Brouwer's Theorem on the Invariance of Domain Luukas Hallamaa

Brouwer’s Theorem on the Invariance of Domain

Luukas Hallamaa

Department of Mathematics and Statistics August 2019

HELSINGIN YLIOPISTO — HELSINGFORS UNIVERSITET — UNIVERSITY OF HELSINKI Tiedekunta/Osasto — Fakultet/Sektion — Faculty Laitos — Institution — Department

Faculty of Science Department of Mathematics and Statistics Tekijä — Författare — Author Luukas Hallamaa Työn nimi — Arbetets titel — Title

Brouwer’s Theorem on the Invariance of Domain Oppiaine — Läroämne — Subject Mathematics Työn laji — Arbetets art — Level Aika — Datum — Month and year Sivumäärä — Sidoantal — Number of pages Master’s thesis August 2019 63 pages Tiivistelmä — Referat — Abstract

The purpose of this thesis is to present some dimension theory of separable metric spaces, and with the theory developed, prove Brouwer’s Theorem on the Invariance of Domain. This theorem states, that if we embed a subset of the n-dimensional Euclidean space into the aforementioned space, this embedding is an open map. We begin by revising some elementary theory of point-set topology, that should be familiar to any graduate student in mathematics. Drawing from these rudiments, we move on to the concept of dimension. The dimension theory presented is based on the notion of the small inductive dimension. We define this dimension function for regular spaces and state and prove various results that hold for this function. Although this dimension function is defined on regular spaces, we mainly focus on separable metric spaces. Among other things, we prove that the small inductive dimension of the Euclidean n-space is exactly n. This proof makes use of the famous Brouwer Fixed-Point Theorem, which we naturally also prove. We give a combinatorial proof of the Fixed-Point Theorem, which relies on Sperner’s lemma. We move on to develop some theory regarding the extensions of functions. These various results on extensions allow us to finally prove the theorem that lent its name to this thesis: Brouwer’s Theorem on the Invariance of Domain.

Avainsanat — Nyckelord — Keywords Dimension theory, Invariance of domain, Separable metric spaces, Topology Säilytyspaikka — Förvaringsställe — Where deposited Kumpula Science Library

Muita tietoja — Övriga uppgifter — Additional information

Contents

1 Introduction 7

2 Preliminaries 9 2.1 Topological spaces ...... 9 2.2 Metric spaces ...... 17

3 Small Inductive Dimension 21 3.1 Dimension 0 ...... 21 3.2 Dimensionn...... 27

4 Brouwer Fixed-Point Theorem 35 4.1 Some Theory of Simplexes ...... 35 4.2 Some Properties ofR n ...... 46

5 Invariance of Domain 49

Bibliography 65

1 Introduction

The aim of this thesis is to present rudimentary dimension theory, in particular dimension theory regarding separable metric spaces, and with the tools developed, prove Brouwer’s Theorem on the Invariance of Domain. This theorem states, that ifA is a subset of the Euclidean spaceR n, an embedding h:A R n is an open map. This result is simple in the way, that anyone → familiar with elementary topology can understand the meaning of it, and yet as we shall see, the proof is not so simple. There are several notions of dimension in topology, but our focus in this thesis is only on the small inductive dimension. Should the reader be so in- clined, Professor Ryszard Engelking’s Dimension Theory [1] contains a com- prehensive exposition – among small inductive dimension – of other types of dimension, i.e. large inductive dimension and covering dimension. We begin by a revision of topological concepts that should be familiar to any graduate student of mathematics. This chapter relies mostly on the two books by Professor Jussi Väisälä, [6] and [7]. The book Topology by Professor James Munkres [5] has served not so much as a mathematical reference, rather as a stylistic guide on language. The deﬁnition of a separator between two subsets of some space is from [1]. The proof of Urysohn’s lemma is from Engelking’s General Topology [2]. Chapter 3 introduces the concept of small inductive dimension, or simply dimension as we refer to it in this thesis. We begin by deﬁning the notion of 0-dimensionality before moving on to the general notion of dimension. We cover many basic results of the dimension theory of separable metric spaces, although some results hold for more general spaces, namely regular spaces. The main source for this chapter is [4]. The book by Engelking [1] has also contributed as a source for this chapter. In Chapter 4 we develop some theory of simplexes in order to prove the

7 8 Chapter 1. Introduction famous Brouwer Fixed-Point Theorem. With the use of this theorem, we are able to show that the dimension of the Euclidean spaceR n is exactly n, as one should expect. The section on simplexes is almost entirely based on [2]. The lecture notes of a combinatorics course taught at Princeton University by Jacob Fox [3] provided help in the proof of Sperner’s lemma. From Proposition 4.18 to the end of Chapter 4 we rely on [4]. In Chapter 5 we develop some theory concerning extensions of functions and use this theory to prove the Brouwer’s Theorem on the Invariance of Domain, which lent its name to this thesis. This chapter is based on [4], with the exception of the Tietze Extension Theorem, the proof of which is from [2]. I would like to express my gratitude to my thesis supervisors, Drs. Erik Elfving and Pekka Pankka, for all the help and guidance they have so kindly oﬀered to me. 2 Preliminaries

Topological spaces

Deﬁnition 2.1 (Topology). LetX be some set andT a collection of subsets ofX, i.e.T P(X), whereP(X) is the power set ofX. The collectionT is ⊂ called a topology onX if the following conditions hold:

(T1) Any union of sets inT is an element ofT. • (T2) Anyﬁnite intersection of sets inT is an element ofT. • (T3) The empty set and the whole setX are members ofT. • ∅

An ordered pair (X,T), whereT is some topology on the setX is called a topological space. The sets inT are the so called open sets inX. A set inX is said to be closed, if its complement is open inX. It might seem intuitive to think that a set being open implies that it is not closed, and vice versa. This is not the case. For an easy counterexample, one can see from (T3) of Deﬁnition 2.1 that the whole spaceX and the empty set are always both ∅ open and closed.

Deﬁnition 2.2 (Subspace topology). Let (X,T) be a topological space. If Y is a subset ofX, the collection

T = Y U:U T Y { ∩ ∈ } is a topology onY called the subspace topology.

9 10 Chapter 2. Preliminaries

Deﬁnition 2.3 (Product topology). Suppose we have some indexed collection of topological spaces (X ,T ) wherej J andJ is some index set. The j j ∈ topology, called the product topology on the Cartesian product

X := Xj j J �∈ is the coarsest topology for which the projection maps pr :X X are j → j continuous. We recall, that ifU is a non-empty open set inX, then pr jU= Xj except for afinite number of indexesj (see Väisälä [7][Lause 7.6, p. 49]). IfX =Y for allj J, we writeX=Y J . j ∈ Definition 2.4 (Neighbourhood). Let (X,T) be a topological space. Letx be an element ofX andA a subset ofX. Ifx U T, thenU is called a ∈ ∈ neighbourhood ofx. The analogous definition holds for sets: ifA U T, ⊂ ∈ thenU is a neighbourhood ofA. Proposition 2.5. A subsetA of a topological space(X,T) is open, if and only if every elementx A has a neighbourhoodU , which is included inA. ∈ x Proof. SupposeA is open. Now we can chooseA=U for everyx A. x ∈ Suppose then that every elementx A has a neighbourhoodU , which is ∈ x included inA. Now we may writeA= x A Ux, whence it follows thatA is ∈ open as a union of open sets. � Oftentimes it is difficult to specify the topology onX by describing the whole collectionT of open sets. In most cases we can specify a smaller collection of subsets ofX, and define the topology using this smaller collection. Definition 2.6 (Basis). SupposeT is a topology on a spaceX. We call a collectionB P(X)a basis for the topology onX, if ⊂ B T • ⊂ Every open setU= can be expressed as a union of some sets inB. • � ∅ Proposition 2.7. Suppose(X,T) is a topological space. The collectionB ⊂ P(X) is a basis forT, if and only if (1)B T ⊂ (2) Ifx U T, there exists aB B such thatx B U. ∈ ∈ ∈ ∈ ⊂ Proof. SupposeB is a basis forT. Then by definition (1) holds. Suppose x U T. Since the setU can be expressed as a union of some sets inB, ∈ ∈ there exists some basis setB x inB such thatB U. � ⊂ Suppose (1) and (2) hold. LetU be a non-empty open set. Now for every x U, wefind some setB x from the collectionB, that is included inU. ∈ x � Now we haveU= x U Bx. This completes the proof. ∈ � Brouwer’s Theorem on the Invariance of Domain 11

Of particular interest in theﬁeld of topology are functions that preserve topological properties. These functions are called homeomorphisms. We recall, that a functionf:X Y , whereX andY are topological spaces, → 1 is said to be continuous, if for every open setU Y , the preimagef − U is ⊂ open inX.

Deﬁnition 2.8 (Homeomorphism, embedding). Let (X,T X ) and (Y,T Y ) be topological spaces. A continuous bijectionf:X Y is called a homeomor- 1 → phism, if the inverse functionf − :Y X is also continuous. If a function → g:X gX, where gX Y is a homeomorphism, the functiong:X Y → ⊂ → is called an embedding. Iff is a homeomorphism between the spacesX and Y , we say that these spaces are homeomorphic and we writef:X Y. ≈ Deﬁnition 2.9 (Closure). Let (X,T) be a topological space. The closure of the subsetA X is the set ⊂

A := x X:U A= , for every neighbourhoodU ofx . { ∈ ∩ � ∅ } When dealing with subspaces of some topological spaceX, we may denote the closure operation as clA, whereA is some subset ofX. Now ifY is a subspace of the spaceX, cl X A denotes the closure of the setA in the whole spaceX, whereas cl Y A denotes the closure ofA in the subspaceY. A pointx X is called a limit point ofA, if every neighbourhoodV ∈ ofx contains some point ofA distinct fromx. If, on the other hand, there exists some neighbourhoodU ofx such thatU A= x , thenx is called ∩ { } an isolated point ofA.

Deﬁnition 2.10 (Interior, exterior, boundary). LetX be a topological space, and letx X andA X. The pointx is called an interior point ofA, ∈ ⊂ ifx has some neighbourhoodU A. Ifx has a neighbourhoodU A, then ⊂ ⊂� x is an exterior point ofA. Ifx is neither an interior nor an exterior point ofA, we say thatx is a boundary point ofA. The sets of interior, exterior, and boundary points ofA are denoted as intA, extA, and∂A respectively.

Proposition 2.11. SupposeA andB are subsets of a topological space (X,T).

1.A A. ⊂ 2. A is always closed.

3. IfB is closed inX, andA B, then A B. ⊂ ⊂ 12 Chapter 2. Preliminaries

4. A is the smallest closed subset ofX containingA.

5. IfA B, then A B. ⊂ ⊂ 6.A is closed inX, if and only ifA= A.

7. intA A, extA A. ⊂ ⊂� 8.A is open, if and only ifA = intA.

9. extA= A, intA= A, A = intA ∂A=A ∂A. � � � ∪ ∪ 10.∂A= A A= A intA, and∂A is always closed. ∩ � \ 11.∂A=∂�A. 12. IfA is open, then∂A= A A. \ Proof. See Väisälä [6, Lause 6.8, p. 48 and Lause 8.3, p. 60].

Proposition 2.12. A subsetA of a topological spaceX is both open and closed, if and only ifA has an empty boundary.

Proof. Suppose, that =A=X, since the result is evident otherwise. ∅� � SupposeA X is open and closed. From Proposition 2.11 it follows that ⊂ ∂A= A A, sinceA is open, andA= A, sinceA is closed. Hence∂A= . \ ∅ Suppose then, that∂A= . Now from Proposition 2.11 we get that ∅ A=A ∂A=A, henceA is closed. Also∂A= A intA= . Hence ∪ \ ∅ A=A = intA, which impliesA is open.

Deﬁnition 2.13(F σ sets,G δ sets). A countable union of closed sets is called anF σ set. The complement of anF σ set, i.e. a countable intersection of open sets, is called aG δ set.

Clearly every closed set is anF σ set and similarly every open set is a Gδ set. Since the countable union of a countable union is again a countable union, a countable union ofF σ sets is anF σ set. Also, it is quite easy to verify, that aﬁnite intersection ofF σ sets is also anF σ set.

Deﬁnition 2.14 (Separation axioms). Letj 0,1,2,3,4 . We say that ∈{ } a spaceX is aT j space, or more succinctly, thatX isT j, if it satisﬁes the condition (Tj) below.

(T ) If a, b X are distinct points, then at least one of the points a, b • 0 ∈ has a neighbourhood not containing the other point. Brouwer’s Theorem on the Invariance of Domain 13

(T ) If a, b X are distinct points, then both of the points a, b have a • 1 ∈ neighbourhood not containing the other point.

(T ) If a, b X are distinct points, the pointsa andb have disjoint • 2 ∈ neighbourhoods.

(T ) Ifa X, andB is a closed subset ofX not containing the point • 3 ∈ a, then the pointa and the setB have disjoint neighbourhoods.

(T ) If the closed setsA,B X are disjoint, the setsA andB have • 4 ⊂ disjoint neighbourhoods.

Spaces satisfying the conditionT 2 are called Hausdorff spaces. We call a topological spaceX regular, if it satisfies the separation axiomsT 1 andT 3.A space is normal, if it satisfies axiomsT 1 andT 4.A completely normal space is a space whose every subspace is normal. Since the propertyT 1 is clearly hereditary, in order to prove that a normal space is completely normal, it suffices to show that every subspace of that space isT 4. Proposition 2.15. A space(X,T) isT , if and only if for everyx X and 3 ∈ for every neighbourhoodU ofx, there exists a neighbourhoodV ofx such that V U. ⊂ Proof. SupposeX is aT -space. Letx X andU X be a neighbourhood 3 ∈ ⊂ ofx. Now the set�U is closed, and does not contain the pointx. Since X isT 3,x and�U have disjoint neighbourhoods, that we shall callV and W , respectively. Now the set�W is closed inX and containsV , whence it follows that V W U=U. ⊂� ⊂ �� Suppose the condition of the proposition holds. Letx X andB X ∈ ⊂ be a closed set not containing the pointx. Now�B is a neighbourhood ofx, and thus there exists some neighbourhoodV ofx such that V B. Now we ⊂� have found disjoint neighbourhoods for the pointx and closed setB, namely V and� V , respectively.

