The Infinite-Dimensional Topology of Function Spaces

By J. van Mill

In this book we study function spaces of low Borel complexity.
Techniques from general topology, infinite-dimensional topology, functional analysis and descriptive set theory
are primarily used for the study of these spaces. The mix of
methods from several disciplines makes the subject
particularly interesting. Among other things, a complete and self-contained proof of the Dobrowolski-Marciszewski-Mogilski Theorem that all function spaces of low Borel complexity are topologically homeomorphic, is presented.

In order to understand what is going on, a solid background in
infinite-dimensional topology is needed. And for that a fair amount of knowledge of dimension theory as well as ANR theory is needed. The necessary material was partially covered in our previous book `Infinite-dimensional topology, prerequisites and introduction'. A selection of what was done there can be found here as well, but completely revised and at many places expanded with recent results. A `scenic' route has been chosen towards the
Dobrowolski-Marciszewski-Mogilski Theorem, linking the
results needed for its 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, Chapters 2 and 3 and part of Chapter 1 should be covered. For a course in infinite-dimensional topology, Chapters 1, 4 and 5. In Chapter 6, which deals with function spaces, recent research results are discussed. It could also be used for a graduate course in topology but its flavor is more that of a research monograph than of a textbook; it is therefore
more suitable as a text for a research seminar. The book
consequently has the character of both textbook and 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 for easy reference and some basic facts that are important in the book have been collected. The book is not intended as a basis for a course in topology; its purpose is to collect knowledge about general topology.

The exercises in the book serve three purposes: 1) to test the reader's understanding of the material 2) to supply proofs of statements that are used in the text, but are not proven there
3) to provide additional information not covered by the text.
Solutions to selected exercises have been included in Appendix B.
These exercises are important or difficult.

• Language: English
• Publisher: Elsevier Science
• Release date: May 24, 2002
• ISBN: 9780080929774

