 be such that

Observe that 0, 1 € /[-B] since A 7^ 0 which implies that diam fn [B] > %. In addition, f[A] = {0} so that diam fn [A] < i/4. From this we conclude that diam fn [A] < diamfn[B]. Also, diam ^ [A] < diam fl [B] for every i since A is a subset of B.

1.11. HYPERSPACES

105

This shows that

71—1

i=l

i=n+l

n— 1

oo 2

^ ~* i=l

diam

2

/i[#] + ~

n diam

/n[£] + XI z=n+l

which is as required. Since clearly u({x}) = 0 for every x G X, this completes the proof.

D

Let uj\ 2X -> [0,oo) be a Whitney function. For each t G [0,o;(A")] the collections {A £ 2X : u(A) = t} and {A € G(X) : w(A) = *}

are called Whitney levels. Lemma 1.11.16. Let X be a continuum, and w: 2X —>• [0,oo) a Whitney function. Then for every t G [0, w(X)] the Whitney level £ = {A G C(X) : u(A) = t}

is closed in Q(X} and moreover covers X. Proof. The first statement is trivial since £ is equal to

which is an intersection of two closed subsets of 2X (use that cj is continuous and Lemma 1.11.14). To check that £ covers X, let us consider any x E X: and let

Observe that 3C ^ 0 since Ct>({x}) = 0 and hence {x} G 3C. Since by continuity of a; and Exercise 1.11.2 the collection OC is closed in C(X) (and hence compact), there is an E G DC with : AT G 3C}.

To see that ui(E) = t, let us fix any e > 0, and let £/ be an open neighborhood of E with u(C) < u(E) + e for any E C C C 17 (Exercise 1.11.22). The

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component H of U which contains x intersects U \ U by Corollary A. 10. 5. So E is a proper subcontimmm of the continuum C = E U H . Hence

By (*), C g X for otherwise u(C) < u(E). Hence u(C) > t and so

u(E) >t-e. Since e > 0 was arbitrary, this shows that, indeed, x 6 E 6 £.

D

Corollary 1.11.17. Let JC be a hereditarily indecomposable continuum, and oj: 2X —>• [0, oo) a Whitney function. Then for every t G the Whitney Jevel £= 4 is an upper semi-continuous decomposition X . Proof. Suppose C and D are intersecting continua in X with o;(C) = Since ^C is hereditarily indecomposable, C C D or D C C (Exercise 1.10.21). If e.g., C C D then C = D for otherwise uj(C] < u(D}. This shows by Lemma 1.11.16 that £ is a partition of X. To prove that £ is upper semi-continuous, let A C X be an arbitrary closed set. Then

({A, X}} n £ = {E e £ : E n A ^ 0} X

is closed in 2 since both ({A,X}) Lemma 1.11.16). As a consequence,

and £ are (use Exercise 1.11.2 and

is closed in X by Lemma 1.11.6.

D

Exercises for §1.11.

1. Let X be a space and let E and F be compact nonempty subsets of X. Prove that gH(E,F) e [0, oo). 2. Let A" be a space and let F C X be closed. Prove that the sets {A € 2X : A n F ^ 0},

{A £ 2 X : .4 C F}

are both closed in 2A . 3. Prove that yn(X) is closed in 2X for every n e N. 4. Prove that 2 X is separable. 5. Let (A n ) n be a decreasing sequence of nonempty compact subsets of X. Prove that An ->• f L ^^ in 2X .

1.11. HYPERSPACES

107

6. Let X be a space containing more than one point and n G N. Prove that the function fn : Xn —>• 3~n(X) denned by

is continuous, and a homeomorphism if and only if n = 1. 7. Let A^ be a subspace of Y. Prove that the natural inclusion 2X <—>• 2y is an imbedding. 8. Let A be a space. In addition, let (An)n and (Bn)n be sequences in 2A converging to A and B, respectively. Prove that if An C Bn for every n then A C B. 9. Let X be a space and let (An)n be a sequence in 2 X converging to an element A 6 2X . Prove that if an G An for every n then there is a subsequence of (a n ) n converging to an element a G A 10. Let A <E 2 X , and for every n let An e 2* be such that A C An C D(A, l/n}. Prove that (A n ) n converges to A in 2 X . 11. Prove that 6(1) is homeomorphic to I 2 . 12. Let X be a space. Prove that the function diam: 2 X —> [0, oo) sending each element of 2X onto its diameter, is continuous. 13. Let X be compact and let E,F G 1X . Prove that , F) = maxjsup g(x, F), sup ^(x, E)}. 14. Prove that I" can be imbedded in 21 for every n. 15. Let X be compact and let 33 = {Bn : n e N} be an open base for it such that B\ = X. Let (A,){ be a sequence in 2 X . (1) Prove that there exists a decreasing sequence (Xn)n of closed subsets of X with the following properties: 1. Xi=X, 2. every neighborhood of Xn contains infinitely many Aj's, and 3. if Xn r\Bn ^ 0 then Xn\Bn has a neighborhood F containing at most finitely many A;'s. (2) Put D = Pl^Li Xn- Prove that there is a subsequence of the sequence (Ai)i converging to D. (So this is an alternative proof of the second part of Corollary 1.11.4.) 16. Let X Prove (1) (2)

and Y be compact, and let /: X —> Y be a continuous surjection. that the following statements are equivalent: / is open. The function F: Y ->• 2X defined by

is continuous.

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17. Let (X, £>) be a compact space and let /: X —> Y be a surjection which is both continuous and open. Prove that Y is homeomorphic to the subspace

of 2X . Conclude that formula d(yi,yi} =

QH(f~l(yi},f~l(y2}}

defines an admissible metric on Y. 18. Prove that the functions s,i: 21 —> I defined by s ( A ) = m i n A and £(/!) = max A are continuous. ^•19. Let Q be an admissible convex metric on the compact space X. Prove that the function e:2Xx [0,oo) -)• 2X defined by e(A,t) = Dt(A) is continuous. ^•20. Let X be a space and let k £ N. In addition, let A = {Al : i £ N} be a sequence of subsets of X, each of cardinality k. Prove by induction on k that there are an infinite subset E C N and a (possibly empty) subset A° of X and for every i € E a partition A®, A} of Ai such that that one of the following statements is true: (A) Every x £ X has a neighborhood meeting finitely many terms of the sequence (Ai)i^E only. (B) A V 0 and (1) the sequence (A°); £j E converges to A° in the Vietoris topology. (2) Every x € X has a neighborhood meeting finitely many terms of the sequence (A\}i^E only. 21. Let X be a compact space with disjoint closed subsets A and B. Define C = {C G 2X : C is a continuum from A to B}. Prove that C is a closed subspace of 2X . 22. Let X be a space, and let a; be a Whitney function for 2X . Let e > 0 be arbitrary. Prove that for every E € 2X there is a neighborhood U of E in X such that for every C e 2X with E C C C Z7 we have w(C) < u;(£) + e. . Let X be a hereditarily indecomposable continuum, and u: 2X —>• [0, oo) a Whitney function. For t e [0, u;(X)] consider the Whitney level £ = {A €

Prove that the natural quotient map TT : X —>• JC/£ is open.

1.11. HYPERSPACES

109

24. Let X be a compact space and let uj: 2X —> [0, oo) be a Whitney map. Prove that if t G (0,o;(X)] is such that its corresponding Whitney level & = {A£2X :u(A) = t} is nonempty then inf{diam(£) : E € 8} > 0. 25. Let X be a compact space and let u: 2X —>• [0, oo) be a Whitney map. Prove that for every e > 0 there is an element t £ (0, oo) such that if A e 2X is such that u>(A) < £ then diam(A) < s. 26. Let T and X be spaces and let ip: T —>• 2 X be continuous. In addition, let TT : X x T —> T be the projection. Prove that if E = {(x,t) : x <E
CXxT

then -n \ E: E —> T is a closed surjection.

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CHAPTER 2

Basic combinatorial topology In this chapter we present some elementary combinatorial results and apply these to get nontrivial information about the topology of the Euclidean spaces R n , n e N. The main result is the Brouwer Fixed-Point Theorem. 2.1. Affine notions

In this section we will introduce simplicial complexes and present some basic results on them. So-called subdivisions of simplicial complexes are important in the proof of Brouwer's Fixed-Point Theorem as well as in ANRtheory. For completeness sake, we begin by reviewing some elementary Linear Algebra. The proofs of these results are left to the reader. Let V denote a fixed vector space. Let F be a finite subset of V, say F = {vi,... ,vn}. An affine combination of v i , . . . , vn is a vector v that can be written in the form J^ILi tivi with ti,..., tn G E and X^=i ^ = •*•• Such a v is called geometrically dependent on F. A linear (affine) subspace of V is a subset of V closed under the formation of linear (affine) combinations. Theorem 2.1.1. Let A be a subset ofV and let a 6 A. Then A is an affine subspace if and only if A — a is a linear subspace. If 5 C V, then the intersection of all affine (linear) subspaces of V containing 5 is the smallest affine (linear) subspace of V containing 5. This subset is called affine hull (linear hull) of 5 and is denoted by aff(S) (lin(5)). Theorem 2.1.2. aff(S) is the set of all affine combinations of elements of S. Moreover, for every a 6 5 the following equality holds: aff(S) — a — lin(5 — a). We see that if 5 C V then aff(5) is a translated linear subspace of V, since for every a £ S the equality aff (S) = a + lin(5 - a) 111

112

2. BASIC COMBINATORIAL TOPOLOGY

holds. We say that S spans the affine subspace aff(S'). Let v\,..., vn € V. Then (1) the vectors vi,..., vn are said to be geometrically independent if for all elements t i , . . . , tn G M with ^"=1 ^ = 0 and SILi ^^ = 0 we have t\ = • • • = tn = 0, (2) a subset 5 C V is linearly (geometrically) independent if and only if every finite subset is. Theorem 2.1.3. Let 5 C V. The following statements are equivalent: (1) (2) (3) (4)

5 is geometrically independent, no x G 5 is geometrically dependent on a finite subset F C 5 \ {x}, for every s G 5, {x — s : x E 5, x ^ s} is linearly independent, for some s e 51, {x — s : x 6 5, x 7^ s} is linearly independent.

Corollary 2.1.4. Let 5 C Rn be geometrically independent. Then S contains at most n + 1 points. Consider the affine subspace aff(S) and fix an arbitrary a G S. We saw that aff (5) = a + lin(5 — a). So there exists a subset T C S such that lin(5 — a) = lin(T — a) while moreover T — a is linearly independent. By Theorem 2.1.3 it follows that T' = {a} U T is geometrically independent. Then aff(T') = o + lin(T' - a) = a + lin(T - a) = a + lin(5 - a) = aff(S). So we conclude that 5 and its geometrically independent subset T' span the same affine subspace. An affine subspace of a linear space is called m-dimensional if it is spanned by m + 1 geometrically independent vectors. Proposition 2.1.5. Let S C V be geometrically independent. If A,B C 5 then aff(A) n aff (B) = aff (A n B}. The following result is a nice test for proving that subsets of V are geometrically independent. Theorem 2.1.6. Let OQ, o > i , . . . , an be elements ofV such that di+i & aff({a 0 ,... ,a»}) for every i < n. Then {O,Q. . . . , an} is geometrically independent. A function between affine subspaces of linear spaces is called affine if it preserves affine combinations. Images and preimages of affine sets under affine functions are again affine.

2.1. AFFINE NOTIONS

113

Theorem 2.1.7. Let Vi and V2 be linear spaces, let A\ C Vi and A? C V? be affine subspaces and let f : A\ —* A2 be a function. Then the following statements are equivalent: (1) f is affine, (2) the composition Al - ai ^4 Al —> A-2 ^ AI - a?

(ai 6 AI, a2

i.e., the function £: AI — a\ -» A2 — a2 defined by

£(x) = f(x + ai) -a 2 , is linear. From Theorem 2.1.7 it follows that an affine function, the domain and range of which are both contained in a finite dimensional normed linear space, is continuous. This can be seen as follows. We first claim that for arbitrary n,m (E N, linear functions

are continuous. First observe that each linear function /: M —> M is a multiplication and hence is continuous. Assume that every linear function /: W —> R is continuous for i < n, and let F: Rn+l -> M be linear. If fl = F \ W1 x {0} and /2 = F f {0} x {0} x • • • x {0} xM. n-l

then they are continuous by assumption. But then F is continuous by linearity, being the sum of the continuous functions

and

( x i , . . . , x n + i ) ^ / 2 (0, . . . , 0 , x n + i ) . Consequently, each linear function /: Mn —>• RTO is continuous since Wn is endowed with the product topology. By Exercise 1.1.18 it follows that each finite dimensional normed linear space is topologically isomorphic to some Mn . So we conclude that a linear function between finite dimensional normed linear spaces is indeed continuous. This gives us what we want since by Theorem 2.1.7 each affine function /, the domain and range of which are both contained in a finite dimensional normed linear space, is of the form

where both £ and 77 are translations and hence homeomorphisms (Exercise 1.1.3), and F is linear.

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2. BASIC COMBINATORIAL TOPOLOGY

Simplices. Let V be a linear space. An n-simplex in V is a geometrically independent subset of V having precisely n + 1 points. Simplices are denoted by Greek letters. If a and r are simplices and a C T then a is called a face of r- to indicate that a is a face of r we shall also use the notation: a ^ r. An n-simplex in V is sometimes also called an n-dimensional simplex. Theorem 2.1.8. Let a be a simplex in V and let A = aff(a). Then every element x G A can be written uniquely as an affine combination ^v£a. tv • v of a. In addition, the functions av : A —> R denned by av(x] = tv are affine. Corollary 2.1.9. Let a and T be simplices in V such that r ^ a and let A = aff(cr),

B =

aff(r).

If x G B then av(x) — 0 for every v G a \ r. The real numbers av(x) for v G a are called the affine coordinates of x with respect to a. We call the av the coordinate functions of aff(cr). This notation will remain in force throughout the book. Observe that by the above remarks and Theorem 2.1.8 it follows that the coordinate functions of aff(cr) are continuous on aff(cr). A geometric simplex is the convex hull of a simplex. We use |cr as an abbreviation for conv(
2.1. AFFINE NOTIONS

115

Proof. We prove that diam(|cr|) < diam(<7). Take arbitrary p, q £ cr|. Then since £veo. av(p) = 1 we get

from which it follows by using standard properties of the norm that

= max||i; — q\ . As a consequence we obtain for every w G a that \\q-i So

<

t> — w ; .

\\P ~ Q\\ < max \\w — q\\ < max \\v — D

as required.

Figure 6. For each 0 < i < k let di 6 R/c+1 be the vector all coordinates of which are 0, except the (i + l)-th coordinate which equals 1. The vectors d0,..., dk are clearly linearly independent (hence geometrically independent). For x in aff ({do,..., dk}) we have that the affine coordinates of x = (XQ, ..., Xk) with respect to do,... ,dk are equal to XQ, . . . , X k , respectively. Put rk =

116

2. BASIC COMBINATORIAL TOPOLOGY

Observe that \Tk = {x G I fc+1 : J^ILi ^i = 1} is a closed and bounded subset of R/c+1 and hence is compact (alternatively, use that \Tk\ is a geometric simplex). Theorem 2.1.12. Let (V, || • ||) be a normed linear space, and let a — {v0,...,vk} C V be a k-simplex. Then the function f : \a\ —»• |r^| denned by /(a) = (a w o (a),...,a! w f c (a)) is a homeomorphism. Proof. First observe that the functions aVi are affine (Theorem 2.1.8). Consequently, / is continuous by the remark following Theorem 2.1.7, and oneto-one by the definition of affine coordinates (Theorem 2.1.8). Moreover, / is clearly surjective since if y G \r^\ then f ( x ) = y, where

By compactness of \cr\ it therefore follows that / is a homeomorphism (Exercise A. 5. 9). D Triangulation. A simplicial complex is a countable collection § of nonempty, finite sets such that: (SC)abstact

if
The elements of the set

s = Us

are called the vertices of the simplicial complex S. Although S need not be a subset of a normed linear space, without loss of generality we may assume that this is the case. In fact, we can think of S as being a linearly independent subset of some normed linear space. This can be seen as follows. For every n G N let xn G I2 be the vector all whose coordinates are 0 except for the n-th coordinate which equals 1. Observe that the set D = {xn : n G N} is linearly independent. Let /: S —> D be an injection, and identify every simplex a G S with /[
(sc)

MOM - ^ n ^ .

In the sequel, when dealing with a simplicial complex S, we shall always require that S = (J § is contained in a normed linear space L, and that (SC) holds. The set |S = [JIM : cr ^ §} is called the geometric realization of S

2.1. AFFINE NOTIONS

117

and |S| is said to be triangulated by S (or S is a triangulation of |8|). Observe that the collection Sy = {|a| : a G §} consists of geometric simplices in L and has the following properties: (SC)geometric

(1) if cr| £ §|| and r is a face of a then |r| 6 Sy, (2) for all |<Ji|, (72 1 € §>\\ with <TI fl <72| 7^ 0, |<TI| H |a2 is a face of <TI as well as |<7 2 |-

Let cr\ be a geometric simplex in L. The collection of all faces of |cr| is clearly a simplicial complex which triangulates \a\. It is called the standard triangulation of |cr| and is denoted by 3(0). Topologizing a simplicial complex. From this moment on, we will no longer distinguish between an abstract simplicial complex and its geometric realization. So a simplicial complex in a normed linear space L is a collection S of geometric simplices (which by abuse of terminology we will call simplices for short in the sequel) in L such that for all a\ , 0*2 G S we have (1) if a G S and r is a face of a then r € S, (2) ai n 02 / 0 =>• 0"i n <72 is a face of a\ as well as 02A subcollection T of 8 which is also a simplicial complex is called a subcomplex ofS. Let T be a simplicial complex in L. For each m > 0 define <j(m) = |s £ 'J : <j is at most m-dimensional}. So T 0 is the set of vertices of all simplices in T, T^1) is the collection of all at most one-dimensional faces of all simplices in T, etc. The elements of T'°) are called the vertices of T. It is a triviality that 7^m^ is a subcomplex of T; we call it the m- skeleton of T. Let L be a normed linear space and let T be a simplicial complex in L. We are interested in useful topologies on the set X = (J T. There are several natural candidates for such a topology. First, X is a subset of L and therefore carries the subspace topology inherited from L. It turns out that this is in general an interesting but not a useful topology. We shall now describe a better topology. To this end, let us first agree that there cannot be ambiguity about the topology that each simplex of T should carry. By Theorem 2.1.12, each ^-dimensional simplex in L with its subspace topology is naturally homeomorphic to a standard fc-dimensional simplex in Rfe+1 ; each simplex in L from this moment on is endowed with its subspace topology. It would be very unnatural if the useful topology we want to define on X should induce a different topology on one of the simplices of T. So there is a natural candidate for a topology on X, namely, the largest topology that induces the 'right' topology on every simplex of T, i.e., U C X is open iff for every simplex T e T, U D T is open in T.

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This is called the Whitehead topology on X and X with this topology is usually denoted by \7\. Lemma 2.1.13. Let T be a simplicial complex in L and let X = \J 7. Then (1) The Whitehead topology on X is a topology and is finer than the topology that X inherits from L, i.e., the inclusion \7\ <-» X C L is continuous, (2) On every simplex of 7 the Whitehead topology and the topology on L induce the same subspace topology. Proof. The proof of (1) is a triviality the verification of which we leave to the reader. For (2), let r be a simplex of T. It suffices to prove that for every relatively open subset U C T there exists a Whitehead-open V C |T| such that V fl T = U. We claim that V — U U (|T| \ T) is as required. It is clear that V fl r = U, and so it remains to prove that V is Whitehead-open. To this end, let a 6 7 and put V = V D (r U a}. Then V is open in r U a since its complement is equal to r \ U which is compact and hence closed in r U a. Consequently, V fl a = V n a is open in cr, as required. D Observe that a subset A of |T| is closed if and only if A n r is closed for every r G 7. This immediately implies that |T| is a T\-space, i.e., a space in which every singleton is closed (this follows also from Lemma 2.1.13(1)). Due to our self-chosen (sometimes unpleasant) restriction to deal with separable metrizable spaces exclusively for the time being, there is a small problem with the Whitehead topology since it need not be metrizable. However, there is a simple combinatorial condition on 7 that ensures that |T| is separable and metrizable; call a simplicial complex T locally finite if every vertex of T is contained in at most finitely many simplices of T. In turns out that for our considerations it always suffices to consider locally finite simplicial complexes and these are precisely the simplicial complexes that are separable metrizable when given the Whitehead topology, see Proposition 2.1.21 below. For the moment, we shall not worry about the metrizability of |T| but we shall prove a few easy but important lemmas from which metrizability in the locally finite case will follow rather easily later. Lemma 2.1.14. Let 7 be a simplicial complex with subcollection S. Then (1) U S is a closed subspace of |T|, (2) if S is a subcomplex then the topology that |J § inherits from Tj coincides with the Whitehead topology on |JS.

2.1. AFFINE NOTIONS

119

Proof. For (1), for convenience, put Y — (J S. Since each simplex of T has finitely many faces only, and T is a simplicial complex, it is clear that for every r G T, Y D T is closed in r. By definition of the Whitehead topology, we therefore conclude that Y is closed in |T|. For (2) observe that since S C T, the topology that (J S inherits from |T| is coarser than its Whitehead topology. Now let A C |S| be closed. We shall prove that A is a closed subset of |T|. To this end, let r be an arbitrary simplex in T. Since T has finitely many faces only, there is a finite subcollection y of 8 such that

\j3~r\r = |J S n r. Consequently, A n T = (A n U S) n r = (A n U 5") n r. Now since A C |8| is closed, for every a G 3", A n a is closed in cr, and consequently, A n
and

carx = ( ] { T 6 T : x 6 r}, respectively. We shall prove that carx G T and it is therefore the smallest simplex of T that contains x. Corollary 2.1.17. Let T be a simplicial complex. Then for every x G |T|, (1) Six is open in T|,

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2. BASIC COMBINATORIAL TOPOLOGY

(2) carx belongs to T. In addition, the family {Six : x G T^} covers |T|. Proof. For (1), let § = {r G T : x €" T}. From Lemma 2.1.14 we obtain that y 8 is closed in |T|, hence Sta: is open. For (2), first take T G 0" such that x G T. Since for every a G T such that x G cr we have cr n r =
(*)

xn G (Knrn+l)\ y rn. i=l

Since K is compact, and metrizable, we may by Theorem A.5.1 assume without loss of generality that x = limn_>.00 xn exists (and belongs to K}. Now let r G T be an arbitrary simplex. Since rn n r is either empty or a face of r, there exists m G N such that oo

m T

U n n=l

n T

= IJ Tn H T, n=l

from which it follows that {xn : n G N} O r C {XI,...,X T O }. By the definition of the Whitehead topology, this implies that {xn : n G N} is closed in |T (recall that |T| is TI). Since the sequence (xn)n converges, it therefore has to be eventually constant, which contradicts (*). D One of the reasons that the Whitehead topology on a simplicial complex is relatively simple to deal with is that it is easy to check that certain functions are continuous.

2.1. AFFINE NOTIONS

121

Lemma 2.1.19. Let 7 be a simplicial complex and let X be a space. A function f : \7\ —>• X is continuous if and only if the restriction of f to every simplex r € 7 is continuous on T. Proof. Suppose that /: |T| —>• X is such that the restriction of / to every simplex r G T is continuous. Let U C X be open. For each r G T, is open in T. By the definition of the Whitehead topology, it therefore follows that f~l[U] is open in |T|. D Corollary 2.1.20. Let T be a simplicial complex. Then \7\ is a normal topological space, i.e., for every pair of disjoint closed subsets A and B of\7\ there exists a continuous function f : 7 —> I such that f \A = 0 and f\B = l. Proof. Since 7 is countable, we can enumerate it as {rn : n G N}. By Lemma A. 4.1 there exists a continuous function /i : r\ —> I such that By induction on n, we shall construct continuous functions fn : UlLi Ti ~^ I such that

(1) /nKAnu: = 1 r,) = o, (2) / n t ( B n U r = i r O = l, (3) if n > 1 then fn extends / n -iSince we already defined /i, assume fi to be constructed for 1 • I by (a; e A'),

Then 5f is clearly continuous and extends fn. Since UF^i Ti '1S compact, it is metrizable by Lemma 2.1.18. By Corollary 1.2.5 there consequently exists a continuous extension /n+i : UlLi T{ —>• I of ^. It is clear that fn+i is as required. Now put / = Ui^i fn- Then / is a well-defined function from |T| into I having the property that / \ A = 0 and / \ B = 1. Moreover, / is continuous by Lemma 2.1.19. D These simple results enable us to prove the following Proposition 2.1.21. Let 7 be a locally finite simplicial complex in the linear space L. Then \7\ is a locally compact separable metrizable space.

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Proof. We shall first prove that each point of |T| has an open neighborhood which is separable, metrizable, and has compact closure in |T|. To this end, take an arbitrary x G |T|. By Corollary 2.1.17(1), Stx is open. Since 7 is locally finite, St x is contained in the union of finitely many simplices of T. This can be seen as follows. First observe that Six is contained in the union of all simplices that contains x. But, as we will show, there are only finitely many of such simplices. Striving for a contradiction, assume that S is an infinite sub collection of T such that x G P| S. Fix
2.1. AFFINE NOTIONS

123

Now let L and E be norrned linear spaces and let 7 and 8 be simplicial complexes in L and E, respectively. We say that 7 and § are combinatorially equivalent if there exists a bijection / : 7^ —>• S^0) such that F C 7^ spans a simplex in 7 if and only if f[F] spans a simplex in 8. Let us also say that |T| and S| are simplicially homeomorphic if there exists a homeomorphism h: \7\ —> §| such that for all r E T and a E §, we have h[r] 6 § as well as h~l[a] e T. Proposition 2.1.23. Let L and E be normed linear spaces and let 7 and 8 be simplicial complexes in L and E, respectively. The following statements are equivalent: (1) T and 8 are combinatorially equivalent, (2) |T| and |S| are simplicially homeomorphic. Proof. For (1) => (2), let /: 7^ -»• S<°) be a bijection such that F C T<°) spans a simplex in 7 if and only if f[F] spans a simplex in 8. So for every simplex T = |{x0, . . . ,Xk}\ € T, the set {/(x 0 ), . - . ,/(£&)} spans a simplex cr of 8. Define a homeomorphism fr : T —> a by

i=0

i=0

here ao(^), • • • ,(*k(x) denote the affine coordinates with respect to T of an arbitrarily chosen point x E T of course. It is easily seen that jr is indeed a homeomorphism, cf. Theorem 2.1.12. Now define /: |T| —>• |S| by riET

By unicity of affine coordinates (Theorem 2.1.8), / is well-defined. In addition, / is continuous since the restriction of / to every simplex of 7 is continuous (Lemma 2.1.19). It is easily seen that / is one-to-one and surjective and that f~l is continuous for the same reason / is. We conclude that / is a homeomorphism. Clearly both / and f~l are 'simplex preserving'. Conversely, let /: |T| —>• |S| be a homeomorphism such as given by (2). The function is clearly as required.

D

From the above proposition we conclude that the topology of a simplicial complex depends only on the combinatorial properties of its vertex set. For that reason when discussing a simplicial complex, it is no longer necessary to mention the normed linear space it is a subset of. Let X be a space. We say that X is a polytope if there exists a locally finite simplicial complex T such that X and T| are homeomorphic. In case 7

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is finite we say that X is a polyhedron. Observe that each polyhedron is compact. Polytopes are very important since they are the 'bridge' between abstract topological spaces and concrete ones. If X is a polytope, say X is homeomorphic to |T|, then we will find it sometimes convenient to not distinguish between X and |T|. This will never cause confusion because the triangulation under consideration will always be explicitly defined but the reader should keep in mind that there usually are many different triangulations of the same polytope. If |T| is a polytope then a subset Y C |T| is called a subpolytope if there is a subcomplex 8 C T such that Y = |S|. Similarly for subpolyhedron. Theorem 2.1.24. Each polyhedron is an ANR. Proof. Let T be a finite simplicial complex. By induction on the cardinality of T we shall prove that |T| is an ANR. If T consists of one simplex only then |T| is an AR by Theorems 2.1.12 and 1.2.9. Suppose that for every simplicial complex 7 of cardinality at most n — I > 1, |T| is an ANR, and let T be a simplicial complex of cardinality n. Since T is finite, there exists a simplex T 6 7 such that for every u E 7 with r ^ cr, a = T (let T be any simplex of maximal dimension). Put S = T \ {T}. Then § is a simplicial complex and |T| = r U |S|. In addition, clearly T n |8| = \{T n cr : a G S}\. So by our inductive assumptions and by Theorem 1.2.16(1) it follows that |T| is an ANR. D In Corollary 4.3.6 we shall prove that every polytope is an A N R , which generalizes Theorem 2.1.24. Exercises for §2.1. 1. Prove Theorem 2.1.1. 2. Prove Theorem 2.1.2. 3. Prove Theorem 2.1.3. 4. Prove Proposition 2.1.5. 5. Prove Theorem 2.1.6. 6. Prove Theorem 2.1.8. 7. Prove Theorem 2.1.10. 8. Let A be an affine subspace of a finite dimensional linear space L. Prove that A is closed in L. ^•9. Prove that if a is an n-simplex in W1 then Int(|cr|) is equal to \a\°. 10. Let a be a simplex and let p G a. Prove that a \ {p} is convex if and only if p is one of the vertices.

2.2. BARYCENTERS AND SUBDIVISIONS

125

11. Let a be a simplex and let C be a convex set which is contained in da. Prove that C is contained in a proper face of a. 12. Let X be a set triangulated by the simplicial complex S. Prove that the collection of all geometric interiors of elements of |S| is a partition of X . 13. Let 11 be a countable collection of sets. Prove that the collection (J : y C U is finite and P| 3 ^ 0} is a simplicial complex. 14. Let T be a simplicial complex. Prove that |T| is paracompact (we do not assume that T is locally finite). 15. Let T be a simplicial complex that is not locally finite. Prove that |T| is not metrizable (this is the converse to Proposition 2.1.21). ^•16. Let T be a simplicial complex and S a subcomplex of T. Prove that the topological interior and topological boundary of |S| in |T| are given by {x e |S| : x e a £ T => a G §} and

|JCr\s)n|S|, respectively.

2.2. Barycenters and subdivisions Througout this section, L is a fixed normed linear space. All simplices under consideration are subsets of L. If a is a simplex in L then a denotes the set of its vertices. Barycenters. Let a be a simplex. The barycenter ba of a is the point in a whose affine coordinates (with respect to ci) are all equal: if a - {V0,. . . , f n }

then

Observe that ba belongs to the geometric interior a° of a. Lemma 2.2.1. Let S be a simplicial complex. Then if cr, a' G § and ba G a' then a is a face of a' . Proof. By (SC), a Pi a' is a face of a. This face contains bff so it must be a itself. D Lemma 2.2.2. Let r\ C r2 C • • • C Tk be a strictly increasing collection of faces of a simplex a. Then the barycenters bTl , bT2 , . . . , 6Tfc are geometrically independent.

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Proof. By Theorem 2.1.6 it suffices to show that for i < k, bTi+1 £aff({6 Tl A 2 ,...,& Ti }). To this end, take an arbitrary vertex v 6 f^i \ T;. Then

Since {6 T1 ,6 T2 ,...,6 T J C aff(f.) we get

aff({6 T l ,6 T 2 ,...,6 T J) C aff(fi). This and Corollary 2.1.9 imply that for every x 6 aff (|&TI , 6T2 , . . . , bTi }) we have av(x) = 0. This is clearly as required. D Let a be a simplex and let 3C = (TI, T2, . . . , r/t} be a chain of faces of a (by a chain we mean of course a chain with respect to inclusion) with corresponding chain of vertices 3C = {ri,f 2 , . . . , fjt}. If 3C is a maximal chain of faces then clearly <7 € 3C and consequently |J JC = o\ This rather trivial remark will be used without explicit reference a few times in the remaining part of this section. For every v 6 (J 3C define its height ht(v) to be the first i < k with v 6 f;. Define a quasi-order '= M(v) < ht(w). Lemma 2.2.3. Let a be an n-simplex and let 9C be a maximal chain of faces of a with corresponding quasi-order =<;. Then (1) OC has cardinality n + 1 and tie corresponding quasi-order =<; is a linear order on a = (J 3C. (2) If/? is the geometrically independent set of all barycenters of elements of 3C then it is geometrically independent and \^\ — {x £ a : (Vv,w E cr)(v ^ w =>• Q W (X) > aw(a:)}. (3) If v ^, w and for certain x G |/?| we have av(x] = aw(x] then for every T G OC with v 6 f and w 0 f we have that the affine coordinate of x in \f3 1 with respect to bT is equal to 0. Proof. We claim that if a is the direct successor of /? in 3C then a \ /? is a singleton: if it contains two distinct points v and w then

would be a chain extending 5C; likewise one sees that the first element of 3C is a singleton. Thus OC = {TO,TI, . . . ,rn} where f^ has i + 1 points. Now write a = {VQ,VI, . . . ,vn} such that for every 0 < i < n, TV = {^0,^1, • • • , ^}Then Vi =
2.2. BARYCENTERS AND SUBDIVISIONS

127

For (2), observe that /3 is geometrically independent by Lemma 2.2.2. An arbitrary element x 6 \(3\ can be written as n

n

x — 2_. ti • bTi where ti > 0 for every i and i=0

(Lemma 1.1.1). Observe that

=0

j=0

i=0

j=i

j +1

J

For every 0 si > • • • > sn. Conversely, let p be a point in a whose affine coordinates SQ, ..., sn have the property that SQ > si > • • • > sn. Then the numbers /L

n

— \^ld („

l-\\e

T^ ±JOn

are non-negative and have the property that n

2_^ ^i ' brt = (SQ — i=0

Si

i=0

= p. In addition, J^ILo ^ ~ Sr=o si ~ ^- ^e ^ are consequently the affine coordinates of p in \j3\ (Theorem 2.1.8). This proves (2). For (3), simply observe that if for p above we have that if Si = Sj for certain i < j < n then all tk for i < k < j are equal to 0. D

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2. BASIC COMBINATORIAL TOPOLOGY

Lemma 2.2.4. Let a be a simplex and let 3Ci and OC? be chains of faces of a. Let @i and fa denote the sets of barycenters of elements of 3Ci and 3C2, respectively. Then

Proof. Since |/?i| n \fa\ is convex and contains fli n/3 2 , it is clear that

\Pinfol = conv(/?i n &) c |ft| n |/?2|. For the reverse inclusion, take an arbitrary x G |/?i | fl \fa\- Since (3Ci\3C2)n(3C2\3Ci)=0, without loss of generality we may assume that <7 0 UCi \ 3^2- We assume without loss of generality that XL \ 3C2 ^ 0 for otherwise we get what we want from Lemma 2.2.2. It is clear that there exists a linear order ^ on the set a such that for every element of T G 3Ci we have that f is an initial =^isegment. This order corresponds to a maximal chain of faces of a. Pick an arbitrary element r G KI \ 3C2; observe that T ^ a. The collection 3C2 can easily be extended to a maximal chain -C of simplices such that r $ £ (Exercise 2.2.1). Consequently, r is not an initial segment with respect to the corresponding linear order =<(2 (Lemma 2.2.3(1)), i.e., there exist distinct elements v , w G cr such that v G r, w; ^ r and w; =^2 ^; consequently, o^Oc) > a w (a;) (Lemma 2.2.3(2)). Observe that v =^i w from which it follows similarly that

av(x) > aw(x). By Lemma 2.2.3(3) we therefore conclude that the affine coordinate of x in the simplex |/?i | with respect to bT is equal to 0. D Subdivisions. Let X be a subset of L which is triangulated by S. A triangulation T of X is called a subdivision of S if for every simplex cr G S we have that the collection

7 (a) = {r € T : T C cr} is finite and is a triangulation of cr. Lemma 2.2.5. Let S and T be triangulations of X. If 7 is a subdivision of § then for every r G T and cr G S sucJi tAat bT G a we Lave r C cr. Proof. Take an element T' G T(cr) such that bT G T'. Then by Lemma 2.2.1 it follows that r C r' C cr, as required. D Theorem 2.2.6. Let S, T and $ be simplicial complexes. If § is a subdivision of 7 and 7 is a subdivision of^R then §> is a subdivision ofR.

2.2. BARYCENTERS AND SUBDIVISIONS

129

Proof. We have to prove that S(p) is a finite triangulation of p G ft. That the collection S(p) is a simplicial complex is clear since condition (SC) holds for all simplices in 8. Let a G S(p). We claim that there exists an element ra G T(p) such that <j C Tff. This is easy. Indeed, since IJT(p) = p there exists Tff G T(p) such that ba G ra. So by Lemma 2.2.5 it follows that ra is as required. Since T(p) is finite and for every r G T(p), S(r) is finite, we obtain that the collection S(p) is finite. To conclude the proof, observe that \JS(p) — p is trivial since (J T(p) = p and \J §(r) = r for every r G T(p). D Let a be a geometric simplex in L. We shall define a special triangulation of cr, the so-called barycentric triangulation. Let -B(cr) denote the set of all barycenters effaces of a. We shall define a simplicial complex 23(cr) consisting of subsets of B(a] as follows: A nonempty subset /3 of B(a) belongs to 13(0") if and only if the faces of a the barycenters of which belong to /3 form a chain. It is clear that if j3 G 23 (cr) and 7 C J3 then 7 G 23 (cr). So 23 (cr) is an abstract simplicial complex. By Lemma 2.2.4 it follows that the collection

satisfies condition (SC) (see Page 116). So the geometric realization of 23 ( is |23(cr) and, in particular, is contained in a. In order to prove that |23(a is a subdivision of cr, it therefore remains to prove that

Let x G a. Without loss of generality we have a = {VQ, . . . , vn} and ttvn (X}

<

a

Vn - 1 (X)

< • • • < ^f 0 (X} •

This ordering corresponds to a (maximal) chain of faces in cr, which in turn corresponds to a (maximal) simplex /? G 23(0"). By Lemma 2.2.3(2), x G |/?|. Let X be a set in L which is triangulated by the simplicial complex §. As above, for every a G S let 23 (cr) be the barycentric triangulation of cr. The collection ®(§) = {|r| G 23(a)| : c r G § } is called the bary 'centric subdivision of §. That 23(8) is a collection geometric simplices with union X is clear. We claim that it is a simplicial complex and is a subdivision of 8. For that we only need to verify condition (SC) (see Page 116) because for every a G 8, { | r | e | ® ( S ) | : | r | C < r } = 23(
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Let fa G S(cTj) for <Ji G 8, i = 1, 2. If <TI D 02 = 0 then there is nothing to prove. If cri n o"2 ^ 0 then a = <TI n cr2 is a face of <TI as well as cr2. Now put Then 71 and 72 belong to 25 (
M = IA|n<7 (. = 1,2). That |7i| C |/3; | n a is clear. Take an arbitrary x G |/3;| n a. There exists an element /3 G 23 (o7) such that a: G |/?|. Observe that

Consequently,

By Lemma 2.2.4 we therefore conclude that

\/3i\ n \/32\ = |&| n |/32| n a = |7i| n |72| = |7l n 72 | = [ft n /?2|, which is as required. Sometimes it will be convenient to denote S by sd^S and 23 (S) by We define the second barycentric subdivision sd^S of S by !B ($(§)). Similarly, one defines sd^S, the n-th barycentric subdivision of S. Notice that by Theorem 2.2.6, sd^S is a subdivision of § for all n. Let S be a triangulation of a subset of L. The mesh of S, denoted by mesh(S), is defined to be the number supdiam(er). ff£§

We allow mesh(S) to be equal to oo. Theorem 2.2.7. Let a be an n-simplex in the normed linear space L and let *B(cr) be the barycentric triangulation of a. Then mesh(|-B(o-)|) < —^— -diam(cr). Proof. First observe that for every /? G *B((7) we have diam(/?) = diam(|/?|) (Theorem 2.1.11). Fix (3 G B(
of (T such that c is the barycenter of {^0,^1, • • • , Vk} and d is the barycenter of {^QJ^IJ ••• ,vm}, respectively. We may assume without loss of generality

2.2. BARYCENTERS AND SUBDIVISIONS

131

that k < m. Then \\d-c\ i=0

TTL ~T~ J-

m

1 k +l

j=0 ,

/v

;+1 + ltl

V i i "

fc / I V 1( 1 k + l j =0 \ V m + 1

M

3=0

max

0<j
Moreover,

m

Vi

-

Vn

m +l

i=0

i=0

m+l m diam(
In the last inequality we used that one of the terms (if i = j) is equal to 0. We conclude that diam(/3) <

ra

m+l

diam(a).

Finally observe that m < n.

D

Corollary 2.2.8. Let § be a triangulation of a set X in a normed linear space L such that every a G § is at most n-dimensional. IfB is the barycentric subdivision of S then mesh(S) <

mesh(S).

Corollary 2.2.9. Let \T\ be a geometric simplex in a normed linear space L. Then for every e > 0 there is an m 6 N such that

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2. BASIC COMBINATORIAL TOPOLOGY

Proof. Assume that r is an n-simplex. By Corollary 2.2.8 it follows that for every m G N we have i

\

/

mesh (sd (m) J(r)) < (

^ iv

Ti I

\m

)

• diam(r).

-L *

We conclude that for a sufficiently large m, mesh(sd^ TO ^5'(r)) <e.

D

We shall now formulate a very important property of polyhedra. Theorem 2.2.10. Let T be a finite simplicial complex. Then for every open cover U of |T| there exists m G N such that sd^T refines U, i.e., for every simplex u G sd^ m 'T there exists an element U G U with a C U. Proof. Fix r 6 T for a moment. By Corollary 2.2.9 and Lemma A.5.3 there exists n(r) G N such that every a G sd^T''y(r) is contained in an element of U. Let m = max{n(r) : r G T}. Now take an arbitrary K G sd^T. By the definition of sd (m) T it is clear that there exists r G T with K G sd (m) 3~(r). Since every element of sd'm'5F(r) is contained in an element of sd^ n ^ T ^9 r (r), the simplex K is contained in an element of U. D The above theorem can be generalized as follows: for every polytope |T| and for every open cover U of |T| there exists a subdivision § of T such that each simplex a G § is contained in an element of U. However, for noncompact polytopes, S generally cannot be chosen to be an iterated barycentric subdivision. For details, see WHITEHEAD [408].

Exercise for §2.2.

^•1. Let a be a simplex and let 3C be a chain of faces of a. Prove that if r is a proper face of a such that r £ 3C then there is a maximal chain of faces L such that 3C C L but r 0 £. 2.3. The nerve of an open covering

Let X be a space and let U be a countable open cover of X. We say that a simplicial complex T is a nerve of U if 7^ can be indexed as {x(U} : U G U} such that for every n > 0, n

(*)

x(U0),... , x(Un) spans a simplex in T iff Q Ui ^ 0. i=0

If T is a nerve of U then it is convenient to adopt the notation {x(U} : U G U} for T(°\ where it is implicitly assumed that the 'indexing' is such that (*) holds.

2.3. THE NERVE OF AN OPEN COVERING

133

Proposition 2.3.1. Let X be a space. Each countable open cover of X has a nerve.

ur\w uni

u n v nw

unv

cover

V

nerve Figure 7.

Proof. Let U = {Un : n G N} be a countable open cover of X. For each n let xn G I2 be the vector all coordinates of which are 0, except for the n-th coordinate which equals 1. The sequence (xn)n is linearly independent, hence geometrically independent. Define 3 = {F C N : F is finite and |°| Un / 0}n<EF

2

For F 6 y let r(F) be the simplex in I spanned by {xn : n G F}. Put It is easy to verify that T is a simplicial complex and that it moreover is a nerve for U. D Clearly, any two nerves of the same open cover are combinatorially equivalent (hence 'isomorphic'), so by Proposition 2.1.23 we can now speak of the nerve N(U) of the cover U. If L is a normed linear space containing a simplicial complex T which is combinatorially equivalent to the nerve N(U) then we say that N(U) can be realized in L. Let N(U) be the nerve of the locally finite open cover U (recall that U is countable, see Page 486). We shall use the Ac-functions with respect to U to

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2. BASIC COMBINATORIAL TOPOLOGY

define a canonical continuous function K: X —>• N(U)\. This function is the 'bridge' between the 'abstract' space X and the 'concrete' space |JV(U)| and is called the n-function of the cover U. Indeed, define K: X -»• \N(U)\ by

K(X) We claim that K is well-defined. To check this, for x G A" put

g~(x) = {C/ e U : x e U}. Observe that by the local finiteness condition, 3*(x) is finite, that KU(X) = 0 if U $ y(x), and that ICt/eg-r^) KU(X) = 1. This implies that K

u(x) • x(U).

Hence the infinite sum in the definition of K reduces to a finite sum. In addition, the {Ku(x) : U G ^(x)} are the affine coordinates of K(X) in the simplex r(x) spanned by the vertices {x(U} G U : U e ?(x)}

of 7V(U). This shows that K(X) is well-defined, and also that K(X) G r(x). Lemma 2.3.2. Let X be a space and let U be a locally finite open cover of X. Then for each x G X, r(x] is the carrier of K(X) in N(U). Proof. As remarked above, r(x) is a simplex of N(li) that contains K(X). As remarked above, the Kjj(xys are the affine coordinates of K(X) with respect to the simplex in N(U) spanned by the vertices {x(U) : U G y(x)}. Now since for every U G 3(x) we have that KU(X] ^ 0, T(X) is the smallest simplex containing K(X), i.e., r(x) = carK(x). D Theorem 2.3.3. Let X be a space and let U. be a locally finite open cover of X. Then the K-function K: X —> 1^(11)1 -has the following properties: (1) AC is continuous; (2) for every U G U: K~I [Stx(U)] = U. Proof. We shall first prove (2). Observe that for every x G X and U Eli, x gU «=> U g 3(x) ^=> x(U] g T(X) - carAt(x) (Lemma 2.3.2). Now if x(U) $ r(x] then since K(X) G r(x), K(X) g Stx(C7). This proves that K~1[Stx(U}] C U. Conversely, if K(X) g Stx(U) then there is a simplex T G N(U) which contains K(X) but not x(U). Since car K (x} C r, this gives us that x(U) 0 car/c(x) = T(X) and hence that x 0 U. This proves that K[U] C Stx(U).

2.3. THE NERVE OF AN OPEN COVERING

135

For (1), let L be a normed linear space which realizes 7V(U) (e.g., £ 2 , see the proof of Proposition 2.3.1). Take an arbitrary x € X. There is a neighborhood W of x meeting only finitely many elements of U. Since the infinite sum in the definition of K for points of W clearly reduces to a finite sum, and the functions KU are all continuous, by the continuity of the algebraic operations on L it follows that K regarded as a mapping from X into L is continuous. Now l e t W = { t / e l t : t / n V ^ ^ 0 } and let Y be the union of all simplices of N(U) the vertices of which correspond to elements of the collection W. Observe that Y is a finite union of elements of N(11) and hence is compact. Since x e W it follows that 3~(x) C W and so

K(X) G T(X) C Y. Since Y is compact, by Lemma 2.1.13(1) it follows that the topology that Y inherits from L is the same as the topology that Y inherits from |Ar(U)|. We conclude that K: X —>• |-/V(U)| is continuous at all points of W, and hence at the point x. D Let X and Y be spaces and let U be an open cover of X. A continuous function /: X —>• Y is called a It-mapping if there is an open cover V of Y such that /^[V] < U. This concept is related to the concept of a e-mapping on Page 34. For let /: X —> Y be an gr-map, where X is compact, and let U be the open cover of X consisting of all open sets of diameter less than e. We claim that / is a It-mapping. To this end, first observe that if

yeV = Y\f[X] then V is an open neighborhood of y such that /-1|V] = 0 is contained in every U € U. Next, if y € f[X] then let U be an open neighborhood of f ~ l (y) such that diam(C/) < e. Since by compactness of X the function / is closed (Exercise A.5.5), there is a neighborhood V of y such that /~ 1 [V r ] C U (Exercise A. 1.15). So we conclude that / is a It-mapping. We see that the concept of an e-map is indeed the 'compact' version of the concept of a It-map. Corollary 2.3.4. Let X be a space and let It be a locally finite open cover o f X . Then K: X -> \N(U)\ is a It-mapping. Proof. This follows from Theorem 2.3.3(2) and Corollary 2.1.17.

D

It is unfortunately not the case that a cover U of a space X is locally finite if and only if the simplicial complex N(U) is locally finite. Since, as we pointed out above, our main interest is in locally finite simplicial complexes, it is natural to wonder when JV(U) is locally finite. First a definition. An open cover U of X is said to be star-finite if for every U 6 U the set

{v e it: v n u + 0}

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2. BASIC COMBINATORIAL TOPOLOGY

is finite. Observe that a star-finite open cover is locally finite. Consequently, every star-finite cover is countable. This definition leads us to the following Theorem 2.3.5. Let X be a space and let U be an open cover of X. Then (1) there exists an open refinement V of U such that V is star-finite, (2) N(W) is locally finite if and only if U is star-finite. Proof. For (1), we may assume without loss of generality that A" is a subspace of the Hilbert cube Q (Corollary A.4.4). For every U 6 U let U C Q be open such that U n X = U and put U = {U : U G U}. Then V = (J U is an open subset of Q and hence is locally compact and <j-compact. There are compacta Fn C V for n 6 N such that F! C Int F2 C F2 C Int F3 C • • • C F n _i C Int Fn C Fn C • • •

while moreover V — U^Li Fn- Put F0 — 0. For every n let £ n be a finite subcollection of U with Fn C (J £ n . Now put W = {U n (Q \ F n _i) n Int Fn+1 : U G £ n , n 6 N}.