Proposition 2.16. A space(X,T) isT 4, if and only if for every closed subset A X and for every neighbourhoodU ofA, there exists a neighbourhoodV ⊂ ofA such that V U. ⊂ Proof. The argument is essentially the same as in the proof of Proposition 2.15. We just replace the pointx with the setA. Proposition 2.17. A space(X,T) is completely normal, if and only if there exist disjoint neighbourhoods for any two subsetsA,B X, for which it ⊂ holds that A B=A B= . ∩ ∩ ∅ 14 Chapter 2. Preliminaries

Proof. Suppose (X,T) is completely normal and thatA,B X are subsets ⊂ ofX such that A B=A B= . LetY= (cl A cl B). Now cl A ∩ ∩ ∅ � X ∩ X Y and clY B are disjoint closed subsets of the spaceY . SinceY is normal and thusT , there exist disjoint neighbourhoodsU ,U Y for the sets cl A 4 A B ⊂ Y and clY B, respectively. SinceY is open inX, so are the setsU A andU B. Since Y= (cl A cl B)= (cl A) (cl B), � X ∩ X � X ∪� X bothA (cl B) andB (cl A) are subsets ofY . Hence we have ⊂� X ⊂� X found two disjoint neighbourhoods forA andB inX, namelyU A andU B, respectively. Suppose the condition of the proposition holds, and thatY is a subspace of the space (X,T). LetA,B Y be disjoint and closed inY . Since ⊂ the topology onY is the subspace topology, we haveA=Y cl A and ∩ X B=Y cl B. Thus ∩ X =A B=(Y cl A) B = (cl A) B, ∅ ∩ ∩ X ∩ X ∩ and similarly =A B=A (Y cl B)=A (cl B). ∅ ∩ ∩ ∩ X ∩ X By assumption, there exist disjoint open subsetsU A andU B ofX such thatA U andB U . TakingU Y andU Y , we have disjoint ⊂ A ⊂ B A ∩ B ∩ neighbourhoods in the subspace topology onY forA andB, respectively. This shows thatY isT 4, completing the proof. Proposition 2.18 (Urysohn’s lemma). For every pairA,B of disjoint closed subsets of a normal spaceX, there exists a continuous functionf:X [0, 1] → such thatf(x) = 0 for everyx A andf(x) = 1 for everyx B. ∈ ∈ Proof. For every rational numberq on the interval [0, 1] we shall deﬁne an open setV q with the conditions

V V whenever q

are thus satisﬁed fork = 2. Suppose the setsV satisfying (3 ) are deﬁned fori n, wheren 2. qi n ≤ ≥ Let us denote byq l andq r, respectively, those of the numbersq 1, q2, . . . , qn that are closest to the numberq n+1 from the left and from the right. Since q < q it follows from (3 ) that V V . LetU � be an open set such l r n ql ⊂ qr that V U � U V . Again the existence of such a setU � follows ql ⊂ ⊂ � ⊂ qr from Proposition 2.16. TakingV qn+1 =U �, we obtain setsV q1 ,Vq2 ,...,Vqn+1

which satisfy (3n+1). The sequenceV q1 ,Vq2 ,... obtained this way satifies contiditions (1) and (2). Consider the functionf:X [0, 1] defined by the formula → inf q:x V , forx V f(x) = { ∈ q} ∈ 1 1, forx X V  ∈ \ 1 Since (2) impliesfA 0 andfB 1 , we only need to show thatf is ⊂{ } ⊂{ } continuous. To show this, it suffices to show that inverse images of intervals of the form [0, a[ and ]b, 1], wherea 1 andb 0, are open. The inequality ≤ ≥ f(x)b holds, if and only if there exists aq � > b such thatx V . Hence by (1) this means that there �∈ q� exists aq>b such thatx V . Thus the preimage �∈ q 1 f − ]b, 1] = X V :q>b =X V :q>b \ q \ q � � � � � � is open as a complement of a closed set.

Since the unit interval [0, 1] is homeomorphic to any interval of the form [a, b], wherea

Deﬁnition 2.19 (Countability axioms). A spaceX is said to have a countable basis at the pointx, if there exists a countable collectionB x of neighbourhoods ofx such that each neighbourhood ofx contains some element of Bx. A space that has a countable basis at each of its points is said to be ﬁrst-countable. If a spaceX has a countable basis for its topology, thenX is said to be second-countable.

Deﬁnition 2.20 (Compact space). A spaceX is compact, if every open covering ofX contains aﬁnite subcovering. 16 Chapter 2. Preliminaries

We recall that the setA is countable, if there exists a surjectionf:N A. → Deﬁnition 2.21 (Lindelöf space). A spaceX is Lindelöf, if every open covering ofX contains a countable subcovering. Proposition 2.22. Every second-countable space is Lindelöf.

Proof. Let (X,T) be a topological space and letB= B n :n N be a basis { ∈ } for the topologyT. SupposeD is an open covering ofX. For everyB B n ∈ that is a subset of some element ofD, we shall choose someU n such that B U D. If no such member ofD exists for some basis setB , the n ⊂ n ∈ n setU n is not defined. Let us denote the collection of the setsU n asA. The collectionA is a countable subcollection ofD. Supposex X. BecauseD is a covering ofX, there exists someU D ∈ ∈ such thatx U. BecauseB is a basis forT, Proposition 2.7 implies the ∈ existence of someB B such thatx B U. Thus there exists a setU n ∈ ∈ n ⊂ n for which it holds thatx B U A. ThusA is a covering ofX. ∈ n ⊂ n ∈ Definition 2.23 (Separation, Connected space). Let (X,T) be a topological space. A separation ofX is a pairU,V of non-empty subsetsX such that U V=U V= , andX=U V . The spaceX is said to be connected, ∩ ∩ ∅ ∪ if no separation ofX exists. The definition of connectedness can also be formulated in the following way: the spaceX is connected, if and only if there exist no other subsets ofX, which are both open and closed inX, except the empty set and the whole spaceX itself.

Deﬁnition 2.24 (Separator between subsets). IfA 1 andA 2 are disjoint subsets of the space (X,T), we say that a subsetB X is a separator ⊂ betweenA andA , ifX B can be split into two disjoint setsA � andA � , 1 2 \ 1 2 open inX B, and containingA andA , respectively. In other words, it \ 1 2 holds that X B=A � A � , \ 1 ∪ 2 A A � ,A A � , 1 ⊂ 1 2 ⊂ 2 A� A � = , 1 ∩ 2 ∅ withA � andA � both open inX B, or equivalently both closed inX B. 1 2 \ \ Proposition 2.25. The empty set is a separator between the subsetsA 1 and A2 of a topological space(X,T), if and only if there exists a setA 1� such that

A A � , 1 ⊂ 1 A� A = , 1 ∩ 2 ∅ andA 1� is both open and closed, or equivalently, has an empty boundary. Brouwer’s Theorem on the Invariance of Domain 17

Proof. TakeA � =X A � . 2 \ 1 Deﬁnition 2.26 (Dense set). A subsetD of a topological space (X,T) is called dense inX, if for every =U T it holds that the intersectionD U ∅� ∈ ∩ is not empty.

For example the set of rational numbersQ is dense in the real numbersR, whenR is endowed with the usual topology. As the set of rational numbers is countable, we get thatR is in fact a separable space, which leads us to our next deﬁnition.

Deﬁnition 2.27 (Separable space). A spaceX is called separable, if it contains a countable dense set.

Metric spaces

Definition 2.28 (Metric). LetX be some set. A functiond:X X R 0 × → ≥ is a metric inX, if the following conditions hold for all x, y, z X: ∈ (M1)d(x, z) d(x, y)+d(y, z), • ≤ (M2)d(x, y)=d(y, x), • (M3)d(x, y) = 0, if and only ifx=y. • A metric space is an ordered pair (X, d), whereX is some set andda metric inX. The following definition gives us a tool to define a topology on a metric space.

Deﬁnition 2.29 (Ball). Supposea is an element of a metric space (X, d), andr> 0. We deﬁne

B(a, r) := x X:d(x, a)

Definition 2.30 (Metric topology). Let (X, d) be a metric space. The collection of balls B(x, r):x X,r>0 is a basis for a topology onX, called { ∈ } the metric topology induced byd. In a metric space (X, d) a setU X is open, if for everyx U we can ⊂ ∈ find an open ballB(x, r), which is included inU. Definition 2.31 (Diameter). The diameter of a subsetA of a metric space (X, d) is defined to be

d(A) := sup d(x, y): x, y A . { ∈ } The distance of a pointx X from the setA is ∈ d(x, A) := inf d(x, y):y A . { ∈ } Proposition 2.32. If(X, d) is a separable metric space, it is second-countable.

Proof. Let (X, d) be a metric space andA= a j j N be a countable dense { } ∈ set inX. We claim that the collection

B := B(a, r):a A, r Q >0 { ∈ ∈ } is a countable basis for the topology onX. LetU be a non-empty open set inX. Now for everyx U, there exists ∈ someε> 0 such that the ballB(x,ε) is included inU. SinceA is dense inX, there exists somea A such thatd(x, a )<ε/2. Hence,a B(x,ε) U. j ∈ j j ∈ ⊂ It holds thatB(a ,ε/2) B(x,ε). For lety B(a ,ε/2). Now from the j ⊂ ∈ j triangle inequality we get that

d(y, x) d(y, a ) +d(a , x)<ε/2 +ε/2<ε. ≤ j j SinceQ is dense inR, for everyε> 0 we canﬁnd a positive rational numberr such thatr<ε. Thus for everyx U, we canﬁnd an open ball ∈ with centre at somea A and a rational radiusr such that j ∈ x B(a , r) B(x,ε) U. ∈ j ⊂ ⊂ Hence we have the inclusion

U B(a, ra), ⊂ a U A ∈�∩ wherer a is a rational radius depending on the pointa, such that the ball B(a, r ) is included inU. The condition that the ballsB(a, r ), wherea A a a ∈ andr Q >0, are included inU gives us the other inclusion ∈ Brouwer’s Theorem on the Invariance of Domain 19

B(a, ra) U. a U A ⊂ ∈�∩ We have shown that every non-empty open subset ofX can be represented as a union of sets in the collectionB. This completes the proof, sinceB is countable.

Proposition 2.33 (The Lebesgue Covering Theorem). For every open cov- eringD of a compact metric space(X, d), there exists anε>0 such that the covering B(x,ε) x X is a refinement ofD. In other words, every set in { } ∈ B(x,ε) x X is contained in some set inD. { } ∈ Proof. For everyx X, we shall choose someε > 0 such that the ball ∈ x B(x,2ε x) is contained in a member ofD. SinceX is compact, the open covering B(x,ε) x X has afinite subcovering, i.e. there exists afinite set { } ∈ x , x , . . . , x X such that { 1 2 k}⊂ k

X= B(xi,ε xi ). i�=1 The numberε := min ε ,ε ,...,ε has the required property. { x1 x2 xk }

Deﬁnition 2.34 (Hilbert cube). The spaceI ω := i N[ 1/i,1/i] is called ∈ − the Hilbert cube. Since [ 1/i,1/i] is compact for alli N, the Hilbert cube − �∈ is compact by Väisälä [7, Lause 18.1, p. 136]. It is often convenient to think of the Hilbert cube as a metric space. For this purpose, the Hilbert cube is 2 2 considered to be a subspace of the separable metric spacel ∗. Thel -norm induces the product topology on the Hilbert cube.

Deﬁnition 2.35 (Metrizable space). A space (X,T) is called metrizable, if there exists a metricd, such that the topologyT d induced by this metric is equal toT.

We recall, that every metric space is Hausdorﬀ, regular and normal, see for example Väisälä [7, Havaintoja, p. 87]. In fact, every metric space is completely normal, since every subspace of a metric space is itself a metric space.

∗Here we mean the space consisting of square-summable sequences of real numbers. 2 2 1/2 Thel -norm is naturally deﬁned as x 2 := i xi . � � ∈N | | �� 

3 Small Inductive Dimension

We shall begin this chapter by examining 0-dimensional spaces and their properties.

Dimension0

Definition 3.1 (Dimension 0). A non-empty regular spaceX has dimension 0, if for every pointx X and for every neighbourhoodU ofx, there exists ∈ a neighbourhoodV ofx such thatV U and∂V= . ⊂ ∅ Proposition 3.2. A 0-dimensional space can also be defined as a non-empty space, in which we can define a basis consisting of sets that are both open and closed. Proof. LetX be some non-empty regular space, which has a basisB consisting of open and closed sets. Letx X andU be a neighbourhood of ∈ x. Now Proposition 2.7 guarantees, that we canfind a basis elementB B ∈ such thatx B U. This basis setB is a neighbourhood ofx, which is ∈ ⊂ both open and closed, and by Proposition 2.12 has an empty boundary. Suppose then, thatX is a 0-dimensional space. Now for every pointx X ∈ and for every neighbourhoodV ofx, we canfind a new neighbourhoodU V ⊂ ofx, whose boundary is empty, or equivalently is both open and closed. Since we canfind arbitrarily small open and closed neighbourhoods for each point inX, Proposition 2.7 implies the collection of open and closed sets inX form a basis for the topology onX. Example 3.3. Every non-empty countable metric space (X, d) is 0-dimensional. For suppose thatx X andU is some neighbourhood ofx. Let ∈ r > 0 be such, that the ballB(x, r) is contained inU. Letx i, wherei N ∈ 21 22 Chapter 3. Small Inductive Dimension

be an enumeration ofX. Letr � be a positive real number less thanr such thatr � =d(x i, x) for alli N. Now the ballB(x, r �) is contained inU, and � ∈ has an empty boundary. HenceX is 0-dimensional.

