[problem books in mathematics] exercises in analysis || topological spaces

Chapter 2

Topological Spaces

2.1 Introduction

2.1.1 Basic Definitions and Notation

Definition 2.1A topological space is a pair (X, τ) where X is a nonempty set andτ is a family of subsets of X called open sets, which satisfies thefollowing three requirements:(a) ∅,X ∈ τ ;(b) τ is closed under arbitrary unions, i.e.,

if {Ui}i∈I ⊆ τ, then⋃

i∈IUi ∈ τ ;

(c) τ is closed under finite intersections, i.e.,

if {Ui}i∈I ⊆ τ and I is finite, then⋂

i∈IUi ∈ τ.

The complements of open sets are called closed sets,

Remark 2.2By the De Morgan law, if {Ci}i∈I is any family of closed sets, theset⋂i∈I

Ci is a closed set. Evidently many different topologies can be

defined on the same set X. The smallest of all topologies on X (i.e.,the one with the fewest open sets) is the topology τ = {∅,X} and itis called the indiscrete topology . The biggest of all topologies on

L. Gasinski and N.S. Papageorgiou, Exercises in Analysis: Part 1,Problem Books in Mathematics, DOI 10.1007/978-3-319-06176-4 2,© Springer International Publishing Switzerland 2014

193

194 Chapter 2. Topological Spaces

X (i.e., the one with the most open sets) is the topology consistingof all subsets of X and it is called the discrete topology . The spaceX with the discrete topology is a discrete space. If τ1, τ2 are twotopologies on X and τ1 ⊆ τ2 then we say that τ1 is weaker than τ2or that τ2 is stronger than τ1. Finally note that a subset of X maybe both open and closed (for example ∅ and X are open and closed inany topology on X or in a discrete space every set is open and closed).A set which is both open and closed is usually called clopen .

If (X, dX) is a metric space then the family τ of all open sets (in

the sense of Definition 1.8) is a topology. We call it the topologyinduced by the metric dX . So every metric space is in a natural waya topological space.

Finally, if for the topological space (X, τ) there exists a metric dsuch that τ is induced by the metric d , then we say that the topologicalspace (X, τ) is metrizable .

Definition 2.3Let (X, τ) be a topological space. A neighbourhood of x is a set U ∈ τsuch that x ∈ U . The family of neighbourhoods of x ∈ X is denotedby N (x). Similarly, a neighbourhood of a set C ⊆ X is an open setU such that C ⊆ U . If U ∈ N (x), then U \ {x} is called the deletedneighbourhood of x.

Definition 2.4Let (X, τ) be a topological space.(a) X is a T0-space (or Kolmogorov space) if for every pair ofdistinct points x, y ∈ X, we can find U ∈ τ such that x ∈ U , y �∈ U .(b) X is a T1-space (or Hausdorff or separated space) if for everypair of distinct points x, y ∈ X, we can find U, V ∈ τ such that x ∈ U ,y �∈ U and x �∈ V , y ∈ V .(c) X is a T2-space if for every pair of distinct points x, y ∈ X, wecan find U, V ∈ τ such that x ∈ U , y ∈ V and U ∩ V = ∅.(d) X is a T3-space (or regular or Vietoris space) if for everyx ∈ X and every closed set C ⊆ X not containing x, we can findU, V ∈ τ such that x ∈ U , C ⊆ V and U ∩ V = ∅.(e) X is a T4-space (or normal or Tietze space) if for every pair ofdisjoint closed sets C,D ⊆ X, we can find U, V ∈ τ such that C ⊆ U ,D ⊆ V and U ∩ V = ∅.

2.1. Introduction 195

Remark 2.5In general a T1-space is always a T0-space, a T2-space is a T1-space,but a T3 space does not need to be a T2-space. A T4 space needs to beneither T3 nor T2 (considerX = {x, y} with topology τ =

{∅, {x},X}).A discrete topological space (see Remark 2.2) is T0, T1, T2, T3 and T4,while the indiscrete topological space X (where X has at least twoelements) is T3 and T4 but not T0, T1 and T2 (in particular it is not aHausdorff space).

Remark 2.6Strictly speaking, for a space X to be non-Hausdorff, seems ratherunreasonable. However, it turns out that there exist useful topologies(such as the Zariski topology in algebraic geometry), which are non-Hausdorff. Nevertheless, for the purpose of Analysis, the Hausdorffseparation axiom suffices. For this reason in this book

all topological spaces are assumed to be Hausdorff.

Every metric space is Hausdorff and in a Hausdorff space singletonsare closed sets.

Definition 2.7Let (X, τ) be a topological space and let {xn}n�1 ⊆ X be a sequence.We say that x ∈ X is the limit of the sequence {xn}n�1, if for everyU ∈ N (x), we can find n0 = n0(U) � 1 such that xn ∈ U for alln � n0.

Remark 2.8In a Hausdorff space a sequence can have at most one limit.

Definition 2.9Let (X, τ) be a topological space and let A ⊆ X be a nonempty set.(a) A point x ∈ X is an interior point of A, if we can find U ∈ N (x)such that U ⊆ A. The set of interior points of A is called the interior

of A and is denoted by intA (or◦A).

(b) A point x ∈ X is an exterior point of A, if x ∈ intAc (recallthat Ac = X \A). The set of exterior points of A is called the exteriorof A and is denoted by extA.(c) A point x ∈ X is a boundary point of A, if for every U ∈ N (x),we have A ∩ U �= ∅ and Ac ∩ U �= ∅. The set of boundary points of Ais called the boundary of A and is denoted by ∂A (or bdA).


(d) A point x ∈ X is a limit point (or accumulation point orcluster point) of A, if every U ∈ N (x) satisfies A ∩ (U \ {x}) �= ∅.The set of limit points of A is called the derived set of A and isdenoted by A′. The set A∪A′ is called the closure of A and is denotedby A (or clA). A point x ∈ A is an isolated point of A if x ∈ A \ A′

(i.e., there exists U ∈ N (x) such that A ∩ (U \ {x} = ∅).(e) We say that a set A is dense in X if A = X. We say that a setA is nowhere dense in X if intA = ∅.(f) We say that the space X is separable if it has a countable densesubset.

Proposition 2.10If (X, τ) is a topological space and A,B ⊆ X are nonempty sets,then(a) A is open if and only if A = intA;(b) intA is the biggest open set contained in A, i.e., if U is the familyof all open sets contained in A, then intA =

⋃U∈U

U ;

(c) int (A ∩B) = intA ∩ intB;(d) int (A ∪B) ⊇ intA ∪ intB.

Proposition 2.11If (X, τ) is a topological space and A,B ⊆ X are nonempty sets,then(a) A is closed if and only if A = A;(b) A is closed if and only if A′ ⊆ A;(c) x ∈ A if and only if for all V ∈ N (x), we have V ∩A �= ∅;(d) A is the smallest closed set which contains A, i.e., if F is thefamily of all closed set containing A, then A =

⋂C∈F

C;

(e) A ∪B = A ∪ B;(f) A ∩B ⊆ A ∩ B.

Proposition 2.12If (X, τ) is a topological space and {Ai}i∈I ⊆ X is a family of sets,then

(a)⋃

i∈IAi ⊆

⋃

i∈IAi;

(b)⋂

i∈IAi ⊆

⋂

i∈IAi..


Proposition 2.13Let (X, τ) be a topological space. The following statements are equiv-alent:(a) D is dense in X;(b) if C is a closed subset of X and D ⊆ C, then C = X;(c) intDc = ∅;(d) for every U ∈ τ , we have U ∩D �= ∅.

Definition 2.14Let (X, τ) be a topological space and let C ⊆ X be a nonempty set. Thesubspace topology (or relative topology or induced topology) τCon C is defined by τC =

{C ∩ U : U ∈ τ

}. With these open sets C

becomes a topological space in its own right and we refer to C as asubspace of X.

Proposition 2.15Let (X, τ) be a topological space and let C ⊆ X be a nonempty sub-space. Every open (respectively, closed) set in C is open (respectively,closed) in X if and only if C is open (respectively, closed) in X.

Proposition 2.16Suppose that (X, τ) is a topological space, C ⊆ X is a nonempty sub-space and D is a nonempty subspace of C. Then the subspace topologieson D induced by X and C coincide.

Proposition 2.17The Hausdorff property and the regularity property (see Definition 2.4)are hereditary, i.e., if (X, τ) is Hausdorff (respectively, regular) andC ⊆ X is nonempty, then (C, τC) is Hausdorff (respectively, regular)too.

Definition 2.18Let (X, τ) be a topological space.(a) A set E ⊆ X is said to be a Gδ-set if E =

⋂n�1

Un with Un ⊆ X

open for all n � 1.(b) A set D ⊆ X is said to be a Fσ-set if D =

⋃n�1

Cn with Cn ⊆ X

closed for all n � 1.


2.1.2 Topological Basis and Subbasis

Definition 2.19Let (X, τ) be a topological space.(a) A family B ⊆ τ is a basis for τ if every U ∈ τ is a union ofelements in B.(b) A family Y ⊆ τ is a subbasis for τ if the collection of all finiteintersections of sets from Y form a basis for τ .

Proposition 2.20If (X, τ) is a topological space and B ⊆ τ ,then B is a basis if and only if for every x ∈ X and every U ∈ N (x),we can find V ∈ B such that x ∈ V ⊆ U .

Proposition 2.21If (X, τ) is a topological space and B is a basis for τ ,then U is open (i.e., U ∈ τ) if and only if for every x ∈ U , we canfind V ∈ B such that x ∈ V ⊆ U .

Definition 2.22Let A be a family of subsets of a set X. The topology on X generatedby A (denoted by τ(A)), is the smallest topology on X containing A.This topology can be obtained from A by taking finite intersections ofelements of A and creating this way a basis for a topology which isunique. Clearly τ(A) is the intersection of all topologies containing A.

Definition 2.23Let (X, τ) be a topological space and let x ∈ X. We say that D ⊆ N (x)is a local basis at x (or a neighbourhood basis at x) if for everyU ∈ N (x), we can find V ∈ D such that x ∈ V ⊆ U .

Definition 2.24A topological space (X, τ) is said to be:(a) first countable if there is a countable local basis at every x ∈ X;(b) second countable if it has a countable basis.

Remark 2.25Every metric space is first countable, but need not be second count-able. From Proposition 1.24 we know that a metric space is secondcountable if and only if it is separable. For example the open intervals


with rational endpoints form a countable basis for the separable spaceX = R (recall Q = R). Clearly every second countable space is firstcountable. The converse is not in general true.

Definition 2.26Let (X, τ) be a topological space.(a) Let Y ⊆ τ . We say that Y is an open cover of X if X =

⋃U∈Y

U .

A subfamily Y ′ ⊆ Y which is also a cover is called a subcover of Y.(b) X is said to be a Lindelof space if every open cover of X has acountable subcover.

Theorem 2.27Every second countable space is Lindelof and separable.

Remark 2.28While for metric spaces separability (or the Lindelof property) is equiv-alent to second countability (see Theorem 2.27), for general spaces thisis not true. Consider X = R and

A ={(λ,+∞) : λ ∈ R

} ∪ {(−∞, μ] : μ ∈ R}.

We consider the topology τ(A) (see Definition 2.22). The topologyhas a basis consisting of the intervals (λ, μ], λ � μ. Note that τ(A) isnot the usual (Euclidean) topology on R since the sets (λ, μ] are notopen sets in the Euclidean topology. In fact the Euclidean topologyis strictly weaker than τ(A) (usually called the upper limit topol-ogy). Then

(R, τ(A)

)is Lindelof and separable (Q is a countable

dense subset), but it is not second countable.

2.1.3 Nets

Sequences which were very useful in the study of metric spaces are oflimited interest in general topological spaces which need not be firstcountable. Instead we use nets, which are a suitable generalization ofsequences. In the indexing of a net the set of positive integers (usedin sequences) is replaced by the more general notion of directed set.


Definition 2.29

(a) A relation � on a set I is a partial ordering if it is reflexive

(i.e., x � x for all x ∈ I), antisymmetric (i.e., x � y and y � x implyx = y for all x, y ∈ I) and transitive (i.e., x � y and y � z implyx � z for all x, y, z ∈ I).(b) A pair (I,�) consisting of a set I and a partial ordering � is saidto be a directed set if for any x, y ∈ I, we can find z ∈ I such thatx � z and y � z.

Example 2.30

(a) If I = N and � is the usual ordering in N, then (N,�) is a directedset.(b) Let X be a nonempty set and let I be the family of finite subsetsof X and let � be the relation on I defined by F1 � F2 (for F1, F2 ∈ I)if and only if F1 ⊆ F2. Then (I,�) is a directed set.(c) Let (X, τ) be a topological space, x ∈ X and let I = N (x). OnI we consider the relation � defined by U � V (for U, V ∈ N (x)) ifand only if U ⊇ V (the partial order is the reverse inclusion). Then(N (x),�

)is a directed set.

(d) Let (I1,�1) and (I2,�2) be two directed sets. If on I = I1 × I2we consider the partial order � defined by (x1, x2) � (u1, u2) if andonly if x1 �1 u1 and x2 �2 u2, then (I,�) is a directed set.

Definition 2.31Let X be any set. A net {xi}i∈I ⊆ X is any function x : I −→ X with(I,�) being a directed set. In particular, a sequence is a net definedon the directed set N with the usual ordering. If (X, τ) is a topologicalspace, then a net {xi}i∈I is said to converge to x, written xi −→ x ifand only if for every U ∈ N (x) there is i0 ∈ I such that xi ∈ U for alli0 � i. This convergence can also be expressed by saying that {xi}i∈Iis eventually in U for every U ∈ N (x). We say that the net {xi}i∈Iis frequently in a set C ⊆ X, if for any i ∈ I, there is a j ∈ I suchthat i � j and xj ∈ C. Note that the negation of the statement “thenet is eventually in C” is the statement “the net is frequently in Cc”Evidently if a net {xi}i∈I is eventually in C1 ⊆ X and eventually inC2 ⊆ X, then it is eventually in C1 ∩ C2.


Proposition 2.32The Hausdorff property of (X, τ) is equivalent to saying that everyconvergent net has a unique limit.

Proposition 2.33If (X, τ) is a topological space and C ⊆ X is nonempty, then(a) x is a limit point of C if and only if there exists a net {xi}i∈I ⊆C \ {x} such that xi −→ x.(b) C is closed if and only if every convergent net {xi}i∈I ⊆ C has itslimit in C.

In Theorem 1.71 we saw that in metric spaces compactness can becharacterized using convergent subsequences. We shall see that thesame can be done for general topological spaces provided we replacesequences by nets and subsequences by subnets, which we define next.

Definition 2.34Let I and J be directed sets and let {xi}i∈I and {yj}j∈J be nets in a setX. We say that {yj}j∈J is a subnet of {xi}i∈I , if there is a functionψ : J −→ I such that:(a) yj = xψ(j) for each j ∈ J ;(b) for every i0 ∈ I, there is a j0 ∈ J such that for every j ∈ J , j0 � jimplies i0 � ψ(j).

Proposition 2.35If (X, τ) is a topological space and {xi}i∈I is a net in X,then x ∈ X is a limit point of {xi}i∈I if and only if x is the limit ofsome subnet of {xi}i∈I .

Corollary 2.36If (X, τ) is a topological space,then a net {xi}i∈I converges to x ∈ X if and only if every subnet of{xi}i∈I converges to x.

2.1.4 Continuous and Semicontinuous Functions

The notion of “continuous function”, which we are about to introduce,is in the core of Topology. We give both the local and global definitionsof continuity.


Definition 2.37Let (X, τX) and (Y, τY ) be two topological spaces and let f : X −→ Ybe a function.(a) We say that f is continuous at x ∈ X if for every V ∈ N (f(x)),we can find U ∈ N (x) such that f(U) ⊆ V . If f is not continuous atx ∈ X, then we say that f is discontinuous at x.(b) We say that f is continuous if for every V ∈ τY we havef−1(V ) ∈ τX .

Proposition 2.38If (X, τX) and (Y, τY ) are two topological spaces and f : X −→ Y isa function,then f is continuous if and only if f is continuous at every x ∈ X.

Recalling that inverse images preserve all set theoretic operations,we have the following equivalence.

Proposition 2.39If (X, τX) and (Y, τY ) are two topological spaces and f : X −→ Y isa function,then f is continuous if and only if for all C ⊆ Y closed, f−1(C) ⊆ Xis closed.

Proposition 2.40If (X, τX) and (Y, τY ) are two topological spaces and f : X −→ Y isa function,then the following statements are equivalent:(a) f is continuous;(b) the inverse image of every basis element of τY is in τX ;(c) the inverse image of every subbasis element of τY is in τX ;(d) for every A ⊆ X, we have f(A) ⊆ f(A);(e) for every B ⊆ Y , we have f−1(B) ⊆ f−1(B);(f) for every net {xi}i∈I , if xi −→ x, then f(xi) −→ f(x).

Proposition 2.41If X,Y,Z are topological spaces and f : X −→ Y is continuous at x0and g : Y −→ Z is continuous at y0 = f(x0),then g ◦ f : X −→ Z is continuous at x0.


If Y is a metric space, the notion of continuity of a functionf : X −→ Y can be expressed in terms of oscillation (see Problem 2.23).The following definition extends Definition 1.37 to a more general no-tion.

Definition 2.42Suppose that X is a topological space, (Y, d

Y) is a metric space and

f : X −→ Y is a function. For x ∈ X, the oscillation of f at x,denoted by ωf (x), is defined by

ωf (x) = infU∈N (x)

diam f(U).

Next we will see how we can “glue” together continuous functions(piecewise definition of functions). First a definition.

Definition 2.43Let X be a topological space and let {Cα}α∈A be a family of subsetsof X. We say that {Cα}α∈A is locally finite (or neighbourhoodfinite) if for every x ∈ X, we can find U ∈ N (x) such that U ∩Cα �= ∅for only finite number of indices α ∈ A.

Using this notion, we can have the following “gluing theorem”.

Theorem 2.44If X is a topological space, {Cα}α∈A is a cover of X, for all α ∈ A,functions fα : Cα −→ Y are continuous, we have

fα∣∣Cα∩Cβ

= fβ∣∣Cα∩Cβ

∀ (α, β) ∈ A×A

and at least one of the following conditions holds:(i) all sets Cα are open; or(ii) all sets Cα are closed and the family {Cα}α∈A is locally finite,then there exists a unique continuous function f : X −→ Y such that

f∣∣Cα

= fα ∀ α ∈ A

(i.e., f extends fα for all α ∈ A).

For real functions (i.e., functions into R or R∗ = R ∪ {±∞}), we

can exploit the fact that half lines{(a,+∞)

}a∈R∪

{(−∞, a)

}a∈R form

a subbasis for the usual topology on R, to have the following result.


Proposition 2.45If X is a topological space and f : X −→ R is a function,then f is continuous if and only if for every λ, μ ∈ R both sets

{x ∈ X : f(x) < λ

}and

{x ∈ X : f(x) > μ

}

are open in X.

If we require openness only for one of the two sets in the aboveproposition, then we have the notion of semicontinuous function (upperor lower).

Definition 2.46Let X be a topological space and let f : X −→ R

∗ = R ∪ {±∞} be afunction.(a) We say that f is lower semicontinuous at x ∈ X, if for everyλ ∈ R with λ < f(x), we can find U ∈ N (x) such that λ < f(y) forall y ∈ U . We say that f is lower semicontinuous, if f is lowersemicontinuous at each x ∈ X.(b) We say that f is upper semicontinuous at x ∈ X, if for everyλ ∈ R with λ > f(x), we can find U ∈ N (x) such that λ > f(y) forall y ∈ U . We say that f is upper semicontinuous, if f is uppersemicontinuous at each x ∈ X.

Remark 2.47According to the above definition f : X −→ R

∗ is lower semicontinuousif and only if −f is upper semicontinuous.

Also f : X −→ R∗ is lower semicontinuous at x ∈ X if and only if

f(x) � supU∈N (x)

infy∈U

f(y).

Sinceinfy∈U

f(y) � f(x) ∀ U ∈ N (x),

we infer that f is lower semicontinuous at x ∈ X if and only if

f(x) = supU∈N (x)

infy∈U

f(y).

Similarly f : X −→ R∗ is upper semicontinuous at x ∈ X if and only

iff(x) = inf

U∈N (x)supy∈U

f(y).


It follows immediately that if f is lower semicontinuous (respectively,upper semicontinuous) at x, then

f(x) � lim infn→∞ f(xn)

(respectively lim sup

n→∞f(xn) � f(x)

)

for every sequence {xn}n�1 converging to x in X. The converse is trueif we assume additional structure on the topological space X.

Proposition 2.48If X is a first countable topological space, f : X −→ R

∗ is a functionand x ∈ X,then the following statements are equivalent:(a) f is lower semicontinuous at x ∈ X (respectively, upper semicon-tinuous at x ∈ X);(b) f(x) � lim inf

n→+∞ f(xn) (respectively, lim supn→+∞

f(xn) � f(x)) for every

sequence {xn}n�1 ⊆ X such that xn −→ x ∈ X;(c) f(x) � lim

n→+∞ f(xn) (respectively, limn→+∞ f(xn) � f(x)) for every

sequence {xn}n�1 ⊆ X such that xn −→ x ∈ X and limn→+∞ f(xn) <

+∞ (respectively, limn→+∞ f(xn) > −∞).

Definition 2.49Let X,Y be two topological spaces.(a) A function f : X −→ Y is said to be sequentially continuousat x ∈ X if for every sequence {xn}n�1 ⊆ X such that xn −→ x inX, we have that f(xn) −→ f(x) in Y . We say that f is sequentiallycontinuous if it is sequentially continuous at every x ∈ X;(b) A function f : X −→ R

∗ is said to be sequentially lower semi-continuous at x ∈ X (respectively, sequentially upper semicon-tinuous at x ∈ X) if for every sequence {xn}n�1 ⊆ X such thatxn −→ x in X, we have that

f(x) � lim infn→+∞ f(xn)

(respectively lim sup

n→+∞f(xn) � f(x)

).

We say that f is sequentially lower semicontinuous (respectively,sequentially upper semicontinuous) if it is sequentially lowersemicontinuous at every x ∈ X (respectively, sequentially upper semi-continuous at every x ∈ X).


Proposition 2.50If X,Y are topological spaces, then(a) f : X −→ Y is continuous (at x ∈ X) implies that f is sequentiallycontinuous (at x ∈ X) and the converse is true if X is first countable;(b) f : X −→ R

∗ is lower semicontinuous (at x ∈ X) (respectively,upper semicontinuous (at x ∈ X)) implies that f is sequentially lowersemicontinuous (at x ∈ X) (respectively, upper semicontinuous (atx ∈ X)).

Definition 2.51Let (X, τ) be a topological space and let A ⊆ X be a set. We say thatA is sequentially open in X, if for every x ∈ A and every sequence{xn}n�1 ⊆ X, such that xn −→ x in X, there exists n0 � 1 suchthat xn ∈ A for all n � n0. We say that A is sequentially closedin X, if for every sequence {xn}n�1 ⊆ X such that xn −→ x in X,we have x ∈ A. Evidently A is sequentially open if and only if Ac issequentially closed.

Remark 2.52The sequentially open sets form a topology τseq. This is the strongesttopology on X for which the converging sequences are the τ -convergingsequences. We have τ ⊆ τseq and the inclusion can be strict (forexample an infinite dimensional Banach space endowed with the weaktopology, see Chap. 5). We have τ = τseq if and only if X is firstcountable.

Proposition 2.53If X is a topological space and f : X −→ R

∗ is a function,then the following statements are equivalent:(a) f is lower semicontinuous (respectively, sequentially lower semi-continuous);(b) for every λ ∈ R the set

{x ∈ X : f(x) > λ

}is open (respectively,

sequentially open);(c) for every λ ∈ R the set

{x ∈ X : f(x) � λ

}is closed (respectively,

sequentially closed);

Proposition 2.54If X is a topological space and f : X −→ R

∗ is a function,then the following statements are equivalent:


(a) f is upper semicontinuous (respectively, sequentially upper semi-continuous);(b) for every λ ∈ R the set

{x ∈ X : f(x) < λ

}is open (respectively,

sequentially open);(c) for every λ ∈ R the set

{x ∈ X : f(x) � λ

}is closed (respectively,

sequentially closed);

Proposition 2.55If X is a topological space, fi : X −→ R

∗ are functions for i ∈ I,then(a) if each fi is lower semicontinuous (respectively, sequentially lowersemicontinuous), then sup

i∈Ifi(x) is lower semicontinuous (respectively,

sequentially lower semicontinuous);(b) if I is finite and each fi is lower semicontinuous (respectively,sequentially lower semicontinuous), then inf

i∈Ifi(x) is lower semicontin-

uous (respectively, sequentially lower semicontinuous);(c) if each fi is upper semicontinuous (respectively, sequentially uppersemicontinuous), then inf

i∈Ifi(x) is upper semicontinuous (respectively,

sequentially upper semicontinuous);(d) if I is finite and each fi is upper semicontinuous (respectively,sequentially upper semicontinuous), then sup

i∈Ifi(x) is upper semicon-

tinuous (respectively, sequentially upper semicontinuous).

Corollary 2.56If X is a topological space and fi : X −→ R

∗ for i ∈ I are continuous(respectively, sequentially continuous) functions,then f = sup

i∈Ifi is lower semicontinuous (respectively, sequentially

lower semicontinuous) and f = infi∈I

fi is upper semicontinuous

(respectively, sequentially upper semicontinuous).

Definition 2.57Suppose that X is a topological space and f : X −→ R

∗ is a func-tion. The relaxed function of f , denoted by f , is the biggest lowersemicontinuous function majorized by f , i.e., f = sup

g∈L(f)g, where

L(f) ={g : X −→ R

∗ : g is lower semicontinuous and g � f}.


Proposition 2.58Let X be a topological space. The set of all functions f : X −→ R =R ∪ {+∞} which are lower semicontinuous (respectively, sequentiallylower semicontinuous) is a cone, i.e., it is closed under addition andscalar multiplication with λ � 0. Similarly for the set of functionsf : X −→ R

∗= R ∪ {−∞} which are upper semicontinuous (respec-

tively, sequentially upper semicontinuous).

2.1.5 Open and Closed Maps: Homeomorphisms

Definition 2.59Let (X, τX) and (Y, τY ) be two topological spaces and let f : X −→ Ybe a function.(a) We say that f is an open function if for every U ∈ τX ,f(U) ∈ τY .(b) We say that f is a closed function if for every C ⊆ X closed,f(C) is closed in Y .

The notion of homeomorphism that we are about to introduce, isof fundamental importance in Topology.

Definition 2.60Let (X, τX) and (Y, τY ) be two topological spaces and let f : X −→ Ybe a function. We say that f is homeomorphism if f is bijectiveand U ∈ τX if and only if f(U) ∈ τY . Evidently f−1 : Y −→ Xis a homeomorphism too and we say that the spaces X and Y arehomeomorphic. We say that f : X −→ Y is a local homeomorphismif each x ∈ X has a neighbourhood which is mapped homeomorphicallyby f onto an open subset of Y .

Remark 2.61The relation “X homeomorphic to Y ” is an equivalence relation in thefamily of topological spaces. It is the fundamental equivalence relationin Topology. A homeomorphism preserves the topological structure (itis an isomorphism of topological structures). From a purely topologicalpoint of view there is no difference between two homeomorphic topo-logical spaces. They can be considered as two representatives of thesame geometric entity. When a space X has several different structures(such as algebraic, metric, linear etc.), we say that a property of X is


“topological”, if it is preserved by homeomorphisms. Roughly speak-ing, a property which can be stated in terms of open sets and derivednotions such as closed sets, limit points, dense sets, etc., is topological.For example, the Hausdorff property is a topological property, but ina metric space completeness is not a topological property.

2.1.6 Weak (or Initial) and Strong (or Final)Topologies

From the definition of continuity (see Definition 2.37), we see that thecontinuity of a function f : X −→ Y is preserved if we strengthen thetopology on X or weaken the topology on Y . More generally, let Xbe a set and let

{(Yi, τi)

}i∈I be a family of topological spaces (finite

or infinite). For a given family of functions{fi : X −→ Yi

}i∈I , we are

looking for those topologies on X which ensure the continuity of allfunctions fi, i ∈ I.

Definition 2.62Let X be a set and let

{(Yi, τi)

}i∈I be a family of topological spaces.

Also for each i ∈ I, let fi : X −→ Yi be a function. The weakesttopology on X which makes all the functions fi, i ∈ I continuous iscalled the weak topology (or initial topology) on X. This is thetopology w

({fi}i∈I)generated by the subbasis

Y ={f−1i (U) : i ∈ I, U ∈ τi

}.

Remark 2.63If I is a singleton, then the defining subbasis Y =

{f−1(U) : U ∈ τY

}

is in fact the weak topology w(f).

Since we are interested on Hausdorff topological spaces, we wantto know when this topology w

({fi}i∈I)is Hausdorff. To check this we

need the following definition.

Definition 2.64Let X be a set and let {fi}i∈I be a family of functions defined on X.We say that the family {fi}i∈I is separating (or total) if for each pairof points x, u ∈ X, x �= u we can find i ∈ I such that fi(x) �= fi(u).

Proposition 2.65If X is a set endowed with the weak topology w

({fi}i∈I)determined by

a separating family of functions{fi : X −→ Yi

}i∈I (where

{(Yi, τi)

}i∈I

is a family of topological spaces),then

(X,w

({fi}i∈I))

is Hausdorff.


Proposition 2.66If X is a set endowed with the weak topology w

({fi}i∈I)determined by

a separating family of functions{fi : X −→ Yi

}i∈I (where

{(Yi, τi)

}i∈I

is a family of topological spaces) and {xα}α∈A ⊆ X is a net,then xα −→ x in X if and only if fi(xα) −→ fi(x) for all i ∈ I.