Book preview

North-Holland Mathematical Library

The Infinite-Dimensional Topology of Function Spaces

Jan van Mill

Faculteit der Exacte Wetenschappen, Amsterdam, The Netherlands

ISSN  0924-6509

Volume 64 • Suppl. (C) • 2002

Table of Contents

Cover image

Title page

North-Holland Mathematical Library

Copyright page

Dedication

Introduction

Chapter 1: Basic topology

1.1 Linear spaces

1.2 Extending continuous functions

1.3 Function spaces

1.4 The Borsuk homotopy extension theorem

1.5 Topological characterization of some familiar spaces

1.6 The inductive convergence criterion and applications

1.7 Bing’s shrinking criterion

1.8 Isotopies

1.9 Homogeneous zero-dimensional spaces

1.10 Inverse limits

1.11 Hyperspace

Chapter 2: Basic combinatorial topology

2.1 Affine notions

2.2 Barycenters and subdivisions

2.3 The nerve of an open covering

2.4 Simplices in n

2.5 The Lusternik-Schnirelman-Borsuk theorem

Chapter 3: Basic dimension theory

3.1 The covering dimension

3.2 Translation into open covers

3.3 The imbedding theorem

3.4 The inductive dimension functions ind and Ind

3.5 Dimensional properties of compactifications

3.6 Mappings into spheres

3.7 Dimension of subsets of n and certain generalizations

3.8 Higher-dimensional hereditarily indecomposable continua

3.9 Totally disconnected spaces

3.10 The origins of dimension theory

3.11 The dimensional kernel of a space

3.12 Colorings of maps

3.13 Various kinds of infinite-dimensionality

3.14 The Brouwer fixed-point theorem revisited

Chapter 4: Basic ANR theory

4.1 Some properties of ANR’s

4.2 A characterization of ANR’s and AR’s

4.3 Open subspaces of ANR’s

Chapter 5: Basic infinite-dimensional topology

5.1 Z-sets

5.2 Extending homeomorphisms in s

5.3 The estimated homeomorphism extension theorem

5.4 The compact absorption property

5.5 Absorbing systems

Chapter 6: Function spaces

6.1 Notation

6.2 The spaces Cp(X): introductory remarks

6.3 The Borel complexity of function spaces

6.4 The Baire property in function spaces

6.5 Filters and the Baire property in Cp(NF)

6.6 Extenders

6.7 The topological dual of Cp(X)

6.8 The support function

6.9 Nonexistence of linear homeomorphisms

6.10 Bounded functions

6.11 Nonexistence of homeomorphisms

6.12 Topological equivalence of certain function spaces

6.13 Examples

Appendix A: Preliminaries

A.1 Prerequisites and notation

Appendix B: Answers to selected exercises

Appendix C: Notes and comments

Bibliography

Special Symbols

Author Index

Subject Index

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. Hörmander, J.H.B. Kemperman, W.A.J. Luxemburg, F. Peterson, I.M. Singer and A.C. Zaanen

Board of Advisory Editors:

A. Björner, R.H. Dijkgraaf, A. Dimca, A.S. Dow, J.J. Duistermaat, E. Looijenga, J.P. May, I. Moerdijk, S.M. Mori, J.P. Palis, A. Schrijver, J. Sjöstrand, J.H.M. Steenbrink, F. Takens and J. van Mill

VOLUME 64

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Printed in The Netherlands.

Dedication

Aan de nagedachtenis van Maarten Maurice.

Introduction

Jan van Mill,     Bussum

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 infinite-dimensional 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 Chapter 2 and 3 and part of Chapter 1. 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 Chapter 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 .

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.

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.1.

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 we have αx + (1 − α)y ∈ A.

Let L be a linear space. If A ⊆ 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

, let convn(A) denote the set of vectors x ∈ L that can be written as a convex combination of at most n vectors from A. Also, put

and observe that conv∞(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) = 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 ∪ {∞} with their usual product topologies are locally convex linear spaces under coordinatewise defined addition and scalar multiplication.

Let L be a vector space. A norm on L is a function ||·||: L → [0, ∞) having the following properties:

(1) 

(2) 

,

If ||·|| is a norm on L then the function

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

and

denote its unit ball and unit sphere, respectively.

A subset A of a normed linear space L is called bounded if there exists an ε ∈ [0, ∞) such that ||a|| ≤ ε 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 .

is the Euclidean norm which is defined by

As noticed on Page 461, Bn 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 endowed with the Tychonoff product topology. It is a classical object in both 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:

(See Exercise 1.1.1 for the verification.)

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

It is clear that σ is a linear subspace of s.

Lemma 1.1.4

If L is a linear subspace of s with σ ⊆ 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

is an open neighborhood of the origin of L. By definition of the product topology on s there are an open neighborhood V such that

(*)

where Vi = V for i ≤ n for i > n. Let y ∈ s be defined by yi = 0 if i ≠ n + 1 and yn+1 = 1. Since y ∈ σ ⊆ L and y ≠ 0 it follows that ε = ||y|| > 0. By (*), t y ∈ U . In particular, ||y/ε|| < 1 but also ||y/ε|| = ε/ε = 1, which is a contradiction.

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 .

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 f ∈ C(X) then define its norm, ||f||, by

(Observe that by compactness of X this supremum is attained. That is: there is an element x ∈ X such that ||f|| = |f(x)|.) It is easily seen that ||·||: C(X) → [0, ∞) is indeed a norm; it is called the sup-norm on C(X). Consequently, the function

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 (fn)n -Cauchy sequence. Then (fnexists 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.