Example 3.4. Any non-empty subspace of the real lineR containing no interval is 0-dimensional. For suppose thatP is such a set. The complement R P is dense inR, for otherwise there would be some non-empty open set \ U which does not intersectR P . Hence we would haveU P . SinceU is \ ⊂ open inR, it contains an interval, a contradiction. Thus sets of the form ]a, b[ P, ( ) ∩ ∗ where a, b R P anda

Example 3.5. The subspaceJ ω of the Hilbert cubeI ω consisting of points, all of whose coordinates are irrational is 0-dimensional. Supposea J , and ∈ ω letU be a neighbourhood ofa inI ω. Since the Hilbert cube has the product topology, there exists a numberN N such that pr iU=[ 1/i,1/i] for ∈ − i > N. Thus we canfind a new neighbourhood fora contained inU, which consists of the pointsx=(x 1, x2,... ) inI ω whosefirstN coordinates are restricted byp i < xi < qi, wherep i < ai < qi, andp i, qi are sufficiently close toa , and the rest of the coordinates ofx are restricted only by x 1/i. i | i|≤ By takingp i andq i rational, we get a neighbourhoodV ofa, each of whose boundary point inI ω has at least one rational coordinate. ThusV has an empty boundary inJ ω, which proves thatJ ω is 0-dimensional. Proposition 3.6. A non-empty subspace of a 0-dimensional space is 0- dimensional.

Proof. SupposeX is a 0-dimensional space and letX � be a non-empty subspace ofX. Letx X and letU � be a neighbourhood of the pointx in the ∈ subspaceX �. SinceX � has the subspace topology, there exists someU open inX such thatU � =U X �. ∩ SinceX is 0-dimensional, there exists a neighbourhoodV ofx, both open and closed inX such thatV U. LetV � =V X �. NowV � is both open ⊂ ∩ and closed inX �, andx V � U �. HenceX � is 0-dimensional. ∈ ⊂ We shall prove that Deﬁnition 3.1 is equivalent to Brouwer’s Theorem on the Invariance of Domain 23

Proposition 3.7. A non-empty regular spaceX has dimension 0, if and only if the empty set is a separator between every singleton set x X and { }⊂ every closed setC not containing the pointx.

Proof. SupposeX is 0-dimensional. Supposex X and letC X bea ∈ ⊂ non-empty closed subset ofX not containing the pointx. SinceX C is a \ neighbourhood ofx, there exists a neighbourhoodV X C ofx, which is ⊂ \ both open and closed. FromV C= and Proposition 2.25, it follows that ∩ ∅ the empty set is a separator between x andC. { } Suppose then, that the condition of the proposition holds. Supposex X ∈ and thatV is a neighbourhood ofx. Now the set�V is a closed subset ofX that does not contain the pointx. Since the empty set is a separator between x and V , Proposition 2.25 implies that there exists an open and closed set { } � U such thatx U, andU V= . Hence we have found a neighbourhood ∈ ∩� ∅ U V ofx, whose boundary is empty. This proves the claim. ⊂ From Proposition 3.7 it follows that a metric space is 0-dimensional, if the empty set is a separator between any two disjoint closed subsets, since in metric spaces singleton sets are closed. The converse also holds:

Proposition 3.8. If the separable metric spaceX is 0-dimensional, the empty set is a separator between any two disjoint closed subsets ofX.

Proof. Suppose (X,T) is 0-dimensional. Proposition 3.7 implies that the empty set is a separator between any singleton set x X and any closed { }⊂ set not containingx. LetC andK be disjoint closed subsets ofX. We shall demonstrate that the empty set is a separator between these two sets inX. For every pointx X, at most one of the conditionsx C andx K ∈ ∈ ∈ holds. SinceX isT 3, there exist open and closed neighbourhoodsU x for each pointx such that again at most one of the following hold:U C= , or x ∩ � ∅ U K= . By Proposition 2.32X is second-countable and by Proposition x ∩ � ∅ 2.22X is Lindelöf. Hence there exists a countable collection U xn :n N of { ∈ } these neighbourhoodsU x that forms an open covering ofX. We shall deﬁne a new sequence of sets as follows:

V1 :=U x1 ,

i 1 − Vi :=U xi Uxk , i=2,3,4,.... \ � � k�=1 Now we have ∞ X= Vi, i�=1 24 Chapter 3. Small Inductive Dimension and V V = , ifi= j. i ∩ j ∅ � i 1 Since k− Uxk is closed for everyi N >1 as theﬁnite union of closed sets, =1 ∈ its complement� is open, and thusV i is open as the intersection of two open sets. We also have eitherV C= orV K= , (or both). i ∩ ∅ i ∩ ∅ LetC � be the union of allV for whichV K= , andK � the union of i i ∩ ∅ the remainingV i. Now X=C � K �, ∪ C� K � = , ∩ ∅ bothC � andK � are open as unions of open sets, and

C� K=K � C= . ∩ ∩ ∅

It follows thatC C � andK K �. We have shown that the empty set is a ⊂ ⊂ separator betweenC andK.

Proposition 3.9. IfC 1 andC 2 are disjoint closed subsets of a separable metric spaceX, andA is a 0-dimensional subspace ofX, there exists a closed separatorB between the setsC andC such thatA B= . 1 2 ∩ ∅

Proof. Let the setsC 1,C 2 andA be as described in the proposition. Since X isT 4, Proposition 2.16 implies there exist open setsU 1 andU 2 such that

C U ,C U , (1) 1 ⊂ 1 2 ⊂ 2 and U U = . 1 ∩ 2 ∅ The disjoint sets U A and U A are closed inA, and by Proposition 3.8 1 ∩ 2 ∩ the empty set is a separator between these sets inA, sinceA is 0-dimensional.

Thus there exist disjoint setsC 1� andC 2� , both open and closed inA, satisfying

A=C � C � 1 ∪ 2 and U A C � , U A C � . 1 ∩ ⊂ 1 2 ∩ ⊂ 2 Thus C� U =C � U = , (2) 1 ∩ 2 2 ∩ 1 ∅ and C C = C C � = . (3) 1 ∩ �2 �1 ∩ 2 ∅ From (1) and (2) we get

C� C =C � C = . (4) 1 ∩ 2 2 ∩ 1 ∅ Brouwer’s Theorem on the Invariance of Domain 25

Furthermore, sinceU 1 andU 2 are open sets, (2) implies

C U = C U = , �1 ∩ 2 �2 ∩ 1 ∅ and hence by (1) C C = C C = . (5) �1 ∩ 2 �2 ∩ 1 ∅ From (3), (4), and (5) together with C C =C C = it follows 1 ∩ 2 1 ∩ 2 ∅ that C C � (C C � ) = (C C � ) C C � = . SinceX is completely 1 ∪ 1 ∩ 2 ∪ 2 1 ∪ 1 ∩ 2 ∪ 2 ∅ normal, Proposition 2.17 implies there exists an open setV such that

C C � V, 1 ∪ 1 ⊂ and V (C C � ) = . ∩ 2 ∪ 2 ∅ Now we have X ∂V=V V ∂V , \ ∪ � \ � � C V,C V ∂V , 1 ⊂ 2 ⊂ � \ � � the setsV and V ∂V are open and closed inX ∂V , and the boundary∂V � \ \ is disjoint fromC � C � =A. We have shown that we can chooseB=∂V, 1 ∪ 2 which completes the proof. Theorem 3.10 (The Union Theorem for Dimension 0). If the separable metric spaceX is the countable union of closed 0-dimensional subsets ofX, then the spaceX is itself 0-dimensional. Proof. Suppose X= Ci, i N �∈ where eachC i is closed and 0-dimensional. LetK andL be disjoint closed subsets ofX. We will show that the empty set is a separator between these sets. NowK C andL C are disjoint closed subsets of the 0-dimensional ∩ 1 ∩ 1 spaceC 1. Thus by Proposition 3.8 there exist subsetsA 1 andB 1 ofC 1, which are closed inC 1, and thus also closed inX, such that

K C A ,L C B , ∩ 1 ⊂ 1 ∩ 1 ⊂ 1 A B =C , andA B = . 1 ∪ 1 1 1 ∩ 1 ∅ The setsK A andL B are closed and disjoint inX. SinceX is normal, ∪ 1 ∪ 1 there exist open setsG 1 andH 1 such that

K A G ,L B H , ∪ 1 ⊂ 1 ∪ 1 ⊂ 1 26 Chapter 3. Small Inductive Dimension and G H = . 1 ∩ 1 ∅ Hence it holds that C G H , 1 ⊂ 1 ∪ 1 and K G ,L H . ⊂ 1 ⊂ 1 We shall repeat this same process, replacingC 1 byC 2, and replacingK andL by G1 and H1, respectively. We get open setsG 2 andH 2 for which

C G H , 2 ⊂ 2 ∪ 2 G G , H H , 1 ⊂ 2 1 ⊂ 2 and G H = . 2 ∩ 2 ∅ We shall continue inductively, constructing sequences G i i N and H i i N { } ∈ { } ∈ of sets open inX for which it holds that

C G H , i ⊂ i ∪ i

Gi 1 G i, Hi 1 H i, − ⊂ − ⊂ and G H = . i ∩ i ∅ Put

G := Gi andH := Hi. i N i N �∈ �∈ NowG andH are disjoint open sets such that

X= C G H i ⊂ ∪ i N �∈ and K G,L H. ⊂ ⊂ This shows that the empty set is a separator betweenK andL, which completes the proof.

Corollary 3.11. If a separable metric spaceX can be represented as a countable union of 0-dimensionalF σ sets, thenX is itself 0-dimensional. Brouwer’s Theorem on the Invariance of Domain 27

Proof. SupposeX= i NCi, where eachC i is a 0-dimensionalF σ set. Now ∪ ∈ each of the setsC i is a countable union of closed sets, which are at most 0-dimensional by Theorem 3.6. The countable union of a countable union is itself a countable union, whence it follows thatX can be represented as the countable union of 0-dimensional closed subsets. Hence by Theorem 3.10X is 0-dimensional.

m Example 3.12. Supposem 0 and denote byQ ω the set of points of the ≥ m Hilbert cube exactlym of whose coordinates are rational. The setQ ω is 0-dimensional. For each selection ofm indexesi 1, i2, . . . , im, and for each selection of m rational numbersr 1, r2, . . . , rm we have a subspace of the Hilbert cube deﬁned by the equations

xi1 =r 1, xi2 =r 2, . . . , xim =r m.

There are countably many of these subspaces. Denote byC i the set consisting of the points satisfying the above equations, and all of whose remaining coordinates are irrational. The setsC i contain all of their limit points and m are thus closed inQ ω . EveryC i can be embedded intoJ ω (Example 3.5). Therefore the setsC i are 0-dimensional by Theorem 3.15, which is proved m below. ThusQ ω is the countable union of 0-dimensional spaces, and by Theorem 3.10 is itself 0-dimensional.

Dimensionn

Deﬁnition 3.13 (Small inductive dimension). LetX be a regular space. The small inductive dimension ofX, denoted by ind(X), is either an integer larger than 2, or . The deﬁnition of the dimension function ind consists − ∞ of the following conditions:

(SID1) ind(X)= 1, if and only ifX= , • − ∅

(SID2) ind(X) n, wheren Z 0, if for every pointx X and for • ≤ ∈ ≥ ∈ each neighbourhoodV X of the pointx there exists a neighbourhood ⊂ U V ofx such that ⊂ ind(∂U) n 1, ≤ − (SID3) ind(X)=n if ind(X) n and ind(X)>n 1, • ≤ −

(SID4) ind(X)= if ind(X)>n for alln Z 1 • ∞ ∈ ≥− 28 Chapter 3. Small Inductive Dimension

We shall refer to the small inductive dimension simply as dimension throughout this thesis. When we talk about the dimension of some space X, it is understood thatX is regular.

Proposition 3.14. Equivalent to the condition that ind(X) n is the ex- ≤ istence of a basis for the topology onX consisting of sets whose boundaries have dimension n 1. ≤ − Proof. The argument is essentially the same, as in the proof of Proposition 3.2.

Theorem 3.15. Dimension is a topological invariant.