Example 2.67(a) Let Y be a topological space, X ⊆ Y and consider the identityfunction iX : X −→ Y (i.e., iX(x) = x for all x ∈ X). Then the weaktopology w(iX ) is nothing else but the subspace topology on X (seeDefinition 2.14). So, we see that the subspace topology is a particularcase of a weak (initial) topology.(b) Let V be a set and let X be a separating family of functionsf : V −→ R. For every v ∈ V , we define eV : X −→ R by

eV (f) = f(v) ∀ f ∈ X .

Then the family {ev}v∈V induces a weak topology on X which is Haus-dorff. Moreover fα −→ f in X with this weak topology if and onlyif fα(v) −→ f(v) for all v ∈ V (i.e., weak convergence coincides withpointwise convergence).

Proposition 2.68If X is a set,

{fi : X −→ R

}i∈I is a separating family of functions

and C ⊆ X,

then w({fi}i∈I

)∣∣∣∣C

= w({fi|C}i∈I

).

The notion of weak or initial topology leads easily to the notion ofproduct topology.

Definition 2.69Let

{(Xi, τi)

}i∈I be a family (finite or infinite) of topological spaces

and let

X =∏

i∈IXi = XI .


We consider the canonical projections pi : X −→ Xi for i ∈ I, definedby

pi((xj)j∈I

)= xi.

Then the weak topology w({pi}i∈I

)on X is called the product topol-

ogy on X. The space(X,w

({pi}i∈I))

is called the product space ofthe family {Xi}i∈I .

Remark 2.70Based on Definition 2.62, we see that a basis for the product topology(obtained by taking finite intersections of subbasis elements) is thecollection of all sets of the form

∏i∈I

Ui which satisfy two conditions:

(a) each Ui is open in Xi;(b) only for a finite number of indices i ∈ I, we have Ui �= Xi.

When I is infinite, if requirement (b) is not satisfied and we permitin the collection set

∏i∈I

Ui, where Ui �= Xi for an infinite number of

indices, we get a stronger topology, known as the box topology . Inwhat follows on product sets

∏i∈I

Xi we will always consider the product

topology.

Proposition 2.71

(a)∏i∈I

Xi is Hausdorff (respectively, regular; see Definition 2.4) if and

only if each Xi if Hausdorff (respectively, regular).(b) The product

∏i∈I

Xi of normal spaces Xi need not be normal.

(c) The product∏n�1

Xn is second countable (see Definition 2.24) if

and only if each Xn is second countable.(d) The product

∏n�1

Xn is separable if and only if each Xn is separable.

(e) The product∏i∈I

Xn of Lindelof spaces need not be Lindelof (see

Definition 2.26).

Next we consider the opposite case from the one that led to theweak (initial) topology. So, now we are given a family of functions{fi : Xi −→ Y

}i∈I , with {Xi}i∈I being a family of topological spaces

and we are looking for the strongest topology which can be placed onY for which all functions {fi}i∈I are continuous.


Definition 2.72Suppose that

{(Xi, τi)

}i∈I is a family of topological spaces, Y is a

set and{fi : Xi −→ Y

}i∈I is a family of functions. The strongest

topology on Y which makes all functions {fi}i∈I continuous, is calledthe strong topology (or final topology) on X. It is given by

{U ∈ Y : f−1

i (U) ∈ τi for all i ∈ I}

and it is denoted by s({fi}i∈I

).

Proposition 2.73If Y is endowed with the strong topology s

({fi}i∈I)determined by a

family of functions{fi : Xi −→ Y

}i∈I , V is a topological space and

g : Y −→ V ,then g is continuous if and only if for every i ∈ I, g ◦ fi : Xi −→ V iscontinuous.

We concentrate our attention on the case of a single functionf : X −→ Y . Let Y have the strong topology s(f). Let Y0 = Y \f(X).If y ∈ Y0, then f

−1(y) = ∅ and so {y} is clopen in Y . Also, f(X) isclosed in Y . Therefore Y is the disjoint union of f(X) with a discretespace. Hence we may assume that f is surjective.

Definition 2.74Let X,Y be two topological spaces and let f : X −→ Y be a function.We say that f is an identification function if the following tworequirements are satisfied:(a) f is a surjective; and(b) Y is endowed with the strong topology s(f).Then s(f) is called the identification topology with respect to f andY is an identification space of f . A set A ⊆ X is said to be f-saturated if f−1

(f(A)

)= A.

Proposition 2.75If f : X −→ Y is an identification function,then

s(f) ={f(V ) : for all V ⊆ X open, f -saturated

}

={f(C) : for all C ⊆ X closed, f -saturated

}.

A special case of the identification topology is the quotienttopology.


Definition 2.76Let X,Y be topological spaces and let ∼ be an equivalence relationon X. Let p : X −→ X/∼ be the quotient map. Then

s(p) ={U ⊆ X/∼ : p−1(U) is open in X

}

is called the quotient topology on X/∼ and(X/∼, s(p)

)is called the

quotient space of X by ∼.

Remark 2.77Quotients of Hausdorff spaces need not be Hausdorff. However, ifGr ∼⊆ X ×X is closed, then X/∼ is Hausdorff.

2.1.7 Compact Topological Spaces

One of the most important topological properties is compactness.

Definition 2.78Let (X, τ) be a topological space.(a) We say that X is compact if every open cover of X admits afinite subcover (see Definition 2.26).(b) A set C ⊆ X is compact if every open cover of C in X (i.e., afamily {Ui}i∈I ⊆ τ such that C ⊆ ⋃

i∈IUi) admits a finite subcover.

Proposition 2.79If X is a topological space and C ⊆ X,then C is a compact subset of X if and only if C with the subspacetopology is a compact topological space.

Definition 2.80Let X be a set and let D ⊆ 2X \ {∅}. We say that D has the finiteintersection property if for every nonempty finite subset F ⊆ D,we have

⋂A∈F

A �= ∅.

The next theorem provides important alternative characterizationof compactness.

Theorem 2.81Let X be a topological space. The following statements are equiva-lent:


(a) X is compact;(b) Every nonempty family of closed subsets of X with the finite in-tersection property has a nonempty intersection;(c) Every net in X has a convergent subnet.

Proposition 2.82If X,Y are two topological spaces, f : X −→ Y is a continuous func-tion and C ⊆ X is a compact set,then f(C) ⊆ Y is compact.

In the next proposition we have gathered some properties of com-pact spaces which are useful in Analysis.

Proposition 2.83

(a) Every closed subset of a compact set is compact.(b) Every compact set in a topological space is closed.(c) If C1 and C2 are disjoint compact sets in a topological space X,then there exist disjoint open sets U1 and U2 such that C1 ⊆ U1 andC2 ⊆ U2.(d) A compact topological space is normal (see (c)).(e) A quotient space of a compact space is compact.

Theorem 2.84If X,Y are two topological spaces, X is compact and f : X −→ Y isa continuous bijection,then f is homeomorphism.

Theorem 2.85 (Heine–Borel Theorem)A set C ⊆ R

N is compact if and only if C is closed and bounded.

Theorem 2.86If X is a compact topological space and f : X −→ R = R ∪ {+∞} is

lower semicontinuous (respectively, f : X −→ R∗= R∪{−∞} is upper

semicontinuous),then f attains its infimum (respectively, supremum).

Corollary 2.87If X is a compact topological space and f : X −→ R is continuous,then f attains both its infimum and supremum.


Definition 2.88Let X be a topological space and let C ⊆ X.(a) We say that C is sequentially compact if every sequence in Chas a subsequence which converges to a point in C.(b) We say that C is countably compact (or Frechet compact ora set with Bolzano–Weierstrass property) if every infinite setD ⊆ C has a limit point in C.

Remark 2.89From Theorem 1.71, we know that in metrizable spaces the above twonotions coincide with compactness. For general topological spaces thisis no longer true.

Proposition 2.90

(a) A compact topological space is countably compact, but the converseis not in general true.(b) A sequentially compact topological space is countably compact, butthe converse is not in general true.(c) A first countable space is sequentially compact if and only if it iscountably compact.(d) A first countable compact space is sequentially compact.(e) A topological space is countably compact if and only if every count-able cover has a finite subcover.(f) A topological space is countably compact if and only if every se-quence has a limit point.

Theorem 2.91 (Tichonov Theorem)If {Xi}i∈I is a family of topological spaces,then X =

∏i∈I

Xi with the product topology (see Definition 2.69) is

compact if and only if for each i ∈ I, Xi is compact.

Definition 2.92A topological space X is locally compact if for every point x ∈ Xthere exists U ∈ N (x) such that U is compact.

Remark 2.93Compact spaces are locally compact, but the converse is not true (con-sider X = R

N , N � 1). Also discrete spaces (see Remark 2.2) are


locally compact and as will see in Chap. 5, a normed space is locallycompact if and only if it is finite dimensional.

Proposition 2.94If (X, τ) is a locally compact topological space, C ⊆ X is a nonemptycompact subset and U ∈ τ is such that C ⊆ U ,then we can find V ∈ τ which is relatively compact (i.e., V is com-pact in X) and C ⊆ V ⊆ V ⊆ U .

Corollary 2.95X is a locally compact topological space if and only if every x ∈ X hasa local basis consisting of relatively compact open sets.

Definition 2.96Let X be a topological space. A compact topological space Y is saidto be a compactification of X, if X is homeomorphic to a densesubspace of Y .

Remark 2.97Such a compactification always exists. The simplest example is theso called Alexandrov one-point compactification. Let (X, τ) be a non-compact topological space and define a new set X∗ by X∗ = X ∪{∞},where ∞ is not already an element of X. We define

τ∗ = τ ∪ {Kc ∪ {∞} : K ⊆ X compact}.

Then τ∗ is a topology on X∗ and (X∗, τ∗) is the Alexandrov one-point compactification of X. Since for us only Hausdorff spacesmatter, the following theorem is important.

Theorem 2.98If (X, τ) is a topological space and (X∗, τ∗) is its Alexandrov one-pointcompactification,then (X∗, τ∗) is Hausdorff if and only if X is locally compact.

Definition 2.99A topological space X is said to be σ-compact if there is a sequence{Cn}n�1 of compact subsets of X such that

X =⋃

n�1

Cn.


Proposition 2.100If X is a locally compact, σ-compact topological space,then there exists an increasing sequence {Cn}n�1 of compact subsetsof X such that

X =⋃

n�1

Cn and Cn ⊆ intCn+1 ∀ n � 1.

Proposition 2.101If X is a locally compact topological space,then X is σ-compact if and only if it is Lindelof (see Definition 2.26).

Corollary 2.102A locally compact, σ-compact topological space is normal.

Definition 2.103Let X be a topological space and let f : X −→ R

∗ be a function. Wesay that f is coercive (respectively, sequentially coercive) on Xif for every λ ∈ R, the set {x ∈ X : f(x) � λ} is countably compact(respectively, sequentially compact; see Definition 2.88).

2.1.8 Connectedness

Connectedness is another important topological property of a space.

Definition 2.104Let X be a topological space. We say that X is disconnected if wecan find two disjoint, nonempty, open sets U, V such that X = U ∪ V(i.e., the pair {U, V } forms a partition of X). The pair {U, V } iscalled a disconnection of X. We say that the topological space X isconnected if it is not disconnected. A subset C of X is connectedif it is connected as a topological space with the subspace topology (seeDefinition 2.14).

Theorem 2.105Let X be a topological space. The following statements are equiva-lent:(a) X is connected;(b) The only clopen subsets of X are ∅ and X;(c) There is no continuous surjection f : X −→ {0, 1}.


Proposition 2.106If X is a topological space and C ⊆ X,then C is connected if and only if for every U, V open subsets of Xsuch that C ⊆ U ∪V and U ∩V = ∅, we have C ∩U = ∅ or C ∩V = ∅.

Corollary 2.107If X is a topological space and C ⊆ X is a connected set,

then every D satisfying C ⊆ D ⊆ C is connected. In particular theclosure of a connected set is connected.

Proposition 2.108If X,Y are two topological spaces, f : X −→ Y is continuous andC ⊆ X is a connected set,then f(C) is connected.

Proposition 2.109If X is a topological space and {Ci}i∈I is a family of connected subsetsof X such that there exists i0 ∈ I with Ci ∩ Ci0 �= ∅ for all i ∈ I orCi ∩ Cj �= ∅ for all i, j ∈ I,

then the set⋃

i∈ICi is connected.

Proposition 2.110If {Xi}i∈I is a family of topological spaces,then X =

∏i∈I

Xi with the product topology (see Definition 2.69) is

connected if and only if for every i ∈ I the space Xi is connected.

Every topological space can be decomposed uniquely into “con-nected components”, which are the maximal connected subsets of thespace. Of course, if X is connected, then it has just one connectedcomponent (X itself).

Definition 2.111Let X be a topological space. A connected component of X is aconnected subset of X which is not properly contained in another con-nected subset of X.

Remark 2.112Here is another way to define the connected components of X. Letx ∈ X and let C(x) be the union of all connected subsets of X which


contain x ∈ X. By Proposition 2.109, C(x) is connected and it ismaximal with respect to inclusion among all connected subsets of Xcontaining x ∈ X. This is the connected component of X containingx ∈ X.

Proposition 2.113If X is a topological space,then the connected components of X are closed and form a partitionof X.

Remark 2.114The connected components of a topological space need not be opensubsets of X. For example, let X = Q with the subspace topologyas a subset of R. Then the connected components of X = Q arethe singletons. Indeed, suppose that A ⊆ X and cardA > 1. Letx, y ∈ A with x < y. We can find u ∈ Q such that x < u < y. Then{U = A ∩ (−∞, u), V = A ∩ (u,+∞)

}is a disconnection of A ⊆ Q.

So, every subset of X = Q with more than one point is disconnected.The connected components {x}, x ∈ X = Q are closed but not open.This example leads to the following definition.

Definition 2.115A topological space X is said to be totally disconnected if its con-nected components are its singletons.

Sometimes it is more important that the space exhibits connectiv-ity “locally”. This leads to the following definition.

Definition 2.116A topological space X is said to be locally connected at x ∈ X ifevery U ∈ N (x) contains a connected open set belonging in N (x). Wesay that X is locally connected if it is locally connected at each pointx ∈ X.

Remark 2.117Another equivalent way to define local connectedness for a topologicalspace X is to say that X has a basis containing of connected sets.

Proposition 2.118A topological space X is locally connected if and only if the con-nected components of open sets are open sets (see Definitions 2.111and 2.116).


Remark 2.119A connected space need not be locally connected. To see this, letY =

{(x, y) : y = sin 1

x , x ∈ (0, 1]}and let X = Y ∪{(0, 0)}. Then X

is connected (note that Y is connected being the image of (0, 1] underthe action of the continuous function ψ(x) = sin 1

x , x ∈ (0, 1]; seeProposition 2.108, note thatX ⊆ Y and use Corollary 2.107). HoweverX is not locally connected since the connected components of X ∩{(x, y) : y < 1

2

}are not open in X; see Proposition 2.118. A discrete

space is a totally disconnected and locally connected topological space.

Let X,Y be two topological spaces and let f : X −→ Y be a home-omorphism. If C is a connected component and x ∈ C, then f(C) is aconnected component of Y containing f(x). Consequently f induces abijection from the connected components of X to the connected com-ponents of Y . Therefore the number of topological components of X isa topological invariant of X. The drawback of this topological invari-ant is that it fails to distinguish between different connected spaces.For this reason we introduce the following notion.

Definition 2.120Let X be a connected topological space and let k ∈ N ∪ {∞}. A pointx ∈ X is a cut point of order k if X \ {x} has k connected compo-nents (see Definition 2.111). The number of cut points of order k is atopological invariant of X.

Refining Definition 2.120, we introduce the following notion.

Definition 2.121Let X be a connected topological space and let x, y ∈ X. We saythat {x, y} is a cut pair of order k if X \ {x, y} has k connectedcomponents (see Definition 2.111).

In many cases in Analysis, only connected spaces of special typeare used. For example, in Global Analysis we deal with spaces whichlocally look like (i.e., are homeomorphic to) R

N . For this reason weintroduce the following notions.

Definition 2.122Let X be a topological space and x, u ∈ X.(a) A path from x to u in X is a continuous functionf : I = [0, 1] −→ X such that f(0) = x and f(1) = u. We say thatx is the initial point and u is the finale point.


(b) We say that X is path-connected (or arcwise-connected) ifevery pair of points x, u ∈ X can be connected by a path.

Proposition 2.123If X is a topological space and x0 ∈ X,then X is path-connected if and only if each x ∈ X can be connectedto x0 be a path (see Definition 2.122).

Proposition 2.124If X,Y are two topological spaces, X is path-connected (see Defini-tion 2.122) and f : X −→ Y is a continuous function,then f(X) ⊆ Y is path-connected. In particular path-connectedness isa topological invariant.

Proposition 2.125A path-connected topological space is connected (see Definitions 2.122and 2.104). The converse is not in general true.

Example 2.126Let

C ={(x, u) : u = sin 1

x , 0 < x � 1}.

Evidently X is connected being the image of the connected set (0, 1]under the continuous function ψ(x) =

(x, sin 1

x

). Hence C is connected

(see Corollary 2.107) and

C = C ∪ ({0} × [−1, 1]).

However, C is not path-connected since there is no path joining (0, 0)to(1π , 0)(see Definition 2.122).

To produce conditions which guarantee that the converse of Propo-sition 2.125 holds, we need the following analog of Definition 2.116.

Definition 2.127A topological space X is said to be locally path-connected at x ∈ X,if every U ∈ N (x) contains a path-connected open set belonging inN (x). We say that X is locally path-connected if it is locally path-connected at each point x ∈ X.


Theorem 2.128A topological space X is path-connected if and only if it is connectedand locally path-connected (see Definitions 2.122 and 2.127).

Proposition 2.129If X is a topological space and {Ai}i∈I is a family of path-connectedsubspaces of X (see Definition 2.122) such that there exists i0 ∈ I withAi ∩ Ai0 �= ∅ for all i ∈ I or Ai ∩Aj �= ∅ for all i, j ∈ I,

then⋃

i∈IAi is path-connected too.

Because of this proposition, we can define path-connected compo-nents of a topological space X, as maximal path-connected subsetsof X.

Definition 2.130Let X be a topological space. A path-connected component of X,is a path-connected subset of X which is not properly contained inanother path-connected subset of X.

Remark 2.131The path-connected components of X form a partition of X (as didthe connected components; see Remark 2.112). However, in contrastto the connected components (see Definition 2.111), path-connectedcomponents of X need not to be closed subsets of X. For example,

C ={(x, u) : u = sin 1

x , 0 < x � 1}

is a path-connected component of C (see Example 2.126), but it is notclosed.

Proposition 2.132If X is a topological space,then X is locally path-connected if and only if the path-connectedcomponents of every open set of X are open (see Definitions 2.122and 2.130).

Corollary 2.133X is locally path-connected (see Definition 2.127) if and only if its con-nected components and path-connected components coincide (see Def-initions 2.111 and 2.130) (and are both open and closed, i.e., clopen).


Corollary 2.134An open set in R

N (or in SN ={x ∈ R

N+1 : ‖x‖ = 1}) is connected

if and only if it is path-connected.

Proposition 2.135If {Xi}i∈I are topological spaces,

then X =∏

i∈IXi with the product topology (see Definition 2.69) is

path-connected if and only if for every i ∈ I, Xi is path-connected (seeDefinition 2.122).

2.1.9 Urysohn and Tietze Theorems

The next two theorems provide alternative characterization of normal-ity (see Definition 2.4).

Theorem 2.136 (Urysohn Lemma)If X is a topological space,then the following two statements are equivalent:(a) X is normal (see Definition 2.4);(b) if C and D are two disjoint closed sets in X, then there exists acontinuous function f : X −→ [0, 1] such that f |

C= 0 and f |

D= 1.

Definition 2.137We say that a topological space X is perfectly normal if for everypair of disjoint closed sets C,D ⊆ X, we can find a continuous functionf : X −→ [0, 1] such that C = f−1

({0}) and D = f−1({1}).

Theorem 2.138 (Tietze Extension Theorem)If X is a topological space,then the following two statements are equivalent:(a) X is normal (see Definition 2.4);(b) for every closed set C ⊆ X and every continuous functionf : C −→ R, we can find a continuous extension f : X −→ R of f(i.e., f |

C= f); moreover, if |f(x)| � M for all x ∈ C and some

M > 0, then we can choose f such that |f(x)| �M for all x ∈ X.

One of the main problems in point set topology is the so called“metrization problem”. Namely, for a given topological space (X, τ),when is it possible to define a metric d

Xon X such that the metric


topology induced by dX coincide with τ? In this direction we have thefollowing important result.

Theorem 2.139 (Urysohn Metrization Theorem)A second countable topological (see Definition 2.24) space is metrizableif and only if it is regular (see Definition 2.4).

In fact a second countable regular space is homeomorphic to asubset of the Hilbert cube [0, 1]N.

2.1.10 Paracompact and Baire Spaces

One of the most important generalizations of the notion of compactnessis paracompactness. Before introducing it, we need some additionalinformation about coverings.

Definition 2.140Let {Ui}i∈I and {Vj}j∈J be two coverings of a set X. We say that{Ui}i∈I is a refinement of {Vj}j∈J , if for each i ∈ I there is a j ∈ Jsuch that Ui ⊆ Vj . We often write {Ui}i∈I ≺ {Vj}j∈J to denote re-finement. A refinement of a covering may contain more sets than thegiven covering. We say that a refinement is precise if I = J andUj ⊆ Vj for all j ∈ J .

Proposition 2.141If the covering {Vj}j∈J has a locally finite refinement {Ui}i∈I (seeDefinition 2.43),then it also has a precise locally finite refinement {Uj}j∈J . Moreover,

if for each j ∈ J , the set Uj is open, then each Uj can be chosen to beopen too.

Now we can introduce the notion of paracompactness.

Definition 2.142A topological space X is paracompact if every open covering of Xhas a locally finite refinement.

Proposition 2.143If X is a regular topological space (see Definition 2.4),then X is paracompact if and only if every open covering of X has aclosed locally finite refinement.


Theorem 2.144Compact spaces and metrizable spaces are paracompact. Every para-compact space is normal (see Definition 2.4).

Proposition 2.145

(a) Paracompactness is invariant under continuous closed surjections.(b) Paracompactness is not a hereditary property; however every Fσ-subset of a paracompact space is paracompact.

One of the main reasons that paracompactness is important is thatparacompact spaces admit arbitrarily fine partitions of unity and thelatter is a major tool in modern analysis.

Definition 2.146Let X be a topological space.(a) For any function f , the support of f is the following closed set

supp f ={x ∈ X : f(x) �= 0

}.

(b) A family {ψi}i∈I of continuous functions ψi : X −→ [0, 1], i ∈ Iis a partition of unity on X if

{suppψi

}i∈I form a closed, locally

finite covering of X and∑i∈I

ψi(x) = 1 for all x ∈ X (in this sum only

finitely many terms are nonzero since{suppψi

}i∈I is locally finite).

(c) If {Ui}i∈I is an open covering of X, then we say that the partitionof unity {ψi}i∈I is subordinated to {Ui}i∈I , if suppψi ⊆ Ui for everyi ∈ I.

Theorem 2.147If X is a paracompact space and {Ui}i∈I is an open covering of X,then there is a partition of unity {ψi}i∈I subordinated to {Ui}i∈I .

Definition 2.148A topological space X is said to be a Baire space if every nontrivialcountable intersection of dense open sets in X is dense too.

From the Baire category theorem (Theorem 1.26), we know thatcomplete metric spaces are Baire spaces. Next we add one more classof spaces to Baire spaces.

Theorem 2.149Locally compact spaces are Baire spaces.


2.1.11 Polish and Souslin Sets

Next we pass to a generalization of separable metric spaces.

Definition 2.150A topological space X is a Polish space if it is separable and metriz-able by means of a complete metric space.

Remark 2.151There are many topological spaces which are Polish, but have no com-plete metric which is particularly natural or simple. However, manyconstructions and results in Analysis depend on the existence of acomplete metric but not on a specific choice of it. So, working withPolish spaces instead of separable complete metric spaces provides amore general framework of Analysis.

Proposition 2.152Every closed or open subset of a Polish space is Polish.

Proposition 2.153

(a) Any finite or countable product of Polish spaces with the producttopology (see Definition 2.69) is Polish.(b) Countable intersection of Polish spaces is Polish.(c) A locally compact, σ-compact metrizable space is Polish.(d) The set of irrationals R \Q with the subspace topology induced byR is a Polish space.

Theorem 2.154

(a) A subspace of a Polish space X is Polish if and only if it is Gδin X.(b) A topological space X is Polish if and only if it is homeomorphicto a Gδ-subset of a Hilbert cube [0, 1]N.

Proposition 2.155Every Polish space X is a continuous image of N∞.


Definition 2.156A topological space (X, τ) is said to be Souslin space if there exists aPolish space Y and a continuous surjection from Y to X. Equivalently,we can say that (X, τ) is a Souslin space, if there exists a topology τon X stronger than τ such that (X, τ ) is homeomorphic to a quotientof a Polish space.

Remark 2.157A Souslin space is separable and Lindelof (see Definitions 2.156and 2.26) but need not be metrizable (for example an infinite dimen-sional separable Banach space with the weak topology; see Chap. 5).A Souslin subset of a Polish space is called in the literature an analyticset .

Proposition 2.158If X is a Souslin space (see Definition 2.156) and C ⊆ X is nonempty,then C has a countable subset D which is sequentially dense in C, i.e.,every x ∈ C is the limit of a sequence in D.

Proposition 2.159

(a) Closed and open subsets of a Souslin space are Souslin spaces (seeDefinition 2.156).(b) Countable products of Souslin spaces with the product topology (seeDefinition 2.69), are Souslin spaces.(c) In a topological space, countable intersections or unions of Souslinsubspaces are Souslin spaces.

Theorem 2.160If X is a locally compact topological space,then the following statements are equivalent:(a) X is Polish (see Definition 2.150);(b) X is Souslin (see Definition 2.156);(c) X is second countable (see Definition 2.24);(d) X is separable metrizable (see Definition 2.9);(e) the Alexandrov one-point compactification X∗ of X has any of theabove properties (see Remark 2.97).


Definition 2.161

(a) A topological space X is said to be dispersed if it is Polish and

every point has a local basis consisting of clopen sets.(b) A topological space is said to be dispersible if it is the continuousimage of a dispersed space.

Proposition 2.162

(a) Countable products of dispersible spaces are dispersible.

(b) Countable intersections of dispersible spaces are dispersible.(c) A subspace of a dispersible space is dispersible if and only if it isa Gδ-set. Also disjoint unions of dispersible spaces are dispersible.

We introduce the following stronger version of a Lindelof space (seeDefinition 2.26).

Definition 2.163A topological space X is said to be strongly Lindelof if every opencover of any open set of X has a countable subcover.

Proposition 2.164

(a) Arbitrary intersections and countable unions of strongly Lindelofspaces are strongly Lindelof.(b) A subspace of a strongly Lindelof space is strongly Lindelof.(c) If (X, τ) is strongly Lindelof, then for any topology τ ′ ⊆ τ , wehave that (X, τ ′) is strongly Lindelof.(d) A second countable topological space (see Definition 2.24) isstrongly Lindelof.(e) The continuous image of a strongly Lindelof space is stronglyLindelof.

Corollary 2.165Every Souslin space is strongly Lindelof.

Remark 2.166The cartesian product of two strongly Lindelof spaces need not bestrongly Lindelof. However, if {Xn}n�1 are second countable (seeDefinition 2.24) or Souslin (see Definition 2.156), then so is

∏Xn

and therefore it is strongly Lindelof (see Proposition 2.164(d) andCorollary 2.165).


Definition 2.167We say that a topological space (X, τ) is a Lusin space if there existsa Polish space Y and a continuous bijection from Y to X.

Remark 2.168The above definition of a Lusin space (X, τ) is equivalent to sayingthat there is a topology τ∗ on X stronger than τ such that (X, τ∗) isa Polish space.

2.1.12 Michael Selection Theorem

In this section, we introduce some continuity notions from the theory ofmultifunctions (set valued functions) and state the celebrated selectiontheorem of Michael.

Definition 2.169Let X,Y be two topological spaces and let F : X −→ 2Y be a multi-function.(a) We say that F is upper semicontinuous at x0 ∈ X (usc at x0,for short), if for any open set V ⊆ Y such that F (x0) ⊆ V , we canfind U ∈ N (x0) such that F (U) ⊆ V . If this is true at every x0 ∈ X,then we say that F is upper semicontinuous (usc, for short).(b) We say that F is lower semicontinuous at x0 ∈ X (lsc at x0,for short), if for any open set V ⊆ Y such that F (x0)∩ V �= ∅, we canfind U ∈ N (x0) such that F (x) ∩ V �= ∅ for all x ∈ U . If this is trueat every x0 ∈ X, then we say that F is lower semicontinuous (lsc,for short). In fact in the above definition V can be taken to be a basisopen set.(c) We say that F is continuous at x0 ∈ X (or Vietoris con-tinuous at x0 ∈ X), if it is both upper semicontinuous and lowersemicontinuous at x0. If this is true at every x0 ∈ X, then we say thatF is continuous (or Vietoris continuous).

Theorem 2.170 (Michael Selection Theorem)If X is paracompact (see Definition 2.142), Y is a Banach space

and F : X −→ 2Y is a lower semicontinuous multifunction with valueswhich are nonempty, closed and convex subsets in Y (a set C ⊆ Y isconvex if for every u, v ∈ C and t ∈ [0, 1], we have tu+ (1− t)v ∈ C;see Chap. 5),


then there exist a continuous function f : X −→ Y such that

f(x) ∈ F (x) ∀ x ∈ X

(we call f , a continuous selector of F ).

Theorem 2.171If X is a metric space, Y is a Banach space and F : X −→ 2Y is alower semicontinuous multifunction with nonempty, closed and convexvalues,then there exists a sequence

{fn : X −→ Y

}n�1

of continuous selectors

of F such that F (x) ={fn(x)

}n�1

for all x ∈ X.