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 is 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 x0 ∉ Xbe an admissible metric for Y. Let A be the subspace of C(Ysuch that f(x0) = 0. Observe that if f ∈ A then

This is clear since if y ∈ Y then

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 be a finite open cover of [–diam(Y), diam(Y)] consisting of open sets with diameter at most 2–n. This cover has a Lebesgue number, say λn > 0 (Lemma A.5.3). By compactness of X n of X such that

Now fix any f ∈ A and V n. Since f is Lipschitz, the diameter of f[V] is at most λn. Hence f[V] is contained in an element U . So for each f ∈ A there is a function

such that for every V n we have

Let n such that ξ1(fm) = ξ1(fk) for all integers m, k ∈ N1. Let m and k be two arbitrary elements of N1, and pick an arbitrary x ∈ X. Pick an element V 1 such that x ∈ V. Then both fm(x) and fk(x) belong to the same element of U1, i.e., |fm(x) − fk(x)| < 2−1. Since x was arbitrary, this shows that ||fm − fk|| ≤ 2−1. So now it is clear how to proceed. Let n1 = min N1 and consider the infinite set N1 \ {n1}. There is an infinite subset N2 ⊆ N1 \ {n1} such that for all m and k in N2 the functions ξ2(fm) and ξ2(fk) agree. Then by a similar argument as the one above, ||fm − fk|| ≤ 2−2 for all m, k ∈ N2. Let n2 = min N2. Continuing in this way resursively, we can construct an infinite sequence

in C(Y). Since C(Y) is a Banach space, this sequence has a limit and since A is closed, this limit belongs to Ais the desired convergent subsequence of (fn)n.

Define i: X → C(A) by the following formula:

Observe that i is well-defined. For i(x) should be an element of C(A. But an element of A is a function f from Y . So the formula tells us that i(x) sends the function f onto its evaluation in the point x ∈ X ⊆ Y.

Claim 2

i: X → i[X] is an isometry.

Proof. Let x1, x2 ∈ X. Then

Observe that the last inequality follows from the fact that f by

If y, y′ ∈ Y are arbitrary then

Also, g(x0) = 0. Hence g ∈ A and

Hence by (1) we get ||i(x1) − i(x(x1, x2), as required.

It remains to prove that i[X] is a linearly independent subset of C(A). To this end, let x1, …, xn+1 be distinct elements of X. We claim that i(xn+1) is not a linear combination of the i(x1), …, i(xn).

by

Claim 3

.h ∈ A.

Figure 1

Proof. We shall first prove that h is Lipschitz. To this end, pick arbitrary a, b ∈ Y. We may assume without loss of generality that h(b) ≤ h(a). Fix i and j such that h(a(a, xi) and h(b(b, xj). Since h(a(a, xi) we get by the definition of h (a, xi(a, xj). From this we conclude that

as required.

Since it is trivial that h(x0) = 0, this completes the proof. ◊

Observe that i(xk)(h) = h(xk) = 0 for every k ≤ n, but since the x1, …, xn+1 are distinct elements of Y,

So i(xn+1) is not a linear combination of the i(x1), …, i(xn), and this is what we had to prove.

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 ∩ i[aX] = i[X]. Hence i[X] is a closed linearly independent subset of the normed linear space L.

Since ||·||: C(A) → [0, ∞) is continuous, and aX is compact, it is clear that i[aX] is a bounded subset of C(A), hence so is i[X].

If X is not compact then C(X) contains unbounded functions (see Exercise A.5.14), and so the formula

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.

Example 1.1.9

The space c0.

, and endow it with the norm

It follows by straightforward calculations that this is indeed a norm compatible with the linear structure on c0. There is however another way of proving this. Let S be a nontrivial convergent sequence including its limit t, and consider

Then L is a closed linear subspace of C(S) which clearly can be identified with c0. So c0 is a closed linear subspace of the function space C(S).

The set c0 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 ².

We saw that the topology on s which behaves better (in this respect).

given by

:

. For every x If x, y From this it follows that

and so

since all infinite series considered are convergent. We conclude that for every x, y ² we have x + y ². If x then trivially tx .

for all x, y ,

which is easily seen to be an inner product. Consequently, ||x|| = p(x² and the metric derived from this norm is:

² with this topology as Hilbert space.

² and so ² inherits from s ² which we just defined. We will comment on the precise relation between these topologies later.