Proof. Throughout the proof suppose thatX andY are regular spaces and thatf:X Y. ≈ Suppose thatX is 0-dimensional. Supposey Y and letV be a neigh- 1 ∈ 1 bourhood ofy. Nowf − V is a neighbourhood of the pointf − (y). Since 1 1 X is 0-dimensional, there exists a neighbourhoodU f − V off − (y) with ⊂ an empty boundary. Since homeomorphisms are open and closed maps, the image ofU underf is open and closed inY . We have thus found a neigh- bourhoodfU V ofy with an empty boundary. HenceY is 0-dimensional. ⊂ Suppose that dimension is a topological invariant for spaces whose dimension does not exceed somen 0. Suppose ind(X)=n + 1. Now there ≥ exists an elementx X such that there exists a neighbourhoodW ofx ∈ such that all neighbourhoodsU W ofx have boundaries with dimension ⊂ ind(∂U) n. Thus there exists a pointy=f(x) Y with a neighbour- ≥ ∈ hoodfW such that every neighbourhoodV fW ofy has boundaries with ⊂ dimension ind(∂V) n by the induction hypothesis. Hence ind(Y) n + 1. ≥ ≥ 1 Suppose thaty Y and letV � be a neighbourhood ofy. Nowf − V � ∈ 1 is a neighbourhood of the pointf − (y) X. Since ind(X) n + 1 there 1 ∈1 ≤ exists a neighbourhoodU � f − V � off − (y) whose boundary has dimension ⊂ ind(∂U �) n. The setfU � V � is a neighbourhood of the pointy. From ≤ ⊂ the induction hypothesis we get that ind(f∂U �) = ind(∂fU �) n. Thus ≤ ind(Y) n + 1, which proves the claim. ≤ Proposition 3.16. For every subspaceM of a regular spaceX, we have ind(M) ind(X). ≤ Proof. If ind(X)= , the proposition is evident, so we may suppose ind(X)< ∞ . The inequality holds clearly, if ind(X)= 1. Suppose the theorem holds ∞ − for all regular spaces, whose dimension does not exceed somen 1. Sup- ≥− poseX is a regular space with ind(X)=n + 1 and thatM is a non-empty subspace ofX. Letx M andV a neighbourhood ofx inM. Because ∈ M Brouwer’s Theorem on the Invariance of Domain 29

the topology onM is the subspace topology, there exists someV X open in X such thatV =M V . Since ind(X) n + 1, there exists an open set M ∩ X ≤ U X such thatx U V and ind(∂U ) n. X ⊂ ∈ X ⊂ X X ≤ The intersectionU =M U is open inM and is included inV . The M ∩ X M boundary∂ M UM in the spaceM is equal to

∂ U =M cl (M U ) cl (M U ). M M ∩ X ∩ X ∩ X \ X

The boundary∂ M UM is a subspace of the space∂U X . Hence by the induction hypothesis we get that ind(∂ U ) n, and thus we have ind(M) n+1= M M ≤ ≤ ind(X). Proposition 3.17. Suppose ind(X)=n. ThenX contains anm-dimensional subspace for everym n. ≤ Proof. It suﬃces to show thatX contains a subspace with dimensionn 1. − Since ind(X)>n 1, there exists a pointx X and a neighbourhoodV X − ∈ ⊂ ofx such that for every open setU satisfying the conditionx U V , we ∈ ⊂ have ind(∂U) n 1. On the other hand, since ind(X) n, there exists an ≥ − ≤ open setU X satisfying the above condition with ind(∂U) n 1. This ⊂ ≤ − proves the claim.

Proposition 3.18. A subspaceX � of a separable metric spaceX satisﬁes ind(X�) n, if and only if every point of the subspaceX � has arbitrarily ≤ small neighbourhoods inX, whose boundaries have intersections withX � of dimension n 1. ≤ − Proof. SupposeX � is a subspace of a separable metric space with dimension ind(X�) n. Letx be a point inX � andU a neighbourhood ofx inX. Now ≤ U � =U X � ∩ is a neighbourhood ofx inX �. Hence there exists a neighbourhoodV � X � ⊂ ofx such that x V � U �, ∈ ⊂ and ind(∂ V �) n 1, X� ≤ − where∂ X� V � denotes the boundary ofV � in the subspaceX �. Now it holds that cl V � (X � cl V �) =V � cl (X� cl V �) = . SinceX is completely X ∩ \ X ∩ X \ X ∅ normal as a metric space, Proposition 2.17 implies that there exists a setW open inX satisfying

V � W and cl W (X � cl V �) = . ⊂ X ∩ \ X ∅ 30 Chapter 3. Small Inductive Dimension

We may assume thatW U, since otherwise we can replaceW by the ⊂ intersectionW U. The boundary∂ W = cl W W contains no point of ∩ X X \ V � and no point ofX � cl V �. Hence we have the inclusion∂ W X � ∂ V � \ X X ∩ ⊂ X� and by Proposition 3.16 we have ind(∂ W X �) n 1, so that the condition X ∩ ≤ − of the proposition is satisﬁed. Suppose then, thatX � is a subspace of a separable metric spaceX for which the condition of the proposition holds. Supposex X � and letU � ∈ be a neighbourhood ofx in the subspaceX �. Since the topology onX � is the subspace topology, there exists a neighbourhoodU ofx inX such that U � =U X �. Thus there exists a setV , open inX, such that ∩ x V U, ∈ ⊂ and ind(∂ V X �) n 1. X ∩ ≤ − LetV � :=V X �. NowV � is open inX �, andx V � U �. We have the ∩ ∈ ⊂ inclusion∂ V � ∂ V X � from which it follows that ind(∂ V �) n 1 X� ⊂ X ∩ X� ≤ − by Proposition 3.16, so that ind(X �) n. ≤ Proposition 3.19. For any two subspacesA andB of a separable metric spaceX, we have

ind(A B) ind(A) + ind(B)+1. ∪ ≤ Proof. The proposition is clear, if

ind(A) = ind(B)= 1, − i.e. the setsA andB are empty. Let ind(A) =m and ind(B) =n, and suppose the proposition holds for the cases

ind(A) m, ind(B) n 1, (6) ≤ ≤ − and ind(A) m 1, ind(B) n. (7) ≤ − ≤ Letx A B. Without loss of generality, we may assumex A. LetU be ∈ ∪ ∈ a neighbourhood ofx inX. Now by Proposition 3.18 there exists a setV, open inX, such that x V U ∈ ⊂ and ind(∂ V A) m 1. X ∩ ≤ − Brouwer’s Theorem on the Invariance of Domain 31

As∂ V B is a subset ofB, by Proposition 3.16 we have X ∩ ind(∂ V B) n. X ∩ ≤ By the induction hypotheses (6) and (7) we get

ind(∂ V (A B)) m+ n. X ∩ ∪ ≤ Hence, from Proposition 3.18 we have

ind(A B) m+n+1. ∪ ≤ This completes the proof. Example 3.20. In Example 3.4 we showed that a subspace of the real line containing no interval is 0-dimensional. Thus both the rational numbersQ, and the set of irrational numbersR Q are 0-dimensional. We will show \ later that, in fact, ind(R) = 1. This shows that Proposition 3.19 gives us the most accurate upper bound for the dimension of the union of two subspaces. In case the reader wonders, we have not contradicted Theorem 3.10, since neitherQ, norR Q is a closed subset ofR. \ Example 3.21. Supposem 0. Denote byM m the set of points of ≥ ω the Hilbert cubeI ω at mostm of whose coordinates are rational. Then ind (Mm) m. This follows by repeatedly applying Proposition 3.19 since ω ≤ m m i Mω = Qω, i�=0 i where eachQ ω is 0-dimensional by Example 3.12. Theorem 3.22 (The Union Theorem for Dimensionn). Suppose the separable metric spaceX is the countable union ofF sets of dimension n. Then σ ≤ ind(X) n. ≤ Suppose the theorem holds for somen 1 1. We shallﬁrst prove − ≥− that this implies the following: Corollary 3.23. Any separable metric spaceY with dimension ind(Y) n ≤ is the union of a subspace of dimension n 1 and a subspace of dimension ≤ − 0. ≤ Proof. LetY be a separable metric space of dimension n. Let B i :i N ≤ { ∈ } be a basis for the topology onY , consisting of sets such that the boundaries have ind(∂Bi) n 1 for alli N. The existence of such a basis is ≤ − ∈ 32 Chapter 3. Small Inductive Dimension

guaranteed by Propositions 2.32 and 3.14. Since the boundaries∂B i are closed sets by Proposition 2.11, and are thusF σ sets, and since Theorem 3.22 was assumed to hold forn 1, we get that −

L := ∂Bi i N �∈ has dimension n 1. Since the boundaries∂B do not meetY L, the ≤ − i \ condition of Proposition 3.18 (withn = 0 andX � =Y L) is satisﬁed. Thus \ ind(Y L) 0. The claim follows from the equationY=L (Y L). \ ≤ ∪ \ Proof of Theorem 3.22. The casen= 1 is clear. Suppose the theorem − holds for somen 1 1. Suppose that − ≥−

X= Ci, i N �∈ ind(C ) n, i ≤ andC i is aF σ set for alli N. We want to show that ind(X) n. Let ∈ ≤

K1 :=C 1, and let

i 1 i 1 − − K :=C C =C X C , i=2,3,4.... i i \ j i ∩  \ j j�=1 j�=1   Now we have X= Ki, (8) i N �∈ and K K = , ifi= j. i ∩ j ∅ � Since i 1 − ∞ X C = C \ j j j�=1 j�=i is anF σ set,K i is also anF σ set inX as the intersection of twoF σ sets. As Ki is a subset ofC i, Proposition 3.16 implies

ind(K ) n. (9) i ≤

Because of (9), we can apply Corollary 3.23 to eachK i: we have

K =M N , i i ∪ i Brouwer’s Theorem on the Invariance of Domain 33 such that ind(M ) n 1, ind(N ) 0. i ≤ − i ≤ PutM := M andN := N . From (8) we get that ∪ i ∪ i X=M N. ∪

EachM i is anF σ inM. For

M =M K =M K , i i ∩ i ∩ i sinceM K , andK K = , wheneveri=j. ThusM is anF set i ⊂ i i ∩ j ∅ � i σ inM as the intersection of the setK i, which isF σ inX, and the subspace M. Therefore, since Theorem 3.22 was assumed to hold forn 1, we can − conclude that ind(M) n 1. By a similar argument, everyN is anF set ≤ − i σ inN and thus ind(N) 0 by Corollary 3.11. ≤ We thus haveX=M N with ind(M) n 1 and ind(N) 0. ∪ ≤ − ≤ From Proposition 3.19 we conclude, that ind(X) n, which completes the ≤ proof.

Proposition 3.24. IfC 1 andC 2 are disjoint closed subsets of a separable metric spaceX, andA X is a subset of dimension n, wheren 0, ⊂ ≤ ≥ there exists a closed separatorB X between the setsC andC such that ⊂ 1 2 ind(A B) n 1. ∩ ≤ − Proof. Ifn = 0, then we have two options. Theﬁrst option is ind(A) = 1. − Now the proposition holds clearly, sinceX isT 4, and thus there exists a neighbourhoodV ofC 1 disjoint from some neighbourhoodU ofC 2 and we can choose the boundary∂V as the separator. The second option is ind(A) = 0, which has already been demonstrated in Proposition 3.9. Suppose then thatn> 0. Applying Corollary 3.23 we haveA=D E, ∪ with ind(D) n 1 and ind(E) 0. Let us use Proposition 3.9 to obtain a ≤ − ≤ closed separatorB between the setsC andC such thatB E= . Thus 1 2 ∩ ∅ A B D. ∩ ⊂ Since ind(D) n 1, Proposition 3.16 implies ind(A B) n 1, which ≤ − ∩ ≤ − completes the proof.

Corollary 3.25. If a separable metric spaceX has ind(X) n, then there ≤ exists a closed separator with dimension n 1 between any two disjoint ≤ − closed subsets of the spaceX.

Proof. This follows from Proposition 3.24 by choosingA=X. 34 Chapter 3. Small Inductive Dimension

Proposition 3.26. LetX be a separable metric space with dimension ≤ n 1, and letC ,C �,i=1, . . . , n ben pairs of closed subsets ofX such that − i i

C C � = . i ∩ i ∅

Then there existn closed setsB i that act as separators between the setsC i andC i�, and n B = . i ∅ i�=1

Proof. From Corollary 3.25 we get a closed separatorB 1 between the setsC 1 andC � with ind(B ) n 2. By Proposition 3.24 we get a closed separator 1 1 ≤ − B2 betweenC 2 andC 2� with

ind(B B ) n 3. 1 ∩ 2 ≤ − By repeatedly applying Proposition 3.24 we arrive atk sets of closed sepa-

ratorsB i between the setsC i andC i� with

k ind Bi n k 1. � � ≤ − − i�=1 Fork=n we conclude that n B = . i=1 i ∅ � 4 Brouwer Fixed-Point Theorem

The focus of this chapter is on a very famous result in topology: the Brouwer Fixed-Point Theorem. This theorem allows us to determine the exact dimension of a Euclidean space. We shall give a combinatorial proof of this theorem, which relies on Sperner’s lemma. The proof also relies somewhat on the theory of simplexes, the rudiments of which are presented in this chapter. We shall begin by stating a familiar result, which we will need in some proofs.

Proposition 4.1. SupposeX is a compact topological space, andY is a Hausdorﬀ space. Any continuous bijectionf:X Y is a homeomorphism. → Proof. See Väisälä [6, Lause 13.25 and Seuraus 13.26, p. 102].

Some Theory of Simplexes

Deﬁnition 4.2(m-simplex, face, vertex, barycentric coordinates). Suppose n a0, a1 . . . , am arem + 1 linearly independent points of the spaceR . The subset ofR n consisting of all the points of the form

m x= λiai, (1) �i=0 where m λ = 1, andλ 0 for alli 0,1, . . . , m (2) i i ≥ ∈{ } �i=0 is called anm-simplex spanned by the pointsa 0, a1, . . . , am and is denoted bya 0a1 . . . am. It is evident that anm-simplex depends only on the points that span it, not on the order of these points.

35 36 Chapter 4. Brouwer Fixed-Point Theorem

n Consider a simplexa 0a1 . . . am R . For anyk + 1 distinct non-negative ⊂ integersj 0, j1, . . . , jk not larger thanm the pointsa j0 , aj1 , . . . , ajk are linearly independent, so that thek-simplexa j0 aj1 . . . ajk is well deﬁned. Every simplex of the aforementioned form is called ak-face of the simplexa 0a1 . . . am. The 0-faces – that is the pointsa 0,a 1 and so forth – ofa 0a1 . . . am are called vertices ofa 0a1 . . . am. The whole simplexa 0a1 . . . am is also considered to be one of its faces.

Thek-facea j0 aj1 . . . ajk consists of all the points of the form (1) satisfying (2) such that λ = 0 wheneveri j , j , . . . , j . i �∈{ 0 1 k} Since the vertices are linearly independent, every pointx a a . . . a can ∈ 0 1 m be represented uniquely in the form (1) under the conditions (2). The coeﬃ- cientsλ 0,λ 1,...,λ m in (1) are called the barycentric coordinates of the point x. The barycentric coordinates ofx are denoted byλ 0(x),λ 1(x),...,λ m(x).