It is clear that W covers V', is star-finite and that W < U. As a consequence, the collection V = W \ X is as required. For (2), first assume that N(li) is locally finite. If there exists U G U such that the set {V G U : V n t / ^ 0 } is infinite then the vertex x(U) of N(IV) is contained in infinitely many onedimensional simplices of TV(U), which contradicts the local finiteness condition on TV (It). The verification of the reverse implication is a triviality which we leave to the reader. D We shall now prove that every space can be 'approximated arbitrarily closely' by a polytope. Corollary 2.3.6. Let X be a space. For every open cover U of X there exists a polytope P and a U-mapping f : X — > P . If X is compact then P can be chosen to be a polyhedron. Proof. Let U be an open cover of X. By Theorem 2.3.5(1), there exists a star-finite open cover V of X which refines U. Let

K: X ^ \N(V)\ be the ^-mapping of V. Since V is star-finite, |./V(V)| is locally finite by Theorem 2.3.5(2). By Corollary 2.3.4, K is a V-mapping. Since V < U, we are done. If X is compact then V is a finite subcover of U and proceed as above. D

2.3. THE NERVE OF AN OPEN COVERING

137

Freudenthal's Approximation Theorem. We now come to an interesting 'Approximation Theorem'. See §1.10 for results that are in the same spirit. Freudenthal's Approximation Theorem 2.3.7. Every compact space is homeomorphic to the inverse limit of an inverse sequence consisting of polyhedra. Proof. We start as in Example 1.10.5. By Corollary A.4.4 we may restrict ourselves to a compact subspace X of Q. For every n let Xn = 7Tn[X\, n

where rn: Q —> JJ is the projection. In addition, let fn: Xn+i —>• Xn be defined by fn(xi,...,xn+i) = ( x i , . . . , a ; n ) . Then ]^m(XnJn)n = X (Example 1.10.5). Unfortunately, the sets Xn need not be polyhedra. It will require a little extra work to take care of this problem. Claim 1. Let X be a compact subspace of JP for some n. Then for every neighborhood U of X there exists a polyhedron P such that X C P C U. Proof. By Corollary A.5.4 there exists 6 > 0 such that if A C J n , diam A < 6 and AnX / 0 then A C U. Since J n is a polyhedron, it has by Theorem 2.2.10 a triangulation § such that for every cr E S we have diamcr < 6. Let P be the union of all simplices in S that intersect X. An easy check show that P is as required. 0 Now for every n G N we shall define a polyhedron Pn in JP with Xn C Pn, as follows. Let PL = JT. By Claim 1 there exists a polyhedron P2 in ]J2 such that X2 C P2 C B(X2, V 2 ). By the Dugundji Theorem 1.2.2, we can extend the function /i: X? —>• X\ C P! to a continuous function £1: P2 —> PI . For PS we have to do a little extra work. Since P2 is an A N R (Theorem 2.1.24), there is a neighborhood U of X% such that the function /2 : X3 —> X? ^ P2 can be extended to a continuous function 77: U —>• P2. Pick a neighborhood F C C7 of Xs such that r)[v] c B(x2, y3), & o

138

2. BASIC COMBINATORIAL TOPOLOGY

We think of Lhn(P n , f n ) and lim(Xn, / n ) = JC as subspaces of claim that Um(P n , £ n ) = ]im(Xn, fn) = X. It is clear that )jm(Xn,fn)

C h'm(P n ,£ n ).

Now take an arbitrary x E Hm(P n ,£ n ) and fix TV e N. Then,

m=N

and hence XN € XN by (*). Since £n extends fn for every n, this easily implies that x £ Hm(JC n ,/ n ). D

Exercises for §2.3. 1. Let X be a space and let It and V be open covers of X such that V < U. For every V £ V pick J7(V) e U such that V C f/(V). This defines a function / 0 : A^(V) (0) -+ N(U)W. Extend this function 'linearly' over each simplex of N(V) and prove that the resulting function from |W(V)| -^ \N(U)\ is continuous. 2. Prove that the following statements for a space X are equivalent: (1) X is compact. (2) Every locally finite open cover of X is star-finite. 2.4. Simplices in En In this section we shall formulate and prove some fundamental properties of simplices in W1 . Among other things, we shall present a proof of the Brouwer Fixed-Point Theorem and some of its applications. Lemma 2 A.I. Let a\ and a-2 be two (geometric) n-simplices in W1 which intersect in a common (n — l)-face T. Then a\ U cr2 is a neighborhood of the barycenter of T. Proof. Observe that the lemma is trivial if n — 1; we may therefore assume that n > 1. Let d\ = {a 0 ,ai,. . . ,a n _i,a^} and a2 = {a 0 ,ai, . . . ,a n _i,a^}, where The points {ao5 &i, . . . , a n-i; a n} are geometrically independent, from which it follows by Theorem 2.1.3 that the points — ao, 02 —

2.4. SIMPLICES IN Rn

139

are linearly independent. Now consider the standard basis ei, 62, . . . , en in W1 and a linear isomorphism gl : W1 —> W1 with the properties: gl(ei) = di - a0 (i < n), g1(en) = a1n-a0. This isomorphism followed by the translation x H-» x + ao is an affine isomorphism (Theorem 2.1.7) and hence a homeomorphism (see Page 113) which we denote by f1. Observe that f l has the following properties: '/1(Q)=a0,

In the standard basis we now change en into — en. By a similar argumentation as above we obtain an affine isomorphism / 2 : Mn —> Mn such that /2(Q)=a0,

Observe that fl and /2 agree on the plane P = {x £ Mn : xn = 0}. Also observe that /* f F = / 2 \ P is an affine isomorphism between P and the hyperplane spanned by the elements of r = (a 0 , ai, . . . , a n -i}- Now define a homeomorphism / : ffin —>• Mn as follows

Let b be the barycenter of r and let c = (Y n , Yn, . . . , Yn, 0). Then /(c) = 6 and the set is a neighborhood of c of which of which it is geometrically obvious that it is mapped by / into a\ U a? (see Exercise 2.4.1). We conclude that a\ U a^ is a neighborhood of b. D The next result is the key in the proof of the Brouwer Fixed-Point Theorem. Theorem 2.4.2. Let r be an n-simplex in W1 , and let S be a finite triangulation of r which subdivides its standard triangulation. Finally, let [i £ S be an (n — 1) -simplex and let b be the barycenter of//. (1) Ifb^dr then p, C dr and fj, is a face of precisely one n-simplex in §. (2) Ifb^dr then // is a face of precisely two n-simplices in §. Proof. Assume that b E dr. We will show that p, C dr. Since S subdivides the standard triangulation of r there is a simplex a G § with b € a such

140

2. BASIC COMBINATORIAL TOPOLOGY

that a is contained in a proper face of r, i.e., a C dr. By Lemma 2.2.1 we obtain that ju is a face of a so that // C dr. Now we shall prove that depending on whether b £ da or not, w is a face of at most one or at most two n-simplices in S. To this end, first assume that b G dr. If there exist two n-simplices in § having // as a common face then the union of these two simplices is a neighborhood of b (Lemma 2.1.5). But this contradicts the fact that b G dr which is equal to the topological boundary of r (Exercise 2.1.9). Ifb^dr and there exist three distinct n-simplices, say <Ji, <72 and a% G S, having IJL as a common face then since S is a simplicial complex, (<TI U cr2) H (cr2 U cr3) D (<73 U <TI) = (<TI H cr2) U (<72 n cr3) U (cr3 n <TI)

For the last equality observe e.g., that IJL is an (n — l)-simplex contained in the simplex a\ H <72. If a\ Pi a2 is an n-simplex then <j\ = <J2 which is a contradiction. So a\ fl cr2 is an (n — l)-simplex since it contains n as a face, hence it must be equal to fi. By Lemma 2.4.1 the above intersection is also a neighborhood of b in W1 . Consequently, // is a neighborhood of b in Mn which is impossible because JJL is contained in an (n — l)-dimensional hyperplane. For the remaining part of the proof, put

Observe that U is a nonempty relatively open subset of r. We shall now prove that // is a face of at least one n-simplices of 8. Striving for a contradiction, assume that there does not exist an n-simplex in S having IJL as a face. According to Lemma 2.2.1 this implies that IJL is the only simplex in S containing b which implies that U C IJL. Now if b 6 dr then we arrive at the desired contradiction because then U C ^ C dr and dr has empty interior in r. If b 0 dr then U n Int(r) is a neighborhood of b € Kn which is contained in yu. But this contradicts the fact that // is contained in an (n — l)-dimensional hyperplane of W1 . All there remains to prove is that if b ^ dr then there are at least two n-simplices having IJL as a face. Striving again for a contradiction, assume that this is not the case. We already know from the above that there is an n-simplex a e S having p as a face. Pick an arbitrary p e U and let 7 G S be such that p G 7. Then & G 7 by the definition of U and so by the same argumentation as above, JJL is a face of 7. But then either 7 = p or 7 = a. In either case, p £ a. We conclude that U C a. So Int(r) n C/ is a neighborhood of & G Mn which is

2.4. SIMPLICES IN R"

141

contained in a. Since )U is a proper face of a we also have b £ 5cr. This is a contradiction. D Let T — \{XQ, . . . ,Xk}\ be an arbitrary fc-simplex in E n . An m-Sperner map for r is a function from the vertices of sd^ m ^(r) to {0, . . . , k} such that if v is such a vertex and v € \{xio,...,xit}\ then /i(u) € {i 0 ,...,«*}. We call a fc-dimensional simplex a in sd^ m ^(r) full if

There are always full simplices as the following result shows. As to be expected, its proof is of a combinatorial nature. Sperner's Lemma 2.4.3. Let T = \{XQ, . . . ,Xk}\ be a k-simplex in Rfc and let h be an m-Sperner map for T. The number of full simplices in sd'm^(r) is odd and hence non-zero. Proof. We shall prove the lemma by induction on k. If k = 0 then T — \ {XQ}\ , and there is one full simplex. Assume that the theorem is true for all (k — 1)simplices, k — I > 0. We shall prove the theorem for T. Put /3 = {x0,...,xk-i}, 3>i = {cr e sd(m}5(T) : a is (k - l)-dimensional, h[cr] = {0, . . . , k - 1}}, ?2 = {a e sd (m) 3 r (r) : a isfc-dimensional,{0, . . . , k - 1} C h[a}},

respectively. Observe that C

By the definition of an m-Sperner map it therefore follows that the restriction of h to the vertices of sd^ m ^9 r (^) is an m-Sperner map for ft. As a consequence, the collection J*i (0) is the set of full simplices in the m-th barycentric subdivision of |/?|, so |Ti(0)| is odd by our inductive assumptions. Also, ^(O) is the set of full simplices in the ?n-th barycentric subdivision of r; so we have to prove that ^(O)! is an odd number. Put R = {(«, ju) € 3*1 x y2 '• K is a face of p,}.

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2. BASIC COMBINATORIAL TOPOLOGY

We compute the cardinality of ft twice. For each K G CPi put #[«;] = {/JG ?2 : ( K ,/x) G ft}. Claim 1. |ft| - |0>i ( 0 ) | + 2 -|Ti(l)|. Proof. Clearly, jft| = £ |ftM|. «£?!

We claim that if K E ^i(O) then |ft[ft]| = 1. Simply observe that such K is a face of precisely one fc-dimensional simplex in sd^ m ^9 r (r) by Theorem 2.4.2, and this simplex clearly belongs to J*2. We next claim that if K G ^1(1) then |ft|V|| = 2. For take an arbitrary K G ?i(l) and assume first that it is contained in a (k — l)-dimensional face of r, say K C |{#;0, . . . , X i k _ 1 } \ . Then Ji[«] C { z o , . . . , « & - i } which implies that {^o, . . . , i/t-i} = {0, . . . , k — 1} and so K C |/?|, contradiction. So such a AC is not contained in a (k — 1) -dimensional face of r, and therefore, again by Theorem 2.4.2, is a (k — 1) -dimensional face of precisely twofc-simplicesin sd^S^r), and these simplices clearly belong to 3V ^ For each /z G O^ put ft- 1 ^] = { K e O ) 1 : (K,H) eft}. Claim 2. |fl| = | y 2 ( 0 ) | + 2 - | ? 2 ( l ) | . Proof. Clearly,

Take /z G T 2 (0). We claim that l^"1^]! = 1. Indeed, h[fi] = {0, . . . , k}, i.e., the function h is one-to-one on fi. So p, has exactly one (k — 1) -dimensional face K such that { 0 , . . . , f c — !} = /&[«] and this face is clearly the only element of Ift"1^]!- If V e y 2 (l) then /i[/t] = (0, . . . , A; - 1}. Since /i has size fc + 1, there exist precisely two subsets E and F of fi of cardinality fc with the property that Consequently, there exist precisely two (fc — l)-dimensional faces K± and ^2 of IJL with /I[KI] = {0, . . . , k — 1} = h[kz\. These simplices are obviously the only elements of Ift" 1 ^] | . 0 We find that

is even. As was remarked at the beginning of the proof, |CPi(0)| is odd, hence sois|T 2 (0)|. D

2.4. SIMPLICES IN R™

143

This result has interesting consequences. Lemma 2.4.4. Let T be a k-simplex in Rk , say r = \{XQ, . . . , Xk}\- In addition, let {Fi : 0 0 be a Lebesgue number for this covering (use that T is compact and apply Lemma A. 5. 3). By Corollary 2.2.9 we can choose ra so large that mesh(sd^ m ^3 r (r)) < e. Let V be the set of vertices of elements of sd^ m ^(r). Let v € V and put Iv = (0 0} (here a0 (v) , • • • , o.k (v ) are the affine coordinates of v with respect to x0 , . . . , Xk of course).

Figure 8. Let Iv = { I Q , . . . , i i } . Then v G \{xltt,...,xit}\CFi0\J---\JFit. So for v G V we can pick h(v) £ {0,.... A;} such that a / l ( u ) ( v ) > 0 , veFh(v). Then h is an m-Sperner map for T. By Theorem 2.4.3 there exists a full simplex • ln be continuous. Then f has a fixed-point.

144

2. BASIC COMBINATORIAL TOPOLOGY

Proof. Let T = \{XQ, . . . , xn}\ be an n-simplex in W1 . Then r is homeomorphic to I n (Exercise 1.1.24) and so it suffices to prove the Theorem for T. To this end, let / : T —> T be continuous. For i — 0, 1, . . . , n let Fl = {xer:ai(f(x))

(here (XQ(X), . . . , an(x) are the affine coordinates of x with respect to XQ, . . . , xn of course) . Let | {xi0 ,...,Xit}\ be a face of r and x € | {xi0 , . . . , Xit } \ . Then for an element i i0 . . . ^

so that "zo (&) + • • • + QV (x) = l> alo (f(x))

+ • • • + a», (/(a;))

and hence there must be 0 < j aii(f(x)).

Consequently, x € Fij . We conclude that \{xi0 , . . . , x^ }\ C F;0 U • • • U Fit . Each Fi is closed by continuity of / and the functions a; (Theorem 2.1.8). So by Lemma 2.4.4 there exists x 6 fllLi -^- Then for every 0 • Q be continuous. For every n G N define J\ — JT*£^I/*ITI-I nr \ — i T i T M i T T T* 1 \\ -i\n — \-*' ^ ^ • V"^l' • ' • i -^n) — \J V x ) i i • ' • •> J V 1 ^ ) n ) ) •

It is clear that for every n the set Kn is closed in Q and that Kn+i C Kn. Fix n 6 N, let pn: Q —> J n denote the projection and define a continuous function fn: J n -> J n by By Theorem 2.4.5, fn has a fixed-point, say ( x i , . . . , x n ), from which it follows that ( x i , . . . , a ; n , 0 , 0 , . . . ) G ATn.

2.4. SIMPLICES IN Rn

145

We conclude that the Kn's form a decreasing collection of nonempty closed subsets of Q so that by compactness of Q,

n=l

It is clear that every point in K is a fixed-point of /.

D

Corollary 2.4.7. Let X be a compact ANR. If/: X —> X is nullhomotopic then f has a fixed-point. In particular, every compact AR has the fixed-point property. Remark 2.4.8. In view of this result, the following question naturally arises. Let X be a compact ANR and let / : X —> X be continuous. Assume that / is homotopic to a continuous function g: X —>• X such that g has a fixed-point. Does / have a fixed-point? The answer to this question is in the negative, see Exercise 2.4.7. So in some sense, Corollary 2.4.7 is 'best possible'. Proof. We may assume that X is a subspace of Q (Theorem A. 4. 4). Since every constant function X —> X is extendable to a constant function Q —> X, by Corollary 1.4.3 it follows that / can be extended to a continuous function Since by Corollary 2.4.6, Q has the fixed-point property, / has a fixed-point, say x. Since f[Q] C X, x belongs to X, and since / extends /, we conclude that x is a fixed-point of /. Since by Theorem A. 12.4 every compact AR is contractible, the second part of the Corollary immediately follows from the first part and Proposition A. 12. 2. D Remark 2.4.9. In §3.14 we will show by elementary but remote methods that Brouwer's Fixed-Point Theorem for I3, implies the Brouwer Fixed-Point Theorem for all dimensions. This seems to show that the 'real' power of the Brouwer Fixed-Point Theorem is already attained at dimension three. This is a rather curious phenomenon and it is not clear whether there is a direct route from Brouwer in I 3 to Brouwer in all dimensions. This is of course only of interest for 'philosophical' reasons. The proof of Sperner's Lemma 2.4.3 is by induction, so it is equally complicated in all dimensions (except for the dimensions 1 and 2). Similar remarks can be made for the various other proofs that exist of Brouwrer's Theorem in the literature. This seems to indicate that even if one finds a more direct route from Brouwer in I 3 to Brouwer in all dimensions, this will not result in a simpler proof of Brouwer's Theorem. But this does not mean that clarifying the rather curious phenomenon that we will encounter in §3.14 would not be interesting. We finish this section by presenting three applications of the techniques developed in this section.

146

2. BASIC COMBINATORIAL TOPOLOGY

Application 1: The No-Retraction Theorem. It is clear that the function r: JJn —> Bn defined by r(x)

~

1

-

"|o;|| < 1).

n

is a retraction. Consequently, B has the fixed-point property by Theorem 2.4.5 and Exercise A.12.10(2) (observe that in fact I n and Bn are homeomorphic, cf. Exercise 1.1.24). Theorem 2.4.10 (No-Retraction Theorem). For every n E N, §n-1 is not a retract of Bn. Proof. To the contrary, suppose that Sn-1 is a retract of Bn. As was remarked above, Bn has the fixed-point property. Consequently, §™-1 has the fixed-point property by Exercise A.12.10(2). However, the antipodal mapping on S™^ 1 clearly demonstrates that §n-1 does not have the fixed-point property. Contradiction. D Corollary 2.4.11. No Sn is contractible. Proof. Suppose that §n is contractible. Then it is an AR by Corollaries 1.2.13 and 1.4.5. Consequently, there exists a retraction r: Bn —> S71"1, which contradicts Theorem 2.4.10. D Application 2: The Theorem on Partitions. The following result is the basis for dimension theory (see Chapter 3). The Theorem on Partitions 2.4.12. Consider J n and for i • JIn by /(x) = (£i(x), • . . ,^ n (a;)). Then / is continuous and the point 0 does not belong to its range. For every x G J n \ {0} the ray from 0 through x intersects the 'boundary' B — UlLi ^ u UlLi ^ °^ ^ n m precisely one point, say r(x). The function r: JIn \ {0} -^ B is easily seen to be continuous. The function g = r o f : $n —> B has the following properties:

2.4. SIMPLICES IN R rl

147

and for every i < n,

C 5i, 0[£i] C A,. Therefore, g has no fixed-point, which contradicts Theorem 2.4.5.

D

Corollary 2.4.13. Consider the Hilbert cube Q and its opposite faces W-1 ={xEQ:Xi = -1},

W} = {x 6 Q : Xl = 1}.

1

If Ci is a partition between W^ and W^ for every i then HiLi ^ ^ $• Proof. For every m, define /m : JIm —> Q by j m \-Eli---i-Em) — \'E'i i • • • i • E m i ' J i U ) . . . ) .

Then /TO is clearly an imbedding. It is easily seen that f^i^i] is a partition between A; and £; for every « < m. Consequently, by Theorem 2.4.12,

By compactness of Q we therefore obtain that oo

c 0

n **'

i=l

which is as desired.

D

Application 3: The Non-Homogeneity Theorem. Theorem 2.4.14. Let n 6 N. TLen (1) if A C S71"1 then Bn \ A is contractible. (2) if A C 5n \ $n~l is nonempty then Bn\A is not contractible. Proof. (1) Define H: (Bn \ A) x I ->• Bn \ A byfl"(a;,*)= (1 - ^)x. It is easily seen that H is a contraction. (Alternatively, apply Exercise 1.2.5 and Theorem 1.4.4.) (2) Without loss of generality, 0 6 A. Assume to the contrary that H:(Bn\A)xl^Bn\A is a contraction. Define F: S71"1 x I ->• §n-1 by

Then F contracts §n~1 to a point which contradicts Corollary 2.4.11. Since JJn and 5n are homeomorphic, cf. Exercise 1.1.24, this yields Corollary 2.4.15. Let n 6 N. Then ln is not homogeneous.

D

148

2. BASIC COMBINATORIAL TOPOLOGY

Exercises for §2.4. Let X be a space. We say that a closed set A in X separates X if X \ A is not connected. ^•1. Make the 'it is geometrically obvious' part in the proof of Lemma 2.4.1 precise. 2. Prove that the 'No-Retraction Theorem', the 'Brouwer Fixed-Point Theorem' and the fact that no S n is contractible are equivalent statements, in the sense that they are easily deduced from each other. ^•3. Let C be a compact convex subset of a norrned linear space L. Prove that for every e > 0 there exists a map fe: C —>• C such that (1) 0 ( / e , l ) < e , (2) fs\C] is contained in a finite dimensional linear space subspace of L. 4. Let X (1) (2) Prove

be a space and let /: X —>• X be a map with the following properties: the closure of f[X] is compact, for each e > 0 there exists x 6 X with Q(X, f(x}] < e. that / has a fixed-point.

5. Let C be a compact convex subset of a normed linear space L. Prove that C has the fixed-point property. (This is called the Schauder FixedPoint Theorem.) 6. Prove that the antipodal map S™ —>• Sn is not nullhomotopic. 7. Give an example of a compact ANR X having a homotopy H: X x I —>• X such that HQ is the identity and HI is fixed-point free. ^•8. Let A C Rn be compact, where n > 1. Prove that if A separates M71 then there is a continuous function /: A —>• S n-1 which is not nullhomotopic. 9. Let A C Sn be compact, where n > I . Prove that if A separates Sra then there is a continuous function /: A —> §n-1 which cannot be extended over S™. 2.5. The Lusternik-Schnirelman-Borsuk theorem The Brouwer Fixed-Point Theorem which was proved in the previous section, is one of the most central results in topology. We presented a detailed proof of it for two reasons. First of all, its proof uses subdivisions of triangulations of simplices. Getting used to this is important because that will be used extensively in ANR-theory later. Secondly, it is the basis for dimension theory. A more powerful result is the so-called Lusternik-Schnirelman-Borsuk Theorem which can be proved along the same lines using a somewhat more complicated combinatorial lemma than the one of Sperner (Lemma 2.4.3). We will not present a proof of the Lusternik-Schnirelman-Borsuk Theorem here since it is used only twice in this book and for our purposes its proof does not give more insight than the proof of Sperner's Lemma.

2.5. THE LUSTERNIK-SCHNIRELMAN-BORSUK THEOREM

149

The Lusternik-Schnirelman-Borsuk Theorem 2.5.1. If M is an arbitrary closed covering of S n such that |M| = n + 1 then at least one element M € M contains a pair of antipodal points. For a nice combinatorial proof, see DUGUNDJI and GRANAS [143, Theorem 4.4]. The Borsuk-Lusternik-Schnirelman Theorem implies many deep theorems about the topology of the Euclidean spaces E n . Exercises for §2.5. A map /: Sm —> S™ is called antipodal-preserving provided that f(—x) = —f(x) for every x G STO. 1. Prove that the Lusternik-Schnirelman-Borsuk Theorem is sharp by showing that there exists a closed covering M of Sra such that |M = n + 2 while no M G M contains a pair of antipodal points. ^•2. Prove that the following statements are equivalent in the sense that they are easily deduced from each other: (1) The Lusternik-Schnirelman-Borsuk Theorem for S n . (2) There is no antipodal-preserving map /: S^-^S"" 1 . (3) (Borsuk Antipodal Theorem) An antipodal-preserving map /: S^^S71'1 is not nullhomotopic. (4) (Borsuk-Ulam Theorem) Every continuous /: Sn ->• Rn

sends at least one pair of antipodal points to the same point. 3. Prove that the Lusternik-Schnirelman-Borsuk Theorem implies the Brouwer Fixed-Point Theorem.

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CHAPTER 3

Basic dimension theory Dimension theory enables us to assign to every topological space X a number, dimX, in {-l,0,l,...}U{oo} having, among other things, the following properties: (1) if X and Y are homeomorphic spaces then dim X = dim Y, (2) dimR n = n for every n € N. So dimX is a topological invariant of X, and by (2) it distinguishes between the euclidean spaces M n , n G N. A space X for which dim X < oo is called finite dimensional, and a space is in finite- dimensional if it is not finite dimensional. The aim of this chapter is to present some basic results from dimension theory. Some of the presented results are classical: they mostly concern finite dimensional spaces. However, during the last decades significant contributions were made concerning the topology of infinite-dimensional spaces and the topology of hereditarily indecomposable continua. We shall also present some of these results in detail. 3.1. The covering dimension Let X be a space and let F be an index set. The Theorem on Partitions 2.4.12 motivates the following definition: a family of pairs of disjoint closed sets r = {(Ai,Bi) : i £ F} of X is called essential if for every family

{Liiier}, where L z is an arbitrary partition between Ai and Bi for every i, we have

if r is not essential then it is called inessential. So Theorem 2.4.12 and Corollary 2.4.13 show that I n has an essential family of n pairs of disjoint closed sets and Q has essential families of size n for every n < NO- It follows easily that every subfamily of an essential family is again essential (Exercise 3.1.1). 151

152

3. BASIC DIMENSION THEORY

Theorem 3.1.1. I n has an essential family consisting of n pairs of disjoint closed sets, but every family consisting of at least n + l pairs of disjoint closed sets is inessential. Proof. By the above remarks we need only to consider the second part of the theorem. Let A and B be disjoint closed subsets of I n and let E C I be dense. Claim 1. There is a partition D between A and B such that

D C {x e I n : (3i < n)(xi € E}}. Proof. This is easy. Every point x € A has a neighborhood of the form n IJfatj&i], i—l

with o,i,bi E E U {0, 1} for every i < n, which misses B. There is a finite family 3 of these neighborhoods whose union covers A, and the boundary D of this union is contained in the union of the boundaries of the elements of f. We conclude that D is the required partition between A and B. <0 Now let r = {(Ai,Bi) : i < n + 1} be an arbitrary family consisting of n+l pairs of disjoint closed subsets of ln. We will prove that T is inessential. Indeed, there exist n + l pairwise disjoint dense subsets of I. One can take for example

Q) nl, E2 = ( > 3 + Q ) n l , E3 = ( \ 5 + Q) nl, . . . , respectively. By the above there exist partitions Di between Ai and Bi such that DiC{x<Eln: ( 3 j < H}(XJ G Er)} (i < n + 1). Since the Ei are pairwise disjoint, clearly HlLi Di = 0.

D

Observe that in Theorem 3.1.1 we formulated a topological property of I n shared by no ITO for m ^ n. In particular we obtain: Corollary 3.1.2. Let n,m € N. Ifn^m

then In ^ lm.

These remarks suggest the following definition: for a space X define its covering dimension, dimX 6 { — 1,0, 1, . . .} U {oo}, by

dim X < n

<&

dim X = n dim X = oo

<£> <£>

every family of n + 1 pairs of disjoint closed subsets of X is inessential, dim X < n and dim X ^ n — I , dim X / n for every n > — 1 .

3.1. THE COVERING DIMENSION

153

A space X with dimJY" < oo is called finite dimensional] if it is not finite dimensional then it is called infinite-dimensional. Observe that dim X > n if and only if there is an essential family consisting of n pairs of disjoint closed subsets of X. So in a sense, there are n different 'directions' in X. Also observe that it is not clear at all why we call dimX the covering dimension of X: the term partition degree seems to be more appropriate. We will explain our terminology later. It is easy to see that if X and Y are homeomorphic spaces then X and Y have the same covering dimension. In our new terminology, Theorem 3.1.1 reads as follows: Theorem 3.1.3. diml n = n for every n. In §1.5 we called a space zero-dimensional if it is nonempty and has a base consisting of clopen sets. In Corollary 1.5.2 it was shown that a space X is zero-dimensional if and only if 0 is a partition between any pair of disjoint closed subsets of X. So, in our new terminology, a space X is zero-dimensional in the sense of §1.5 if and only if dimX = 0. It will turn out that zero-dimensional spaces are, in a sense, the 'building blocks' of all the other finite dimensional spaces. Without too much trouble, the proof of Theorem 3.1.1 can be adapted to show that dimR" = n for every n. This equality however will turn out to follow trivially from Theorem 3.1.1 and results to be derived in §3.2. For that reason we will not verify it here. For later use we shall now study some elementary properties of essential families of pairs of disjoint closed sets. The following simple result is fundamental for dimension theory since it allows to extend partitions in subspaces to partitions in the whole space. Lemma 3.1.4. Let Y be a subspace of a space X. In addition, let A and B be disjoint closed subsets of X. If U and V are open neighborhoods of A and B, respectively, having disjoint closures, and if S is a partition in Y between Yr\U and Y Pi V, then there is a partition T in X between A and B such that T n Y C S. Proof. Write Y \ 5 as the disjoint union of two open (in Y) sets E and F such that

YnucE, Ynv CF. Since E n V = 0 we obtain E n B = 0, and similarly F n A = 0. From this it follows easily that A U E and B U F are separated. By Corollary A.8.2 there exist disjoint open neighborhoods U' and V of A\JE and B(JF, respectively. Clearly, T = X \ (U1 U V) is as required. D For a closed subspace we can do a little better.

154

3. BASIC DIMENSION THEORY

Corollary 3.1.5. Let Y be a closed subspace of a space X. In addition, let A and B be disjoint closed subsets of X. If S is a partition in Y between the sets Y n A and Y n B, then there is a partition T in X between A and B such that T n Y C S. Proof. Write Y \ S as the disjoint union of two open (in Y) sets E and F such that YnACE,

YHB CF.

Observe that S U F U B is closed in X and that A n (S U F U B) = 0. By Corollary A. 4. 3 there is an open neighborhood U of A in X such that £7n(SuFu£) = 0. By a similar argument it is possible to find an open neighborhood V of B in X such that

Vn(SuEuu) =0. By construction, S is a partition between U C\Y and F n Y. Now apply Lemma 3.1.4. D Corollary 3.1.6. Let Y be a zero- dimensional subspace of X . Then for all disjoint closed subsets A and B of X there exists a partition S between A and B in X such that S n Y = 0. Proof. Let A and B be disjoint closed subsets of X and let U and V be disjoint neighborhoods of A and B, respectively, such that UnV = 0 (Corollary A. 4. 3). Since Y is zero-dimensional, 0 is in Y a partition between the sets U n Y and V n Y. Now apply Lemma 3.1.4. D Corollary 3.1.7. If a space X is the union of at mostn + l zero-dimensional subspaces then dim X < n. Remark 3.1.8. We will prove later in Corollary 3.3.9 that the converse of this result is also true; so a space is at most n-dimensional if and only if it is the union of a family of n + I (not necessarily pairwise distinct) zerodimensional subspaces. Proof. Let X = \J^ Xl with dimXi < 0 for i < n + I . Let be an arbitrary family of n + 1 pairs of disjoint closed subsets of X. By Corollary 3.1.6, for each i < n + I there exists a partition L% between Ai and Bi in X such that LiCiXi = 0. Then f}™=i Li = 9 so that T is inessential. We conclude that dim X
3.1. THE COVERING DIMENSION

155

Theorem 3.1.9. Let X be a space, let {(Al^Bi) : i € F} be essential in X, and let F0 be a proper subset of T. If for every i £ F 0 , Li is a partition between Ai and Bi and L= p| Lz ier 0 then is essential in L. Proof. For i (£ TO let £^ be a partition in L between LnAi and L n Bi. By Corollary 3.1.5, for i £ F0 we can find a partition F; in X between Ai and ^ such that Fl H L C E^ Then

,n

e L n Pi E,= f|

which is as required.

D

Corollary 3.1.10. Let X be a space and let n > 0. If dim X > n then there exist disjoint closed subsets A and B of X such that if L is a partition between A and B then dim L > n — 1. Proof. If n — 0 then there is nothing to prove. So assume that n > 1. Since dim X > n there exists an essential family in X. We claim that AI and BI are as required. Indeed, let L be a partition between AI and BI. Observe that L ^ 0 since T is essential. We consider two cases. If n = 1 then since L / 0, dim!/ > 0 and so we are done. If n > 1 then by Theorem 3.1.9, L has an essential family of cardinality n — 1, namely the collection {(L n A^ L n Bi) : i = 2, . . . , n}. And so dim L > n — 1 . D Exercises for §3.1. A space X is called countable dimensional if it can be written as the union of countably many zero-dimensional subspaces (cf. Corollary 3.1.7). In addition, a space X is called strongly infinite-dimensional if it has an infinite essential family. See e.g., §3.13 for more information on these notions. Let X be a space with disjoint closed subsets A and B. We say that a closed subset S of X is an irreducible partition between A and B if S is a partition between A and B while no proper closed subset of S shares this property. 1. Prove every subfamily of an essential family is again essential.

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2. Let r = {(Ai, Bi) : i < n} be a family of pairs of disjoint closed subsets in the space X. Prove that if there exist different indices i-,j
{(AinE,Bir\E) :ier} is inessential in E. ^•9. Let X be a compact space and let r = {(Ai,Bi) : i 6 F} be an essential family of pairs of disjoint closed subsets of X. Prove that there is a component C of X such that the family r \C = {(AlnC,BinC)

:i€r}

is essential in C. Conclude that every strongly infinite-dimensional compact space has a strongly infinite-dimensional component. 10. Prove that 21, the hyperspace of I, is infinite-dimensional. 11. Let /: X —>• Y be continuous, and let r = {(Ai, Bi} : i 6 /} be a collection of pairs of disjoint closed sets in X. Assume that the collection consists of pairs of disjoint sets and is inessential. Prove that r is inessential. 12. Let /: X —> Y be continuous, and let T = {(Ai,ft) : i € /} be an inessential collection of pairs of disjoint closed sets in Y. Prove that the collection f = {(/-1[Ai],/-1[ft]):t€/} is inessential in X.

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157

13. Let X be a space and let {(Ai, Bi) : i £ F} be essential in X. Suppose that the set y C X is such that y fl P|i£r L; 7^ 0 for any choice of partitions Li of Ai and Bi in X. For each i E F let f/i and Vi be disjoint closed neighborhoods of Ai and Bi, respectively. Prove that

{(t/iny,v;ny): i e r} is essential in y. 14. Let X be a zero-dimensional space containing the pairwise disjoint closed subsets A, B and C'. Prove that C is a partition between A and B. (So in a zero-dimensional space every irreducible partition is empty.) ^•15. Give an example of a space X containing two disjoint closed subsets A and B such that no partition between A and B is irreducible. ^•16. Prove that the Brouwer Fixed-Point Theorem and the fact that there is a space X with dim X = oo are equivalent statements, in the sense that they are easily deduced from each other. (This complements the list in Exercise 2.4.2 with an additional statement.) 3.2. Translation into open covers We defined the covering dimension dim X in terms of properties of the family of all closed subsets of X. In this section we see that dimX is also describable in terms of properties of the family of all open covers of X. As an application of the results derived here we present a simple proof that W1 is n-dimensional for every n. We conclude from this that W1 and RTO are homeomorphic if and only if n = m. Let X be a space and let A and 13 = {B(A) : A e A} be families of subsets of X. We say that 13 is a swelling of A if (1) for every A e -A, A C B(A), (2) for every finite J C A, f| 3~ = 0 iff [}A^B(A] = 0. Observe that if B(Ao),B(Ai) G 13 are distinct then so are A0 and AI, but the converse need not hold. Before we state the following proposition, let us recall that a locally finite collection is countable (see the remark on Page 486 following the definition of locally finite collection). Proposition 3.2.1. Let 3 be a locally finite collection of closed subsets of a space X. Then 3 has a swelling consisting of open subsets of X. Proof. We can enumerate 3 as {F; : i £ N}. Without loss of generality, assume that F\ = 0. By induction on i £ N we shall construct an open neighborhood U% of F; such that

158

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is a swelling of 3. Put U\ = 0, and assume that for some i the sets U\ , . . . , have been constructed. Observe that 3i is locally finite. Put finite subcollection of 3l and ( P) £) n Fl+l = 01. Observe that 23 is locally finite as well and put B = \J eS>. Then B is closed by Exercise A. 7.1 and clearly B D Fi+\ — 0. By Corollary A. 4. 3 there consequently exists an open neighborhood Ui+i of FJ+I the closure of which misses B. It is clear that the set Ui+\ is as required. We claim that {Ui : i G N} is a swelling of 3. To this end, take arbitrary «'(!)) . . . ,i(n) G N and assume that D?=i ^i(j) = ®- Let m = max{z(l), . . . , i(n)}. Then {Ui, . . . , Um} is a swelling of {Fi, . . . , Fm} by our construction. We conclude that f|™=1 C/i(j) = 0 . D Corollary 3.2.2. Let 3 be a locally finite family of closed subsets of a space X. Also, for every F G 3 let V(F) be a neighborhood of F. Then there exists a swelling {U(F) : F 6 3} of 3 consisting of open subsets of X such that for every F e 3 we have U(F) CV(F). Proof. By Proposition 3.2.1 there exists an 'open' swelling {W(F) : F 6 3} of 3. By Corollary A. 4. 3 there exists for every F G 3 an open neighborhood U(F) such that U(F) C V(F) n W(F). We claim that [U(F) : F e 3} is as required. To this end, let £5 C 3 be finite such that H S = 0- Then D clearly f) F€S W(F] = 0 from which it follows that fl^es U(F) = 0 " We now show how finite closed collections can be modified. Lemma 3.2.3. Let A,Ai,...,An be closed subsets of a space X. Moreover, let V\ , . . . , Vn be open subsets of X with A n Ai C Vi for every i < n. If

1=1 for i < n then

(i) A u U I U ^ c i u U I U ^

(2) innr=i^cnr =1 Fr\/ z .

Proof. Observe that (1) is trivial. For (2), simply observe that for i < n,

AnAz = ( A \ v l } n ( A l u v l ) = [(An ^ \Vi\\j [An (Vi \Viil So we are done.

D

3.2. TRANSLATION INTO OPEN COVERS

159

One should think of the previous triviality in the following way. The closed sets A are changed into the closed sets A. But this is not done in an arbitrary way because (1) tells us that what is covered by the ^4's is also covered by the A's, and (2) tells us where the intersection of the A's can be found. The results obtained so far have nice applications in dimension theory. Corollary 3.2.4. Let X be a space such that dimX < n < oo. Then for every countable open cover U of X and for every subcollection 3 C U of cardinality n + 2 there exists an open shrinking V — {V(U) : U G U} o f U having the following properties: (1) for every U G U, V(U] C V(U) C U, (2) r\U€
y= {c/i,..., un+2}. By Proposition A.7.1 there exists a closed shrinking {Bl : i 6 N} of U. For every i < n + 1 define Ai = X \ Ui. Then {(Bi,Ai) : i < n + 1} is a family of n + 1 pairs of disjoint closed subsets of X. Since dimX < n, there exist open sets Vi C X for i < n + 1 such that (3) B1CV1CV1CX\A1 = Ui, Now consider the closed sets B\, . . . , Bn+2- For i < n + I put Bi = Vi and let Bn+2 = Bn+2 \ ljr=i ^- Observe that the B's are the 'improved' sets we get from Lemma 3.2.3 by considering the collections of closed sets

and open sets

Vi,.. . , So we get by (4) that n+2

(5)

n+1

f l ^ C p| i=l

i-l

Observe that ^ — {B\ , .62 , . . . , Bn+2 , Bn+3 , -Bn+4 , • • • } covers X since \J™=i Bi C Ul^i ^- ^n addition, the first n + 2 members of 23 have empty intersection by (5). Since Bi C Ui for every i < n + 2, the desired result now follows from Corollary 3.2.2. D

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Let U be a cover of a space X and let n > 0 (we do not assume U to be open). We say that the order of U is at most n, ord(U) < n, if for every x 6 X, \{U £U:x£ U}\ 0. The following statements are equivalent: (1) dimX < n, (2) for every open cover U of X there exists a locally finite closed refinement V of U such that ord(V) < n, (3) for every open cover U of X there exists an open refinement VofU with ord(V) < n, (4) for every open cover U of X there exists a closed shrinking VofU with ord(V) < n, (5) for every open cover U of X there exists an open shrinking VofU with ord(V) < n, (6) for every finite open cover U of X there exists a closed shrinking V of 1C such that ord(V) < n, (7) for every finite open cover U of X there exists an open shrinking V of It such that ord(V) < n. Remark 3.2.6. This theorem explains the earlier terminology covering dimension. The vague idea behind dimension theory is that one wishes to cover a given space with finitely many 'small' open sets in such a way that as few of the open sets as possible have points in common. If dim X < n then it is possible to do that in such a way that at most n + 1 elements intersect. If dimX > n then no matter how the 'small' cover is chosen, there are always at least n + I elements having a point in common. It is interesting and intriguing that these ideas can be used to distinguish between various spaces topologically. Remark 3.2.7. Missing in the list of statements in Theorem 3.2.5 is: (8) for every open cover U of X there exists a dosed refinement V of U with ord(V) < n. This is not a very interesting statement, since every open cover U of a space X is refined by the closed cover {{x} : x € X} of X which has order 0. The following statement seems to be more interesting: (8') for every open cover U of X there exists a countable closed refinement V o f U with ord(V) < n. It is not true that (8') characterizes all at most n-dimensional spaces. We will show in Exercises 3.2.7 and 3.2.8 that Erdos' space described in Example 1.5.18 has the property that every open cover U of it has a countable

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161

disjoint closed refinement, yet E is one-dimensional (as will be shown in Example 3.4.13). We do not know whether spaces that satisfy condition (8') are at most (n + l)-dimensional. Proof. We prove (1) =>• (2). Since a refinement of a refinement is a refinement, by Corollary A. 7. 3 we may assume that U is locally finite. Enumerate U as {£/o,i : i & N}. In addition, let {F(j) : j G N} enumerate the collection of all subsets of N of cardinality precisely n + 2. By Corollary 3.2.4, for j G N there exists an open cover Vj = {Uj^ : i G N} of X having the following properties: (8) for each i, Uj i C Uj i C t/j-i ;, (9) Now for each i 6 N define

Claim 1. The collection V = {Si : i G N} covers X. Proof. Take an arbitrary x G A". Since U is locally finite, x is contained in finitely many elements of U only. Pick ra G N such that x ^ Ui> m ^o,i- We conclude that for every j' G N there is an index k(j) < m such that x G f^j,fc(j) • So there exists k < m such that a; belongs to infinitely many of the C/j,fe. By (8) this implies that x G Sk, which is as required. <> We conclude that V is a closed shrinking of U and hence is locally finite. Moreover, ord(V) < n by (8) and (9). We prove (2) => (3). By (2) there exists a locally finite closed refinement § of U such that ord(S) < n. For each S G 8 pick U(S) G U containing 5. By Corollary 3.2.2 there exists an open swelling V = {V(S) : S G S} of 8 such that for every element 5 G 8 we have V(S] C U(S). Since ord(S) < n we get ord(V) < n, and so V is as required. We prove (3) =>• (5). By (3) there exists an open refinement W of U such that ord(W) < n. For each W G W pick U(W) G U be such that W C U(W). Now for each U G U define

V(U) = \J{W G W : U(W) = U}. Clearly, V = {V(U) : U G U} is an open shrinking of U and it therefore suffices to prove that ord(V) < n. Take pairwise distinct V(U1),...,V(Un+2)eV

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3. BASIC DIMENSION THEORY

and assume that x G C\i=i V(Ui)- For each i < n+2 there exists Wi € W such that U(Wi) = Ui and x G Wi. Since the [7; are pair wise distinct, this implies that the Wi are pairwise distinct. But ord(W) < n, and so x G [^^ Wi = 0. This is a contradiction. We prove (5) =>• (4). Observe that if A is a cover of X such that ord(,A) < n and I> is a shrinking of .A then ord(!B) < n. So (4) follows directly from (5) and Corollary 3.2.4. We prove (4) =3- (6). This is a triviality. We prove (6) => (7). This follows immediately from Corollary 3.2.2. We prove (7) => (1). Let {(A;,.E?i) : i < n + 1} be a family of n + 1 pairs of disjoint closed subsets of X. Observe that n+l

n+l

p| Ai n U 5, = 0 i=l

i=l

so that the collection n+l

{x\A1,X\A2,...,X\An+l,X\ covers X. By (7) there consequently exists an open cover V = {Vi : i < n + 2} of X such that (10) ViCX\Ai

(i
(12) ord(V) < n. Let J = {Fx : i < n + 2} be a closed shrinking of V = {Vi : i < n + 2} (Proposition A. 7.1). Now for i < n + 1, define Ai and Bi by

Ai = Ai\J(Fn+2\Vi),

Claim 2. For every i < n + 1, Ai n B» = 0. Proof. Simply observe that A% n Bi — 0, that F; C V^ from which it follows that Ai n Fi = 0 and (Fn+2 \ Vi) n F» = 0 and that Fn+2 n 5» = 0. 0

Claim 3.

1

u

& = X.

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163

Proof. Since 3~ covers X, we are done if we show that (JF^i -^ - ^=i &i and Fn+2 C (J^/ AY The first inclusion is a triviality. For the second one observe that Fn+2 C Vn+2 and that by (12), Vn+2C\^\^ Vi = 0. We conclude that n+l

n+1

i—l

i=l

Fn+2 = Fn+2 \ p| Vi c y A, which proves the claim.

Q

By Corollary A. 4. 3 there exists for every i < n + I a partition Li between Al and Bi. Since A^ C A; and Bi C B; for every z, L; is also a partition between the sets A; and Z?;. By Claim 3 we obtain n+l

n+l

as required.

D

Naturally, the reader wonders about the relation between the dimension of a space X and the various dimensions of its subspaces. We finish this section by deriving two results about that relation. The Countable Closed Sum Theorem 3.2.8. If X can be covered by countably many closed and at most n-dimensional sets then X is at most n- dimensional. Proof. It is clear that without loss of generality we may assume that n < oo. Enumerate the closed cover "3 as {Fi : i e N}, and put FQ — 0. Let U be a finite open cover of X. By induction on i > 0 we shall construct an open cover U(i) of X such that the following conditions are satisfied: (1) U(0) = U, (2) if 0 < j < i then U(i) is a shrinking of U(j'), (3) ord(U(i) r Fi) < n.

Observe that conditions (2) and (3) are satisfied for i = 0. Now assume that we completed the construction for all j with 0 < j < i. Since dim .Ft < n, by assumption, the open cover U(i — 1) \ Fi of Fi has an open shrinking V = {V(U} : U € U(i - 1)}

of order at most n (Theorem 3.2.5). For each U 6 U(i — 1) put W(U) - (U\Fi)\JV(U). It is clear that W = {VT(C7) : t / e U ( i - l ) } is an open shrinking of U(i — 1) such that ord(W \ Fi) < n. An easy application of Proposition A.7.1 and Corollary 3.2.2 now shows that there exists

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3. BASIC DIMENSION THEORY

an open cover U(i) of X satisfying (2) and (3). This completes the inductive construction. Let the cardinality of U be k < oo. It is clear that for i e N we can enumerate li(i) as {Umji : m < k} such that for 0 < i < j and ra < k we have Umj C Um^. Now for each x £ X there exists m(x] < k such that x belongs to infinitely many of the t/ m ( x ),;. So by construction we have

By (2) and (3) this implies that oo

oo

Um,i : m < A;} - { f| t7m,i : m < fc}

is a closed shrinking of U and has order at most n. Consequently, dim X < n by Theorem 3.2.5. D Since Mn is a countable union of homeomorphs of I n , Theorems 3.1.1 and 3.2.8 imply that dimR n < n. In the remaining part of this section we shall present, among other things, two additional proofs of this inequality which are interesting in their own rights. The Subspace Theorem 3.2.9. Let A be a subspace of a space X. Then dim A < dim.Y. Proof. If dim.Y = oo then there is nothing to prove. So without loss of generality assume that n = dim X < oo. First assume that A is closed. Observe that if r is an essential family in A then it is also an essential family in X. From this it follows immediately that dim A < n. Next, assume that A is open. Observe that A is an F^-subset of X by Exercise A. 2. 3. That dim A < n therefore follows from the above and the Countable Closed Sum Theorem 3.2.8. Finally, assume that A is an arbitrary subspace of X. In addition, let U be a cover of A by sets that are open in A. For each U 6 U pick an open subset V(U] of X with V(U] C\A = U. Put V = \JUeli V(U). Then V is an open subset of X and hence, by the above, dim V < n. Since V = {V(U) : U G U}

covers V, Theorem 3.2.5 yields the existence of an open refinement W of V with ord(W) < n. Then X = W \ A is an open refinement of U with order at most n. Another application of Theorem 3.2.5 now gives us that dim A < n, as required. D

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165

A different proof of Theorem 3.2.9 shall be given in Corollary 3.4.14(1)2. Observe that since W1 is homeomorphic to (0, l) n we have by Theorems 3.1.1 and 3.2.9 that dirnlP < n. The results in this section can also be used to prove that various spaces are zero-dimensional. We shall demonstrate this by the following example. For all n € N and 0 < m < n let y\.n>m — {x G W1 : exactly m coordinates of x are rational}. Proposition 3.2.10. Let n G N. Then W1 = U^ =0 ^Vm dimensional for all 0 < m < n.

and

^n,m is zero-

Proof. The first part of this proposition is trivial. For the second part, observe that if m = 0 then there is nothing to prove since 9^n,o is the product of n copies of IP. So let m > 1 and let A = |i(l), . . - , i(m)} be a set of m indices in {1, 2, . . . , n} and let q\ , . . . , qm £ Q- Put X — {x € W1 : x^) = qj for every j < m}. Observe that X is a closed subspace of R™ and that X n *Rn,m — [x 6 W1 : £;(j) = QJ for every j < m and x% G P for i £ A}. Consequently, Xr\9tnjTn is a closed subspace of 9^n,m which is homeomorphic to the product of n — m copies of F. We conclude that X n ytn,m is zerodimensional. Since lHn;TO is the union of countably many sets of the above form X n 9ln,m, we conclude that dimlH njm = 0 by Theorem 3.2.8. D Corollary 3.2.11. For all n £ N and 0 < m < n, dimja: G IRn : at most m coordinates of x are rational} < m. Proof. Observe that the set of all x E Mn for which at most m coordinates are rational coincides with So we are done by Proposition 3.2.10 and Corollary 3.1.7.