Lemma 1.1.11

(Exercise 1.1.25). The metric on ² defined above is complete.

² is also a Banach space. But there is a difference with the spaces C(X² is derived from an inner product, while no inner product on C) yields its standard norm (Exercise 1.1.9).

The topology on s is the topology of ‘coordinatewise convergence’, see 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))n is a sequence in ², and x ². The following statements are equivalent:

(1) limn→∞ x(n) = x (in ²),

(2) limn→∞||x(n)|| = ||x|| and for every i , limn→∞ x(n)i = xi.

From ² inherits from s. However, more can be concluded. For example, consider the unit sphere S = {x ² : ||x|| = 1}. Since all points in S have the same norm, the topology that S ² is precisely the same as the topology that S inherits from s² and s are topologically homeomorphic (see also VAN MILL [297, Chapter 6] for a complete proof of this result).

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 α > 0 such that such that

Proof. For each α > 0 put Bα = {x ∈ E : ||x|| ≤ α}. Since

and F is a Baire space (see Page 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. In addition, Bm is symmetric, i.e., –Bm = Bm. Again since T is linear, it follows that T[Bmis symmetric as well. Now choose y ∈ F . Let z ∈ F be such that ||z|| ≤ β. Then ||(z + y)− y. Similarly, ||(y − z) − yby convexity we get

from which it follows easily that

where α = m/β.

In the remaining part of the proof we adopt the notation introduced in Claim 1.

Claim 2

{y ∈ F : ||y|| ≤ 1} ⊆ T[B2α].

Proof. Let y ∈ F with ||y|| ≤ 1. We shall define recursively a sequence (yn)n in T[Bα] such that for all n,

By Claim 1, there exists y1 ∈ T[Bα] such that ||y - y1|| ≤ 1/2. Suppose that y1, …, yn are chosen properly. Then

and therefore, again by Claim 1, there exists yn+1 ∈ T[Bα] such that

It is clear that yn+1 is as required.

From (*) it easily follows that

For every n choose a point xn ∈ Bα with T(xn) = yn. Since ||xn|| ≤ α < ∞ for every n, it is easily seen that the sequence

is Cauchy, and hence that

exists. Since ||·||: E is continuous,

By observing that T is continuous and linear we obtain

from which we conclude that y ∈ T[B2α].

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 ε > 0 with D(x, ε) ⊆ U. Consequently, Bε ⊆ U − x. It follows by Claim 2 that

Consequently, since T[Bε] ⊆ T[U − x] = T[U] − T(x), we have

So T[U] is a neighborhood of T(x) and since x was an arbitrarily chosen point from U, we are done.

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.

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

be linear. If the graph

of T is a closed subset of E x F then T is continuous.

Proof. Observe that Γ is a closed linear subspace of the Banach space E x F (Exercise A.1.8) and hence is a Banach space itself. In addition, the function S1: Γ → E defined by S1(x, Tx) = x is clearly continuous (since it is the restriction of a projection). As a consequence, S1 is a linear homeomorphism by Corollary 1.1.14. Observe that the function S2: Γ → F defined by S2(x, T(x)) = T(x) is continuous as well. So the function E → F defined by x (x, TxTx is continuous, being a composition of continuous functions. So T is continuous.

Corollary 1.1.16

Let E and F be Banach spaces additionally endowed with weaker vector space topologies. If 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. For this simply observe that the product of the weaker topologies is weaker than the product of the Banach space topologies.

The Michael selection theorem. We shall prove that certain set-valued 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 (Y) \ {∅}, i.e., F: X (Y) \ {∅}. By the symbol

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 ⊆ Y we put

We say that F is lower semi-continuous (abbreviated LSC) provided that for every open subset U of Y, F⇐[U] is open in Xis a covering of Y then

covers X since for every x ∈ X, F(x) ≠ ∅.