Proposition 4.3. For anym+1 of linearly independent pointsa 0, a1, . . . , am n n ofR , the simplexΔ :=a 0a1 . . . am is a compact subspace ofR and the barycentric coordinatesλ 0,λ 1,...,λ m are continuous functions fromΔ to [0, 1].

m+1 th Proof. Let us denote bye i R , the unit vector whosei coordinate ∈ is 1, and the rest of the coordinates are 0. The pointse 1, e2, . . . , em+1 are m+1 linearly independent, so that the simplexS :=e 1e2 . . . em+1 R is well ⊂ deﬁned. The barycentric coordinates of the points inS coincide with their coordinates inR m+1 and by (2)S is bounded.S is closed as the intersection m+1 k+1 of two closed sets, namely the hyperplane x R : i=1 xi = 1 and the m+1 { ∈ } subset ofR consisting of the points whose every coordinate� is at least 0. ThereforeS is compact as a closed and bounded subset ofR m+1 by Väisälä [6, Lause 13.14, p. 99]. The functionf:S Δ deﬁned by the rule → f(x) = pr (x) a + pr (x) a + + pr (x) a 1 · 0 2 · 1 ··· m+1 · m is continuous, since the projection maps are continuous by Väisälä [6, Lause 5.6, p. 43]. The functionf is injective, since the pointsa 0, a1, . . . , am are linearly independent. Sincef(e i) =a i 1 fori=1,2, . . . , m + 1, we have − fS=Δ. In other wordsf is surjective. Thusf is a homeomorphism by Proposition 4.1. HenceΔ is a compact subspace ofR n and the barycentric coordinatesλ i, fori=0,1, . . . , m are continuous as the composition of two 1 continuous functions, sinceλ (x) = (pr f − )(x). i i+1 ◦ Corollary 4.4. Any twom-simplexes are homeomorphic. Brouwer’s Theorem on the Invariance of Domain 37

Proof. From the proof of Proposition 4.3 we get for any twom-simplexesΔ 1 andΔ , thatΔ S Δ , from which the claim follows. 2 1 ≈ ≈ 2 Deﬁnition 4.5 (Simplical subdivision, mesh).A simplical subdivision of a n k n simplexΔ R is a familyS := Δ i i of simplexes inR satisfying the ⊂ { } =1 following three conditions The familyS coversΔ, that is,Δ= k Δ . • i=1 i For any i, j k he intersectionΔ �Δ is either empty or a face of • ≤ i ∩ j bothΔ i andΔ j. Fori=1,2, . . . , k all faces ofΔ are members ofS. • i The mesh of a simplical subdivision Δ k of a simplexΔ is the largest of { i}i=1 the numbersd(Δ 1), d(Δ2), . . . , d(Δk). Deﬁnition 4.6 (Barycenter). The barycenter of a simplexΔ :=a a . . . a 0 1 m ⊂ Rn is the point 1 1 1 b(Δ) = a + a + + a . m+1 0 m+1 1 ··· m+1 m Clearlyb(Δ) Δ andb(Δ) does not belong to anyk-face, wherek

n Proposition 4.7. LetΔ :=a 0a1 . . . am R be a simplex. For every de- ⊂ creasing sequenceΔ Δ Δ of distinct faces of the simplexΔ, 0 ⊃ 1 ⊃···⊃ k the pointsb(Δ 0), b(Δ1), . . . , b(Δk) are linearly independent. The familyS of all simplexes of the formb(Δ 0)b(Δ1)...b(Δ k) is a simplical subdivision of the simplexΔ. Every(m 1)-simplexS S is a face of one or twom- − ∈ simplexes ofS, depending on whether or not the simplexS is contained in an (m 1)-face ofΔ. − Proof. Every decreasing sequence of distinct faces ofΔ can be completed to a sequenceΔ Δ Δ consisting ofm + 1 faces ofΔ such that 0 ⊃ 1 ⊃···⊃ m

Δ0 =a i0 ai1 . . . aim ,Δ 1 =a i1 ai2 . . . aim ,...,Δ m =a im , (3) wherei 0, i1, . . . , im is a permutation of 0,1, . . . , m. To prove theﬁrst part of the proposition, it suﬃces to show that the pointsb(Δ 0), b(Δ1), . . . , b(Δm) are linearly independent. Consider a linear combination m µjb(Δj). (4) j�=0 38 Chapter 4. Brouwer Fixed-Point Theorem

Using the deﬁnition of the barycenter, we can represent (4) as a linear combination of the pointsa i0 , ai1 , . . . , aim . This linear combination is of the form

λij aij , (5) j�=0 where j 1 λ = µ (6) ij m+1 k k k�=0 − and m m j 1 m m 1 m λ = µ = µ = µ . (7) ij m+1 k k m+1 i i i j�=0 j�=0 k�=0 − �i=0 �j=i − �i=0

Suppose the linear combination (4) is equal to 0. As the pointsa i0 , ai1 , . . . , aim are linearly independent, from (5) we get thatλ = 0 for allj 0,1, . . . , m . ij ∈{ } Now from (6) it follows thatµ = 0 for allk 0,1, . . . , m . Thus the points k ∈{ } b(Δ0), b(Δ1), . . . , b(Δm) are linearly independent and the simplex spanned by the barycenters is well-deﬁned. Every simplex inS is a face of anm-simplex of the formb(Δ 0)b(Δ1)...b(Δ m). From (4)–(7) it follows that the simplexb(Δ 0)b(Δ1)...b(Δ m) is a subset of Δ, when the coeﬃcientsµ j in (4) are chosen in such a way that they satisfy the conditions in (2). We shall show thatb(Δ 0)b(Δ1)...b(Δ m) coincides with the set

F := x Δ:λ (x) λ (x) λ (x) . (8) { ∈ i0 ≤ i1 ≤···≤ im } By (6), it suﬃces to show that every pointx F can be represented in the ∈ form (4) with m µ = 1 andµ 0 forj 0,1, . . . , m . We get such a j=0 j j ≥ ∈{ } representation� by choosing

µ0(x) = (m + 1)λi0 (x) andµ j(x) = (m+1 j)(λ ij (x) λ ij 1 (x)) (9) − − − m m forj=1,2, . . . , m, since j=0 µj(x) = j=0 λij (x) = 1 and clearly every µj(x) is non-negative. � � From (9) it follows that the faces of the simplexb(Δ 0)b(Δ1)...b(Δ m) can be described by adding to condition (8) a number of conditions of the form

λij (x) =λ ij 1 (x), where 1 j m and possibly the conditionλ i0 (x) = 0. As − ≤ ≤ the intersection of a face determined by such conditions with a face of another simplexb(Δ 0� )b(Δ1� )...b(Δ m� ) corresponging to a different permutation of 0,1, . . . , m still satisfies the aforementioned conditions, or is empty, the familyS satifies the second condition of Definition 4.5. Thefirst condition of Definition 4.5 is also satisfied, since any point inΔ belongs to a set of the Brouwer’s Theorem on the Invariance of Domain 39

form (8). By the deﬁnition ofS, every face of a simplex inS belongs toS, so that the third condition of Deﬁnition 4.5 holds. ThereforeS is a simplical subdivision ofΔ. LetS :=b(Δ 0)b(Δ1)...b(Δ m 1) be an (m 1)-simplex inS. If the simplex − − S is contained in an (m 1)-face ofΔ, in other words, ifΔ =Δ, thenS is a − 0 � face of exactly onem-simplex ofS, namelyb(Δ)b(Δ 0)b(Δ1)...b(Δ m 1). If, − on the other handS is not contained in any (m 1)-face ofΔ, i.e.Δ =Δ, − 0 then eitherΔ m 1 is a 2-simplex, or there exists aj

m m m d(x, y)= x y = λ a λ y = λ (a y) � j j j � � j j � � − � �j=0 − j=0 � �j=0 − � �� m � � m � � � � � � λj aj y max� �aj y λj � ≤ � − �≤ j m � − � j�=0 ≤ j�=0 = max aj y = max d(aj, y). j m j m ≤ � − � ≤

n Lemma 4.10. The diameter of a simplexa 0a1 . . . am R is equal to the ⊂ diameter of the set a , a , . . . , a . { 0 1 m} Proof. Let x, y a a . . . a . From Lemma 4.9 we get that ∈ 0 1 m d(x, y) max d(aj, y). j m ≤ ≤ 40 Chapter 4. Brouwer Fixed-Point Theorem

Applying Lemma 4.9 again, we get

d(aj, y) max d(aj, ai). i m ≤ ≤

Thusd(x, y) max i,j m d(ai, aj), which proves the claim. ≤ ≤ Lemma 4.11. The mesh of the barycentric subdivision of them-simplex Δ :=a 0a1 . . . am is at most(m/(m + 1))d(Δ).

Proof. By virtue of Lemma 4.10 it suﬃces to show that the distance between any two points of the form

1 1 b(Δ ) = (a +a + +a ) andb(Δ ) = (a +a + +a ), j j+1 i0 i1 ··· ij k k+1 i0 i1 ··· ik

wherek

d(b(Δ ), b(Δ )) d(a , b(Δ )) k j ≤ il j for some 0 l m. Thus ≤ ≤ d(b(Δ ), b(Δ )) d(a , b(Δ )) k j ≤ il j = b(Δ ) a � j − il � 1 = (ai +a i + +a ij ) a i �j+1 0 1 ··· − l � � � � � � 1 j � = � (a a ) � � ih il � j+1 �h=0 − � �� 1 �j � � a a � ≤ j+1 � ih − il � h�=0 j d(Δ) ≤ j+1 m d(Δ). ≤ m+1

Corollary 4.12. For every simplexΔ and for everyε>0 there exists a natural numberl such that the mesh of thel th barycentric subdivision is less thanε. Brouwer’s Theorem on the Invariance of Domain 41

Proof. By Lemma 4.11 the mesh of the second barycentric subdivision of the m-simplex is at most m 2 d(Δ) �m+1 � and similarly the mesh of thel th barycentric subdivision is at most

m l d(Δ). �m+1 � Ifε> 0, by choosingl large enough, we get that

m l d(Δ)<ε. �m+1 �

Proposition 4.13 (Sperner’s lemma). LetS be thel th barycentric subdivision of anm-simplexa 0a1 . . . am and letV be the set of all vertices of simplexes inS. If a functionh:V 0,1, . . . , m satisﬁes the condition →{ } h(v) i , i , . . . , i wheneverv a a . . . a , ∈{ 0 1 k} ∈ i0 i1 ik then the number of simplexes inS, on the vertices of whichh assumes all values from0 tom, is odd.

Proof. We apply induction with respect tom. Sperner’s lemma holds for m = 0, sinceS= a , andh(a ) = 0. Suppose the lemma holds for some { 0} 0 m=n 1. Consider ann-simplexΔ :=a a . . . a , thel th barycentric − 0 1 n subdivisionS ofΔ and a functionh satisfying the condition in Sperner’s lemma. We shall consider the number of (n 1)-simplexes inS, the vertices of − which map onto 0,1, . . . , n 1 underh. Denote the set consisting of these { − } (n 1)-simplexes byS. − Letr be the number ofn-simplexes inS, whose vertices receive all the values from 0 ton underh. Each of these simplexes contributes exactly one (n 1)-face to the setS. Letq denote the number ofn-simplexes inS, the − vertices of which map onto 0,1, . . . , n 1 underh. That is to say, two { − } vertices map to the same value underh. These simplexes contribute two (n 1)-faces each to the setS. We arrive at theﬁgurer+2q. Note that this − number is greater than #S, since we have counted some of the (n 1)-faces − twice; some of these faces are faces of twon-simplexes. Letx denote the number of elements ofS, which lie on an (n 1)-face of − Δ. It follows from the deﬁnition of the functionh, that the only face ofΔ on 42 Chapter 4. Brouwer Fixed-Point Theorem

which the elements ofS can lie, is the facea 0a1 . . . an 1. By the induction − hypothesis,x is an odd number. Lety be the number of the rest of the elements ofS, i.e.y := #S x. − From the last part of Proposition 4.7, it follows that every (n 1)-simplex − is a face of one or twon-simplexes: one if the simplex is contained in an (n 1)-face ofΔ and two if this is not the case. Thus in theﬁgurer+2q we − have counted the simplexes that contribute to the numbery twice, and the simplexes that contribute to the numberx only once. Thus, counting in two diﬀerent ways, we arrive at the equation

r+2q=x+2y, whence it follows thatr is an odd number, sincex is odd. This completes the proof.

The next result wasﬁrst proved by Knaster, Kuratowski and Mazurkiewicz. This is why it is known as the KKM lemma.

Proposition 4.14 (The KKM lemma). Let F m be a family of closed { i}i=0 subsets of a simplexΔ :=a 0a1 . . . am. If for every facea i0 ai1 . . . aik ofΔ it holds that k a a . . . a F , i0 i1 ik ⊂ ir r�=0 then the intersection m Fi i�=0 is non-empty.

Proof. Suppose, for a contradiction, that m F = . The collection U k , i=0 i ∅ { i}i=0 whereU :=Δ F is an open covering ofΔ. SinceΔ is compact, by i \ i � Proposition 2.33 there exists anε> 0 such that every subset ofΔ, whose diameter is less thanε is contained in a setU i, i.e. disjoint fromF i. By Corollary 4.12 we can choose the natural numberl to be such, that the mesh of thel th barycentric subdivisionS ofΔ is less thanε. Denote byV the set of vertices of simplexes inS. We construct a functionh:V → 0,1, . . . , m as follows. For everyv V the intersection of all faces ofΔ { } ∈ that containv is a face ofΔ. In other words, this intersection is of the form a a . . . a . Sincev a a . . . a , by the assumption of the proposition i0 i1 ik ∈ i0 i1 ik there exists a numberj k such thatv F . We seth(v) =i . By ≤ ∈ ij j constructionh satisﬁes the assumptions of Sperner’s lemma, so that there exists at least one simplexS :=v v . . . v S such thath(v ) =i, for 0 1 m ∈ i Brouwer’s Theorem on the Invariance of Domain 43

i=0,1, . . . , m. Thus fori=0,1, . . . , m we havev F , which in turn i ∈ i implies thatS F = fori=0,1, . . . , m. This is a contradiction, since the ∩ i � ∅ diameter ofS is less thanε.