Theorem 2.172If X is a metric space, Y is a Banach space and F : X −→ 2Y isa lower semicontinuous multifunction with nonempty, convex valuessuch that intF (x) �= ∅ for all x ∈ X,then F admits a continuous selector.

Definition 2.173Let X and Y be two topological spaces and let F : X −→ 2Y be amultifunction. The graph of F is the set GrF ⊆ X × Y , defined by

GrF ={(x, y) ∈ X × Y : y ∈ F (x)

}.

2.1.13 The Space C(X; Y )

Next we will topologize the space of continuous functions between twotopological spaces using a topology that corresponds to the concept ofuniform convergence on compacta.

Definition 2.174Let X and Y be two topological spaces and let C(X;Y ) denote thespace of all continuous functions from X into Y . The compact-opentopology on C(X;Y ) is the one having as subbasis all sets of the form:

(K,V ) ={f ∈ C(X;Y ) : f(K) ⊆ V

}

with any compact K ⊆ X, and open V ⊆ Y . In what follows thecompact-open topology is abbreviated as the c-topology and is denotedby c.


Proposition 2.175

(a) C(X;Y ) with the c-topology is Hausdorff (in fact(C(X;Y ), c

)is

Hausdorff if and only if Y is Hausdorff).(b) C(X;Y ) with the c-topology is regular (see Definition 2.4) if andonly if Y is regular.

Definition 2.176Let X,Y,Z be three topological spaces. We introduce the following twofunctions.(a) The composition function τ : C(X;Y )×C(Y ;Z) −→ C(X;Z),defined by

τ(f, g) = g ◦ f ∀ f ∈ C(X;Y ), g ∈ C(Y ;Z).

(b) The evaluation function e : C(X,Y )×X −→ Y , defined by

e(f, x) = f(x) ∀ f ∈ C(X;Y ), x ∈ X.

Theorem 2.177

(a) The composition function τ : C(X;Y ) × C(Y ;Z) −→ C(X;Z) iscontinuous on each argument separately, when C(X;Y ), C(Y ;Z) andC(X;Z) are endowed with the respective c-topology.(b) If Y is locally compact, then τ(·, ·) is jointly continuous, whenC(X;Y ), C(Y ;Z) and C(X;Z) are endowed with the respectivec-topology.

Theorem 2.178

(a) For fixed x0 ∈ X, the function ex0 : C(X;Y ) −→ Y , defined by

ex0(f) = e(f, x0) = f(x0)

is continuous, when C(X;Y ) is endowed with the c-topology.(b) If Y is additionally locally compact, then the evaluation functione : C(X;Y )×X −→ Y is jointly continuous, when C(X;Y ) is endowedwith the c-topology.


Theorem 2.179

(a) If f : X×Y −→ Z is continuous and f : X −→ C(Y ;Z) is definedby

f(x)(·) = f(x, ·),then f is continuous, when C(Y ;Z) is endowed with the c-topology.(b) If f : X −→ C(Y ;Z) is continuous, when C(Y ;Z) is endowedwith the c-topology and Y is locally compact, then f : X × Y −→ Z isjointly continuous.

Assuming that Y is a metric space, we will identify the compactsubsets of

(C(X;Y ), c

)(the Arzela–Ascoli theorem). We start with a

definition.

Definition 2.180Let X be a topological space and let (Y, d) be a metric space. A subsetA ⊆ C(X;Y ) is said to be equicontinuous at x0 ∈ X, if for everyε > 0, we can find U ∈ N (x0) such that f(U) ⊆ Bε

(f(x0)

)for

all f ∈ A. The set U ∈ N (x0) is called a neighbourhood of ε-continuity for A. We say that A is equicontinuous if it is equicon-tinuous at every x0 ∈ X.

Theorem 2.181 (Arzela–Ascoli Theorem)If X is a topological space, (Y, d) is a metric space, A ⊆ C(X;Y )and(i) A is equicontinuous;(ii) for every x ∈ X, ex(A) ⊆ Y is compact (i.e.,

{f(x) : f ∈ A

}is

relatively compact in Y ),then A

c(c denotes the c-topology on C(X;Y )) is compact and equicon-

tinuous.

Next we have a closer look at the sequential convergence in thec-topology.

Definition 2.182Let X be a topological space and let (Y, d

Y) be a metric space. A

sequence {fn}n�1 ⊆ C(X;Y ) is said to converge to f ∈ C(X;Y ) uni-formly on compacta if for every compact set K ⊆ X and everyε > 0, we can find n0 = n0(K, ε) � 1 such that

dY

(fn(x), f(x)

)< ε ∀ n � n0, x ∈ K.


Theorem 2.183If X is a topological space, (Y, d

Y) is a metric space, {fn}n�1 ⊆

C(X;Y ) and f ∈ C(X;Y ),then

[fn −→ f uniformly on compacta

] ⇐⇒ [fn

c−→ f in C(X;Y )].

Definition 2.184Let X,Y be two topological space, {fn}n�1 ⊆ C(X;Y ) and f ∈C(X;Y ). We say that the sequence {fn}n�1 converges continu-ously to f , if for every x ∈ X and every sequence {xn}n�1 ⊆ X suchthat xn −→ x ∈ X, we have fn(xn) −→ f(x) in Y .

Theorem 2.185If X is first countable, (Y, dY ) is a metric space, {fn}n�1 ⊆ C(X;Y )is a sequence and f ∈ C(X;Y ),then

[fn −→ f continuously

] ⇐⇒ [fn −→ f uniformly on compacta

].

2.1.14 Elements of Algebraic Topology I: Homotopy

Themain problem of topology is the classification of topological spaces.Given two topological spaces X and Y , are they homeomorphic? Ingeneral, this is a very difficult question to answer, without the use oftools from other fields such as algebra. This then leads us to the realmof “algebraic topology”. In the rest of this chapter we present someintroductory notions and results from this field, which can be usefulto the analyst.

Definition 2.186

(a) A topological pair (X,C) consists of a topological space X and asubspace C ⊆ X. If C = ∅, then the pair (X, ∅) stands for the space X.A subpair (X ′, C ′) ⊆ (X,C) is a topological pair such that X ′ ⊆ Xand C ′ ⊆ C. A function f : (X,C) −→ (Y,D) between topologicalpairs, is a continuous function f : X −→ Y such that f(C) ⊆ D. IfC = {x} with x ∈ X, then the topological pair (X,C) is said to be apointed topological space and is denoted by (X,x).


(b) For a topological pair (X,C) and I = [0, 1], we set

I × (X,C) =(I ×X, I × C

).

Let V ⊆ X and suppose that f0, f1 : (X,C) −→ (Y,D) are functionsbetween topological pairs such that f0|V = f1|V . We say that f0 is

homotopic to f1 relative to V (denoted by f0V� f1), if there exists

a continuous function h : I × (X,C) −→ (Y,D) such that

h(0, x) = f0(x) ∀ x ∈ X,

h(1, x) = f1(x) ∀ x ∈ X,

h(t, x) = f0(x) = f1(x) ∀ t ∈ I, x ∈ V.

The function h is a homotopy relative to V from f0 to f1. If V = ∅,then we omit the expression relative to V and write � instead of

V�.(c) A continuous function f : X −→ Y is said to be nullhomotopicif it is homotopic to a constant function. In this case we write f � 0.(d) Two paths f, g : [0, 1] −→ X are said to be path homotopic iff(0) = g(0) = x0, f(1) = g(1) = x1 (i.e., the two paths have the sameinitial and finale points) and there exists a continuous map h : [0, 1] ×[0, 1] −→ X such that

h(s, 0) = f(s), h(s, 1) = g(s) s ∈ [0, 1]

h(0, t) = x0, h(1, t) = x1 t ∈ [0, 1].

We call h a path homotopy between f and g and we write f �p g.

Proposition 2.187

If W ⊆ V ⊆ X and f0V� f1,

then f0W� f1.

Proposition 2.188The homotopy relation is an equivalence relation.

Definition 2.189For a continuous function f : X −→ Y the equivalence class under �is denoted by [f ] and is called the homotopy class of f .

The next proposition shows that the equivalence relation � (seeProposition 2.188) is compatible with composition.


Proposition 2.190If X,Y,Z are topological spaces and f, f : X −→ Y , g, g : Y −→ Z are

continuous functions such that f � f and g � g,then g ◦ f � g ◦ f .Therefore, we can define the composition of homotopy classes, by

[g] ◦ [f ] = [g ◦ f ].

Recall that two topological spaces X,Y are said to be homeomor-phic, if there exist continuous functions f : X −→ Y and g : Y −→ Xsuch that g ◦ f = idX and f ◦ g = idY . We can extend this notion byreplacing “=” by “�”. This leads to the following definition extendingthe idea of homeomorphic spaces.

Definition 2.191Two topological pairs (X,C) and (Y,D) are said to be homo-topy equivalent if there exist functions f : (X,C) −→ (Y,D) andg : (Y,D) −→ (X,C) such that g◦f � id

(X,C)(i.e., the homotopy keeps

the image of C in C) and f ◦ g � id(Y,D)

(i.e., the homotopy keeps theimage of D in D). We write (X,C) � (Y,D). If C = D = ∅, theng ◦ f � id

X, f ◦ g � id

Yand we write X � Y . Clearly homotopy

equivalence between topological spaces is an equivalence relation. Theequivalence classes are called homotopy types.

The simplest topological spaces are the singletons. We characterizethe homotopy type of a singleton as follows.

Definition 2.192A topological space X is said to be contractible if it is homotopyequivalent to a singleton. This is equivalent to saying that the identityfunction on X is nullhomotopic (i.e., id

X� 0). Another equivalent

definition is that there exists a continuous function h : I × X −→ Xsuch that

h(0, ·) = idX , and h(1, x) = u0 for some u0 ∈ X and all x ∈ X.

Remark 2.193In other words contractible is a topological space which can be con-tinuously shrunk to a point. Some obvious examples of contractiblespaces are convex sets in R

N and more generally, any star-shaped set


C ⊆ RN . This means that there is a point u0 ∈ C such that for every

x ∈ C, the line segment

[u0, x]def={u = (1− t)u0 + tx : 0 � t � 1

}

is contained in C.

Proposition 2.194A topological space X is contractible (see Definition 2.192) if andonly if for any topological space Y any two continuous functionsf, g : Y −→ X are homotopic.

The next theorem establishes an important relation between null-homotopy and extendability of the special class of functions definedon spheres.

Theorem 2.195If SN =

{x ∈ R

N+1 : ‖x‖ = 1}, u0 ∈ SN , Y is a topological space

and f : SN −→ Y is a continuous function,then the following statements are equivalent:(a) f � 0;

(b) f{u0}� 0;

(c) there exists a continuous function

f : BN+11 =

{x ∈ R

N+1 : ‖x‖ � 1} −→ X

such that f |SN

= f .

Definition 2.196Let X be a topological space and let C ⊆ X be a subset.(a) We say that C is a retract of X if there is a continuous func-tion r : X −→ C such that r|C = id

C. The function r is called a

retraction.(b) We say that C is a deformation retract of X if there exists aretraction r : X −→ C which is homotopic (as a function into X) tothe identity function id

X. Therefore C is a deformation retract of X,

if there exists a continuous function h : I ×X −→ X such that

h(0, ·) = idX, h(1, x) ∈ C ∀ x ∈ X and h(1, ·)|

C= id

C.

This homotopy h(t, x) is called deformation retraction.


(c) We say that C is a strong deformation retract of X if thereexists a retraction r : X −→ C which is homotopic (as a function intoX) relative to C to the identity function id

X. Therefore C is a strong

deformation retract of X, if there exists a continuous function h : I ×X −→ X such that

h(0, ·) = idX,

h(1, x) ∈ C ∀ x ∈ X,

h(t, ·)|C= id

C∀ t ∈ [0, 1].

This homotopy h(t, x) is called strong deformation retraction.

Proposition 2.197A topological space X is contractible if and only if any singleton inX is a deformation retract of X (see Definitions 2.192 and 2.196).Hence a contractible space is path-connected (see Definition 2.122).

Theorem 2.198If X is a topological space and C is a deformation retract of X (seeDefinition 2.196),then C and X are homotopy equivalent (i.e., of the same homotopytype).

Next we will briefly discuss the fundamental group, an algebraicgroup, which in some sense measures the number of “holes” in a space.Consider for example the plane and the punctured plane, which areboth path-connected. Using the standard topological properties, wecannot distinguish between the two and this may suggest that theyare homeomorphic. Yet intuition suggests that due to the “hole” inthe punctured plane, the two cannot be homeomorphic. To see thatthis can be detected topologically, observe that every closed curve inR2 can be continuously shrunk to a point, while by contrast it is clear

that this cannot be done in the punctured plane, without leaving it.This observation is formalized by using the fundamental group.

Definition 2.199Let X be a topological space and let x0 ∈ X. We say that a con-tinuous function f : I = [0, 1] −→ X is a loop based at x0 iff(0) = f(1) = x0. Two loops f, g : I −→ X are said to be homotopic


if f{0,1}� g. The homotopy relation between loops is an equivalence re-

lation and it partitions the set of all loops based at x0 into equivalenceclasses. This set of equivalence classes is denoted by π1(X,x0) and iff is a loop based at x0, then [f ] denotes its equivalence class.

On π1(X,x0) we can define an operation “�” as follows. For anytwo loops f, g based at x0, we set

(f � g)(t) =

{f(2t) if t ∈ ∣∣0, 12

∣∣,g(2t− 1) if t ∈ ∣∣12 , 1

∣∣.

Evidently (f � g)(·) is again a loop based at x0. We can lift thisoperation to the set π1(X,x0) by setting

[f ] � [g] = [f � g].

This operation “�” is well defined on π1(X,x0), namely, if f{0,1}� g

and f{0,1}� g, then f � f

{0,1}� g � g.

Theorem 2.200(π1(X,x0), �

)is a group (not necessarily abelian).

Definition 2.201The group π1(X,x0) is called the fundamental group of X at x0.

Remark 2.202In general the dependence of the fundamental group on the base pointcannot be omitted. However, for an important special class of spacesit can be ignored.

Theorem 2.203If X is path-connected (see Definition 2.122) and x0, x1 ∈ X,then π1(X,x0) and π1(X,x1) are isomorphic.

Remark 2.204So, for path-connected topological spaces we can speak of the funda-mental group π1(X) without any reference to a base point.

The power of the fundamental group lies on the fact that contin-uous functions between pointed topological spaces induce homomor-phisms between the corresponding algebraic structures.


Theorem 2.205Every continuous function h : (X,x0) −→ (Y, y0) induces a homomor-phism h∗ : π1(X,x0) −→ π1(Y, y0), defined by

h∗([f ])def= [h ◦ f ].

The function h∗ is called the homomorphism induced by h.

The induced homomorphism introduced above has two “functionalproperties”, which are crucial in applications.

Proposition 2.206

(a) If h : (X,x0) −→ (Y, y0) and g : (Y, y0) −→ (Z, z0) are continuousfunctions,then (g ◦ h)∗ = g∗ ◦ h∗.

(b) If i : (X,x0) −→ (X,x0) is the identity function,then i∗ : π1(X,x0) −→ π1(X,x0) is the identity homomorphism.

Corollary 2.207If h : (X,x0) −→ (Y, y0) is a homotopy equivalence,then h∗ : π1(X,x0) −→ π1(Y, y0) is an isomorphism.

Definition 2.208A topological space S is simply connected if it is path-connected (seeDefinition 2.122) and π1(X) is trivial.

Theorem 2.209A contractible topological space is simply connected (recall that a con-tractible space is automatically path-connected; see Proposition 2.197).

Proposition 2.210If h, g : (X,x0) −→ (Y, y0) are homotopic,then h∗ = g∗.

Next let us have an alternative description of the fundamentalgroup which motivates the introduction of higher order homotopygroups. Consider the loop ε : I = [0, 1] −→ S1 =

{x ∈ R

2 : ‖x‖ = 1},

defined by

ε(t)def= e2πit ∀ t ∈ I.


This function, known as the exponential function , wraps the inter-val once around the circle counterclockwise and functions 0 and 1 tothe base point 1 of S1. From Problem 2.51, we know that ε is an iden-tification function (see Definition 2.74). If X is a topological space andf : I = [0, 1] −→ X is a loop, then there exists a unique continuousfunction f : S1 −→ X such that

f ◦ ε = f,

i.e., we have the commutative diagram

I

S1 X

εf

f

Definition 2.211We call f the circle representative of the loop f .

The next proposition gives a convenient criterion for detectingnullhomotopic loops using their circle representatives (see also Theo-rem 2.195).

Proposition 2.212A loop in a topological space X is nullhomotopic if and only if itscircle representative f : S1 −→ X extends to a continuous functionf0 : B1 −→ X, where

B1def={x ∈ R

2 : ‖x‖ � 1}

and f = f0∣∣S1.

Here by identifying loops with their circle representatives, we canview the fundamental group π1(X,x) as the set of equivalence classesof functions from S1 into X sending 1 to x modulo homotopy relativeto the base point 1. Generalizing this, we have the following definition.

Definition 2.213For every integer n � 1, we define πn(X,x) to be the set of all equiva-lence classes of continuous functions from Sn =

{x ∈ R

n+1 : ‖x‖ = 1}


into X sending (1, 0, . . . , 0) to x modulo homotopy relative to the basepoint (1, 0, . . . , 0). For given such elements f, g, we define f#g, by

f#g(x1, . . . , xn) =

{g(x1, . . . , xn−1, 2xn) if xn � 1

2 ,

f(x1, . . . , xn−1, 2xn − 1) if xn � 12 .

Using #, we introduce the operation “+” on πn(X,x), defined by

[f ] + [g] = [f#g].

We have that(πn(X,x),+

)is a group, which is abelian if n � 2.

We call(πn(X,x),+

)the nth homotopy group of the pointed space

(X,x).

Remark 2.214For every n � 1, πn(X,x) is a topological invariant. The higher ho-motopy groups (n > 1) are notoriously difficult to compute. If X,Yare pointed topological spaces, then

πn(X × Y ) = πn(X)× πn(Y ) ∀ n � 1.

Theorem 2.215

(a) π1(S1) � Z.

(b) If n � 2, then πk(Sn) = 0 for all k < n and πn(S

n) � Z. So, forn � 2, Sn is simply connected (see Definition 2.208).

Remark 2.216However, if k > n, then πk(S

n) can be nontrivial. For example, it canbe shown that π3(S

2) � Z.

Definition 2.217Let X and X be two topological spaces and let p : X −→ X be a con-tinuous function. We say that a set U ⊆ X is evenly covered by pif U is open, connected and each component of p−1(U) is an open setwhich is mapped homeomorphically onto U .

The next definition introduces an important tool in computingcritical groups.


Definition 2.218Let X and X be two topological spaces with X being path-connectedand locally path-connected (see Definitions 2.122 and 2.127) and letp : X −→ X be a continuous surjection. If every x ∈ X has a neigh-bourhood U which is evenly covered by p, then p is called a coveringfunction and X is a covering space of X.

Example 2.219

(a) The exponential function ε : R −→ S1, defined by ε(x) = e2πix,i.e.,

ε(x) =(cos 2πx, sin 2πx

) ∀ x ∈ R

is a covering function.(b) The function pn : S

1 −→ S1, given by

pn(z) = zn ∀ z ∈ S1

(here we consider S1 as a subset of C with |z| = 1, z ∈ C), is a coveringfunction for every n � 1.(c) Consider the n-torus T n = S1 × . . .× S1

︸︷︷︸n−times

(if n = 2 we have the

usual torus T = S1×S1, i.e., the doughnut) and let εn : Rn −→ T n be

the function, defined by

εn(x) =(ε(xk)

)nk=1

∀ x = (xk)nk=1 ∈ R

n.

(d) Let Pn be the n-dimensional projective plane (this is the set of

one-dimensional linear subspaces (lines through the origin) in Rn+1).

There is a natural function p : Rn+1 \ {0} −→ Pn, defined by sending

a point x ∈ Rn+1 \ {0} to its span. We topologize P

n be giving it theidentification (quotient) topology (see Definition 2.74) with respect top (an alternative way to define Pn is to say that it is the space obtainedfrom Sn =

{x ∈ R

n+1 : ‖x‖ = 1}by identifying antipodal points, i.e.,

the points where every line through the origin intersects Sn). Considerthe function p : Sn −→ P

n (n � 1) which sends each x ∈ Sn to the linethrough the origin and x, thought as an element of Pn. Then p is acovering function.

Definition 2.220Let X,X and Y be three topological spaces, let p : X −→ X be a cov-ering function and let f : Y −→ X be a continuous function. A lift


of f is a continuous function f : Y −→ X such that p ◦ f = f . So wehave the following commutative diagram

X

Y X

pf

f

The next theorem establishes a crucial link between covering spacesand the fundamental groups.

Theorem 2.221If p : X −→ X is a covering function, γ0, γ1 : I = [0, 1] −→ X are

path homotopic and γ0, γ1 : I −→ X are lifts of γ0 and γ1 such thatγ0(0) = γ1(0),then γ0 and γ1 have the same end point and are path homotopic.

The next theorem characterizes the fundamental group homomor-phism induced by a covering function.

Theorem 2.222If p : X −→ X is a covering function and x ∈ X,

then p∗ : π1(X, x) −→ π1(X, p(x)) is injective.

We still need to determine when a given continuous functionf : Y −→ X admits a lift. The theorem that follows transforms thistopological problem to an equivalent algebraic one.

Theorem 2.223If p : X −→ X is a covering function, Y is a connected and locallypath-connected topological space (see Definitions 2.104 and 2.127) andf : Y −→ X is a continuous function,then for a given y0 ∈ Y and x0 ∈ X such that p(x0) = f(y0), f hasa lift f : Y −→ X such that f(y0) = x0 if and only if the subgroupf∗(π1(Y, y0)

)of π1

(X, f(y0)

)is contained in p∗

(π1(X, x0)

).

Two immediate consequences of this theorem are useful inapplications.


Corollary 2.224If p : X −→ X is a covering function and Y is a simply connected andlocally path-connected topological space (see Definitions 2.104, 2.208and 2.127),then every continuous function f : Y −→ X admits a lift.Moreover, for a given y0 ∈ Y , we can choose the lift to map y0 to anypoint in the fibre over f(y0).

Corollary 2.225If p : X −→ X is a covering function, X is simply connected, Yis connected and locally path-connected (see Definitions 2.104, 2.208and 2.127) and f : Y −→ X is a continuous function,then f admits a lift to X if and only if f∗ is the trivial homomorphismfor any base point y0 ∈ Y and the lift can be chosen to map y0 to anypoint in the fiber above f(y0).

Theorem 2.226If p : (X, x0) −→ (X,x0) is a covering function,then there is a surjection ξ : π1(X,x0) −→ p−1(x0).Moreover, if X is also simply connected, then ξ is a bijection.

Definition 2.227If X is simply connected covering space of X (see Definition 2.208),then we say that X is a universal covering space of X.

For easy reference let us recall the fundamental groups of some wellknown spaces.

Example 2.228

(a) π1(S1) = Z and πk(S

n) = 0 for all n � 2 and k < n, whileπn(S

n) = Z (see Theorem 2.215).(b) π1

(Rn \ {0}) = π1(S

1) (here Rn \ {0} is simply connected; Defi-

nition 2.208).(c) π1(T ) = Z × Z (here we use the fact that π1

(X × Y, (x0, y0)

)=

π1(X,x0)× π2(Y, y0)).(d) π1(P

2) = Z2.(e) The “double torus” T2 is the surface obtained by taking two copiesof the torus T deleting a small open disk from each of them and pastingthe remaining pieces. The fundamental group of T2 is not abelian.


Before passing to homology theory, we state an important theorem,which is a useful tool in the computation of fundamental groups. Itdescribes the fundamental group of a space in terms of the fundamentalgroups of its subspaces, subject to certain conditions.

Theorem 2.229 (Van Kampen Theorem)If X = U ∪V where U, V ⊆ X are open, U ∩V is path-connected (seeDefinition 2.122) and x0 ∈ U ∩ V ,then every element ϑ ∈ π1(X,x0) can be written as a sum

ϑ = η1 + . . .+ ηn,

where for each k � 1, we have

ηk ∈ π1(U, x0) or ηk ∈ π1(V, x0).

2.1.15 Elements of Algebraic Topology II: Homology

Algebraic topology came into existence by the efforts to solve topo-logical problems using algebraic tools. Homology theory is the coresubject in algebraic topology. As usual, we start with the geometricapproach to the subject which is called “simplicial homology theory”and then pass to the abstract approach which has two branches: “sin-gular homology theory” and “Cech homology theory”.

Definition 2.230Let {xk}nk=0 ⊆ R

N with n � N . We say that these vectors are affinelyindependent if the vectors {xk − x0}nk=1 ⊆ R

N are linearly indepen-dent. The convex hull of n+1 affinely independent vectors in R

N (withn � N) is called an n-simplex and is denoted by σn. All these n+ 1vectors are called vertices of σn and n is the dimension of σn.

Remark 2.231

If u ∈ σn, then u =n∑k=0

λkxk with λk � 0 for all k ∈ {0, 1, . . . , n}

andn∑k=0

λk = 1. The numbers {λk}nk=0 are called the barycentric

coordinates of u. Every u ∈ σn admits a unique set of barycentriccoordinates. The boundary of σn consists of (n − 1)-simplices.

Definition 2.232An m-face of an n-simplex σn is the convex hull of any set of mvertices of σn, where m < n.


Definition 2.233K is a geometric simplicial complex if it is a finite set of simplexesin some Euclidean space RN satisfying the following requirements:(a) If σn ∈ K and σm is a face of σn, then σm ∈ K.(b) If σn, σm ∈ K, then σn∩σm = ∅ or otherwise σn∩σm is a commonface of both simplexes and so by (a) belongs in K.The dimension of K, denoted by dimK, is the maximum of the di-mensions of its simplexes. A subcomplex L of K is a subfamily ofsimplexes of K satisfying (a).

Definition 2.234The union of all simplexes of a geometric simplicial complex K inRN , topologized as a subset of RN is called the polyhedron of K and

is denoted by |K|. If L is a subcomplex of K, then |L| is called asubpolyhedron of K.

Proposition 2.235A subset C of |K| is closed if and only if C ∩σ is closed in σ for everysimplex σ ∈ K.

Definition 2.236For given two simplicial complexes K and L, a function f : |K| −→ |L|is a simplicial function if it satisfies the following requirements:(a) If x is a vertex of a simplex in K, then f(x) is a vertex of asimplex in L.(b) If σ is a simplex of K with vertices {xk}nk=0, then τ =conv

{f(xk)

}nk=1

is a simplex of L (possibly of lower dimension).

(c) If x =n∑k=0

λkxk is in a simplex σ of K with vertices {xk}nk=0, then

f(x) =n∑k=0

λkf(xk), i.e., f is “linear on each simplex”.

A simplicial function of simplicial pairs f : (|K|, |L|) −→(|M |, |N |) is a simplicial function f : |K| −→ |M | such thatf(|L|) ⊆ |N |.

Proposition 2.237A simplicial function f : |K| −→ |L| is continuous.

There is a slight difficulty in the use of polyhedra. Namely, notevery topological space which is homeomorphic to a polyhedron is itselfa polyhedron. We rectify this problem with the following definition.


Definition 2.238If X is a topological space, a triangulation of X is a pair (K, f),where K is a simplicial complex and f : |K| −→ X is a homeomor-phism. A topological space X with a triangulation is called a triangu-lated space. Similarly, if (X,A) is a pair of topological spaces, a tri-angulation consists of a simplicial pair (K,L) and a homeomorphismof pairs f : (|K|, |L|) −→ (X,A). A pair (X,A) with a triangulationis called a triangulated pair.

Remark 2.239In general the particular homeomorphism f involved does not matterand so we often mention onlyK when referring to a triangulation of X.

Definition 2.240Let K be a simplicial complex and x ∈ |K|.(a) The simplicial neighbourhood of x, denoted by NK(x), is theset of all simplexes of K which contains x together with all their faces.(b) The link of x, denoted by LkK(x), is the subset of simplexes ofNK(x), which do not contain x.(c) For every simplex σ ∈ K, the star of σ, denoted by stK(σ), is theunion of the interiors of the simplexes of K that have σ as a face.

Remark 2.241We usually omit the suffix K, when it is clear from the context whichcomplex is used.

Proposition 2.242For every simplex σ ∈ K, st(σ) is an open set and if x is a point inthe interior of σ, then st(σ) = |N(x)| − |Lk(x)|.

Theorem 2.243If K and L are simplicial complexes and f : |K| −→ |L| is a homeo-morphism,then for every x ∈ |K|, ∣∣LkK(x)

∣∣ and∣∣LkL

(f(x)

)∣∣ are homeomorphic.

Definition 2.244Let K and L be simplicial complexes and let f : |K| −→ |L| be acontinuous function. A simplicial function g : |K| −→ |L| is a sim-plicial approximation of f , if for every vertex u of K, we havef(stK(u)

) ⊆ stL(g(x)

).


Remark 2.245A simplicial function is a simplicial approximation of itself.

Theorem 2.246If K and L are simplicial complexes and f : |K| −→ |L| is a continu-ous function,then there exists a simplicial approximation g of f and every suchsimplicial approximation is homotopic to f .

Orientation of a simplex means an ordering of its vertices.

Definition 2.247Two orderings of the vertices are said to determine the same orienta-tion of the simplex if every even perturbation of the vertices transformsone ordering to the other. If the permutation is odd, then we say thatthe orientations are opposite. Hence there are basically two orienta-tions of a simplex. An oriented simplex is a simplex together witha choice of orientation. So, for a k-simplex σ ordered by the linearordering (u0, . . . , uk) of its vertices, by −σ we will denote the samesimplex but with the opposite orientation.

This definition allows us to produce an abelian group out of asimplicial complex.