A basic example of an LSC set-valued function is the following one. If X and Y are spaces and f: X → Y is an open surjection then we define

by F(y) = f−1(y). The function F is LSC since for every open U ⊆ Y we have

Other examples of LSC set-valued mappings will be presented later.

Let X and Y be sets and let F: X ⇒ Y. A function f: X → Y is called a selection for F if for all x ∈ X, f(x) ∈ F(x). Since for all x ∈ X 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 Fby

There does not exist a continuous function fthe graph of which is contained in the ‘graph’ of F This is left as an exercise to the reader, as well as the proof 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(xfor all 1/3 < x < 2/3 then F does admit a continuous selection.

Although not every LSC map has a continuous selection, any map with sufficiently many continuous selections must be LSC. This is shown in our next result.

Proposition 1.1.18

Let X and Y be spaces and let F: X ⇒ Y be a set-valued function with the following property:

for all x ∈ X and y ∈ F(x) there exists a continuous selection f for F such that f(x) = y.

Then F is LSC.

Proof. Let U ⊆ Y . Pick y ∈ F(x) ∩ U. By assumption there exists a continuous selection f: X → Y for F such that f(x) = y. Put V = f−1[U]. By continuity of f, V is a neighborhood of xsince if x′ ∈ V then

is open.

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 defined by is LSC,

(2) if f: X → Y is continuous and is an admissible metric for Y and the number r > 0 is such that

then the function G: X ⇒ Y defined by

is LSC.

Proof. For (1), simply observe that for all x ∈ X and open U ⊆ Y we have that F(x) ∩ U ≠ ∅ . Consequently, for every open subset U ⊆ Y holds.

defined by

is LSC. Let V ⊆ Y . There exists y ∈ Y such that

Let ε = r (y, f(x)) and choose 0 < δ < ε such that

Since F(x)⊆B(y, δ/2) ≠ ∅ and F is a neighborhood of x. In addition, since f is continuous, U1 = f−1[B(f(x), δ/2)] is also a neighborhood of x. Put U = UU.

To this end, take an arbitrary x′ ∈ U. Since x′ ∈ U0, there exists a point y′ ∈ F(xB(y, δ/2). In addition, f(x′) ∈ B(f(x), δ/2) since x′ ∈ U1.

Consequently,

and therefore,

. We conclude that the point x.

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 ∈ X. Then for every r > 0 there exists a continuous function f : X → L such that for every x ∈ X we have (f(x), F(x)) < r.

Proof. Put

By on X . Consequently, for each p there exists bp ∈ L such that

Define f: X → L by

For each x ∈ X there are a neighborhood Ux and a finite subset G(xsuch that Ux p−1 [(0,1]] ≠ ∅ if and only if p ∈ G(x).

Now fix an arbitrary x ∈ X. Observe that the restriction of f to Ux is given by

which is a continuous expression in y since G(x) is finite. From this we conclude that f is well-defined and continuous at x.

Put

we have

and so there exists yp ∈ F(xB(bp, r). Observe that

This implies that

Since F(x) is convex and Σp∈g(x) p(x) = 1 we have

((f(x), F(x)) < r.

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 ⊆ X is closed and that f : A → Y is a continuous selection for the function F A: A ⇒ Y. Define G: X ⇒ Y by

Then G is LSC.

Proof. Let U ⊆ Y . Indeed, take an arbitrary x ∈ f−1[U]. Then f(x) ∈ F(x) ⊆ (U, hence x ∈ F⇐[U]. Observe that by continuity of f, the set f−1[U] is open in A. So there is an open V ⊆ X such that V ⊆ A = f−1[Uand F⇐[U. Consequently,

is open.

Let (L, ||·||) be a normed linear space. A subset A of L is called complete with respect to (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

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 = ∅.

By induction on n we shall construct a sequence (fn)n in C(X, L) such that

(2) 

Apply Lemma 1.1.20 with r = ½ to find f1: X → L such that for every x ∈ X,

Suppose that fn has been defined. Define Fn: X ⇒ L by

Then Fn is LSC by Lemma 1.1.19. By another appeal to Lemma 1.1.20 we find h: X → L (h(x), Fn(x)) < 2−n for every x ∈ X. It is easy to see that fn+1 = h is as required.