Lemma 4.15. Every compact convex subsetA R n, with intA= is ⊂ � ∅ homeomorphic to the closed unit ball Bn, and the boundary∂A is a homeo- n 1 morph of the unit sphereS − .

Proof. Without loss of generality, we may assume that 0 intA. Every ray ∈ that emanates from the origin intersects with the boundary∂A exactly once: should some ray intersect the boundary more than once,A would not be convex, and should the intersection be empty,A would not be bounded and thus would not be compact. Hence every point ofA 0 lies on one, and \{ } only one ray emanating from the origin. n 1 The mapf:∂A S − , defined with the rule → x f(x) = x � � is a continuous bijection from a compact space to a Hausdorff space and is n 1 thus a homeomorphism by Proposition 4.1. Hencef:∂A S − . ≈ Letg: Bn A be a function defined by the rule → 1 x x f − x , forx= 0 g(x) = � � � � � .  � � 0, forx= 0 The functiong is continuous except maybe at the origin. SinceA is compact the norms of the elements ofA are bounded. PutM := sup x :x A . {� � ∈ } Now for allx Bn we have g(x) M x . To prove thatg is continuous ∈ � � ≤ � � at the origin, letε> 0. Now g(x) <ε, whenever x <δ=ε/M. Thus � � � � g is a continuous function. Sinceg is bijective, it is a homeomorphism by Proposition 4.1. This proves the claim.

Theorem 4.16 (Brouwer Fixed-Point Theorem). Iff: Bn Bn is a con- → tinuous function, there exists a pointx Bn such thatf(x) =x. ∈ Proof. By virtue of Lemma 4.15 we can replace the ball Bn by ann-simplex Δ :=a 0a1 . . . an, with a non-empty interior and investigate the behaviour of a continuous functionf:Δ Δ. → For every pointx Δ, we have ∈ x=λ (x)a +λ (x)a + +λ (x)a , (10) 0 0 1 1 ··· n n 44 Chapter 4. Brouwer Fixed-Point Theorem where n λi(x) = 1 (11) �i=0 andλ (x) 0 fori=0,1, . . . , n. i ≥ The image of the pointx Δ under the functionf can be written as ∈ f(x) =λ (f(x))a +λ (f(x))a + +λ (f(x))a , (12) 0 0 1 1 ··· n n where n λi(f(x)) = 1 (13) �i=0 andλ (f(x)) 0 fori=0,1, . . . , n. i ≥ Fori=0,1, . . . , n the set

F := x Δ:λ (f(x)) λ (x) (14) i { ∈ i ≤ i } is closed as the preimage of a closed set under a continuous function. We shall next show, that the collection F n satisﬁes the assumptions of Proposition { i}i=0 4.14. Leta a . . . a be a face ofΔ. Consider a pointx a a . . . a . We i0 i1 ik ∈ i0 i1 ik have k

λij (x) = 1, j�=0 so that by (13) we get

k k λ (f(x)) λ (x). ij ≤ ij j�=0 j�=0 Henceλ (f(x)) λ (x) for at least onej k, whence it follows that ij ≤ ij ≤ x F . Hence we have shown, that ∈ ij k a a . . . a F . i0 i1 ik ⊂ ij j�=0 Now, by Proposition 4.14 there exists a pointx n F . From (14) it fol- ∈ i=0 i lows thatλ (f(x)) λ (x) fori=0,1, . . . , n. However, the strict inequality i ≤ i � cannot hold for anyi because of (11) and (13). Therefore,λ i(f(x)) =λ i(x) fori=0,1, . . . , n, whence it follows thatf(x) =x, because of (10) and (12). This proves the claim.

n n 1 Corollary 4.17. There exists no continuous functionr: B S − which n 1 → keeps each point of the boundaryS − ﬁxed. Brouwer’s Theorem on the Invariance of Domain 45

n n 1 Proof. Suppose a continuous functionr: B S − which keeps the points n 1 →n on the boundaryﬁxed exists. Letf:S − B be the function which maps n 1 → each pointx S − to its antipode, i.e.f(x) = x. Now the composition ∈ − f r: Bn Bn is continuous, and contains noﬁxed-point, which contradicts ◦ → the Brouwer Fixed-Point Theorem.

n n n + Proposition 4.18. Consider the cubeI := [ 1, 1] R . LetF i be the − ⊂ th face of the cube determined by the equationx i = 1, wherex i denotes thei n coordinate of a point ofI , and letF i− be the opposite face, i.e.x i = 1. + − LetC i be a closed separator betweenF i andF i−. Then n C = . i � ∅ i�=1 + Proof. SupposeC i is a closed separator betweenF i andF i−. Hence there + n exists setsU ,U − I such that i i ⊂ n + I C =U U − \ i i ∪ i + + F U ,F − U − i ⊂ i i ⊂ i and + U U − = , i ∩ i ∅ + n n n with bothU andU − open inI C and thus open inI , sinceI C is i i \ i \ i open inI n. For each pointx I n letv be the point whosei th component ∈ x has the value d(x, C ), ± i + where the sign is positive, ifx U −, and negative, ifx U . We deﬁne a ∈ i ∈ i functionf:I n I n with the rule →

f(x) =x+v x. This function is well-deﬁned, since the way the sign of the coordinates of n the vectorv x is determined, the image ofx underf is inI . The function f is continuous, since the distance function is continuous. We can thus apply the Brouwer Fixed-Point Theorem (Theorem 4.16), for Bn andI n are homeomorphic. Thus there exists a pointx 0 such that

f(x 0) =x 0. This means thatd(x ,C ) = 0 for alli 1,2, . . . , n , and since eachC 0 i ∈{ } i is closed, we havex C . Thus the intersection of the setsC , where 0 ∈ i i i 1,2, . . . , n , is non-empty. This completes the proof. ∈{ } 46 Chapter 4. Brouwer Fixed-Point Theorem

Some Properties ofR n

We have not yet demonstrated that sets of dimension> 0 even exist. We shallﬁx this shortcoming in this section. We shall prove a very important result, namely that the dimension of the Euclidean spaceR n is exactlyn. Our theory of dimension would violate intuition quite seriously, should this not be the case.

Theorem 4.19. The real lineR has ind(R) = 1.

Proof. SinceR is connected, no real number has arbitrarily small neighbourhoods, whose boundaries are empty. Hence we get ind(R) 1. ≥ Supposex R and letU be a neighbourhood ofx. NowU contains ∈ some open interval ]α,β[ x. The topology on the subspace∂]α,β[ = � α,β is the subspace topology, which is discrete. Now the open and closed { } neighbourhood α of the pointα is contained in every neighbourhood ofα. { } The analogous is true for the other pointβ. Thus ind(∂]α,β[) = 0. Now it follows from (SID2) of Deﬁnition 3.13 that ind(R) 1, which proves the ≤ claim.

Corollary 4.20. Any non-empty intervalJ R has ind(R) = 1. ⊂ Proof. The intervalJ is connected, thus no element of the interval has arbitrarily small neighbourhoods with empty boundaries. Hence ind(J) 1. As ≥ J R, Proposition 3.16 and Theorem 4.19 give us ind(J) 1. ⊂ ≤ By a similar argument we can show that the sphereS 1 also has ind(S1) = 1. Every point in the sphere has arbitrarily small neighbourhoods, whose boundaries are two-point sets. Hence ind(S1) 1. SinceS 1 is connected, no ≤ point inS 1 has arbitrarily small open and closed neighbourhoods, whence it follows that ind(S1)> 0.

Proposition 4.21. The spacesR n andS n are at mostn-dimensional. Proof. The base casen = 1 follows from Theorem 4.19 and the discussion below it. Suppose the proposition holds for spaces whose dimensions do not exceed somen 1. For every pointx in the Euclidean spaceR n+1, or the ≥ sphereS n+1 and for each neighbourhoodV of the pointx there exists a neighbourhoodU V ofx, whose boundary is homeomorphic toS n. Hence, ⊂ by Theorem 3.15 and the induction hypothesis, we have ind(∂U) n, which ≤ implies ind(Rn+1) n + 1, and ind(S n+1) n + 1. ≤ ≤ Corollary 4.22. Then-cubeI n := [ 1, 1]n has ind(In) n. − ≤ Brouwer’s Theorem on the Invariance of Domain 47

Proof. SinceI n R n, the claim follows from Propositions 3.16 and 4.21. ⊂ Proposition 4.23. ind(I n) n. ≥ Proof. Suppose, that ind(In) n 1. Then by Proposition 3.26 there exists ≤ − n closed subsetsB I n, fori=1,2, . . . , n, with eachB being a sepa- i ⊂ i rator between two opposite faces ofI n, and n B = . This contradicts i=1 i ∅ Proposition 4.18. � Corollary 4.24. ind(I n) =n.

Proof. This follows from Corollary 4.22 and Proposition 4.23.

Proposition 4.25. ind(Rn) n. ≥ Proof. Since by Proposition 4.23 ind(I n) n, and sinceI n R n, the claim ≥ ⊂ follows from Proposition 3.16.

Corollary 4.26. The sphereS n R n+1 has ind(Sn) =n. ⊂ Proof. Any pointx R n+1 has arbitrarily small neighbourhoods, whose ∈ boundaries are homeomorphic toS n. By Proposition 4.25 we have ind(Rn+1) ≥ n + 1, whence it follows that ind(Sn) n, which together with Proposition ≥ 4.21 proves the claim.

n 1 The above corollary explains the notationS − for the boundary of a ball Bn. The dimension of the boundary is strictly smaller than the dimension of the ball itself. Now we areﬁnally ready to determine the exact dimension of a Euclidean space.

Theorem 4.27. The Euclidean spaceR n has dimensionn. Proof. This follows from Propositions 4.21 and 4.25.

Theorem 4.28. The spacesR n andR m are homeomorphic, if and only if n=m.

Proof. It is evident, thatR n R m, ifn=m. The converse follows from ≈ Theorems 3.15 and 4.27.

5 Invariance of Domain

We are almost suﬃciently prepared to prove Brouwer’s Theorem on the Invariance of Domain. The proof of this theorem, which we shall present shortly, relies on results concerning the extensions of continuous functions. Thus most of this chapter is devoted to developing this theory. Having these new tools at hand, we conclude this chapter, and hence the whole thesis, by proving the Invariance of Domain.

Deﬁnition 5.1 (Stable and unstable values). Suppose (X, dX ) and (Y, dY ) are metric spaces andf:X Y a continuous function. A pointy fX is → ∈ called an unstable value of the functionf, if for everyε> 0 there exists a continuous functiong:X Y satisfying → d (f(x), g(x))<ε, for everyx X, Y ∈ gX Y y . ⊂ \{ } If a point offX is not an unstable value off, it is a stable value.

Example 5.2. SupposeX is a topological space andf:X I n a continuous → function. Now every point on the boundary ofI n is unstable. For given any 0<ε< 1, the functions

g (x) = (1 ε)f (x), i=1,2, . . . , n i − i deﬁne a continuous function whose image does not contain any boundary points ofI n.

Proposition 5.3. Suppose(X, d) is a separable metric space with ind(X)< n, and thatf:X I n is a continuous map. Now all values off are unstable. → 49 50 Chapter 5. Invariance of Domain

Proof. Example 5.2 shows that it suﬃces to show, that there are no stable interior points ofI n. To prove this, it is enough to show that the origin is not a stable value off. We consider the coordinate functionsf i, where + i=1,2, . . . , n, off. Suppose 0<ε< 1. LetC i be the subset ofX, for the points of which it holds that f (x) ε, i ≥ and letC i− be the set of points ofX for which f (x) ε. i ≤− + For eachi, the setsC i andC i− are closed and disjoint. Hence by Propo- sition 3.26 there exist closed setsB 1,B2,...,Bn such thatB i is a separator + between the setsC i andC i−. In other words, there exist disjoint open sets + + U C andU − C − such that i ⊃ i i ⊃ i + X B =U U −, \ i i ∪ i and n B = . (1) i ∅ i�=1 We shall deﬁne functionsg , g , . . . , g :X [ 1, 1] with the following rules: 1 2 n → −

+ g (x) =f (x), ifx C C − i i ∈ i ∪ i d(x, Bi) + + gi(x) =ε + , ifx cl X (Ui C i ) d(x, Ci ) +d(x, B i) ∈ \ d(x, Bi) gi(x) = ε , ifx cl X (Ui− C i−) − d(x, Ci−) +d(x, B i) ∈ \ g (x) = 0, ifx B . i ∈ i

We shall show that the functionsg i are continuous. It is clear that the subfunctions that formg i are continuous. Since all of the domains of the subfunctions are closed inX, if the subfunctions whose domains intersect attain the same values in this intersection, then the whole piecewise deﬁned functiong i is continuous. + + + + In the intersection (C C −) cl (U C ) =∂C , we havef (x) =ε, i ∪ i ∩ X i \ i i i and

d(x, Bi) d(x, Bi) ε + =ε =ε. d(x, Ci ) +d(x, B i) 0 +d(x, B i) + Similarly in (C C −) cl (U − C −) =∂C −, we havef (x) = ε, and i ∪ i ∩ X i \ i i i − Brouwer’s Theorem on the Invariance of Domain 51

d(x, B ) d(x, B ) ε i = ε i = ε. − d(x, Ci−) +d(x, B i) − 0 +d(x, B i) −

The subfunctions, whose domain has a non-empty intersection withB i vanish in this intersection, as can be seen from the deﬁnitions of the sub- + + functions. If the sets cl (U C ) and cl (U − C −) meet, they meet inB , X i \ i X i \ i i and thus the relevant subfunctions attain the value zero in this intersection. We have checked all non-empty intersections. Thus the functionsg i are continuous, and we have

g (x) f (x) 2ε (2) | i − i |≤ for allx X. By (1) there is no point inX such that all the functionsg ∈ i vanish simultaneously, sinceg (x) = 0 only whenx B . Hence the origin is i ∈ i not an image point of the continuous functiong := (g1, g2, . . . , gn), and with (2) this shows that the origin is not a stable value off.