Definition 2.248Let K be a simplicial complex. A k-chain is a function g from the setof all oriented k-simplexes of K to the integers such that:(a) g(σ) = −g(τ), if the k-simplexes σ, τ have opposite orientation;(b) g(σ) = 0 for all but finitely many oriented k-simplexes σ.

We add k-chains in the usual way by adding their values. Similarlyfor the multiplication of k-chains with c ∈ Z. The resulting group iscalled the group of (oriented) k-chains and is denoted by Ck(K).If k < 0 or k > dimK, then Ck(K) = 0.

If σ is an oriented simplex, the elementary chain g correspond-ing to σ is the function g defined by g(σ) = 1, g(−σ) = −1 andg(τ) = 0 for all other oriented simplexes.


Proposition 2.249Ck(K) is a free abelian group. A basis for Ck(K) can be obtained byorienting each k-simplex and the using of corresponding elementarychains as a basis.

Remark 2.250The group C0(K) differs from the others, since it has a natural basis(note that the 0-simplex, being a point, has only one orientation). Incontrast, the group Ck(K), for k � 1, has no “natural basis”. We needto orient the k-simplexes of K in some arbitrary fashion in order toproduce a basis.

Proposition 2.251A function f from the oriented k-simplexes of K to an abelian groupG extends uniquely to a homomorphism from Ck(K) to G providedf(−σ) = −f(σ) for every oriented k-simplex σ.

Definition 2.252We define a homomorphism ∂k : Ck(K) −→ Ck−1(K) called theboundary operator as follows. If σ = [u0, . . . , uk] is an orientedk-simplex (k � 1), then

∂kσ =

n∑

k=0

(−1)n[u0, . . . , un, . . . , uk

],

where the symbol un means that the vertex un is deleted from the array.Since Ck(K) = 0 for k < 0, the operator ∂k is the trivial homomor-phism for k � 0.

Proposition 2.253∂k−1 ◦ ∂k = 0.

This fundamental equation leads to the definition of the (simplicial)homology groups.

Definition 2.254Consider the boundary operator ∂k : Ck(K) −→ Ck−1(C). The kernelof ∂k is an abelian subgroup of Ck(K) and is denoted by Zk(K). It isthe group of k-cycles. The image of ∂k+1 : Ck+1(K) −→ Ck(K) is an


abelian subgroup of Ck(K) and is denoted by Bk(K). It is the groupof k-boundaries. By Proposition 2.253, we have

Bk(K) ⊆ Zk(K).

We can define the quotient (factor) group

Hk(K) = Zk(K)/Bk(K),

which is called kth (simplicial) homology group of K.

Theorem 2.255H0(K) � Z if and only if |K| is connected.

Theorem 2.256

If |K| =n⋃i=1

|Ki|, where |Ki| are pairwise disjoint and connected,

then

Hk(K) =n∑

i=1

Hk(Ki) ∀ k � 0.

Sometimes it is more convenient to consider an alternative versionof the 0-dimensional homology.

Definition 2.257Let ε : C0(K) −→ Z be the surjective homomorphism, defined by

ε(u) = 1 for every vertex u of K.

Then, if g is a 0-chain, ε(g) equals the sum of the values of g on thevertices of K. We call ε the augmentation function for C0(K).Note that ε(∂1c) = 0 if c is a 1-chain. The reduced homology groupof K in dimension 0, denoted by H0(K), is defined by

H0(K) = ker ε/im ∂1.

If k � 1, then Hk(K) = Hk(K).

Proposition 2.258H0(K) is a free abelian group and

H0(K) � H0(K)⊕ Z.

Hence, if |K| is connected, then

H0(K) = 0.


Definition 2.259If L is a subcomplex of K, the quotient group Ck(K)/Ck(L)

is called

the group of relative chains of KmodL and is denoted by Ck(K,L).

Remark 2.260The group Ck(K,L) is a free abelian group. It has as basis all cosetsof the form [σi] = σi + Ck(L), where {σi} is a basis for Ck(K) andσi �∈ L.

Proposition 2.261The boundary operator ∂k : Ck(L) −→ Ck−1(L) is the restriction of theboundary operator ∂k : Ck(K) −→ Ck−1(K) (we use the same sym-bol for both). Then this homomorphism induces a homomorphism∂k : Ck(K,L) −→ Ck−1(K,L) (again the same symbol is used). Wehave ∂k−1 ◦ ∂k = 0. We set

Zk(K,L) = ker ∂k,

Bk(K,L) = im ∂k+1,

Hk(K,L) = Zk(K,L)/Bk(K,L).

We call Zk(K,L) the group of relative k-cycles, Bk(K,L) thegroup of relative k-boundaries and Hk(K,L) the kth relative(simplicial) homology group of KmodL.

Remark 2.262A relative k-chain is a coset c + Ck(K). This is a relative k-cycle ifand only if ∂kc is carried by L. It is a relative k-boundary if and onlyif there is a (k + 1)-chain gk+1 of K such that c− ∂k+1gk+1 is carriedby L. If L = {v} (a singleton), the

H0(K,L) � H0(K).

If K consists of an n-simplex and its faces and L is the set of all properfaces of K, then

Hk(K,L) = δk,nZ ∀ k � 0,

where n is the dimension of the simplex.

One of the greatest advances in the algebraic topology has beenthe extension of homology theory to general topological spaces. So far


homology groups have been defined for a special kind of spaces, namelycompact polyhedra and complexes resulting from them. Singular ho-mology theory extends the notion of homology groups to general topo-logical spaces.

Definition 2.263Let en = (0, . . . , 0, 1) ∈ R

n (n � 1). By means of the standard identifi-cation of Rn with the subspace R

n×{0} of Rn×Rm = R

n+m, en can beviewed as an element of Rn+m for any m � 0. Let e0 = (0, . . . , 0) ∈ R

n

(n � 1). Clearly {e0, . . . , en} are affinely independent and so they maybe taken as vertices of an n-simplex denoted by Δn. We call Δn thestandard n-simplex. If X is a topological space, a singular n-simplex, is a continuous function T : Δn −→ X.

Definition 2.264The free abelian group with the singular n-simplexes as generators andcoefficients in Z is called the nth singular chain group and is de-noted by Sn(X). For n < 0, Sn(X) = 0. If c ∈ Sn(X), then c is calleda singular n-chain.

Definition 2.265Let X,Y be two topological spaces and let f : X −→ Y be a continuousfunction. If T : Δn −→ X is a singular n-simplex in X, then thecomposition f ◦ T : Δn −→ Y is a singular n-simplex in Y denoted by

fT . If c =n∑i=1

aiTi, ai ∈ Z, belongs in Cn(X) (an n-chain in X), then

f∗(c) =n∑

i=1

ai fTi ∈ Cn(Y )

(an n-chain in Y ). The homomorphism f∗ : Cn(X) −→ Cn(Y ) is calledthe homomorphism induced by f . Clearly

(f2 ◦ f1)∗ = (f2)∗ ◦ (f1)∗.

Definition 2.266For each i ∈ {1, . . . , n}, let di : Δn−1 −→ Δn be the affine function,defined by

di(x0, . . . , xn−1) = (x0, . . . , xi−1, 0, xi, . . . , xn−1),


called the i-face function. It maps Δn−1 onto Δin (the ith face

of Δn opposite to the vertex ei). For a given singular n-simplexT : Δn −→ X, then T ◦ di : Δn−1 −→ X are n + 1 different singu-lar (n − 1)-simplexes, which can be thought of as the boundary of T .So, for each n � 1, the boundary operator ∂n : Cn(X) −→ Cn−1(X)is the homomorphism, defined by

∂nT =

n∑

i=0

(−1)iT ◦ di.

Proposition 2.267For all n � 1, we have ∂n ◦ ∂n+1 = 0.

Definition 2.268The collection C∗(X) =

{Cn(X), ∂n

}n�0

of abelian groups and bound-ary operators is called a singular chain complex for X. The re-mainder of the construction of singular homology copies that of sim-plicial homology. So, we set

Zn(X) = ker ∂n ∀ n � 1

and

Z0(X) = C0(X).

These are abelian subgroups of Cn(X) and their elements are calledsingular (n)-cycles. Also we set

Bn(X) = im ∂n+1 ∀ n � 0.

These are abelian subgroups of Cn(X) too and their elements arecalled singular (n)-boundaries. From Proposition 2.267 we see thatBn(X) ⊆ Zn(X) and so we can define the quotient groups

Hn(X) = Zn(X)/Bn(X) =

{ker ∂n/im ∂n+1

if n � 1,

C0(X)/im ∂1if n = 0.

This is the nth singular homology group of X. The singularhomology of X is the collection

H∗(X) ={Hn(X)

}n�0

.


Remark 2.269The elements of Hn(X) are called singular homology classes, the cosetu + Bn(X) being the class for the singular n-cycle u. If two singularn-cycles u and u′ belong to the same singular homology class, theyare said to be homologous. Evidently u and u′ are homologous if andonly if u − u′ = ∂n+1c for some singular (n + 1)-chain. If Hn(X) isfinitely generated, then rankHn(X) is the nth Betti number of X.Note that Zn(X) and Bn(X) being subgroups of an abelian group arenormal. Finally note that in contrast to the simplicial homology, ifX �= ∅, then

Cn(X) �= 0 ∀ n � 0.

Proposition 2.270

(a) If X = {�} (a singleton), then H0(�) � Z and

Hn(�) = 0 ∀ n � 1.

(b) If X is path-connected, then H0(X) � Z and if X has path-connected components {Xi}i∈I (see Definitions 2.122 and 2.130), then

Hn(X) =∑

i∈IHn(Xi) ∀ n � 0.

(b) If X ⊆ RN is convex, then H0(X) � Z and

Hn(X) = 0 ∀ n � 1.

Remark 2.271As with simplicial homology (see Theorems 2.255 and 2.256), H0

counts the number of path-connected components ofX, thus giving thesame information as the zero homotopy group π0(X) (in fact H0(X)is isomorphic to the abelianization Z

[π0(X)

]of π0(X)). However,

unlike higher homotopy groups, the singular homology groups giveinformation about all the path-connected components of X (see Defi-nition 2.130).

Proposition 2.272If B

n1 =

{x ∈ R

n : ‖x‖ � 1}

and Sn = ∂Bn+11 =

{x ∈ R

n+1 :

‖x‖ = 1}, then


(a) we haveHk(B

n) � δk,0Z ∀ k � 0,

where

δk,0 =

{1 if k = 0,0 if k �= 0.

(b) we have

Hk(Sn) �

{Z if k = 0 or k = n,0 otherwise.

We can also define relative singular homology. In this theory chainson a certain subspace A ⊂ X are identified with 0. That is, for a chainto be a singular cycle of XmodA, it must have a boundary which isa singular chain on A rather than zero. So, we make the followingdefinition.

Definition 2.273Let (X,A) be a topological pair. We set

Cn(X,A) = Cn(X)/Cn(A)∀ n � 1.

This is the relative n-singular chain group of XmodA. Itis a free abelian group with generators those singular n-simplexesT : Δn −→ X whose image is not completely contained in A. Theelements of Cn(X,A) are called relative singular n-chains ofXmodA. Since ∂n : Cn(X) −→ Cn−1(X) (n � 1) is a homomor-phism and ∂

(Cn(A)

) ⊆ Cn−1(A), there exists a unique homomorphism∂n : Cn(X,A) −→ Cn−1(X,A) (for notational economy it is denotedby the same symbol). This is the boundary operator for the relativesingular chain groups. Again

∂n−1 ◦ ∂n = 0 ∀ n � 1.

As before (see Definition 2.268), we set

Zn(X,A) = ker ∂n and Bn(X,A) = im ∂n+1 ∀ n � 0.

Then Zn(X,A) is the group of relative singular n-cycles ofXmodA and Bn(X,A) is the group of relative singular n-boundaries of XmodA. Both are abelian subgroups of Cn(X,A)and

Bn(X,A) ⊆ Zn(X,A) ∀ n � 0.


We set

Hn(X,A) = Zn(X,A)/Bn(X,A)∀ n � 0.

This is the nth relative singular homology group of X modA andit is a free abelian group. If it is finitely generated, then rankHn(X,A)is the nth Betti number of the topological pair (X,A).

Remark 2.274Note that

Zn(X,A) =

{ {c ∈ Cn(X) : ∂nc ∈ Cn−1(A)

}if n � 1,

C0(X) if n � 0,

andBn(X,A) = Bn(X) +Cn(A)

(the subgroup of Cn(X) generated by Bn(X) and Cn(A)). If A = ∅,the

Hn(X, ∅) = Hn(X).

One of the characteristic features of homology theory is a long exactsequence, which relates the homologies of X,A and XmodA. Todescribe it we will need the following proposition.

Proposition 2.275If f : (X,A) −→ (Y,B) is a continuous function between topologicalpairs,then f induces a homomorphism f∗ : Hn(X,A) −→ Hn(Y,B), n � 1such that:(a) if f is the identity function i : (X,A) −→ (X,A), then f∗ is theidentity isomorphism;(b) for another continuous function between topological pairsg : (Y,B) −→ (Z,C), we have (g ◦ f)∗ = g∗ ◦ f∗.

Let (X,A) be a topological pair and let i : A −→ X andj : (X, ∅) −→ (X,A) be the inclusion functions. They induce the ho-momorphisms i∗ : Hn(A) −→ Hn(X) and j∗ : Hn(X) −→ Hn(X,A)(see Proposition 2.275). Also, there is a unique homomorphism∂∗ : Hn(X,A) −→ Hn−1(A), defined by

∂∗([c+Bn(X) + Cn(A)]

)= [∂nc+Bn−1(A)]

(see Remark 2.274).


Theorem 2.276If (X,A) is a topological space,then the following infinite homology sequence is exact

. . . −→ Hn(A)i∗−→ Hn(X)

j∗−→ Hn(X,A)∂∗−→ Hn−1(A) −→ . . .

. . . −→ H0(A)i∗−→ H0(X)

j∗−→ H0(X,A)∂∗−→ 0.

Moreover, a continuous function f : (X,A) −→ (Y,D) between topo-logical pairs induces a homomorphism of the infinite exact sequenceof (X,A) into that of (Y,D), namely in the following diagram everysquare is commutative.

Hn(A) Hn(X) Hn(X,A) Hn−1(A))

Hn(D) Hn(Y ) Hn(Y,D) Hn−1(D)

i∗

(f|A)∗

j∗

f∗

∂∗

f∗ (f|A)∗

i∗ j∗ ∂∗

Theorem 2.277If A ⊆ D ⊆ X is a triple of topological spaces,then there is an exact sequence

. . . −→ Hn(D,A) −→ Hn(X,A) −→ Hn(X,D)∂∗−→ Hn−1(D,A) −→ . . .

. . . −→ H0(D,A) −→ 0,

where the unlabelled homomorphisms result from the correspondinginclusion functions and ∂∗ = j∗ ◦ ∂∗, i.e., the composition

Hn(D,A)∂∗−→ Hn(D)

j∗−→ Hn−1(D,A).

Moreover, a continuous function f : (X,D,A) −→ (X ′,D′, A′) betweentopological triples induces a homomorphism of the exact sequence of(X,D,A) into that of (X ′,D′, A′) (see Theorem 2.276).

In the relative singular homology of X modA, the singular chainson A are identified with 0. This suggests that removal of a set E ⊆ Ashould not affect the homology. This is true under the stronger con-dition E ⊆ intA and is known as “excision property” of singularhomology.


Theorem 2.278 (Excision Property)If (X,A) is a topological pair and E ⊆ intA,then Hn(X \ E,A \ E) � Hn(X,A) for all n � 0.

This property leads to the following definition.

Definition 2.279Let X be a topological space and let A,D ⊆ X be two sets such thatX = A ∪D. We say that the pair (A,D) is excisive if the inclusionfunction i : (A,A ∩ D) −→ (A ∪ D,D) = (X,D) induces an isomor-phism of the singular homology groups.

Theorem 2.280If X is a topological space and A,D ⊆ X are such that X = A∪D =(intA) ∪ (intD),then the pair (A,D) is excisive.

Theorem 2.281If X is a topological space and (A,D) is an excisive pair,then there is a homomorphism ∂n∗ : Hn(X) −→ Hn−1(A∩D) such thatthe following infinite homology sequence is exact:

. . . −→ Hn(A ∩D)(j1∗,−j2∗)−→ Hn(A)⊕Hn(D)

i1∗+i2∗−→ Hn(X)∂n∗−→ Hn−1(A ∩D) −→ . . .

. . . −→ H1(X)∂1∗−→ H0(A ∩D)

(j1∗,−j2∗)−→ H0(A)⊕H0(D)i1∗+i2∗−→ H0(X) −→ 0,

where i1 : A −→ X, i2 : D −→ X, j1 : A ∩D −→ A, j2 : A ∩D −→ Xare the inclusion functions. This exact sequence is known as theMayer–Vietoris sequence.

Theorem 2.282If (X,A) is a topological pair and � ∈ A,then the following infinite singular homology sequence is exact:

. . . −→ Hn(A, �) −→ Hn(X, �) −→ Hn(X,A) −→ Hn−1(A, �) −→ . . . −→ 0,

where the homomorphisms are induced by inclusions or boundary op-erators (see Theorem 2.277). This sequence is known as the reducedexact homology sequence.


Theorem 2.283Let f : (X,A) −→ (Y,D) be a continuous function between topologicalpairs. If f : X −→ Y and f |A : A −→ D are homotopy equivalences(see Definition 2.191),then f∗ is an isomorphism in relative singular homology.

First we defined homology groups for a particular class of spaces,namely polyhedra. Then we passed to general topological spaces, bymeans of the singular homology theory. In fact there is a plethoraof homology theories, a fact that led to an axiomatic unification byEilenberg–Steenrod.

Definition 2.284Let P be a class of topological pairs such that:(a) If (X,A) ∈ P, then (X,X), (X, ∅), (A,A), (A, ∅) ∈ P.(b) If (X,A) ∈ P, then (I ×X, I ×A) ∈ P (where I = [0, 1]).(c) There is a singleton {�} such that ({�}, ∅) ∈ P.

Then we call P an admissible class of spaces for a homology theory

Using this notion we can produce an axiomatic definition of a ho-mology theory.

Definition 2.285Let P be an admissible class of spaces. A homology theory on Pconsists of the following three “functions”:(a) A function H from Z × P into the family of abelian groups withthe image of (n, (X,A)) denoted by Hn(X,A).(b) A function which to every n ∈ Z and every continuous functionf : (X,A) −→ (Y,D) with (X,A), (Y,D) ∈ P, assigns a homomor-phism (f∗)n : Hn(X,A) −→ Hn(Y,D).(c) A function which to every n ∈ Z and every (X,A) ∈ P, assignsa homomorphism (∂∗)n : Hn(X,A) −→ Hn(A) (where A denotes theelement (A, ∅) ∈ P).

These three functions should satisfy the following axioms:Axiom 1. If i is the identity, then i∗ is the identity homomorphism.Axiom 2. (g ◦ f) = g∗ ◦ f∗.Axiom 3. If f : (X,A) −→ (Y,D) is a continuous function betweentopological pairs, then the following diagram commutes:


Hn(X, A) Hn(Y, D)

Hn−1(A) Hn−1(D)

f∗

∂∗ (∂∗)n

(f|A)∗

Axiom 4. If i : A −→ X and j : X = (X, ∅) −→ (X,A) are theinclusion functions, then the sequence

. . . −→ Hn(A)i∗−→ Hn(X)

j∗−→ Hn(X,A)∂∗−→ Hn−1(A) −→ . . .

is exact (this is known as the exactness axiom).Axiom 5. If f and g are homotopic, then f∗ = g∗ (this is known asthe homotopy axiom).Axiom 6. Let (X,A) ∈ P and let U be an open subset of X such thatU ⊆ intA. If (X \ U,A \ U) ∈ P, then

Hn(X \ U,A \ U) � Hn(X,A) ∀ n � 0

(this is known as the excision axiom).Axiom 7. If {�} is the singleton space in P, then

Hn({�}) = δn,0Z ∀ n � 0

(this is known as the dimension axiom).

Theorem 2.286

(a) Simplicial homology theory on the class of triangulated pairs (seeDefinition 2.238) satisfies the Eilenberg–Steenrod axioms of Defini-tion 2.285.(b) Singular homology theory on the class of topological pairs satisfiesthe Eilenberg–Steenrod axioms.

Theorem 2.287If X is a topological space, K is a simplicial complex and X, |K| arehomeomorphic,then for every integer Hn(X) (singular homology group) is isomorphicto Hn(K) (simplicial homology group).


Cohomology theory can be axiomatized in the same way as homol-ogy theory.

Definition 2.288Let P be an admissible class of spaces and let G be an abelian group(the group of coefficients). A cohomology theory on P consists ofthe following three “functions”:(a) A function H from G × P into the family of abelian groups withthe image of (n, (X,A)) denoted by Hn(X,A;G).(b) A function which to every n ∈ Z and every continuous functionf : (X,A) −→ (Y,D) with (X,A), (Y,D) ∈ P assigns a homomorphism(f∗)n : Hn(Y,D;G) −→ Hn(X,A;G).(c) A function which to every n ∈ Z and every (X,A) ∈ P assigns ahomomorphism (δ∗)n : Hn−1(A;G) −→ Hn(X,A;G).

These three functions should satisfy the following axioms:Axiom 1. If i is the identity, then i∗ is the identity homomorphism.Axiom 2. (g ◦ f) = f∗ ◦ g∗.Axiom 3. If f : (X,A) −→ (Y,D) is a continuous function betweentopological pairs, then the following diagram commutes:

Hn(X,A; G) Hn(Y, D; G)

Hn−1(A; G) Hn−1(D; G)

f∗

(f|A)∗

(δ∗)n(δ∗)n

Axiom 4. If i : A −→ X and j : X = (X, ∅) −→ (X,A) are theinclusion functions, then the sequence

. . .←− Hn(A;G)i∗←− Hn(X;G)

j∗←− Hn(X,A;G)(∂∗)n←− Hn−1(A;G)←− . . .

is exact.Axiom 5. If f and g are homotopic, then f∗ = g∗.Axiom 6. Let (X,A) ∈ P and let U be an open subset of X such thatU ⊆ intA. If (X \ U,A \ U) ∈ P, then

Hn(X \ U,A \ U) � Hn(X,A) ∀ n ∈ Z.

Axiom 7. If {�} is the singleton space in P, then

Hn({�}) = δn,0Z ∀ n � 0.


Definition 2.289The singular n-cochain is defined to be the homomorphismc : Cn(X;G) −→ G. We use the bracket notation [·, c] and we have

[T1 + T2, c] = [T1, c] + [T2, c] ∀ T1, T2 ∈ Cn(X;G)

and

[gT, c] = g[T, c] ∀ g ∈ G, T ∈ Cn(X;G).

The set of all singular n-cochains Hom(Cn(X;G), G

)is denoted by

Cn(X;G). We have

[T, c1 + c2] = [T, c1] + [T, c2] ∀ T ∈ Cn(X;G), c1, c2 ∈ Cn(X;G)

and

[T, gc] = g[T, c] ∀ g ∈ G, T ∈ Cn(X;G), c ∈ Cn(X;G).

So the duality brackets [·, ·] is a bilinear form on Cn(X;G) ×Cn(X;G). Then the dual operator of the boundary operator (∂∗)n withrespect to [·, ·] is called the coboundary operator and is denoted by(δ∗)n. We have

[(∂∗)nT, c] = [T, (δ∗)nc] ∀ T ∈ Cn(X;G), c ∈ Cn(X;G).

Hence (δ∗)n : Cn−1(X;G) −→ Cn(X;G) is a homomorphism and(∂∗)n−1 ◦ (∂∗)n = 0 implies (δ∗)n ◦ (δ∗)n−1 = 0. Then singular co-homology is defined as follows. Let (X,A) be a topological pair andlet

Cn(X,A;G) = Hom((Cn(X;G)

)/Cn(A;G)

, G)

and

(δ∗)n : Cn−1(X,A) −→ Cn(X,A)

is the dual operator of the boundary operator

(∂∗)n : Cn(X,A;G) −→ Cn−1(X,A;G).

We define

Hn(X,A) = ker (δ∗)n/im (δ∗)n−1.

This is the relative nth singular cohomology group of X modA.


Remark 2.290Note that

Cn(X,A;G) � {c ∈ Cn(X;G) : [T, c] = 0 for all T ∈ Cn(A;G)}.

The isomorphism is realized by the dual homomorphism

ξ∗ : Cn(X,A;G) −→ Cn(X;G)

of the homomorphism ξ : Cn(X;G) −→ Cn(X;A;G). Therefore

ker (δ∗)n = Zn(X,A;G)

={c ∈ Cn(X,A;G) : [T, c] = 0 for all T ∈ Bn(X,A;G)

}

and

im (δ∗)n−1 = Bn(X,A;G)

={c ∈ Cn(X,A;G) : [T, c] = 0 for all T ∈ Zn(X,A;G)

}.

In general we have a canonical homomorphism

β : Hn(X,A;G) −→ Hn(X,A;G)∗.

If G is a field, then β is surjective. Note that in this case Hn(X,A;G)is a vector space and Hn(X,A;G) is its dual.

Proposition 2.291The singular cohomology theory satisfies the Eilenberg–Steenrod ax-ioms.

Definition 2.292Let{Cn(X), (∂∗)n, ε

}n�1

be an augmented chain complex (see Defini-

tion 2.257). Let ε : Hom (Z, G) −→ C0(X;G) be the dual to the aug-mentation function. We define the reduced singular cohomologygroups by setting

˜Hn

(X ;G) = Hn(X ;G) if n � 1, and ˜H0

(X ;G) = ker (δ∗)1/im ε.

The next theorem relates homology and cohomology groups.

Theorem 2.293 (Alexander Duality Theorem)If A is a proper nonempty subset of Sn =

{x ∈ R

n+1 : ‖x‖ = 1}and

(Sn, A) is a triangulated pair,

then˜Hk

(A) � Hn−k−1(Sn \ A).


We conclude this section with the diagram summarizing the rela-tions between the various types of spaces introduced in the theory andin the problems (arrows stand for inclusions).

Hausdorff

regular

completely regular Baire

completely normal normal locally compact

perfectly normal paracompact

metric regular Lindelöf

separable separable metric σ-compact

stronglyLindelöf Borel compact complete metric

second countable Souslin Polish

compact metric

2.2. Problems 265

2.2 Problems

Reminder: All topological spaces are taken to be Hausdorff.

Problem 2.1

Suppose that X is a topological space and U ⊆ X is an open set. Showthat for every set A ⊆ X, we have U ∩ A ⊆ U ∩A and the inclusioncan be proper.

Problem 2.2

Suppose that X is a topological space and A,C ⊆ X. Show that:(a) if ∂C ⊆ A ⊆ C, then ∂C ⊆ ∂A;(b) ∂A need not be equal to ∂A;(c) ∂(A ∪ C) ⊆ ∂A ∪ ∂C and the inclusion can be proper;(d) ∂(A ∩ C) ⊆ ∂A ∪ ∂C and the inclusion can be proper.

Problem 2.3

Let A ⊆ C[0, 1] be the subset of Lipschitz functions. Furnish C[0, 1]with the supremum metric topology (see Example 1.3(d)). Show thatintA = ∅.

Problem 2.4

Suppose that X is a topological space and C ⊆ X be a closed set.Show that C is nowhere dense in X if and only if X \C is dense in X.

Problem 2.5

Suppose that X is a topological space, A ⊆ X is a set furnished with

the subspace topology, C ⊆ A and let CA

(respectively, C) denotethe closure of C as a subspace of A (respectively, of X). Similarlylet intAC (respectively, intC) be the interior of C as a subspace of A(respectively, of X). Show that:

(a) CA= C ∩A.

(b) intC ⊆ intAC and the inclusion can be strict.

Problem 2.6

Suppose that X is a topological space and A,C,D are three subsets ofX such that D ⊆ A∩C. Show that if D is open (respectively, closed)in both A and C (with their respective subspace topologies), then D isalso open (respectively, closed) in A∪C (with the subspace topology).


Problem 2.7

Let X be a topological space and let C ⊆ X be a nonempty subspace.Show that C is closed if and only if every x ∈ C has a neighbourhoodU such that C ∩ U is closed in U .

Problem 2.8

Let X be a topological space and let C,D ⊆ X be two nonemptysubsets. Show that if there exists E ⊆ X such that E \D is closed inX and contains C \D, then we have

C ∩D = C ∩D.

Problem 2.9

Let X be a topological space with a countable subbasis. Show that Xis second countable.

Problem 2.10

Show that a closed subspace of a Lindelof space is also a Lindelof space.

Problem 2.11

Suppose that X is a topological space and B is a basis for the topology.Show that there exists a dense subset D ⊆ X such that #D � #B(where # denotes the cardinality of a set).

Problem 2.12

Suppose that X is a topological space and {Ui}i∈I is an open cover ofX. Show that C ⊆ X is closed if and only if Ui ∩ C is closed in Ui(with the subspace topology) for all i ∈ I.

Problem 2.13

Can R be the continuous image of [0, 1)? Justify the answer.

Problem 2.14

Show that the uniform limit of a net of continuous R-valued functionson a topological space X is continuous.

Problem 2.15

Suppose that X is a topological space and f : X −→ R∗ = R∪ {±∞}.

Show that f is lower semicontinuous (respectively, sequentially lowersemicontinuous) if and only if epi f =

{(x, λ) ∈ X × R : f(x) � λ

}is

closed (respectively, sequentially closed).

2.2. Problems 267

Problem 2.16

Suppose that X is a topological space, f : X −→ R∗ = R ∪ {±∞} is a

function and f : X −→ R∗ = R ∪ {±∞} is the relaxed function of f .

Show that for every λ ∈ R, we have

{x ∈ X : f(x) � λ

}=⋂

μ>λ

{x ∈ X : f(x) � μ

}

and

epi f = epi f in X × R.