Claim 1

exists and belongs to F(x).

Proof. Take an arbitrary x ∈ Xa point an ∈ F(x(fn(x), an) < 2−n. Consequently, by (1) we have

We conclude that the sequence (an)n (fn(x), an) < 2−n for every nalso exists and is equal to a ∈ F(x).

By Lemma A.3.1 we consequently conclude that the function g is the required continuous selection.

Now if A ≠ ∅, the above special case and Lemma 1.1.21 yield the desired result.

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 of T. Then there exists a continuous function f : F → E such that T f = 1F. The function h: E → KerT x F defined by

is a homeomorphism.

Proof. Define H : F ⇒ E by H(y) = T−1(y). Then for each open U ⊆ E , 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 f follows directly from Theorem 1.1.22.

The easy proof that h is a homeomorphism is left as an exercise to the reader.

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) 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 [268].

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 ∈ L \ {0} can be written uniquely in the form

with xi ∈ B \ {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.

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.,

Let X be a space. For a compact subset K in X and an open U define

Topologize C(X) by taking the collection

as an open subbase. This topology is called the compact-open topology on C(X).

1. Prove that the formula

defines an admissible complete metric on s.

2. Let L be a linear space. Prove that for every x ∈ L and every convex 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 ∈ Lbe a local base at the origin of L. Show that x has arbitrarily small neighborhoods of the form x + W, where W .

5. Prove Lemma 1.1.1.

6. Prove that every normable linear space is locally convex.

by |||x||| = max{|xi| : 1 ≤ i ≤ nand that it generates the Euclidean topology.

8. Let L be a linear space and let B be a Hamel basis for L. If

with xi ∈ B for every i ≤ n, then put

Prove that ||·|| defines a norm on L.

9. Prove that the sup-norm on C) cannot be derived from an inner product by showing that it violates the parallelogram law.

10. Prove that for every compact space (X) there exists an isometry

such that

(1) for every subset Y ⊆ X the image set i[Y] is closed in conv(i|Y]),

(2) i[X] ⊆ {f ∈ C(X) : ||f|| ≤ diam X}.

11. Prove that there does not exist a homeomorphism h² → s which is linear, i.e., has the property that

for all x, y .

12. Let A be a bounded subset of a normed linear space L. Prove that if (tn)n such that tn → 0 and (an)n is any sequence of elements in A then tnan → 0.

13. Let 0 < p p is a linear subspace of s p:

p 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 ||·||.

² and s are separable.

are linearly homeomorphic.

17. Let

. Prove that X and Q are homeomorphic.

and ε 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.

and let fbe linear but not identically 0. Prove that there exists t > 0 such that

(1) f−1 (tB(0, 1) = ø,

(2) f−1 (tD(0, 1) ≠ ø.

21. Let V be an infinite-dimensional normed linear space. Prove that there is a sequence (en)n in V such that

(1) {en } is linearly independent,

.

22. Let V be an infinite-dimensional normed linear space. Prove that the unit sphere S = {x ∈ V : ||x|| = 1} is not compact.

and let

Prove that for every α ∈ [0,1) there exists ε > 0 such that B(αx, ε) ⊆ K.

with nonempty interior is homeomorphic to Bn .

² defined on Page 9 is complete.

26. Prove Lemma 1.1.12.

27. Let X be a space and let f, g: X with f 1sc and g use. Prove that if g ≤ f, i.e., g(x) ≤ f(x) for every x ∈ X, then the function F: X defined by F(x) = [g(x), f(xsuch 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 Δ(X) is homeomorphic to the subspace

of L .

30. Let E and F be Banach spaces and let φ: E → F such that

for every x ∈ E.