D

It therefore again follows that dimM n < n. We now come to what is sometimes called the 'Fundamental Theorem of Dimension Theory'. Theorem 3.2.12. For all n e N, dim IT = diml n = dim§ n = n. Proof. By Theorem 3.1.1, dim I™ = n. By Theorem 3.2.9 this implies that dimM n > n. Since dimM n < n, we obtain dimM n = n. Also, §n is the union of two homeomorphs of I n , so an appeal to Theorem 3.2.8 gives us that dimS n < n. It now follows easily that dim§ n = n as well. D

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3. BASIC DIMENSION THEORY

Theorem 3.2.12 is of fundamental importance since it confirms our geometric intuition that M n , ln and S n are n-dimensional. Since the covering dimension is a topological notion, we also obtain the following Corollary 3.2.13. Ifn^m then W1 and Mm are not homeomorphic. We finish this section with three characterizations of dimension, two of which are in the spirit of Theorem 3.4.4 and one of which is 'combinatorial'. Let r be an n-dimensional simplex in a linear space L. Exercise 1.1.24 gives us that T is homeomorphic to E n , hence dim r — n. Now let X = \7\ be a polytope. Since T is countable, by the Countable Closed Sum Theorem 3.2.8 and the above remark we get dim X = supjdim r : r e T} = supjA; : r is afc-simplex,r e T}. We conclude that the topological dimension of X equals its 'combinatorial' dimension. Theorem 3.2.14. Let X be a nonempty space and let n > 0. The following statements are equivalent: (1) dimX < n, (2) for every open cover li of X there exists a locally finite open refinement V of U such that such that ord(V) < n, (3) for every open cover U of X there exists a star-finite open refinement VofU such that such that ord(V) < n, (4) for every open cover U of X there exist a polytope P such that dim P < n and a ll-mapping f : X — > P . Proof. We prove (1) =» (3). Without loss of generality, U is star-finite (Theorem 2.3.5(1)). Now by Theorem 3.2.5 there exists an open shrinking V of U such that ord(V) < n. Since each shrinking of a star-finite cover is clearly star-finite, we are done. We prove (3) => (2) => (I). That (3) => (2) is a triviality and that (2) =>> (1) follows from Theorem 3.2.5. We prove (3) => (4). Let U be an open cover of X. By (3) there exists a (countable) star-finite open refinement V of U such that ord(V)
is a V-mapping by Corollary 2.3.4, and hence a U-mapping. Also, P is a polytope by Theorem 2.3.5. Finally, since ord(V) < n, N(V) has no simplices

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167

of dimension greater than n. By the Countable Closed Sum Theorem 3.2.8 it follows that dim P < n. We prove (4) =>• (1). Let U be an open cover of X. By (4) there exists a poly tope P such that dimP < n and a U-mapping f : X -> P. For each U € U let E(U) be the set of all y G P with an open neighborhood Vy such that /~ 1 [T / y ] C U. Then the collection £ = {E(U) : U £ 11} is an open cover of P while moreover f-l[E(U)]CU for every U G U. Since dimF < n, Theorem 3.2.5 implies the existence of an open shrinking V = {V(U) : U G U} of £ = {E(U) : £7 e U} such that ord(V) < n. It is clear that the collection W = {f~l[V(U)] : U G U} is an open shrinking of U such that ord(W) < n. By Theorem 3.2.5 we therefore conclude that dim A7" < n. D Exercises for §3.2. A space X is called almost zero-dimensional if it has an open base 23 such that every B e "B has the property that X \ B is the union of clopen subsets of X. 1. Let X be a space containing more than one point and let p £ X. Prove that dim(A \ {p}) = dimX. (So the dimension cannot be raised by the adjunction of a single point.) 2. Let (Xn, f n ) n be an inverse sequence of compact spaces such that

dim An < ra < oo for all n. Prove that the inverse limit of the sequence is at most mdimensional. (The compactness condition is superfluous in this result. But we are not in the position yet to prove this. See Exercise 3.5.2 for more information.) ^•3. Let A' be a dense-in-itself space such that dim X < n < oo. Prove that for every cover {Ui,...,Um} consisting of nonempty open sets there is a collection {Fi,...,F m } consisting of closed sets having the following properties: (1) Q^FiCUt for every i < m, (2) U™! K = X, (3) if i\ < i-2 < • • • < in+2 < m are arbitrary then P|n=i2 Fij — $• ^•4. Prove that every n-dimensional compact space has a component which is also n-dimensional. 5. Let X be a space. Prove that X is almost zero-dimensional if and only if X has a base !B such that for each pair G, H of elements of 3 with disjoint closures there is a clopen set W in A" with G C W C X \ H. 6. Prove that every almost zero-dimensional space is totally disconnected. ^•7. Prove that Erdos' space E is almost zero-dimensional.

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8. Prove that if X is almost zero-dimensional then every open cover of X has a countable refinement by pairwise disjoint closed sets. 9. Prove that C(I 2 ) is infinite-dimensional. 10. Let X be a homogeneous space. Prove that every nonempty open subspace of X has the same dimension as X. 3.3. The imbedding theorem The aim of this section is to prove that every space A" with dim X < n can be imbedded in E2n+1 . This is a result of fundamental importance. For n > 0. define yin = {x G E2n+1 : at most n coordinates of x are rational}.

So fc=0

where the sets D^n+i.fc ^ M2n+1 are as in the previous section. Observe that 9to is the space of irrational numbers P. The space 9tn is called Nobeling's universal n-dimensional space. For later use, we do not aim at imbeddings in R2n+1 but at imbeddings in 9tn. This does not require extra work and explains our terminology. Lemma 3.3.1. For n > 0, dim 9^ = n and 9Tn is the union of n + I zerodimensional subspaces. Proof. By Corollary 3.2.11 we get dimWn < n. To see that dim9T n > n, we claim that 91n contains a homeomorph of I n . This is easy. Define an imbedding i : I" -» 9?n by i(xi,...,xn) = ( x i , . . . , x n , \ / 2 , \/2, . ..,-s/2). By Theorems 3.2.9 and 3.2.12 we now obtain dim01n > n. Since, as observed above, 9Tn = !H2n+i,o U ^n+i,! U • • • U ^n+i.n? an application of Proposition 3.2.10 gives us that 9tn is the union of n + I zerodimensional subspaces. D A finite subset F = {xi, . . . , xn} of Mm is said to be in general position if for every collection i(0)
is geometrically independent, cf. §2.1.

3.3. THE IMBEDDING THEOREM

169

Lemma 3.3.2. Let G = {xi,... ,xn} and F = {?/i,... ,y/t} be subsets of Em such that G is in general position, and let e > 0. Then for each i < k there exists a point xn+i G Km such that (1) ||j/i -2; n+ i|| < £ >

(2) {xi,...,xn,xn+i,...,xn+k} is in general position. Proof. It is clear that the lemma follows by induction once we have established it for k — I . So we assume without loss of generality that k — 1. For every nonempty subset F C { l , . . . , n } of cardinality at most m, let v(F)=aS(F), the affine hull of F. Each v(F) is a nowhere dense closed subset of Mm since \F\ < m. Since G = {xi,... ,x n } is finite, there consequently exists a point

xn+l € 5T \ \J{v(F) : F C {1,..., n}, \F\ < m} such that g(y\, xn+\) < e. We claim that {x\,..., x n +i} is in general position. Since G is, we need only prove that if F C {!,..., n} has cardinality at most m then {xi : i 6 F} U {xn+i} is geometrically independent. This is clear however since by construction, xn+\ 0 v(F] (Theorem 2.1.6). D Now for the remaining part of this section, let X be a fixed space such that 0 < dim^ < n. Proposition 3.3.3. Let f 6 C(X,R2n+l; Q), let e > 0, and let H be an at most n-dimensional affine subspace of M 2n+1 . For every finite open cover li ofX there exists g e C(X, E 2n+1 ; g) such that (1) g ( f , g ) < e,

(2) g[X] n H = 0, (3) g is a U-mapping.

Proof. Since / is bounded, f[X] is compact. There consequently exists a finite open cover V of it such that mesh(V) < £/4. Let W be the common refinement of U and /-1[V]. Observe that W is finite. Since dim X < n, there exists by Theorem 3.2.5 a finite open shrinking £ of W of order at most n. The essential properties of the cover £ are the following ones: (4) £ has order at most n, (5) £ < U, (6) for each E e £, diam(/[£]) < e/4We assume without loss of generality that every member from £ is nonempty. Let £ = {-Bi, • • • , Em}. Now for each i < m, pick an arbitrary point

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There is a geometrically independent set 5 = {hi, . . . , hk] C R 2n+1 of cardinality at most n + 1 such that S spans H. By Lemma 3.3.2, for each i < m there exists yi e M2n+1 such that (7) g ( y l , z l ) < e/4, (8) {hi , . . . , hk , yi , • - • , ym } is in general position. Let KI, . . . , Aem : X —)• I denote the ^-functions with respect to the cover £. Define 5: X -»• R2n+1 by

Then g is clearly continuous and we claim that it is as required. First observe that g is bounded because its range is contained in the simplex spanned by the vectors {yi,... , y ^ } which is compact by Lemma 1.1.1(2). For F C {1, 2 , . . . , m} we let F denote {j/» : z 6 F}. For every x G X let Fx — {i < m : x G Ei}. Observe that Ki(x) > 0 if and only if i G Fx so that Y^ierx Ki(x} = 1It follows by (4) and (8) that Fx is geometrically independent. Consequently, the Ki(x), i G Fx, are the affine coordinates of g(x) with respect to Fx. In particular, g(x) G aS(Fx). Claim 1. 0(f,g) < e. Proof. Fix an arbitrary x G X. It follows by (6) and (7) that \\f(x)—yi\ for every i G Fx. Consequently, \\f(x) — 9(x)\

=

\\ / . Ki(x) ' f ( x ) — /

J

< e/2

KI(X) • yi

We conclude that g(f,g) < £/2 < £, which is as required.

•(>

Claim 2. g[X] C g[X] C \JX&X ztt(Fx) C M2^1 \ H. Proof. Let x e X. By (4), |FZ| < n + 1 and so if we put E = Fx U 5 then \E\ < (n + 1) + (n + 1) = 2n + 2.

By (8) this implies that E1 is geometrically independent. Now observe that by Proposition 2.1.5 we have aff(Fx)

n H = afi(Fx) n aff(S) = aff(F x n 5) = aff(0) = 0.

3.3. THE IMBEDDING THEOREM

171

Since g(x) G aff(F x ) and x is arbitrary it therefore follows that g[X] C |J aS(Fx) C R2n+l \ H. But since aS(Fx) is closed by Exercise 2.1.8 and there are only finitely many F^'s, we even have that g[X] C g\x] C |J aff(F x ) C R2n+l \ H, x£X

as desired.

§

It remains to prove that g is a U-mapping. Let p G
U A| < (n + 1) + (n + 1) = 2n + 2.

By (8) this implies that Gp U A is geometrically independent. So by Proposition 2.1.5 we get p G aff (A) n aff (Gp) = aff (A n Gp). Since i0 ^ A and the affine coordinate of p corresponding to y^ is non-zero, by unicity of affine coordinates (Theorem 2.1.8), this is a contradiction. <0> Now let B be the complement of

and for some i G Gp,yi $ A}. By Claim 3, B is an open neighborhood of p in M2n+1 . Claim 4. g-i[B]C^GpEl. Proof. Pick an arbitrary z G g~~l[B] and let IQ G G p . By the definition of g we have g(z) G aff(F z ). Since |FZ| < n + 1 by (4), if i0 ^ F2 it would follow that which is a contradiction. So io G F2 or, equivalently, z G £^i() .

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3. BASIC DIMENSION THEORY

If p E U = W1 \ g[X] then U is a neighborhood of p and g~l[U] = 0. Since by (5) we have £ < U, this completes the proof.

D

Proposition 3.3.4. Let X be a space 0 < dim^Y" < n. In addition, let

{Ui-.ie N} be a sequence of finite open covers of X such that mesh(Ui) < V; for every i. Then the function space C(X, R 2n+1 ; £>) contains a dense GSsubset G having the following properties: (1) (2) (3) (4)

Every f 6 C is an imbedding. Every f 6 G is a Ui-map for every i. The range of every f £ C is contained in 9Tn. The range of every / E C has compact closure in 9tn.

Proof. Let A = [i(l),... ,i(n + l)} C {1, 2 , . . . , 2n + l} and g i , . . . ,g n+1 G Q be arbitrary. Put # = (re G M2n+1 : xi(j) = q3 for every j < n + 1}. Observe that H is an n-dimensional affine subspace of M2n+1 and that

H n 9?n = 0. 2n+1

In addition, each point of R \ 9In has at least n + 1 rational coordinates and is therefore contained in an affine subspace of the form H. We conclude that the complement of 9tn is the union of countably many n-dimensional affine subspaces, say {Li : i G N}. Now consider the function space C(X, R 2n+1 ; Q). For every i G N, put C, = {/ 6 C(X, M 2n+1 ; Q) : f is a U z -map and f[X] n Li = 0}. It follows from Proposition 3.3.3 that Gz is dense in C(X, R 2n+1 ; Q) for every i. Claim 1. C, is open in C(X,R2n+1; Q) for every i. Proof. Take an arbitrary / 6 C^. Since / is bounded, f[X] is compact. Therefore, since f[X] fl Li = 0, there exists £ > 0 such that (5)

{yeE 2 ^ 1 :g(yJ[X])<e}nLi

=

(Corollary A.5.4). Now since / is a U^-map, every y in the compact set f[X] has a neighborhood Vy in R2n+1 such that f~l\Vy\ is contained in an element of Ui. Put V = {Vy : y e /[A"]}. By Lemma A.5.3 there exists 8 > 0 such that every A C M2n+1 with diam(A) < 6 and which moreover intersects f[X] is contained in an element of V. Let 7 — min{ 5/3, e}. We claim that the open

3.3. THE IMBEDDING THEOREM

173

ball about / with radius 7 is contained in C;. To this end, take an arbitrary function g G C(X, R 2n+1 ; g) such that 0(f,g) < 7. We shall first prove that

g[X] n L, = 0. This is easy. Take an arbitrary x G X. Then g(f(x),g(x)} that g(x) is contained in the compact set

< £, which implies

:Q(y,\X}}<e}. So by (5), g[X]r\Li = 0. We next prove that g is a Ut-map. Take an arbitrary element p G g[X}. There exists x G X such that g(x) G B(p, %). Consider the open neighborhood of x. We claim that B is contained in an element of U^. If z G B then clearly g(g(x),g(z))

< %.

Since g(f,g} < %, we also get

Q(g(x),f(x))<%, From these inequalities it follows that g(f(z)J(x))

<

g(g(z)J(z})<%.

e(f(z),g(z))+g(g(z),g(x))+e(g(x)J(x))

< % + % +% =8 and so

(6)

g-l[B(g(x),%}}Cf^[B(f(x),S})].

Since by the special choice of 6 the set B(f(x),6) is contained in an element of V, f~l[B(f(x),6)] is contained in an element of IL1. By (6) it therefore follows that g ~ l [ B ( g ( x ) , %)] is contained in an element of Hi. But this is as desired since B(p, %) C B(g(x), %). 0 By Corollary 1.3.6, C(X, M2n+1 ; g} is a topologically complete space, and is therefore a Baire space (Theorem A. 6. 6). Consequently, Claim 1 implies that

is a dense G^-subset of C(X, R2n+1 ; g). We shall prove that C is as desired. Observe that (2) follows by construction. Take any function / G G. Then f[X] n (JSi ^ = 0, i.e., f[X] C Wn. This proves (3) and (4) for /. So it suffices to prove that / is an imbedding. We shall prove that for every x G X and every neighborhood W of x there exists a neighborhood V of } ( x ) such that / -1 [V^] C W. From this it follows that / is one-to-one and that /: X —>• f[X] is open, i.e., / is an

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3. BASIC DIMENSION THEORY

imbedding. Take an arbitrary x e X and a neighborhood W of x. Let e > 0 be such that B(x,e) C W. There exists i G N with l/i < e. Since / is a Uimap, there exists a neighborhood V of /(x) such that /~ 1 [T / ] is contained in an element U of Hi. So x G C/ and since diam(t/) < y^ < e, we conclude that f~l[V] CUC B(x,e) C W, as required. D We now come to the main result in this section. The Imbedding Theorem 3.3.5. Let X be a space with 0 < dim AT < n. Then there exists an imbedding i: X —>• 9ln such that i[X] has compact closure in 9ln. Proof. By Exercise A.6.5, X has an admissible metric Q for which there exists a sequence of finite open covers {Hi : i G N} such that for every z, mesh(Ui) < l/i (i G N). So we are in a position to apply Proposition 3.3.4 to get what we want. D Remark 3.3.6. Notice that we proved in fact the stronger statement that for a space X with 0 < dim^C < n, any bounded map /: X -> R2n+1 can be approximated arbitrarily closely by an imbedding in yin the range of which has compact closure in 9tn. Remark 3.3.7. Theorem 3.3.5 is 'best possible' for there exist n-dimensional spaces that cannot be imbedded in M 2n . An example of such a space is the union X of all n-faces of a (2n + 2)-dimensional simplex. This X is obviously n-dimensional and that it cannot be imbedded in R2n is due to FLORES [164]. The proof of this fact for n — I is simple and is based on the Jordan Curve Theorem for R 2 . For n > 1, the proof is more complicated. Its main ingredient is the Borsuk-Ulam Antipodal Theorem in Exercise 2.5.2(4), which is equivalent to the Lusternik-Schnirelman-Borsuk Theorem 2.5.1. See ENGELKING [155, p. 106] for details and additional information. Some applications. We now present some applications of the results in this section. The proofs are based on Theorem 3.3.5 which explains our interest in imbeddings in 9tn instead of imbeddings in R 2n+1 . Corollary 3.3.8. If dim X < n, n > 0, then X has a compactification *yX such that dim 7^ < n. Proof. Let i: X -» 9Tn be an imbedding such that jX = i[X] is a compact subset of yin (Theorem 3.3.5). Then dim7^Y < n by Lemma 3.3.1 and the Subspace Theorem 3.2.9. D Corollary 3.3.9. Let X be a nonempty space and let n > 0. The following statements are equivalent: (1) dim X
3.3. THE IMBEDDING THEOREM

175

Proof. Suppose that dim X < n. By Theorem 3.3.5 we may assume that A is contained in 9ln. By Lemma 3.3.1, 9tn is the union of n+1 zero-dimensional subspaces. Now apply the Subspace Theorem 3.2.9. Conversely, assume that X is the union of at most n + 1 zero-dimensional subspaces. Then apply Corollary 3.1.7 to conclude that dim A" < n. D We now prove a generalization of Corollary 3.1.6 to n-dimensional spaces. Corollary 3.3.10. Let 0 < n < oo and let X be an n-dimensional subspace of the space Y. Then for all disjoint closed subsets A and B ofY there exists a partition S between A and B in Y such that dim(S n A") < n — 1. Proof. Write A = (J^i ^> where dimA^ < 0 for every i (Corollary 3.3.9). Let A and B be disjoint closed subsets of A". By Corollary 3.1.6 there is a partition S between A and B such that 5 n X \ = 0 . But then SnX is contained in (J™J2 Xi and hence is at most (n — l)-dimensional by Corollary 3.3.9. D By a repeated application of the previous corollary, we get the following result. Corollary 3.3.11. Let A be a space with subspace Y such that dimY < n, where 0 < n < oo. For each collection T — {(F^Gi) : i < n + 1} of pairs of disjoint closed subsets of X there exists for every i < n + 1 a partition Si between Fi and Gi such that n+1

Y n n S, = 0. The following result is known as the G$ -Enlargement Theorem. Corollary 3.3.12. Let M be a subspace of a space X. If dim M < n then there exists a G$-subset S C X with M C S and dim S < n. Proof. We first prove the corollary for the special case dim M = 0. Let £ be a countable open base for M. For every pair (Eo,Ei) of elements of £ such that EX) C EI there exists by Corollary 3.1.6 an open set B C M such that E0 C B C B C EI

while moreover M n Frf? = 0. The collection I> of Z?'s obtained in this way is clearly a countable base for M. In addition,

F= \J FrB is an Fcr-subset of M which misses M. Let S = M\F. Then 5 is a G^-subset of the G^-subset M and hence is a G^-subset of A. It is clear that "B \ S consists of (relative) clopen sets, and so S is zero-dimensional.

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3. BASIC DIMENSION THEORY

We will now use the just proved special case to prove the general case. To this end, write M = M0U---UMn, where Mi is zero-dimensional for every 0 < i < n (Corollary 3.3.9). By the above there exists for every 0 < i < n a zero-dimensional G$-subset M* of X with Mi C M*. The union of the M*'s is clearly a G^-subset of X and is at most n-dimensional by Corollary 3.3.9. D Exercises for §3.3. 1. Give an example of a space X, an open cover U of X, and an imbedding /: X —t X such that / is not a U-mapping. 2. Let X be a space. Prove that every continuous function /: X —>• Q can be approximated arbitrarily closely by an imbedding in Q. 3. Prove that for every n > 0 there exists a compact space Xn such that (1) dimXn = n, (2) every space Y with dimy < n can be imbedded in Xn. >4. Let X be a compact space without isolated points such that 0 < dimX < n < oo. Prove that there exists a continuous surjection /: C —>• X such that each fiber of / has cardinality at most n + 1 5. Let X be a compact space and let /: C —» X be a continuous surjection such that each fiber of / has cardinality at most n + 1. Prove that X is at most n-dimensional. 6. Let X be a compact space with 0 < n = dimX < oo. Prove that every open cover U of X has a closed refinement 7 such that for every F £ 7 the collection {F1 € 5 : F' n F ^ 0} has cardinality at most 32"+1. ^•7. Prove that every (n+l)-dimensional compactum contains an n-dimensional compactum (and hence an n-dimensional continuum by Exercise 3.2.4). 3.4.

The inductive dimension functions ind and Ind

There are two additional dimension functions that are important in dimension theory, namely, the small and the large inductive dimension function, abbreviated ind and Ind, respectively. It turns out that for a given space these functions and the dimension function dim take the same value. The functions ind and Ind are important because in certain situations it is easier to deal with them than with dim. For example, it is a triviality to verify that for all X and y, ind(X xY) < i

3.4. THE INDUCTIVE DIMENSION FUNCTIONS ind AND Ind

177

by equality of ind and dim it therefore follows that dim(X x Y) < dim A + dimF. However, to verify this straight from the definition of dim is unpleasant. The aim of this section is to study basic properties of the dimension functions ind and Ind and to prove equality of all three dimension functions. We shall now give the definition of ind, which differs from the definition of dim in the sense that it is an inductive definition. Indeed, for a space X define its small inductive dimension ind X £ { — 1 , 0 , 1 , . . . } U {00} as follows: indA = -l <£> A = 0, ind X < n <4> for every x G X and every closed subset A of X not containing x there exists a partition L between {x} and A such that indL < n — 1, ind X = n <& ind X < n and ind X <£. n — 1, ind X = oo <{=> ind X ^ n for every n > — I. It is clear that if X and Y are homeomorphic spaces then ind A" = indl'. The property of having small inductive dimension at most n can be expressed in terms of a special countable base. Lemma 3.4.1. A space X satisfies 0 < ind A" < n if and only if X has a countable base *B such that ind Fr B < n — 1 for all B e !B. Proof. If £ X < n then A" has clearly a base U such that ind Fr U < n — 1 for every U £ U. This base has a countable subcollection !B which is still a base (Exercise A.2.12). So !B is as required. The converse of the lemma is trivial. D A consequence of this lemma is that for a space X we have ind A" = 0 if and only if X is nonempty and has a base consisting of clopen sets. This means that X is zero-dimensional if and only if ind X — 0. We shall now derive a few additional properties of the dimension function ind. The following triviality shall be used several times in the forthcoming: if X is a space, Y C X. y 6 Y and A is a (relatively) closed subset of Y not containing y. then y does not belong to the closure of A in X. Proposition 3.4.2. Let X be a space. Then dim A" = 0 iff ind A = 0. Proof. This is a triviality since, as observed above, ind A = 0 iff the clopen subsets of X form a base iff dim A = 0 (see Page 41). D Lemma 3.4.3. Let X be a space, and let A be a subspace of X. ind A < ind X.

Then

Proof. There is nothing to prove if ind A = oo, so we assume that ind A is finite. Again, there is nothing to prove if ind A = — 1. So assume the lemma

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3. BASIC DIMENSION THEORY

to be true for all spaces Y with 0 < ind Y < n — 1, and assume that ind X < n. Take x E A and let C be a closed subset of A not containing x. Since x $ C there is a partition L in X between x and C such that indL < n — 1. Then L n A is a partition between x and C in A so that, by our inductive assumption, ind(L n A)
The Addition Theorem 3.4.4. If A and B are subspaces of a space X then 'md(AuB) < ind A + ind B + I . Proof. If ind A — oo or ind B = oo then there is nothing to prove. So we assume that ind A and indL? are both finite. We induct on ind A + indB. If A and B are both empty, i.e., if ind A = indB = — 1, then there is again nothing to prove. So assume that the theorem is true for all subspaces E and F of X with

ind E + ind F < n - 1, where n > — 1. Let a = ind A and 0 — ind B and assume that a + fi = n. We shall prove that ind(Auf?)
ind (5 n (A U B))
and Y = F then indX = 1 and

This shows that Theorem 3.4.4 is 'best possible'.

Corollary 3.4.6. If X can be written as the union ofn + 1 zero-dimensional subspaces then indX < n. Proof. Since for every space F, dimF = 0 iff indl" = 0 (Proposition 3.4.2) the result follows immediately from Theorem 3.4.4. D

3.4. THE INDUCTIVE DIMENSION FUNCTIONS ind AND Ind

179

We want to prove the equality of dim and ind. Before being able to do that, we need to derive a preliminary lemma. Lemma 3.4.7. Let X be space and let n > 0. The following two statements are equivalent: (1) For every x G X and for every open neighborhood U ofx there exists a partition L between {x} and X \U such that dim L < n — 1. (2) For every pair A and B of disjoint closed subsets of X there exists a partition L between A and B such that dimL < n — 1. Proof. First observe that (2) => (1) is a triviality. For (1) => (2), let A and B be disjoint closed subsets of X. By assumption, for each x G X there exist open sets U(x] and V(x] such that (3) x G t/0), U(x)r\V(x) = 0, (4) if L(x) = X\ (U(x) U V(x)) then dim L(x) < n - 1, (5) U(x) n A = 0 or U(x) n B = 0. The cover {U(x) : x G X} has a countable subcover, say U (Corollary A.2.3). Put L = \J{L(x) : U(x) G U}. Observe that by (4) and the Countable Closed Sum Theorem 3.2.8 we have dim L < n - 1. Let U0 = {U G U : U n A ^ 0} and Ui = U \ U 0 . Enumerate U0 as {U0ii : i € N} and Ui as {E7M : i G N}, respectively. For each i G N put (6) Ei = U0ti \ \Jj hence y G t/o,i \ ^0,1 ^ L, and we are done. Suppose therefore that y $ UQ^. If y G t/i^ then y £ F^ which is also not the case. Consequently, y £ C/i^, which implies that in this case y G L as well. D Theorem 3.4.8. For every space X, dim X = ind^C. Proof. We shall first prove that dim^C < indX. If ind AT = CXD then there is nothing to prove. So assume that ind^C < oo. We induct on indX. If indJ^C = 0 then apply Proposition 3.4.2. Now assume that the inequality holds for all spaces Y with indF < n — 1, n > 1, and let ind^sT = n. Then for each point x G X and for every open neighborhood U of x there exists a partition L between {x} and X\U such that ind L < n — l. By our inductive

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3. BASIC DIMENSION THEORY

hypothesis, all these partitions have covering dimension at most n — 1. We conclude that dim A < n by Lemma 3.4.7 and Corollary 3.1.10. We shall now prove that ind X < dimX. If dim A = — 1 or dim A^ = oo then this is a triviality. So assume that 0 < dim X — n. By Corollary 3.3.9, X is the union of at most n+1 zero-dimensional subspaces. So by Corollary 3.4.6 we get ind A Ind X ^ n for every n > — 1. As in the case of ind, it is easy to see that if X and Y are homeomorphic spaces then Ind A = IndF. We shall prove that Ind = dim, thereby establishing the announced equality. Lemma 3.4.9. Let A be a space. Then ind A < Ind A. Proof. If Ind A = oc or Ind A = — 1 then there is nothing to prove, so assume that the lemma is true for all spaces Y with — 1 < Ind Y < n — 1 and 0 < n < oo, and assume that Ind A = n. Let x G A and let U be an open neighborhood of x. Since {x} and X \U are disjoint closed sets in A, there is a partition L between them such that IndL < n — 1. By our inductive assumption we have indL < IndL < n — 1. From this we consequently conclude that ind A
3.4. THE INDUCTIVE DIMENSION FUNCTIONS ind AND Ind

181

subsets A and B of X there exists a partition L between A and B such that dimL < n — 1. By our inductive hypothesis it follows that these partitions all have large inductive dimension at most n — I. From this we conclude that indeed Ind X < n = dim X. D Remark 3.4.11. From now on we will no longer formally distinguish between dim, ind and Ind. Theorems proved for dim will be used for ind and vice versa, etc. Let X and Y be spaces. It is natural to wonder about the relation between dim A, dimF and dim(A x Y}. Since dimIR71 = n for every n, one would expect the relation dim(A r x Y} — dim AT + dimF to hold for all nonempty X and Y. Unfortunately, this is not true, see Example 3.4.13. We shall now prove 'half of the expected equality.

Theorem 3.4.12. Let X and Y be nonempty spaces. Then dim(A x Y) < dimX + dimY. Proof. By the Coincidence Theorem 3.4.10, it suffices to prove that ind(A x y) < ind A" + indY. First observe that without loss of generality, 0 < ind X, ind Y < oo. We shall prove the theorem by induction on ind X + ind Y. If ind X + ind Y = 0 then both X and Y are zero-dimensional, and so is X x Y. Assume that the theorem is true for all spaces X and Y with ind A + ind 1" < n- 1,

where n > 1, and let X and Y be spaces such that ind A = a, indF = /3 and a + (3 = n. By Lemma 3.4.1 there is a countable base 23 for A such that dimFrS < a - I for every B 6 B. Put B' = \JBe^ FrE. Then B' is an FO--subset of X and hence by the Countable Closed Sum Theorem 3.2.8 it follows that dim B' < a - 1. Since Fr B n (A \ B') = 0 for every B e £ it also follows that *B \(X\B') consists of relatively clopen sets, i.e., dim(A\.B) < 0. Similarly, construct an F^-subset E C Y such that dimE < (3 — 1 and dim(y \ E) < 0. Now observe that by our inductive hypothesis and the Countable Closed Sum Theorem 3.2.8, B' x Y is an at most (a — 1) + /?dimensional Fff-subset of A" x y. It follows similarly that X x E is an at most a + (/3 ~ 1)-dimensional F^-subset of A x Y. Again by the Countable Closed Sum Theorem 3.2.8 we get dim ((B' x y) U (X x E))
Since the complement of (B' x Y} U (X x E} is zero-dimensional, being the product (A \ B') x (Y \ E), we get by Theorem 3.4.4 that dim(A xY)<(a + 0-I) + Q + l = a + /3, as required.

D

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3. BASIC DIMENSION THEORY

The following example shows that the other 'half of the expected equality dim(X xY] = dimX + dimY need not hold. See Theorem 3.9.5 for a (strong) generalization of this result to all dimensions. Example 3.4.13. There exists a one-dimensional space E such that E x E is one-dimensional as well. We will show that the space E in Example 1.5.18, i.e., Erdos' space, is the required example. From Exercise 1.5.14 we know that E « E x E. So it suffices to prove that dim.E = 1. In Example 1.5.18 we showed that dim E > 1. We shall prove here that ind E < 1. First observe that E is a subgroup of I2, and hence is a topological group. By homogeneity, it therefore suffices to prove that indE < 1 at the zeroelement of E. We claim that for every e > 0, the sphere S£ consisting of all points in E with norm e, is zero-dimensional. Indeed, since all elements of S£ have the same norm, Lemma 1.1.12 shows that S£ is homeomorphic to a subspace of the countably infinite product of rational numbers, which is zerodimensional by Proposition 1.5.3 and the trivial observation that products of zero-dimensional spaces are again zero-dimensional. So S£ is indeed zerodimensional, being a subspace of a zero-dimensional space. We now summarize the results obtained so far in the following Corollary 3.4.14. (1) Let X be a nonempty space and let n > 0. The following statements are equivalent: 1. dimX < n, 2. for every subspace AofX, dim A < n, 3. X has a compactification 'yX such that dim^yX < n, 4. for every pair of disjoint closed subsets A and B of X there exists a partition L between A and B such that dim L < n — I , 5. X is the union of at most n + 1 zero-dimensional subspaces. (2) If X is a space and if A and B are subspaces of X then dim(AuB} < dimA + dimB + 1, dim(A x B) < dim A + dimB. Exercises for §3.4. Let X be a space. Define the compactness degree cmpX of X, as follows: cmpX = — 1 <=>• X is compact, cmp X < n <&• for every x G X and every closed subset A of X not containing x there exists a partition L between {x} and A such that cmp L < n — 1, cmpX = n «=>• cmpX < n and cmpX •£ n — 1, cmp X = oo <$ cmp X ^ n for every n > — 1.

3.5. DIMENSIONAL PROPERTIES OF COMPACTIFICATIONS

183

So the compactness degree of a space is intuitively its small inductive dimension modulo the class of all compact spaces. It is clear that if X and Y are homeomorphic spaces then they have the same compactness degree. Let *yX be a compactification of X. We call the space ^X \ X the remainder of the compactification ^X of X. Let the compactness deficiency be the least integer n G [ — 1, oo] such that X has a compactification with remainder of dimension n. Observe that homeomorphic spaces have the same compactness deficiency. Given a space X and n £ N, we shall denote by X(n) the set of all points in X that have arbitrarily small neighborhoods with at most (n — l)-dimensional boundaries. 1. Prove that Erdos' space E is not topologically complete. Give an example of a one-dimensional totally disconnected topologically complete space F such that F and F x F are one-dimensional 2. Let X be a space. Prove that cmpX < def X < 3. Let X be a totally disconnected space which is not compact and let ^X be a compactification of X. Prove that dim^X \ X > dimX, i.e., X cannot be compactified by adding a set of smaller dimension than the dimension of X. ^•4. Let X be a compact space such that 0 < dimX = n < oo, and let Y be the product Q x X. (1) Prove that if S is a topologically complete space containing Y then S\Y contains a homeomorph of X . As a consequence, the dimension of S \ Y is at least n. (2) Prove that def Y = n. 5. Let X be a space and n E N. Prove that there is a countable family open sets U such that (1) U is a local base at every point of X( n ), (2) dim(I7 \ C7) < n - 1 for every U e U. Prove that dimX( Tl ) < n. 3.5. Dimensional properties of compactifications In this section we collect some results on compactifications preserving certain dimensional properties. We saw in Corollary 3.3.8 that every m-dimensional space X has an mdimensional compactification. Such a compactification is called dimension preserving. The following result generalizes this and is central in this section. Theorem 3.5.1. Let X be a space, and let {(Fn,Gn) : n € N} be a countable collection of pairs of disjoint closed subsets of X. Then there is a dimension preserving compactification aX of X such that for every n the sets Fn and Gn have disjoint closures in aX .

184

3. BASIC DIMENSION THEORY

Proof. We will prove this in the case of finite dimensional X only, the infinite-dimensional case being simpler. Let Q be an admissible metric on X for which there exists a sequence {Un : n G N} of finite open covers of X such that mesh(Un) < l/n for every n (Exercise A.6.5). Since An = {X \ Fn,X \ Gn} is an open cover of X, we can replace lin by the common refinement of Vtn and An. So we may assume without loss of generality that for every n and U E Un we have that U n Fn — 0 or U n Gn = 0. Let ra = dimX. By Proposition 3.3.4 there is an imbedding i: X —> Otm such that (1) i[X] has compact closure in ytm. (2) i is a U n -map for every n. Put aX = i[X]. Then aX is a dimension preserving compactification of X (cf. Corollary 3.3.8). Fix n E N. We claim that i[Fn] and i[Gn] have disjoint closures in aX. Since i is a Un-map, there is an open cover 9 of M2m+1 such that

{rMGJrGeS} refines Un. Striving for a contradiction, assume that

pei[Fn]ni[Gn]. There is an element G E S such that p E G. Pick an element £/ E Un such that i~1[G] C U. We may assume without loss of generality that UC\Fn = 0. But then i~l[G] n ,Fn = 0 and so G n z[Fn] = 0. Since M2m+1 \ G is closed, this implies that p £ i[Fn], which is a contradiction. D Corollary 3.5.2. Let X be a space, and let bX be a compactincation of X. Then there is a compactincation aX ofX which is dimension preserving while moreover aX > bX. Proof. By Lemma A.5.6 there is a collection {(A'n,B'n) : n E N} of pairs of disjoint closed subsets of bX such that for any pair (E, F} of disjoint closed subsets in bX there is an ra E N such that E C A'm and F C B'm. For every n, let An = A'n C\X and Bn = B'n n JC, respectively. By Theorem 3.5.1 there is a compactification aX of X such that for every n the closures in aX of An and Bn are disjoint, while moreover dimaX = dimX. But this implies by Lemma A.8.3 that the identity function X —> X can be extended to a continuous function aX —> bX, i.e., aX > bX. D We now show that compactifications can also preserve the dimensions of countably many closures of closed sets.

3.5. DIMENSIONAL PROPERTIES OF COMPACTIFICATIONS

185

Theorem 3.5.3. Let X be a space, and let "H be a countable family closed subsets of X. Then there is a Wallman base 3 of X such that (1) :KC_J, (2) dimH = dimH for every H 6 "K (here closure means closure in u(X,f)). Proof. List 3~C as {Hi : i e N} such that every H 6 JC is listed infinitely often. For every H 6 !K put HH = dimH, and if H = Hl then m = HHLet 3~i be an arbitrary Wallman base for X (Lemma A. 9.1) and let ai X be the compactification cj(X, 3"i) of X. Consider the set H\. The closure of HI in a\X is a compactification of HI, say PiHi. By Corollary 3.5.2 there is a compactification /32Hi > ftiHi such that dim ,$2 HI = n\. By an application of Exercise A. 11. 8, there is a compactification £2-^ of X for which there is a continuous function /i : t2X ->• aiX

which extends the identity on X while moreover the closure of HI in t2X is ^2 HI . Hence the closure of HI in t2X has dimension n\ . By Corollary A. 9. 8 there is a Wallman base TI of X such that t\X and w(X, TI) are equivalent compactifications. Let S"2 be a Wallman base of X containing the collection 9~iUTiU{Hi} (Lemma A. 9.1), and let a2X be u(X,y2}. By Exercise A. 9. 4 there is a continuous function gi : a2X —>• iiX which extends the identity on X. Observe that at this stage we have the following compactifications and continuous functions: The Wallman bases fi and 32 have the property that 3"i C y2, and the compactification t\X has the property that the closure of HI in it has the 'right' dimension, namely n\. This is the first step in an inductive process. We now proceed recursively as above, dealing with the set Hi in the i-ih step of our process. We thus obtain an inverse sequence 1

9l

2

9

2

3

„ V j f + V j 4 V , V , f a\X
(3) 3"i_i c y^ (4) Hitfi, (5) u(X,?i) = aiX,

186

3. BASIC DIMENSION THEORY

(6) fa: tiX —> aiX extends the identity on X, (7) gi : di+iX -» tiX extends the identity on X, (8) the closure of Hi in tiX has dimension U{. Put y = Ui^i ^i- We claim that 3~ is the desired Wallman base of X (that it is a Wallman base follows from Exercise 1.10.11). Let a^X be the inverse limit of our inverse sequence. We already know by Exercise 1.10.11 that So docX is a compactification of X and every projection restricts to the identity on X. Now let H G "K be arbitrary. The closure of H in tiX is mapped by fa onto the closure of H in aiX. Similarly, the closure of H in di+iX is mapped by gi onto the closure of H in tiX. Also, the closure of H in a^X is mapped by the various projections onto the closures of H in the corresponding compactifications. This is all true because of the compactness of the spaces involved and the fact that every function in our inverse sequence restricts to the identity on X. So all the closures of H in the various compactifications form an inverse system of which the limit is the closure of H in a^X (Exercise 1.10.7). Since H is listed infinitely often, by (8) this sequence has a subsequence all elements of which have dimension UH- But then, since every subsequence of an inverse system has the same limit (Lemma 1.10.6), by Exercise 3.2.2 it consequently follows that the closure of H in a^X has dimension at most riH- But since H is a subspace of its own closure in a^X , by the Subspace Theorem 3.2.9 we get that this closure is at least HH -dimensional. So we are done. D Corollary 3.5.4. Let X be a space, and let 3 be a countable family closed subsets of X. Then there is a compactification aX of X such that dim F = dim F

for every F € J. Remark 3.5.5. An inspection of the proof of Theorem 3.5.3 shows that for this corollary it is not necessary to deal with Wallman bases. The proof of it can be completed entirely within the framework of inverse limits. The following corollary to Theorem 3.5.3, which will be important in §3.11, does however not follow from Corollary 3.5.4. An inspection of its proof will show that we need to control the dimensions of various partitions between points and closed sets, and for that control over closures alone does not suffice. It turns out that via the use of Wallman bases, one gets the desired control automatically. One simply has to put the right collection of closed sets in a suitable Wallman base, and the Wallman compactification theory does the rest.

3.5. DIMENSIONAL PROPERTIES OF COMPACTIFICATIONS

187

Corollary 3.5.6. Let X be a space with countable family closed subsets 9Then X has a compactification aX having the following properties: (1) dimG = dimG for all G G g, (2) X(n) C (aX)(n} for every n > 0. Proof. For n > 0 there is by Exercise 3.4.5 a countable collection of open sets Un in X such that (3) Un is a local base at every point of X(n), (4) dim(Z7 \ U) < n - I for every U G Un. Let IK be the collection {U : U G U n , n > 0} U {X \ U : U G U n ,n > 0} U

U{!7\[/:£/GUn,n>0}ug. Then IK is a countable collection closed subsets of X. By Theorem 3.5.3 there is a Wallman base f of X such that

(5) IKCJF, (6) dimH = dimH for every H G IK (here closure means closure in

u(x,y}}. Put aX = a;(X, IF). Observe that (1) follows from (6) since S C IK. So we need only worry about (2). To this end, let x G -X"(n) for certain n > 0 and let V be an arbitrary open neighborhood of x in aX. Let W" be a closed neighborhood of x in aX such that W C V. There is an element U G Un such that x 6 C7 C U C WnX (closure means closure in X here). Observe that

{u,u\u,x\u} cy. This means that

u*u(x\uy = u(x,?)

and

C/*n(.Y\C/)* = (C7\C7)*. (Lemma A. 9.4). Observe that U* is the closure of U in u}(X,3) (Exercise A.9.1). Similarly, (U\U)* is the closure of U \ U in u(X, J). Since W is a closed neighborhood of x, it consequently contains U* . Hence (U \ U)* is a partition between {x} and aX\W which is at most (n — l)-dimensional by (4) and (6). Since V was arbitrary, this proves that x G (aX)^, as required. D Let X be a space, and let / : X -> X be continuous. A set A C X such that f[A] n A — 0 or, equivalently, A n /-1[A] = 0, is called a color of /. A coloring of / is a finite collection of colors of / which covers X. We say

188

3. BASIC DIMENSION THEORY

that / is colorable if there is a finite cover of X consisting of closed colors of/Observe that every colorable map is fixed-point free. There is an easy to describe example of a fixed-point free map which is not colorable. The verification of this is difficult. See §3.12 for details. Let /: X —>• X be fixed-point free. Then every x E X has a closed neighborhood Vx such that Vx Pi f[Vx] = 0. By Corollary A. 2. 3, the cover {Vx:xe X} has a countable subcover. There consequently exists for every fixed-point free map / : X —>• X a countable and potentially infinite closed cover V of X such that every V G V misses its own image. So a fixed-point free map on X is colorable if such a cover can be chosen to be finite. Corollary 3.5.7. Let X be a space and let 3 be a countable collection of continuous functions from X to X. Then there is a compactification aX of X such that

(1) (2) Every f 6 y can be extended to a continuous function f : aX -> aX. If f is a homeomorphism then so is f. In addition, if f £ 3~ is colorable then f is fixed-point free. Proof. We assume without loss of generality that the identity on X belongs to 3~ and that for each homeomorphism / £ J we also have that f~l G 3~. Assume that / € 3 is colorable. Then there is a finite closed cover 9/ ofX such that Gr\f~l[G] = 0 for every G e S/. Put Kf = Sf U f~l[9f]Let
U r^n]. Observe that since the identity belongs to 3~ we have An C An+\ . In addition, since X G An we have dimo;(X, A n ) = dim JC. As in the proof of Theorem 3.5.1, oo

A = |^J .An

n=l

is a Wallman base such that aX = u)(X,A) is the inverse limit of an inverse sequence of compact spaces having the same dimension as X, hence aX is dimension preserving.

3.5. DIMENSIONAL PROPERTIES OF COMPACTIFICATIONS

Since

189

r^cyt

for every / G y we have that / can be extended to a continuous function /: aX —>• aX by Exercise A. 9. 4. If / is a homeomorphism then so is / because f~l G 5F can be extended as well and this extension is the inverse of /. Now assume that there exists an / G 9" which is colorable. The collection

is a sub collection of A. We claim that /: aX —>• aX is fixed-point free. To see this, observe that the finite collection 9 covers X and so does [f~l[G] : G G 5}- For a given p € aX there consequently exists G G 9/ such that p G f ~ l [ G ] . The continuity of /: aX —> aX therefore implies that

Since G and f ~ l [ G ] are disjoint and both belong to A, they have disjoint closures in aX (Lemma A. 9.4(2)). So f(p) ^ p. D The dimension of almost zero-dimensional spaces. We will use compactifications to prove that every almost zero- dimensional space is at most one-dimensional. A subset X of a compactum K is L-imbedded in K if for every open cover U of K there is a neighborhood V of X in K such that every subcontinuum of K which is a subset of V is contained in an element of U. Let A and B be disjoint closed sets in a space X, and let e > 0. A triple

is an L£ -enlargement for the pair (A, B} if (1) G and H are open neighborhoods of A and .B, respectively,

(2) G n # = 0, (3) y is a discrete collection of closed sets of diameter at most e such that X \ (G U H) = U ?• Theorem 3.5.8. Let X be an L-imbedded subspace of a compact space (K,g). If A and B are disjoint closed subsets of X having disjoint closures in K then for each e > 0 there is in X an L£ -enlargement (G, H, J) for the pair A, B such that for every F G J the sets

G D F,

~H n F

have disjoint closures in K (here closure means closure in K).

190

3. BASIC DIMENSION THEORY

Proof. Let 8 = Q(A,B). Our assumptions give us that 6 > 0. Pick an arbitrary e > 0. We will construct in X an Z/e-enlargement for the pair (A, B}. There is by assumption an open neighborhood U of X in K such that all continua in U have diameter at most y 2 min{c, 6}. Let (p: K —>• I be continuous such that U = (p~l [(0, 1]] (Lemma A. 4.1). For each n, let ^» = y" 1 [[ 1 /(n+l), 1 /n]].