Problem 2.17

Suppose that X is a topological space and f : X −→ R∗ = R ∪ {±∞}

is a function. Let f be the relaxed function of f (see Definition 2.57)Corollary 2.56 guarantees that f is lower semicontinuous. Show that

f(x) = supU∈N (x)

infy∈U

f(y) ∀ x ∈ X.

Problem 2.18

Suppose that X is a first countable topological space, f : X −→ R =R ∪ {+∞} is a function and f : X −→ R

∗ = R ∪ {±∞} is the relaxedfunction of f . Show that for every x ∈ X, f(x) is characterized by thefollowing two properties:(a) For every sequence xn −→ x, we have f(x) � lim inf

n→+∞ f(xn).

(b) There exists a sequence yn −→ x in X such that lim infn→+∞ f(yn) �

f(x).

Problem 2.19

Suppose that (X, dX) is a metric space, f : X −→ R+ = [0,+∞] is a

function and f : X −→ R∗ = R ∪ {±∞} is the relaxed function of f .

For every λ > 0, we define

fλ(x)def= inf

y∈X(f(y) + λd

X(x, y)

).

Show that:(a) fλ is λ-Lipschitz;(b) lim

λ→+∞fλ(x) = f(x) for all x ∈ X.


Problem 2.20

Let (X, dX) be a metric space and let f : X −→ R

∗ = R ∪ {±∞} be afunction. Show that:(a) f is lower semicontinuous and bounded below if and only if there isa sequence {fn}n�1 of bounded continuous functions such that fn ↗ fand n→ +∞.(b) f is upper semicontinuous and bounded above if and only ifthere is a sequence {fn}n�1 of bounded continuous functions such thatfn ↘ f and n→ +∞.

Problem 2.21

Is the pointwise limit of upper semicontinuous functions, an uppersemicontinuous function too? How about the uniform limit? Justifyyour answer.

Problem 2.22

Suppose that X and Y are two topological spaces, D ⊆ X is a densesubset and f, g : X −→ Y are two continuous functions such thatf(x) = g(x) for all x ∈ D. Show that f(x) = g(x) for all x ∈ X.

Problem 2.23

Suppose that X is a topological space, (Y, dY) is a metric space,

f : X −→ Y is a function and ωf is the oscillation function of f (seeDefinition 2.42). Show that f is continuous at x if and only if ωf (x) = 0(this problem extends to general topological spaces Problem 1.46).

Problem 2.24

Suppose that X is a topological space, (Y, dY) is a metric space,

f : X −→ Y is a function and cont f ={x ∈ X : f is continuous at x

}.

Show that cont f is a Gδ-set in X.

Problem 2.25

Suppose that X is a topological space and f : X −→ R is a function.Show that f is continuous if and only if for every λ ∈ R, the sets{x ∈ X : f(x) � λ

}and

{x ∈ X : f(x) > λ

}are closed and open

respectively.

Problem 2.26

Suppose that X is a topological space and {fn : X −→ R}n�1 is a se-quence of continuous functions such that fn(x) −→ f(x) for all x ∈ X.Show that the following two statements are equivalent:

2.2. Problems 269

(a) f is continuous on X;(b) For every ε > 0 and k � 1, we can find n � k such that the set{x ∈ X :

∣∣f(x)− fn(x)∣∣ < ε

}is open in X.

Problem 2.27

Let X be a normal space and let Y be a locally finite open cover of X.Show thatX has a locally finite open cover D∗ such that

{V : V ∈ D∗

}

is a refinement of Y.

Problem 2.28

Let X be a topological space and let Y = {Uα}α∈J be an open coverof X which is locally finite. Show that for any K ⊆ J , the set

⋃x∈K

Uα

is closed.

Problem 2.29

Suppose that X,Y,Z are three topological spaces and f : X −→ Y ,g : Y −→ Z are two functions. Show that:(a) If f, g are open (respectively, closed), then g ◦ f is open (respec-tively, closed) too.(b) If g ◦ f is open (respectively, closed) and g is bijective and con-tinuous, then f is open (respectively, closed).(c) If g ◦ f is open (respectively, closed) and f is surjective and con-tinuous, then g is open (respectively, closed).

Problem 2.30

Suppose that X is a topological space, A ⊆ X is a set and iA : A −→ Xis the canonical injection function (i.e., iA(x) = x for all x ∈ A).Show that iA is open (respectively, closed) if and only if A is open(respectively, closed).

Problem 2.31

Suppose that X and Y are two topological spaces and f : X −→ Y isa surjection. Show that f is a homeomorphism if and only if

A = f−1(f(A)

) ∀ A ⊆ X.

Problem 2.32

Suppose that X and Y are two topological spaces, X is normal andf : X −→ Y is a continuous closed surjection. Show that Y is normaltoo.


Problem 2.33

Suppose that X and Y are two topological spaces, Y is second count-able and E ⊆ X × Y is a nowhere dense set. For every x ∈ X, we set

Exdef={y ∈ Y : (x, y) ∈ E

}. Show that Ex is nowhere dense for all

x ∈ X except on a set of first category.

Remark. The result of this problem is known in the literature asthe Kuratowski–Ulam theorem and can be found in Kuratowski[11, p. 249] and Oxtoby [16, p. 56] The result can be viewed as atopological analogue of the Fubini theorem (see Chap. 3).

Problem 2.34

Suppose that X and Y are two topological spaces with Y being secondcountable. Let C ⊆ X andD ⊆ Y be nonempty sets. Show that C×Dis of first category in X × Y if and only if at least one of the sets C orD is of first category.

Problem 2.35

Suppose that X and Y are two topological spaces and f : X −→ Y isa continuous surjection. Show that if f is open or closed, then f is anidentification function.

Problem 2.36

Suppose that X and Y are two topological spaces and f : X −→ Yis a function. Show that f is continuous and closed if and only ifthe projections proj

X: Gr f −→ X and proj

Y: Gr f −→ Y are closed

functions (recall that Gr f ={(x, y) ∈ X × Y : y = f(x)

}and on it

we consider the subspace product topology).

Problem 2.37

Suppose that X and Y are two topological spaces and A ⊆ X, C ⊆ Yare two nonempty sets. Show that A×C is closed (respectively, open,dense) if and only if both sets A and C are closed (respectively, open,dense).

Problem 2.38

Suppose that I is an uncountable set and{(Xi, dXi

)}i∈I is a family of

metric spaces. Assume that for each i ∈ I, the metric space (Xi, dXi)

is nontrivial (i.e., Xi is not a singleton). Consider the set Xdef=∏i∈I

Xi

with the product topology. Show that X is not metrizable.

2.2. Problems 271

Problem 2.39

Suppose that (X, τX) and (Y, τY ) are two topological spaces andf : X −→ Y is a continuous open function. Show that(a) if f is a surjection, then τY = s(f).(b) if f is an injection, then τX = w(f).

Problem 2.40

Suppose that X and Y are two topological spaces and f : X −→ Yis a bijection. Suppose that either the topology of X is w(f) or thetopology of Y is s(f). Show that f is a homeomorphism.

Problem 2.41

Suppose that X is a topological space, � is an equivalence relation onX and p : X −→ X/� is the quotient function. We say that � is open(respectively, closed) if for every open (respectively, closed) set A,its p-saturation p−1

(p(A)

)(see Definition 2.74) is open (respectively,

closed). Show that � is open (respectively, closed) if and only if thequotient function is open (respectively, closed).

Problem 2.42

A topological space X is said to be completely regular if for ev-ery nonempty closed set C ⊆ X such that x �∈ C, we can finda continuous function f : X −→ [0, 1] such that f(x) = 0 andf |

C= 1. Let C(X) (respectively, Cb(X)) be the space of all continu-

ous (respectively, bounded continuous) functions f : X −→ R. Showthat a topological space (X, τ) is completely regular if and only ifτ = w

(C(X)

)= w

(Cb(X)

).

Problem 2.43

Let K be a nonempty compact set in a completely regular topologicalspace X (see Problem 2.42) and let U be an open set containing K.Show that there exists a continuous function f : X −→ [0, 1] such thatf |

K= 1 and f |

X\U = 0.

Problem 2.44

Let X be a topological space and let f : X −→ R∗ = R ∪ {±∞}

be a function. Suppose that f is coercive and lower semicontinuous(respectively, sequentially coercive and sequentially lower semicontin-uous). Show that:(a) There exists x0 ∈ X such that f(x0) = inf

Xf .


(b) If {xn}n�1 ⊆ X is a minimizing sequence for f (i.e., f(xn) −→infXf) and x is a limit point of {xn}n�1 (respectively, x is the limit

point of a subsequence of {xn}n�1), then x is a minimizer of f .(c) If f is not identically +∞, then every minimizing sequence for fhas a limit point (respectively, convergent subsequence).

Problem 2.45

Is a convergent net necessarily compact? Justify the answer.

Problem 2.46

Suppose that X is a topological space and assume that{fn : X −→ R

∗ = R ∪ {±∞}}n�1 is a sequence of functions. We saythat the sequence {fn}n�1 is equicoercive if for every λ ∈ R, thereexists a closed, countably compact set Kλ such that

⋃

n�1

{x ∈ X : fn(x) � λ

} ⊆ Kλ.

Show that the sequence {fn}n�1 is equicoercive if and only if thereexists a lower semicontinuous, coercive (see Definition 2.103) functionψ : X −→ R

∗ = R ∪ {±∞} such that

ψ(x) � fn(x) ∀ x ∈ X, n � 1.

Problem 2.47

Suppose that X is a topological space, f : X −→ R = R ∪ {+∞} is acoercive function. Show that the relaxed function f is coercive too.

Problem 2.48

Suppose that X is a topological space, f : X −→ R∗ = R ∪ {±∞} is a

coercive function and f : X −→ R∗ is the relaxed function of f . Show

the following statements:(a) f admits a minimizer in X and min

Xf = inf

Xf .

(b) Every limit point of a minimizing sequence for f is a minimizerfor f .(c) If X is first countable, then every minimizer of f is the limit pointof a minimizing sequence for f .

Problem 2.49

Suppose that X and Y are two topological spaces with Y being com-pact and C is a closed subset of X × Y . Show that proj

X(C) is closed

2.2. Problems 273

in X (here projX: X × Y −→ X is the natural projection on the first

factor X).

Problem 2.50

Show that a topological space X is compact if and only if every opencover consisting of basic open sets has a finite subcover.

Problem 2.51

Suppose that X and Y are two topological spaces and f : X −→ Yis a continuous surjection. Show that if X is compact, then f is anidentification function.

Problem 2.52

Suppose that X and Y are two topological spaces and f : X −→ Y isa continuous function. On X we introduce the equivalence relation �,defined by

x � u ⇐⇒ f(x) = f(u)

Let X = X/�. Show that if X is compact, then X is homeomorphicto f(X).

Problem 2.53

(a) Show that two comparable compact Hausdorff topologies on a setX are equal.(b) Give an example of two topologies on a set which are both com-pact Hausdorff but not comparable.

Problem 2.54

Suppose that X is a topological space, Y is a compact topologicalspace and f : X −→ Y is a function. Show that f is continuous if andonly if Gr f ⊆ X × Y is closed (see Definition 1.132). Is any of theimplications false if we do not assume the compactness of Y ?

Problem 2.55

Let X be a topological space. Show that a locally compact densesubset of X is open.


Problem 2.56

(a) Suppose that X is a compact topological space and assume that{fi : X −→ R}i∈I is a net of continuous functions converging to acontinuous function f : X −→ R. Suppose that the net {fi}i∈I ismonotone (increasing or decreasing). Show that fi ⇒ f (this is thetopological version of the so called Dini theorem ; see Problem 1.104).(b) Show that the above result fails if we drop the assumption on thecompactness of the space X.(c) Show that the above result fails if we drop the assumption on thecontinuity of the limit function f .

Problem 2.57

Show that, if I = [0, 1], then I/(0∼1) is homeomorphic to S1, whereS1 =

{x ∈ R

2 : ‖x‖ = 1}.

Problem 2.58

Show that the Alexandrov one-point compactification of RN is home-omorphic to SN =

{x ∈ R

N+1 : ‖x‖ = 1}.

Problem 2.59

Suppose that X is a topological space and {xn}n�1 is a sequence in X

such that xn −→ x in X. Show that the set K ={xn : n � 1

} ∪ {x}is compact in X. (Compare with Problem 2.45.)

Problem 2.60

Suppose thatX is a topological space and C,D ⊆ X are two nonempty,compact and disjoint sets. Show that we can find two open setsU, V ⊆ X such that C ⊆ U , D ⊆ V and U ∩ V = ∅.

Problem 2.61

Suppose that X is a topological space, K ⊆ X is a nonempty com-pact set and U1, U2 ⊆ X are two nonempty open sets such thatK ⊆ U1 ∪ U2. Show that we can find two compact sets K1,K2 ⊆ Xsuch that K1 ⊆ U1, K2 ⊆ U2 and K = K1 ∪K2.

Problem 2.62

Suppose that X and Y are two topological spaces and f : X −→ Y isa function with a closed graph Gr f . Show that for every compact setK ⊆ Y , the set f−1(K) is closed in X.

2.2. Problems 275

Problem 2.63

Suppose that X and Y are two topological spaces and f : X −→ Y is aclosed function such that for every y ∈ Y , the set f−1(y) ⊆ X is com-pact. Show that for every compact set K ⊆ Y , the set f−1(K) ⊆ X iscompact.

Problem 2.64

Suppose that X is a compact topological space, f : X −→ R is afunction and for every λ ∈ R, f−1

([λ,+∞)

)is closed. Show that we

can find x0 ∈ X such that f(x0) = supXf < +∞.

Problem 2.65

Suppose that X and Y are two topological spaces, with X being com-pact and p

Y: X × Y −→ Y is the canonical projection on the second

factor Y . Show that pYis continuous, closed and for all y ∈ Y , the set

p−1Y

(y) is compact.

Problem 2.66

Suppose that X is a Lindelof topological space in which every sequencehas a limit point. Show that X is compact.

Problem 2.67

Suppose that I is an infinite index set and {Xi}i∈I is a family oftopological spaces. Assume that an infinite number of these spacesare noncompact and let K ⊆ ∏

i∈IXi be a compact set (for the product

topology see Definition 2.69). Show that intK = ∅ (i.e., K is nowheredense).

Problem 2.68

A topological space X is said to be a k-space , if the following is true:“A ⊆ X is closed if and only if for every compact set K ⊆ X, the setA ∩K is closed”. Suppose that X and Y are topological spaces, withX being a k-space and f : X −→ Y is a function such that for everycompact set K ⊆ X, f |K is continuous. Show that f is continuous.

Problem 2.69

Suppose that X is a locally compact topological space and A ⊆ X.Show that if A is open or closed in X, then A with the subspacetopology is locally compact.


Problem 2.70

Suppose that X is a compact topological space and x ∈ X. Show thatX \ {x} is locally compact.

Problem 2.71

Let K be a compact, convex subset of RN with nonempty interior.

Show that K is homeomorphic to the closed unit ball BN1 by homeo-

morphism which sends SN−1 = ∂BN1 to ∂K.

Problem 2.72

Suppose that X and Y are two topological spaces with Y being locallycompact and f : X −→ Y is a continuous bijection which returns com-pact sets in Y to compact sets in X (i.e., if K ⊆ Y is compact, thenf−1(K) ⊆ X is compact). Show that f is closed.

Problem 2.73

Suppose that X is a compact topological space and there is a countableseparating family Φ of continuous functions fromX into a metric space(Y, d

Y). Show that the topology of X is metrizable.

Problem 2.74

Suppose that X is a locally compact topological space and X∗ is theAlexander one-point compactification ofX. Show that X is σ-compactif and only if the point ∞ ∈ X∗ has a countable local basis.

Problem 2.75

(a) Suppose that X is a compact topological space and f : X −→ X isa continuous function. Show that there exists a nonempty and closedsubset A ⊆ X such that f(A) = A.(b) Show that the result fails if X is not compact.

Problem 2.76

Suppose that X is a compact topological space, f, g ∈ C(X), f(x) � 0for all x ∈ X and if f(x) = 0, then g(x) > 0. Show that there existsλ > 0 such that λf(x) + g(x) > 0 for all x ∈ X.

Problem 2.77

Let X be a topological space and let {Ci}i∈I be a family of compactsubsets of X. Let U ⊆ X be an open set such that

⋂i∈I

Ci ⊆ U . Show

that there exists a finite set F ⊆ I such that⋂i∈F

Ci ⊆ U .

2.2. Problems 277

Problem 2.78

Let X be a topological space and let f : X −→ R be a functional. Wesay that f is locally bounded if for each x ∈ X, there exist U ∈ N (u)and M =M(U) > 0 such that

∣∣f(x)∣∣ � M ∀ x ∈ U.

Show that, if X is compact, then every locally bounded function isbounded.

Problem 2.79

Let D be a locally compact dense subset of a metrizable space X.Show that D is open in X.

Problem 2.80

Show that every locally compact metrizable space X is completelymetrizable.

Problem 2.81

Is the circle S1 ={x ∈ R

2 : ‖x‖ = 1}homeomorphic to the closed

interval [0, 1]?

Problem 2.82

Suppose that X is a topological space and K ⊆ X is a nonempty,compact and disconnected set. Show that we can find two disjointopen sets U, V ⊆ X such that K ⊆ U ∪ V , K ∩U �= ∅ and K ∩ V �= ∅.

Problem 2.83

Suppose that X is a topological space and {Kn}n�1 is a decreasing se-quence of compact sets in X. Let K =

⋂n�1

Kn. Show the following:

(a) K ⊆ X is nonempty and compact.(b) If U is open and K ⊆ U , then we can find n0 � 1 such thatKn ⊆ U for all n � n0.(c) If for every n � 1, the set Kn is also connected, then K is con-nected too.

Problem 2.84

Is it true that, if X is a topological space and {Cn}n�1 is a decreas-ing sequence of nonempty closed connected sets in X, then

⋂n�1

Cn is

connected too? Justify your answer.


Problem 2.85

Let f : [0, 1] −→ A×C be a homeomorphism. Show that one of A andC is a singleton.

Problem 2.86

(a) Let U ⊆ RN be an open and connected set. Show that U is

path-connected.(b) Give an example showing that the above is no longer true if wereplace R

N by any topological space.

Problem 2.87

Are the intervals (x, u) and [y, z) in R homeomorphic? Justify jouranswer.

Problem 2.88

Suppose that X is a topological space, {Ai}i∈I is a family of connectedsubsets of X and for every pair (i, j) ∈ I×I, there exists a finite family{ik}nk=0 ⊆ I such that i0 = i and in = j and

Aik−1∩Aik �= ∅ ∀ k ∈ {1, . . . , n}.

Show that the set⋃i∈I

Ai is connected.

Problem 2.89

Suppose that X and Y are two topological spaces with X being com-pact and f : X −→ Y is a continuous function. Suppose that for everyy ∈ Y , the set f−1(y) ⊆ X is connected. Show that the inverse imageof every connected set in Y is connected in X.

Problem 2.90

Suppose that X is a topological space, A ⊆ X is a nonempty set andC ⊆ X is a connected set such that C ∩ A �= ∅ and C ∩ (X \ A) �= ∅.Show that C ∩ ∂A �= ∅.

Problem 2.91

Let X and Y be two topological spaces such that X is path-connectedand Y is not. Show that there can be no continuous surjectionf : X −→ Y .

2.2. Problems 279

Problem 2.92

Show that for every N � 1, the sphere SNdef={x ∈ R

N+1 : ‖x‖ = 1}

is path-connected.

Problem 2.93

Suppose that X and Y are two topological spaces and f : X −→ Y isan open, continuous surjection. Show that, if X is locally connected(respectively, locally compact), then so is Y .

Problem 2.94

Assume that X is a disconnected topological space and assume thatX = C ∪D, where both C and D are nonempty and connected. Showthat C and D are the connected components of X.

Problem 2.95

Suppose that X is a topological space and C and D are two nonemptyconnected subsets of X such that C ∩ D �= ∅. Show that C ∪ D isconnected.

Problem 2.96

Let X be a connected topological space and let A ⊆ X be a nonemptysubset. Show that, if ∂A is connected, then A is connected too.

Problem 2.97

Suppose that X is a topological space, A ⊆ X is a nonempty subset,x ∈ A, u ∈ X \A and γ : [0, 1] −→ X is a path such that γ(0) = x andγ(1) = u. Show that γ

([0, 1]

) ∩ ∂A �= ∅.

Problem 2.98

Consider the following subsets of R2:(a) the set of points with both coordinates rational;(b) the set of points with at least one coordinate rational;(c) the set of points with coordinates which are either both rationalor both irrational.Determine which of the above sets is connected. Justify your answer.

Problem 2.99

Suppose that X is a connected normal topological space. Show thatX is either a singleton or uncountable.


Problem 2.100

Suppose that X is a normal topological space and Y is locally finiteopen cover of X. Show that for every U ∈ Y, we can find a continuousfunction fU : X −→ [0, 1] such that f |

X\U = 0 and∑U∈Y

fU (x) = 1 for

all x ∈ X.

Problem 2.101

Suppose that X is a normal topological space and A,C are twononempty, disjoint closed subsets in X. Show that there exist opensets U and V such that A ⊆ U , C ⊆ V and U ∩ V = ∅.

Problem 2.102

Suppose that X is a normal topological space and A ⊆ X is anonempty closed set. Show the following:(a) There exists a continuous function f : X −→ [0, 1] such that

A = f−1({0}) ⇐⇒ A is a Gδ-subset of X.

(b) If A is also a Gδ-set and C ⊆ X is nonempty, closed and disjointfrom A, then we can find a continuous function f : X −→ [0, 1] suchthat A = f−1

({0}) and f |C= 1.

Problem 2.103

Suppose that X is a locally compact topological space, K ⊆ X is acompact set and U ⊆ X is an open set such that K ⊆ U . Show thatthere is a continuous function f : X −→ [0, 1] such that f |

K= 0 and

fX\U = 1.

Problem 2.104

Suppose that X is a locally compact topological space and K ⊆ X isa nonempty, compact set. Show that K is a Gδ-set if and only if thereexists a continuous function f : X −→ [0, 1] such that K = f−1

({0}).

Problem 2.105

Show that a normal topological space X is perfectly normal (see Def-inition 2.137) if and only if every closed subset of X is a Gδ-set.

Problem 2.106

Suppose that X is a second countable, regular topological space andU ⊆ X is a nonempty and open set. Show that there exists a contin-uous function f : X −→ [0, 1] such that f |

U> 0 and f

X\U = 0.

2.2. Problems 281

Problem 2.107

Let X be a locally compact topological space. Show that X is secondcountable if and only if X is separable and metrizable.

Problem 2.108

Let X be a locally compact and paracompact topological space andlet Y be a locally finite open cover of X. Show that X has a locallyfinite open cover D such that for every V ∈ D, the set V is compactand

{V : V ∈ D} is a refinement of Y.

Problem 2.109

Let X be a topological space in which the union of every countablefamily of closed nowhere dense sets is nowhere dense. Show that X isa Baire space. Is the converse true? Justify your answer.

Problem 2.110

Suppose that X is a Baire space and {An}n�1 is a sequence of nowheredense sets. Show that

⋃n�1

An �= X.

Problem 2.111

Suppose thatX is a Baire space andD1,D2 ⊆ X are two dense subsets.Is D = D1 ∩D2 necessarily dense in X? Justify your answer.

Problem 2.112

Suppose that X is a topological space in which every point has an openneighbourhood, which is a Baire space (with the subspace topology).Show that X is a Baire space.

Problem 2.113

Let X be a Baire topological space and let f : X −→ R∗ = R ∪ {±∞}

be a lower semicontinuous function. Suppose that the set{x ∈ X :

f(x) < +∞} is not of the first category. Show that we can find anopen set U ⊆ X and M > 0 such that f(x) �M for all x ∈ U .

Problem 2.114

Suppose that X is a Baire topological space and {Cn}n�1 is a sequenceof closed subsets of X such that X =

⋃n�1

Cn. Show that the set⋃n�1

intCn is dense in X.


Problem 2.115

Suppose that X is a Baire topological space, Y is a separable metricspace and f : X −→ Y is a function such that the inverse image ofevery open set is an Fσ-set. Show that f is continuous at every pointof a dense Gδ-set.

Problem 2.116

Suppose that X is a Baire topological space in which every open setis Fδ (for example, a complete metric space; see Theorem 1.26 or aperfectly normal Baire space; see Problem 2.105). Show that a lowersemicontinuous (or an upper semicontinuous) function f : X −→ R iscontinuous at every point of a dense Gδ-set.

Problem 2.117

Show that a closed subset C of a paracompact spaceX is paracompact.

Remark. In fact something more general is true. Every Fσ-subset ofa paracompact space is paracompact (see Proposition 2.145(b)).

Problem 2.118

Let X be a Baire space. Show the following:(a) Every open subset of X with the subspace topology is Baire too.(b) If ∼ is an equivalence relation, X/∼ is the quotient space with thequotient topology and if U ⊆ X/∼ is open, then U is Baire too.

Problem 2.119

Let X be a separable Baire space and let U ⊆ X be a nonempty openset. Consider a function f : U×X −→ R such that for every u ∈ X, thefunction x �−→ f(x, u) is lower semicontinuous and for every x ∈ U , thefunction u �−→ f(x, u) is continuous. Show that there exists a densesubset D of U such that for all u ∈ X, the function x �−→ f(x, u) iscontinuous on D.

Problem 2.120

Suppose that X is a paracompact space and C,D ⊆ X are two disjointclosed sets. Assume that the set C is covered by open sets {Uα}α∈Iwith Uα ∩ D = ∅. Show that the sets C and D have disjoint neigh-bourhoods.

2.2. Problems 283

Problem 2.121

Let X be a normal space and let Y be a locally finite open cover of X.Show that there exists a partition of unity {ψi}i∈I subordinate to Y.Problem 2.122

Show that every open or closed subset of a Lusin space is itself Lusin.

Problem 2.123

Let {Xn}n�1 be a sequence of Lusin spaces. Show that the productspace X =

∏n�1

Xn is a Lusin space too.

Problem 2.124

Show that a Souslin space is dispersible.

Problem 2.125

Suppose that X is a strongly Lindelof topological space andassume that {fi : X −→ R}i∈I is a family of lower semicontinuousfunctions. Show that there is a countable subfamily J ⊆ I such thatf = sup

i∈Ifi = sup

j∈Jfj.

Problem 2.126

Let X be a regular strongly Lindelof topological space. Show thatevery open set in X is a Fσ-set and every closed set in X is a Gδ-set.

Problem 2.127

Suppose thatX and Y are two topological spaces andX×X is stronglyLindelof. Suppose that {fi : X −→ Y }i∈I is a separating family ofcontinuous functions. Show that there is a countable subset J ⊆ Isuch that the sequence {fj : X −→ Y }j∈J is separating too.

Problem 2.128

Suppose that X is a Souslin space, Y is a topological space andf : X −→ Y is a function such that Gr f is a Souslin subspace ofX × Y . Show that the inverse image of every Souslin subspace of Y isa Souslin subspace of X.

Problem 2.129

Suppose that X and Y is two topological spaces with Y being regularand F : X −→ 2Y \ {∅} is a multifunction which is upper semicontinu-ous. Show that GrF is closed in X × Y . Is the converse true? Justifyyour answer.


Problem 2.130

Let X and Y be two topological spaces. Show that a multifunctionF : X −→ 2Y \ {∅} is compact valued (i.e., F (x) ⊆ Y is a compact setfor every x ∈ X) and upper semicontinuous if and only if for every net{(xi, yi)}i∈I ⊆ GrF such that xi −→ x in X, {yi}i∈I has a limit pointin F (x).

Problem 2.131

Suppose that X and Y are two topological spaces andF : X −→ 2Y \ {∅} is a compact valued multifunction (i.e., F (x) ⊆ Yis compact for every x ∈ X) which is upper semicontinuous. Showthat:(a) for every compact set K ⊆ X, the set F (K) ⊆ Y is compact;(b) GrF ⊆ X×Y is closed (note that no regularity on Y is assumed;cf. Problem 2.129).

Problem 2.132

Suppose that X is a paracompact topological space and f : X −→ R isan upper semicontinuous function, g : X −→ R a lower semicontinuousfunction such that

f(x) < g(x) ∀ x ∈ X.

Show that there exists a continuous function h : X −→ R such that

f(x) < h(x) < g(x) ∀ x ∈ X.

Problem 2.133

Suppose that X is a paracompact space, Y is a Banach space andF : X −→ 2Y \ {∅} is a lower semicontinuous function with closed andconvex values. Suppose that (x, y) ∈ GrF . Show that we can find acontinuous function g : X −→ Y such that g(x) ∈ F (x) for all x ∈ Xand g(x) = y.

Problem 2.134

Suppose thatX is a paracompact space, Y is a Banach space, C ⊆ X isa nonempty, closed set and F : X −→ 2Y \{∅} is a lower semicontinuousmultifunction with closed, convex values. Show that every continuousfunction f : C −→ Y such that f(x) ∈ F (x) for all x ∈ C can beextended to a continuous function f : X −→ Y such that f(x) ∈ F (x)for all x ∈ X.

2.2. Problems 285

Problem 2.135

Suppose that X is a compact topological space, Y is a topologicalvector space and F : X −→ 2Y \ {∅} is a multifunction with convexvalues such that for every y ∈ Y , the set

F−({y}) def={x ∈ X : y ∈ F (x)

}

is open. Show that there is a continuous function f : X −→ Y suchthat f(x) ∈ F (x) for all x ∈ X.

Problem 2.136

Suppose that X and Y are topological spaces and F : X −→ 2Y \{∅} isa multifunction which is lower semicontinuous or upper semicontinuouswith connected values. Show that F maps connected sets to connectedsets.