Proposition 5.4. SupposeX is a separable metric space and supposef:X → In is continuous. If there exists a pointy I n such that ∈ fX I n y , ⊂ \{ } then for everyε>0 there exists a continuous functiong:X I n such that → f(x) g(x) <ε for everyx X � − � ∈ gX I n y . ⊂ \{ } Proof. If the pointy lies on the boundary ofI n, the function defined in Example 5.2 satisfies the desired conditions. Suppose thaty is an interior n point ofI . Letf �(x) be the projection of the pointf(x) fromy on the n boundary∂I . If the length of the segment joiningf(x) andf �(x) is greater than, or equal toε/2, we define the pointg(x) to be the point on this segment at distanceε/2 fromf(x). If the length of this line segment is less thanε/2, n we putg(x) =f �(x). The functiong:X I is as wanted. → Proposition 5.5. Suppose(Y, d) is a compact metric space and letX be any space. Then the space(C(X,Y), ) ∗ is complete. �·� ∞ ∗Here we mean the space consisting of all the continuous functions fromX toY with the sup-metric which is naturally defined as f g := supx X d(f(x), g(x)) . � − � ∞ ∈ { } 52 Chapter 5. Invariance of Domain

Proof. Let (f ) be a Cauchy sequence inC(X,Y ). Now for eachx X the n ∈ sequence (fn(x)) is Cauchy inY , sinced(f m(x), fn(x)) f n f m . Since ≤� − �∞ Y is a compact metric space and therefore complete by Väisälä [6, Lause 13.28, p. 102], the sequence (fn(x)) converges to a pointf(x). The sequence (f ) converges uniformly tof, and thusf C(X,Y ) by Väisälä [7, Lause n ∈ 10.13, p. 81]. Proposition 5.5 allows us to apply the Baire Category Theorem to the space (C(X,Y), ), whereY is a compact metric space. To refresh our �·� ∞ memory, we shall state the theorem here. Proposition 5.6 (Baire Category Theorem). Suppose(X, d) is a complete metric space, and suppose(G j)j N is a sequence of open dense subsets ofX. ∈ Then the intersection i N Gi is dense inX. ∈ Proof. Väisälä [7, Lause� 10.8, p. 78]

Corollary 5.7. The countable intersection of denseG δ sets in a complete metric space is a denseG δ set.

Proof. Suppose (X, d) is a complete metric space, and suppose (Gj)j N is a ∈ sequence of denseG δ subsets ofX. EachG j is a countable intersection of dense open sets, and since the countable intersection of countable intersections is again a countable intersection, by the Baire Category Theorem the intersection i N Gi is a denseG δ subset ofX. ∈ Proposition� 5.8. SupposeX is a separable metric space. ThenX can be embedded in the Hilbert cubeI ω. Moreover the set of embeddings ofX inI ω contain a dense∗ Gδ set in the function space(C(X,I ω), ). �·� ∞ Before we prove Proposition 5.8, we shallfirst define some new concepts and prove some auxiliary results, of which we will make use. Definition 5.9(A-map). SupposeX andY are topological spaces, and thatA is afinite open covering ofX. We say that the continuous function g:X Y is anA-map, if every point ofY has a neighbourhood inY such → that the preimage of this neighbourhood underg is entirely contained in some member ofA. Definition 5.10 (Basic sequence of coverings). LetAbeafinite open covering of the topological spaceX. Denote byS A(x) the open set which is the union of all the members ofA that contain the pointx X. A countable ∈ ∗In other words, making arbitrarily small modifications to a continuous function suffice to make it an embedding. Brouwer’s Theorem on the Invariance of Domain 53

collectionA 1,A 2,... ofﬁnite open coverings is called a basic sequence of coverings, if given a pointx X and a neighbourhoodU ofx, at least one ∈ of the open setsS A1 (x),S A2 (x),... is contained inU. Proposition 5.11. For every separable metric spaceX there exists a basic sequence of coverings.

Proof. Let U ∞ be any countable basis ofX. We consider pairsU ,U { i}i=1 m n of members of this basis for which it holds that

U U . n ⊂ m

Denote byA n,m the covering ofX that consists of two members, namely the setsX U andU . The collection consisting of these coveringsA is \ n m n,m countable. Moreover,x U impliesS A (x) =U . Thus the collection ∈ n n,m m of sets S A (x) for a given pointx includes the collection of all theU { n,m } m containingx, which proves that A is a basic sequence of coverings. { n,m}

Proposition 5.12. SupposeX andY are metric spaces, and thatA 1,A 2,... is a basic sequence of coverings ofX. If the functiong:X Y is anA -map → i for everyi N, theng is an embedding. ∈ Proof. We shall show, that ifx is any point ofX andU a neighbourhood of x, there exists a neighbourhoodV ofg(x) such that the preimage ofV under g is contained inU. From this follows the injectivity ofg and the continuity 1 of the inverseg − :gX X, whereg :X gX is the function deﬁned by 1 → 1 → g. By the deﬁnition of a basic sequence of coverings, there exists anA i for which SA (x) U. i ⊂ Sinceg is anA -map there exists a neighbourhoodV ofg(x) and a setU i A i 0 ∈ i for which

1 i g− V U . ⊂ 0 1 i Sincex g − V U , we have ∈ ⊂ 0 i U S A (x). 0 ⊂ i 1 Thusg − V U, which proves the claim. ⊂ Proposition 5.13. LetX be a separable metric space andY a compact metric space. For eachﬁnite open coveringA ofX, the setG A of allA- maps fromX toY is open in the function space(C(X,Y), ). �·� ∞ 54 Chapter 5. Invariance of Domain

Proof. Supposeg:X Y is anA-map. Hence every point ofY has a neigh- → bourhood whose preimage underg is contained in some member ofA. Since Y is compact, there is aﬁnite subcollection of these neighbourhoods which form a coveringC ofY . From the Lebesgue Covering Theorem (Proposition 2.33) one derives a numberλ> 0 with the property that any set inY whose diameter is less thanλ is contained in some member ofC, and thus has its preimage underg entirely contained in some member ofA. Letf:X Y → be a continuous function satisfying 1 f g < λ. � − � ∞ 3

1 Take a spherical neighbourhood of diameter 3 λ around every point ofY. Denote byV the preimage of one of these spherical neighbourhoods under 1 1 f, that isV :=f − B(y, λ) for somey Y . The diameter of the setgV 3 ∈ is less thanλ, which� is a result� of the inequality above. Thus the setV is contained in some member ofA, which in turn implies thatf is anA-map. This completes the proof.

Proof of Proposition 5.8. Consider the function space (C(X,Iω), ). Let �·� ∞ A1,A 2,... be a basic sequence of coverings ofX, which exists by Proposition

5.11, letG Ai be the set ofA i-maps ofX intoI ω and let

∞ H := GAi . i�=1 By Proposition 5.12 each element ofH is an embedding. By Proposition 5.13 eachG Ai is open inC(X,I ω) and thusH is aG δ set. Thus by Corollary 5.7 we only need to show the following:

For eachﬁnite open coveringA ofX denote byG A the set of A-maps fromX toI ω. ThenG A is dense inC(X,I ω). Supposef C(X,I ) and letε> 0. We shall construct a continuous ∈ ω functiong:X I such that → ω f g <ε (3) � − � ∞

g G A. (4) ∈ As a compact space,I ω has afinite open coveringK of mesh less than 1 ε. Let U r be the covering ofX made up of the non-empty sets of the 2 { i}i=1 form 1 A f − K, ∩ Brouwer’s Theorem on the Invariance of Domain 55 whereA A andK K. Thus for everyU it holds thatd(fU )< 1 ε. ∈ ∈ i i 2 We shall select linearly independent pointsp , p , . . . , p I for which it 1 2 r ∈ ω holds that 1 d(p , fU )< ε, i=1,2, . . . , r. (5) i i 2 For eachx X setw (x) =d(x, X U ), wherei 1,2, . . . , r with the ∈ i \ i ∈{ } understanding, thatw (x) = 1, ifU =X. Noww (x)> 0 ifx U and i i i ∈ i w (x) = 0 ifx U . For everyx X at least onw (x) is positive, since i �∈ i ∈ i U r is a covering ofX. Thus the functiong:X I defined with the { i}i=1 → ω rule 1 r g(x) = r wi(x)pi i=1 wi(x) �i=1 is well defined, and evidently continuous.� We shall next show, thatg satisfies (3) and (4). Letx X and suppose ∈ that the collection U r is so numbered, thatU ,U ,...,U are the sets { i}i=1 1 2 s that contain the pointx. Thenw (x)> 0 fori s, andw (x) = 0 fori>s. i ≤ i Hence when examining the pointg(x), we only need to consider the points p , p , . . . , p . Sincex U , wheni s, and from the fact thatd(fU )< 1 ε, 1 2 s ∈ i ≤ i 2 together with (5) we get

d(p , f(x))<ε, i s. i ≤ Thus the pointg(x) satisﬁes

d(g(x), f(x))<ε.

SupposeU ,U ,...,U are all of the members of U r that contain i1 i2 ik { i}i=1 a given pointx X. Consider the aﬃne subspaceA(x) of the Hilbert ∈ cube spanned by the pointsp i1 , pi2 , . . . , pik . In other words,A(x) consists of exactly those points of the Hilbert cube, that can be represented in the form

λij pij , j�=1 k where j=1 λij = 1. Clearly� the pointg(x) is contained inA(x). Letx � be another point of the spaceX. We claim, that ifA(x) A(x �)= , these aﬃne subspaces contain ∩ � ∅ some common pointp i, and thusx andx � are contained in a common member of U r . Otherwise, from the equation { i}i=1 k m

αij pij = βlj plj , j�=1 j�=1 56 Chapter 5. Invariance of Domain

k m where j=1 αij = j=1 βlj = 1, it would follow that

� � k m α p β p = 0 ij ij − lj lj j�=1 j�=1 and thus the pointsp 1, p2, . . . , pr we chose before would not be linearly independent, a contradiction. Since there are only afinite number of these affine subspacesA(x), there exists a numberδ> 0 such that any two of these affine subspacesA(x) and A(x�) either meet, or have a distance δ from each other. If ≥ d(g(x), g(x�))<δ, it certainly holds thatd(A(x),A(x �))<δ, and as was noted above, this shows r thatx andx � are contained in a common member of U . This shows, { i}i=1 thatg is anA-map, and thus the setG A is dense inC(X,I ω). Proposition 5.14. SupposeX is a separable metric space with ind(X) n, ≥ wheren 1. Then there exists a continuous functionf:X I n such that ≥ → f has at least one stable value. Proof. Suppose that no continuous functionf:X I n has stable values. → Now, by the definition of unstable values, for each pointy fX, the function ∈ f can be approximated arbitrarily closely by a continuous functiong � :X n → I , for whichy g �X. Proposition 5.4 gives us in turn an arbitrarily close �∈ n approximation ofg �, namely a functiong:X I , with the propertyy → �∈ gX. Let us consider the function spaceC(X,I ω) with the metric induced by the sup-norm. LetM=M(i 1, i2, . . . , in;c 1, c2, . . . , cn) be the subspace of the Hilbert cube defined by then equations

xi1 =c 1, xi2 =c 2, . . . , xin =c n. (6)

We shall denote byG M the subset ofC(X,I ω), consisting of functionsg with the property gX I M. ⊂ ω \ The set of functionsG is dense inC(X,I ). To show this, letf C(X,I ). M ω ∈ ω Now we have

f(x) = (f 1(x), f2(x),...), f i(x) 1/i, wherei N. (7) | |≤ ∈ The functionsf , f , . . . , f deﬁne a continuous function f¯:X I n, and ii i2 in → as we remarked above, there exists an arbitrarily close approximationf � of f¯, such that (c , c , . . . , c ) f X. Thus,G is dense inC(X,I ). 1 2 n �∈ � M ω Brouwer’s Theorem on the Invariance of Domain 57

To show thatG is open inC(X,I ) consider a functiong G . M ω ∈ M Now the distanced :=d(g(X),M) is positive, and thus every function f C(X,I ω) with f g < d belongs toG M . ∈ � − � ∞ Let us now focus on the functionsg C(X,I ) with the property ∈ ω n 1 gX M − , (8) ⊂ ω n 1 whereM ω− denotes the points in the Hilbert cube at mostn 1 whose − n 1 coordinates are rational (see Example 3.21). The complement ofM ω− is the countable union of hyperplanesM 1,M2,... of type (6), namely those corre- sponding to all possible combinations ofn indexesi j and rational numbers cj. Thus (8) is equivalent to

g G Mi , for everyi N. ∈ ∈

By Proposition 5.8 the spaceC(X,I ω) contains a denseG δ setE, each member of which is an embedding. The set