Observe that every Kn is compact. Fix n for a moment. If C is a component of Kn then C is not a continuum from A n Kn to 5 n K"n . This is so because diam(C) < l/28 since C C U and 0(.A,.B) = 6. Hence Kn can be split into two disjoint closed sets Pn and Qn containing Ar\Kn and Br\Kn, respectively (Exercise A. 11. 5). Observe that the sets oo

oo

s= |J A n U p n t / ) , r = U Q 2 n U ( £ n £ / ) 71=1

n=l

are disjoint closed subsets of U. Let a : U —> I be a Urysohn function such that a[S] C {0} and a[T] C {!}. Put F = a-1 [[0,i/0],

W = a'1 [(3/4,1]].

Clearly, Vflt/ and WC\U are disjoint. Observe that U\(V(JW) is contained in the union of the discrete collection % = {K2n_! : n G N} in [/. Each K2n-i is a compactum whose components have all diameter at most i/2£. So by Corollary A. 10. 2 we can split K2n-i into a finite pairwise disjoint collection compacta y2n-i of diameter at most e. Observe that the traces of V and W on any of these compact pieces are disjoint. We conclude that the triple n—l

is the required //^-enlargement for (A, B).

D

Corollary 3.5.9. Let X be an L-imbedded subspace of a compact space K. Then dim X < I . Proof. In this proof A stands for the closure of A in K.

3.5. DIMENSIONAL PROPERTIES OF COMPACTIFICATIONS

191

Let F be a closed subset of A r , and let p G X \ F. We will prove that there are disjoint open subsets V, W in X with p E V, F C W and dim ((A" \(V\J W)) < 0. By Theorem 3.4.10, this suffices. We shall construct two increasing sequences of open sets

p G GI C G2 C • • • ,

F C HI C H2 C • • •

and for every n G N a collection Jn of closed subsets of X such that (1) G n n # n n A = 0, (2) the triple (Gn+i,Hn+i,yn) is an L^-enlargement for the pair

(Gnnx,Hnnx), (3) if F € 3~n is arbitrary then

F n Gn n F n #n = 0. Let GI , HI be any pair of open sets containing p and F, respectively, having disjoint closures in K. Suppose Gn,Hn have been defined. For n = 1, we get the triple (G2,#2j3~i) directly from Theorem 3.5.8. Assume n > 2. For each F 6 3>i-i consider the disjoint closed sets A(F] = Gn n F,

£(F) =Hn^F

in A. These sets have disjoint closures in K by (3). By Theorem 3.5.8 there is an L^-enlargement of the pair (A(F), -B(F)) , such that for every R 6 ft^ the sets (4)

G(F) n fl, fT(F) n R

have disjoint closures in K. For each F 6 J n _i let C/(F) D F be an open set in A" such that the collection (5)

is discrete in A (Exercise A.7.3). Put

Gn+l =GnU \J{G(F) n U(F) : F e ^.J, Hn+1 =HnU \J{H(F) n t/(F) : F 6 ^_i}. It is clear that Gn+i and Hn+i are open, and by (5) and (1) it follows that

G n + 1 n H n + i n A = 0. Put

Jn = {FnR:F e ^.RG^F}. Then 3"n is a discrete collection of closed sets with diameter at most l/n. It is easy to see that the triple (Gn+i,Hn+i, Jn) is an L^-enlargement for the pair

192

3. BASIC DIMENSION THEORY

This completes the inductive construction. Now put oo

oo

V = U Gn,

W=\jHn.

n=l

n=l

Observe that for each e > 0 the complement X \ (V U W) is the union of a discrete collection of closed sets of diameter at most e. As a consequence, the partition X \ (V U W) is at most zero-dimensional (Exercise 1.5.3). D We are now in a position to present the proof of the following interesting result.

Theorem 3.5.10. If X is almost zero-dimensional then dimX < 1. Proof. Let X be nonempty and almost zero-dimensional. In addition, let !B be a base for X with the properties stated in Exercise 3.2.5. We may assume that 1$ is countable (Exercise A.2.12). For every pair G, H G 23 such that G n # = 0,

pick a clopen set C in X with

G C C C H. Since 23 is countable, the collection C of clopen sets picked in this way is countable as well. There is a compactification ^X of X such that for every C E C we have that C and X\C have disjoint closures in jX (Exercise A.9.3). That is, the closure of C in ^X is clopen in jX. For every B E 23 let B* be an open subset of ^X such that B* n X = B. We claim that X is L-imbedded in ~/X. To this end, let e > 0, and put J{ = {B E 23 : diamtf* < Y3e}, and

U = \jK, respectively. If x G X then there is an element B 6 23 such that

B CB(x, y7e) n X. Since 5 is dense in B*, the set 5* is contained in the closed ball D(x, VT^), hence B* C U. We conclude that U is an open neighborhood of X in ~fX. To conclude the proof, let K C [7 be an arbitrary continuum. Pick two elements BQ,BI G ^K such that

(i)

B*nK^®^KnB*.

If .Bo and #1 have disjoint closures in X then there is an element C G C such that B0CC CBl.

3.6. MAPPINGS INTO SPHERES

193

The closure of C in jX is clopen. By (1), K intersects this closure, as well as its complement. But this contradicts the connectivity of K. So the closures of BQ and BI intersect. Since there is a subfamily £ of IK such that K C \J{B* :B e £}

this proves that the diameter of K is at most E. So an appeal to Theorem 3.5.9 finishes the job.

D

Exercises for §3.5.

1. Let X be a space and let Y be a compact space for which there is a continuous function /: X —> Y. Prove that there is a dimension preserving compactification ^X of X such that the function / can be extended to a continuous function g: ^X —>• Y. ^•2. Let ( X n , fn)n be an inverse sequence consisting of at most m-dimensional spaces. Prove that its inverse limit is at most m-dimensional. 3. Let X be a space which is not compact. Prove that X has a compactification *yX such that dim'jX = oo. >4. Give an example of a compact space X containing two disjoint closed subsets A and B such that no partition between A and B is irreducible. 5. Let X be a subspace of the compact space Y. Prove that the following statements are equivalent: (1) X is L-imbedded in Y. (2) For every e > 0 there is a neighborhood U of Y in X such that every continuum in U has diameter less than e. >-6. Let A and B be L-imbedded subspaces of the compacta X and Y, respectively. Prove that A x B is L-imbedded in X x Y. Conclude that A x B is at most one-dimensional. 3.6. Mappings into spheres In the previous sections we presented several characterizations of dimension, some 'internal' (e.g., Theorem 3.2.14(2) and (3)) and some 'external' (e.g., Theorem 3.2.14(4)). In this section we shall prove that a space X is at most n-dimensional if and only if every continuous function /: A —>• §n, where A C X is closed, can be extended over X, thereby deducing another important 'external' characterization of dimension. As an application of this result we shall present a proof of the Brouwer Invariance of Domain Theorem. We first formulate and prove two lemmas that are needed in the proof of the main result in this section. Lemma 3.6.1. Let X be a space and let A\ and A? be disjoint closed subsets of X such that 0 < dim (X \ (A\ U ^2))
194

3. BASIC DIMENSION THEORY

Proof. Put A = A\ U A 2 . By Corollary A. 4. 3 there exist open subsets U and V_ of X such that AI C U, A2 C V and UnV = 0. Since dim (A" \ A) < n and U H (X \ A) and V C\ (X \ A) are disjoint closed subsets of X \ A, by Corollary 3.4.14 there exist disjoint open sets E and F in X \ A such that

(1) u_n(X\A)cE, (2) VH(X\A}CF, (3) if L = (X \ A) \ (E U F) then dim L < n - 1. Since AiUE = UUE, and J5 is open in X being open in the subspace X\A, we conclude that AI U E is open in X. Similarly, A2 U F is open in X. Consequently, L is closed in X and is a partition between AI and A2 in X with dimL < n — 1 by (3). So L is as required. D Corollary 3.6.2. Let X be a space and let AI and A2 be closed subspaces of X such that 0 < dim (X \ (Ai U A 2 )) < n. Then there exist closed subspaces Xi and X2 of X such that (1) (2) (3) (4)

Ai=Xin(AiUA2), A2 = X 2 n ( A i U A 2 ) , dim ((Xi n X2) \ (Al U A 2 )) < n - 1, ^!UX 2 = X.

Proof. Put F = JC \ (Ai n A 2 ). By Lemma 3.6.1 there exists a closed set L in Y with dim L < n — 1, such that F \ L can be written as the disjoint union of two open sets E and F such that AI D Y C E and A2 n Y C F. It is easy to see that Xi = AI U E U L and JC2 = A2 U F U I/ are as required. D These simple results enable us to derive the following: Theorem 3.6.3. Let X be a space and let A be a closed subspace of X such that 0 < dim(X \ A) < n. Then every continuous function g: A -> §>n can be extended to a continuous function g: X —> §n. Proof. We shall prove the theorem by induction on n > 0. Suppose first that n = 0 and recall that S° = { — 1, 1}. By Corollary 1.5.2 there exists a clopen set C C X such that g'1^!) C C and g~l(l) C X \ C. Define the function g : X -t §° by

-i

xeC

An easy check shows that g is as required. Now assume that the theorem is true for n — 1, n > 1, and let g: A -» §n. Define §? = {x € §n : xi > 0} and §£ = {x G §n : Xi < 0}, respectively. Observe that §^ and §2 are both homeomorphic to I n and that S^ n Sg is

3.6. MAPPINGS INTO SPHERES

195

homeomorphic to ^n~1. Now put AI — g~v^\ and A2 = g~l[$2], respectively. Let Xi and X2 be such as in Corollary 3.6.2 for AI and A2. By our inductive assumption, we can extend g \ A\ n A2 to a continuous function

Now for i = 1,2 define &: AlU(X1nX2) -> S? by gl = (g\Ax)Uh. Since S? is homeomorphic to I n it is an AR (Corollary 1.2.12) and so we can extend gi for i = 1, 2 to a continuous function #; : Xi —> §". Clearly,

is a continuous extension of g.

D

Corollary 3.6.4. Let X be a space and ]et /, g : X —> §n be continuous. If

is at most (n — 1) -dimensional then f and g are homotopic. Proof. Consider the product X x I and put

Then A is closed in X xl (Exercise A. 1.9) and the function £: A -> §" defined by f /(*) (* = 0), ^(x,*) = «{ f ( x ) = g ( x ) (xeX\D(f,g)), { 9(x] (t = 1), is clearly continuous. In addition, (X x I) \ A is contained in

and hence is at most ((n — 1) + l) -dimensional by Theorem 3.4.12. So Theorem 3.6.3 implies that £ can be extended to a continuous function £: X x l - > § n , which means that / and g are homotopic.

D

Theorem 3.6.5. Let X be a space and let n > 0. TLe following statements are equivalent: (1) dimX • §n can be continuously extended over X. Proof. For (1) =^ (2) observe that dim A < n and apply Theorem 3.6.3. For (2) => (1), let T = {(Ai, Bi) : i < n + I} be a family of pairs of disjoint closed subsets of X. We shall prove that T is inessential. To this end, for

196

3. BASIC DIMENSION THEORY

every i < n + I let oti : X —»• I be a Urysohn function such that cti \ Ai = 0 and al \ Bl = 1 (Corollary A.4.1). Define a: X -> I n+1 by a(x) = ( a i ( x ) , . . . ,a n + i(a;)). Put A = \J*=i(Ai U Bi) and (3 = a \ A. Then j3[A] is contained in the boundary B of I n+1 which is homeomorphic to S n by Exercise 1.1.24. By assumption, there exists a continuous extension 7 : X —> B of J3. Now for every i < n + 1 put E^ = (^i ° ^}~l(lli}- Then clearly E^ is a partition between Ai and Bi for every « such that HlLi Ei = ®. Consequently, r is inessential. D Application 1: The Brouwer Invariance of Domain Theorem. To begin with, we shall first present an 'internal' characterization of the boundary points of an arbitrary closed subset of a fixed Rn . Proposition 3.6.6. Let n 6 N and let X be a closed subspace ofW1. Then a point x in X belongs to the boundary Fr(X) of X in W1 if and only if x has arbitrarily small neighborhoods U in X such that every continuous function g : X \ U -> Sn~1 can be extended continuously over X. Proof. First assume that x € Fi(X). Without loss of generality, x = 0. If V is a neighborhood of 0 in X then there is e > 0 such that B(x,e)r\X C V. Consequently, it suffices to consider for some e > 0 a 'spherical' neighborhood .0(0, e) of 0 and a continuous function g : X\B(Q,e") -» S71"1. Without loss of generality assume that e = 1 which means that the boundary of 5(0, e] is $n-1. Since 0 belongs to the boundary of X, there is a point q e B(Q, 1) \ X. For each p e Bn \ {q} let r(p) denote the 'projection' of p on S71"1 from q. Put X0 = X n Bn and Xi =X\B(Q,1), respectively. Then X0 and X1 are closed in X, cover X, and have the following additional property:

X0 n S71-1 = Xi n S71"1. Since dimS n-1 = n - 1 (Theorem 3.2.12), the function g \ Y can be extended to a continuous function g \ Y : S71"1 —>• S71"1 by Theorem 3.6.5. In addition, define h: XQ —>• S71"1 by h — g \ Y o T. Observe that r(y) = y for y 6 y so that This means that the functions 5 and /i agree on Y from which it follows that the function g = h Ug is the desired continuous extension of g.

3.6. MAPPINGS INTO SPHERES

197

Conversely, let x e X and assume that x is an interior point of X. There is e > 0 such that B = D ( x , e ] C X. Let U be an open neighborhood of x in X such that U C B(x,e). We identify Bn~l and the boundary of B. For each p G X \ U let r(p] denote the 'projection' of p on Sn-1 from x. Then r cannot be extended over X since this would yield a retraction from J5 onto its boundary, which is impossible by Theorem 2.4.10. D Corollary 3.6.7. Let n € N and Jet X and Y be closed subspaces ofW1. If f : X —> Y is a homeomorphism then f[Fr(X)] — Fr(F). We are now in a position to present a proof of the following interesting: Brouwer Invariance of Domain Theorem 3.6.8. Let n G N and let U be an open subset of M n . If/: U —> W1 is injective and continuous then the following statements hold: (1) f[U] is open in W1, (2) / : £ / — ) • f[U] is a homeomorphism. Proof. We shall first prove (1). Take an arbitrary x G U. We shall prove that f ( x ) belongs to the interior of f[U]. There exists e > 0 such that D(x,e) = {y€Wl

:\\x-y\\<e}CU.

Since B is compact, by Exercise A. 5. 9, / \ B is a homeomorphism. By Corollary 3.6.7 we conclude that f(x] belongs to the interior of f[B] and hence to the interior of f[U]. We shall now prove (2). This is easy. If V is an open subset of U then (1) applied to / f V shows that f[V] is open in W1 and hence in f[U]. We conclude that / : £ / — » f[U] is an open mapping and therefore by injectivity is a homeomorphism. D The question naturally arises whether something like Theorem 3.6.8 can also be derived for the class consisting of all infinite- dimensional linear spaces. As to be expected, this is not possible. For the first part of the theorem, this can be demonstrated quite easily by considering Hilbert space I2. The function / : I2 —> I2 defined by

is an imbedding with nowhere dense range. We now turn to the second part of Theorem 3.6.8. Theorem 3.6.9. For an infinite-dimensional normed linear space L there exists a bijective continuous function f : L —> L such that f is not a homeomorphism.

198

3. BASIC DIMENSION THEORY

Proof. Let S be the unit sphere in L. Since L is infinite-dimensional, S is not compact by Exercise 1.1.22. Consequently, by Exercise A. 5. 13 there exists a continuous function A: S —>• (0,1] such that infA(S) = 0. Now define f:L-^Lby the formula

/(O) = Q. Claim 1. If x £ L and a E [0, oo) then f ( a x ] = a f ( x ) . Proof. This is a triviality. If x = 0 or a = 0 then there is nothing to prove. In addition, if x ^ 0 and a ^ 0 then OLX \ —-J -ax = a f ( x ) , \ax\. as required. 0

(

We shall now prove that / is continuous. It is clear that this need only be checked at the origin. To this end, let (xn}n be a sequence in L \ {0} such that limn^.00 xn = 0. Then II £( Nil \ l *^n 1 I I l l ^ - r l l I I n—too n ||/(a:n)|| = A f Tj-^ J ' \\Xn\\ 0. H^nll '

We conclude that limn_j.oo f ( x n ) = 0. We shall next prove that / is injective. To this end, take x,y € L and assume that f ( x ) = f ( y ) . Without loss of generality, x ^ 0. Then f ( x ) ^ 0 so that f ( y ) ^ 0 from which it follows that y ^ 0. So f ( x ) = f(y) gives us that y

MM

Ml'

-y,

and since A(5) C (0,1] this implies that there exists a > 0 such that x = ay. By the claim, this yields f ( x ) = a f ( y ] , i.e., a = I. We conclude that x = y. Now take y € L \ {0}. Put

x=

y

An easy application of the claim gives us that f ( x ) = y. We shall now prove that / is not a homeomorphism by showing that /-1 is not continuous. Since inf A(5) = 0, we can choose a sequence (xn)n in S such that lira A(x n ) — 0. n—¥oo

It follows that \\f(xn)\\=X(xn}-

\\Xn\\ ^ 0 - 1 = 0 ,

so f(xn} —> 0. However, \\xn\\ — 1 for every n, so xn -ft 0.

D

3.6. MAPPINGS INTO SPHERES

199

Application 2: Dimension and mappings. We shall now discuss the question how continuous mappings can lower or raise dimension. The results obtained earlier in this section are important in the proof of Theorem 3.6.10 below. They are not used in the proof of the subsequent Theorem 3.6.14. However, Theorem 3.6.14 is motivated by Theorem 3.6.10 and so it is natural to discuss it here. We already know that every compact space is a continuous image of C (Theorem 1.5.10). So continuous closed surjections, can arbitrarily raise dimension. In general it is therefore not possible to conclude anything. Certain conditions have to be imposed. It is natural to look at the fibres of the map under consideration. If something can be said about their dimensions, then our intuition tells us that there should be a relation between the dimensions of the domain and the range of the map. This is what the next result is about. Theorem 3.6.10. Let X and Y be spaces and let f : X —>> Y be a continuous closed surjection. If there is an integer k > 0 such that dim/~ 1 (y) < k for every y 6 Y then dimX Proof. Wre may assume without loss of generality that dimy < oo. We shall prove the theorem by induction on dim.y. If dimF = — 1 then X and Y are empty and so the theorem trivially holds. Now assume the theorem holds for all spaces X and Y and all continuous closed surjections f : X — t Y such that dimy < n, where n > 0, and assume that we are in the situation that Y is a space such that dim Y = n. Claim 1. If y € F and U is an open neighborhood of f ~ l (y) in X then there is an open neighborhood V of f ~ l ( y ) such that V C U while moreover dim Fr V < n - 1 + k.

Proof. Put W = Y \ f[X \ U]. Since the map / is closed, W is an open neighborhood of y. Let E be an open neighborhood of y in y such that EC E C W

while moreover dimFrE < n — I (Theorem 3.4.10). Then V = f~1[E] is as required since clearly V CU and FrV = Fr/-![£] = f~l[E] \ f~l[E} C rl[E] \ /^[E] = f~l[F,E] and so dim FrV < n — 1 + A; by the Subspace Theorem 3.2.9 and our inductive hypothesis applied to the closed map / \ f'1 [Fr E]: f'^Fr E] -> Fr £. 0 Put m = n + k. Now let A be an arbitrary closed subset of A", and let h: A —>• S m be continuous. We will prove that h can be extended to a

200

3. BASIC DIMENSION THEORY

continuous function h : X —> S m . It will then follow that dim X < m = n + k by Theorem 3.6.5. Let y 6 Y be arbitrary. Then dimf~l(y) < k < n + k and so we can extend h continuously over A U f ~ 1 ( y ) (Theorem 3.6.3). Since STO is an ANR, this function can be extended over a neighborhood Uy of A U f ~ l ( y ) (Corollary 1.2.13). By Claim 1 there is an open neighborhood Vy of f~~l(y) such that Vy C Uy while moreover dim Fr Vy < n — 1 + k. So h can evidently be extended over A\JVy. The open cover V = {Vy : y £ Y} has a countable subcover by Corollary A. 2. 3, say {Vyi : i G N}. We shall inductively define a sequence of continuous functions hi : A U Uj-=i Vy, -^ Sm such that z-l

(*)

hi\A\J \JVyj =hi-!, 3=1

for every i > 1. Let hi be an arbitrary continuous extension of h over AuVyi . Assume that the functions hi, . . . , hi-\ have been defined satisfying (*). Put t-l

i-l

3=1

3=1

respectively. Observe that K U L = A\J Uj=i ^y> • By the above we can extend the function / to a continuous function g: L —»• S m . Observe that i—l

i—l

i—l V

D = {xeLnK: h^(x) ± g ( x ) } C P = (J ^w \ U y, = U Frl ^' j=i j=i j=i Since dimP < n — l + k — m — 1 by the Countable Closed Sum Theorem 3.2.8, we have that dimD < m — 1 by the Subspace Theorem 3.2.9. We conclude that the functions hl-i \Kr\L and g \Kr\L are homotopic by Corollary 3.6.4. The Borsuk Homotopy Theorem 1.4.2 implies that the function h^-i \ Kr\L can be extended to a continuous function g: L -> Sm. Since hk-i and g agree on the closed set K n L it follows that their union is the desired function hk • This completes the inductive construction. Now define h: X ->• §m by

/i(x) = /ij(x)

(x e vyi}.

Then /i is clearly a continuous function extending h.

D

As we saw above, closed continuous surjections can arbitrarily raise dimension. It is natural to ask whether the same can happen for open continuous surjections. But things are slightly different here. For observe that if X is zero-dimensional, and / : X —> Y is an open and closed continuous surjection, then Y is zero-dimensional (Exercise 1.5.5). So open maps defined on

3.6. MAPPINGS INTO SPHERES

201

zero-dimensional compacta do not raise dimension. In contrast, open maps denned on one-dimensional compacta can arbitrarily raise dimension, as we will prove after the completion of some preparatory work. Proposition 3.6.11. For each n > 1 there exists a continuous function fn : Bn+l ->• 2s" such that for every x G S n , fn(x) — {x}. Proof. Define fn : Bn+l ->• 2 s ™ by t (r\ Jn\^J

{ {§-} - \) Lr f J \ I

f _ I _ on . Jn(X) —
Z —

x

i

-

^9 9 || II I 5- Z — ^ • ||X||

(z = 0), f i n\ (^T ^ UJ.

It is easy to visualize this function e.g., for n — 1. It is clear that fn is continuous and that / n (z) = {x} for every x 6 §n. D Lemma 3.6.12. Let X be a compact space with closed sets A,B,E and F such that A n B = 0 and E n F — 0. Then there exist a compact space Y and an open surjection f:Y-$X such that the collection of pairs is inessential in Y. Proof. Put Z = (A U B) U (E U F). In addition, pick Urysohn functions £,77: Z ->• I such that £[A] = Q,£[B] = l,rj[E] = 0 and n[F] = I (Lemma A. 4.1). The function h(x) = (£(x),ri(x)) 2

(xeZ) 2

maps Z into the boundary of I . Let h: X —>• I be a continuous extension of h (Corollary 1.2.12). By the fact that I 2 w B2 under a homeomorphism which (necessarily) maps its boundary W = dl2 onto S1 (Exercise 1.1.24), there exists by Proposition 3.6.11 a continuous function g : I 2 -> 2W such that g(x) — {x} for every x G W. Now define Y= U ( W x 0 ( & ( x ) ) ) C X x W x£X

This is the space Y we are looking for. We first claim that Y is closed in X x W. To this end, suppose that (xn,yn)n is a sequence in Y converging to an element ( x , y ) & X x W. Then xn —>• x and so 51 o h(xn] —> g o /i(x). Since yn£ g° h(xn} for every n, and y n -> y, it therefore follows that y G g o /i(x)

(Lemma 1.11.2(2)). This proves that Y is closed, whence compact.

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3. BASIC DIMENSION THEORY

Now let / = ?TI \ Y —> X , where TTI : X x W —>• A^ is the projection. This is the map / we are looking for. The function F: X -> 2XxW defined by

F(x) = {x} x [0 o h(x)] is continuous because g and /i are. As a consequence, / is open by Exercise 1.11.16. Let 7T2 : X x W —> W denote the projection. Now take arbitrary elements (x,y] G /-1[A] and (p,q) 6 f~l[B\. Then obviously x € A and y G g ( h ( x ) ^ . But since x G Z, it follows that h(z) = h(x) E W and so

g(h(x))={h(x)}. So this implies that ( x , y ) = (x, /&(#)). Similarly it follows that p € B and that (p, g) = (p, h(p)} . Since h(x) ^ h(p) and (x, y) and (p, g) were arbitrarily chosen, this implies 7T2 [/^[A]] n 7T2 [/"^.B]] = 0. It follows similarly that

This implies that the collection of pairs is by Exercise 3.1.11 inessential since dimVF = 1.

D

Corollary 3.6.13. Let X be a compact space. Then there exist a compact space Y and an open surjective map /: Y —» X such that if A,B,E and F are arbitrary closed subsets of X such that A n B = 0 and E n F — 0 then the pairs are inessential. Proof. Let £ = {(En: Fn) : n e N} be a collection of pairs of disjoint closed sets for X such as in Lemma A. 5. 6. That is, if (A, B) is any pair of disjoint closed subsets of X then there exists n e N such that A C En and F C Bn. Let {(rii , mi) : i G N} be an enumeration of all pairs of natural numbers. We aim at applying Lemma 3.6.12 repeatedly. In the first step of the construction we find a compact space Y\ and an open continuous surjection /o : Y\ —> X such that the collection of pairs (**)

is inessential. Then we again apply Lemma 3.6.12 but now with Y\ and the collection of pairs

3.6.

MAPPINGS INTO SPHERES

203

obtaining an open surjection /i : Y-2 -» YI such that the collection of pairs

{ (/r1 1/o"1 [En2]} , /r1 [/o-1 [Fn2]] ) , (/r1 [.fir1 (Em2]\ , /r1 [/o-1 [^2]] ) } ,

is inessential. Observe that by (**) and Exercise 3.1.12 it follows that the collection of pairs is inessential as well. So now it is clear how to proceed. In an infinite process we can deal with all pairs in the collection £, thus obtaining an inverse sequence y

/O

y

fl

Y

_

_

V"

/"

y

with continuous, surjective, open bonding mappings. If YOO denotes its inverse limit, then the projection /oo : Y^ -> X is an open continuous surjection by Exercise 1.10.5. In addition, if A,B,E and F are closed subsets of X such that A n B — 0 and E n F — 0 then there exist n, m e N such that Since by construction the pairs { (&&„], &[Fn]),

(/^[

are inessential, the same holds for the pairs So the space 1^, and the map /^ are as required.

D

This leads us to the following interesting result. Theorem 3.6.14. Let X be a compaction. Then there exists an at most one-dimensional compactum Y which can be mapped onto X by an open map. Proof. By applying Corollary 3.6.13 we obtain an inverse sequence y

_ ~y*

/(I

y

fl

y

_

_

y

/».

of compact spaces, with continuous, open and surjective bonding mappings, having the following additional property: if for some n > 0, A, B,E and F are closed sets in Yn such that A n B = 0 and E n F = 0 then the collection of pairs is inessential in Yn+\. Let Y^ denote the inverse limit of the sequence. By similar arguments as the ones in the proof of Corollary 3.6.13 it follows that for every n > 0 the canonical projection /^ from 1^ onto Yn is an open surjection. So we are done if we can prove that dim YOO < 1. But this follows by a compactness argument. For let A:B,E and F be closed subsets of Y^ such that A n B = 0 and E n F = 0. By Exercise 1.10.12 it follows that

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3. BASIC DIMENSION THEORY

there exists n € N such that f^[A] C\ f£[B] = 0 and /£[£] n /£[F] = 0. By construction we have that the pairs are inessential. So now we are done by using Exercise 3.1.11 and by observing that /"+1 [4 C/- 1 ^ [4], etc. D Exercises for §3.6. ^•1. Let n > 0 and let X be a compact space which can be written as the union of closed sets Fi, i (E N, such that for all distinct i,j £ N, dim(F; n F j ) < n. Prove that every continuous function from F\ into Sn is continuously extendable over X. 2. Observe that the previous exercise is a generalization of Theorem A. 10. 6. 3. Let X be a compact space and let H be the union of all components of X of dimension at most n. Prove that dim H < n. ^•4. Let A be a closed subspace of a space X such that dim(X \ A) < n, let 0 < k < n, and let /: A —> S fc be continuous. Prove that there exists a closed subspace B of X such that A D B — 0 and dim B < n — k — 1, while moreover the function / can be extended over X \ B. 5. Let n > 2 and consider the n-dimensional cube I n . Define En~l = {(xi,...,xn-i,l) : 0 < xl < l,i = l , . . . , n - 1}. Prove that if X = ln \ En~l then def X = n - 1. 6. Let X be an at most (n — l)-dimensional space. Prove that every continuous function /: X —> Sra is nullhomotopic. 7. Prove that S" is unicoherent for every n > 2. 3.7.

Dimension of subsets of W1 and certain generalizations

In this section we prove some interesting dimensional properties of subsets of the Euclidean spaces Rn , n G N. Theorem 3.7.1. Let X C Rn . Then

dimX = n

Proof. First, if Int^T ^ 0 then X contains a homeomorph of In and is therefore n-dimensional by Theorems 3.2.12 and Corollary 3.4.14. Next, assume that Int X = 0. There consequently exists a countable set D C W1 \ X which is dense in W1 .

3.7. DIMENSION OF SUBSETS OF R™ AND CERTAIN GENERALIZATIONS

205

By Proposition 3.2.10, Rn = 9tnj0 U 9tn,i U • • • U
Notice that 9ln,n consists of all points of IRn having rational coordinates only, and hence is a countable dense set. By Proposition 3.2.10 and Corollary 3.1.7 we get that dim Y < n — I . By the countable dense homogeneity of W1 (Corollary 1.6.10), there is an element h G CK (W1} such that h[D] = ftn,n. This implies that h[X] C Y, and hence by Theorem 3.2.9 that dim^C = dim/i[.Y] < n — 1, as required. D Corollary 3.7.2. Let U C W1 be nonempty and open. If U is not dense then dim Fr U = n — I . Proof. Put F = FrU. Then by Theorem 3.7.1 it follows that dim F< Pick an arbitrary point m £ Mn \ U.

n-l.

Striving for a contradiction, assume that dim F < n — 2. We will first prove that we may assume without loss of generality that U is compact. For this we will use a trick on 'exchanging' points at infinity. The one-point compactification X — W1 U {00} is homeomorphic to Sn (Exercise A. 4. 8). The boundary of U in X is F if it is compact and F U {00} otherwise. Since dimF = dim(F U {oo}) (Exercise 3.2.1) we arrive at the conclusion that the boundary of U in X is at most (n — 2)-dimensional. Now observe that Y — X \ {m} is homeomorphic to R n , that the closure of U in Y is compact, and that its boundary is at most (n — 2)-dimensional. So from now on assume that U is compact in Rn and that FrC/ is at most (n — 2)-dimensional. We may assume without loss of generality that the origin 0 of Mn belongs to U. By compactness of U, the collection {r-U:r G (0,1]} is a local base at 0 consisting of compact neighborhoods all of whose boundaries are at most (n — 2)-dimensional. But since W1 is homogeneous, the same is true for all points of E n . By Theorem 3.4.10 (applied twice) this implies that dim W1 < n - 1, which contradicts Theorem 3.2.12. D Corollary 3.7.3. Let L be a closed subset ofW1 with dim L < n-2. Then L does not separate W1 . Proof. Suppose that Mn \L = UUV, where U and V are disjoint nonempty open sets. Then Fr£7 C L and so dimFrL < n — 2 which contradicts Corollary 3.7.2. D

206

3. BASIC DIMENSION THEORY

We now aim at proving in Theorem 3.7.6 a much stronger version of this corollary. First we need to do derive some elementary results about essential families and continua. Proposition 3.7.4. Let X be a compact space, let {(Ai,Bi) : i G F} be an essential family of pairs of disjoint closed subsets of X and let F(0) C F. If for each i E F(0), Li is a partition between Ai and Bi in X and n G F \ F(0) then L = H w ^ contains a continuum from An to Bn. Proof. Suppose that L does not contain any continuum from An to Bn. Let H = An n L,

G = BnnL,

and put

£ = {C : C is a component of L and C n H / 0},

E = (J £,

3" = {C : C is a component of L and C n G ^ 0},

F = |J J,

and

respectively. From Proposition A. 10. 3 it follows that both E and F are closed. Also, by assumption, E D -F = 0. Take an arbitrary C E £. Since L \ F is an open neighborhood of C in L, by Lemma A. 10.1 there exists a clopen (in L) set Uc which contains C but misses F. By compactness, finitely many t/c's cover .E. We conclude that 0 is a partition in L between H and G. By Corollary 3.1.5 there is a partition T in X between An and Bn such that T n L = 0. But this contradicts the fact that is essential (Exercise 3.1.1).

D

Corollary 3.7.5. Let X be a compact space, let be an essential family of pairs of disjoint closed subsets of X and let n G F. Suppose that Y C X is such that Y meets every continuum from An to Bn. For each i G F \ {n} let Ui and Vi be disjoint closed neighborhoods of Ai and Bi, respectively. Then is essential in Y. Proof. By Proposition 3.7.4, if for every i G F \ {n} the set Li is a partition between Ai and Bi then Y^r\i^r\{n} ^i ^ ®- Now apply Exercise 3.1.13. D A space X is called continuum- connected provided that for any pair of points x,y G X there is a subcontinuum C of X containing both x and y. A continuum-connected space is obviously connected, but the converse need not be true (Exercise 3.7.1).

3.7. DIMENSION OF SUBSETS OF R™ AND CERTAIN GENERALIZATIONS

207

Theorem 3.7.6. Let M C G, where G C W1 is open and connected. If M is at most (n — 2)-dimensional then G\M is continuum-connected. Proof. We first prove the special case that G = W1. Take two arbitrary distinct points x, y G Mn \ M and let K denote the closed ball in Rn with center 1/2(x + y] and radius l/2\\x — y\\. Let /: ln —>• K be a continuous surjection which maps a pair of opposite faces A and B of I n to x and y, respectively, and has the property that the restriction g — / \In\ (A\JB] is a homeomorphism from In\ (A\JB) onto K\ {x, y}. Such a function can easily be constructed with the technique used in the solution of Exercise 1.1.24. The set f ~ l [ M r \ K ] is at most (n — 2)-dimensional. Hence by Corollary 3.7.5 there is a continuum C from A to B which misses f~l[Mr\K]. But then f[C] is a continuum containing x and y but missing M. Now consider an arbitrary open and connected set G C W1 and arbitrary points x,y £ G\ M. By Lemma 1.5.21 there exist open balls BI,B<21 • • •,B^ in G such that x E B\, y £ Bk and BiC\Bi+\ ^ 0 for alH < fc — 1 (here we use that G is open and connected). Since M has empty interior by Theorem 3.7.1 there exists for i < k— 1 a point Zi € (BiC\Bi+i)\M. Put ^o = x and 2^ = y. By the above for i I is called an n-dimensional Cantormanifold if for each closed subset L C X with dim L < n — 2

the complement X \ L is connected. The examples of n-dimensional totally disconnected spaces for all n that will be constructed in §3.9 show that a generalization of this concept to noncompact spaces makes no sense; simply observe that the singletons are the only connected subsets of a totally disconnected space. Observe that each Cantor-manifold is a continuum and that the one-dimensional Cantor-manifolds are precisely the one-dimensional continua. Proposition 3.7.7. Every n-dimensional compactification of W1 is an ndimensional Cantor-manifold. In particular, §n and In are n-dimensional Cantor-manifolds. Proof. Let 7Mn be an n-dimensional compactification of R n , and let A in jX be at most (n — 2)-dimensional. Then B = W1 \ A is connected by Corollary 3.7.3. In addition, A n W1 has empty interior in W1 by Theorem 3.7.1

208

3. BASIC DIMENSION THEORY

which implies that B is dense. We conclude that ^W1 \ A contains the dense connected set B and is therefore connected itself. D We will now show that there are 'many' n-dimensional Cantor-manifolds. Theorem 3.7.8. Let n > 1. Every compact n-dimensional space X contains an n-dimensional Cantor-manifold. Proof. By Theorem 3.6.5 there exists a closed subset A C X and a continuous function /: A —> Sn-1 such that / admits no continuous extension over X. Let "B = {Bi : i 6 N} be a countable open base for X. We define a decreasing sequence (Xi)i>o of closed subsets of X, as follows. First, let XQ = X. If Xi has been defined then we consider two cases. If / does not admit a continuous extension over A U (Xi \ Bi+i) then put

otherwise put We claim that M = Di^o -^ *s an ^--dimensional Cantor-manifold. Observe that by construction, for no i the function / admits a continuous extension over A U Xi . Claim 1. / does not admits a continuous extension over A U M. Proof. Striving for a contradiction, assume that f : A\J M —>• S71"1 is a continuous extension of /. By Corollary 1.2.13 it follows that there is an open neighborhood U of M such that / can be extended to a continuous function f:A\JU-> S™"1. Pick N so large that M C XN C U (here we use that X is compact, see Exercise A. 5. 17). Then / \ (Au XN) extends / which is impossible by construction. 0 So by Theorem 3.6.3 it follows that dimM > n. Since dimM < n being a subspace of X we conclude that dimM = n. To complete the proof it clearly suffices to prove that if MI and M2 are proper closed subsets of M with M = MI U M2 then dim (Mi n M2) > n — 1. Striving for a contradiction, assume that there exist proper closed subsets MI and M2 of M with M = M!UM 2 ,

dim(Mi n M2) < n - 2.

Put AI - A U MI and A2 = A U M2, respectively. Claim 2. / admits continuous extensions over AI and A 2 , say to /i and /2, respectively.

3.7. DIMENSION OF SUBSETS OF R n AND CERTAIN GENERALIZATIONS

209

Proof. It suffices to prove this for A\. To this end, pick x G M2 \ MI (here we use the fact that MI and M2 are proper subsets of M). Pick i G N so large that x E -B;+i C X \ M\. Consider the set Xi+i. It contains x and hence it intersects Bi+i. By construction this can only happen when / can be extended continuously over A U (Xl \ Bl+i). Since A\ C A U (Ar; \ -B;+i), we are done. 0 Put 5 = A U (Mi n M 2 ). The functions /i f # and /2 \ B only differ in a subset of MI n M2, and hence on a set of dimension n — 2 at most. We conclude that f i \ B and /2 f Z? are homotopic by Corollary 3.6.4. Since /2 \ B admits a continuous extension over A U M2 , the Borsuk Homotopy Extension Theorem 1.4.2 implies that the same is true for /i \ B, say to the function f[ : A U M2 ->• §n~l. Since B= ( A u M ! ) n ( A u M 2 ) , it follows that the formula

defines a continuous extension of / over AU M. This is a contradiction. D This yields another solution to Exercise 3.2.4. Corollary 3.7.9. If X is compact and dim X = n then one of its components is n- dimensional. Exercises for §3.7. 1. Give an example of a connected space which is not continuum-connected. 2. Give an example of a space X having the property that for every component C C X we have dimC < 3. Let X be compact and let A and B be disjoint closed subsets of X. In addition, let L be a closed subset of X such that no continuum in L intersects both A and B. Prove that there is an open neighborhood U of L such that no continuum in U intersects both A and B. >4. Let X be a compact space, let {(Ai,Bi) : i G F} be an essential family of pairs of disjoint closed subsets of X and let F(0) C F. Suppose that for each i G F(0), Li is a partition between Ai and Bi in X and let

L= ier(o) n L,.

Finally, let m,n 2 G F \ F(0) be distinct. Call a subset T C X small if no continuum in T meets both Ani and Bm .

210

3. BASIC DIMENSION THEORY

Prove that if N is a finite pairwise disjoint collection of small closed subsets of X then there is a continuum in from An2 to Bn2 .

Let X be a compact space such that dimX > 2. Prove that there are real numbers 5 > 0 and e > 0 such that if N is any finite collection of pairwise disjoint closed sets with mesh(!N) < 5 then there is a continuum C Q X \ (J N with diameter at least e. Let X be a compact space with dimX 0 there is a family K of pairwise disjoint closed subsets of X with mesh(N) < e such that each continuum C C X \ (J [NT has diameter less than s. 3.8. Higher-dimensional hereditarily indecomposable continua Indecomposable continua are strange objects. We proved their existence in §1.10. In this section we go one step further, and prove the existence of 'many' hereditarily indecomposable continua of arbitrary large dimension. Higher-dimensional hereditarily indecomposable continua were first constructed in 1951. But only the last decade it was realized that they are fundamental objects that can be used for the solution of various open problems in dimension theory. See the notes for more information. The following result is the heart of the construction of hereditarily indecomposable continua. Theorem 3.8.1. Let X be a compaction and let F0,Fi be disjoint closed sets in X. Then there are disjoint open neighborhoods WQ and W\ of FQ and FI, respectively, such that any two intersecting continua K0, KI in are comparable, i.e., K0 C KI or KI C K0. Remark 3.8.2. A compact space X is called Bing provided that for all subcontinua A and B of X we have Ar\B = ® or A C B or B C A. So a zerodimensional compact space is Bing. Theorem 3.8.1 shows that disjoint closed sets in a compact space can always be partitioned by a Bing compactum. Proof. Let g be an admissible metric on X. The Euclidean metric on I will be denoted by d. Let P be the union of the five open rectangles in I 2 , described in Figure 9. The set I 2 \ P is the union of the disjoint open sets MQ and MI in the picture. Clearly, M0 H MI = 0 . Observe that (1)

(OjxlCMo,

{l}xICMi.

3.8. HIGHER-DIMENSIONAL HEREDITARILY INDECOMPOSABLE CONTINUA 211

Hence P is a partition between {0} x I and {1} x I in I2. Let WQ,O and W1)0 be open neighborhoods of FQ and F\, respectively, such that /O\

Tf7

yZ.)

o TT/

(ft

K^O o I I ''1 0 — W-

5

/6

2

/3

P

Vs

Figure 9. In addition, let F = {/; : z e N} be a dense subset of C(-X", E) (Proposition 1.3.3). Recursively, determine for every i > I open sets WQ^ and W\^ in X and a continuous function M; : X —>• I 2 , by the following formulas: (3)

^(x,Wo,i-i)

f(

Q(X,

(4)

Since Wo,onWo,i = 0 5 it follows that u\ is well-defined and continuous. Notice that by (1) and the fact that the first component of u\ is an appropriate Urysohn function, we have Ui[W0i0] C {0} x I C Mo,

WitVTi.o] C {1} x I C Mi.

By (4) and the fact that MQ fi MI = 0, we consequently obtain that By induction on i this reasoning can easily be extended. So oo

W0 = U W0ti,

oo

= U wi>*> i=0

212

3. BASIC DIMENSION THEORY

are disjoint open neighborhoods of FQ and F\ , respectively. We claim that these neighborhoods are as desired. Let KQ and K\ be arbitrary intersecting continua of X such that KoUKi

CX\(W0UWi).

Striving for a contradiction, assume that there exist points ao£K0\Ki,

ai£Ki\K0.

There is by Lemma A. 4. 2 a Urysohn function u: X —> I such that

u(o 0 ) = 0,

u(ai) = 1,

u[K0] C [0, 1/2],

u[#i] C [i/2, 1].

Since F is dense in C(J\T,I), there exists i such that d(u, fi) < l/6. So we have (5)

/,(a0) <E [0, i/6), £(01) e (%, 1], /i[^o] C [0, %], /^i] C [1/3, 1],

hence

(6)

K 0 n/- 1 (V 3 ,i]-0, K 1 n/- 1 [o,y 3 )=0.

This shows that the continua To = Ui[Ko],

T! =ui[Kl],

have the property that (7)

To C I x [0,2/3],

T0 C l x [1/3,1].

Since KQ and K\ are disjoint from Wo,i U Wi^, the continua TO and T\ are contained in P, cf. (4). By (5), TO intersects the strip E x [0, Ye) and, similarly, T\ intersects I x ( 5 / 6 , 1]. Since TO and T\ are continua, the geometry of P and (7) imply that they are disjoint. But this is a contradiction since K0 and KI intersect, hence T0 fl TI / 0. D Corollary 3.8.3. Every (n + 1) -dimensional compactum X contains an ndimensional hereditarily indecomposable subcontinuum. Proof. Let X be an (n + l)-dimensional compact space with essential family {(Ai,Bi) : i < n + I}. An arbitrary partition T between A\ and B\ must be at least n-dimensional by Theorem 3.1.9. Theorem 3.8.1 shows that for T we may take a closed set having the property that each of its components is hereditarily indecomposable (Exercise 1.10.21). One of those components, say C, must have the same dimension as T by Exercise 3.2.4 or Corollary 3.7.9. So C is a hereditarily indecomposable continuum of dimension n or n + 1. But every (n + l)-dimensional compactum contains an n-dimensional compactum by Exercise 3.3.7, and hence an n-dimensional continuum by Exercise 3.2.4. So X contains an n-dimensional hereditarily indecomposable continuum. D

3.8. HIGHER-DIMENSIONAL HEREDITARILY INDECOMPOSABLE CONTINUA 213

Corollary 3.8.4. Every strongly infinite-dimensional compact space contains a strongly infinite-dimensional hereditarily indecomposable continuum. Proof. If X is a strongly infinite-dimensional compactum with essential family {(Ai, BI) : i E N} then an arbitrary partition between AI and BI must be strongly infinite-dimensional by Theorem 3.1.9. By using Exercise 3.1.9 one of the components of such a partition T must be strongly infinite-dimensional as well. We can therefore use the same method as in the proof of Corollary 3.8.3. D Remark 3.8.5. In view of Corollary 3.8.3 and its proof, the question naturally arises whether there exists for n > 2 a hereditarily indecomposable continuum separating the top from the bottom in ln+1. The answer to this question is in the affirmative. Simply observe that I n+1 is locally connected and by contractibility is unicoherent (Corollary A. 12.10). As a consequence, the Bing partition between the top and the bottom of I 3 considered in the proof of Corollary 3.8.3 contains a connected partition by Exercises 3.10.2 and 3.10.3. This partition is clearly a hereditarily indecomposable continuum. Henderson compacta. A compactum X is called Henderson if it is infinite-dimensional but every finite dimensional closed subspace is zerodimensional. It is not clear at all that such a compactum exists. By using Corollary 3.8.4 we will present a surprisingly simple construction of such a space (for a stronger result, see Theorem 3.13.10). Let X be a compact space. For t > 0 and n > 0 put Xn,t — \\{C C X : C is a continuum with dimC < n and diamC > t}. Lemma 3.8.6. Let X be a Bing compactum. Then for every t > 0 and n > 0 we have dim Xnj < n. Proof. Let F be an arbitrary closed subset of X with diamF < t. Claim 1. The components of F of dimension at most n cover Xn,t n F. As a consequence, dim(Xntt n F) < n. Proof. Indeed, let x G Xn^ H F. There is a continuum A contained in Xnj with dim A < n and diamA > t such that x € A. Let B be the component of x in F. Then diam B < t and B n A 7^ 0. So A g. B and hence since X is Bing it follows that B C A. By the Subspace Theorem 3.2.9 we conclude that dimB < n. Hence x is contained in an at most n-dimensional component of F.

214

3. BASIC DIMENSION THEORY

Let H be the union of all components of F with dimension at most n. Then dimH < n by Exercise 3.6.3. We just observed that H D Xnj n F. So we are done by the Subspace Theorem 3.2.9. <£> Since X is compact, there is a finite closed cover 7 of X with mesh less than t. By the claim, Xn,t n F is at most n-dimensional for every F. Hence dimXn:t < n by the Countable Closed Sum Theorem 3.2.8. D This leads us to the following unexpected result. Theorem 3.8.7. Let X be a Bing compaction. The union of all nondegenerate finite dimensional subcontinua of X is countable dimensional. Proof. Every nondegenerate subcontinuum of X has positive diameter. So the set we are interested in is a subset of

71=1

This means that we are done by Lemma 3.8.6, Corollary 3.3.9 and Theorem 3.2.9. D This is all we need for the construction of a Henderson compactum. Corollary 3.8.8. Every strongly infinite-dimensional compactum contains a strongly infinite-dimensional Henderson compactum. Proof. By Corollary 3.8.4 we may assume without loss of generality that X is a hereditarily indecomposable continuum. Let be essential in X. By Theorem 3.8.7 there is a countable family

{N2i :i e N} of zero-dimensional subspace of X covering all finite dimensional nondegenerate subcontinua of X. For every i e N let Ti be a partition between the sets A-2i and B2i such that Tz n N2i = 0. By Theorem 3.1.9 it follows that the compactum

is strongly infinite-dimensional. In addition, T contains no nondegenerate finite dimensional subcontinua. For such a subcontinuum is contained in \J°^=1 NH and hence misses T. We claim that T is Henderson. For if A is a finite-dimensional closed subspace of T which is not zero-dimensional, then one of its components is not zero-dimensional (Exercise 3.2.4) and hence by construction misses T. So this is impossible. D

3.8. HIGHER-DIMENSIONAL HEREDITARILY INDECOMPOSABLE CONTINUA 215

Homogeneity. As we observed in §1.10, the dyadic solenoid £2 is a topological group and hence is homogeneous. So there is a homogeneous indecomposable continuum. Is there an example of a homogeneous hereditarily indecomposable continuum? Surprisingly, there is such a space which is even planar. It is the so-called pseudo-arc which can be constructed by a slightly more complicated procedure than the one used for the indecomposable continuum P in Figure 5 on Page 95. So the construction of the pseudo-arc is not difficult. The verification that it is hereditarily indecomposable is not difficult as well. But the proof of its homogeneity is not simple. This was established by BING in [55]. The pseudo-arc is clearly one-dimensional. So in view of the existence of higher-dimensional hereditarily indecomposable continua (Corollary 3.8.3), the question arises whether such a space can be homogeneous. The answer to this question is in the negative. Proposition 3.8.9. If X is a Bing compaction such that 2 < dim AT < oo then X is not homogeneous. Proof. Let dim AT = n. Let U be a finite open cover of X of mesh less than 1. Since every element of U is an F^-subset of X (Exercise A.2.3), the Countable closed Sum Theorem 3.2.8 shows that there are an element U G U and a compact subset F C U such that dimF = n. Corollary 3.7.9 shows that we may assume without loss of generality that F is a continuum. So X contains an n-dimensional continuum of diameter less than 1. Observe that besides compactness, we did not use any of the properties of X. So by repeating this reasoning, we can construct by induction a decreasing sequence (F;); of n-dimensional subcontinua of X such that diam Fi < l/i for every i. By compactness, the intersection Hi^i -^ ^ ®- In fact, it is a single point since diam Fi < l/i for every i. Let x be the unique point in that intersection. Claim 1. If C is a nondegenerate subcontinuum of X containing x then dimC = n. Proof. For every i, x G Fi n C. Since X is hereditarily indecomposable, this implies that Fi C C or C C Fi (Exercise 1.10.21). So if there exists i such that Fi C C then we are done since by the Subspace Theorem 3.2.9 (applied twice) we then have n = dim Fi < dim C < dim X = n. If there does not exist such i then C C Fi for every i and hence C f) # = i—l

But this contradicts the fact that C is nondegenerate.