Problem 2.137

Suppose that X is a topological space and f : X −→ R is a function.Show the following:(a) f is lower semicontinuous if and only if the multifunction

X � x �−→ Lf (x) ={λ ∈ R : f(x) � λ

} ∈ 2R

is upper semicontinuous;(b) f is upper semicontinuous if and only if the multifunction

X � x �−→ Lf (x) ∈ 2R

is lower semicontinuous.

Problem 2.138

Suppose that X and Y are topological spaces, f : X × Y −→ R∗ =

R ∪ {±∞} is a lower semicontinuous function and F : Y −→ 2X \ {∅}is a lower semicontinuous multifunction. Let m(y)

def= sup

x∈F (y)f(x, y).

Show that the function m : Y −→ R ∪ {+∞} is lower semicontinuous.

Problem 2.139

Suppose that X and Y are two topological spaces, f : X×Y −→ R∗ =

R∪{±∞} is an upper semicontinuous function and F : Y −→ 2X \{∅}is an upper semicontinuous multifunction with compact values. Let

m(y)def= sup

x∈F (y)f(x, y). Show that the function m : Y −→ R ∪ {+∞}

is upper semicontinuous.


Problem 2.140

Suppose that X is a compact topological space and F : X −→ 2X \{∅}is an upper semicontinuous with compact values. Then there exists acompact set C ⊆ X such that F (C) = C.

Problem 2.141

Let (X, dX) be a compact metric space and let (Y, d

Y) be a complete

separable metric space (a Polish space). Show that C(X;Y ) with thec-topology (see Definition 2.174) is a Polish space.

Problem 2.142

Let X be a topological space and let

Pk(X) def

={C ⊆ X : C is nonempty and compact

}.

The topology on Pk(X)generated by the sets of form

{C ∈ Pk

(X):

C ⊆ U}and

{C ∈ Pk

(X): C ∩ U �= ∅}, where U ⊆ X is open, is

called the Vietoris topology on Pk(X). Show that

(a) The set of all finite nonempty sets in X is dense in Pk(X)with

the Vietoris topology.(b) If X is separable, then so is Pk

(X)with the Vietoris topology.

Problem 2.143

Let (X, dX) be a bounded metric space and let h be the Hausdorff

metric on Pk(X)(see Definition 1.134). Show that the Hausdorff

metric h induces the Vietoris topology on Pk(X)(see Problem 2.142).

Problem 2.144

Suppose that X is a topological space and A is a retract of X. Showthat A is closed. Also, if Y is another topological space and f : A −→ Yis a continuous function, then show that f admits a continuous exten-sion, i.e., there exists a continuous function f : X −→ Y such thatf |A = f .

Problem 2.145

Show that a retract of a normal space is itself normal.

Problem 2.146

Show that SN = ∂BN+11 =

{x ∈ R

N+1 : ‖x‖ = 1}

is a strongdeformation retract of RN+1 \ {0}.

2.2. Problems 287

Problem 2.147

Show the following:(a) A retract of a compact (respectively, (path-)connected) topologi-cal space is itself compact (respectively, (path-)connected).(b) A retract of a simply connected topological space is itself simplyconnected.(c) A retract of a retract is itself a retract (i.e., if C ⊆ A ⊆ X and Ais a retract of X and C is a retract of A, then C is a retract of X).

Problem 2.148

Show that for N � 3, the sets SN−1 = ∂BN1 =

{x ∈ R

N : ‖x‖ = 1}

and RN \ {0} are simply connected.

Problem 2.149

By constructing a homotopy equivalence show that [0, 1] and (0, 1) arehomotopy equivalent to the singleton.

Remark. Since every open interval (a, c) is homeomorphic to (0, 1),it follows that every open interval (a, c) is homotopy equivalent to {0}.This also holds for half-lines (a,+∞) and (−∞, a).

Problem 2.150

Consider the annulus Adef={(x, y) ∈ R

2 : 1 �√x2 + y2 � 2

}. Show

that A is homotopy equivalent to the unit sphere S1 def={(x, y) ∈ R

2 :x2 + y2 = 1

}.

Problem 2.151

Show that the cylinder I × S1 and the circle S1 are homotopy equiva-lent.

Problem 2.152

Suppose that X is a topological space, SNdef={x ∈ R

N+1 : ‖x‖ = 1}

and f : X −→ SN is a continuous function, which is not surjective.Show that f is nullhomotopic, i.e., f � 0.

Problem 2.153

Let X and Y be two topological spaces and let f : X −→ Y be acontinuous function. Show that Gr f is a retract of X × Y .


Problem 2.154

Compute the fundamental group of the following figure (the union oftwo tangent circles):

x0a b

C1 C2

Problem 2.155

Suppose that X is a compact topological space, A ⊆ X is a connected,closed subset of X and f : A −→ A is a continuous function.(a) Show that the set C =

⋂n�1

fn(A) is connected, where fn =

f ◦ . . . ◦ f︸︷︷︸n-times

.

(b) Let g : S1 −→ X \ C be a nullhomotopic function. Show thatthere exists an integer n � 1 such that g(S1) ⊆ X \ fn(A).

Problem 2.156

Consider the torus T = S1 × S1 and let A ⊆ T be defined by

Adef=[S1 × {1}] ∪ [{1} × S1

].

Is A a retract of T ? Justify your answer.Hint: Use the fact that π1(A) is nonabelian with two generators aand b.

Problem 2.157

Suppose that X and Y are two topological spaces, with X being path-connected, f : X −→ Y is a continuous function and x0, x1 ∈ X. Sup-pose that the induced homomorphism f∗ : π1(X,x0) −→ π1

(Y, f(x0)

)

is surjective. Show that f∗ : π1(X,x1) −→ π1(Y, f(x1)

)is surjective.

Problem 2.158

Find the fundamental group of the projective n-space PN (N � 2).

Problem 2.159

Find the fundamental group of P2 × S2.

2.2. Problems 289

Problem 2.160

(a) Suppose that X is a topological space, A is a retract of X,

r : X −→ A is a retraction and i : A −→ X is an inclusion function.Show that for all x ∈ A, (iA)∗ : π1(A, x) −→ π1(X,x) is injective andr∗ : π1(X,x) −→ π1(A, x) is surjective.(b) Show that the fundamental groups of R2 \ {0} and T = S1 × S1

(the torus) have infinite cyclic subgroups.

Problem 2.161

Let X be a path-connected topological space with a compact universalcover and let x ∈ X. Show that π1(X,x) is finite.

Problem 2.162

Let X be a locally path-connected and simply connected topologicalspace. Show that every continuous function f : X −→ S1 is nullhomo-topic. (Compare with Problem 2.173.)

Problem 2.163

Suppose that X is a path-connected and locally path-connected topo-logical space and π1(X) is finite. Show that every continuous functionf : X −→ S1 is nullhomotopic.

Problem 2.164

Show that homotopy equivalent spaces have the same number of path-connected components.

Problem 2.165

Let p : X −→ X be a covering map and suppose that x0, x1 ∈ X belongto the same path-connected component of X. Let u = p(x0) = p(x1).Show that p∗

(π1(X, x0)

)and p∗

(π1(X, x1)

)are conjugate subgroups

of π1(X, u), i.e., there exists v ∈ π1(X, u) such that p∗(π1(X, x1)

)=

v−1p∗(π1(X, x0)

)v.

Problem 2.166

Suppose that X is a topological space and A is a retract of X. Showthat

Hn(X) � Hn(A)⊕Hn(X,A) ∀ n � 0.


Problem 2.167

Suppose that X is a topological space and A ⊆ X is a deformationretract. Show that

Hn(X,A) = 0 ∀ n � 0.

Remark. In particular this implies that Hn(X,X) = 0 for all n � 0.

Problem 2.168

Suppose that X is a topological space and � ∈ X. Show that

Hn(X) = Hn(X, �) ⊕Hn(�) ∀ n � 0.

Problem 2.169

Suppose that X is a topological space and{Ak}mk=1

are disjoint closed

subsets of X such that Xdef=

m⋃k=1

Ak. Show that Hn(X) =m⊕k=1

Hn(Ak)

for all n � 0.

Problem 2.170

Show that, if X is a contractible topological space, then

Hn(X, �) = 0 ∀ n � 0, � ∈ X.

Problem 2.171

Suppose that X is a topological space and A ⊆ X is a contractiblesubspace. Show that

Hn(X,A) � Hn(X, �) ∀ n � 1, � ∈ X.

Problem 2.172

Show that

H0(S1) = H1(S

1) = Z and Hn(S1) = 0 ∀ n � 2.

Then show that

H0(S2) = H2(S

2) = Z and Hn(S2) = 0 ∀ n ∈ N \ {0, 2}.

Problem 2.173

(a) Suppose that N � 2. Is there a continuous function f : SN −→ S1

which is not nullhomotopic? Justify your answer.

2.2. Problems 291

(b) Let T = S1 × S1 be the torus. Is there a continuous functionf : T −→ S1 which is not nullhomotopic? Justify your answer.

Problem 2.174

Let X be a topological space. The suspension ΣX of X is the iden-tification space obtained from X × [−1, 1] by identifying X × {−1}to a point and X × {1} to another point (we call them identifica-tion points). Compute the homology groups of ΣX in terms of thehomology groups of X.

Problem 2.175

Let X be a topological space such that X = U ∪ V , with U, V ⊆ Xbeing open, path-connected with U ∩V nonempty and path-connectedand H1(U) = 0. Let the inclusion j2 : U ∩V −→ V induce a surjectivehomomorphism j2∗ : H1(U ∩ V ) −→ H1(V ). Show that H1(X) = 0.Is the result true if throughout H1 is replaced by H2? Justify youranswer.

Problem 2.176

Let BN1 =

{x ∈ R

N : ‖x‖ � 1}with N � 2 and let f : B

N1 −→ B

N1

be a continuous function such that f |SN−1

is a homeomorphism from

SN−1 to SN−1. Show that f is surjective.

Problem 2.177

Let X ={(x, y, z) ∈ R

3 : xy = 0}. Show that X is not homeomorphic

to R2 but it is of the same homotopy type.

Problem 2.178

Let X ⊆ R3 be the union of S2, of B

21 =

{u ∈ R

2 : ‖u‖ � 1}in the

xy-plane and of C being the portion of the z-axes which is inside S2.Compute the fundamental and homology groups of X.

Problem 2.179

Find the homology groups of a cylinder C = S1 × R.

Problem 2.180

Show that SN is not a retract of BN+11 (N � 1).

Remark. In contrast, in an infinite dimensional Banach space X,the set ∂B1 is a retract of B1 =

{x ∈ X : ‖x‖ � 1

}.


Problem 2.181

Let U ⊆ RN (N � 2) be an open set and let x ∈ U . Show that

HN−1(U \ {x}) �= 0.

Problem 2.182

Let U ⊆ RN , V ⊆ R

M be two nonempty open sets and N �=M . Showthat U and V cannot be homeomorphic.

Problem 2.183

Show that

Hk(SN , �) =

{Hk−N(�) if k � N,0 if k < N,

for � ∈ SN .

Problem 2.184

Let X1 ⊆ X2 ⊆ X3 ⊆ X4 be topological spaces. Show that

rankHk(X3, X2)− rankHk(X4, X1) � rankHk−1(X2, X1) + rankHk+1(X4, X3),

for all k � 1.

2.3. Solutions 293

2.3 Solutions

Solution of Problem 2.1Let x ∈ U ∩A and let V ∈ N (x). Then V ∩U ∈ N (x). Because x ∈ A,we have that

V ∩ (U ∩A) = (V ∩ U) ∩A �= ∅,

which shows that x ∈ U ∩A. Therefore we conclude that U ∩ A ⊆U ∩ A.

Now, let X = R with the natural topology. Let U = (0, 2) andA = (1, 3). Then

U ∩A = [1, 2) � [1, 2] = U ∩A.

Solution of Problem 2.2

(a) Let x ∈ ∂C and U ∈ N (x). Then x ∈ A ∩ U and

∅ �= (X \ C) ∩ U ⊆ (X \A) ∩ U,

which implies that x ∈ ∂A. Hence ∂C ⊆ ∂A.

(b) Let A = BN1 \ {0}, where BN

1 ={x ∈ R

N : ‖x‖ � 1}. Then

0 ∈ ∂A but 0 �∈ ∂BN1 . Therefore ∂A �= ∂A.

(c) From Proposition 2.11(e) and (f), we have

∂(A ∪ C) = (A ∪ C) ∩ (X \ (A ∪ C)) = (A ∪ C) ∩ ((X \A) ∩ (X \ C))

⊆ (A ∪ C) ∩ (X \A) ∩ (X \ C)⊆ (

A ∩ (X \A)) ∪ (C ∩ (X \ C)) = ∂A ∪ ∂C.

To see that this inclusion can be proper, let X = R, A = [0, 1) andC = [1, 2]. Then

∂A = {0, 1}, ∂C = {1, 2} and ∂(A ∪ C) = {0, 2}.

(d) We have

∂(A∩C) = ∂(X \(Ac∪Cc)) = ∂

(Ac∪Cc) ⊆ ∂Ac∪∂Cc = ∂A∪∂C


(see (c) above). Again, if X = R, A = [0, 1) and C = [1, 2]. Then

∂A = {0, 1}, ∂C = {1, 2} and ∂(A ∩C) = ∅

and so we see that the inclusion can be proper.

Solution of Problem 2.3Clearly A is a vector subspace of C[0, 1]. Consider the function f(x) =√x. Then f ∈ C[0, 1] \ A, since

f(x)x −→ +∞ as x→ 0+.

Let g ∈ A. For every ε > 0 the ball

Bε(g) ={h ∈ C[0, 1] : d∞(h, g) < ε

}

contains the function g+ ε2f which is not Lipschitz continuous. There-

fore intA = ∅.

Solution of Problem 2.4We have

intC = ∅ ⇐⇒ intC = ∅ ⇐⇒ X = int (X \ C) ∪ ∂(X \ C) = X \ C

(see Definition 2.9(e)) and so we conclude that C is nowhere dense ifand only if X \ C is dense.


(a) The set C ∩ A is closed in A (see Definition 2.14) and so CA ⊆

C ∩ A. Let x ∈ C ∩ A. Let U be a neighbourhood of x in A. ThenU = V ∩ A, with V being a neighbourhood of x in X. Since x ∈ C,

we have V ∩ C �= ∅. Since C ⊆ A, we have U ∩ C �= ∅, hence x ∈ CA.

Therefore C ∩A ⊆ CAand we conclude that C

A= C ∩A.

2.3. Solutions 295

(b) Note that intC is an open subset of X contained in A (seeProposition 2.15). Hence from the definition of the subspace topol-ogy (see Definition 2.14), we have that intC is also open in A. HenceintC ⊆ intAC.

This inclusion can be strict. To see this let X = R × R = R2,

A = R × {0} ⊆ R × R and C = (0, 1) × {0}. Then intAC = C, whileintC = ∅.

Solution of Problem 2.6Since by hypothesis D is open in both A and C, we can find two opensubsets U, V of X such that

D = U ∩A and D = V ∩A.

Then

(U ∩ V ) ∩ (A ∪ C) = (U ∩ V ∩A) ∪ (U ∩ V ∩ C) = D.

Because U ∩ V is open in X, we conclude that D is open in A ∪ C.

Solution of Problem 2.7“=⇒”: Since the set C is closed, from the definition of the subspacetopology (see Definition 2.14), the set C ∩U is closed in U (in fact forevery subset U ⊆ X).

“⇐=”: Arguing by contradiction, suppose that the set C is not closed.Then C �= C and so we can find x ∈ C \ C. For every neighbourhoodU of x, every open set V ⊆ U with x ∈ V satisfies C ∩ V �= ∅ (recallthat x ∈ C). So, we can find u ∈ C ∩ V = C ∩ U ∩ V and this meansthat x ∈ U \C belongs in the closure of C ∩U , which contradicts ourhypothesis.


Solution of Problem 2.8Evidently, we always have C ∩D ⊇ C ∩D. So, we need to showthat the opposite inclusion also holds. Let K = E \ D. We haveC ⊆ (C ∩D) ∪K, hence

C ⊆ C ∩D ∪K = C ∩D ∪K,so

C ∩D =(C ∩D ∪K) ∩D = C ∩D ∪ (K ∩D) = C ∩D,

thus C ∩D ⊆ C ∩D.

Solution of Problem 2.9Let Y be the countable subbasis (see Definition 2.19) and let F ⊆ 2Y

be the collection of all finite elements. Then F is countable. Everybasic element in X is of the form

⋂F with F ∈ F . Hence the basis B

is countable (see Definition 2.24).

Solution of Problem 2.10Let C be a closed subset of a Lindelof space X (see Definition 2.26)and let Y be an open cover of C. Every V ∈ Y has the form V = U∩C,where U is an open subset of X. Let Y∗ be the set of all such opensets U ⊆ X together with X \ C. Evidently this is an open cover ofX. Because X is a Lindelof space, we can find a countable subcoverof X. Removing X \ C (if it is included in the countable subcover),we see that Y has been reduced to a countable subcover of C. Thisshows that C with the subspace topology is a Lindelof space.

Solution of Problem 2.11Let U ∈ B, U �= ∅ and choose xU ∈ U . We claim that

D = {xU}U∈B\{∅}

is the desired set. First we show that D is dense in X (see Defini-tion 2.9(e)). To this end, let V be an open set in X. If u ∈ V , then

2.3. Solutions 297

we can find U ∈ B such that xU ∈ U ⊆ V (see Definition 2.19). HencexU ∈ V and so V ∩D �= ∅, which proves the density of D in X.

Next we show that #D � #B. For this purpose, we introduce thefunction ξ : B \ {∅} −→ D, defined by ξ(U) = xU . This function issurjective. The Axiom of Choice implies that there exists Y ⊆ B suchthat for every x ∈ D, Y ∩ ξ−1({x}) is a singleton. Then ξ : Y −→ D isa bijection and so we conclude that #D = #Y � #B.

Solution of Problem 2.12Let C ⊆ X be closed. Then from the definition of the subspace topol-ogy (see Definition 2.14), it follows that Ui ∩ C is closed in Ui for alli ∈ I.

Conversely, suppose that for every i ∈ I, the set Ui ∩ C is closedin Ui. Let {xj}j∈J ⊆ C be a net such that xj −→ x in X (see Def-inition 2.31). Because {Uj}j∈J is an open cover of X, we can findi0 ∈ I such that x ∈ Ui0 . Since xj −→ x, we can find j0 ∈ J such thatxj ∈ Ui0 for all j � j0, hence xj ∈ Ui0 ∩ C for all j � j0. But Ui0 ∩ Cis closed in Ui0 and xj −→ x. Hence x ∈ Ui0 ∩C and so x ∈ C, whichproves that C is closed in X.

Solution of Problem 2.13Yes. Consider the function f : [0, 1) −→ R, defined by

f(x) = 11−x sin

(1

1−x).

Evidently f is continuous and f([0, 1)

)= R.

Solution of Problem 2.14Let {fi : X −→ R}i∈I be a net of continuous functions and supposethat

supx∈X

∣∣fi(x)− f(x)∣∣ −→ 0


To show that f : X −→ R is continuous, it suffices to show that, if{xj}j∈J is a net such that xj −→ x, then f(xj) −→ f(x). To this end,let ε > 0 be given and we pick i0 ∈ I such that

∣∣fi(u)− f(u)∣∣ < ε

3 ∀ i � i0, u ∈ X.

Exploiting the continuity of fi0 , we can find j0 ∈ J such that

∣∣fi0(xj)− fi0(x)∣∣ < ε

3 ∀ j � j0.

Therefore for j � j0, we have∣∣f(xj)− f(x)

∣∣ �∣∣f(xj)− fi0(xj)

∣∣+∣∣fi0(xj)− fi0(x)

∣∣ +∣∣fi0(x)− f(x)

∣∣< ε

3 + ε3 + ε

3 = ε,

so f(xj) −→ f(x) and thus f is continuous.

Solution of Problem 2.15Suppose that f is lower semicontinuous (respectively, sequentiallylower semicontinuous; see Definitions 2.46 and 2.49). Let λ ∈ R andconsider the function

h(x, λ) = f(x)− λ.

Evidently h is lower semicontinuous (respectively, sequentially lowersemicontinuous) on X × R. Hence the set

{(x, λ) ∈ X×R : h(x, λ) � 0

}={(x, λ) ∈ X×R : f(x) � λ

}= epi f

is closed (respectively, sequentially closed; see Definition 2.51) inX×R.Now suppose that epi f is closed (respectively, sequentially closed)

in X × R. We need to show that for every λ ∈ R, the set

Lλ ={x ∈ X : f(x) � λ

}

is closed (respectively, sequentially closed) in X. So let {xα}α∈J ⊆ Lλ(respectively, {xn}n�1 ⊆ Lλ) be a net (respectively, sequence) such

that xα −→ x (respectively, xn −→ x) inX. Note that{(xα, λ)

}α∈J ⊆

epi f (respectively,{(xn, λ)

}n�1

⊆ epi f) and (xα, λ) −→ (x, λ) (re-

spectively, (xn, λ) −→ (x, λ)) in X × R. Then (x, λ) ∈ epi f and sof(x) � λ, which proves that Lλ is closed (respectively, sequentially

2.3. Solutions 299

closed), hence f is lower semicontinuous (respectively, sequentiallylower semicontinuous).

Solution of Problem 2.16From the definition of f (see Definition 2.57), we see that

f(x) = min{lim inf f(xi) : {xi}i∈I is a net converging to x

}.

Let x ∈ {x ∈ X : f(x) � λ}. For a given μ > λ, we can find a net

{xi}i∈I ⊆ X such that xi −→ x in X and

f(x) � lim inf f(xi) = supi0∈I

infi�i0

f(xi) < μ.

So, we can find a subnet {xj}j∈J of {xi}i∈I such that

f(xj) −→ lim inf f(xi) and f(xj) � μ.

Hencexj ∈ {

x ∈ X : f(x) � μ} ∀ j ∈ J

and since xj −→ x in X, we have that x ∈ {x ∈ X : f(x) � μ}. But

μ > λ was arbitrary. So, we infer that

{x ∈ X : f(x) � λ

} ⊆⋂

μ>λ

{x ∈ X : f(x) � μ}.

On the other hand let

x ∈⋂

μ>λ

{x ∈ X : f(x) � μ

}.

Thenx ∈ {

x ∈ X : f(x) � μ} ∀ μ > λ.

So, for a given μ > λ, we can find a net {xi}i∈I ⊆{x ∈ X : f(x) � μ

}

such that xi −→ x in X. Then f(xi) � μ and so

f(x) � lim inf f(xi) � lim inf f(xi) � μ.

Because μ > λ was arbitrary, we let μ −→ λ+ and so we have f(x) � λ.Hence we have

⋂

μ>λ

{x ∈ X : f(x) � μ

} ⊆ {x ∈ X : f(x) � λ

},


so {x ∈ X : f(x) � λ

}=⋂

μ>λ

{x ∈ X : f(x) � μ

}.

Since f � f , we have epi f ⊆ epi f . Because f is lower semicontinuous(see Definition 2.46), epi f is closed in X × R (see Problem 2.15) andso epi f ⊆ epi f . Suppose that the inclusion is strict. So, we can find(x, λ) ∈ epi f such that (x, λ) �∈ epi f . Hence there exist U ∈ N (x)and ε > 0 such that

(U × (λ− ε, λ+ ε)

) ∩ epi f = ∅.

Let μ ∈ (λ, λ + ε). We claim that x �∈ {y ∈ X : f(y) � μ}. Indeed,

if {xi}i∈I is a net in X converging to x, then we can find i0 ∈ I suchthat

(xi, μ) ∈ U × (λ− ε, λ+ ε) ∀ i � i0,

hence (xi, μ) �∈ epi f and so

f(xi) > μ ∀ i � i0,

which proves the claim. But then this contradicts the first part of theproblem.

Solution of Problem 2.17Let

g∗(x) = supU∈N (x)

infy∈U

f(y).

According to Remark 2.47, g∗ is lower semicontinuous (see Defini-tion 2.46) and of course g∗ � f . Therefore g∗ � f . On the other hand,let g ∈ L(f) (see Definition 2.57). Then again by Remark 2.47 andsince g � f , we have

g(x) = supU∈N (x)

infy∈U

g(y) � supU∈N (x)

infy∈U

f(y) = g∗(x),

2.3. Solutions 301

so f � g∗. Hence we conclude that

f(x) = supU∈N (x)

infy∈U

f(y) ∀ x ∈ X.

Solution of Problem 2.18Since f � f is lower semicontinuous (see Definitions 2.46 and 2.57),for every sequence xn −→ x in X, we have

f(x) � lim infn→+∞ f(xn) � lim inf

n→+∞ f(xn)

(see Proposition 2.48). This proves property (a). To prove (b), wemay assume that f(x) < +∞. Let {Un}n�1 be a local basis at x ∈ Xsuch that Un+1 ⊆ Un for all n � 1 (recall that by hypothesis X is firstcountable) and let λn −→ f(x) in R

∗ = R∪ {±∞} with λn > f(x) forall n � 1. From Problem 2.17, we have

infy∈Un

f(y) < λn ∀ n � 1

and so there exists yn ∈ Un such that f(yn) < λn. Evidently yn −→ yin X and so

lim supn→+∞

f(yn) � limn→+∞λn = f(x).

This proves property (b).


(a) Let x, u, y ∈ X. Then using the triangle inequality, we have

f(y) + λdX(x, y) � f(y) + λd

X(y, u) + λd

X(u, x),

sofλ(x) � fλ(u) + λd

X(u, x).

Reversing the roles of x and u in the above argument, we also have

fλ(u) � fλ(x) + λdX(u, x)

and so we conclude that fλ is λ-Lipschitz.


(b) Evidently fλ � f and from part (a), fλ is λ-Lipschitz. So fλ � f .Let x ∈ X and assume that

supλ>0

fλ(x) = M < +∞.

Choose xλ ∈ X such that

f(xλ) + λdX(x, xλ) � fλ(x) +

1λ ,

so

dX(x, xλ) � 1

λM + 1λ2

(since f � 0) and thus

dX(x, xλ) −→ 0 as λ→ +∞,

i.e., xλ −→ x in X. Then from the definition of f (see Definition 2.57),we have

f(x) � lim infλ→+∞

f(xλ) � lim infλ→+∞

(f(xλ)+λdX

(x, xλ))

� lim infλ→+∞

fλ(x).

Therefore, we conclude that

limλ→+∞

fλ(x) = f(x) ∀ x ∈ X.


(a) “=⇒”: We may assume that for some x0 ∈ X, f(x0) < +∞. Let

fn(x) = infy∈X

(f(y) + nd

X(x, y)

).

For every x ∈ X, we have

ξ � fn(x) � f(x) and ξ � fn(x) � f(x0)+ndX(x, x0) < +∞.

Therefore, we have

ξ � f1 � f2 � . . . � fn

2.3. Solutions 303

and each fn is R-valued. From triangle inequality, for every x, u, y ∈ X,we have

f(y) + ndX (x, y) � f(y) + ndX (y, u) + ndX (u, x)

sofn(x) � fn(u) + nd

X(u, x).

Reversing the roles of x, u, we also have

fn(u) � fn(x) + ndX(u, x),

so ∣∣fn(u)− fn(x)∣∣ � nd

X(u, x)

(i.e., fn is n-Lipschitz). We have

limn→+∞ fn(x) � f(x) ∀ x ∈ X.

For a given ε > 0, choose yn ∈ X for n � 1 such that

f(yn) + ndX(x, yn) � fn(x) + ε.

As n → +∞, either fn(x) ↗ +∞ and so fn(x) ↗ f(x) = +∞, sincelim

n→+∞ fn � f or else dX(x, yn) −→ 0. Therefore

f(x) � lim infn→+∞ f(yn) � lim

n→+∞ fn(x) + ε.

Since ε > 0 was arbitrary, we let ε↘ 0, to obtain

f(x) � limn→+∞ fn(x),

hence fn ↗ f as n→ +∞. Let us define

fn = min{n, fn}.

Then for every n � 1, the function fn is bounded continuous and

fn ↗ f.

“⇐=”: This part is an immediate consequence of Corollary 2.56.


(b) Consider −f which is lower semicontinuous and bounded belowand use part (a).

Solution of Problem 2.21The pointwise limit of upper semicontinuous functions (see Defini-tion 2.46) need not be an upper semicontinuous function. To see this,consider the following sequence fn : [0, 1] −→ R of continuous functions

fn(x) =

{nx if x ∈ [0, 1n ],1 if x ∈ ( 1n , 1],

∀ n � 1.

For every x ∈ [0, 1], we have fn(x) −→ f(x) as n→ +∞, where

fn(x) =

{0 if x = 0,1 if x ∈ (0, 1],

which is not upper semicontinuous (if xn −→ 0, xn �= 0, thenlim

n→+∞ f(xn) = 1 > 0 = f(0)).

On the other hand, the uniform limit of upper semicontinuous func-tions is an upper semicontinuous function. Let fn : R −→ R, n � 1, bea sequence of upper semicontinuous functions and assume that fn ⇒ f(see Definition 1.59). Let λ ∈ R and x ∈ {y ∈ R : f(y) > λ

}. Since

fn ⇒ f , we can find δ > 0 small enough such that

fn(x) > λ+ δ ∀ n � n1.

The upper semicontinuity of each fn, implies that we can find εn > 0such that

fn(y) > λ+ δ ∀ y ∈ (x− εn, x+ εn).

Because fn ⇒ f , we can find an integer n2 � n1 such that

∣∣fn(u)− f(u)∣∣ < δ ∀ u ∈ R, n � n2.

Hencefn(u) < f(u) + δ ∀ n � n2, u ∈ R

and soλ < f(u) ∀ u ∈ (x− εn, x+ εn),

2.3. Solutions 305

which shows that the set{y ∈ R : λ < f(y)

}is open. This in turn

implies the upper semicontinuity of f .