E� :=E G ∩  Mi  i N �∈   is dense inC(X,I ω) as the countable intersection of denseG δ sets by Corol- lary 5.7. In particular,E � is non-empty. n 1 Thus there exists an embeddingh, which embedsX inM ω− . Hence n 1 by Theorem 3.15 and Proposition 3.16 we have ind(X) ind(M ω− ). In n 1 ≤ Example 3.21 we in fact showed that ind(M − ) n 1, thus we have ω ≤ − contradicted the assumption that ind(X) n. ≥ Proposition 5.15. SupposeX is a metric space and letf:X I n be → continuous. An interior pointy offX is an unstable value off, if and only if for every neighbourhoodU ofy, there exists a continuous mapg:X I n → satisfying g(x) =f(x) iff(x) U (9) �∈ g(x) U iff(x) U (10) ∈ ∈ y gX. (11) �∈ Proof. Suppose that the condition of the proposition holds. From (9) and (10) it follows that for every neighbourhoodU of the pointy there exists a continuous functiong:X I n such that → f(x) g(x) d(U) � − � ≤ 58 Chapter 5. Invariance of Domain

for everyx X, andgX I n y . Thusy is an unstable value off. ∈ ⊂ \{ } Suppose then thaty is an interior point ofI n and thaty is an unstable value of the functionf. Letε> 0. PutU :=B(y,ε). Sincey is an unstable n value off, there exists a continuous functiong � :X I such that → f(x) g �(x) <ε/2 (12) � − � g�(x)=y (13) � for allx X. We shall construct a new functiong as follows: ∈ g(x) =g �(x) if f(x) y ε/2, (14) � − �≤ f(x) y 2 f(x) y g(x) = 2 1 � − � g�(x) 1 � − � f(x) (15) � − ε � − � − ε � ifε/2 f(x) y ε, and ≤� − �≤ g(x) =f(x) if f(x) y ε. (16) � − �≥ The subfunctions that defineg are all continuous and their domains are closed inX, since the domains are preimages of closed sets under the continuous functionf. If in the intersections of the domains of these subfunctions, the subfunctions receive the same values, the piecewise defined functiong is continuous. From the definitions of the subfunctions, it is evident that we only need to check the cases f(x) y =ε/2 and f(x) y =ε. � − � � − � In thefirst caseg(x) =g �(x) and ε/2 2ε/2 g(x) = 2 1 g�(x) 1 f(x) =g �(x). � − ε � − � − ε � In the second case we have ε 2ε g(x) = 2 1 g�(x) 1 f(x) =f(x), � − ε� − � − ε � andg(x) =f(x). We have thus established, that indeedg:X I n is a → continuous function. Condition (9) is the same as (16). Ifε/2 f(x) y ε we get from ≤� − �≤ (12) and (15) that g(x) y f(x) y f(x) g(x) � − �−� − �≤� − � 2 f(x) y = f(x) g �(x) + (f(x) g �(x)) 1 � − � � − − � − ε �� f(x) y � = 2� 1 � − � f(x) g �(x) � � − ε � � − � <ε f(x) y , −� − � Brouwer’s Theorem on the Invariance of Domain 59 and thus 0< g(x) y <ε. (17) � − � By (12), (13) and (14) the inequality (17) also holds when f(x) ε/2. � � ≤ Hence the conditions (10) and (11) hold, completing the proof. Definition 5.16 (Extendable function). If for a continuous functionf:M → Y defined on a subspaceM of a spaceX, there exists a continuous function F:X Y , such thatF(x) =f(x) for allx M, we say thatf is contin- → ∈ uously extendable, or more briefly extendable, and callF an extension off overX. We have already encountered a result concerning extensions, namely Urysohn’s lemma (Proposition 2.18). It states that if a subspaceM of a normal spaceX can be represented as a union of two disjoint closed sets A, B X, the functionf:M [0, 1], defined byf(x) = 0, whenx A and ⊂ → ∈ f(x) = 1 whenx B is continuously extendable overX. Actually a much ∈ more general theorem holds, which is the Tietze Extension Theorem. Before we prove this theorem, wefirst state and prove a small lemma. Lemma 5.17. SupposeX is a normal space andA X is closed,a>0 ⊂ and that the functionf:A [ a, a] is continuous. Now there exists a → − continuous functionh:X [ a/3, a/3] such that f(x) h(x) 2a/3 for → − | − |≤ allx A. ∈ 1 1 Proof. The setsA :=f − [ a, a/3] andA :=f − [a/3, a] are closed and 1 − − 2 disjoint. By Urysohn’s lemma (Proposition 2.18), there exists a continuous functionh:X [ a/3, a/3] such that hA a/3 and hA a/3 . → − 1 ⊂{− } 2 ⊂{ } Nowh is the desired function. Theorem 5.18 (Tietze Extension Theorem). Every continuous funtion from a closed subspaceM of a normal spaceX to a closed interval[a, b] is continuously extendable overX. Proof. Letf:M [a, b] be continuous. Since the closed interval [a, b] is → homeomorphic to [ 1, 1] we may assume [a, b] = [ 1, 1] for simplicity. We − − shall apply Lemma 5.17 to define a sequenceg 1, g2,... of continuous functions fromX to [ 1, 1] such that − i 1 1 2 − gi(x) , forx X (18) | |≤ 3 �3� ∈ and i 2 i f(x) g (x) , forx M. (19) | − j |≤ 3 ∈ j�=1 � � 60 Chapter 5. Invariance of Domain

To obtaing 1, we apply Lemma 5.17 to the functionf. Thus 1 g (x) , forx X | 1 |≤ 3 ∈ and 2 f(x) g (x) , forx M. | − 1 |≤ 3 ∈

Suppose we have deﬁned the functionsg 1, g2, . . . , gi. Applying Lemma 5.17 to the functionf i g M we obtain a functiong satisfying − j=1 j i+1 � � � � � 1� 2 i gi+1(x) , forx X | |≤ 3 �3� ∈ and i+1 2 i+1 f(x) g (x) , forx M. | − j |≤ 3 ∈ j�=1 � � From (18) and the Weierstraß test (see Väisälä [7, Lause 10.14, p. 81]) it

follows that the formulaF(x) := i∞=1 gi(x) deﬁnes a continuous function F:X [ 1, 1] (see Väisälä [7, Lause 10.13, p. 81]), and (19) impliesF(x) = → − � f(x) for allx M, so thatF is an extension off overX. ∈ Corollary 5.19. Every continuous function from a closed subspaceM of a normal spaceX to the cubeI n is continuously extendable overX.

Proof. This follows from applying the Tietze Extension Theorem to each of the coordinate functions.

Corollary 5.20. SupposeM is a closed subset of a normal spaceX, and let f:M S n be a continuous function. Now there exists a neighbourhood of → the setM over whichf can be extended.

Proof. SinceS n has radius 1, we have for the coordinates of the pointf(x) ∈ Rn+1 n+1 2 fi(x) = 1, �i=1 and thus f (x) 1 fori=1,2, . . . , n+1. | i |≤ Hence we can apply the Tietze Extension Theorem (Theorem 5.18) to each of the coordinate functionsf :M [ 1, 1]. Thus for eachf we get extensions i → − i F :X [ 1, 1]. i → − Brouwer’s Theorem on the Invariance of Domain 61

LetU X denote the set of points for which ⊂ n+1 2 Fi(x) >0. �i=1 Thus n+1 1 U= F − ([ 1, 1] 0 ), i − \{ } i�=1 which makesU an open subset ofX as a union of open sets. It is evident that M U. We shall deﬁne the coordinate functions of the functionG:U S n ⊂ → with the rule

Fi(x) Gi(x) = 1/2 fori=1,2, . . . , n+1. n+1 2 i=1 Fi(x) � � The mappingG is the desired� extension off overU. Proposition 5.21. A separable metric spaceX has ind(X) n, if and only ≤ if for each closed setC X and continuous functionf:C S n there exists ⊂ → an extension off overX. Proof. Supposef is a continuous function from a closed subsetC X to ⊂ Sn. SinceS n and∂I n+1 are homeomorphic, we think off as function fromC n+1 n+1 toI . By Corollary 5.19 there exists a continuous functionF � :X I , → which is an extension off. Suppose that ind(X) n. Proposition 5.3 implies, that the origin is ≤ not a stable value ofF �. Proposition 5.15 gives us a continuous function n+1 F �� :X I such that 0¯ F ��X, whileF ��(x) =F �(x) for allx X that → �∈ n+1 ∈ do not map to the interior ofI underF �. In particular, forx C, we ∈ have F ��(x) =F �(x) =f(x). n+1 LetF:X ∂I be the projection of the pointF ��(x) from the origin on → the boundary∂I n+1. NowF is the desired extension off. Suppose the condition of the proposition holds. Now, in order to prove that ind(X) n, it is enough to show, according to Proposition 5.14, that ≤ a continuous functionf:X I n+1 cannot have stable values. Example 5.2 → tells us that a boundary point ofI n+1 is never stable. Hence it suﬃces to show that the interior points ofI n+1 cannot be stable. Lety be an interior point n+1 1 ε ofI and let 0<ε< 1. Denote byC the inverse imagef − ∂B(y, 2 ) . As the functionf is continuous,C is a closed subset ofX. Let � f¯ denote� the restriction off toC. By hypothesis, there exists a continuous function F:X ∂B(y, ε ) such thatF(x) = f¯(x) forx C. → 2 ∈ 62 Chapter 5. Invariance of Domain

We shall construct a new functiong defined in the whole spaceX, taking values inI n+1 with the rules ε g(x) =f(x) iff(x) B y, , �∈ � 2� ε g(x) =F(x) iff(x) B y, . ∈ � 2� 1 n+1 ε In other words,g(x) =f(x), wheneverx f − I B y, and ∈ \ 2 1 ε � � �� g(x) =F(x), whenx f − B y, . The intersection of these two pre- ∈ 2 images is the setC defined above.� � By�� the definition of the functionF , we havef(x) =F(x), ifx C, and thus the functiong is continuous. n+1∈ ε Nowg:X I B y, 2 is a continuous function with f g <ε. → \ � − � ∞ This completes the proof.� � Corollary 5.22. SupposeC is a closed subset of the separable metric space X. If ind(X C) n, then every continuous functionf:C S n can be \ ≤ → extended overX. Proof. Supposef:C S n is continuous. Corollary 5.20 shows that there → exist an open setU C and an extensionf � off overU. SinceX is normal, ⊃ by Proposition 2.16, there exists an open setV X satisfying ⊂ C V V U. ⊂ ⊂ ⊂ Let us consider the restriction

f¯ :=f � V (X C). ∩ \ � � This is a continuous function from� a closed subset of the spaceX C to the \ sphereS n. Now the subspaceX C has dimension n, and by Proposition \ ≤ 5.21 there exists an extensionf �� of f¯ overX C. By putting \ F(x) =f(x) ifx C, ∈ F(x) =f ��(x) ifx X C, ∈ \ we get the desired extensionF:X S n off. → LetA be a subset of an arbitrary topological spaceX. Iff:X X is → a homeomorphism, then all interior points ofA map to the interior points offA, and conversely, iff(x) is an interior point offA, thenx intA. ∈ However, if we consider an embeddingh:A X, it is not generally true, → thath maps the interior points ofA to the interior of hA. If the spaceX is a Euclidean space we in fact have Brouwer’s Theorem on the Invariance of Domain 63

Theorem 5.23 (Brouwer’s Theorem on the Invariance of Domain). Suppose A is a subset of the Euclidean spaceR n and leth:A R n be an embedding. → Then, ifx A is an interior point ofA, the pointh(x) belongs to the interior ∈ of hA. In particular, ifU andV are homeomorphic subsets ofR n andU is open, thenV is open.

We shall prove Brouwer’s Theorem on the Invariance of Domain with the help of the proposition below:

Proposition 5.24. LetA R n be compact. The pointx is a boundary ⊂ point ofA, if and only if the pointx has arbitrarily small neighbourhoods n 1 U A open inA, such that any continuous functionf:A U S − can ⊂ \ → be extended overA.

Proof. Supposex ∂A. Letε> 0, and putB(x) :=B n(x,ε) andU := ∈ A B(x). We will show that the setU has the desired property. Because the ∩ setA U=A U is compact, it is closed in any containing space. Thus by \ ∩� Corollary 5.22 any continuous functionf:A U ∂B(x) can be extended \ → to a functionf � :(A U) ∂B(x) ∂B(x). Letq be a point ofB(x) that is \ ∪ → not inA. For eachy A denote byy � the projection ofy from the pointq ∈ to the boundary∂B(x). We deﬁne

F(y)=f �(y�) ify U, ∈ F(y)=f(y) ify A U. ∈ \ The functionF:A ∂B(x) is the desired extension. → Supposex is an interior point ofA. Letε> 0 be such that Bn(x,ε) A. ⊂ We again denoteB n(x,ε) byB(x). We shall show that for any neighbourhood U B(x) ofx, there exists a continuous functionf:A U ∂B(x), that ⊂ \ → cannot be extended overA. Letf be the projection ofA U on the boundary \ ∂B(x) from the pointx. Suppose we can extendf overA. Let us denote this extension by f¯. Now the restriction f¯ B(x) is a continuous function from a � closed ball to the sphere∂B(x), which� leaves each of the boundary points ﬁxed, thus contradicting Corollary 4.17.� This completes the proof.

Proof of Theorem 5.23. We may assume thatA R n is compact, since an ⊂ interior pointx ofA is also an interior point of some ball Bn(x,ε), where ε> 0. Since the image of a compact set under a continuous function is compact by Väisälä [6, Lause 13.18, p. 100], we can apply Proposition 5.24 to the boundary points of hA. Suppose, thatx intA, and suppose, for a contradiction, thath(x) ∈ ∈ ∂(hA). We denote byh :A hA the homeomorphism deﬁned by the 1 → 64 Chapter 5. Invariance of Domain

embeddingh. LetU hA be some neighbourhood ofh(x) in hA. Now ⊂1 the inverse imageh − U is a neighbourhood of the pointx inA, and by the n 1 normality ofA, there exists a numberε> 0 such that B (x,ε) = : B h − U. ⊂ Letf: hA U ∂ B be a function which maps everyy hA U to the \ → ∈ \ boundary∂ B in such a way, that the pointf(y) corresponds to the projection 1 ofh 1− (y) to the boundary∂ B from the pointx. Clearly,f is continuous as a composition of two continuous functions. We may apply Proposition 5.24. Hence there exists an extension off, namely some function f¯: hA ∂ B. Now the composition f¯ h B is a → ◦ continuous function from a closed ball to its boundary, which keeps� � � each � boundary pointﬁxed thus contradicting Corollary 4.17. � We have shown that the embeddingh maps each interior point ofA to the interior of hA, which completes the proof of Brouwer’s Theorem on the Invariance of Domain. Bibliography

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