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3. BASIC DIMENSION THEORY

By equality of dim and ind (Theorem 3.4.8), there is a closed subspace G of X such that dimG = n — 1. Corollary 3.7.9 shows that we may assume that G is a continuum. Since n — 1 > 1, the set G is a nondegenerate subcontinuum. Let y be an arbitrary element in G. Then by Claim 1 there does not exist a homeomorphism of X which maps x onto y. Hence X is not homogeneous. D Remark 3.8.10. The proof of Proposition 3.8.9 clearly breaks down if X is infinite-dimensional. It would work if every infinite-dimensional compact space would contain a nondegenerate finite dimensional continuum. But this is not true, as was shown in Corollary 3.8.8 (see also §3.13). By a different technique, ROGERS [354] proved that there is no infinite-dimensional hereditarily indecomposable homogeneous continuum. It is unknown however whether there is an infinite-dimensional homogeneous indecomposable continuum (this is a question of ROGERS). Exercise for §3.8.

1. Let X be a space and let FQ, FI be disjoint closed sets in X. Prove that there are disjoint open neighborhoods WQ and W\ of .Fo and F\ > respectively, such that any two intersecting continua Ko, K\ in X \ (Wo U W\) are comparable, i.e., KQ C K\ or K\ C KQ. 3.9. Totally disconnected spaces In this section we shall prove that totally disconnected spaces of every dimension exist, cf. Examples 1.5.18 and 3.4.13 where we constructed an example of a one-dimensional totally disconnected space. Our main tool is the concept of a so-called 'G^-selection'. After some preparatory work, the construction of the examples is surprisingly simple. The lexicographic order. Suppose that X is compact and that

is a continuous surjection. The set-valued function

F: Y ^ X l

defined by F(y) = f ~ ( y ] in general unfortunately does not admit a continuous selection, see Page 13. We shall show however that there always exists a selection s for F such that the range of s is a G^-subset of X. Let n e N U {00} and consider the product JJn. It can be ordered by the so-called lexicographic order ^, as follows:

x Xy

^=>

(3 n € N) (xn < yn and Xi = yi if i < n).

A straightforward verification shows that -< is a linear order on J n .

3.9. TOTALLY DISCONNECTED SPACES

217

Lemma 3.9.1. Let n € N U {00} and let X C JJn be compact. Then X has a smallest element with respect to ^. Proof. Since X is compact and TTI is continuous, TTI[^] is compact as well and hence x\ — minTTipC] exists. Observe that X n Trf l ( x i ] is a nonempty compact subset of X. So x 2 = min^fX n Trf l ( x \ } ] exists as well. Observe that the set

x n^l(xi) n7T2"1(x2)

is again a nonempty compact subset of X. So the minimum of its projection onto the third factor exists. Etc. It is clear that the point x = (xi}i defined in this way is the smallest element of X with respect to ^ (observe that by compactness of X it belongs to X, even if n = oo). D Theorem 3.9.2. Let X be a compact space and let f : X ->• Y be a continuous surjection. There exists a G^-subset S of X which intersects each fiber of f in precisely one point. Proof. By Corollary A. 4. 4 we may assume that X is a subspace of Q. Since / is a surjection, for every y G Y the fiber f ~ l ( y ) is a nonempty compact subset of X. Hence it has a smallest element with respect to the lexicographic ordering ^ on Q (Lemma 3.9.1), say p(y). We claim that the set S = {p(y}

:yeY}

is a GVsubset of X and is therefore as required. Define for n, m G N the following sets: Un^m — {x G X : for all z G X such that Xi — Zi for i < n — 1 and zn < xn — 1/m we have f(x] ^ f ( z ) } . We first prove that the t/n,m's are open by establishing that their complements are closed. To this end, for each j G N take x(j) G X \ Un>m and assume that the sequence (x(j)) . converges to a point x G X. For each j there exists an element z(j) in X such that x(j)i = z(j)i for every i < n — I while moreover z ( j ) n < x(j}n - Vm and f(x(j)) = f ( z ( j ) ) . By compactness, without loss of generality we assume that (z(j)) . converges to a point z in X. By continuity of the projections and / we get Xi = Zi for every i m is closed. We next claim that for all n and m, S C Un,m- To this end, take an arbitrary y G Y and consider the point p(y}. Fix arbitrary n,ra G N. Take a point z E X such that p(y)i — Zi for all i < n — 1 and zn < p(y}n — l/mIf z G f ~ l ( y ) then by construction p(y)n < zn- But this is impossible and so z g. f ~ l ( y ) , as required.

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3. BASIC DIMENSION THEORY

We finally claim that the intersection U of all the t/n,m's is equal to S. Since S intersects every fiber of / in precisely one point, and as we just observed, S C £/, it suffices to prove that U intersects every fiber of / in at most one point. To this end, take an arbitrary y 6 Y and assume that there are distinct points x ( l ) , x ( 2 ) 6 U C \ f ~ 1 ( y ) . Let n 6 N be the first index such that x(l)n ^ x(2)n. We may assume without loss of generality that

x(l}n <x(2)n. There exists i G N such that x(l)n < x(2)n — l/i. Since x(2) 6 Un>i and x(l)m ~ x(2)m for all m < n — 1, we get f ( x ( l } ] ^ f ( x ( 2 ) } . This is a contradiction. D The existence of n-dimensional totally disconnected spaces. We are now in a position to prove the main result of this section. Theorem 3.9.3. For each n > 0 there exists an n-dimensional topologically complete totally disconnected space. There also exists a strongly infinitedimensional topologically complete totally disconnected space. Remark 3.9.4. The existence of a strongly infinite-dimensional topologically complete totally disconnected space turns out to have a surprising application: it is the main ingredient in the solution of a famous problem in dimension theory that remained open for more than 50 years. See Theorem 3.13.8 for details. Proof. We shall construct the examples simultaneously. Consider Y — — JJH X v JJlTn , -A.

where n 6 NU {oo}. Let TT: X —>• JJ be the projection onto its first factor and put A = TT~ I (— 1) and B = 7r~ 1 (l), respectively. Define C = {C € 2X : C is a continuum from A to B}. By Exercise 1.11.21 it follows that C is a closed subspace of 2X . Since 2X is compact by Proposition 1.11.4, this proves that C is compact. Let C be a Cantor set in J and since C / 0 there is a continuous surjection a: C -»> C (Theorem 1.5.10). Now put

Claim 1. Y is compact and 7r[Y] = C. Proof. We shall first prove that ir[Y] = C. Clearly, 7r[F] C C. Take an arbitrary t G C. Then a(t) is a continuum from A to B. Consequently, 7r[o;(i)] is a continuum in J from —1 to 1 and therefore is equal to J. So there exists an element y £ a(t) with TT(T/) = t. Since this y belongs to Y, this proves that C C 7r[y].

3.9. TOTALLY DISCONNECTED SPACES

219

We shall next show that Y is compact. To this end, let (yi)i be a sequence in Y converging to a point y G X. Then t; = 7t(yi) belongs to C for every i and (ti)i converges to t = TT(?/). Consequently, t belongs to C. Since a is continuous, the sequence of continua (a(t;)). converges to a(t) in C. We claim that y e a(t). If this is not the case then we can find e > 0 such that y£D(a(t},e}. Since (cu(£;)). converges to a(t), for all but finitely many i, yi£a(ti) which implies that y G D(a(t),e). that yeY.

CD(a(t),e) This is a contradiction. We conclude 0

Since TT is continuous and Y is compact (Claim 1), by Theorem 3.9.2 there exists a G^-subset 5 of Y which intersects each fiber of TT \Y in precisely one point. The compactness of Y implies that S is topologically complete (Theorem A.6.3). Claim 2. S intersects each continuum from A to B. Proof. Take an arbitrary C € C. Since a is surjective, there exists t G C such that a(t) = C. Observe that (TT r IT1^) = 7r ~ 1 W n r = ^H*) n «(*) = T"1^) n (7 C C. Since 5 intersects each fiber of TT \ Y we consequently get S fi C 7^ 0.

<(>

Observe that TT f S is a one-to-one continuous function from 5 onto the zero-dimensional space C. This implies that S is totally disconnected, cf. Example 1.5.18. Now assume that n is finite. Then dim S > n by Theorem 2.4.12 and Corollary 3.7.5. Also dim 5 < n since 5 is contained in C x JJ n , which is n-dimensional by Theorems 3.2.12 and 3.4.12. We conclude that dim S = n. Now if n — oo then S is strongly infinite-dimensional by Corollaries 2.4.13 and 3.7.5. D Application: The dimension of infinite products. As an unexpected application of the results obtained here we will show that there exists for every n > 0 a topologically complete n-dimensional space whose infinite power is also n-dimensional. This shows that there exists an example such as Example 3.4.13 in every dimension. We will in fact show that the spaces constructed in the proof of Theorem 3.9.3 are as required. We have to be a little careful though. In the proof of Theorem 3.9.3 the precise nature of the 'G^-selection' 5 is irrelevant. We used Theorem 3.9.2 to prove its existence, but how it was created in the proof of Theorem 3.9.2 did play no role. But this is important here. So we assume

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3. BASIC DIMENSION THEORY

in the proof of Theorem 3.9.5 that the 'G^-selection' S was constructed via the method used in the proof of Theorem 3.9.2, i.e., via the lexicographic order on JIn. Theorem 3.9.5. For every n > 0 there exists an n-dimensional topologically complete space whose infinite power is also n-dimensional. Proof. Let S be the space constructed in the proof of Theorem 3.9.3. We claim that S is as required. Observe that S C JJ x JP and that the projection

TT: J J x J n -> JJ maps 5 in a one-to-one way onto the Cantor set C. Define a function

/: S x 5 -> C 2 x Rn as follows: f((x,z),(x',z')) 2

= ((x,x'),z + z').

n

Since dim(C x R ) < n (Theorems 3.2.12 and 3.4.12) it will follow that dim S2 < n if we can prove that / is an imbedding. It is clear that / is oneto-one and continuous. The potential problem is the continuity of the inverse. To check this, let ( f ( ( x i , Z i ) , (X,^)))^ be a sequence in /[52] converging to an element /((x, z), (x', z')) in f[S2]. By the definition of / this implies that Xi —> X,

x'i —> x',

Zi + z'i —¥ Z + z'.

By compactness we may assume without loss of generality that Zi —> a,

z'i —> &,

n

where a,6 e JJ . So a + b = z + z'. Since (xi,Zi) £ a(xj) for every i and Gi(xi) —>• a(x) by continuity of a it follows easily that (re, a) 6 a(x] (Lemma 1.11.2(b)). But this implies by construction that z -< a. It follows similarly that z' -< b. Since z + z' = a + b we get z = a and z' = b. This implies that ((zi,Zi),(z;,z;)) -> ( ( x , z ) , ( x ' , z ' ) ) , as required. So dim S2 < n from which it follows that dim S2 = n since S2 contains a (closed) copy of the n-dimensional space 5 (Theorem 3.2.9). The same method can be be used to show that the dimension of each finite power of S is equal to n. Since the countably infinite power of 5 is the inverse limit of the inverse sequence consisting of all of its finite powers (Example 1.10.3), we are done by an appeal to Exercise 3.5.2. D Remark 3.9.6. It was shown by POL [345] that if X is a compact space which is (n + l)-dimensional then it contains an n-dimensional G^-set 5 such that dim 5°° = n. The methods used in this section show that this is true for the special case X = J n+1 .

3.10. THE ORIGINS OF DIMENSION THEORY

221

Exercises for §3.9.

1. Let X be a compact space such that dimX > n + l. Prove that X contains a totally disconnected G^-subset S such that dim S > n. 2. Let X and Y be totally disconnected. Prove that X x Y is totally disconnected. 3. Prove that there exists a countable dimensional infinite-dimensional topologically complete totally disconnected space. 3.10. The origins of dimension theory The first definition of a dimension function is due to BROUWER [77]. He seems to have been put on the right track by PoiNCARE [337, 338] who had proposed to call a continuum n-dimensional, if it can be dissected in separate pieces by one or more (n — l)-dimensional continua. Brouwer refined this idea and defined a dimensional invariant intended for topologically complete spaces without isolated points (in the terminology of his days, normal sets in the sense of Frechet) which he called Dimensions grad. There is no reason for the restriction to topologically complete spaces without isolated points, so we will state his definition for arbitrary spaces instead. Indeed, define the Dimensionsgrad Dg X of a space X, as follows: 0 < Dg.Y < n

<£>

Dg X = n Dg X = oo

<£> <&•

X is nonempty and for every pair A, B of disjoint closed subsets of X there exists a closed set T C X which is a cut between A and B such that Dg T is at most n — 1, DgX < n and Dg^Y is not smaller than n, Dg X ^ n for every n > — 1 .

So this invariant is very similar to the large inductive dimension, which was defined much later. It is easy to see that if X and Y are homeomorphic spaces then Dg X is equal to DgY. Hence Dg is a topological invariant, as are all dimension functions considered in this chapter. It is clear that DgX = 0 if and only if X ^ 0 and X contains no continuum of cardinality larger than 1 . Such spaces are called punctiform. There is something very peculiar about the definition of Dimensionsgrad, namely, a cut between two closed subsets of an arbitrary space is not necessarily a partition, even in compact spaces. In the sin (llx} -continuum in the plane {(x,sin(V,)) :0<x< i/J U {(0,y) : -1 < y < 1} the point (0,1) cuts between the points (Y^O) and (0, — 1). But the complement of (0, 1) is connected! (See also Exercise A. 10. 7.) This was pointed

222

3. BASIC DIMENSION THEORY

out to Brouwer by Urysohn. As a result of this, Brouwer issued a flood of notes and papers correcting his 'mistake'. For a description of the troubles this caused in the early days of dimension theory, see e.g., JOHNSON [213]. Lemma 3.10.1. For every space X we have Dg X < dim X. Proof. This is easy. If dim X — — 1 then X = 0 and so Dg X = — 1. Suppose that the inequality is true for all spaces X with dimX < n — 1, where n > 0. By the Coincidence Theorem 3.4.10 it follows that for every pair of disjoint closed sets A, B C X there exists a partition T between A and B such that dimT < n — 1. By our inductive assumption, DgT < dimT < n ~ 1. Since a partition is a cut, it follows that DgJ\T < n, as required. D The proof of Lemma 3.10.1 and Urysohn's Example suggest the question whether every cut is a partition in some well-behaved class of spaces. Lemma 3.10.2. Let X be a space which is both tope-logically complete and locally connected. If A and B are closed subsets of X and T cuts between A and B then T is a partition between A and B. Proof. Let U be the family of all components of X \ T. Since X is locally connected, every member of U is open (Exercise A.2.8). Assume that some U G U intersects both A and B. Then U is a connected, locally connected and topologically complete space (Theorem A.6.3). But then by the Mazurkiewicz Theorem 1.5.22 it follows that there is an imbedding $: I —> U such that $[1] meets both A and B. But this contradicts the fact that T cuts between A and B. Now let U be the union of all elements of U that meet A and let V be the union of all the other elements of U. Then U is an open neighborhood of A, V is an open neighborhood of B and U U V = X \T. Hence T is a partition between A and B. D So from Lemma 3.10.2 one would hope to be able to prove that Dg and dim take the same values on locally connected, topologically complete spaces. This was indeed widely believed to be true for a long time. In the book by HUREWICZ and WALLMAN [208], one can read on Page 4 that Dimensionsgrad and dimension agree even in the class of all locally connected spaces. A proof of this assertion was not presented. Similar statements for various classes of locally connected spaces were subsequently made in various other books on dimension theory. None of these claims was supported by a proof. That there are troubles, even for compact locally connected spaces, is easily demonstrated. Let X be a compact space of Dimensionsgrad 0. Then it contains no nontrivial continuum, and hence is zero-dimensional (Corollary 3.7.9). This triviality will be used in the proof of Theorem 3.10.3 below. If moreover X is locally connected and DgX < 1 then by Lemma 3.10.2 every two disjoint closed sets can be partitioned by a zero-dimensional set,

3.10. THE ORIGINS OF DIMENSION THEORY

223

i.e., the large inductive dimension of X is at most 1. But if X is compact and locally connected and has Dimensionsgrad at most 2 then all there can be concluded from the above is that every pair of disjoint closed subsets can be partitioned by a closed subset with Dimensionsgrad at most 1. But that closed set does not need to be locally connected, so we are at a dead end. Brouwer always claimed that his Dimensionsgrad could serve as the basis of an equally important theory of dimension as ordinary dimension. But one would guess from the above that for a sufficiently 'bad' compact space dim and Dg can differ. The following surprising result shows that this is not true and that Brouwer, as usual, was right after all with his claim at least for compact spaces. Theorem 3.10.3. Let X be a compact space. Then dimX = DgX. Proof. By Lemma 3.10.1 we only need to prove that dim^C < DgX for all compact spaces X. If DgX = 0 then by the above remarks we have that dim X = 0. So now suppose that for every compact space X such that 0

we have dim X < Dg X.

Assume that X is a compact space such that Dg X = n. Let 13 be a countable base for X. For every pair (So, BI) of elements of 23 such that BQ C B\ there is a cut between BQ and X \ BI with Dimensionsgrad at most n — I . So by our inductive assumption, every such cut has dimension at most n — 1. By the Countable Closed Sum Theorem 3.2.8, the union of these cuts is at most (n — l)-dimensional. By Corollary 3.3.12 this union is contained in an at most (n — 1) -dimensional G$ -subset 5 of X. Then T = X \ S is an Fa -subset of X. We claim that T contains no nontrivial continuum. Striving for a contradiction, assume that C is a nontrivial continuum contained in T, and pick arbitrary distinct points a, b £ C. There exist elements BQ,BI G "B such that a e 5o C £0 C #! \ {b}. By construction, 5 contains a cut between BQ and X\Bi. But C is a continuum meeting both B0 and X \ BI but missing 5. This is a contradiction. So we conclude that T contains no nontrivial continua. But then every compact subset of T is zero-dimensional by the same reasons as before. So T is a countable union of compact zero-dimensional subspaces, and hence is zerodimensional itself (again by the Countable Closed Sum Theorem 3.2.8). So by Theorem 3.4.4 we conclude that dim X = dim(S U T) < dim S + dim T + l = ( n - l ) + 0 + l = n, as required.

D

224

3. BASIC DIMENSION THEORY

It is clear that there are spaces X such that DgX ^ dimX. For let A" be a totally disconnected n-dimensional space (Theorem 3.9.3). Then X is punctiform and so DgX = 0. The gap between Dimensionsgrad and dimension can therefore be arbitrarily large. In view of the claim in the book by HUREWICZ and WALLMAN [208], the question naturally arises whether there are similar examples within the class of locally connected spaces. Observe that Lemma 3.10.2 draws specific attention to the class of all topologically complete locally connected spaces, for there the notions of a cut and a partition coincide. C

00,01,10,11

0,1

Figure 10. Theorem 3.10.4. For each n — 2 , 3 , . . . , oo there exists a locally connected topologically complete space Xn such that DgXn = 1 < dimXn = n. Proof. Represent the set S, the set consisting of all finite sequences consisting of O's and 1's (including the empty sequence), by dots in the plane as in Figure 10. These dots are compactified in a natural way by a copy C of the Cantor set. We now connect every dot with its two 'successor' dots by an interval (see Figure 11). In this way we obtain a connected space which we call T. By U we denote one of the basic neighborhoods of a point of C. The set U is clearly connected, so T is locally connected. But it has a stronger property which we will now describe. Claim 1. For an arbitrary subset CQ C C the space TQ U CQ is locally connected. Proof. In fact, a basic neighborhood of an arbitrary point c G CQ C TO U CQ has the form

(C/nTo)u([/nC 0 ). This set is connected, since it contains the dense connected set U n TQ.

<0>

3.10. THE ORIGINS OF DIMENSION THEORY

225

Now let / : C —> Q be a continuous surjection from C onto the Hilbert cube Q (Theorem 1.5.10). Let Z = T(Jf Q and let g: T ->• Z be the quotient map. So Z is the disjoint union of TO and the Hilbert cube Q. For every n let Yn C Q be an n-dimensional totally disconnected Gs-set (Theorem 3.9.3) and let Xn = T0UYn. Claim 2. Xn is a topologically complete space. Proof. This is clear since the complement of Xn in Q is an TV -subset of Q by Theorem A. 6. 3. So the complement of Xn in the compact space Z is an TV -subset of Z as well. Claim 3. Xn is locally connected. Proof. In fact, Xn = /[/^[Xn]]. The set f ~ 1 [ X n ] is locally connected by Claim 1. Hence Xn is locally connected by Exercise A. 2. 11. <0>

C

Figure 11.

Claim 4. dimXn = n. Proof. It is clear that Yn is a closed n-dimensional subset of Xn. So Xn is at least n-dimensional by the Subspace Theorem 3.2.9. On the other hand, Xn \ Yn is a countable union of one-dimensional compacta. Hence dimXn < n by the Countable Closed Sum Theorem 3.2.8. 0

Claim 5. DgXn = 1. Proof. Let A and B be disjoint closed subsets of Xn. Take an arbitrary partition So in Yn between A n Yn and B n Yn. Corollary 3.3.10 gives us a partition S in Xn between A and B such that SC\Yn C S0 and dim(S\S 0 ) < 0.

226

3. BASIC DIMENSION THEORY

We claim that S is totally disconnected. So we must show that 0 is a partition between any pair of distinct points a and 6 of 5. Since S \ Yn is an open zero-dimensional subspace of S, a moments reflection shows that we may assume without loss of generality that both a and b belong to the totally disconnected closed space Yn of Xn. So 0 is a partition between a and b in 5 n Yn. By another appeal to Corollary 3.3.10 it follows that 0 can be extended to a partition between a and b in S such that the dimension of its intersection with S \ Yn is — 1. We conclude that 0 is a partition in S between a and b. Since a totally disconnected space is punctiform, it has Dimensionsgrad 0. We conclude that DgXn < I . That is clear since Xn contains copies of the unit interval.

0

This concludes the proof of the theorem.

D

Let us remark that there also is an infinite-dimensional locally connected topologically complete space with Dimensionsgrad 1. It suffices to take the topological sum of the spaces Xn in Theorem 3.10.4. This space is countable dimensional. There is also a strongly infinite-dimensional example since there is a strongly infinite-dimensional totally disconnected G<$-subset in Q (Theorem 3.9.3).

Exercises for §3.10.

1. Let X be a cr-compact space. Prove that Dg X = dimA^. ^•2. Let A be a space which is both topologically complete and locally connected. Suppose that A and B are disjoint closed subsets of A. Prove that every partition S between A and B contains an irreducible partition T between A and B. ^•3. Let A be a unicoherent Peano continuum containing two disjoint continua A and B. Prove that if S is an irreducible partition between A and B then S is connected. 4. Show that the fact that A and B are continua in the previous exercise is essential. ^•5. Prove that Sn for n > 2 cannot be imbedded in a product of the form ^\ 1 X

' * *

X

Y\ -ft ,

such that every Xi is at most one-dimensional. 6. Prove that the local connectivity assumption is essential in Exercise 3.10.2, even for compact spaces.

3.11. THE DIMENSIONAL KERNEL OF A SPACE

227

3.11. The dimensional kernel of a space For a space X and a point x 6 X, indx X denotes the dimension of X at the point x, and is defined as follows: mdxX = -l 0 < mdx X < n

<£> <£>

mdx X = n indj; X = oo

<£> <£>

X = 0, for every closed subset A of X not containing x there exists a partition L between {x} and A such that dimL < n — 1, indx X < n and ind z X ^ n — 1 , indx X ^ n for every n > — 1 .

Observe that dim X = ind X < n if and only if indx X < n for every x 6 X. If X is n-dimensional then its dimensional kernel A.(X) is the set {x £ X : indxX = n}. Lemma 3.11.1. Let X be an n-dimensional space, n>l. (1) The dimensional kernel of X is an Fff -subset of X. (2) The dimensional kernel of X has dimension > n — 1. Proof. Let E denote the complement of the dimensional kernel of X. For every i e N and for every x 6 E there is an open neighborhood Ulx of x such that indFr(C7*) < n - 2, diam(L^) < i/iThe cover {[/.£ : x G E} of E has a countable subcover, say Uz (Corollary A. 2. 3); put Ui = \JUi. Then Ui is open, and oo

E C U = p| t/i. i=l

We claim that L7" C E as well. Pick arbitrary x € C/ and e > 0. Let z be so large that l/i < e. Let V G Ui be such that x G F. Then diam(V ) = diam V < l/i < e which implies that V C B(x,e). Since indFr(y) < n — 2 and e > 0 was arbitrarily chosen, this implies that ind^ X < n — 1, i.e., x & E. We conclude that U C E and hence that E = U. As a consequence, £" is a G^-subset of JC, which proves (1).

For (2), let CO

F=U U Then F is the union of countably many closed subsets of X with dimension at most n — 2. By the above, the collection U = U^i ^i '1S a local base at every point of E (cf. Exercise A.2.15). So it follows that the collection U\(E\F]

228

3. BASIC DIMENSION THEORY

consists of open and closed subsets of E \ F and is in E \ F a local base at any point. From this we see that ind(£' \ F) < 0. Now let K = X \E be the dimensional kernel of A", and suppose that ind K < n — 2. We will derive a contradiction. By (1) and the above, K U F is an F0.-subset of X and by Theorems 3.4.10 and 3.2.8 it follows that md(K\JF) < n - 2 . But then by Theorem 3.4.4, ind X = ind ((K U F) U (E \ F ) ) < n - 2 + 0 + 1 = n - 1,

which is a contradiction.

D

For compact spaces, we can do a little better. Theorem 3.11.2. Let X be an n-dimensional compact space. Then the dimensional kernel of M is n-dimensional and a-compact. Proof. The dimensional kernel is cr-compact by Lemma 3.11.1(1). To prove that it is n-dimensional, first observe that if n = 0 then there is nothing to prove. So let n > 1. By Theorem 3.7.8, X contains an n-dimensional Cantor-manifold Y. We claim that Y is contained in the dimensional kernel of X. If so then we are done. To this end, let y G Y be arbitrary. Since dimF > 1 there exists a point z G Y \ {y}. If indy X < n then there is a partition F between {y} and {z} such that dimF < n — 2. But then F D Y separates Y and is at most (n — 2)-dimensional, which is a contradiction since Y is an n-dimensional Cantor-manifold. D Let X be zero-dimensional, and let x E X. It is clear that indx X — 0. So the dimensional kernel of X is X itself. This explains why in the following definition we require X to be at least one-dimensional. A space is called weakly n-dimensional, where n > 1, if it is n-dimensional, but its dimensional kernel is of dimension n — l. By Theorem 3.11.2, no compact subspace of a weakly n-dimensional space is n-dimensional. It is not clear at all whether weakly n-dimensional spaces exist. But in Theorem 3.11.8 we will construct for every n > 1 such a space. We have to so some preparatory work first. Lemma 3.11.3. Let X be weakly n-dimensional with subspace F. Then F is at most (n — I)-dimensional or weakly n-dimensional. Proof. Assume that dim F = n. We claim that if x belongs to the dimensional kernel of F then x belongs to the dimensional kernel of X. For assume that x £ F does not belong to the dimensional kernel of X. Then x has arbitrarily small neighborhoods U in X such that indFrt/ < n — 2. By Lemma 3.4.3 and Exercise A.1.16 it follows that the boundary of U n F in F

3.11. THE DIMENSIONAL KERNEL OF A SPACE

229

has dimension at most n — 2. So this shows that x does not belong to the dimensional kernel of F. So the dimensional kernel of F is a subspace of the dimensional kernel of X and hence is at most (n — l)-dimensional by the Subspace Theorem 3.2.9 (or by Lemma 3.4.3). D Lemma 3.11.4. Let n > 1. In addition, let X be compact such that there exists a continuous function f : X —t I such that for every q € Q D (0,1) we have dim f ~ l ( q ] < n. Then there is a G5-subset S C (0,1) with Qn(0,1) C S such that dim/" 1 (z) < n for every z € S. Proof. By the Countable Closed Sum Theorem 3.2.8 it follows that

U r'fo)

g€Q

is at most n-dimensional and hence can be enlarged to a G^-subset S of X with dim S < n (Corollary 3.3.12). Then T = f[X \ S] is by compactness of X an FCT-subset of I which misses Q. It is clear that for every z in the GS~ subset I \ T of I we have that f ~ l ( z } C 5. Hence f ~ l ( z ] is at most ndimensional. D Lemma 3.11.5. Let K be an (n + I)-dimensional compact space with disjoint closed subsets A and B. There exist a continuous function f : K —> I and a Cantor set A in [i/3, 2/3] such that f[A] C {0}, f[B] C {1} and dim ( f ~ l ( t ) ) < n for each t e A. Proof. Let a: K —>• I be a Urysohn function such that

a[A] C {0},

a[B] C {1}

(Lemma A.4.1). Identify I and the set I x {0} x ••• x {0} ^m^Mmv^B^H*. jtm^—^^^^^mi/

2n+2 2n

3

in R + . By Remark 3.3.6 there is an imbedding g : K -> Wn+i such that g ( a , g ) < Ys- It follows that we may think of K as subspace of yin+i having the additional property that Ki[A] C (—00, Ys) and ^2 [-6] C ( 2 / 3 , oo). Now consider an arbitrary element q G Q n [V3, 2/3]. The set 5 = {x € M 2n+3 : xl = q} can be identified with M 2n+2 , and the set K n S with a subspace of {x G M 2n+2 : at most n coordinates of x are rational}. But this implies that dim(K D 5) < n by Corollary 3.2.11. So the function £ = 7T! \K: K ->R

has the property that for every q £ Qn [Y3, 2/3] we have dim^" 1 ^) < n. By Lemma 3.11.4 it follows that we may enlarge Qn[Y3, 2/3] to a G^-subset S having the same property. Observe that by the Baire Category Theorem A.6.6

230

3. BASIC DIMENSION THEORY

and the fact that Q H [Y3, 2/3] is dense in [1/3, 2/3], it follows that S is uncountable. So 5 contains a Cantor set A by Theorem 1.5.12, and all we need to do is to adjust £ slightly so as to satisfy the additional property that it sends A to 0 and B to 1. This adjustment is left as Exercise 3.11.1 to the reader. D Lemma 3.11.6. If X is an n- dimensional space, where n > 0, then there exists a zero- dimensional F^-subset N C X such that dim(A" \ N) < n — 1. Proof. If dim X = 0 then N = X is as required. Suppose therefore that the lemma is true for all spaces Z with dim Z < n — 1, where n > I. Assume that dim A = n. Let $ be a countable base for X such that dimFr(-B) < n — 1 for every B € £ (Lemma 3.4.1). Put

Y = |J Then Y is an Fa-subset of X since every Fr(B) is closed in A, and so Y is at most (n — l)-dimensional by the Countable Closed Sum Theorem 3.2.8. In addition, dim(X \ Y} < 0 since T> \ X \ Y is a clopen base for A \ Y . By our inductive assumption there exists a zero-dimensional F^-subset N C Y such that dim(y \ N) < n — 2. Observe that N is an F^-subset of A as well. Since dim(A \ F) < 0, it follows by Theorem 3.4.4 that dim(A \ N) < dim(Y \N} + dim(A \ Y) + 1 < (n - 2) + 0 + 1

= n — 1, as required.

D

Lemma 3.11.7. Let X and Y be compact spaces, and let f : X —>• Y be a continuous surjection. IfY is zero-dimensional at y £ Y and if f ~ l ( y ) is zero- dimensional then X is zero- dimensional at every point of f ~ l ( y ) . Proof. Pick an arbitrary x £ f ~ l ( y ) , and let U be a neighborhood of x in X. Since dimf^1(y) = 0 there is a clopen subset C of f ~ l ( y ] such that x G C C U . Pick disjoint open subsets E and F in A such that

It is clear that without loss of generality we may assume that ECU. Observe that E U F is a neighborhood of f ~ l ( y ) in A. Consequently, since / is a closed map, there is a neighborhood V of y in Y such that f ~ l [ V ] C E U F (Exercise A. 1.1 5). Since Y is zero-dimensional at y, we may assume without loss of generality that V is clopen. Then f~l[V]r\E is a clopen neighborhood of x in A which is contained in U. D We now come to the main result in this section.

3.11. THE DIMENSIONAL KERNEL OF A SPACE

231

Theorem 3.11.8. Let n > 1 and Jet K be an (n + l)-dimensionaJ compactum. Then K contains a weakly n-dimensional G§ -subset X (hence X is topologically complete). Proof. Let r = {(A0, Z?0), . . . , (A n ,5 n )} be an essential family in K. By Lemma 3.11.5 there exist a continuous function /: K —>• I and a Cantor set A C (0, 1) such that f[A0] = {0}, f[B0] = {1} and dim ( f ~ l ( t ) ) < n for every t G A. We now closely follow the construction in Theorem 3.9.3. Put C = {C 6 2K : C is a continuum from A0 to B0}. Then C is closed in 2K and hence is a compact space (see the proof of Claim 1 of Theorem 3.9.3). There consequently is a continuous surjection 0: A —>• C (Theorem 1.5.10). Put

Then Z is compact, and f[Z] = A (see the proof of Claim 2 of Theorem 3.9.3). Now dim Z > n because Z intersects every continuum from AQ to BQ (Corollary 3.7.5). In addition, the fibers of g = f \ Z are at most n-dimensional and A is zero- dimensional. We conclude that dimZ < n by Theorem 3.6.10. Now let A denote the dimensional kernel of Z. Then because Z is compact, A is an n-dimensional (T-compact subset of Z (Theorem 3.11.2). There consequently exists a a-compact zero-dimensional subset N of A such that dim(A\AT) < n- 1 (Lemma 3.11.6). Now put X = Z\N. Then X is clearly a G<$-subset of K and we first claim that it is n-dimensional. Let C e 6. We will prove that C meets X. Pick t € A such that (£) = C. Observe that f ~ l ( t ) nC C Z. If dimf~l(t} n C > 0 then f ~ l ( t } n c intersects X because the complement of X in Z is zero-dimensional. We may therefore assume that dimf~l(t) n C = 0. But since A is zero-dimensional, we get that Z is zero-dimensional at all points of f ~ l ( t ] nC (Lemma 3.11.7). Consequently, /-1(*) n C n A = 0, i.e., 0 ^ f ~ l ( t ) n C C X. So we conclude from Corollary 3.7.5, that dimX > n. However, because X is a subspace of Z, we also have dimX < n. Consequently, dimX — n, as required. We finally claim that X is weakly n-dimensional. This is however a triviality (cf. the proof of Lemma 3.11.3). Simply observe that if x is a point of X at which X is n-dimensional, then Z is n-dimensional at x, which implies that x E X n A. Since by construction, dim(X n A) < n — 1, we are done. D Remark 3.11.9. If one takes K = In+l in the above theorem then there is no need to use Lemma 3.11.5 in the proof because we can use for / the projection onto the first coordinate. It seems that this gives the easiest known examples of weakly n-dimensional spaces.

232

3. BASIC DIMENSION THEORY

A universal weakly n-dimensional space. In §3.3 we showed that for every n there is a universal space for n-dimensional spaces. That is, there is an n-dimensional space X which contains a homeomorphic copy of any n-dimensional space. We will prove now that a similar result holds for weakly n-dimensional spaces. Theorem 3.11.10. For each n > 1 there exists a weakly n-dimensional space E such that any weakly n-dimensional space imbeds in E. Proof. Let p: (0, 1}°° x Q ->• {0, 1}°° be the projection. It will be convenient in this proof to denote the hyperspace of Q, i.e., the space 2^, by 3C. For a fixed natural number n, let JVC be the subspace of the product X x OC°° x (DC x K)00, consisting of all elements of the form (1)

(K^Lfr^dtDfr)

satisfying the following conditions: (2)

LiCL2C---CK,

dimLi
(i e N),

(3)

K = Cl U Di:

dim(Cl n Dx] < n - 2

(z e N).

There are a subspace T of {0, 1}°° and a continuous mapping (p from T onto the space M (use Exercise 1.5.10 and Corollary 1.5.7). It will be convenient to denote (p as follows:

(4)

t ^ ( K ( t ) , (1^)^(0^}, n^t}):).

For any t € T, let

(5)

H(t) = {z € K(t) \ U~! Li(t) : for any neighborhood U of z in Q there is an 2 e N with 2 e C;£ C £7 and 2 £

For * e r, put

(6) i—l

Finally, let E C {0, 1}°° x Q be the subspace of all points of the form (t, z), where t e T and z £ E(t). We shall show that (7)

E is weakly n-dimensional,

and

(8)

any weakly n-dimensional space imbeds in E.

3.11. THE DIMENSIONAL KERNEL OF A SPACE

233

To this end let, us consider the sets K*={(t,z):t£T,ze #(*)},

LI = {(t,z}:t(ET,ze

Claim 1. For every i 6 N we have (9)

dim LI < n - 1, EC C* U £>*,

dim(Cf n £>*) < n - 2.

Proof. It follows from Exercise 1.11.26 that the projection TT = p \L* is closed. Observe that the fibers of TT are of the form

Hence by (2) the fibers of TT are at most (n — l)-dimensional. Since dim T = 0, the desired result follows from Theorem 3.6.10. By (2) and (3) the other statements follow by similar considerations. •(> By (9) and Theorem 3.2.8, dim(|J2i Ll )
(10)

E\\jL*CE(n_1).

This implies among other things that the dimension of the set E \ -E( n _i) is at most n — I . So if dim.E > n then we contradict Lemma 3.11.1. This implies that dimE < n and also by the same argument that E is weakly n-dimensional provided that E is n-dimensional. The latter fact will follow once we established (8). Simply observe that there are weakly n-dimensional spaces (Theorem 3.11.8) which imbed into E by (8). Hence E is at least n-dimensional since it contains n-dimensional subsets (Theorem 3.2.9). Proof. Let us consider any point a ~ (i, z] from the set on the left hand side of the inclusion (10), i.e., z G H(t) by (6). Let W be a neighborhood of a in {0,1}°° x Q, and let

U = {xeQ: ( t , x ) £W}. Then U is a neighborhood of z\ let i 6 N be an index given by (5). The projection p[C* \ W] is closed in T (cf. the proof of Claim 1). It also misses t since Ci(t) C U. Let V be a clopen neighborhood of t in T, disjoint from p[C* \ W]. Then C? np~l[V]CW and therefore

is an (n — 2)-dimensional partition between the point a and the set E \ W , cf. (9). 0

234

3. BASIC DIMENSION THEORY

Let us consider now an arbitrary weakly n-dimensional space X. We shall find an element t E T such that X embeds in E(t). The dimensional kernel of X is an Jf^-set in X (Lemma 3.11.1). Hence there exist closed sets Zi C Z2 C • • •

in X such that dimZi < n — 1 for every i and X \ U^i ^ f= -^"(n-i)Corollary 3.5.6 and the fact that every compact space imbeds in Q (Corollary A.4.4), provide an imbedding h: X —>• Q such that dim/?,[Zj] < n — 1, for i = 1, 2 , . . . , and h(x) G (h[X])(n_ij for any x G X \ (J^i ^- Let

K = h[X],

(11)

Ll =

Then oo

(12)

h[X]\\jLlCK(n_1}. i=l

By (12) there exist pairs (Ci,Di), i G N, of compact sets in K, with K = d\JDi,

dim(ClHDi)
such that for any z G /&LX"] \ Ui^i ^ an<^ any neighborhood C/ of 2 in Q, there is an i € N with 2 G C; C C7 and z $ Di. The sets K and I/; in (11), together with the pairs (Ci,Di) determine a point in the set M, cf. (1), (2) and (3). So there exists t G T such that (13)

K

Let us consider the section 2£(£) of the set .E, cf. (6). According to formula (5),

and hence by (13), h[X] C E(t). This completes the proof of (8) and ends the proof of the theorem.

D

Dimension of products of weakly one-dimensional spaces. In his paper [284], MENGER asked whether there exists a weakly one-dimensional space X such that dimXn — n for every n. We will prove here among other things that the product of an arbitrary family weakly one-dimensional spaces is at most one-dimensional, thereby giving a strong negative answer to MENGER'S question. Theorem 3.11.11. If X is weakly one-dimensional then X has a compactification 'jX in which it is L-imbedded. Proof. We put K = A.(X). In addition, let C be a countable collection clopen subsets of X having the property that it is a local base at every point ofX\K (Exercise A.2. 12).

3.11. THE DIMENSIONAL KERNEL OF A SPACE

235

There is a compactification ^X of X such that for every C G C we have that C and X\C have disjoint closures in ^X (Exercise A. 9. 3). That is, the closure of C in ^X is clopen in jX. We claim that X is L-imbedded in ~fX. To this end, let U be a finite open cover of X. For every point x G X\ K, pick elements Cx G C and Ux G U such that (1)

X 6 C X CCX CUX.

Put A = X \ \JX£x\i
= U Cx u U WA-

E

Then E is an open neighborhood of X in *yX. Let Z C E be an arbitrary continuum. If Z fl Cx ^ 0 for some x E X \ K then Z C Cz C Ux by connectivity of Z and (1). So we may assume without loss of generality that Z n Cx — 0 for every x E X \ K. But then

zc\JwA so that by (3) and the connectivity of Z there is an A G A such that Z C WA • So this shows that Z C UAd Corollary 3.11.12. The product of a finite family of weakly one-dimensional spaces is one-dimensional. Proof. Such a product if L-imbedded in one of its compactifications by Exercise 3.5.6 and Theorem 3.11.11. So the desired result follows by an application of Corollary 3.5.9. D In the light of this result, the question arises whether the dimension of products of weakly n-dimensional spaces can be estimated more accurately

236

3. BASIC DIMENSION THEORY

than the bounds we get from Theorem 3.4.12. It is unclear whether this is possible. Exercises for §3.11. 1. Let X be a space and let A and B be disjoint closed subsets of X. In addition, let /: X —> I be continuous such that /[A] C [0, V 3 ), g [ B ] C ( % , ! } . Prove that there is a continuous function /: X —>• I having the following properties: (1) /(x) = / ( x ) i f x € / - 1 [ [ 1 / 3 , 2 / 3 ] ] ,

(2) f[A] C {0} and f[B] C {!}. 2. For every n € N let Xn be weakly one-dimensional. Prove that O^Li Xn is one-dimensional.

We now present a different proof of the fact that the product of two weakly one-dimensional spaces is one-dimensional. Let X and Y be weakly one-dimensional. Pick an arbitrary point (x, y) G (X x Y) \ (A(AT) x A.(Y)). We claim that X x Y is at most one-dimensional at ( x , y ) . We first show that this suffices. Striving for a contradiction, assume that X x Y is two-dimensional. Then by our claim, A(AT x Y) C A(X) x A(y) hence dim (A.(X) x A(F)) < 0. This contradicts the fact that A.(X x F) is at least one-dimensional (Lemma 3.11.1(2)). We assume without loss of generality that y £ A(F). If in addition x £ A.(X) then X x Y is zero-dimensional at (x,y). So we assume further that x G A(AT). Let U and V be arbitrary open subsets of X and Y, respectively, such that x & U and y G F. We will construct an open subset E1 C X x F such that (x,y)£ ECU xV 1

while moreover indFrE < 0. Since y 0 A(Y) we may assume without loss of generality that V is clopen. Let U' be an open neighborhood of x such that C^CC/,

FrU' C X\\(X).

It is possible to pick U' since ind (A" \ A(A)) < 0 (Corollary 3.1.6). If dim V < 0 then Fr(U' xV}= Fr(U") x V is at most zero-dimensional and so we are done. Assume therefore that dim V = 1. Then V is weakly one-dimensional (Lemma 3.11.3), and so we may assume without loss of generality that V = Y. Put A = FrU' and let U" be an open subset of X such that A C U" C U" C U. 3. Prove that for every n € N there exist pairwise disjoint clopen subsets Uin, Uzn,... of X such that (1) Uin n A ^ 0 for every i,

3.12. COLORINGS OF MAPS

237

(2) diam Uin < l/n f°r every z, (3) A C U^! Uin C (J* i tt« C tf", (4) Fr(U~ i tfi«) C A(X). Now for every n e N let £ n denote the collection of all clopen subsets of Y of diameter at most l/n and put

Observe that if b e Y is such that for every n there exists a clopen subset Cn of Y containing 6 and of diameter at most l/n then Y is zero-dimensional at b. So if 6 € A(Y) then there exists n e N such that b £ Bn. 4. Prove that for every n there is a decreasing sequence Vin , i G N, of clopen subsets of Y such that H^i ^« = ^«5. Let 00

w

00

= U LK^"*^")n=l i=l

Prove that A x A(Y) C W. 6. Prove that FrW C (A(X) x A(Y)) U (A x Y). Put

E = W\J(U' x Y). Then E is an open neighborhood of ( x , y ) and by Exercise 3.11.2(3) it follows that 'ECU xY.

We claim that Fr E is zero-dimensional. First observe that (*)

FrE C FrWU(FrU' x Y) = FrW(J (A x Y).

Put Bo = Fr £ n (A x Y) and BI = Fr£ \ S0, respectively. Since A x A(Y) C W (Exercise 3.12.4) and Fr E n W = 0, it follows that B0CAx ( Y \ A ( Y ) ) , which is zero-dimensional. We conclude that BQ is a zero-dimensional closed subspace of Fr E. In addition, Exercise 3.12.5 and (*) imply that Bi = Fr E \ Bo C Fr E \ (A x Y) C Fr W \ (A x Y) C A(X) x A(Y),

which is also zero-dimensional. So B\ is zero-dimensional as well. Since BQ is closed, the Countable Closed Sum Theorem 3.2.8 now easily gives us that dimE < 0, as desired. 3.12. Colorings of maps Let X be a space and let /: X —>• X be a fixed-point free continuous function. On Page 188 we defined / to be colorable if there is a finite closed cover A of X such that A n f[A] — 0 for every A € A. Since finite open covers can be shrunk to closed covers, and finite closed covers can be swelled to open covers, finite open covers do equally well (see §3.2).

238

3. BASIC DIMENSION THEORY

There are two natural questions we consider here. First of all the question whether every fixed-point free continuous function can be colored. And secondly, whether something can be said about the number of colors needed for a colorable function. Negative results.

Theorem 3.12.1. There is a space X with a fixed-point free involution which admits no coloring. Proof. If an: S n —>• Bn is the antipodal map, then every closed cover 3~ of S n such that F n ct[F] = 0 for every F G 3~ has at least n + 2 elements by Theorem 2.5.1. Let X be the topological sum of the S™ (n e N), and let h be the involution on X which is defined by the requirement that h f S n = an for every n. Now assume that there is a finite closed cover 3 such that F n h[F] = 0 for every F <E 3~. Pick m so large that |3~| - 2 < m. Then 9 = 5 \ §m is a closed cover of STO no element of which contains an antipodal pair. This implies that m + 2 < |S| < |3*|, which is a contradiction. D Observe that the space X in this theorem is infinite-dimensional. It is therefore natural to ask whether there is also a finite dimensional example. The answer to this question is quite interesting. For continuous functions there is such an example, but for homeomorphisms this is not true. Before we can answer it, we need to make a very interesting detour which will lead us via topological Ramsey theory to the desired result. We consider the Cantor set {0,1}°° with its usual Tychonoff product topology. As was remarked on Page 458, the function /: J'(N) -> {0,1}°° defined by f ( A ) = XA, the characteristic function of A, is a bijection. The product topology on {0,1}°° therefore induces a topology on T(N). With this topology, !P(N) is homeomorphic to {0,1}00 and hence is a Cantor set. The collection of all sets of the form ( N ) : (a C A] A (A n T = 0)}, where a and T are arbitrary finite subsets of N, is a clopen base for J"(N). For this simply observe that if a and r are disjoint finite subsets of N then the basic open subset {x G {0,1}°° :(ie0=> Xi = l ) and (i e T => xx = 0)} of {0,1}°° corresponds under this function to (cr, T). So the finite subsets of N form a countable dense subset of T(N). The complement, i.e., the collection of all infinite subsets of N, is a G^-subset which is also dense by the Baire Category Theorem A.6.6.