Solution of Problem 2.22Since D = X, for a given x ∈ X, we can find a net {xi}i∈I ⊆ D suchthat xi −→ x (see Proposition 2.33(a)). By hypothesis f(xi) = g(xi)for all i ∈ I. Also since f and g are continuous, from Proposition 2.40,we have

f(xi) −→ f(x) and g(xi) −→ g(x) in Y.

Therefore f(x) = g(x) (see Proposition 2.32).

Solution of Problem 2.23“=⇒”: Since f is continuous, for a given ε > 0, we can find U ∈ N (x)such that

dY

(f(u), f(x)

)< ε

2 ∀ u ∈ U.

Hence, if u, v ∈ U , then

dY

(f(u), f(v)

)� d

Y

(f(u), f(x)

)+ d

Y

(f(x), f(v)

),

so0 � ωf (x) � diam f(U) � ε.

Because ε > 0 was arbitrary, we let ε↘ 0 and obtain ωf (x) = 0.

“⇐=”: Since ωf (x) = 0, from the definition of the oscillation, wesee that for a given ε > 0, we can find U ∈ N (x) such that

diam f(U) < ε,

sodY

(f(u), f(x)

)< ε ∀ u, x ∈ U

and thus f is continuous at x ∈ X.


Solution of Problem 2.24Consider the oscillation function ωf : X −→ R+ = [0,+∞) (see Defi-nition 2.42). From Problem 2.23, we know that

cont f ={x ∈ X : ωf (x) = 0

}.

For every λ > 0, let

Lλ ={x ∈ X : ωf (x) < λ

}.

Let x ∈ Lλ, then we can find U ∈ N (x) such that

diam f(U) < λ

(from the definition of ωf ). So, if u ∈ U , then U ∈ N (u) and we haveωf (u) < λ, which proves that Lλ is an open set in X (hence ωf isupper semicontinuous; see Definition 2.46). But

cont f =⋂

n�1

{x ∈ X : ωf (x) <

1n

}

and each set{x ∈ X : ωf (x) <

1n

}is open. Therefore, cont f is a

Gδ-set in X (see Definition 2.18).

Solution of Problem 2.25We know that open half-lines (−∞, λ) and (λ,+∞) with λ ∈ R forma subbasis for the usual topology on R. Then by Proposition 2.40,the function f is continuous if and only if for every λ ∈ R, the setsf−1((−∞, λ)

)and f−1

((λ,+∞)

)are open in X. Therefore, the func-

tion f is continuous if and only if the sets{x ∈ X : f(x) < λ

}and{

x ∈ X : f(x) > λ}

are open. Hence we conclude that the func-tion f is continuous if and only if the sets

{x ∈ X : f(x) � λ

}and{

x ∈ X : f(x) > λ}are closed and open respectively.

2.3. Solutions 307

Solution of Problem 2.26“(a) =⇒ (b)”: For any n � 1, the function x �−→ (f − fn)(x) is con-tinuous and so for every ε > 0, the set

{x ∈ X :

∣∣f(x)− fn(x)∣∣ < ε

}

is open in X.

“(b) =⇒ (a)”: Let ε > 0 be given. By hypothesis, we can find strictlyincreasing sequence {nk}k�1 such that sets

Ck ={x ∈ X :

∣∣f(x)− fnk(x)∣∣ < ε

3

}

are all open in X. Since

fn(x) −→ f(x) ∀ x ∈ X,we have

X =⋃

k�1

Ck.

Let x0 ∈ X. Then, we can find k0 � 1 such that x0 ∈ Ck0 . Letx ∈ Ck0 . Then∣∣f(x)− f(x0)

∣∣ �∣∣f(x)− fnk0

(x)∣∣+∣∣fnk0

(x)− fnk0(x0)

∣∣+∣∣fnk0

(x0)− f(x0)∣∣

< 2ε3 +

∣∣fnk0(x)− fnk0

(x0)∣∣.

The continuity of fnk0implies that there exists U ∈ N (x0) such that

U ⊆ Ck0 and∣∣fnk0

(x)− fnk0(x0)∣∣ < ε

3 ∀ x ∈ U,

so ∣∣f(x)− f(x0)∣∣ < ε ∀ x ∈ U,

which establishes the continuity of f at x0 ∈ X. Since x0 ∈ X wasarbitrary, we conclude that f is continuous.

Solution of Problem 2.27Let L be the set of all open covers D of X such that for every V ∈ D,we have

V ∈ Y or V ⊆ U for some U ∈ Y.For each D ∈ L, let

ΓD ={V ∈ D : V ⊆ U with U ∈ Y}.


We introduce the order by inclusion on{ΓD : D ∈ L}. Then every

chain has an upper bound. Thus by the Kuratowski–Zorn lemma,there exists a maximal element ΓD∗ which coincides with Y. Then D∗is the desired locally finite open cover of X (see Definition 2.43).

Solution of Problem 2.28Let x �∈ ⋃

x∈KUα. Let V be an open neighbourhood of u such that V

intersects only Uα1 , . . . , Uαn (see Definition 2.43). Since u �∈ Uαi fori ∈ {1, . . . , n}, we can find Wi ∈ N (u) such that

Wi ∩ U i = ∅ ∀ i ∈ {1, . . . , n}.

We set

W = V ∩W1 ∩ . . . ∩Wn.

Then W ∈ N (x) and

W ∩ (⋃

α∈KUα)

= ∅,

so the set ⋃

x∈KUα

is closed.


(a) For every set D ⊆ X, we have

(g ◦ f)(D) = g(f(D)),

from which we immediately infer that, if f and g are open (respec-tively, closed), then g ◦ f is open (respectively, closed) too (see Defini-tion 2.59).

(b) Since g is bijective, then for every D ⊆ X, we have

f(D) =(g−1 ◦ (g ◦ f))(D).

2.3. Solutions 309

By hypothesis, we know that g◦f is open (respectively, closed). So, if Ais open (respectively, closed), then (g◦f)(A) ⊆ Z is open (respectively,closed). The continuity of g implies that

g−1((g ◦ f)(A)) = f(A)

is open (respectively, closed), which proves that f is an open (respec-tively, closed) function.(c) Because f is surjective, for every D ⊆ Y , we have

g(D) = (g ◦ f)(f−1(D)).

By hypothesis f is continuous and so, if D ⊆ Y is open (respectively,closed), then f−1(D) ⊆ X is open (respectively, closed). Since g ◦ fis assumed to be open (respectively, closed), we have that the set(g ◦ f)(f−1(D)

)= g(D) is open (respectively, closed). Therefore g is

open (respectively, closed).

Solution of Problem 2.30First suppose that A is open. Then the open subsets of A (with thesubspace topology) are the open subsets of X contained in A. There-fore their images under i are open in X and so i is an open function(see Definition 2.59). Next suppose that iA : A −→ X is an open func-tion. Since A is an open subset of itself, iA(A) = A ⊆ X is open.Similarly for the “closed case”.

Solution of Problem 2.31“=⇒”: Since f is bijection, we have

f−1(f(A)

) ⊇ f−1(f(A)

)= A ∀ A ⊆ X.

The set f−1(f(A)

)is closed and so we have

A ⊆ f−1(f(A)

) ∀ A ⊆ X.

On the other hand, since f is surjective, we have

f(f−1(f(A)

))= f(A) ∀ A ⊆ X


and because f is closed (see Definition 2.59), we have

f(A) ⊆ f(A) ∀ A ⊆ X.

Hencef(f−1(f(A)

)) ⊆ f(A) ∀ A ⊆ X.

Acting with f−1 and using the injectivity of f , we obtain that

f−1(f(A)

) ⊆ A ∀ A ⊆ X.

We conclude that

f−1(f(A)

)= A ∀ A ⊆ X.

“⇐=”: We show that f is also injective. So, suppose that f(x) = f(y).Then

{x} = {x} = f−1({f(x)}) = f−1

({f(y)}) = {y} = {y}.Now let A ⊆ X be closed. Then, since f−1

(f(A)

)= A = A and f is

surjective, we obtain f(A) = f(A), which proves that f is continuousand closed, hence a homeomorphism.

Solution of Problem 2.32Let C and D be disjoint closed subsets of Y . Because f is continuous,we have that f−1(C) and f−1(D) are disjoint closed subsets of X.Exploiting the normality of X (see Definition 2.4), we can find twodisjoint open sets U, V ⊆ X such that

U ⊇ f−1(C) and V ⊇ f−1(D).

Let us set E = Y \f(X \U). Since by hypothesis f is a closed function(see Definition 2.59), the set f(X \U) ⊆ Y is closed and so E is openin Y . Moreover, using the fact that f−1(C) ⊆ U and the surjectivityof f , we have

E = Y \ f(X \ U) ⊇ Y \ f(X \ f−1(C))

= Y \ f(f−1(Y \ C))

= Y \ (Y \ C) = C.

2.3. Solutions 311

Similarly, if G = Y \ f(X \ V ), then we show that G is open andG ⊇ D. Evidently E and G are disjoint open neighbourhoods of Cand D respectively and we have proved that Y is normal.

Solution of Problem 2.33Note that E is nowhere dense too and Ex is nowhere dense wheneverEx is of the first category (see Definition 1.25). So, we may assumethat E is closed.

Let {Vn}n�1 be a countable basis of Y (see Definition 2.24) and letD = (X×Y )\E. Then D is an open dense subset of X×Y . For eachn � 1, let

Dn = projX

((X × Vn) ∩D

),

where projX

is the projection from X × Y onto X. Hence

Dn ={x ∈ X : (x, y) ∈ D and y ∈ Vn

}.

Recalling that the projections are open functions (see Definition 2.59),we have that Dn ⊆ X is an open set. For any nonempty and open setU ⊆ X, we have

D ∩ (U × Vn) �= ∅(since D is dense) and so Dn ∩U �= ∅, which proves the density of Dn

in X. The set⋂n�1

Dn is the complement of a set of first category in

X. Let x ∈ ⋂n�1

Dn. Then

Dx ∩ Vn �= ∅ ∀ n � 1.

Hence Dx is an open dense subset of V and so Ex = Y \Dx is nowheredense. So for all x ∈ X \ C, with C of first category, Ex is a nowheredense set.

Solution of Problem 2.34If S is an open dense subset of X, then S×Y is open dense in X ×Y .So, if C ⊆ X is nowhere dense (see Definition 2.9(e)), then so is C×Yin X × Y . Since

( ⋃

k�1

Dk

)× Y =⋃

k�1

(Dk × Y ),


it follows that C ×D is of the first category in X × Y whenever C isin X. Similarly for D in Y .

Conversely, suppose that C ×D is of first category in X × Y andC is not of first category in X. As Y is second countable (see Defini-tion 2.24), then by Problem 2.33, we can find x ∈ C such that the set(C ×D)x is of first category in Y . But note that (C ×D)x = D. So,D is of first category in Y .

Solution of Problem 2.35Suppose that f is open (see Definition 2.59). Let τX and τY be thetopologies of X and Y respectively. We need to show that for U ⊆ Y ,f−1(U) ∈ τX if and only if U ∈ τY (see Definition 2.74). From the con-tinuity of f , we know that, if U ∈ τY , then f

−1(U) ∈ τX . Conversely,suppose that f−1(U) ∈ τX . Then since by hypothesis f is an openfunction, we have f

(f−1(U)

) ∈ τY . Because f is surjective, we havef(f−1(U)

)= U , hence U ∈ τY . This proves that f is an identification

function.

If f is closed, then the argument is similar by replacing open setswith closed ones and open functions with closed functions.

Solution of Problem 2.36“=⇒”: First note that the continuity of f implies thatproj

X: Gr f −→ X is a homeomorphism. Indeed, proj

Xis continuous,

injective and proj−1X

= ϕ, where ϕ : X −→ Gr f is defined by

ϕ(x) =(x, f(x)

),

which is continuous because each component function is continuous.Therefore proj

Xis a closed function (see Definition 2.59). Because f

is continuous, Gr f ⊆ X × Y is closed (see Problem 2.54). We endowGr f with the subspace product topology (see Definition 2.69). SinceGr f is closed in X × Y , every closed subset C of Gr f is also closedin X × Y . Then proj

Y(C) = f

(proj

X(C)). But since f and proj

Xare

both closed, we infer that projY(C) is closed in Y and so we conclude

that projYis a closed function.

2.3. Solutions 313

“⇐=”: Now we assume that projX: Gr f −→ X and

projY: Gr f −→ Y are both closed functions. We have

proj−1X

(x) =(x, f(x)

)and proj

Y

(proj−1

X(x))

= f(x),

i.e., projY◦ proj−1

X= f . Let C ⊆ X be closed. Then

f(C) = projY

(proj−1

X(C)).

But projX

is continuous, hence proj−1X

(C) is closed. By hypothesisproj

Yis a closed function, hence proj

Y

(proj−1

X(C))is closed in Y . This

proves that projY◦ proj−1

X= f is a closed function. Also, let D ⊆ Y

be a closed set. Then f−1(D) = projX

(proj−1

Y(D))(since f = proj

Y◦

proj−1X

). But projY

is continuous on Gr f and so proj−1Y

(D) ⊆ Gr fis closed. By hypothesis proj

Xis closed, hence proj

X

(proj−1

Y(D))=

f−1(D) is closed in X and this proves the continuity of f .

Solution of Problem 2.37First we show that

A× C = A× C.

Evidently, A×C ⊆ A×C and because A×C is closed, we have that

A× C ⊆ A× C.

On the other hand, let (x, y) ∈ A× C and let W be a neighbourhoodof (x, y). We know that we can find U ∈ N (x) and V ∈ N (y) suchthat (x, y) ∈ U × V ⊆ W (see Remark 2.70). Since x ∈ A and y ∈ C,we have

A ∩ U �= ∅ and C ∩ V �= ∅,hence

(A× C) ∩ (U × V ) �= ∅and so

(A×C) ∩W �= ∅.Since W was an arbitrary neighbourhood of (x, y), we conclude that(x, y) ∈ A× C and so, finally

A× C = A× C.

From this we infer that A × C is closed if and only if A ⊆ X andC ⊆ Y are both closed.


Next we show that

int (A× C) = intA× intC.

Evidently, intA× intC ⊆ A×C and because intA× intC is open, wehave intA × intC ⊆ int (A × C). Next, let (x, y) ∈ int (A × C). Wecan find U ∈ N (x) and V ∈ N (y) such that U × V ⊆ A × C. HenceU ⊆ A and V ⊆ C and these two inclusions imply that x ∈ intAand y ∈ intC, hence (x, y) ∈ intA × intC. So, we conclude thatint (A × C) = intA× intC. From this we infer that A× C is open ifand only if A ⊆ X and C ⊆ Y are both open.

Finally, note that

A× C = X × Y

�A× C = X × Y

�A = X and C = Y.

So, we conclude that A×C is dense in X ×Y if and only if A is densein X and C is dense in Y .

Solution of Problem 2.38Let x = (xi)i∈I ∈ X. For every finite set F ⊆ I and r > 0, we set

UF,r =∏

i∈IUF,r,i,

where

UF,R,i =

{Bir(xi) if i ∈ F,

Xi if i �∈ F,

withBir(xi) =

{u ∈ Xi : dXi

(u, xi) < r}.

We know that{UF,r

}F,r

form a local basis at the point x = (xi)i∈I ∈ X

in the product topology (see Definition 2.69 and Remark 2.70). Pro-ceeding by contradiction, suppose that X is metrizable. So, thereexists a metric d

Xon X generating the product topology. The balls

B2−n(x) ={u ∈ X : d

X(u, x) < 2−n

}, n � 1,

2.3. Solutions 315

form a local basis at the point x ∈ X. Therefore, for every n ∈ N, wecan find a finite set Fn ⊆ I and rn > 0 such that UFn,rn ⊆ B2−n(x).Let

S =⋃

n�1

Fn.

This set being a countable union of finite sets is countable. Thereforethe inclusion S ⊆ I is strict. Let y = (yi)i∈I ⊆ X be such that yi = xifor all i ∈ S. Then

yn ∈ UFn,rn ⊆ B2−n(x) ∀ n � 1.

Hence

y ∈⋂

n�1

B2−n(x) = {x}

and so y = x. This implies that yi = xi for all i ∈ I \ S. So we haveshown that for all i ∈ I \ S, Xi is a singleton, a contradiction to ourhypothesis. Therefore X equipped with the product topology is notmetrizable.


(a) By Definition 2.72, s(f) is the strongest topology on Y for which

f is continuous. Hence τY ⊆ s(f). On the other hand, if U ∈ s(f),then f−1(U) ∈ τX (from the definition of s(f)). The openness of f(see Definition 2.59) implies that f

(f−1(U)

) ∈ τY . But because f issurjective, we have

U = f(f−1(U)

) ∈ τY

and so we conclude that τY = s(f).

(b) By Definition 2.62, w(f) is the weakest topology on X for whichf is continuous. Hence w(f) ⊆ τX . On the other hand, let U ∈ τX .Because f is open, we have f(U) ∈ τY . The injectivity of f implies


that f−1(f(U)

)= U and from definition of the weak (initial) topology,

we have

f−1(f(U)

)= U ∈ w(f).

Therefore, we conclude that τX = w(f).

Solution of Problem 2.40From the definitions of w(f) and s(f) (see Definitions 2.62 and 2.72respectively), we see that in both cases f is continuous. Moreover, ifX is furnished with the weak (initial) topology w(f), then given V ∈w(f), we know that V = f−1(U) with U ⊆ Y open (see Remark 2.63).Hence due to surjectivity of f , we have f(V ) = f

(f−1(U)

)= U and

so we infer that f is an open function (see Definition 2.59). Thus f isa homeomorphism.

On the other hand, if Y is furnished with the strong (final) topologys(f), then for a given V ⊆ X open, we have f−1

(f(V )

)= V (since

f is injective) and so from the definition of s(f) (see Definition 2.72),we have that f(V ) ∈ s(f) and so f is an open function. Thus f is ahomeomorphism.

Solution of Problem 2.41For every open (respectively, closed) set A ⊆ X, we have that the setAs = p−1

(p(A)

)is open (respectively, closed) if and only if p(A) is

open (respectively, closed) (see the definition of the quotient topol-ogy; Definition 2.76) if and only if the quotient function p is open(respectively, closed).

Solution of Problem 2.42First note that directly from Definition 2.62, we have thatw(C(X)

) ⊆ τ . Also, we have

w(Cb(X)

) ⊆ w(C(X)

).

2.3. Solutions 317

On the other hand, let f ∈ C(X), x ∈ X and ε > 0 and consider thesubbasic element

U(x, f, ε) ={u ∈ X :

∣∣f(u)− f(x)∣∣ < ε

}.

We set

h(u) = min{f(x) + ε,max

{f(x)− ε, f(u)

}}.

Evidently h ∈ Cb(X) and

U(x, f, ε) = U(x, h, ε).

It follows that w(C(X)

) ⊆ w(Cb(X)

), hence w

(C(X)

)= w

(Cb(X)

).

So, the two weak topologies w(C(X)

)and w

(Cb(X)

)are equal for any

topological space.Now assume that (X, τ) is completely regular. Let U ∈ τ and

x ∈ U . Because of the complete regularity ofX, we can find f ∈ Cb(X)such that

f(x) = 0 and f |X\U = 1.

ThenV (f, x, 1) =

{u ∈ X : f(u) < 1

}

is a w(C(X)

)-neighbourhood of x and V (f, x, 1) ⊆ U . Therefore

τ ⊆ w(C(X)

)

and so from the observation of the beginning of proof, we concludethat

τ = w(C(X)

)= w

(Cb(X)

).

Conversely, suppose that τ = w(C(X)

)= w

(Cb(X)

). Let C ⊆ X be a

nonempty closed set and let x �∈ C. ThenX\C = U is w(C(X)

)-open,

contains x and we can find

V =

n⋂

k=1

{u ∈ X :

∣∣fk(u)− fk(x)∣∣ < 1

},

with fk ∈ C(X), k = 1, . . . , n such that x ∈ V ⊆ U . For everyk ∈ {1, . . . , n}, let

hk(u) = min{1,∣∣fk(u)− fk(x)

∣∣} ∀ u ∈ X

andh(u) = max

khk(u) ∀ u ∈ X.


Evidently h : X −→ [0, 1] is a continuous function satisfying

h(x) = 0 and h|C

= 1.

This proves that (X, τ) is completely regular.

Solution of Problem 2.43From the definition of complete regularity (see Problem 2.42), for everyx ∈ K, we can find a continuous function fx : X −→ [0, 1] such that

fx(x) = 1 and fx|X\U = 0.

Let

Ux ={y ∈ X : fx(y) >

12

}.

Then Ux is open, x ∈ Ux and {Ux}x∈K is an open cover of K. By thecompactness of K, we can find a finite subcover {Uxk}Nk=1. Let

g(x) = 1N

N∑

k=1

fxk(x) ∀ x ∈ X.

Then the function g : X −→ [0, 1] is continuous,

g|X\U = 0 and g(x) � 1

2N ∀ x ∈ K.

Let ξ : [0, 1] −→ [0, 1] be a continuous function such that

ξ(0) = 0 and ξ|[ 12N

,1]= 1.

Then f = ξ ◦ g is the desired continuous function.

Solution of Problem 2.44Clearly for (a) and (b) also, we may assume that f is not identically+∞.

Let f be coercive (respectively, sequentially coercive; see Defini-tion 2.103) and lower semicontinuous (respectively, sequentially lower

2.3. Solutions 319

semicontinuous; see Definitions 2.46 and 2.49). Let {xn}n�1 ⊆ X be aminimizing sequence for f . So

f(xn) −→ m = infx∈X

f(x) < +∞.

Let us consider the set

L =

{ {x ∈ X : f(x) � m+ 1

}if m ∈ R,{

x ∈ X : f(x) � 0}

if m = −∞.

Because f is coercive (respectively, sequentially coercive), the set Lis countably compact (respectively, sequentially compact; see Defini-tion 2.88). Note that we can find an integer n0 � 1 such that

{xn}n�n0 ⊆ L.

So the sequence {xn}n�1 must have a limit point (respectively, con-vergent subsequence); see Definition 2.88. This proves statement (c).Due to the lower semicontinuity (respectively, sequential lower semi-continuity; see Definitions 2.46 and 2.49) of f , we have

f(x) � lim infn→+∞ f(xn) = m

(see Remark 2.47) and so

m = f(x).

This proves statements (a) and (b).

Solution of Problem 2.45No. Let I be the set of all rational numbers in (0, 1) equipped withthe usual ordering of R. We introduce a net {xi}i∈I ⊆ [0, 1], definedby xi = i for all i ∈ I. Evidently xi −→ 1 in [0, 1] (see Definition 2.31).If i0 ∈ I, then

{xi}i�i0 ∪ {1} ={g ∈ [i0, 1] : g is a rational number

}

and the latter fails to be compact for any i0 ∈ I.


Solution of Problem 2.46“=⇒”: We assume that the sequence {fn}n�1 is equicoercive. Thenfor every λ ∈ R, we can find a closed, countably compact set Kλ (seeDefinition 2.88) such that

{x ∈ X : fn(x) � λ

} ⊆ Kλ ∀ n � 1.

Let ψ : X −→ R∗ = R ∪ {±∞} be defined by

ψ(x) = inf{μ ∈ R : x ∈ Kλ for all λ > μ

},

with the usual convention that inf ∅ = +∞. If fn(x) � μ for all n � 1,the x ∈ Kλ for all λ > μ, hence ψ(x) � μ. Therefore ψ � fn for alln � 1. Note that

{x ∈ X : ψ(x) � μ

}=⋂

λ>μ

Kλ,

so the set{x ∈ X : ψ(x) � μ

}is closed and countably compact

and thus ψ is lower semicontinuous (see Definition 2.46), coercive (seeDefinition 2.103).

“⇐=”: Suppose that there is a lower semicontinuous, coercive functionψ : X −→ R

∗ = R ∪ {±∞} such that ψ � fn for all n � 1. Then forall n � 1 and all λ ∈ R, we have

{x ∈ X : fn(x) � λ

} ⊆ {x ∈ X : ψ(x) � λ

}= Kλ

and the set Kλ is closed and countably compact. This shows that thesequence {fn}n�1 is equicoercive.

Solution of Problem 2.47From Problem 2.16, we know that

{x ∈ X : f(x) � λ

}=⋂

μ>λ

{x ∈ X : f(x) � μ

} ∀ λ ∈ R

(see Definition 2.57). Since f is coercive (see Definition 2.103), the

set{x ∈ X : f(x) � μ

}is countably compact (see Definition 2.88),

2.3. Solutions 321

hence the set{x ∈ X : f(x) � λ

}is countably compact too, which

establishes the coercivity of f .


(a) Note that f is coercive (see Definition 2.103 and Problem 2.47)

and lower semicontinuous (see Definition 2.46). So, we can find x0 ∈ Xsuch that f(x0) = inf

Xf (see Problem 2.44(a)). Clearly the constant

function infXf is lower semicontinuous and so inf

Xf � f . Hence

infXf � min

Xf.

Because f � f , we also have minX f � infX f and so we conclude thatminX f = infX f .

(b) Let x be a limit point of {xn}n�1 ⊆ X which is a minimizingsequence for f . Then from Remark 2.47, we have

f(x) � lim infn→+∞ f(xn) � lim inf

n→+∞ f(xn) = infXf = min

Xf

(see (a)), so

f(x) = minX

f.

(c) Suppose that X is first countable and let x ∈ X be a minimizer off (see (a)). Then by Problem 2.18(b), we can find a sequence {yn}n�1

such that yn −→ x and

lim supn→+∞

f(yn) � f(x) = minX

f = infXf.

Since f � f , we conclude that

limn→+∞ f(yn) = min

Xf = inf

Xf.


Solution of Problem 2.49Let {xi}i∈I be a net in proj

X(C) and assume that xi −→ x in X. Then

we can find yi ∈ Y such that (xi, yi) ∈ C. Because of the compactnessof Y , we can find {yj}j∈J a subnet of {yi}i∈I such that yj −→ y ∈ Y(see Theorem 2.81). We consider the corresponding subnet {xj}j∈J of{xi}i∈I and we have

(xi, yi) −→ (x, y) in X × Y.

Because{(xj , yj)

}j∈J ⊆ C and the latter set is closed, we have

(x, y) ∈ C. Moreover, from the continuity of the natural projectionproj

X(see Definition 2.69), we have

projX(xj , yj) −→ proj

X(x, y).

Hence x ∈ projX(C) and so we have proved that proj

X(C) is closed

in X.

Solution of Problem 2.50“=⇒”: It follows from the definition of compactness (see Defini-tion 2.78).

“⇐=”: Let Y = {Uα}α∈I be an open cover of X. We know thateach open set Uα is a union of basic open sets. Let Mα be the familyof basic open sets whose union is Uα. Then the collection of all basicopen sets V such that V belongs in some Mα is an open cover of X.By hypothesis this open cover has a finite subcover {Vk}nk=1. For eachVk, let αk ∈ I be the index such that Vk ∈ Mαk

. Then {Uαk}nk=1 is

the desired finite subcover of X and so X is compact.

Solution of Problem 2.51Let C ⊆ X be a closed set. Then C is compact and so f(C) ⊆ Yis compact too (see Proposition 2.82), in particular then closed (see

2.3. Solutions 323

Proposition 2.83). This proves that f is a closed function. ThenProblem 2.35 implies that f is an identification function (see Defini-tion 2.74).

Solution of Problem 2.52From Problem 2.51, we know that the function f : X −→ f(X) ⊆ Yis an identification function (see Definition 2.74). The points of X arethe sets

{f−1(y)

}, y ∈ Y . Define h : X −→ Y , by

h(f−1(y)

)= y ∀ y ∈ Y.

Then h is a bijection and if p : X −→ X = X/� is the quotient function(see Definition 2.76), then h ◦ p = f . So, by Proposition 2.73, h iscontinuous. Because X is compact, X is compact (as the image of Xunder the action of the continuous quotient function p). Therefore wecan apply Theorem 2.84 and conclude that h : X −→ f(X) ⊆ Y .


(a) Let τ and τ be two comparable compact Hausdorff topologieson a set X. Suppose that τ ⊆ τ . Then the identity functioni : (X, τ ) −→ (X, τ) is continuous. Then Theorem 2.84 implies that iis a homeomorphism, hence τ = τ .

(b) Let X be any countable set, let a, b ∈ X with a �= b. Let

A ={

1n : n � 1

} ∪ {0}

be the topological space with the natural topology (induced from R).Let f, g : A −→ X be two bijections such that f(a) = 0 and g(b) = 0.On X we consider two topologies τ = s(f) and τ = s(g) (see Def-inition 2.72). Then both topologies are Hausdorff compact, but notcomparable (as {a} ∈ τ \ τ and {b} ∈ τ \ τ).


Solution of Problem 2.54“=⇒”: Let

{(xi, yi)

}i∈I ⊆ Gr f be a net such that (xi, yi) −→ (x, y)

in X × Y (see Definition 2.31). Then xi −→ x in X and yi −→ yin Y . The continuity of f implies that f(xi) −→ f(x) in Y . Hencey = f(x) and so (x, y) ∈ Gr f , which proves that Gr f is closed inX×Y (see Proposition 2.40). For this implication we do not need thecompactness of Y (we need only the continuity of f).

“⇐=”: We argue by contradiction. So, suppose that f is not con-tinuous. Then we can find a net {xi}i∈I such that xi −→ x in Xand f(xi) �−→ f(x) (see Proposition 2.40). So, there is U ∈ N (f(x))and a subnet of

{f(xi)

}i∈I (which after renaming, we also denote by{

f(xi)}i∈I) such that f(xi) �∈ U for all i ∈ I. The compactness of Y

guarantees that a subnet of{f(xi)

}i∈I (still denoted by

{f(xi)

}i∈I)

converges to y ∈ Y . Then(xi, f(xi)

) −→ (x, y) in X × Y

and so y = f(x), which contradicts the fact that f(xi) �∈ U for alli ∈ I. Therefore f is continuous.