31. Let E and F be Banach spaces and let φ: E → F such that for every x ∈ E we have

1.2 Extending continuous functions

Suppose that X, Y and Z are topological spaces with Y a subspace of X and let f: Y → Z be continuous. In topology it is often of interest to know whether f is the restriction to Y : X → Z. Easy examples show that in general this need not be the case. If f is the restriction to Y : X → Z then we say that f is continuously extendable over X is a continuous extension of f. 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 f: Y → Z is continuously extendable over X (respectively, over some neighborhood of Y).

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 of X. If E = A ∪ F then by Lemma A.4.1 it follows that the continuous function f: E defined by f A ≡ 0 and f B : X . The function f: E 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.

Lemma 1.2.1

Let X be a space, A a closed subset of X, and let A′ ⊆ A be dense in A. Then there exist a locally finite open cover of X \ A and a sequence of points {aU : U } in A′ such that

(1) for all U and x ∈ U(x, aU(x, A),

(2) if Un for every n and

then

Proof. Let

Since A is an open cover of X \ A. By paracompactness of X \ A (of X \ A , for each U there exists xU ∈ X | A with

In addition, since A′ is dense in A, for each U there exists aU ∈ A′ with

We claim that the U’s and the aU’s are as required.

Claim 1

For every U and x ∈ U the following inequalities hold:

Proof. The first inequality is easy since

Also,

from which it follows that

as required.

So it remains to verify (2). To this end, assume that Un for all n and that

For each n pick pn ∈ Un . By the claim we obtain

for all n. So we are done.

and a sequence of points {aU : 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 ⊆ 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 : X → C.

Remark 1.2.3

For Part 2 of The Dugundji Theorem, see Page 394.

of X \ A be a Dugundji system for X and A. In addition, let κU : X \ A for U .

: X → L by

is well-defined and continuous at all points of X \ A. To this end, fix an arbitrary x ∈ X \ Ais locally finite, there is a neighborhood W of x in X \ A only, say U1, …, Un. For U missing W we clearly have

As a consequence, for every y ∈ W we have

(y) is a convex combination of points in C and therefore belongs to C. By continuity of the κW is continuous.

at the points of A. Pick an arbitrary element a ∈ A. An arbitrary neighborhood of f(a) is without loss of generality of the form f(a) + W, where W is a convex neighborhood of 0 (Exercise 1.1.4). So let f(a) + W be such a neighborhood. The continuity of f at a implies that there exists δ > 0 such that B(a, δ) ⊆ A ⊆ f−1[f(a) + W].

Claim 1

[B(a, δ/3)] ⊆ f(a) + W.

Proof. Pick an arbitrary x ∈ B(a, δ/3). If x ∈ A then there is nothing to prove. So assume without loss of generality that x ∉ A. Then

as a consequence, if x ∈ U then

by (1) of Lemma 1.2.1. From this we conclude that if x ∈ U and so f(aU) ∈ f(a) + W. Consequently,

Since f(aU) – f(a) ∈ W for every U (x(a) is a convex combination of elements of W. Hence by convexity of W (x(a) ∈ W, as required.

is continuous at a.

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 [139] appeared in 1951. It was finally solved in 1994 by CAUTY [87] in the negative. His construction used in an essential way DRANIŠNIKOV’s result in [138] about the existence of an infinite-dimensional 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 or 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 f : X → Y is a homeomorphism then Y is an AR as well. Consequently, X is an AR 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→ X. Notice that every AR is an ANR.

As above, Theorem 1.2.7 below easily implies that X is an ANR 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 ANR is again an ANR (Proposition 1.2.10).

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 f: A → X can be extended over Y (over a neighborhood (depending on f) of A in Y). 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.

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 ⊆ Y be closed, and f: 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 f. 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−1[U]. Since G[A] ⊆ X, V is clearly a neighborhood of A in Y. Let h denote the restriction of r G to V (observe that h is well-defined since G[V] ⊆ U). Then h is continuous and extends f since for every y ∈ A we have h(y) = r(G(y)) = f(y).

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.

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.

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