3.12. COLORINGS OF MAPS

239

If H C N is infinite then [H]u denotes the collection of all infinite subsets ofH. The usual topology on [N]w is the subspace topology that [N]w inherits from 5)(N) (with the just defined topology). If A, B C N then we write A < B provided that max(A) < min(B}. So this makes sense only if A is finite. For a < A, where a C N is (necessarily) finite and A C N is infinite, let (*)

[a,A] = {Se [Nf

:aCSCa(JA}.

u

Observe that [0, A] is simply [A] . The Ellentuck topology on [N]w has as basic open sets all sets of the form (*). It is easy to verify that they satisfy the axioms for a base for a topology. There is a problem though with the Ellentuck topology since it does not have a countable base. So it violates our assumption that all topological spaces are separable and metrizable. This causes no problem, since the Ellentuck topology is used merely as a tool here to prove interesting theorems on [N]w endowed with its usual topology, and the language of topology which we are used to is convenient for that. Lemma 3.12.2. If H C N is infinite then the subspace [H]w of [N|w with the Ellentuck topology is homeomorphic to [N\u. Proof. Let /: H —> N be any bijection. It is easy to see that the function F: y(H) ->> T(N) defined by F(A) = f[A]

is a homeomorphism.

D w

Lemma 3.12.3. The Ellentuck topology on [N] contains the usual topology on mu. Proof. It suffices to prove that if a and T are arbitrary finite subsets of N then

<<7,r)n[Nr is open in the Ellentuck topology. To this end, let A C N be infinite such that A e (cr, T}. Put B = A \
or

[0,A] C [N]w \X.

It is called completely Ramsey if for every o, A C N with a finite and A infinite and a < A there is an infinite set B C A with [a, B] C X

or

[a, B] C [N]w \ X.

240

3. BASIC DIMENSION THEORY

We introduce some additional notation. Let U C [N]w be open in the Ellentuck topology. Call [a, A] good if for some B C A we have [a, B] C U. If [a, A] is not good then it is called bad. We shall call [a, A] very bad if it is bad and for every n G A, [a U {n}, A/n] is bad, where A/n = {m 6 A : m > n}. Notice that if [a, A] is (very) bad and B C A is infinite then [a, B] is (very) bad. Lemma 3.12.4. If [a, A] is bad then there is an infinite subset B C A such that [a, B] is very bad. Proof. Indeed, suppose that this is not true. Then [a, A] is bad but not very bad and so there exists no 6 A such that [a U {no}, A/HQ] is good. There consequently exists an infinite B0 C A/HO such that [a U {no},#o] ^ ~U. Since [a, BQ] is not very bad, there exists ni G B such that [a U {HI}, BO/HI] is good. Observe that no < n\. So there exists an infinite B\ C BO/HI with [a U {ni},Bi] C £7, etc. Let B = {no,ni,...}. Pick an arbitrary element C e [a, 5]. Then a C (7 C a U B. Let n^ be the first integer in (C n B) \ a. Since [a U {r^}, £?j] C U and C G [a U {n^}, #;] we conclude that C 6 U. Since C was arbitrary, it therefore follows that [a,B] C [/, so [a, A] is good, which is a contradiction. D Proposition 3.12.5. Let U C [N]w be open in the Ellentuck topology. Then U is completely Ramsey. Proof. Suppose that [a, A] is given. We may assume without loss of generality that it is bad. Claim 1. There is a decreasing sequence A D B0 D BI D • • • ,

with Hi — min(.B;) strictly increasing, such that for any b C {no,... , n^_i} we have that [a U b, Bi] is very bad. Proof. By Lemma 3.12.4 there exists an infinite BQ C A such that [a,-Bo] is very bad. Let n0 = min(E). Since [a U {no},-B/{no}] is bad, again by Lemma 3.12.4 there exists an infinite BI C B/HQ such that [a U {no}, .Si] is very bad. Then [a, Si] is very bad since BI C B and [a U {no}, BI] is very bad by construction. Now assume that for i > 1

A ^B0 D BI D ••• D Bi, are as we want. So n;_i < ni = min(.B;) and if b C { n o , . . . , n;_i} is arbitrary then [a U 6, Bi] is very bad. So [a U b U {n;}, B/rii] is bad and we can use Lemma 3.12.4 repeatedly (finitely often) to get an infinite set Bi+\ C Bi such that for every b C {no,..., n^_i} we have that

[a\Jb(J{ni},Bi+i]

3.12. COLORINGS OF MAPS

is very bad. Then Bi+i is clearly as required.

241

<0

Let B = {n0, rai,... }. We claim that [a, B] C [N]w \ U. If not then [a, B] H £7 / 0

and so since C7 is open there is [a', B'] C [a,B] such that [a',B'] C U. By Exercise 3.12.1 for some i, a' — aUb with b C { n o , . . . , nx} and B'/rii C Bi/rii. So since [a U 6, B'/Hi] C £/, we have that [a U 6, -Bz/n] is good. This is a contradiction. D Corollary 3.12.6. Let [N]w = P0 U • • • U Pk, where each Px is open in the natural topology. Then there are an infinite H C N and an i < k such that every infinite subset of H is contained in Pi. Proof. It is clear that we may restrict ourselves to the case that k = I (Lemma 3.12.2). Since PQ is open in the Ellentuck topology by Lemma 3.12.3 it is completely Ramsey by Proposition 3.12.5. So we are done. D We are now ready to present our second example of a map which cannot be colored. Theorem 3.12.7. There are a zero-dimensional space X and a fixed-point free continuous function f : X -+ X such that (1) f is closed, (2) / is finite-to-one, (3) / cannot be colored. Proof. Put A" = [N]w endowed with its natural topology. Define f : X — > X

by f ( A ) = A\{mmA} (AeX). It is clear that / is fixed-point free. We claim that it is continuous. To this end, let (An)n be a sequence in X converging to an element A 6 X. Let p = mmA,

a = {p},

/3 = {1, 2 , . . . ,p - 1},

respectively. Then A 6 (a, j3] and since An ->• A we may therefore assume without loss of generality that min An = minyl for all n. Now assume that A \ {min A} 6 (a, T)

for certain finite cr, T C N. Then if a' = a U {min A} and r' = T \ {min A}, it follows that A € U, where U — {cr', r'}. There exists N € N such that An € U for all n > N. But since minAn = min .4 it follows that a C An \ {mmAn} and T n (An \ {min An}} = 0. We conclude that f(An) € (a, T} for all n > N. To prove that / is closed, let A C X be closed, and assume that B g f[A]. Let p = min B. Observe that

242

3. BASIC DIMENSION THEORY

Since A is closed, for every i < p — I there exist finite sets a^Ti C N such that (**)

Bu{i}e(vl,Tl)C[N\"\A. {p}U(UfJ11

Put a = &i\{i}) and r = Uf=i Tii respectively, and let V — (a, r}. Then B 6 V . We claim that V n f[A] = 0. Striving for a contradiction, assume that there exists A G A such that A \ {min A} G F. Then since

a C A\ {min A} it follows that p € A \ {min A} and so min A < p. Put i — min A. Observe that Ti n (A\ {i}) = 0 and that i $ n. So A e {CT^T;} which contradicts (**). We already implicitly proved that / is finite-to-one. To show that it cannot be colored, let U be an arbitrary finite open cover of X. By Corollary 3.12.6 there are an element U 6 U and an infinite subset H C N such that [H]u C U. But then f ( H ] 6 U so that f[U] H t/ / 0, i.e., U does not color /. D Positive results. We will now present the main results on the existence of colorings. If /: X —> X is a homeomorphism then A C X is called /-invariant Theorem 3.12.8. Let X be finite dimensional, and let f be a fixed-point free homeomorphism. Then f can be colored. Proof. We will first prove that if dimJY" = 0 then / can be colored with at most three pairwise disjoint colors. By the remarks on Page 188 there exists an open cover U of X such that f [ U ] r \ U = 0 for all U e It. By Proposition 1.5.1 we may assume without loss of generality that U is a clopen partition. So U is countable and we may list it as {Un : n e N}, where U\ — 0. By induction on n we shall now construct a clopen partition {Vj71, V^1, V3n} of Un such that

a) i—i

i=l

for j < 3. Put Vj- — 0 for j < 3, and assume that the V^'s have been constructed for i < n and j; < 3. For j < 3 put

Claim 1.

3.12. COLORINGS OF MAPS

243

Proof. First observe that H 3 =i Ej ~ 0 from which it follows that both the sets /[f>)3:=1 Ej] and f~1[T\3-i Ej] are empty since / is one-to-one. So if the claim is not true then we may assume without loss of generality that e.g., But this is impossible since E\ n E<2 = 0. So the claim implies that is an open cover of Un , which consequently can be refined by a clopen partition (Proposition 1.5.1). So there exists a clopen partition n

T/ +l T/ 0/n+l 1 ' 1 i 3 V

V

n

+1\ /

of t/n+i such that

(2)

v/ +1 n(/- 1 [s j .]u/[£; j ]) = 0

for j < 3. Claim 2. The V^+^s are as required. Proof. If (1) fails for n + 1 then since it holds for n for some j < 3 either

which is impossible by (2), or

/[T/7+1] n V/+1 / 0 which is impossible since f[Un+i] n Un+i = 0, or

which is also impossible by (2).

<0

This completes the inductive construction. So Vj = Ui^i ^7 ^or J ^ 3 is the desired coloring of / with three colors. Now assume that dim X = n + 1 and that the theorem has been proved for all spaces of dimension at most n, where n > 0. Let D be a countable dense subset of X and let !B be a countable base for X such that dim Fr B < n for all B <E !B. Put 5 = £> U Uses Fr5 > and T=

\Jfn[S], nGZ

respectively. It is clear that T is a dense /-invariant _FCT-subset of X which by the Countable Closed Sum Theorem 3.2.8 has dimension at most n. In

244

3. BASIC DIMENSION THEORY

addition, Z = X \ T is also /-invariant and is of dimension at most 0 since Fr B n Z = 0 for all B € 33 (so 23 \ Z is a countable base for Z consisting of relatively clopen sets). There exists by our inductive assumption a finite cover U of T consisting of relatively open sets such that f[U] n U — 0 for all U 6 U. For every U € U let U' C X be open such that U' n T = U. Then /[£/'] H [/' = 0 since / is a homeomorphism and T is dense in X. We conclude that U' = [Ur : U £ U} is a finite collection of open subsets of X which 'colors' T. Now consider the zero-dimensional /-invariant set Z. We would like to apply the same technique as the one we just used. But Z need not be dense in X. But we know that a strong form of the theorem is true for zero-dimensional spaces, so we will proceed from there. There exist relatively open pairwise disjoint sets Vi, V% and V3 in Z that 'color' Z. Observe that if e.g., x 6 V\ then f(x) 6 Vz U V3. By Lemma A. 8.1 there are pairwise disjoint open sets V/, V2' and V3' in X such that VI n Z = Vi for z < 3. Now put V/' = V/ r\f~l[V^\J V3']. Then V/' is an open subset of X and clearly f[V"] D V/' = 0. We conclude by proving that V\ C V/'. Indeed, take an arbitrary element x E Vi. Then /(x) G V"2 U V3 from which it follows that x e f ~ l [ V 2 U V3] C f~l\yi U V3']. We conclude that x e V/'. In other words, V/' 'colors' all the points of Vi. Define the sets V2" and V3" similarly. So U U {V0", Vi", V3"} is the desired coloring of X. D Remark 3.12.9. Observe that the proof of Theorem 3.12.8 shows that if X is n-dimensional and / is a fixed-point free homeomorphism of X then / can be colored with at most 3n + 3 colors. The number of colors. It is a natural to try to reduce the number of colors needed to color a given function. This is sometimes possible and we will come to that after some preparatory work. Our main tool is the following theorem, which is interesting in its own right. Theorem 3.12.10. Let X be a space with dimX < n and let U be a finite open cover of X with |Uj > n + 3. In addition, let f : X —> X be a, homeomorphism. Then there is an open shrinking V = {Vu : U G U} ofU such that for any 3 C U with |J| = n + 3 we have C\u^(vu u f~^\Yu\) = 0It is tempting to think of the shrinking V in this result as a cover for which Jr[Vu} :UeU}
3.12. COLORINGS OF MAPS

245

In addition, let V = {{0,4}, {1,4}, {2,4}, {3,4}}. Then the collection W = [VUf-^V]

:V e V}

has precisely two elements, namely, the sets {0,1,4} and {2,3,4}. So the order of W is 1. But since P) W ^ 0, the cover V does not have the property stated in Theorem 3.12.10. Before presenting the proof of Theorem 3.12.10, we will first show how we can use it to reduce the number of colors of a homeomorphism. Theorem 3.12.12. Let X be a space with dim.Y < n, and let /: X —> X be a fixed-point free homeomorphism. If f is colorable, then it can be colored with n + 3 colors. Remark 3.12.13. The space X — {0, 1, 2} and the homeomorphism / = {(0,1), (1,2), (2,0)} of X show that Theorem 3.12.12 is sharp for n — 0. By results of STEINLEIN [380, 379] it follows that Theorem 3.12.12 is sharp for every n. Proof. Let k be the minimum cardinality of a coloring of /, and let be a coloring. Assume that k > n + 3. We will derive a contradiction. Observe that Ui / Uj if i ^ j. By Theorem 3.12.10 we may without loss of generality assume that for any 3~ C li of cardinality n + 3 we have

Since / is a homeomorphism, this means that for any 7 C U of cardinality n + 3 we have

(i) Let y = {FI, . . . , Ffc} be a closed shrinking of U (Proposition A. 7.1). For every i < k — 1, define

B, = F, u (Fk n f[x \ U,} n rl[x \ u,}) . We claim that £ = {£?i, . . . , Bk-i} is a coloring of /. To see that f[Bi] n B, - 0

observe that

/[Bi] - f[F,} u (f[Fk] n f2[x \ u,} n (X \ [/,)) . This gives us what we want since

Fx n /[F,] = 0,

F, n (X \ U,} = 0,

Fk n f[Fk] = 0.

246

3. BASIC DIMENSION THEORY

To see that !B covers, first note that \j*ll Fi C (J*!/ Bi. Hence it suffices to show that Fk is covered. To this end, pick an arbitrary x G Fk and consider the collection Since k — 1 > n + 2, (1) implies that there exists i < k — I such that But then x 6 Bi. This contradiction completes the proof of the theorem.

D

Before turning to the proof of Theorem 3.12.10, we make the following remarks. Recall that we wish to shrink an open cover U to an open cover V={Vu:U€ U} such that for any subfamily W C U of cardinality n + 3 we have

PI (uu A moments reflection shows that our task is to construct V in such a way that for any subfamily W of V of cardinality n + 3 and any partition 3" U S of W we have

Lemma 3.12.14. Let X be a space with dim .AT < n. Let Fi C Ui fori < m, with Fl closed and Ui open. In addition, Let Gj C U'- for j < k, with Gj closed and £/• open. I f m + k>n + 3 then there exist closed subsets

FI, . • . ,Fm,

GI, . . . , Gk

such that

(1) Fl C Ui for i < m, and G3 C C/j for j < k, (2) (J™! Ft C U^ x F,, and \Jkj=1 G3 C \Jk.=1 Gjt

(3) n^i^nn* =1 G,- = 0.

Proof. We assume without loss of generality that m + k = n + 3. Suppose first that m > n + 2. Since dim^ < n, there exist open sets Vi, . . . , Vn+i such that

(4) F, C Vl C V, C Ui for i < n + I , So we get what we want by a direct application of Lemma 3.2.3 (let Ai = Fi for i < n + 1 and A = Fn+z}. We may therefore assume without loss of generality that m < n + 1 and, similarly, that k 2.

3.12. COLORINGS OF MAPS

247

Observe that for i < m — I and j < k — 1 we have Fi n Fm C Ui,

GJ n Gk C U'j.

Since dim X < n and (m — 1) + (k — 1) = n + 1, there exist open sets Vi,...,Vm_i,

Wi,...,Wfc_i

such that (6) F, n Fm C y, C Vi_^ Uz for i < m - 1, (7) Gj n G fc C Wj C Wj C tfj for j < k - 1, (8) nS'F^nn^FrW, =0.

So we again get what we want by a direct application of Lemma 3.2.3.

D

Proof of Theorem 3.12.10. Let 5F be a closed shrinking of li (Proposition A. 7.1). Fix n + 3 different elements of U, say,

g = {c/ 1 ,...,[/ m ,c/{,...,t/a, where m + k = n + 3. Let

be the elements of y corresponding to the elements in 9. So Fi C Ui for i < m and f - l [ G j ] C f - l [ U f i for j < k. Since m + k = n + 3, by Lemma 3.12.14, there exist closed sets A\ , . . . , ATO, BI , . . . , Bk such that (1) Ai C Ui for i < m, and Bj C f

[U',] for j < k,

Put Aj = f[Bj] for j < fc. In J replace Fl by A; for i < m and GJ by A'- for j < k. The other elements of 3r are not being replaced. We claim that the collection 3*' obtained in this way covers X, and hence is a closed shrinking of U. To see that it covers, pick an arbitrary x E X. If x E UI^i Fi then we are done by the first part of (2). If x E Uj=i GJ then f ~ l ( x ] E Uj=i f~1[Gj] Q Uj=i Bj by the second part of (2), As a consequence, for some j < fc, x E A'-. Finally, if x does not belong to any element of 'K then the element of 3" that contains x belongs to 3"'. Observe next that f|™ i A» n f|*U /"M^] = 0Since / is a homeomorphism, there exists an open swelling V of 3"' which is simultaneously a shrinking of U (Corollary 3.2.2) and which is as required. So we arrive at the conclusion that U has an open shrinking V such that the Vs and /-1[V]'s corresponding to 9 have empty intersection. The same procedure can be repeated with any subfamily of V of cardinality n + 3. So after finitely many steps, we arrive at the required shrinking of U. D

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3. BASIC DIMENSION THEORY

Corollary 3.12.15. Let X be a space with dimX < n and let f be a fixedpoint free homeomorphism of X. Then f can be colored with at most n + 3 colors. Proof. By Theorem 3.12.8, / is colorable, and by Theorem 3.12.12 the number of colors can be reduced to n + 3. D Coloring continuous functions. An inspection of the proof of Theorem 3.12.12 shows that it very strongly depends on the fact that we are dealing with homeomorphisms. In fact, as Theorem 3.12.7 shows, homeomorphisms are essential for the basic coloring result Theorem 3.12.8 and they are also essential for the process of reducing the number of colors in Theorem 3.12.12. We will investigate now the natural question whether the results obtained so far can be generalized. In the light of Theorem 3.12.7, the only reasonable question is whether colorable maps can be colored by a color of small size. So we are aiming at a generalization of Theorem 3.12.12. The only place in the proof of Theorem 3.12.12 where we used that the map under consideration is a homeomorphism, is in the proof of Theorem 3.12.10. So we will first think about the question whether Theorem 3.12.10 is also true for continuous maps instead of homeomorphisms. For n = 0 there are no problems. To see this, assume that X is zerodimensional, U is a finite open cover of X, and /: X —> X is continuous. Since dim^f = 0, there is an open shrinking V = {Vy : U £ U} of U with ord(V) < 0, i.e., V is pairwise disjoint (Theorem 3.4.4). But then V is as required. For let x e X, and observe that there is precisely one element of V that contains x. Similarly, there is precisely one element of /^[V] that contains x. But for larger n, Theorem 3.12.10 does not hold for continuous functions, as the following simple example shows. Example 3.12.16. Let X be the topological sum of the spaces I n , n e N, where each I n is a copy of I. In addition, let {di : i > 2} be a countable dense subset of Ii. Now define /: X —>• X as follows. / f l i is a homeomorphism from Ii onto I 2 . In addition, / f Ii is the function with constant value di for every i > 2. Then / is clearly continuous (and has no fixed-point). For every i e N let Ei and Fi be open subsets of Ii such that

Ei\JFi= I, and Define

3.12. COLORINGS OF MAPS

249

Then U — {C/i, . . . , C/4} is an open cover of X. Let V = {Vi,...,^} be an arbitrary open shrinking of U. Then V\_ and V?, cover Ii, and are clearly proper subsets of Ii. So by connectivity of Ii, V\ D V? ^ 0. Pick i > 2 with di 6 Vi n V2- Since Vs and V4 cover \Ji>2 Ii, the same argument shows that V3 O 1/4 n Ii 7^ 0. We conclude that as required. This example shows that if one wishes to color continuous functions, the method used so far does not work. Fortunately, there is another way to obtain what we want. Theorem 3.12.17. Let X be a space and let f : X -> X be fixed-point free and continuous. If f is colorable and dimX < n, then f can be colored with at most n + 3 colors. Proof. We claim that by Corollary 3.5.7 we may assume without loss of generality that X is compact. For let aX be an at most n-dimensional compactification of X such that / can be extended to a fixed-point free continuous function /: aX —> aX. By compactness of aX the function / can be colored. So if the theorem for compact spaces is true then / can be colored with at most n + 3 colors. It suffices therefore to observe that if y colors / then 2F \ X colors /. So assume that X is compact. We also assume for the time being that / is surjective. The general case will be derived from this special case and will be dealt with later. Consider the dynamical system (X, /) and the inverse system \r

j

/_

~\r

.

J_

j

J_

~\r

.

J

_

It was shown on Page 92 that the shift function a : X^ -> X^ defined by a(xi,X2,...) = (/(zi), £1,2:2,...). is a homeomorphism. Moreover, for each n the diagram

commutes and a is fixed-point free (Lemma 1.10.24). Now observe that dim ^00 < n by Exercise 3.2.2. So by Theorem 3.12.10 there is a closed cover y of X^ such that a[F] fi F — 0 for every F e J

250

3. BASIC DIMENSION THEORY

while moreover |3"| < n + 3. Now for each F 6 3~ by compactness there exists an NF e N such that f ^ [ F ] n /^[cr[F]] = 0 for every m > NF (Exercise 1.10.12). So if n - max{7VF : F e 3~} then /~[F] n /£° [
K = n /TO oo

i—l

Then f[K] = K by Exercise 1.10.18. By the result just proved there is a cover C = {Ci : i < n + 3} of K consisting of relatively open sets such that f[Ci] n Ci — 0 for every i. Let {Di : i < n + 3} be a closed shrinking of C (Proposition A. 7.1). Then f[Di] n Di = 0 for every i. Let Ul C X be an open neighborhood of Di for i < n + 3 such that f \Uj\r\Ui = 0. The union U of the t/i's is an open neighborhood of K. By compactness, for a sufficiently large k we have fk[X] C [7. So

is an open cover of X . We claim that it is a coloring of /. To prove this, take an arbitrary i < n and assume that there exists Then f k ( x ) e C/; and there exists an element z 6 f~k[Ui] such that /(z) = x. Then on the one hand fk (x) 6 Ui while on the other hand This shows that f k ( x ) G C/i fl /[E/;], which is a contradiction.

D

Exercises for §3.12. Let /: X —> X be a fixed-point free and continuous. If / is colorable then its color number C(f) is the minimal cardinality of a coloring of /. If / cannot be colored then we put C(f) = oo. Suppose that / : X —> X is a fixed-point free and continuous. We say that a subset A of X is invariant if f[A] C A. Observe that a color of / \ A is a subset of A 1. Let a,b,A,B C N be such that a and b are finite, while A and B are infinite. Prove that [a, A] C [b, B] iff a D 6, a \ 6 C B and A C B. 2. Prove that the collection in (*) on Page 239 is a base for a topology on [N]'" .

3.13. VARIOUS KINDS OF INFINITE-DIMENSIONALITY

251

3. Let A be a countable collection infinite subsets of N. Prove that there is an infinite subset X C N such that for every A G A we have XnA^®^

An(N\X).

4. Prove that the Ellentuck topology on [N]" does not have a countable base. We now present an interesting and unexpected addition theorem for the color number of a fixed-point free map. Theorem 3.12.18. Let X be a space and let /: X —> X be a continuous fixedpoint free function such that X — A U B for invariant subsets A and B of X . Then

The following two simple exercises will be useful. 5. Let f : X —>• X be a fixed-point free map with invariant subspace A. Suppose that the subset C C A is a closed color of / \ A. Prove that there exists an open subset W of X such that (/ \ A)~l[C] C W and W is a color of /. 6. Suppose that / : X —> X is a fixed-point free map and that C is a collection of n colors of / each of which is open or closed. Prove that there exists an open coloring of / of size at most n. From Exercise 3.13.6 it follows that the color number of a map /: X —> X is the minimal cardinality of a cover of X consisting of colors that are open or closed. The following exercise relates an open coloring of an invariant subspace to a collection of open colors of the ambient space. 7. Suppose that / : X —>• X is a fixed-point free map and that the subset A of X is invariant. Let II be an open coloring of / \ A with |1I| = in < oo. Prove that there exists a collection V consisting of at most ra open colors of / such that A C \J V. We are now in a position to finish the proof of Theorem 3.12.18. 8. Prove Theorem 3.12.18. 3.13. Various kinds of infinite-dimensionality Recall that a space X is strongly infinite- dimensional if it has an infinite essential family of pairs of disjoint closed sets. By Corollary 2.4.13, the Hilbert cube Q is an example of a strongly infinite-dimensional space. A space X is called weakly infinite- dimensional if it is not strongly infinitedimensional, i.e., if for every family

of pairs of disjoint closed subsets of X there exist partitions Di between and Bi such that A = 0-

252

3. BASIC DIMENSION THEORY

Proposition 3.13.1. Let X C Q be weakly infinite-dimensional. Then for every sequence {(A;, Bi) : i G N} of pairs of disjoint closed subsets ofQ there exist partitions Di between Ai and Bi in Q such that

Proof. For every i 6 N let Ui and Vi be disjoint closed neighborhoods of Ai and Bi, respectively (Corollary A. 4. 3). Since X is weakly infinitedimensional, the sequence {(Ui n X, Vi D X),i 6 N} is inessential. Consequently, there exist partitions Si between UiC\X and Vi n X in X such that the intersection of the 5;'s is empty. By Lemma 3.1.4 there exists for every i a partition Di between Ai and Bi in Q such that Di n X C Si. Then clearly oo

oo

fl Di n x c p| s, = 0, i=l

i=l

as required.

D

Finally, recall that a space X is called countable dimensional if it can be written as the union of countably many zero-dimensional subspaces. So every finite dimensional space is countable dimensional by Corollary 3.3.9, but not conversely. An important example of a countable dimensional space which is infinite-dimensional is given in the following example. Example 3.13.2. As in §1.1, let a = {x e s : (3n e N)(Vm > n)(xm = 0)}.

For each n G N put

an = {x E s : (Vm > n)(xm = 0)}. It is clear that an is homeomorphic to R n , n e N, and hence is the union of n + 1 zero-dimensional subspaces (Corollary 3.3.9). Since a is equal to the union of the crn's, it follows that a is countable dimensional. Observe that a is infinite-dimensional by Theorem 3.2.12 and Corollary 3.4.14(1). Observe that every subspace of a countable dimensional space is again countable dimensional. Example 3.13.3. There exists an infinite-dimensional compact space which is countable dimensional. This is a triviality. Let X be the one-point compactification of the topological sum of the spaces I, I2, I3, • • • . Then X is infinite-dimensional for the same reason a is. In addition, X is countable dimensional by Corollary 3.4.14(1). Proposition 3.13.4. Every countable dimensional space is weakly infinitedimensional.

3.13. VARIOUS KINDS OF INFINITE-DIMENSIONALITY

253

Proof. Let X be countable dimensional. Then X is the union of countably many zero-dimensional subspaces, say Xi, i G N. Let {(Ai,Bi) : i 6 N} be a family of pairs of disjoint closed subsets of X. By Corollary 3.1.6 for every i there exists a partition Di between Ai and £?; such that Di D Xi = 0. Since the union of the X^s equals X, the intersection of the IV s is empty, which is as required. D Corollary 3.13.5. No strongly infinite- dimensional space is countable dimensional. Corollary 3.13.6. The Hilbert cube cannot be written as the union of countably many zero- dimensional subspaces. In view of the above results, the question naturally arises whether the converse to Proposition 3.13.4 holds. This question was known as Alexandroff's problem for decades, [8, §4, Hypothesis]. We shall answer this problem in the negative. However, we need to derive the following result first. Theorem 3.13.7. Every topologically complete space X has a compactification 'jX such that the remainder ^X \ X is countable dimensional. Proof. We may assume that X is a subspace of the Hilbert cube Q (Corollary A. 4. 4). Since X is topologically complete, there is a decreasing sequence (Un)n of open sets in Q such that

n—l

(Theorem A. 6. 3). Since Q is compact, for each e > 0 there is a finite open cover of Q with mesh less than e. Consequently, for each n there exist finitely many open sets £/n,i, . . . , Un,mn C Q such that (1) £7n,i U • • • U Unjmn = Un, (2) mesh({t/ n ) 1 ,...,*7 n > m n })< i/n. For each n G N and m < ran, define / n>m : Q —> I by fn,m(x)

= 1/2&(X, Q \ Un,m)

and define / : Q —> Q coordinate wise as follows: f ( x ) = (fi,i(x), • • • , f i , m i ( x ) , f 2 , i ( x ) , . . . , /2,m 2 (a:),/3,i(a;), . . . , - . . ) • By (2) it follows easily that / \ X is an imbedding. We claim that f[Q]\f[X]Ca. Take an arbitrary point f(q) E f[Q] \ f[X]. Since q $. X, by (1) there exists an N € N such that q £ UN. Consequently, since the C/n's decrease, for every n > N and m < mn we have fn,m(q] = 0,

254

3. BASIC DIMENSION THEORY

as required. Now let jX be the closure of f[X] in f[Q]. Then jX is a compactification of X the remainder 'jX \ X of which is contained in the countable dimensional space a (Example 3.13.2). So 7 A7" \ X is countable dimensional as well. D The topological completeness of the spaces under discussion in the above theorem is essential. It can be shown that a is an example of a countable dimensional space the remainder of each compactification of which is strongly infinite-dimensional, see Exercise 3.13.4. We shall now present the solution to Alexandroff's problem. Theorem 3.13.8. There exists a weakly in finite- dimensional compact space which is not countable dimensional. Proof. By Theorem 3.9.3 there exists a strongly infinite-dimensional totally disconnected and topologically complete space, say S. Let X — jS be a compactification of S with countable dimensional remainder (Theorem 3.13.7). We claim that X is the required example. To this end, first observe that X is not countable dimensional since by Corollary 3.13.5 its subspace 5 is not countable dimensional. Now let T = {(A;, BI) : i > 0} be a sequence of pairs of disjoint closed subsets of X . Write X \ S as the union of countably many zero-dimensional subspaces, say Xi, i € N. By Corollary 3.1.6 for each i e N there exists a partition Di between Ai and Bi such that Di Pi Xi = 0. We conclude that the compact set D — Hi^i Di is contained in 5, and S being totally disconnected, therefore has to be zero-dimensional (Exercise 1.5.12). By another appeal to Corollary 3.13.5 there exists a partition DO between AQ D and BQ such that D0 n D = 0. We conclude that f£0 D* = 0An infinite-dimensional space X is called hereditarily infinite- dimensional if for every nonempty subspace A of X either dim A = 0 or dim A = oc. A hereditarily infinite-dimensional compact space is clearly Henderson, but the converse need not be true. See the notes for more information. We finish this section by presenting an example of a hereditarily infinitedimensional space. Lemma 3.13.9. Let K C JJ be a Cantor set, let £ 6 N and Jet K C N\{£} be an infinite set the complement of which is also infinite. For each j; € K there exists a partition Sj between the opposite faces W~l and Wj ofQ such that JS each subset X C S w^n K infinite-dimensional. Proof. Without loss of generality we may assume that £ = I and K is the set of even natural numbers. For each i e N let Ci = vr-^-l, - 1/2] and A = ^[1/2, 1],

3.13. VARIOUS KINDS OF INFINITE-DIMENSIONALITY

255

respectively. Now put A = {/ e C(Q,Q) : (Vi 6 M)(/-1[I^-1] = C2t and f-^Wft

= D2l)}.

Then A is a separable metric space since by Proposition 1.3.3 C(Q, Q) is. By Exercise 1.5.10 there exists a subspace T of K such that T can be mapped onto A, say by the map T. Let £=TxJ2xJ3x---cg and define 4> : E —> Q by $0) =T(XI)(X). Then 4> is clearly continuous. Claim 1. For each i, (7TiO$)~ 1 (0) is a partition in £ between the sets C and D2i n E. Proof. Fix i 6 N. Then

and similarly, ^>~1[VFi1] = £^ n D 2 i- Now since 7rl~1(0) is a partition between W~l and VF/, the claim follows. 0> Since W^1 and Wg^ are in the interior of C^i and D 2 i, respectively, there exists by Lemma 3.1.4 a partition 6*2 i between W^1 and l/t^i m Q such that (*)

^ a . H ^ C (TTioS)- 1 ^).

We claim that

s

is as required. Claim 2. SnEC ^(O). Proof. This immediately follows from (*).

<0

Now let M be a subspace of S such that K C 7Ti[M]. We shall prove that M is strongly infinite-dimensional. Striving for a contradiction, suppose that this is not the case, i.e., that M is weakly infinite-dimensional. By Proposition 3.13.1 for i 6 N there exist partitions Ml in Q between C 2j and D-zi such that P l ^ 1 M i n M = 0. For i € N there exists a continuous function fi : Q —>• JJ; such that

256

3. BASIC DIMENSION THEORY

(Exercise A.4.9). Define /: Q -> Q by /(x) = (/!(x),/ 2 (x),...). Claim 3. / 6 A and M n /-1(0) = 0Proof. That / is continuous is trivial. Now take i G N. Then by the definition of /;. Similarly one proves that /^[W/] = D-U. We conclude that / G A. Finally, oo

f - l ( o ) n M = Pi Mi n M = 0, i=l

as required.

<)

Now since r is onto, there exists p G T such that r(p) = /. Since T C 7Ti[M], there exists x G M such that #1 = p. Then x £ S n E from which it follows by Claim 2 that f ( x ) = r(p)(x) = T(XI)(X)

= $(x) = 0.

But this contradicts Claim 3.

D

This result enables us to construct our final example. Theorem 3.13.10. There exists a hereditarily infinite-dimensional compact space. Proof. By Exercise 3.13.6 there exists a collection {Ki : i G N} of pairwise disjoint Cantor sets in JJ such that each nondegenerate subinterval of JJ contains one of the Ki. So for each i choose pairwise disjoint such Cantor sets Kn,Ki2, . . . in Ji = [— 1, l]i. In addition, let Kik (i, k G N) be a collection of pairwise disjoint infinite subsets of N \ {1} such that

i 0 Kik for all i and k. Fix i, k E M. By Lemma 3.13.9, for every j G Kik we can choose a partition Sj in Q between W~l and Wj such that each subset in the intersection

whose projection onto the i-fh axis contains Kik, is infinite-dimensional. Now put

We claim that 5 is hereditarily infinite-dimensional. Clearly 5 is a compact subspace of Q. Since 1 & \JikKik, by Corollary 2.4.13 S intersects every partition between W±l and W±. By Exercise 3.1.13 it therefore follows

3.14. THE BROUWER FIXED-POINT THEOREM REVISITED

257

that S is not zero-dimensional (for otherwise there would exist a partition P between W^1 and W^ such that P fi S — 0, contradicting the fact that the family of opposite faces of Q is essential). Now let M be an arbitrary nonempty subspace of S. There are two cases to consider. First assume that for each i e N, 7Tj[M] is zero-dimensional. Then M is contained in the set oo

Jj7T z [M],

which is clearly zero-dimensional. So then dim M = 0 as well. Next assume that there exists i 6 N such that 7T;[M] is not zero-dimensional. Then 7T;[M] contains a nondegenerate subinterval of JJ^ by Proposition 1.5.3. Consequently, by construction, TT^fM] contains some Kij which implies that M is (strongly) infinite-dimensional because M is contained in Sij. D Exercises for §3.13. A real-valued function / on a space X is countably continuous provided that X can be partitioned into countably many sets EI, #2, • • • such that for every i the restriction / \ Ei is continuous. 1. Prove that if a space X is the union of countably many weakly infinitedimensional subspaces then X is weakly infinite-dimensional. Conclude that the Hilbert cube is not the union of countably many weakly infinitedimensional subspaces. 2. Let X be a strongly infinite-dimensional compact space. Prove that X contains a hereditarily infinite-dimensional closed subspace. ^•3. Prove that if G is a GVsubset of s containing a then G\o contains a copy of the Hilbert cube. 4. Prove that if 70- is a compactification of a then the remainder 70- \ a contains a copy of the Hilbert cube and is therefore strongly infinitedimensional. ^•5. Give an 'explicit' example of an upper semi-continuous function that is not countably continuous. 6. Prove that there exists a collection {Ki : i e N} of pairwise disjoint Cantor sets in J such that each nondegenerate subinterval of JT contains one of the Ki. 3.14. The Brouwer fixed-point theorem revisited In this section we show that the existence of a two-dimensional hereditarily indecomposable continuum, which uses Brouwer's Fixed-Point Theorem for I3 only, implies the Brouwer Fixed-Point Theorem for all dimensions. This seems to show that the 'real' power of the Brouwer Fixed-Point Theorem is already attained at dimension three.

258

3. BASIC DIMENSION THEORY

Kelley's construction. Let X be a hereditarily indecomposable continuum with dimX > 2. The existence from such a space follows from Corollary 3.8.3. An inspection of the proof shows that for the construction of X one needs Brouwer's Fixed-Point Theorem for I 3 only. We will show that from X we can build an infinite-dimensional space. Our strategy will be to 'blow up' the points of X and to topologize them in such a way that we get an infinite-dimensional hereditarily indecomposable continuum. Since dimX > 2, by Exercise 3.7.5 the following holds for X: (1) there are real numbers 6 > 0 and e > 0 such that if X is any finite collection of pairwise disjoint closed sets with mesh(N) < 26 then there is a continuum C C X \ [J N with diameter at least e. Let us: 2X —>• I be a Whitney map (Proposition 1.11.15). By Exercise 1.11.25 there is an element t 6 (0, oo) such that if A E 2X is such that uj(A) < t then diam(A) < e. Consider the Whitney level

£ = [A e G(X) : u(A) = t}. Then £ is an upper semi-continuous decomposition of X by Corollary 1.11.17. Here we use for the first time that X is hereditarily indecomposable. By Exercise 1.11.24 we find that (2)

7 = inf{diam(E) : E £ £} > 0.

This property of the collection £ is crucial. So for the diameter of an arbitrary element E G £ we have (3)

0 < 7 < diam£<£.

Let Y be the space X/£, and let TT : X —>• Y be the natural quotient map. Since TT is monotone, we already know from general considerations that Y is a hereditarily indecomposable continuum (Exercise 1.10.22). Observe that TT is open by Exercise 1.11.23. So Y is homeomorphic to the subspace £ of 2X, and the formula

d(tfi,y2) = ^(r 1 (yi) J /~ 1 (!/2)) defines an admissible metric on Y (Exercise 1.11.17). This is the metric on Y that we will use in the sequel. Observe that since £ consists of continua, the space Y is in fact homeomorphic to a subspace of G(X). This observation will be used in the solution to Exercise 3.14.1. Having found the collection £ of "blown up" points, we fix any n > 2 and set (4)

6n

3.14. THE BROUWER FIXED-POINT THEOREM REVISITED

259

Observe that by (1) and (2), 6n > 0. Proposition 3.14.1. For any finite closed cover J ofY of mesh less than 6n there is an element A E y intersecting at least n elements of 9r Proof. It will be convenient to introduce the following notation. For A C Y we put Striving for a contradiction, let 9" be a finite closed cover of Y with mesh less than 6n such that each element of A 6 y meets at most n — I elements of y \ {A}. We shall associate to each element A G J a compact set yMCTr-1^]

(5) such that

(6) y(A] H E ^ 0 for any E G £(A), (7) diamip(A) < 26n, (8) if A, B e y are distinct then (p(A) n (p(B] = 0.

Assume that (p is already defined on 9 C 2F, and let A e 9" \ 9 be arbitrary. The set JC of all elements of 9 hitting A has by assumption cardinality less than or equal to n — 1. For each D e 9, let B(D) be the closed 5 about tp(D}. Observe that by (7) we have (9)

diamB(D) < 4Jn.

Let us fix an arbitrary element E(A) e £(^4.). We shall show that (10)

E(A) \ {]{B(D} :D € K] / 0.

Otherwise, ^(A) is covered by the collection {B(D} -.D^'K} and hence, -E(A) being a continuum, by Exercise A. 10. 3 and (9) we get < \3<\ • max diamB(D] < (n - 1) • 45n, which contradicts (4). Let a be any element given by (10) and let B be the closed 5n-ball about a. We set

(11)

(p(A) =BnK-1[A\.

We claim that
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3. BASIC DIMENSION THEORY

If D E !K then a does not belong to the closed Jn-ball about (p(D}. So the closed Jn-ball about a misses tp(D], i.e., (p(A) r\(p(D] = 0. Moreover, if D belongs to $ \ 'H then A n D = 0 and hence tp(A) n ^>(D) = 0 since (p(A) Cn-^A],

(p(D) CTT-^D],

TT1^] n Tr'1^] = 0 -

So (8) is also satisfied (for g U {A}}. We conclude that we can extend (p over 9 U {A} without creating problems. By the fact that 3~ is finite, we may therefore assume that (p is defined on all of y. The collection (12)

N = {v?(A) : A e ?}

is disjoint and has by (7) mesh less than or equal to 26n < 26. So it can serve as input for (1). To reach a contradiction, let C be an arbitrary continuum in X \ (J N. Let E e £ intersect C and pick A 6 J such that E G £(A). By (6),

Since C does not intersect \J N, this shows that E1 is not a subset of C. But then since X is hereditarily indecomposable, it must follows that C is a subset of E. This implies by (3) that diam C < diam E < e. By (1) there is such a continuum C with diameter at least e. This contradiction establishes the proof. D By Exercise 3.3.6, we now come to the following surprising conclusion. Corollary 3.14.2. The existence of an at least two-dimensional hereditarily indecomposable continuum implies the existence of an hereditarily indecomposable in finite- dimensional continuum. Dimension theory. By Exercise 3.1.16 we know that the Brouwer FixedPoint Theorem and the fact that there is a space X with dim^Y = oo are equivalent statements, in the sense that they are easily deduced from each other. So from Brouwer's Fixed-Point Theorem for I3, implying the existence of a two-dimensional hereditarily indecomposable continuum, we derived the Brouwer Fixed-Point Theorem for all dimensions. This is a rather curious phenomenon and it is not clear whether there is a more direct route from Brouwer in I3 to Brouwer in all dimensions. This is of course only of interest for 'philosophical' reasons. The proof of Sperner's Lemma 2.4.3 is by induction, so it is equally complicated in all dimensions (except for the dimensions one and two). Similar remarks can be made for the various other known proofs of Brouwer's Theorem. This seems to indicate that even if one finds a

3.14. THE BROUWER FIXED-POINT THEOREM REVISITED

261

more direct route from Brouwer in I 3 to Brouwer in all dimensions, this will not result in a simpler proof of Brouwer's Theorem. But this does not mean that clarifying the rather curious phenomenon that we encountered in this section would not be interesting. Exercises for §3.14.

1. Let Z be a compact space with dimZ > 3. Prove that dimC(-Z) = oo. 2. Give an example of a continuum X such that dimC(A') = 2.

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CHAPTER 4

Basic ANR theory In this chapter we shall present several basic results from A N R theory. We introduce and characterize ANR-pairs. They enable us to prove that the hyperspace of a Peano continuum is an AR, which is an important step in the proof of Curtis-Schori-West Hyperspace Theorem that 2X Kt Q if and only if X is a Peano continuum, and that there are deformations of hyperspaces through various interesting subspaces. A curious important result that we will prove is that for a Peano continuum X there are arbitrarily close to the identity maps from 2X into the finite subsets of X. Our concept of an ANRpair, which was motivated by TORUNCZYK's concept of a locally homotopy negligible set from [390], turns out to be a convenient framework for obtaining such results.

4.1. Some properties of AN R's

In this section we shall derive a few useful properties of AN R's. We have to introduce some terminology first. Let Y be a space and let U be an open cover of Y. Two continuous functions f,g:X-*Y are called U-close if for every x £ X there exists an element U G U such that {f(x),g(x)} CU. A homotopy H: X x I ->• Y is said to be limited by U provided that for any x G X there exists U G U such that H[{x} x I] C U.

Two continuous functions f , g : X -> Y are U-homotopic, if there is a homotopy H: X x I —>• Y which is limited by U and connects / and g, i.e., HQ = f and HI = g. Small functions and small homotopies. Here is our first property of ANR's.

Theorem 4.1.1. Let X be an A N R . Then for every open cover U of X there exists an open refinement V of U such that for every space Y, any two V-close maps f,g:Y-+X are U-homotopic. 263

264

4. BASIC ANR THEORY

Proof. By Corollary 1.1.8 we may assume that X is a closed subspace of a normed linear space L. Observe that L is locally convex (Exercise 1.1.6). Since X is an A N R , there are a neighborhood V of X in L and a retraction r: V -> X. Put £ = {r~l[U} : U E U}. Then £ is an open cover of V, and since V is open in L, there is an open cover If of V having the following properties:

(i) if < £ , (2) If consists of convex sets. Let V = y \ X. We claim that V is the required open cover of X. To this end, let Y be a space, and assume that f,g:Y—*X are continuous and V-close. Define a homotopy G : Y x I —> L by

Then G is continuous and clearly connects / and g. Since by (2) the elements of If are convex, by the special definition of G it follows that for every y E Y there exists F G If such that (3)

G({y} xl]CF.

We conclude that G can be considered to be a mapping from Y x I into V and that moreover G is an If-homotopy. Now define H : Y x I —>• X by

H = roG. Then H connects / and g since r is a retraction. Moreover, by (3) and (1) it follows easily that H is limited by U. D Remark 4.1.2. It is a natural problem whether the property of ANR's stated in Theorem 4.1.1 in fact characterizes the class of all ANR's. This was a difficult and fundamental problem which remained unanswered for decades. It was finally solved in 1994 by CAUTY [87] in the negative by his example of linear space which is not an ANR (Remark 1.2.4). To see that Cauty's Example really solves the problem, let L be a linear space. In addition, let U be an open cover of L. The function A : I/ x L x E —> L defined by X ( x , y , t ) = (l-t)-x + t-y, is defined in terms of the algebraic operations on L and is therefore continuous. For every x e L pick an element Ux 6 U containing x. Since A is continuous and X[{x} x {x} x I] = {x}, by compactness of I there exists for every x E X a neighborhood Vx of x such that X[VX x Vx x I] C Ux (Exercise A. 5. 7). Put V = {Vx : x £ L}. We claim that V is as required. To this end, let X be a space and let /, g : X —>• L be continuous V-close functions. Define a homotopy H : X x I —>• L in the obvious way by the formula

4.1. SOME PROPERTIES OF ANR'S

265

Then clearly H0 = f and HI = g. Fix an arbitrary x 6 X. Since / and g are V-close, there exists an element p 6 L such that f ( x ) , g ( x ) G Vp. But then [f(x)} x {g(x}} x I C V p X V p X l from which it follows that H(x,t) = \ ( f ( x ) , g ( x ) , t ) 6 Up for every t 6 I. So this indeed proves that / and g are U-homotopic. Homotopies with control. In many applications one needs homotopies 'with control'. We will therefore present a 'controlled' version of the Borsuk Homotopy Extension Theorem, cf. Theorem 1.4.2. Theorem 4.1.3. Let X be an ANR and let U be an open cover of X. Then for every closed subset A of a space Y and for every homotopy H: A x E —>• X such that (1) H is limited by U, (2) H0 can be extended to a continuous function h0 : Y ->• X there exists a homotopy H: Y x E —>• X such that (3) H is limited by U, (4) H0 = ho and H\(Axl} = H. Proof. Let H, A and Y be as in the formulation of the theorem. By Theorem 1.4.2 there exists a homotopy F: Y x I —> X such that FQ — ho and For each a G A there exists Ua 6 U containing H[{a} x E]. There consequently exists by compactness of I and continuity of H a neighborhood Ea of a in Y such that

F[Ea xl]CUa (Exercise A. 5. 7). Put E = (JaeA Ea . Then E is a neighborhood of A in Y. By Corollary A. 4.1 there is a Urysohn function X:Y-tI such that A f A = 1 and A \ Y \ E = 0. Now define H : Y x I -> X as follows: Then H is clearly continuous and H \ (A x I) = H. In addition, H(y,0)=F(y,Q)=ho(y) for every y 6 Y so ^o — ^o- It remains to prove that H is limited by U. This is a triviality. Take an arbitrary y G Y. If y £ E then there exists a G A such that y e Ea. Consequently, H({y} x I) C F[{y} x E] C t/a.

Now if y $ E then A(y) = 0 from which it follows that

266

4. BASIC ANR THEORY

consists of precisely one point and is therefore also contained in an element of U. D ANR-pairs. We shall now derive another important property of ANR's. For later use, we will state it in terms of ANR-pairs. Let X be a space with subspace Y. We say that the pair (X,Y) is an ANR-paz'r provided that for every space Z and every closed set A C Z and every continuous function / : A —> X there exist a neighborhood U of A in Z and a continuous extension /: U —> X of / having the additional property that f[U\A] C Y. Observe that if ( X , Y ) is an ANR-pair then so is (E,Y), where E is any subspace of X containing Y . Define the notion of an (\R-pair similarly. Observe that if (X, Y) is an A(N)R-pair then both X and Y are A(N)R's (see Exercise 4.1.7). Moreover, if X is an ANR then ( X , X ) is an ANR-pair. Also, Y is dense in X. For if Y is not dense in X then there exists a point x e X such that for some e > 0 we have that B(x,e) D Y = 0. Now let Z — I, A — {0} and /: A —>• X the constant function with value x. Then obviously / cannot be extended over a neighborhood of A with 'new' values iny. Let X be a space, and let Y C X. A deformation of X through Y is a homotopy H : X x I —>• X such that HQ is the identity on X, and Ht[X] C Y for every t 6 (0, 1}. Lemma 4.1.4. Let X be a space and let U be a neighborhood of X x {0}

in X xl. Then there is a continuous function a: X ->• (0, 1] such that {(x,t) :t (0, 1] defined by a(z) =

i/2d((x,Q),(Xxl)\U)

is continuous (and clearly well-defined since Q is bounded by 1). We claim that it is as required. To this end, consider for some x G X a point of the form (x,t), where t < a(x). If ( x , t ) € (X x I) \ U then a(x) =

So we conclude that a(x) — 0, which is a contradiction.

D

4.1. SOME PROPERTIES OF ANR'S

267

Corollary 4.1.5. Let X be a space with subspace Y. If X can be deformed through Y then for every open cover U of X there exists a homotopy H: X x I -> X

such that (1) H deforms X through Y, (2) H is limited by U. Proof. Let F: X x I —> X be a deformation through Y. For every x 6 X pick Ux G U containing x. Since F~1[[/x] is a neighborhood of (x, 0) in X x I, there are an open neighborhood Vx of x and 0 < tx < 1 such that F[Vxx[0,tx]]

CUX.

Put

V = |J Vxx[0,tx). x£X

By Lemma 4.1.4 there is a continuous function a: X —> (0,1] such that {(x,t) :x e X,t< a(x}} C V. Now define H: X x I ->• X by H(x,t) = F(x,t-a(x)). Since a(x) > 0 for every x, we clearly have that H deforms X through Y. Now if x € JC then (x,a(x)) G V. So there exists x' G X such that (z,a(z)) 6 Vx, x [O,^)But then { x } x [ 0 , a ( x } ] CVX, x [0,^) and so

H[{x] x I] C F[{x} x [0,a(x)]] C C/^, as required.

D

These results enable us to characterize A(N)R-pairs, as follows: Theorem 4.1.6. Let X be a space, with subspace Y. The following statements are equivalent: (1) ( X , Y ) is an A(N)R-pair. (2) X is an A ( N ) R and there exists a deformation from X through Y. Proof. We will show this for ANR-pairs only, the proof for AR-pairs being entirely similar. For (1) => (2), consider .Y x 3, its subspace X x {0} and the function /: X x {0} —>• X defined by /(x,0) = x. There exist a neighborhood t / o f ^ L x { 0 } i n ^ C x I and a continuous extension /: U —> X of / such

268

4. BASIC ANR THEORY

that f[U\(X x {0})] C Y. Let a: X -> (0, 1] be the function in Lemma 4.1.4 for U . Now define H : X x I —> X by the formula

Then H is a deformation through Y. Since, as observed earlier, X is an ANR, we are done. For (2) => (1), let X be an ANR and H : X x I ->• X be a deformation through Y. In addition, let Z be a space, A C Z be closed and f : A -> X continuous. Since X is an A N R , there are a neighborhood U of A in Z and a continuous extension f:U-*X. Let Q be an admissible metric for Z which is bounded by 1 (Exercise A. 1.6), and define g: Z -> X by the formula

Then g is clearly as required.

D

As we saw, it is sometimes possible to replace a homotopy by one with better properties. Another example of such a phenomenon is the following result, which shall be used frequently in the sequel. Proposition 4.1.7. Let (X, g) be a space with subspace Y and let H be a deformation of X through Y. Then H can be replaced by a deformation F of X through Y such that

g(F(x,t},x) < t for all x E X and t el. Proof. For every x 6 X and t G I put G(x,t) = {s e I : d i a m ( H [ { x } x [0,s]]) > *}. Observe that possibly G(x,t) = 0. Claim 1. G(x,t) is closed in I for every x and t. Proof. This follows easily from the continuity of the functions H and diam (Exercise 1.11.12). <> Now define f : X x I ->• I by J ^'

J

[

By Claim 1, £ is well-defined. Claim 2.

is Isc.

4.1. SOME PROPERTIES OF AIMR'S

269

Proof. Suppose that for some v G [0, 1), x G X and t G I we have £(x, i) > v. Suppose that there are sequences (x n ) n in X and (£ n ) n in I such that

while moreover £(xn,tn) < v for all n. We may assume without loss of generality that £(x n , t n ) —> w < t>. Observe that since v < 1, diam (H({xn} x [0, £(x n ,£ n )]) > tn for every n so that by continuity of H and diam (Exercise 1.11.12) we get

This shows that £(x,t) < w < v, which is a contradiction.

Q

Let j] : X xl —>• I be the constant function with values 0. Then r\ is continuous, and hence use. By Corollary A. 7. 6 there is a continuous function S : X xl —>• I such that for every ( x , t ) G X x I we have 0 < £(z,t) <£(£,*) and if 0 < ^(x, i) then 0 < 6 ( x , t ) <£(x,t). Now define the homotopy F : X x I —>• X by the formula

Then F is clearly continuous. We claim that it is as required. To this end, pick arbitrary x 6 X and t G I. Then g(x,F(x,t))

=g(x,H(x,6(x,t)))

is less than t if £(x, t} > 0 and 0 < £ otherwise. It therefore suffices to prove that F deforms X through Y' . It is clear that 0(z, F(x, 0)) = 6(x, H(x, e(ar, 0))) = 0(ar, ff (x, 0)) = 0,

i.e., F(x,0) = x. Finally, take arbitrary x e X and t € (0, 1]. If f(a:,£) = ° then 0 = di&m(H[{x} x {0}]) > t > 0 which is impossible. So £ ( x , t ) > 0 from which it follows that 6(x,t] > 0 and consequently F(x,t) = H(x,6(x,t)) as required.

EY, D

Corollary 4.1.8. Let Z C Y C X. If (Y, Z) and ( X , Y ) are A(N)R-pairs then so is (X, Z) .

270

4. BASIC ANR THEORY

Proof. We prove this for ANR-pairs only, the proof for AR-pairs being entirely similar. Let F: X x I —> X be a, deformation through Y (Theorem 4.1.6). We may assume by Proposition 4.1.7 that g ( H t ( x ) , t ) < t for all x 6 X and t £ I. Similarly, let G: Y x I —>• Y be a deformation through Z having the property that g ( G t ( y ) , t ) < t for all y e Y and t e I. Define the homotopy H: X x I —>• .Y by V

'

\G(F(x,t),t)

(*>0).

We only need to check that H is continuous at points of the form (x, 0). So let (xn)n be a sequence in X converging to x and let (tn)n be a sequence in (0,1] such that tn \ 0. Then

< 0(x,xn) + g ( x n , F t n ( x n ) ) + g ( F t n ( x n ) , G t n ( F t n ( x n ) ) _ Q\3-"> -En) i ^n

So we are done by another application of Theorem 4.1.6.

D

Partial realization. The following interesting property of ANR's is not very appealing at first glance. It turns out however that it is of crucial importance in the process of finding a usable topological characterization of ANR's. We need to introduce some terminology first. Let X be a space and let U be an open cover of X. In addition, let T be a locally finite simplicial complex and let § be a subcomplex of T containing all the vertices of T. A partial realization of T in X relative to (S, U) is a continuous function such that for every a £ T there exists U 6 U such that In case S = T we say that / is a full realization of T in X relative to U. Theorem 4.1.9. Let X be an ANR. Then for every open cover U of X there exists an open refinement V of U such that for every locally finite simplicial complex T and every subcomplex S of T containing all the vertices of 7, every partial realization of T in X relative to (S, V) can be extended to a full realization of 7 in X relative to U. Proof. Recall that by Corollary 1.1.8 we may think of X as a closed subspace of a normed linear space L. Since X is an ANR, there are a neighborhood V of X in L and a retraction r: V ->• X. Put £ = {r~l[U] : U G U}. Then £ is

4.1. SOME PROPERTIES OF ANR'S

271

an open cover of V, and since V is open in L, there is an open cover 3 of V having the following properties: (1) ? < £ , (2) J consists of convex sets.

As in the proof of Theorem 4.1.1, put V = 3 \ X. We claim that V is the required open cover of X. To this end, let T be a locally finite simplicial complex, let S be a subcomplex of T containing all the vertices of T, and let / : |S| -> X be a partial realization of T in X relative to (S, V). For each a G T let a* denote the convex hull of f[cr D |S|] in L; observe that by (2) a* is contained in an element of y. For each n > 0, let

Kn = 7(n) U § (here T^ n ^ denotes the n-skeleton of T of course, see §2.1). Then OCn is a subcomplex of T. By induction on n > 0 we shall construct a continuous function fn: \%n\ —>• V such that the following conditions are satisfied: (3) /o = /, (4) if n > 1 then fn extends / n _i, (5) for every a G T, fn[a n \3Cn\] C a*.

Observe that /o is well-defined since T^0) C §. Now assume that for certain n > 0 we constructed the function fn-i- We will define fn on every simplex a G OCn separately and will conclude by applying Lemma 2.1.19 that the union of all the constructed functions is continuous. We have to be careful of course since we want the union of all the functions to be well-defined. So take an arbitrary simplex a G OCn — T^ U §. We want to define fn on a. There are two cases to consider. First assume that a G 3Cn-i- Then put

\ a. Suppose therefore that a £ 3C n _i, i.e., a G 7^ \ 3C n _i. Then a* over a. Since a* is convex, this can be done by Theorem 1.2.9. This defines fn on a. Observe that the union of all constructed functions is a function since the geometric interiors of all simplexes in "Kn form a pairwise disjoint collection (Exercise 2.1.12) and that it is continuous by Lemma 2.1.19. Since (5) is easily seen to be satisfied, this completes the inductive construction. Now define (p: |T| —>• V by
(x G T' n >).

272

4. BASIC ANR THEORY

Observe that (p is clearly well-defined by (4) , that the range of (p is contained in V by (5) and that (p extends / by (3). In addition, by another appeal to Lemma 2.1.19 it follows that

g = ro(f. Since r is a retraction and (p extends /, g extends / as well. Now take an arbitrary a E T and an element W E V such that There exists n > 0 such that a E 7^n\ By (5) and the definition of (p we obtain (p[u\ C a*. There exists F E 3~ such that F D X = VF; observe that a* C F. By (1) there exists t/ E U such that F C r"1 [£/"]. We conclude that sM=r[
is finite. By Exercise A. 7.1 it therefore follows that Wx =

p| Un \ ( |J

n£Ax

Fn)

n£Ax

is open. Put W = {Wx : x G X}. Then W is an open cover of X. We claim that ,W) :x G X} < U .

To this end, take x G X and fix an arbitrary n G Ax. Let Wy e W be such that x G Wy. Then clearly Ax C Ay from which it follows that Wy C Un. We conclude that St({x}, W) C Un, as required.

4.1. SOME PROPERTIES OF ANR'S

273

Repeating this argument for W we get an open cover Wi of X such that for every x € X there is W € W with St({x},Wi) C VF. Observe that this implies that Wi refines W. Now let W\ G Wi be arbitrary. We are interested in computing St(Wi, Wi). To this end, pick an arbitrary x 6 Wi and let W{ e W be such that W{ C\Wi ^ 0, say y e W[ n Wi . There exists by construction an element W G W such that St({y},Wi) C W. We conclude that x e Wi C Wi U W[ C St({y}, Wi) C W and so ^ C St({x},W). This shows that St(^i,Wi) C St({x},W) from which it follows that St(VKi, Wi) C C7 for some Z7 e U. By Theorem 2.3.5(1) there is a star-finite refinement V of Wi. Then V is clearly as required. Now assume that U is finite. Then there is no need to refine U by a locally finite open cover. Observe that the cover W is finite in that case and that the same is true for Wi . So Wi is a star-refinement of U which is obviously star-finite, being finite. D Corollary 4.1.11. Let (X, Y) be an ANR-pair. Then for every open cover U of X there exists an open refinement VofU such that for every locally finite simplicial complex T and every subcomplex § of T, containing all the vertices of 7, and every partial realization f : |S| —> Y of 7 in Y relative to (S, V \Y) can be extended to a full realization g: |T| —> Y of 7 in Y relative to U \Y. Proof. Let U' be a star-refinement of U (Lemma 4.1.10). Since X is an ANR, we let V be the refinement of U' we get from Theorem 4.1.9. We claim that this V also works here. By Corollary 4.1.5 there exists a deformation H from X through Y which is limited by U'. Now let a locally finite simplicial complex T be given, let § be a subcomplex of T, containing all the vertices of T, and let /: |S| —> Y be a partial realization of T in Y relative to (§, V \Y). Since / is also a partial realization in X relative to (S, V), / can be extended to a full realization g of T in X relative to U'. We now use H to push g into Y, as follows. First observe that |T| is a metrizable space, and that |S| is a closed subspace of |T| (Lemma 2.1.14 and Proposition 2.1.21). Let Q be an admissible metric for |T| which is bounded by 1 (Exercise A. 1.6), and define /: |T| —> X by the formula

Then / is clearly continuous. If x € |S| then g(x, |S|) = 0, and so

274

4. BASIC ANR THEORY

Moreover, if x $ |8| then g(x, |S|) > 0 and so J ( x ) G Y. We conclude that the range of / is contained in Y. Moreover, since H is a U'-homotopy, the functions / and g are U'-close. Pick an arbitrary a G T. There is an element UQ G U' such that g[a] is contained in UQ. In addition, g and / are U'-close. For every x € a there consequently exists an element U'x G U' containing g(x] as well as J(x). We conclude that f[a] is contained in St([/o,U'). So we are done since U' is a star-refinement of U. D Dominating polytopes. Let X and Y be spaces. We say that X dominates Y if there are two maps g:Y-*X,

f-.X^Y,

such that the composition / o g \ Y —>• Y is homotopic to the identity on Y. The space X is called a dominating space for Y. Lemma 4.1.12. A space X is contractible if and only a one-point space dominates X . Proof. Suppose first that X is contractible and let H: X x I —> X be a homotopy contracting X to a point, say p. Now let g : X —>• {p} be the obvious function and let / : {p} —>• X be the inclusion. Then H connects the identity on X with f o g , i.e., {p} dominates X. Conversely, assume that there exists a point p and functions g : X —> {p} and /: {p} —> X such that f o g is homotopic to the identity on X. Since f o g is clearly a constant function we conclude that X is contractible. D As remarked above, one sometimes needs homotopies 'with control'. For that reason we define a 'controlled version' of the concept of domination. Again let X and Y be spaces and let U be an open cover of Y. We say that X U-dominates Y if there are two maps such that the composition f o g : Y —> Y is U-homotopic to the identity on Y. The space X is called a U- dominating space for Y. Theorem 4.1.13. Let X be a (compact) ANR. Then for every open cover U of X there exists a (polyhedron) polytope P such that P U-dominates X. Proof. Let V be an open refinement of U with the properties stated in Theorem 4.1.1. Let V be a star-refinement of V (Lemma 4.1.10). By Theorem 4.1.9 there exists an open refinement W of V such that for every locally finite simplicial complex T, for every subcomplex S containing all the vertices of T, and for every partial realization / : |S| —> X of T in X relative to (S, W) there exists a full realization g: |T| ->• X of T in X relative to V such that g extends /.

4.1. SOME PROPERTIES OF ANR'S

275

Claim 1. Such a full realization g has the following property: For every a e T and for every W € W with /[a n S|] C W there exists V € V such that #[0-] U W C V. Proof. Let cr e T and W e W be such that /[cr n |S|] C W. There exists an element V 6 V such that 0[ P be the ^-mapping of A (see Page 488 for the definition of a ^-mapping). Now for every vertex x(A) in N(A)(°) choose an arbitrary point f ( x ( A ) } G A. This defines a function By Corollary 2.1.15, |A/"(.A)( 0 )| is a discrete topological space, hence / is continuous. Claim 2. / is a partial realization of N(A) in X relative to (N(A^), W). Proof. This is easy. Let a — \{x(Ao),x(Ai), . . . ,x(An)}\ be a simplex in N(A). Then

A0 n AI n • • • n An ^ 0. Since A is a star-refinement of W there exists W 6 W such that AQ U AI U • • • U An C

W.

Consequently, f[a n N(A)W] = f [ { x ( A o ) , x(Al], . . . , x(An)}] C AQ U AI U • • • U An

cw. So we are done.

<$•

Now by the special choice of W, the function / can be extended to a full realization g: \N(A)\ —> X relative to V. Claim 3. The identity function lx on X and the function go K are V-close. Proof. Take an arbitrary x G X. Since A is star-finite, there are only finitely many elements of A that contain x, say AQ, AI , . . . , An. Then a = \{x(A0),...,x(An)}\ is a simplex in N(A), and by Lemma 2.3.2, K(X) € a. There exists W G W such that AQ U AI U • • • U An C

W.

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4. BASIC ANR THEORY

Then x G W and since f(x(Ai)}

G A{ for every 0 |] C VK By Claim 1 there exists V G V such that g[a] U PF C I7. Since K(X) G cr we have 0 («(#)) € V and since W C V we have x 6 V. So {x,g(K,(x))} CV.Q By the special choice of V we now conclude that the indetity function lx and g o K are U-homotopic, i.e., P U-dominates X. D

Exercises for §4.1. Let X be a space. An equiconnecting function on X is a continuous function A : X x X x I —>• X such that A(x, x, i) = x for every £ £ I. The spaces X and Y have the same homotopy type if there are continuous functions /: X —> Y and g: Y —> X such that / o g is homotopic to the identity function on Y and g o f is homotopic to the identity function on X. 1. Present a different proof of Lemma 4.1.4 than the one in the text by using Corollary A.7.6. 2. Let r: X —>• Y be a retraction. Prove that X dominates Y. 3. Let X be an A N R with dimX < n. Prove that for every open cover II of X there exists a polytope P such that P U-dominates X and dim P < n. 4. Let X be a compact space with divaX — n < oo. Prove that there is a finite open cover U of X such that for every space Y that U-dominates X we have dimY > n. ^•5. Let X be a compact ANR. Prove that for every e > 0 there exists <5 > 0 such that for every space Y and for every closed subspace A of Y and for all maps /, g: A —> X with £(/, g) < (5, if / has a continuous extension J:Y ->X then g has a continuous extension

g:Y ^X such that g(f,g)

< e.

6. Let X be an AR. Prove that X has an equiconnecting function. 7. Show that if (X, Y) is an A(N)R-pair then both X and Y are A(N)R's. In addition, prove that X and Y have the same homotopy type. 8. Let (X, Y) be an ANR-pair. Assume that for some closed subspace A C X there is a retraction r: X —>• A having the additional property that

r[Y] C Y n A. Prove that (A, Y n A) is an ANR-pair. 9. Let X be a space with subspace Y. Suppose that there is a deformation of X through Y. Let Z be a space with closed subspace K and let e: Z —> I be continuous such that £ -1 (0) = X. Prove that for every continuous function /: Z —> X there is a continuous function g: Z —> X such that

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

277

(1) g ( f ( z ) , g ( z ) ) < e ( z ) for every z £ Z, (2) g\K = f\K, (3) g[Z \ K] C Y.

4.2. A characterization of ANR's and AR's The aim of this section is to present purely topological characterizations of the classes of all ANR's and AR's. We will demonstrate the strength of our characterization by showing that the hyperspace of each Peano continuum, i.e., a locally connected continuum, is an AR. Here is the announced characterization, stated in terms of ANR-pairs. Theorem 4.2.1. Let X be a space and let Y C X be dense. The following statements are equivalent: (1) (X,Y) is an ANR-pair, (2) for every open cover U of X there exists an open refinement VoflL such that for every locally finite simplicial complex 7 and every subcomplex S of T containing all the vertices of T, every partial realization of T in Y relative to (§,V \ Y) can be extended to a full realization of 7 in Y relative toll \Y. Observe that by Corollary 4.1.11 we only need to verify the implication (2) => (1). This will be done by proving a series of lemmas. So let (X, Y) be a pair of spaces having the realization property stated in Theorem 4.2.1(2). By Corollary 1.1.8 we may assume that X is a closed subspace of a normed linear space L. We shall prove that there are a neighborhood U of X in L and a retraction r: U —> X such that r[U \ X] C Y. We first show that this suffices. Since we will not use anything specific about L, we can replace L by the normed space L' = L x M. We identify X and X x { 0 } i n L x I . By our claim, there are a neighborhood U of X in L' and a retraction r: U —> X such that r[U \ X] C Y. This can be used to show that (X, Y) is an ANR-pair, as follows. Let E be a space wdth closed subspace F, and let

be continuous. Since X is an ANR, being a neighborhood retract of L (Theorem 1.2.9), there are a neighborhood V of F in E such that / can be extended to a continuous function g: V —>• X. Since U is a neighborhood of X in I/, the set Ur\ (X x I) is a neighborhood of X x {0} in X x I. There consequently is a continuous function a : X -» (0, 1] such that {(x,t) :0
278

4. BASIC ANR THEORY

(Lemma 4.1.4). Let g be an admissible metric on E which is bounded by 1 (Exercise A. 1.6), and define £: V ->• X x I by t(x) =

(g(x),Q(x,F)-a(g(x))).

Then £[V] CU,£\F = fand£[V\F}CU\(Xx the function /: V -> X by

{0}). Now finally define

f ( x ) = (r°$(x). Then / extends / and by the special property of r, f[V \F] CY. So we are done. For a space X we will let rX denote the family of all open subsets of X. Lemma 4.2.2. Let X be a space with subspace Y. The function a: rY -+rX

defined by a(A) = {x <= X : g(x,A) < g(x,Y\A)} (where, by convention, g(x, 0) = ooj has the following properties: (1) CT(0) = 0, a(Y) = X, (2) a (A] n Y = A for every A 6 rY, (3) if A,B G rY then ACS iff a (A) C a(B], (4) if A, B e r r theno-(AnS) = (7(^)0(7(5).

Proof. Let K: pi" —> pX be as in Lemma A.8.1. Then for every U G rl", cr([7) = X \ / c ( y \ C 7 ) . An easy check shows that a is as required.

D

It will be convenient to fix some notation. Let a: rX —> rL be the function of Lemma 4.2.2. In addition, by applying our assumptions on X and Lemma 4.1.10, by induction on n > 0 it is a triviality to construct two sequences of open (= open in X) covers An and !Bn of X with the following properties: (i) Ao = {X}, (ii) *Bn < An and for every locally finite simplicial complex T and every subcomplex S of T containing all the vertices of T, every partial realization of T in Y relative to (S, Sn \ Y) can be extended to a full realization of T in Y relative to An \ Y, (ni) for n > 1, An is a star-refinement of *B n _i, (iv) mesh(.An) < l/n.

Finally, by Lemma 1.2.1 and the fact that Y is dense in X there exists a countable open cover U of L \ X and a sequence of points (au)ueu m Y with the following properties:

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

279

(v) if p G U € U then g(p,au) < 2g(p,X), (vi) if Un G V for every n and lim

n—>-oo

then lim diam(C/n) = 0.

n—>-oo

By an appeal to Theorem 2.3.5(1) we find that there exists a countable open cover V of L \ X such that (vii) V is star-finite and refines U. So without loss of generality we may assume that U is star-finite. We also assume that every U e U is nonempty. Lemma 4.2.3. Let x 6 X and let E be a neighborhood of x in L. There exists a neighborhood F of x in L such that if U Eli meets F then U C E. Proof. Let E > 0 be such that B(x,e) C E. If the lemma is not true then for every n there exists Un E U such that

(1) un n B(x, */„) ^ 0, (2) Un\E^$. By (1), linin-^oo g(Un,X) — 0 so by (vi), lim.^-^ diam(t/ n ) = 0. This easily contradicts (1) and (2). D This simple result is used in the proof of the following lemma. Lemma 4.2.4. For each n > 0 there exists an open neighborhood Hn of X in L such that H0 = L and for n > I ,

(!)„ E_n C (2) n Hn C Hn-i, _ (3)n if U Eli meets Hn then there exists A 6 An such that

Proof. We shall construct the Hn's by induction on n > 0. Let n > 1 and suppose that Hn-i has been constructed. By Corollary A. 4. 3 there exists a neighborhood V of X in L such that For each x G X there exists Ax € .An such that x E Ax. Then cr(A x ) is a neighborhood of x in L and consequently by Lemma 4.2.3 there exists a neighborhood Bx of x in L such that if U € U intersects Bx then U is contained in a(Ax) D F. Put #n = |J B x .

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4. BASIC ANR THEORY

We claim that Hn is as required. Observe that (l) n and (2) n are trivially satisfied. For the verification of (3) ra , assume that U G U meets Hn. Since U is open, U D #n 7^ 0 and so there exists x such that UC\BX / 0. From this it follows that U C a(Ax}r\V . Since V C Hn-i, we are done. D Fix U G U for a moment and put

Since #o 7^ 0, E 7^ 0. We claim that .E is finite. Striving for a contradiction, assume that E is infinite. Then by Lemma 4.2.4(1), g(U,X) = 0 from which it follows by (vi) that diam(f/) = 0. Since U is nonempty, we find that U consists of a single point which belongs to X since g(U, X) = 0; contradiction. Now for every U G U define n(U) = max{n > 0 : U D Hn ^ 0}. Let K: C \ X ->• |AT(U)| be the ^-function of the cover U, cf. §1.1. For every U G U select a point z\j G C/. By Lemma 4.2.4(3) n (j/) we may pick an element At/ 6 -An(u) such that C7 C cr(Au). Lemma 4.2.5. For each U € U there exists a point j/[/ G A[/ n Y such that

Proof. Let U G U. If aj/ G A{/ then we put ?/[/ = aj/ and are done by (v) and the fact that a\j G Y. So assume that au 0 -4t/- Since zv G Lr C cr(Afy) it follows by the definition of a that

Since au ^ AU, (v) consequently implies that Q(ZU,AU) < g(zu,au) < 2g(zu,X). So any point in AU fl F will do.

D

Recall that by convention the vertex of N(li) corresponding to U G U is denoted by x(U), cf. §2.3. Now we define $: |A^(U)^| -> F as follows:

Since 7V(U)(°)| is a discrete topological space (Corollary 2.1.15) it follows that $ is continuous. We now aim at extending $ over a large part of |JV(U)|. For each m > 0 put Bm — Hm \ HTO-I-I,

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

281

and let

Clearly Tm is a subcomplex of N(IL). Lemma 4.2.6. If\m-n\>2 then \3>n n \fm\ = 0. Proof. This follows directly from Lemma 4.2.4(3).

D

Lemma 4.2.7. For each m > 0, <1> \ |OP m )^| is a partial realization of Tm in Y relative to (CP m )(°>,S m _i \Y). Proof. Let \{x(U^0)), . . . ,x(U^k))}\ be a simplex in 3>m. Then by the definition of lPm, for each 0 < j < k, U^ n Hm ^ 0, so that m < n(Ui^). Consequently, (ii) and (iii) imply that for each 0 < j < k,

so that by Lemma 4.2.4(3) and the fact that a is monotone there exists an element A, 6 Am such that Observe that by the definition of $ this implies that for every 0 < j < k, *(*(Ui(j)) eA^., CAj. Since |{x(t/ i(0 )), . . . ,x(C/ i(fc) )}| is a simplex in N(U), Ui(0) H • - • n C7i(fc) / 0, so that from which it follows by Lemma 4.2.2(1), (4) that A0 n • • • n Ak ^ 0.

Since by (iii) Am is a star-refinement of S m _i, there exists an element of "Bm-i which contains all Aj and hence all ^(x(Ui^. D By (ii) we may extend for every m > 1 the partial realization $ of CP2m in Y relative to ((J^m)^, ^m-i f F) to a full realization *2m: |T 2 m| ~ ^ y ,

relative to A^m-i \ Y. Now for m > 0 consider the subcomplexes O^m+i of A^(U). Observe that for every m the complex l^m+i meets its neighbors |J"2m| and |T2 m +2| only (Lemma 4.2.6). Define ^2ml =

^2ml

( 0 )

U T 2 m 1 H T2m

U ?2ml H T

2 m 2

.

282

4. BASIC ANR THEORY

Clearly, T2m_|_i is a subcomplex of O^m+i? containing every vertex of Define a function $2m+i : |^2m+i| —> Y by (X G |0> 2m

Since the functions *2m and * 2m+2 extend $ f|(T 2 m) ( 0 ) and respectively, and since by Lemma 2.1.14(1) the sets

are closed in |JV(U)|, the function $ 2m +i is well-defined and continuous. Lemma 4.2.8. For each m > 1, $ 2m+ i is a partial realization of 92m+i in relative to (T 2m+ i,!B 2m _ 2 Proof. Let m > 1 and let T G 0 5 2m _ ( _i. By Lemma 4.2.7 there exists an element B 6 !B2m such that

Since by (ii)
Let r' be a face of r such that T' 6 O^m+i n ? 2m . By construction there exists A,-/ G ^L2m-i such that

Observe that A n A T / ^ 0. Now let r" be a face of r such that T" G CP2m+i n J ) 2m+2 . By construction there also exists AT" G ^I2m+i such that

As above, A H A T // ^ 0. Now since .A2m+i < ^,2m, «^2m < ^- 2 m-i, and by (iii), yi 2m _i is a star-refinement of !B2m_2, we are done. D So by Lemma 4.2.8 and (ii), we may extend for each m > 1 the function 3>2m+i to a full realization

relative to the covering ^I 2m _ 2 \ Y. Lemma 4.2.9. For each n > 0, K,[Bn] C |?n .

4.2. A CHARACTERIZATION

283

OF ANR'S AND AR'S

Proof. Take an arbitrary x £ Bn = Hn \ Hn+i. By (vii) there exist finitely many elements of U that contain x only, say t/i(o), • • • , U^)- The simplex T =

\{x(Ui(0)),...,x(Ui(k))}\

is an element of CPn. By Lemma 2.3.2, K(X) belongs to T, and since r is a D subset of \yn\ we are done. For each n > 0 let Pn =

|J

Define * : P2 -> Y by *(ar) = * m (ar)

(x € 3 > m | ; m > 2 ) .

Then * is continuous by Lemma 2.1.19. Notice that for every m > 3, ^ \ |Pm| is a full realization of relative to Am-3 \ Y.

n

Finally, define r : H<2 —>• X by

Then r is well-defined by Lemma 4.2.9. We claim that r is a retraction. It is clear that r[H-2 \ X] C Y. Since by Theorem 2.3.3, K is continuous and ^Y is closed in L, it remains to check the continuity of r at the points of the boundary Fr.Y of X.

Lemma 4.2.10. r is continuous at every point ofFrX. Proof. Let x G FT X and let (tn)n be a sequence in H2 \ X such that limn_>oo tn — x. We shall prove that limn^00 r(tn) = x. To this end, for every n pick Un 6 U containing tn. In two steps we shall prove that lim r(tn) = lim

Claim 1. limn_j.oo Proof. Since limn-^

g(Un,X) = 0, by (vi) it follows that lim diam(f/ n ) = 0.

n—>-oo

From this we conclude that lim z\jn = x.

n—too

By Lemma 4.2.5 we therefore obtain lim g(zun,yun) < 2 lim g(zUn,X) — 0,

284

4. BASIC ANR THEORY

so that by the definition of \I>, lim fy(x(Un)} = lim yu = lim zu = x,

n—too

n—*oo

' ra— >oo

as desired.

As in §2.3 for every y e ~H2 \ X let F(y) = {U € U : y G U} and let r(y) be the simplex of -/V(U) spanned by the (finitely many) vertices {*(*/) : U € F(y)}. By Lemma 2.3.2, r(y) is the carrier of K,(y) in JV(U), so in particular,

«(2/) € r(y). Claim 2. l i m n ^ 0 0 ^ * « t n , * x C / n

=0.

Proof. For every n E N let fc(n) e N be such that t n e Bk(n} (without loss of generality, for every n, k(n) > 3). By Lemma 4.2.4(2) and the fact that the sequence (tn)n converges to x e X, it follows easily that lim^-^ k(n) — oo. Observe that for every n, r(tn) € fk(n)- By construction, for every n there exists An G -A,k(n)-3 such that *[r(t n )] C An, and since {x(f/ n ),«(f n )} C r(tn) we therefore obtain £7

*«*

C*rt

CA .

From limn^.oo fc(n) = oo we find by (iv) that lim diam(A n ) = 0,

n—>oo

so we are done.

<0>

This completes the proof.

D

Remark 4.2.11. Notice that if U in the above proof has order at most n + l, where n > 0, then the dimension of N(W) is at most < n+l. This follows from the Countable Closed Sum Theorem 3.2.8 and the obvious fact that N(U.) has no simplexes of dimension greater than n + 1 . This will be used in the proof of Theorem 4.2.30. Corollary 4.2.12. Let X be a space. The following statements are equivalent: (1) X is an A N R , (2) for every open cover U of X there exists an open refinement V of U such that for every locally finite simplicial complex T and every subcomplex § of T containing all the vertices of 7, every partial realization of 7 in X relative to (§, V) can be extended to a full realization of 7 in X relative to U.

285

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

We shall now derive some applications. Call a space X homotopically trivial if for every n > 0, every continuous function /: §n —>• X can be continuously extended over Bn+l. So every AR is homopically trivial for obvious reasons. Notice that a space X is path-connected iff every function /: §° —>• X can be continuously extended over Bl. So a homotopically trivial space is one which is 'path-connected in every dimension'. Lemma 4.2.13. Every contractible space is homotopically trivial. Proof. Let X be a contractible space and let H: X x I —> X be a homotopy contracting X to a single point, say p. In addition, let n > 0 and let g be a continuous function from Sn to X. Define g: Bn+1 —> X by g(x)=H(g(*/M),l-\\x\

("A Uj,

An easy check shows that g is the required continuous extension of g.

D

There exist elementary examples of homotopically trivial spaces that are not contractible. A well-known example of such a space is the so-called Warsaw circle in the figure below. See Exercise 4.2.1.

Figure 12. We now present the following consequence of Theorem 4.2.1. Theorem 4.2.14. Let X be a locally connected space with dense subspace Y If X has an (open) base !B such that for every finite subfamily 3 ofB: if p| y is nonempty then every component L o property that L n Y is homotopically trivial. Then (X,Y) is an ANR-pair.

has the

286

4. BASIC ANR THEORY

Proof. Let it be an open cover of X. Since 33 is an open base for the topology of X, there exists a refinement V of U consisting entirely of elements of f£>. By Lemma 4.1.10 there exists a star-refinement W of V. Since a refinement of W is also a star-refinement of V, and since by local connectivity of X components of open subsets are open (Exercise A. 2. 8), we may assume without loss of generality that W consists of components of elements of "B. We aim at applying Theorem 4.2.1. To this end, let T be a locally finite simplicial complex, let 8 be a subcomplex of T containing all the vertices of T, and let /: |S| ->• Y be a partial realization of T in Y relative to (S, W \ Y). For each a e T pick Wff € W such that f[cr n |S|] C Wff. We first want to extend / to a continuous function g: \7^ U S| —>• Y such that g is a partial realization of 1 in Y relative to (T^1) U 8, V \ Y). As to be expected, we shall construct the function g 'simplex- wise'. To this end, take an arbitrary We distinguish between two subcases. If a € 8 then / is already defined on all of a: put gff — j \ a. If a £ 8 then a e T^1) \ T^°). Since § contains all the vertices of T, crn|8| = da. So by the fact that Wo-flY is path-connected, there exists an extension ga : a —>• Wff H Y of / \ da. Now define g : \7^ U 8| —> Y as follows: g(x)=ga(x) (xea 6T ( 1 ) US). Since T^ 1 ) U 8 is a simplicial complex, g is well-defined. Also, g is continuous by Lemma 2.1.19. We claim that g is a partial realization of T in Y relative to (T^1) U S, V \ Y). To this end, take an arbitrary simplex a 6 T. First observe that By construction it therefore follows that g[a n |T(1) U S|] C ( i ^ r : r is a one-dimensional face of a} U Wff. Again since S contains all the vertices of T, each of the W T 's intersects Wa. Since W is a star- refinement of V, there consequently exists Va e V such that which is as required. Now let a 6 T be an arbitrary simplex and put £(0-) = {T e T : ( j ^ r } . Since by the local finiteness of T each vertex of a is contained in at most finitely many simplexes in T, it is clear that £(cr) is finite. Define

F(a)=

H VT. r(E£(o-)

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

287

Claim 1. For each a 6 T, every component of F(a] n Y is homotopically trivial and if a ^ T then F(a) C F(r). Moreover, if a € T(1) U S then g[o-]CF(o-). Proof. That every component of F(a) is homotopically trivial follows by the assumptions on !B and the fact that the £(cr)'s are finite. Also, if a ^ r then

£(r) C £( CT ) from which it follows that F(cr) C F(r). Now take an arbitrary a E T^1) U § and T e £(cr). Then

We conclude that #[
0

Now by induction on n > I we shall construct continuous functions / n : |o-< n >us|->r such that the following conditions are satisfied: (1) f i = 9 and /n+1 extends /n, (2) for every a e 7^ U S, /n[a] C F(a}. Assume that for certain n > 2, / n _i has been defined. We shall construct fn 'simplex- wise'. To this end, take an arbitrary a G 7^ U S . l f c r e T^"1) U S then / n _i is already defined on all of a: put fa = fn-i \ a. If a £ T(n-1) U 8 then a € 7^\7^n-^. Observe that
Since n > 2 and cr 0 T^" ), a is at least two-dimensional. Consequently, da is connected from which it follows that fn_i[da] is connected. By Claim 1 it follows that fn-i[da] is contained in a homotopically trivial subset of the intersection F(a] n Y. The function / n _i f 5cr can therefore be extended to a continuous function fff:F((r)nY. Now define /n : |T< n ) U §| -> F as follows: Since T n u S is a simplicial complex, /n is well-defined. Also, fn is continuous by Lemma 2.1.19. We conclude that fn is as required. This completes the inductive construction of the functions fn.

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4. BASIC ANR THEORY

As to be expected, define h: |T| —>• Y by h(x)=fn(x)

(xe|TuS|).

It is clear that by (1), h is well-defined. By another appeal to Lemma 2.1.19 it follows that h is continuous. By construction, h extends / and by (2), it is a full realization of T in Y relative toV\Y and therefore to U \ Y. D Corollary 4.2.15. Let C be a convex subspace of a locally convex linear space. IfCCECC then (E, C) is an AR-pair. Proof. Since convex sets in a linear space are contractible and hence homotopically trivial, and the linear space under consideration is locally convex, and the intersection of an arbitrary family of convex sets is again convex, this follows immediately from Theorem 4.2.14. D Remark 4.2.16. Since the intersection of an arbitrary family of convex sets in a normed linear space is again convex, it seems that Corollary 4.2.15 gives a new proof of Theorem 1.2.9. This is not true however, since Theorem 1.2.9 was used in the proof of Theorem 4.2.1. Corollary 4.2.17. Let 0 < n < oo and let A be an arbitrary subset of§>n. Then Bn+l \ A is an AR. Corollary 4.2.18. Let X be a locally connected space. If X has an (open) base 3 such that for every finite subfamily J of "B: if p| y is nonempty then every component C of P) IF is homotopically trivial. Then X is an ANR. Naturally, Theorem 4.2.1 suggests the question whether there exists a characterization of the class of AR's in the same spirit. In Corollary 1.4.5 we showed that a space X is an AR if and only if X is a contractible ANR. So it seems that our question has already been answered. However, deciding whether a given space is contractible is sometimes quite a complicated task, cf. Corollary 2.4.11. Within the class of ANR's it turns out that contractibility and homotopy triviality are equivalent notions. Since for a given space X, verifying that it is homotopically trivial is usually much easier than verifying that it is contractible, we shall present a proof of our assertion in full detail. We first derive the following lemma. Lemma 4.2.19. Let X be a space. The following statements are equivalent: (1) X is homotopically trivial, (2) for every locally finite simplicial complex T and every subcomplex 8 of T, every continuous function g: |S| —> X can be extended to a continuous function g: |T| —>• X.

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

289

Proof. The implication (2) =>• (1) is a triviality. Simply observe that there is a homeomorphism from Bn onto an n-dimensional simplex a taking §n-1 onto the geometric boundary da of a (Exercise 1.1.24). The implication (1) =>• (2) is also very simple to prove. Let us first show that without loss of generality we may assume that § contains all the vertices of T. Since |T^°^| is a discrete topological space by Corollary 2.1.15, and since both |S| and |T(°)| are closed in |T| by Lemma 2.1.14(1), we can extend g over |S U T(°)| in an arbitrary way. Now, by induction on n > 0 by a standard procedure we shall construct a continuous function gn: |S U 7^\ —>• X such that go = g,

gn+i extends gn.

Since S contains all the vertices of T, we may indeed put go = g. Assume that the function gn-i has been constructed for certain n > 1. As previously in this section, we shall construct gn 'simplex-wise'. To this end, take an arbitrary a€SUT(n). If a 6 S U T^71"1) put ga = gn-i \ a — 9 \ &• So assume that tre7(n)\(S\J7(n-V). Now we use our assumption on X to extend the function gn-i \da: da —> X to a continuous function ga: a —>• X. Define gn : S U 7^\ —I X by 9n(x}=gff(x}

OrEseSuT^).

As in the previous proofs in this section it follows that gn is well-defined and continuous. This completes the inductive construction of the functions gn for n 6 N. Now define g: \7\ —> X as follows: g(x}=gn(x]

(xe
Then g is clearly as required.

D

We shall now present a proof of the following Theorem 4.2.20. Let X be a space. The following statements are equivalent: (1) X is an AR, (2) X is a contractible A N R , (3) X is a homotopically trivial ANR.

Proof. Observe that the implications (1) =>• (2) => (3) are trivialities, see Corollary 1.4.5 and Lemma 4.2.13. So let X be a homotopically trivial ANR. We shall prove that X is an AR. Observe that X has the realization property stated in Corollary 4.2.1(2).

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4. BASIC ANR THEORY

It will be convenient to think of X as the space X in the proof of Theorem 4.2.1. We adopt all notation and terminology introduced in that proof. We shall prove that the retraction r: H^, —> X can be extended over L so that the desired result follows from Corollary 1.2.9 and the obvious fact that a retract of an A R is an AR. Recall that = |J m=2

Define

Q = |iPo|u|a> 1 |u|iP2|. Observe that by Lemma 4.2.6,

P 2 ng = |3>2|, and also that by definition, P2 U Q — PQ. Now since T = y0 U CPi U CP2 is a simplicial complex, and 7 2 is a subcomplex of T, by Lemma 4.2.19 we can extend ^ f l^l to a continuous function

e-.Q->x. By Lemma 4.2.9 it follows that K\C \ H?\ C Q. Consequently, the function f: C ^ X

defined by ~( } rx

=

0OK)(x)

(cec\ff2),

is well-defined if for every x e D = H2\H2 we have r(x) = (^OK)(X). To verify this, take an arbitrary x £ D. Then x E B%. Consequently, Lemma 4.2.9 gives us that K(X) 6 [O^l- By the definition of 9 we therefore have

r(x) = (# o K)(X] = (0o K)(X], which is as required. Finally, f is continuous since the sets H? and C\H<2 are closed in C and the restrictions f \ H^ and f \ C \ H-2 are both continuous. D Corollary 4.2.21. Let (X, F) be an ANR-pair. If Y is homotopically trivial then (X,Y) is an AR-pair. Proof. By the previous theorem it suffices to prove that X is homotopically trivial. By Theorem 4.1.6 there is a deformation H : X x I —>• X through Y. Let /: Sn —>• ^f be continuous, where n > 0. For sufficiently small t > 0, the functions / and Ht ° / are as close as we please, so they are homotopic by Theorem 4.1.1 since X is an ANR. Since Y is homotopically trivial, the function Ht o / can be extended over Bn+l . So the functions Ht o / and / are homotopic functions from S n into X, one of which can be extended over Bn+l . But then the other one can also be extended by the Borsuk Homotopy Extension Theorem 1.4.2. D

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

291

Application to hyperspaces. As in §1.11, for every compact space X we let 2X denote its hyperspace. As announced above, we shall apply the results in this section to prove that if X is a Peano continuum then 2X is an AR. This demonstrates the power of the techniques derived so far. We shall conclude from this by using TORUNCZYK'S Theorem from [391] that the hyperspace of each non-degenerate Peano continuum is homeomorphic to the Hilbert cube; this is the Curtis-Schori-West Hyperspace Theorem. (See also [298, Chapter 8] for details.) Since the Hilbert cube is an AR, at first glance there seems no reason for proving the AR-property for hyperspaces. However, that the hyperspace of each Peano continuum is an A R is a fundamental step in the proof of the Curtis-Schori-West Hyperspace Theorem. Lemma 4.2.22. For each n G N, there exists a map r : Bn+1 -> 5s (§n) such that for every x G S n , r(x] — {x}. Proof. First consider the case n = 1. Parametrize S1 as [0, 2?r], with 0 and 2?r identified. For 9 G [0, 27r], set to — K — \9 — TT|. Define a homotopy H: § x x [0,7r] ->> ^(S1)

by (te < t < TT).

Observe that Ho(9) — {0} for every 9 G S1. In addition, define a homotopy K: § x x [0,7r]-> Js^1) as follows: K(9,t) = {t,2K-t}\J ({0-te,0 + to}r\[t,2n-t]). It is easy to see that the homotopy H followed by K provides a homotopy from S1 into 3s (S1) connecting the inclusion map 9 -> {9} and the constant map 9 ->• {TT}. Now we proceed by induction on n. Assume that for certain n > 2 there exists a map /: Bn -»• ^(S71"1) such that f ( x ) = {x} for every x G S71"1. We identify Bn+l and the two-point compactification of Bn x (— 1, 1). Define a map g: Bn+1 -> J3(§n) as follows: ' g ( x , t ) = {(p,t) :pe f ( x } } ^(+00) = {+00}, -oo) = {-oo},

( ( x , t ) eBnx (-1,1)),

An easy check shows that g is as required.

D

A subcollection £ of3'00(X) is called an expansion hyperspace of X if for every E G £ and F G ^^(X) such that E C F we have F G £. Observe that ^oo(X) is an expansion hyperspace.

292

4. BASIC ANR THEORY

Dense expansion hyperspaces are AR's, as will be shown now. Theorem 4.2.23. Let X be a compact space. The following statements are equivalent: (1) (2) (3) (4)

X is a Peano continuum, 2X is a Peano continuum, 2X is an AR. If £ C 2X is a dense expansion hyperspace then (2 X , £) is an AR-pair.

Proof. We already know that the equivalence (1) <3> (2) is true. See Propositions 1.11.4 and 1.11.13. Since (4) =$• (3) => (2) are trivial (Exercise 1.2.6), it remains to verify the implication (1) & (2) =^ (4). Naturally, we aim at applying Theorem 4.2.14 to prove that (2X , £) is an AR-pair. Claim 1. £ is path-connected and locally path-connected. Proof. Let F 6 £, say F = {xi,...,x n }, where x% ^ Xj if i / j. If £7 is a neighborhood of F in 2X then there is a smaller neighborhood of the form {{Vi, . . . ,yn}}, where Vi is an open neighborhood of Xi for every i (Lemma 1.11.11(2)). We may assume without loss of generality that

Vi n Vj = 0 if i ^ j, and that each Vl is connected since X is locally connected. We claim that A = {{Vi, . . . , Vn}) n £ is path-connected, which will prove that £ is locally path-connected. This will implicitly also show that £ is pathconnected since £ = ({X}} n £ and X is connected. To this end, let G 6 A be arbitrarily chosen. For every i < n let Gi = G n Vi. Then every Gi ^ 0 while moreover G = \J^=1 Gi. Let us concentrate on GI = {ai, . . . , aTO} for a moment. By Theorem 1.5.22, Vi is path-connected. So there exists for every j < m a path Aj : I —> V\ such that Aj(0) = Q.J and Aj(l) = x\. The function fii : I -> {{Vi}) n £ defined by Hi(t) = {xl}\J{Xj(t)

:j<m}

is a path connecting {xi} U GI and {xi} (Corollary 1.11.10). Define the functions [ij for 2 < j < n similarly. Then the function // : I —> A defined by

is a path connecting F and F U G. Using the same technique there exists a path in A connecting FUG and G. Indeed, for j < m define Vi : I -> ({V\ }}n£ by

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

293

Define the functions Vj for 2 < j < n similarly. Then the function v : I —>• A defined by

is a path connecting .F U G and G. The join of these paths /i and v is therefore the required path.

<0>

So we see in particular that £ is locally connected. Hence components of open sets of £ are open in £ (Exercise A. 2. 8). We aim at applying Theorem 4.2.14 to prove that (2 X , £) is an AR-pair. Consider the base 'B(X) for 2X identified in Lemma 1.11.11(2). Since T>(X) is closed under finite intersections (Lemma 1.11.11(2)), it suffices to prove that every component of an element of 'B(X) \ £ is homotopically trivial. To this end, let £/i, . . . , Un C X be nonempty and open, and consider a component C of B = {{t/i, . • . , Un}) n £. Observe that B ^ 0 since £ is dense, hence C ^ 0. Claim 2. C is homotopically trivial. Proof. We already know by Claim 1 that C is open in £. By connectivity of C and the fact that £ is locally path-connected, it therefore follows by Lemma A. 10. 8 that C is path-connected. So for the verification that C is homotopically trivial it suffices to consider a continuous function g : §m —>• C, where m > 1. By Lemma 4.2.22 there exists a map r: Bm+1 -> jF3(STO) such that for every x G S m , r(#) = {x}. By Corollary 1.11.8, the function g: Bm+l ~^2X defined by

g(p) = \Jg[r(p)] is continuous. It is clear that the range of g is contained in £ and that

g\$m = gWe claim that g[Bm+l] C B. To this end, take an arbitrary;? € Bm+l . Notice that g[r(p}} consists of at most three elements of B. So each of these elements meets every Ui and is contained in their union. But then clearly |J^[r(p)] has the same property. By connectivity of Bm+l it now follows that g[Bm+l] C C, which is as required.

<0>

By Theorem 4.2.14 it therefore follows that (2X , £) is an ANR-pair. However, implicitly, we also proved that £ is homotopically trivial itself since clearly £ = {{A"}} fi £ and £ is connected by Claim 1. So we conclude that (2X , £) is an AR-pair by Corollary 4.2.21. D

294

4. BASIC ANR THEORY

Corollary 4.2.24. Let X be a Peano continuum, and £ C 1X a dense expansion hyperspace. Then there is a deformation of2x through £. Proof. Apply the previous theorem and Theorem 4.1.6.

D

Observe that since 3~oo(X) is an expansion hyperspace, this result shows that we can approximate compacta in X by finite sets in a continuous way. This curious result will turn out to be of crucial importance later. Toruiiczyk's Theorem. Let X be a space. We say that X has the disjoint- cells property provided that for every n G N, every continuous function /: I n x {0,1} —> X is approximable (arbitrarily closely) by maps sending I n x {0} and I n x {1} to disjoint sets. Formally, for every n, every continuous function / : E n x {0, 1} —> X and every £ > 0 there is a continuous function g: In x {0, 1} such that (1) Q ( f , g ) < e , (2) 0prx{0}]n0pL n x{l}] = 0. It is clear that Q has this property. To see this, let e > 0 and a continuous function /: I n x {0, 1} —>• Q be given. There exists m e N such that 2-< m - 1 ) < e. Define g: I n x {0, 1} -»• Q as follows: 9(x,i) = (/fo * ) i > " - , / ( z , * ) m - i , M , - - . ) > (« £ {0,1})An easy check shows that g satisfies (1) and (2). Toruiiczyk's Theorem states (among other things) that Q is topologically the only AR with the disjointcells property. Toruiiczyk's Theorem 4.2.25. Let X be a space. The following statements are equivalent: (1) X is homeomorphic to the Hilbert cube Q. (2) X is a compact AR and has the disjoint-cells property. For details, see TORUNCZYK [391] or VAN MILL [298, Chapter 8]. So all there remains to prove for the Curtis-Schori-West Hyperspace Theorem is that 1X has the disjoint-cells property. Lemma 4.2.26. Let X be a Peano continuum and let "X, be a compact subset of TOO (X). Then for every e > 0 there is a continuous function

such that QH(/, 1) < £• Proof. Define £ = {F e 3"oo(X) : F is not contained in an element of OC}.

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

295

Observe that £ is an expansion hyperspace of X and that £ Pi 3C = 0. We claim that £ is dense in 2X . Since yoo(X) is dense in 2X (Lemma 1.11.11(2)), it suffices to prove that for arbitrary F G J'00(X) and e > 0 there exists an element E e £ with gH(F,E] < e. So let F e Joo(^) and e > 0 be given. Pick a point x E F and a sequence (xn)n in X \ {x} converging to x, with Q(xn,x} < £ for each n. Define Observe that for each n, Fn € ^^(X) and gn(F,Fn) < e. We claim that for some n, Fn € £. If not, then there exists a sequence (Kn}n in 3C such that Fn C Kn for each n. By compactness of 3C we assume without loss of generality that Kn —»• K , n —> oo. Claim 1. (xi, £2,0:3,. . . } C ^. Proof. Suppose that for some n, xn £ K. Since Kn —> K, n —> oo, there is an ra > n such that Km C K \ {#„}. But this is impossible since xntFmCKmCK\ {xn}. So we are done.

•<)

From the claim we conclude that K is infinite, which contradicts the fact that K is finite. We see that £ is dense in 2X . By Corollary 4.2.24 there are arbitrarily small maps 2X —> £. Since £ n IK = 0, we are done. D The Curtis-Schori-West Hyperspace Theorem 4.2.27. Let X be compact space. The following statements are equivalent: (1) X is a Peano continuum, (2) 2X is homeomorphic to the Hilbert cube Q. Proof. As observed above, it suffices to prove that for a Peano continuum X, 2X has the disjoint-cells property. Since 2X admits arbitrarily small maps into ^^(X) (Corollary 4.2.24) and since for every compact subset % of jFoopsT) there are arbitrarily small maps from 2X into 2X \ % (Corollary 4.2.26), this is obvious. D Characterization of finite dimensional ANR's and AR's. We shall now characterize the class of all finite dimensional ANR's and AR's in simple topological terms. Let X be a space and let 0 < n < oo. We say that X is connected in dimension n, abbreviated C n , provided that for every 0 < ra < n, every continuous function /: STO —> X can be extended to a continuous function /: Bm+1 ->• X. So the homotopically trivial spaces are precisely those spaces that are connected in every dimension. In addition, we say that X

296

4. BASIC ANR THEORY

is locally connected in dimension n, abbreviated LC n , provided that for every x E X and for every neighborhood U of x and for every 0 < m < n there exists a neighborhood V of x such that every continuous function /: Sm —> V extends to a continuous function /: Bm+1 —>• C7. We shall prove that if dim A" = n < oo then X is an ANR if and only if X is LCn and also that A7" is an AR if and only if X is both LC n and C n . The proof of the following lemma is precisely the same as the proof of Lemma 4.2.13 and is therefore left as an exercise to the reader. Lemma 4.2.28. Let X be a space. IfX is locally contractible then X is LC n for every n. Proposition 4.2.29. Let X be a space and let 0 < n < oo. The following statements are equivalent: (1) X isLC n . (2) for every open cover U of X there exists an open refinement V of U such that for every locally finite simplicial complex T with

dim |T| < n + 1 and every subcomplex 8 of T containing all the vertices of T, every partial realization of T in X relative to (§, V) can be extended to a full realization of 7 in X relative to U. Proof. We prove (2) => (1). Take an arbitrary x G X and a neighborhood U of x. There exists an open neighborhood W of x such that W C U. Consider the open cover U = {U,X\W} of X. By assumption, for the open cover U there exists an open refinement V such as in (2). Pick V G V such that x G V. Fix an integer m < n, and let a be an arbitrary (m + l)-dimensional simplex in M m+1 . It will be convenient to identify Sm and da (Exercise 1.1.24). Now let 3~(cr) be the simplicial complex consisting of all the faces of a and let 8 be the subcomplex consisting of all proper faces. Consider a continuous function /: da —> WnV and observe that / is a partial realization of ^(cr) in X relative to (S, V). By assumption, / can be extended to a full realization /: a = |3~(0")| —> X relative to U. Consequently, there exists U' G U with f[cr] C U'. Since / extends /, U' n W + 0, i.e., U' = U. We prove (1) => (2). Let U be an open cover of X. By downward induction, we shall construct open covers U n , V n , U n _ i , V n _ i , . . . , U Q , V0 of X having the following properties:

(3) Un = U,

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

297

(4) if 0 • V there exist U € Hi with V C U and a continuous extension /: Bi+l ->• C7 of /, (5) if 0 < i < n — 1 then Ui is a star-refinement of Vi+i (Observe that (4) implies Vi < Hi.) The construction of these covers is a triviality. By Lemma 4.1.10 there is never trouble with the construction of the U n 's. Let us construct the cover Vn. Take an arbitrary x € X and pick Ux G lin containing x. Since X is LCn there is an open neighborhood Vx of x such that Fx C Ux while moreover every continuous function / : S n —>• Vx can be extended to a continuous function /: Bn+l —> Ux. The cover Vn = {Vx : x G X ]

is clearly as required. The construction of the other V^'s is similar. Now put V = VQ, let T be a locally finite simplicial complex with dim |T < n + 1, let § be a subcomplex of T containing all the vertices of T, and let /: |S| —> X be a partial realization of T in X relative to (S, V). By induction, we shall construct for every 0 < i < n a continuous function

having the following properties: (6) /0 = / and for 1 < i < n, fi extends /;_i, (7) for 0 < i < n, fi is a partial realization of T in X relative to the pair Since § contains all the vertices of T, our choice /o = / is as required. Now assume that for certain 1 < i < n the function /;_i has been constructed. As in the previous proofs, we shall construct fl 'simplex- wise'. Let a be an element of T^ \ (S U1J^~1)). By (7) there exists an element V e V»_i such that fi[da] C F. So by (4) there exist U £ U z _i with V C. U and a continuous extension /CT : a -)• f/ of the function /;_i f 9cr. Now define £ : |§ U T^ | ->• A" as follows:

By Lemma 2.1.19 fi is continuous. We now claim that fi is a partial realization of T in X relative to (S U T^, V^). To this end, take an arbitrary simplex o 6 T. By (7) there exists V e V,_i such that f^i[(7r\\B\J7^-^\] C V. Let T be a face of a such that r e T^. By construction there exists U E Ui_i such that fi[r] C U. Observe that V intersects U. Consequently,

Now since V;_i < Ui-i and by (5) Ui_i is a star-refinement of Vi, we are done. This completes the inductive construction.

298

4. BASIC ANR THEORY

With the same technique it is clear that we can extend fn to a continuous function / n + 1 :|SUT< n + 1 > -+X such that for every a € T(n+1) there exists U £ Un — U such that fn+i[a]CU. n+1

However, by assumption j( tion of T in X relative to U.

) = T so that fn+i is the required full realizaD

We now come to the announced characterization. Theorem 4.2.30. Let X be a space and let 0 < n < oo. The following statements are equivalent: (1) X isLCn. (2) for every space Y and for every closed subspace AofY with dim(Y\A} • X can be continuously extended over a neighborhood of A, (3) for every x e X and for every neighborhood U of x there exists a neighborhood V of x with V C U such that for every space Y with dim Y < n, for every continuous function f : Y -» V there exists a homotopy H : Y x I —>• U such that HQ = f and HI is a constant function. Proof. We prove (1) => (2). Let Y be a space, let A C Y be closed such that dim(F \A) < n + l and let / : A —> X be continuous. Without loss of generality we assume that X and Y have empty intersection. Claim 1. The set C = (Y \ A] U X can be topologized in such a way that (3) X is a closed subspace of C, (4) Y \ A is a subspace of C, (5) the function f-.Y^-C defined by

-

_ f z

f ( x )}

(zeY\ (zeY\A),

- \ /(*) (z e A).

is continuous. Proof. Let a: rA ->• rY be a function such as in Lemma 4.2.2 and let 'B be a countable open base for X. In addition, let 5" be a countable open base for Y \ A. For every B e 13 and n € N define B(ri) CC by B(n) = {y£(Y\A)n a ( f - l [ B ] ) : g(y, A) < i/n} U B.

4.2. A CHARACTERIZATION OF ANR'S AND AR'S

299

An easy check shows that the collection 3~ U {B(n) : B e 23, n G N} is as a base for the required topology on C. 0 We shall now prove that X is a neighborhood retract of C. Once this has been established, by part (5) of the claim we are done. The proof of our assertion is precisely the same as the proof of Theorem 4.2.1. Only two minor changes should be made. We shall adopt the notation and the terminology introduced there, we shall indicate the required changes and we shall leave the precise verification to the reader. By Proposition 4.2.29 we have a realization property available up to dimension n + 1. So in the construction of the covers An and 23n, one should replace 'locally finite simplicial complex T' by 'locally finite simplicial complex T with dim T| < n + 1' everywhere. The special open cover U of C \ X used in the proof of Theorem 4.2.1 should have the additional property that its order does not exceed n + 1. However, by the fact that dim(C \ X) < n + 1, this can be achieved by Theorem 3.2.5(5). This implies that the nerve N(U) has dimension at most n + 1 (Remark 4.2.11). With these two small changes, the proof of Theorem 4.2.1 can now be followed verbatim. We prove (2) => (3). Suppose that (3) is not true at the point p. Then there exists e > 0 such that for every z € N there exist a space Yi with dim Yi < n and a continuous function /;: Yi —> B(p, e/^) that is not homotopic within B(p, e) to a constant function. Define A and Y by

(j(Y, x n n i i u

^v and

r= i=l

x

Here oo ^ Z^i(^ ^)- Topologize r by requiring that X^i(-^ x ^) ^s an open subspace of Y and a basic neighborhood of oo has the form X^/c(^ x ^) for fc G N. It is easy to see that r is a separable metrizable space. Now define the function /: A —> X as follows

< f(x,V=P

(ser')i

( /(oo) = p Then / is clearly continuous. Observe that by Theorems 3.4.12 and 3.2.9 we have dim(l A \A) < n + l so that by assumption there exists a neighborhood W of A in Y such that / can be extended to a continuous function f:W—>X. Then V = f ~ l [ B ( p , e}] is a neighborhood of oo in Y and therefore contains all but finitely many of the Yi x I, i 6 N. But this implies that for all but

300

4. BASIC AMR THEORY

finitely many i £ N the function /j is homotopic within B(p,e) to a constant function. This is a contradiction. We prove (3) => (1). Since dim§ m = m for all m (Theorem 3.2.12), this is clear. D Theorem 4.2.31. Let X be a space and let 0 < n < oo. The following statements are equivalent: (1) X isLC n andCn. (2) For every space Y and for every closed subspace AofY with dim(Y\A) < n + 1, every continuous function f : A —> X can be continuously extended over Y, (3) X is LCn and for every space Y with dim Y < n, every continuous function f : Y —>• X is nullhomotopic. Proof. The proof of (1) =^ (2) is precisely the same as the proof of the implication (3) => (1) in the proof of Theorem 4.2.20 and is therefore left as an exercise to the reader. We shall now prove (2) =» (3). That X is LCn follows from Theorem 4.2.30. Let Y be a space with dimF < n and assume that /: Y —> X is continuous. Since by Theorems 3.4.12 and 3.2.8 we have dim A(Y~) < n + 1 (here A(F) denotes the cone over Y of course), by (2) we conclude that / can be extended to a continuous function /: A(F) —>• X. The contractibility of A(y) (Exercise A.12.5) implies that / is nullhomotopic. From this it follows easily that / is nullhomotopic as well. The proof of (3) => (1) is a triviality of course since dim§ m = m for every m (Theorem 3.2.12). D These results have the following corollaries. Corollary 4.2.32. Let X be a space. The following statements are equivalent: (1) X is a Peano continuum, (2) there is a continuous surjection f : I —> X. Proof. Assume first that X is a Peano continuum. By Theorem 1.5.10, there exists a continuous surjection g: C —>• X, In addition, by Theorem 1.5.22, X is LC° and C°. Consequently, an easy application of Theorem 4.2.31 yields that g can be extended to a continuous function /: I —> X. The implication (2) =>• (1) follows from Exercises A.5.5 and A.2.11. D

4.3. OPEN SUBSPACES OF ANR'S

301

We can now state our characterization theorem for finite dimensional ANR's and AR's. Theorem 4.2.33. Let X be a space, let 0 < n < oo and let dimX < n. Then (1) X is an ANR iff X is locally contractible iff X is LCn, (2) X is an AR iff X is LCn and Cn. Proof. We shall first present a proof of (1). To this end, first observe that if X is an ANR then X is locally contractible (Exercise 1.2.1) and consequently LCn by Lemma 4.2.28. Conversely, assume that X is LC n . Then X is locally contractible by Theorem 4.2.30(3). Consequently, X is LCm for every m (Lemma 4.2.28). Now let m = 2n + 1. By Theorem 3.3.5, we may assume that X is a subspace of §m. Put Z = {x 6 Bm+1 : \\x\\ < 1} U X. Then Z is an AR by Corollary 4.2.17 and X is clearly closed in Z. Since dimja: 6 Bm+l : \\x\\ < 1} < m + 1, by Theorem 4.2.30 it follows that X is a neighborhood retract of Z. Since Z is an AR, Proposition 1.2.10 implies that X is an ANR. We shall now present a proof of (2). To this end, assume that X is LC n and Cn. By Theorem 4.2.31 we conclude that X is contractible and hence Cm for every m (Lemma 4.2.13). Now define Z as in the proof for (1). By Theorem 4.2.31(2) there is a retraction r: Z —> X from which we conclude that X is an AR, as required. D Remark 4.2.34. Unfortunately, a characterization of infinite-dimensional ANR's and AR's as simple as Theorem 4.2.33 does not seem to be possible. BORSUK [70, p. 124] constructed an example of a contractible and locally contractible compactum which is not an ANR. Exercise for §4.2. 1. Give an example of a homotopically trivial compact space which is not contractible 4.3. Open subspaces of ANR's

In this section we shall prove that every open subspace of an A N R is again an ANR and that every space that admits an open cover by ANR's is an ANR as well. As an application we conclude that every polytope is an ANR. We also show that every A(N)R has an A(N)R completion. Theorem 4.3.1. Let (X, Y) be an ANR-pair and let U be an open subspace o f X . Then (U, U n Y) is an ANR-pair.

302

4. BASIC ANR THEORY

Proof. Let Z be a space, A C Z be closed, and let g : A —> U be continuous. Since (X,Y] is an ANR-pair, there exists an open neighborhood V of A in Z such that g can be extended to a continuous function / : V ->• X while moreover /(V \ A] C F. Since / is continuous and extends g, and since U is open in X, we conclude that W — f~l[U] is an open neighborhood of A. Then g — f \ W : W —>• C7 is a continuous extension of g. In addition, if z belongs to W \ A then g(z) — f(z] G U n F. So we are done. D Corollary 4.3.2. Let X be an ANR and ]et U be an open subspace of X. Then U is an ANR. Observe that [—1,0) U (0,1] is not an AR but is an open subspace of the AR JJ. So an open subspace of an AR need not be an AR. We now aim at proving that a 'local' ANR is an ANR. First we need to derive a few elementary results. In our proofs in this section we make use of cones. As we saw in Exercise A. 12. 6, if A is a (closed) subspace of a space X then A (A) can be thought of as a (closed) subspace of A(X). It will be convenient to do that. We will also use the fact that the cone of an an ANR is an AR (Theorem 1.4.6). Proposition 4.3.3. Let X be a space and assume that X can be written as U U V, where both U and V are open in X . I f U and V are ANR's then so is X. Proof. We shall prove that the cone over X is an AR so that X is an ANR by Theorem 1.4.6. By Proposition A. 7.1 there exist closed sets E and F in X such that ECU,FCV andE\jF = X. Observe that C A(C7),

A(F) C A(V),

A(E) U A(F) =

Now let Y be a space, let A C Y be closed, and let /: A —» A(X) be continuous. Put E' = f~l[A(E)] and F' = f~l[&(F)], respectively. By Lemma A. 8.1 there exist closed sets E" and F" in Y such that Since U n V is an open subspace of U, it is an ANR by Corollary 4.3.2. Consequently, Theorem 1.4.6 implies that the function / \ E' n F': E' n F' -)• A(E n F)

can be extended to a continuous function g: E" n F" —> A(t/ D V). Now define fE:E'U (E" n F") -+ A(t7) by

Then /# is obviously continuous. By another application of Theorem 1.4.6, there exists a continuous extension /i: E" —> A(£7) of JE- Similarly define fp

4.3. OPEN SUBSPACES OF ANR'S

303

and a continuous extension /2 : F" ->• A(V) of /F. Now define f:Y-> in the obvious way, namely,

Then / is the required continuous extension of /.

D

Lemma 4.3.4. Let X be a space having an open cover U by pairwise disjoint ANR's. Then X is an ANR. Proof. This is a triviality. Assume that X is a closed subspace of a space Y. By Lemma 4.2.2, for every U e It there exists an open subset V(U) of Y such that (2) the collection {V(U) : U e U} is pairwise disjoint. Note that U is closed in V(U) for every U e U. So by our assumptions on U, for every U € U there exist an open subset W(U) of V(U) which contains U and a retraction r\j\ W(U) —>• C7. Observe that the W(U)'s are open in Y since the V(C/)'s are. Now put VF = Uc/eu ^(^0- Then VF is a neighborhood of X in F and the function r: W —> X defined by r(x) = rv(x}

(x 6 W(U), U € U)

is a retraction.

D

We now come to the following Theorem 4.3.5. Let X be a space and suppose that X admits an open cover U consisting of ANR's. Then X is an ANR. Proof. Without loss of generality assume that U is countable. Enumerate U as {Un : n 6 N}. Since by Proposition 4.3.3 the union of finitely many elements of U is an ANR, we may assume without loss of generality that for every n, Un C Un+i. For every n, put Vn = {xeX:g(x,X\Un)>

l

/n}.

It is clear that each Vn is open, that Vn C V n +i, and that (J^Li Vn — X. Finally define Vn \ Vn-2 (n>3). Then X — U^Li Rn d for \m — n\ > 2, Rn D -Rm = 0. Also, ,Rn is an open subset of the ANR Un. So by Corollary 4.3.2, we conclude that Rn is an ANR. Consequently, Lemma 4.3.4 implies that an

~ LJ ^2n-!'

n=l

^= LJ

304

4. BASIC ANR THEORY

are open ANR subspaces of X. Since their union equals X, we infer by Proposition 4.3.3 that X is an ANR. D As announced in §2.1, we now get Corollary 4.3.6. Every poly tope is an A N R . Proof. Let 7 be a locally finite simplicial complex and consider P = |T|. Take an arbitrary x € |T(°)|. Since T is locally finite, the collection § = {r 6 T : x € r}

is finite. Consequently, J* = {a e 7 : (3 r 6 S)(<J ^ r)} is a finite subcomplex of T. From Theorem 2.1.24 we conclude that |?| is an ANR. Since Stx, the star of x, is contained in |T| and is open in P by Corollary 2.1.17, we conclude from Corollary 4.3.2 that Star is an ANR. Since by Corollary 2.1.17, is an open cover of P, Theorem 4.3.5 implies that P is an ANR.

D

Enlarging A(N)R's. We now show that every A ( N ) R has a completion which is also an A(N)R. Theorem 4.3.7. Let Y be a space with A ( N ) R subspace X. Then there is a subspace SofY such that X C S C X while moreover (1) 5 is a Gs-subset ofY, (2) ( S , X ) is an A(N)R-pair, Proof. It is clear that we may assume without loss of generality that X is dense in Y . We first claim that the theorem for ANR's implies the theorem for AR's. For let X be an AR. By the ANR-case of the theorem, we may assume that for some 5 with X C S C Y we have that (5, X] is an ANR-pair. Since X is homotopically trivial, being an AR, we conclude by Corollary 4.2.21 that ( S , X ) is an AR-pair. So now assume that X is an ANR. We begin the proof as the proof of Corollary 1.1.8. Let oY be a compactification of Y (Corollary A. 4. 8). By Lemma 1.1.6 we may think of aY as a linearly independent subspace of a Banach space of the form C(A) (here A is a compact space). By linear independence, if EQ is the linear hull of X then EQ D aY — X. Hence AT is a closed subspace of EQ. Since X is an A N R , there are an open neighborhood UQ of X in EQ and a retraction r: UQ —» X. Select an open subset U in C(A) such that U D EQ — UQ. Consider the closure EQ of EQ in C(A). Since X is dense in aY it follows that aY C EQ. Also,

U = UnE0 c E / n E -

4.3. OPEN SUBSPACES OF ANR'S

305

Since aY is topologically complete, being compact, there is a GVsubset T of U C\ EQ which contains UQ such that r can be extended to a continuous function f: T ->• aY (Corollary A.8.4). Observe that T is a GVsubset ofC(A) since Ur\E0 obviously is and T is a GVsubset of UC\EQ. In addition, X C T since X C UQ. Put S — T n Y. Observe that S is obviously a G^-subset of Y. In addition, f(y) — y for every y e S. This is so since X is dense in S (Exercise A. 1.9). Now put S^f-^S}. Then f f Si: Si —> 5 is a retraction. Observe that

C7n£ 0 = c/o c Si c unE0. Since (Eo,Eo) is an ANR-pair by Corollary 4.2.15 it follows that

(Ur\E0,ur\E0) is an ANR-pair as well (Theorem 4.3.1). We conclude from this that (S^UHEo) is an ANR-pair. Since f is a retraction from Si onto S such that

f[u n EQ] = r[U n EQ] = X it follows by Exercise 4.1.8 that that (S,X) is an ANR-pair.

D

Corollary 4.3.8. Let X be an A ( N ) R . Then there exists an A ( N ) R S and an imbedding i: X <-) S such that (1) i[X] is dense in S, (2) (S,i[X]) is an A(N)R-pair, (3) S is topologically complete. Proof. Let aX be a compactification of X (Corollary A.4.8). It follows from Theorem 4.3.7 that there is a G^-subset 5 of aX which contains X while moreover (S,X) is an A(N)R-pair. But then S is an A ( N ) R and moreover is topologically complete since aX is (Theorem A.6.3). D Exercises for §4.3. 1. Give an example of a space X which can be written as EUF, where both E and F are closed ANR subspaces of X, while X is not an ANR. 2. Let X be the union of two open subspaces U and V such that U, V and U n V are AR's. Prove that A' is an AR. ^•3. Let X be a space having a point x such that (1) x has arbitrarily small homotopically trivial neighborhoods, (2) A" \ {x} is an ANR. Prove that X is an ANR. (This generalizes Theorem 1.4.6.)

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CHAPTER 5

Basic infinite-dimensional topology In this chapter we shall prove some elementary results from infinitedimensional topology. The results are elementary in the sense that no powerful apparatus is needed, but the proofs are not always easy. In infinite-dimensional topology one studies the topology of objects such as the Hilbert cube Q, the Hilbert space I2 and its infinite-dimensional linear subspaces, the countably infinite product of lines M°° and its infinitedimensional linear subspaces, and manifolds modeled on them. See the notes for more information. In this chapter we develop basic homeomorphism theory in Q. Much emphasis is on so-called absorbing systems. They are the basis for the topological classification of function spaces of low Borel complexity in the next chapter. 5.1. Z-sets Homeomorphism extension results are useful in infinite-dimensional topology. In this section we shall present a class of subsets of Q, the so-called class of Z-sets, for which an important homeomorphism extension result shall be derived in §5.3. The concept of a Z-set does not seem to be very appealing, yet it is the most central concept in infinite-dimensional topology. It turns out, although this is not immediately clear from the definition, that Z-sets are 'small'. Here smallness has a different meaning than being 'small' with respect to category or measure. It is 'small' in the sense of homotopy. This will be made precise in Exercise 5.1.7 below. Let X be a space. A closed subset A C X is called a Z-set in X provided that for every open cover U of X and every function / e C(Q,X) there is a function g € C(Q,X) such that (1) / and g are U-close, (2) g[Q] n A = 0.

A a Z-set is a countable union of Z-sets. The collection of Z-sets and aZ-sets in X are denoted by Z(X) and Z a ( X ) , respectively. 307

308

5. BASIC INFINITE-DIMENSIONAL TOPOLOGY

It will sometimes be convenient to have a 'metric' translation of the concept of a Z-set. For compact X the following lemma is a triviality since then every open cover has a Lebesgue number. For general X, the proof is only slightly more complicated. Lemma 5.1.1. Let X he a space and let A C X be closed. Then the following statements are equivalent: ( 2 ) V e > O V / e C ( g , X ) 39£C(Q,X\A): Q ( f , g ) < e . Proof. Since (1) =>• (2) is trivial, it suffices to establish the reverse implication. Let U be an open cover of X, and let /: Q —> X be continuous. By compactness of /[Q], there exists 6 > 0 with the property that every A C X with diam(A) < 6 and which moreover intersects f[Q], is contained in an element U E U (Lemma A. 5. 3). By (2) there is a function g G C(Q,X) with £>(/, g} < 6 and g[Q] n A = 0. We claim that / and g are U-close. This is a triviality. Pick an arbitrary x G Q. Then @(f(x),g(x)) < 6 and since f(x) £ /[Q], {f(x),g(x)} is contained in an element of U. D It is also possible to use finite dimensional cubes to characterize Z-sets, see Exercise 5.1.3. We shall now derive some elementary properties of Z-sets. Lemma 5.1.2. Let X be a space. Then (1) If A e Z(X) and B C A is closed in X then B e Z(X}. (2) If A G Z(X) then A has empty interior in X. (3) If (X, g) is complete, A 6 Zo-(X) and A is closed then A e Z(X); in particular, finite unions of Z-sets are again Z-sets. (4) If (X, Q) is complete and A 6 Za(X] then for every e > 0 and for every f E C(Q,X) there exists g 6 C(Q,X) such that g(f,g) < e and g[Q] n A = 0. (5) If A e Z(X) and Y is any space then A x Y e Z(X x Y}. (6) If A e Z(X) and h e ttpf) then h[A] e Z(X). (7) If A C X is closed and for every e > 0 there exists / e C(X, X \ A) such that g(f, !*•) < £ then A e Z(X). Proof. Statement (1) is trivial. For (2), (5) and (6), see Exercise 5.1.1. For (4), write A = \J^=lAn with An 6 Z(X) for every n G N. Using Lemma 5.1.1 and Exercise A. 5. 3, it is clearly possible to construct a sequence of maps fn e C(Q, X \ An), n 6 N, such that (8) e ( / i , / ) < * / 2 ,

5.1. Z-SETS

309

(9) £(/„+!, / n ) < 3 - n - * / 2 ,

(10) 0(/ n +i, /n) < 3~n •min{e(/ i [g],A i ) : 1 

~~ n-»oo ^—' m=0

3~ m • e/2

- 3/2 ' £/2

From this we conclude that g(F, f) < e. It now suffices to prove that F[Q] misses A. This however is an immediate consequence of (10) and Lemma 1.6.1. Observe that (3) is a direct consequence of (4). For (7), let £ > 0 and g G C(Q,X). Find / G C(X,X \ A) such that g(f, lx) < £• Put h = f °g. Then clearly h[Q] r\A = 0, while moreover,

g(h, g) = g(f o g, g) < £(/, lx) <£ (Exercise 1.3.2). We conclude that A G Z(Q) (Lemma 5.1.1).

D

We are interested particularly in Z-sets in the Hilbert cube Q. The following lemma provides us with 'many' Z-subsets of Q. Lemma 5.1.3. Let A C Q be a closed set. Then: (1) A is a Z-set if and only if for every e > 0 there exists f G C(Q, Q) such that g(f, 1Q) < e and f[Q] n A = 0. (2) Ifirn[A] ^ Sn for infinitely many n G N, then A G Z(Q). (3) If 7rn[A] C {-1,1} for certain n G N, then A G Z(Q). Proof. For (1), let A G Z(Q). Since 1Q G C(Q,Q), the definition of a Z-set immediately gives us that for every e > 0 there exists / G C(Q,Q) with g(f, IQ) < e and f[Q] fl A = 0. The converse is a direct consequence of Lemma 5.1.2(7). For (2), choose £ > 0 and find n G N so large that 2~( n-1 ) < e while moreover ifn[A] ^ JJ^. Take an arbitrary t G JJ n \7r n [.4], and define /: Q —> Q by Then clearly f[Q] n A = 0 and g(f, 1Q) < e. So by (1), A G Z(Q). For (3), first observe that { — 1,1} G %(J). We consequently get what we want by first applying (5) and then (1) of Lemma 5.1.2. D

310

5. BASIC INFINITE-DIMENSIONAL TOPOLOGY

Remark 5.1.4. Observe that this lemma implies that each singleton subset of Q is a Z-set. Similarly, every endface of Q is a Z-set. So Q has 'many' Zsubsets, There are also spaces with 'few' Z-sets. It is easy to see for example that the collection £(J) is equal to

At first glance it seems a little surprising that all points in Q are Z-sets. However, we already know that Q is homogeneous (Theorem 1.6.6), so once one singleton Z-subset has been found it follows automatically that all singleton subsets share this property. Corollary 5.1.5. (1) B(Q) 6 Zff(Q), (2) ifKCs is compact then K e Z(Q).

Exercises for §5.1. A closed subset A of Q is called an A-set if for every open U C Q such that U is nonempty and homotopically trivial, the set U \ A is again nonempty and homotopically trivial. 1. Let X (1) (2) (3)

be a space. If A £ Z(X) If A £ Z(X) If A £ Z(X)

Prove the following statements: then A has empty interior in X. and Y is any space then A x Y € Z(X x Y). and h £ ft (A") then /i[A] £

2. Give an example of a space A" which can be covered by a countable union of Z-subsets. (This shows that in Exercise 5.1.1(2) it is not possible to replace Z-set by er Z-set.) 3. Let X be a space and let A C X be closed. Prove that the following statements are equivalent: (1) A £ Z(X), (2) V n £ N V e > 0 V / £ (7(1™, A) 3g E C(I n , X \ A) : g(f.g) < s. 4. Let n > 1 and n

A C Bn = {x € Rn : ^ Xi < l} i=l

n

be closed. Prove that A e 2,(B ) iff ^ C S""1. 5. Prove that an A-set in Q is nowhere dense. •6. Let A C Q be an A-set. Prove that for each polyhedron P, for each s > 0 and for each / £ C(P, Q) there exists 5 £ c7(P, Q) such that £(/, 5) < £ and 0 P n A = 0. Let A C

be closed. Prove that A is an A-set iff A is a Z-set.

5.2. EXTENDING HOMEOMORPHISMS IN s

311

8. For every n let Xn be a space. Assume that a closed set 00

ACX = Yl*n

has the property that 7rn[A] ^ Xn for infinitely many n. Prove that A is a Z-subset of X.

5.2. Extending homeomorphisms in s The aim of this section is to prove that if £", F C s are compact and / is a homeomorphism from E onto F which moves the points less than e then it can be extended to a homeomeomorphism of Q that satisfies the same smallness condition. An element h 6 ^K(Q) is boundary preserving if h[B(Q}] = B(Q), or, equivalently, h[s] — s. Lemma 5.2.1. For every compact subset K C s such that K ^ 0 there is a boundary preserving h 6 IK(Q) such that Tt\h\K] = {0}.

Figure 13.

Proof. Without loss of generality we may assume that

n=l

where — 1 < an < bn < 1 for n 6 N. For n > 1, let hn : Si x J n ->• Ji x J n be a homeomorphism of Ji x JJn having the following properties (see Figure 13): (1) hn does not change the Ji-coordinate of any point, (2) every horizontal line intersects hn [[ai, bi] x [an, bn]] in a set of diameter at most l/n.

312

5. BASIC INFINITE-DIMENSIONAL TOPOLOGY

It is geometrically obvious how to define hn. Let U be the region in Figure 14 below the dotted line. Similarly, V is the region above the dotted line. We will describe hn \ U only, hn \ V being entirely similar. Let t 6 JJ be fixed and let t be the vertical line through the point (t, — 1) in Ji x J n . In Figure 14 on the line i four bullets are drawn, the first one of which is x\ — (t, — 1) and the second one is x-2 = (t,an). The third and fourth bullet will be denoted by x% = (t, 6) and x± = (t,c), respectively. Here an < b < c. Now define hn \ {t} x $n by requiring that the interval from x\ to x2 is mapped linearly onto the interval from x\ to x 3 , and the interval from x 2 to x4 linearly onto the interval from xj, to x 4 . Etc. Then hn is as required. Define / : Q -»> Q by where (xi:yn) = hn(x\,xn] for every n > 1. The composition -K\ o f : Q is equal to vri and is therefore continuous. For n > 2, the composition is equal to the composition pohnO 7T{1>n},

where p: Ji x JJ n ->• J n is the projection.

Figure 14. Hence 7rn o / is continuous, being the composition of three continuous functions. By the definition of the product topology on Q it therefore follows that / is continuous. It is easy to display the continuous inverse of /. Alternatively, one can show that / is a bijection. It then suffices to apply Exercise A.5.9 to conclude that / is a homeomorphism. We shall prove here that / is surjective only. To this end, let y € Q be arbitrary. Put Xi = y\. Since hi: JJi x JJ2 -* Ji x J 2

5.2. EXTENDING HOMEOMORPHISMS IN s

313

is a homeomorphism, there exists a point (xi,a) 6 JJi x ^2 such that

Put X2 = a. Etc. Then clearly f ( x ) = y. That / is boundary preserving follows easily from the fact that all hn are boundary preserving.

Figure 15. Now think of Q as Ji x Qi and take two points in f[K] having the same second coordinates, i.e., points of the form (x,z) and ( y , z ) . If x ^ y then \x - y\ > l/n for certain n > 2. By the definition of / we have that the points (x,zn) and (y,zn) both belong to /i n [[ai,&i] x [a ra ,6 n ]]. But this contradicts (2). So we conclude that the function g: f[K] -> Qi defined by g(x,y) = y is one-to-one and by compactness is therefore an imbedding (Exercise A. 5.9). See Figure 15. Let B = gf[K] and let

be the inverse of g. In addition, let £: B —> [ai,&i] be

[ai,&i]. For every t G [ai,&i] let Ht be the unique homeomorphism of Ji taking [—!,£] linearly onto [—1,0] and [t, 1] linearly onto [0,1]. Then H: JJi x [01,61] -4 Ji defined by H(x,t) — Ht(x) is an isotopy (Exercise 1.8.1). Now put F(x,y) = ( H ( x , X ( y ) ) , y ) . Then F belongs to IK(Q) by Theorem 1.8.2. In addition, F has the property that for every ( x , y ) 6 /[AT], F(x,y) = ( H ( x , X ( y ) ) , y ) = ( H ( x , x ) , y ) = (0,y).

314

5. BASIC INFINITE-DIMENSIONAL TOPOLOGY

Since F is clearly boundary preserving, we find that the function h = F o f is as required. D Corollary 5.2.2. For every compact subset K C s and e > 0 there are an infinite TV C N and a boundary preserving homeomorphism h: Q ->• Q such that (1) the complement of N is infinite and ^n&N 2~ n < e, (2) Q(h,lQ)<e, (3) nnh[K] = {0} for every n e TV. Proof. Choose n e N so large that

Write N\ {1, 2, . . . , n — 1} as the disjoint union of infinitely many infinite sets, say C\ , C<2 , . . . . Recall that for every i

denotes the projection. Let

hi". be a homeomorphism with the following properties: (1) hi is boundary preserving (this has its obvious meaning), (2) hi^d [K] projects onto 0 in the first factor of the product (Lemma 5.2.1). Define h: Q ->• Q by

i

(J
It is clear that h is as required (observe that h is nothing but the product of the hi$ and the identity on the first n — I factors). Finally, observe that the set TV = {mm(Ci) : i € N} has infinite complement in M and TV C {n, n + 1, . . . } so E meJV 2~ m < e. D The following lemma is the key in deriving our main result. The strategy of the proof is similar to the one in Theorem 1.8.4, but is more complicated. It will be convenient to introduce some notation that shall be fixed throughout the remaining part of this section. Let A and B be complementary infinite subsets of N. Put QA = {x € Q : xn = 0 if n g A},

respectively.

QB = [x € Q : xn = 0 if n g B},

5.2. EXTENDING HOMEOMORPHISMS IN s

315

It will be convenient to think of Q as QA x QB- We will specify A and B later. Let 6 > 0 be such that

The 'origins' of Q, QA and Qs, i.e., the points having all coordinates 0, shall be denoted by 0, or, as we think of Q as QA x QB, by (0,0). Let X,Y and Z be compact subsets of s such that X U Y C Q^ and Z C Qs and let p: A^ —>• Z and q: Y —> Z be homeomorphisms such that

Q(Q IP,

(*)

7

for certain 7 > 0. Again we will specify p,q,*y,X,Y and Z later (see Figure 16). We would like to construct a homeomorphism h\ which takes X onto the 'graph' G(p) of p. Similarly, a homeomorphism hi which takes Y onto the 'graph' G(q) of q. And finally, a homeomorphism ^3 mapping G(p) onto G(• F. Before we can make this precise, we define two technical concepts. A homeomorphism /: Q —>• Q is called an yl-homeomorphism (respectively, .B-homeomorphism) if for any x E Q we have

for all n G A (respectively, n e £?). So an A-homeomorphism is a 'vertical action'. Similarly, a S-homeomorphism acts 'horizontally'.

Figure 16. The desired homeomorphisms are constructed in the following lemma. Lemma 5.2.3. (1) There is a boundary preserving A-homeomorphism hi : Q —> Q such that hi(x,0) = ( x , p ( x ) ) for every x 6 X.

316

5. BASIC INFINITE-DIMENSIONAL TOPOLOGY

(2) There is a boundary preserving A-homeomorphism h-2: Q —> Q such that hi(y, 0) = (y, q(y}) for all y € Y. (3) There is a boundary preserving B-homeomorphism h% : Q —> Q such that hs(x,p(x)] = (q~1p(x),p(x)} for every x € X while moreover Q(h3,lQ) < 7. The reader should pause to see what is going on. The homeomorphism hi indeed takes X onto the 'graph' G(p) of p. Similarly, h2 takes Y onto the 'graph' G(q) of q. Finally, /i3 is a 'small' homeomorphism mapping G(p] onto G(q). So h%1 ° /i3 o hi is a 'small' homeomorphism of Q extending the homeomorphism q~l op: X ^ Y (see Figure 17) if QB has 'small' diameter. Proof. Since X U Y U Z is compact we can find for every n an interval such that X U Y U Z C 0^=1 h r n> r n]- Put oo

X^ = {x £ JJ [-rn, r n ] : a;n = 0 if n n=l

and 71=1

respectively. Observe that both KA and KS are products of symmetric intervals (the symmetry of the intervals will be important later), that X U Y C KA ^ QA a n d that Z C K B B .

Figure 17. By applying the Tietze Extension Theorem 1.2.5 to each factor of KB separately, we can extend p: X —>• Z C KB to a continuous function p: QA ^ KB

5.2. EXTENDING HOMEOMORPHISMS IN s

317

(alternatively, apply Corollary 1.2.12). For every t € (—1,1) let (pt : J ->• J be the unique homeomorphism taking the interval [—1,0] linearly onto [— 1, t] and [0, 1] linearly onto [t, 1]. For each n 6 B let the isotopy # n : JJ x [-r n ,r n ] ->• J be defined by Hn(x,i] = (pt(x). The H^'s define a .ftTs-isotopy H:QB xKB^QB as follows:

H(y, x)n = Hn(yn, xn)

(n e B)

(Proposition 1.8.1) (we make an obvious identification here). Now define

hi : QA x QB -* QA * <2s

by hi(x,y) =

(x,H(y,p(x))).

Then /ii is an A-homeomorphism (Theorem 1.8.2). In addition, if x G X then (for the last equality, use the definition of the Hn's for n G -B). We conclude that hi is as required. The construction of h? is precisely the same. The construction of ^3 is similar but slightly more complicated because of the smallness condition involved. Consider p~l : Z -> X C KA and q~l : Z ->• Y C ^. By Exercise 1.3.8 and (*) on Page 315,

e(p~l,g~1} = Q(p~lp,q~lp) = o(^x,q~lp) < 7Claim 1. There exist continuous extensions £, rj: QB —> KA of p~l and g"1, respectively, such that Q(£, 77) < 7. Proof. By applying the Tietze Extension Theorem 1.2.5 to each factor of KA separately (alternatively, apply Corollary 1.2.12), we can extend p~l and q~l to continuous functions f , g : QB —> KA- Put

It is clear that U is an open neighborhood of Z in QB- Let a: QB —>• I be a Urysohn function such that a\Z = l,

a\(QB\U)=0

(Corollary A. 4.1). Now define £,r?: QB ->• KA by £(x)=a(x)-f(x),

r](x}=a(x}-g(x},

318

5. BASIC INFINITE-DIMENSIONAL TOPOLOGY

respectively. Observe that by the special choice of KA these functions are well-defined (and obviously continuous). Now if x 0 U then a(x) — 0, and so £(x) = r)(x). On the other hand, if x £ U then 1~nn/(r\ f(r\ — n(r\ z o^xj • I\J\X)n y\^)n\I n=l

= a(x] < 1 -7

-g(f(x),g(x))

From this we conclude that £(£,17) < 7 (Exercise A. 5. 4). If x G Z then clearly a(x) — 1, so £(x) = 1 • f(x] = f ( x ] . We conclude that £ extends /, and similarly, that T] extends g. Consequently, £ and TJ are as required. <0For all ( x , y ) € (-1,1)2 let

be the unique homeomorphism taking [— l,x] linearly onto [—I,?/] and [x, 1] linearly onto [y, 1]. Observe the following easy: Claim 2. If ( x , y ) e (-1,1)2 then £(<£>( x > y ) ,lj) = \x - y\. We now define a (KA x /d)-isotopy FrQ^x^x/iCO-X^ coordinate wise as follows: F(q, x, y)n = V ( X n , y n ) (qn)

(n £ A),

cf. Proposition 1.8.1. Now define h3: QAxQB ^ QAx QB

by h3(x,y) =

(F(t(y),r,(y))(z),y).

Then h% is a 5-homeomorphism (Theorem 1.8.2). We shall prove that ^3 is as required. Take an arbitrary x E X. Then h3(x,p(x)) = = =

(F(£p(x)^p(x)}(x},p(x)) (F(x,q-ip(x))(x),p(x)) ((1~1P(X):P(X))

(for the last equality, use the definition of F).

5.2. EXTENDING HOMEOMORPHISMS IN s

319

Finally, take an arbitrary (x,y) G QA x QB- Then

<7 (use Claims 1 and 2). We conclude that g(hs, IQ) < 7 (Exercise A. 5. 4).

D

We now come to the main result in this section. Theorem 5.2.4. Let .E, F C s be compact and let f : -E -> F be a homeomorphism such that £(/, Is) < e. Then / can be extended to a boundary preserving homeomorphism f:Q-*Q such that £>(/, IQ) < e. Proof. Let e\ = g(f, 1^) and put 5 = l/6(e — e^) (this specifies £). By Corollary 5.2.2 we find a boundary preserving homeomorphism g: Q -> Q and an infinite subset B C N such that (1) 0(g, 1Q) <S, (2) Kng[E U F] = {0} for every n € 5, (3) A = N \ 5 is infinite and £ n£B 2"n < %

(this specifies A and 5). Put X — g[E], Y ~ g[F] and h = g <=> f o g~l. Observe that Notice that ^C U F C Q^. Since Q^ PS Q, we can find a topological copy Z of X in s such that Z C QB (Corollary A. 4. 4) (Qs is not a subset of s but a smaller infinite-dimensional 'subcube' is a subset of s ) . Let p: X —>• Z be any homeomorphism and let g — p o h~l. Then clearly Put 7 = ei + 2(5. This specifies p, q and 7. Now let hi,h<2 and /is be as in Lemma 5.2.3. By (3) we have e(hi,!Q)<6, Q(h2,lQ)<6. Consequently, if we put 1 T = /i" o /i3 o hi, then Q(T, IQ} < 7 + 26 = ei + 46.

320

5. BASIC INFINITE-DIMENSIONAL TOPOLOGY

Since r clearly extends h, we find that f = g'1 or og

extends / and that £(/, IQ) < e\ + 46 + 26 — e. Since it is clear that / is boundary-preserving, this finishes the proof.

D

Exercise for §5.2. Let e > 0. An isotopy H: Q x I —> Q is called an e-isotopy provided that for every x G Q the diameter of the set H [{x} x I] is less than E. ^•1. Let E, F C s be compact and let /: E —>• F be a homeomorphism such that Q(/,IE) < £• Prove that there is an e-isotopy H: Q x I —>• Q such that Ho = IQ and Hi \ E — f (i.e., HI extends /). (This improves Theorem 5.2.4.) 5.3. The estimated homeomorphism extension theorem The aim of this section is among other things to prove that if E, F F is a homeomorphism such that £(/, IE) < £ then / can be extended to a homeomorphism f'Q^Q such that £(/, IQ) < e. This result is known as the (Estimated) Homeomorphism Extension Theorem and is of fundamental importance in infinite-dimensional topology. The strategy of the proof is that we first push E and F into s by a small motion and then apply Theorem 5.2.4. Extending homeomorphisms. Recall that for n G N and # 6 { —1,1} by W% we mean the endface ir~l({9}). Throughout, let K C s be a fixed compact subset. This set plays no role of importance in this section. But by introducing and dealing with it here we are able to simply our life considerably later. We will first show that endfaces can be pushed into s by a small movement. This is done by pushing endfaces into endfaces with larger and larger indexes. Lemma 5.3.1. For each n e N, 6 6 (—1,1} and e > 0 there are a homeomorphism h: Q —> Q and an m > n such that

(1) fc[W«]nU{Wf:t<m,Aie{-l,l}} (2) h[W°n}CW^ (3) g(h,lQ) <e, (4) h\K = l. Proof. Let e > 0 and choose m > n such that

= 0,

5.3. THE ESTIMATED HOMEOMORPHISM EXTENSION THEOREM

321

We first push W% into W^ and then away from the endfaces in the lower coordinate directions. Without loss of generality we assume that 0=1. It is geometrically obvious that there is a homeomorphism Jn X Jm

having the following properties: (5) ^ [ { l } x J T O ] C J n x { l } , (6) if x < I — 2~( m ~ 1 ) then
Figure 18. This is precisely the same homeomorphism as the one in Figure 4 on Page 61. Let hi 6 J(.(Q) be the homeomorphism of Q that is defined by crossing

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