Finally we show that the second implication does not hold if wedrop the assumption on the compactness of Y . Let X = Y = R. Let

f(x) =

{1x if x > 0,0 if x � 0.

Then Grf ⊆ R× R is closed, but f is not continuous.

Solution of Problem 2.55Let D be a locally compact dense subset of X and let x ∈ D. Dueto the local compactness of D, we can find U open in D (with the

subspace topology) such that K = UD

(the closure in D) is compactin D. We know that U = V ∩D, with V ∈ N (x) and clearly K is alsocompact in X. A fortiori then K is closed in X and so X \K is openin X. Let u ∈ V \ K. We claim that u ∈ D. Indeed, if this is notthe case, because D is dense in X, u is a limit point of D and becauseV \K ∈ N (u), we have

(V \K) ∩D �= ∅ =⇒ U ∩Kc �= ∅,

2.3. Solutions 325

a contradiction. So, we infer that V \K ⊆ D and so V ⊆ D, henceV = U . This shows that every point x ∈ D has an X-open neighbour-hood which remains in D, hence D is open.

Solution of Problem 2.56We will do the proof for a decreasing net {fi}i∈I (the proof beingsimilar if the net is increasing). Replacing fi by fi− f if necessary, wemay assume that f = 0.

Let ε > 0 be given. Then for every x ∈ X we can find ix ∈ I suchthat

0 � fix(x) < ε.

The continuity of fix implies that we can find Ux ∈ N (x) such that

0 � fix(u) < ε ∀ u ∈ Ux.

Since

fi � fix ∀ i � ix,

we see that

0 � fi(u) < ε ∀ u ∈ Ux, i � ix.

Note that {Ux}x∈X is an open cover of X. The compactness of X (seeDefinition 2.78) implies that we can find a finite subcover {Uxk}nk=1

of {Ux}x∈X . Choose i∗ ∈ I such that i∗ � ixk for all k ∈ {1, . . . , n}.Then

0 � fi(u) < ε ∀ u ∈ X, i � i∗,

so

fi ⇒ f

(see Definition 1.59).

(b) The functions

fn : [0, 1) � x �−→ xn

satisfy fn ↘ 0, but the convergence is not uniform. In this case thespace X = [0, 1) is not compact.


(c) The functionsfn : [0, 1] � x �−→ xn

satisfy fn ↘ χ{1} , but the convergence is not uniform. In this case thelimit function f(x) = χ{1} is not continuous.

Solution of Problem 2.57Let σ : [0, 1] −→ S1 be the map, defined by

σ(t) =(cos 2πt, sin 2πt

).

Let P (f) be the partition of [0, 1] determined by the σ-preimages ofthe elements of S1. Then P (f) is the same as the partition inducedby the equivalence relation 0 ∼ 1. Finally note that σ : I/P (f) −→ S1

is a homeomorphism onto, since I/P (f) −→ S1 is compact and σ is a

bijection (see Proposition 2.83(e) and Theorem 2.84).

Solution of Problem 2.58Let en = {δk,N+1}N+1

k=1 be the “north pole” for the sphere SN and lets : SN \ {en} −→ R

N be the stereographic projection, defined by

s(u) = 11−uN+1

(u1, . . . , uN ) ∀ u ∈ SN \ {en}.

Clearly, if U ⊆ SN is an open set containing the north pole en, thens(SN \ U) is compact and s|

SN\U is homeomorphism. Therefore the

Alexandrov one-point compactification of RN (see Remark 2.97) is

homeomorphism to SN .

Solution of Problem 2.59Let {Ui}i∈I be a family of open sets in X whose union contains K.Since x ∈ K, we can find an element Ui∗ of this family such thatx ∈ Ui∗ . Because xn −→ x, we can find n0 � 1 such that xn ∈ Ui∗ forall n � n0. For every 1 � n � n0, we have xn ∈ K and so we can findin ∈ I such that xn ∈ Uin . Then the subfamily {Uin}n0−1

n=1 ∪ {Ui∗}

2.3. Solutions 327

is a finite subcover of K. This proves that K is compact (see Defini-tion 2.78(b)).

Solution of Problem 2.60First assume that D = {x}. Since X is Hausdorff, for every y ∈ C, wecan find Uy ∈ N (y) and Vy ∈ N (x) such that Uy ∩Vy = ∅. The family{Uy}y∈C is an open cover of C and so we can find a finite subfamily

{Uyk}nk=1 such that C ⊆n⋃k=1

Uyk . We set

Ux =

n⋃

k=1

Uyk and Vx =

n⋂

k=1

Vyk .

Both are open sets and

C ⊆ Ux, x ∈ Vx and Ux ∩ Vx = ∅.So, we have established the separation when D is a singleton.

For the general case, we proceed as follows. From the first partof the proof, we know that for every x ∈ D, we can find an open setUx with C ⊆ Ux and Vx ∈ N (x) such that Ux ∩ Vx = ∅. The family{Vx}x∈D is an open cover of D. The compactness of D implies thatwe can find a finite subcover {Vxk}nk=1 of {Vx}x∈D. Let

U =n⋂

k=1

Uxk and V =n⋃

k=1

Vnk.

Then U and V are two open sets in X such that C ⊆ U , D ⊆ V andU ∩ V = ∅.


C1 = K \ U1 and C2 = K \ U2.

Both are closed subsets of K, hence they are compact. Moreover,C1 ∩ C2 = ∅. By Problem 2.60, we can find two open sets V1 and V2such that

C1 ⊆ V1 ⊆ U2, C2 ⊆ V2 ⊆ U1 and V1 ∩ V2 = ∅.


Let

K1 = K \ V1 and K2 = K \ V2.Then both are compact, K1 ⊆ U1, K2 ⊆ U2 and K = K1 ∪K2.

Solution of Problem 2.62Let {xi}i∈I ⊆ f−1(K) be a net such that xi −→ x. Then

f(xi) = yi ∈ K ∀ i ∈ I.

The compactness of K (see Definition 2.78) implies that there is asubnet {yj}j∈J of {yi}i∈I such that yj −→ y ∈ K. Then (xj , yj) ∈ Gr fand (xj , yj) −→ (x, y) in X × Y . The closedness of Gr f implies that(x, y) ∈ Gr f . Hence

x ∈ f−1(y) ⊆ f−1(K)

and this proves that the set f−1(K) ⊆ X is closed.

Solution of Problem 2.63Let K ⊆ Y be a compact set and consider a net {xi}i∈I ⊆ f−1(K).Then

{yi = f(xi) : i ∈ I

} ⊆ K. Because K is compact, we canfind a subnet of {yi}i∈I (which for the sake of notational simplicity, wedenote by the same index) such that yi −→ y ∈ K. For every i0 ∈ I,we define Ai0 =

{xi : i � i0

}. Since f is closed (see Definition 2.59),

we have

f(Ai) ⊆ f(Ai) ∀ i ∈ I

and because yi = f(xi) −→ y in Y , it follows that

{y} =⋂

i∈If(Ai) ⊆

⋂

i∈If(Ai).

Let C = f−1(y). Then for every x ∈ C, we have

f(x) ∈⋂

i∈If(Ai)

2.3. Solutions 329

and so the closed sets Di = C∩Ai ⊆ C are nonempty and clearly havethe finite intersection property. Since by hypothesis C is compact, itfollows that ⋂

i∈IDi = C ∩ (

⋂

i∈IAi) �= ∅.

So, we can find a limit point of {xi}i∈I belonging in C. Thus wecan find a suitable subnet of {xi}i∈I (still labelled by the same indexset I) such that xi −→ x ∈ C ⊆ f−1(K). This proves that f−1(K) iscompact.

Solution of Problem 2.64We claim that f is bounded from above. Indeed, if this is not the case,then for every integer n � 1, we can find xn ∈ X such that f(xn) � n.We consider the decreasing sequence of nonempty closed set

{Cn = f−1

([n,+∞)

)}n�1

.

Evidently this has the finite intersection property. Since X is compact,by Theorem 2.81, we have that

⋂n�1Cn �= ∅. Let y ∈ ⋂

n�1Cn. Then

f(y) � n ∀ n � 1,

hence f(y) = +∞, a contradiction. Therefore supXf =M < +∞. Let

En ={x ∈ X : f(x) �M − 1

n

} ∀ n � 1.

By hypothesis each En, n � 1 is closed and {En}n�1 is decreasing.Once again the compactness of X via Theorem 2.81 implies that

⋂

n�1

En �= ∅.

Then, if x0 ∈⋂n�1

En, we have

f(x0) � M − 1n ∀ n � 1,

hencef(x0) = M = sup

Xf < +∞.


Solution of Problem 2.65The continuity of p

Yfollows from the definition of the product topology

(see Definition 2.69). Let C ⊆ X×Y be closed and let {yi}i∈I ⊆ pY(C)

be a net such that yi −→ y. Then we can find a net {xi}i∈I ⊆ X suchthat (xi, yi) ∈ C for all i ∈ I. The compactness of X implies thatwe can find a subnet of {xi}i∈I (which we denote as the original net)such that xi −→ x ∈ X. Then (xi, yi) −→ (x, y) and because byhypothesis C is closed, we have that (x, y) ∈ C and so y ∈ p

Y(C),

which proves that pY

is closed (see Definition 2.59). Also, if y ∈ Y ,then p−1

Y(y) ⊆ X × {y} and X × {y} ⊆ X × Y is compact (see Theo-

rem 2.91). Because p−1Y

(y) is closed (due to the continuity of pY ), weconclude that p−1

Y(y) ⊆ X × Y is compact.

Solution of Problem 2.66Let {Ui}i∈I be an open cover of X. Because X is Lindelof (see Defi-nition 2.26), we can find a countable subcover {Un}n�1. We need toextract from {Un}n�1, a finite subcover. We proceed by induction.Let V1 = U1 and for every integer n � 2, let Vn be the first set in thesequence {Un}n�1 not covered by U1∪. . .∪Un−1. If such a choice is not

possible, then this means that {Uk}n−1k=1 is a cover of K and so we are

done. Otherwise, for every n � 1, choose xn ∈ Vn such that xn �∈ Vkfor all integers 1 � k < n. By hypothesis the sequence {xn}n�1 hasa limit point x ∈ X. Then x ∈ Vm for some integer m � 1 and sinceit is a limit point of {xn}n�1, we can find n > m such that xn ∈ Vm,a contradiction to the choice of the sequence {xn}n�1. So the cover{Un}n�1 has a finite subcover, which proves that X is compact (seeDefinition 2.78).

Solution of Problem 2.67Arguing indirectly, suppose that intK �= ∅ and let x ∈ intK. Thenwe can find a basic element U such that x ∈ U ⊆ K. We know that

U =⋂

i∈Fp−1i (Vi),

2.3. Solutions 331

with finite set F ⊆ I and open sets Vi ⊆ Xi (see Definition 2.69). Ifj ∈ I \ F , then

pj(K) = Xj

and so Xj is compact being the continuous image of a compact space(see Proposition 2.82). So, all but a finite number of the spaces Xi arecompact, a contradiction to our hypothesis. This prove that intK = ∅(i.e., K is nowhere dense; see Definition 1.25).

Solution of Problem 2.68Let C ⊆ Y be a closed set. Since by hypothesis f |

Kis continuous,

we have that f−1(C) ∩ K is closed in K and because K is compact,it is also closed in X (see Proposition 2.83). But then, because X isa k-space, we infer that f−1(C) is closed in X, which in turn impliesthe continuity of f (see Proposition 2.39).

Remark. It can be shown that locally compact or first countablespaces are k-spaces. In particular metric spaces are k-spaces.

Solution of Problem 2.69Since X is locally compact (see Definition 2.92), for every x ∈ A, wecan find U ∈ N (x) such that U is compact in X.

If A is open and since {x} is compact, invoking Proposition 2.94,we can choose U ∈ N (x) such that

x ∈ U ⊆ UA ⊆ U ⊆ A

and this proves that A with the subspace topology is locally compact.If A is closed, then

U ∩ A ∈ NA(x) and U ∩AA = U ∩A ⊆ U

with the latter being compact. Therefore, A with the subspace topol-ogy is again locally compact.


Solution of Problem 2.70Let y ∈ X \ {x}. We can find U ∈ N (x) and V ∈ N (y) such thatU ∩ V = ∅. Note that X \ U is closed in X, hence compact in X (seeProposition 2.83). Also

V ∈ NX\{x}(y) and VX\{x} ⊆ X \ U,

henceVX\{x}

is compact inX \ {x}(see Proposition 2.79). So, we conclude that X \{x} is locally compact(see Definition 2.92).

Solution of Problem 2.71Let x ∈ intK. By replacing K with its image under the translationy �−→ y − x (which is a homeomorphism of RN with itself), we mayassume that x = 0 ∈ intK. So, we can find ε > 0, small such that

BNε =

{y ∈ R

N : ‖y‖ < ε} ⊆ K.

Then using the dilation x �−→ 1εx (another homeomorphism on R

N

with itself), we see that without any loss of generality we may assumethat BN

1 ⊆ K.Now we show that every ray originating at origin intersects ∂K at

exact one point. Because K is compact, the intersection with everyclosed ray is compact. Denote this compact intersection by D. Sincethe distance function from the origin is continuous, it attains its supre-mum on D. Let x0 ∈ D be a point where it realizes this supremum.Evidently x0 ∈ ∂K. Consider the line segment

[0, x0] ={x ∈ R

N : tx0, t ∈ [0, 1]}.

Since BN1 ⊆ K, every line segment from x0 to u ∈ BN

1 is contained inK. As y moves over all BN

1 , we sweep a set C (which looks like anice-cream cone). For every point tx0, t ∈ [0, 1), we see that

B1−t(tx0) ={y ∈ R

N : ‖y − tx0‖ < 1− t}

is in C, hence in K too. This proves that x0 is unique and so we haveproved that every ray originating at origin intersects ∂K at exact onepoint.

2.3. Solutions 333

Now let f : ∂K −→ SN−1 = ∂BN1 be a function, defined by

f(x) = x‖x‖ ∀ x ∈ ∂K.

Evidently f(x) corresponds to the point where the line segment joining

the origin and x ∈ ∂K intersects the unit sphere SN−1 = BN1 (recall

that BN1 ⊆ K). Clearly f is continuous and a bijection (see the previ-

ous discussion). Because ∂K is compact, we can apply Theorem 2.84

and infer that f is a homeomorphism. Next let h : BN1 −→ K be a

function defined by

h(x) = ‖x‖f−1(x

‖x‖).

Clearly h is continuous and takes every line segment joining the ori-gin with u ∈ SN−1 onto the line segment joining the origin withf−1(u) ∈ ∂K. The convexity of K implies that h has its values inK. Evidently h is injective and onto, therefore a homeomorphism(again Theorem 2.84).

Solution of Problem 2.72Let C ⊆ X be a closed set and let y ∈ ∂f(C). Since Y is locallycompact (see Definition 2.92) we can find U ∈ N (y) such that U iscompact in Y . Because y ∈ ∂f(C), for every V ∈ N (y), we have

f(C) ∩ V �= ∅ and(X \ f(C)

) ∩ V �= ∅.Using this we can easily verify that we also have y ∈ ∂

(f(C) ∩ U

).

From the hypothesis on f , we have that f−1(U) is compact in X,which implies that C ∩ f−1(U) ⊆ X is compact too. Because f is acontinuous bijection we have

f(C ∩ f−1(U)

)= f(C) ∩ f(f−1(U )

)= f(C) ∩ U

is compact (see Proposition 2.82). So, y ∈ f(C) ∩ U ⊆ f(C), whichshows that ∂f(C) ⊆ f(C) and this proves that f(C) is closed. Thusf is closed (see Definition 2.59).


Solution of Problem 2.73First we note that for every f ∈ Φ, the weak topology w({f}) onX (see Definition 2.62) is semimetrizable (i.e., it is generated by asemimetric; see Remark 1.2). To see this, let

df(x, x′) = d

Y

(f(x), f(x′)

) ∀ x, x′ ∈ X.Then w({f}) is generated by the semimetric d

f. To establish this

it is enough to show that the two topologies w({f}) and τdf

have

the same convergent nets (see Definition 2.31). By Proposition 2.66,xi −→ x in w({f}) if and only if dY

(f(xi), f(x)

)= d

f(x1, x) −→ 0.

Therefore w({f}) = τdfand so w({f}) is semimetrizable. Because

Φ is a countable family and the supremum of a countable family ofsemimetrizable topologies is semimetrizable, it follows that

∨

f∈Φw({f}) = w(Φ)

is semimetrizable (∨f∈Φ

w({f}) denoting the supremum of the topolo-

gies{w({f})}

f∈Φ).In fact since by hypothesis Φ is separating (see

Definition 2.64), w(Φ) is metrizable. By the definition of the weaktopology, we have that w(Φ) ⊆ τ , where τ denotes the topology onX. Invoking Theorem 2.84, we conclude that w(Φ) = τ and so τ ismetrizable (see Remark 2.2).

Solution of Problem 2.74“=⇒”: Since X is locally compact (see Definition 2.92) and σ-compact(see Definition 2.99), by Proposition 2.100, we can find an increasingsequence {Cn}n�1 of compact subsets of X such that

X =⋃

n�1

Cn and Cn ⊆ Cn+1 ∀ n � 1.

Then{X∗ \ Cn

}n�1

is a local basis of ∞ in X∗ (see Definition 2.23

and Remark 2.97).

“⇐=”: Let {U∗n}n�1 be a local basis of ∞ in X∗. We have

U∗n = X∗ \Kn, ∀ n � 1,

2.3. Solutions 335

with Kn ⊆ X being compact. Then

X =⋃

n�1

Kn

and so X is σ-compact.

Solution of Problem 2.75(a) Let C1 = f(X) and inductively define

Cn+1 = f(Cn) ∀ n � 1.

This way we produce a decreasing sequence {Cn}n�1 of nonempty andcompact sets (hence closed too; see Proposition 2.83). Due to thecompactness of X,

the set⋂

n�1

Cn = A is nonempty and compact.

Clearly f(A) = A.

(b) Let X = (0, 1] and consider the continuous function

f(x) = x2 ∀ x ∈ X.

Suppose that there is a nonempty and closed subset A ⊆ (0, 1] suchthat f(A) = A. Let x = supA ∈ A. Because f(A) = A, we can findu ∈ A such that f(u) = 1

2u = x. Therefore we have

u = 2x � x,

a contradiction (since x �= 0).

Solution of Problem 2.76Arguing by contradiction, suppose that no such λ > 0 exists. Wedefine the sets

Cn ={x ∈ X : nf(x) + g(x) � 0

} ∀ n � 1.


This is a decreasing sequence of nonempty closed subsets of X. ByTheorem 2.81, we have that

⋂

n�1

Cn �= ∅.

Let u ∈ ⋂n�1

Cn. Then

f(u) � − g(u)n ∀ n � 1,

so

f(u) = 0

(recall that f � 0). Thus, by hypothesis g(u) > 0 and so f(u) < 0, acontradiction.

Solution of Problem 2.77Let i0 ∈ I and consider the set D = Ci0 \ U . Then D is compactand {X \ Ci}i∈I is an open cover of D. Therefore we can find a finitesubcover {X \ Ci}i∈F . Hence

⋂

i∈FCi ⊆ U.

Solution of Problem 2.78Since f is locally bounded, for every x ∈ X, we can find a neighbour-hood Ux of x and Mx > 0 such that

∣∣f(y)∣∣ � Mx ∀ y ∈ Ux.

Then {Ux}x∈X is an open cover of X. Since X is compact, we can finda finite subcover {Uxk}Nk=1 of X (see Definition 2.78). Let

M = max1�k�N

Mxk .

2.3. Solutions 337

Then ∣∣f(x)∣∣ � M ∀ x ∈ X

and so we conclude that f is bounded.

Solution of Problem 2.79Let x ∈ D. Since by hypothesis D is locally compact, we can find anopen set U in D such that x ∈ U and U ∩D is compact, hence closedin X. We have

U ⊆ U ∩D ⊆ D.

Let V be an open subset of X such that U = V ∩D. From the densityof D in X and the fact that V is open in X, we have

V = V ∩D,so

x ∈ V ⊆ V = V ∩D = U ⊆ D.

So, for every x ∈ D, there exists an open set V in X such that x ∈V ⊆ X, which means that D is open in X.

Solution of Problem 2.80Let d

Xbe a metric on X generating the topology of X and let X be

the completion of (X, dX). Then X is a locally compact dense subset

of the metric space X. By Problem 2.79, X is open in X. Invokingthe Alexandrov theorem (see Theorem 1.58), we conclude that X iscompletely metrizable.

Solution of Problem 2.81No. To show this we argue indirectly. So, suppose that

f : S1 −→ [0, 1]

is a homeomorphism. Let h = f−1 : [0, 1] −→ S1. Then consider

X = [0, 12 ) ∪(12 , 1]

and Y = S1 \ h(12).


Hence h(X) = Y and f(Y ) = X. Equip X and Y with the corre-sponding subspace topologies. Then f : Y −→ X and h : X −→ Y arecontinuous bijections and h = f−1. ThereforeX and Y are homeomor-phic. However, note that X is disconnected (U = [0, 12) and V = (12 , 1]is a disconnection of X; see Definition 2.104) while Y is connected(being homeomorphic to (0, 1) by the stereographic projection; cf. thesolution of Problem 2.58). So, X and Y cannot be homeomorphic,a contradiction. This proves that S1 and [0, 1] cannot be homeomor-phic.

Solution of Problem 2.82Since K is disconnected (see Definition 2.104), we can find twononempty disjoint closed sets C,D ⊆ K such that K = C ∪ D. Thesets C and D are compact and disjoint. So, we can use Problem 2.60and obtain two disjoint open sets U, V ⊆ X such that C ⊆ U andD ⊆ V . Then

K = C ∪D ⊆ U ∪ V.Moreover, we have

K ∩ U = C �= ∅ and K ∩ V = D �= ∅.


(a) Note that the sets Kn are closed and Kn ⊆ K1 for all n � 1with K1 being compact. Moreover, the sequence {Kn}n�1 has fi-nite intersection property. Therefore, by Theorem 2.81, we have thatK =

⋂n�1

Kn is nonempty and closed, hence compact.

(b) We argue by contradiction. So, suppose that for every n � 1, wehave Kn ∩ U c �= ∅. Let xn ∈ Kn ∩ U c for all n � 1. Then {xn}n�1 isa sequence in K1, which is compact and so {xn}n�1 has a limit pointx ∈ K1. Note that x ∈ Kn for every n � 1 and so x ∈ K. Also, x ∈ U c,since the latter is closed. Therefore K ∩ U c �= ∅, a contradiction tothe fact that K ⊆ U . This means that there is an integer n0 � 1such that Kn ⊆ U for all n � n0 (recall that the sequence {Kn}n�1 isdecreasing).

2.3. Solutions 339

(c) Arguing indirectly, suppose that K is disconnected (see Defini-tion 2.104). Then according to Problem 2.82, we can find two openand disjoint sets U, V ⊆ X such thatK ⊆ U∪V , K∩U �= ∅, K∩V �= ∅.From part (b) above, we know that there exists an integer n0 � 1 suchthat Kn ⊆ U ∪V for all n � n0. Then Kn ∩U and Kn ∩V is a discon-nection of Kn for all n � 1, a contradiction to the fact that for everyn � 1, the set Kn is connected.

Solution of Problem 2.84No. Consider the following counterexample:

Cn ={(x1, x2) ∈ R

2 : |x1| � n or |x2| � 1} ∀ n � 1.

Evidently {Cn}n�1 is a decreasing sequence of nonempty closed setswhich are connected (in fact path-connected; see Definitions 2.104and 2.122). However

C =⋂

n�1

Cn ={(x1, x2) ∈ R

2 : |x2| � 1}

which is obviously disconnected.

Solution of Problem 2.85Because [0, 1] is compact and connected and f is a homeomorphism,A× C is compact and connected too (see Theorem 2.91 and Proposi-tion 2.110). Then from Theorem 2.91 and Proposition 2.110, it followsthat both sets A and C are compact and connected. If x, u ∈ A, x �= uand y, v ∈ C, y �= v, then since f−1 is a homeomorphism, we have

Ex = f−1({x} × C

), Eu = f−1

({u} × C),

Gy = f−1(A× {y}), Gv = f−1

(A× {v})


are all compact and connected in [0, 1], hence closed subintervals of[0, 1]. Note that

Ex ∩Gy = f−1({x, y}), Ex ∩Gv = f−1

({x, v}),

Eu ∩Gy = f−1({u, y}), Eu ∩Gv = f−1

({u, v})

and

Ex ∩ Eu = ∅, Gy ∩Gv = ∅.Let a ∈ Ex ∩Gy, b ∈ Ex ∩Gv, c ∈ Eu ∩Gy and d ∈ Eu ∩Gv. Withoutany loss of generality assume that a � b � c � d. Then [a, c] ⊆ Gy and[b, d] ⊆ Gv, hence [b, c] ⊆ Gy ∩ Gv �= ∅, a contradiction. This provesthat one of the sets A or C must be a singleton.

Solution of Problem 2.86Let x ∈ U and let V (x) be the path-connected component of U con-taining x (see Definition 2.130). Clearly R

N is locally path-connected(since every open set contains a convex open set; see Definition 2.127).So, by Proposition 2.132, we have that V (x) is open. We claim thatU \ V (x) is open too. Let u ∈ U \ V (x). Since U is open, we can findε > 0 small enough such that

Bε(u) ={y ∈ R

N : ‖y − u‖ < ε} ⊆ U.

All points y ∈ Bε(u) are contained in X \ V (x) since they can beconnected to u by a straight-line path (a line segment) contained inBε(u) (see Definition 2.122). This proves that U \ V (x) is open. Butbecause by hypothesis U is connected (see Definition 2.104), we musthave U \ V (x) = ∅, hence

V (x) = U

and so U is path-connected (see Definition 2.122).

(b) It is enough to take any connected space which is not path-connected (see, e.g., Example 1.100).

2.3. Solutions 341

Solution of Problem 2.87No. The reason being that every point in (x, u) is a cut point of order2 (see Definition 2.120), while in [y, z) the point y is a cut point of or-der 1. Recall that the number of cut points of order k is a topologicalinvariant. This implies that the two intervals cannot be homeomor-phic.


x ∈⋃

i∈IAi

and let ix ∈ I be such that x ∈ Aix . For every other u ∈ ⋃i∈I

Ai, we

can find iu ∈ I such that u ∈ Aiu . Then by hypothesis, we can find afinite set {ik}nk=0 ⊆ I such that i0 = ix and in = iu and

Aik−1∩Aik �= ∅ ∀ k ∈ {1, . . . , n}.

We claim that for every m ∈ {0, 1, . . . , n} the setm⋃k=0

Aik is connected

(see Definition 2.104). To see this note that for m = 0, this is truesince Ai0 = Aix is by hypothesis connected. Suppose that the claim istrue for m = r ∈ {0, 1, . . . , n− 1}. We will show that it is also true for

m = r + 1. The setsr⋃

k=0

Aik and Air+1 are connected with nonempty

intersection (since Air ∩Air+1 �= ∅). Proposition 2.109 implies that the

setr+1⋃k=0

Aik is connected. This proves the claim and so we have shown

that the setn⋃k=0

Aik is connected. This means that the points x and u

are in the same connected component of⋃i∈I

Ai (see Definition 2.111).

But u ∈ ⋃i∈I

Ai was arbitrary. Therefore the set⋃i∈I

Ai has only one

connected component, i.e., it is connected.


Solution of Problem 2.89Let C ⊆ Y be a connected set (see Definition 2.104) and A = f−1(C).We proceed by contradiction. So, suppose that A is not connected.Then we can find two closed sets D1,D2 ⊆ X such that

A = (D1 ∩A) ∪ (D2 ∩A), (D1 ∩A) ∩ (D2 ∩A) = ∅,D1 ∩ A �= ∅ and D2 ∩A �= ∅.

If there exist x1 ∈ D1 ∩A, x2 ∈ D2 ∩A such that f(x1) = f(x2) = y,then

y ∈ C and x1, x2 ∈ f−1(y).

But by hypothesis the set E = f−1(y) is connected,

E = (D1 ∩ E) ∪ (D2 ∩E) and (D1 ∩ E) ∩ (D2 ∩E) = ∅,which contradicts the connectedness of E. Therefore, we must have

C = f(D1 ∩A) ∪ f(D2 ∩A) with f(D1 ∩A) ∩ f(D2 ∩A) = ∅.Recall that C is connected. Since the sets f(D1 ∩ A) and f(D2 ∩ A)are nonempty, if we show that they are closed in C, we would have acontradiction and so we are done. To this end let {yi}i∈I ⊆ f(D1 ∩A)be a net and assume that yi −→ y ∈ C. Then

yi = f(xi) with xi ∈ D1 ∩A ∀ i ∈ I.

Because of the compactness of X, we can find a subnet {xj}j∈J of{xi}i∈I such that xj −→ x in X. Since D1 is closed in X and{xj}j∈J ⊆ D1, we infer that x ∈ D1. Also yj −→ y ∈ C in Y andby the continuity of the function f , we have f(xj) −→ f(x). Hencey = f(x) ∈ C and so x ∈ A = f−1(C). Therefore x ∈ D1 ∩ A andso y = f(x) ∈ f(D1 ∩ A). This proves that f(D1 ∩ A) is closed inC. Similarly, we show that f(D2 ∩ A) is closed in C. So, finally wehave a contradiction to the fact that C is connected. This proves thatf−1(C) = A is connected.

Solution of Problem 2.90If C ∩ ∂A = ∅, then

C ∩ intA �= ∅ and C ∩ extA �= ∅

2.3. Solutions 343

(see Definition 2.9(b)). Then C ∩ intA and C ∩ extA is a discon-nection of C, contradicting the hypothesis that C is connected (seeDefinition 2.104).

Solution of Problem 2.91Since Y is not path-connected (see Definition 2.122), we can findy1, y2 ∈ Y which cannot be joined by a path in Y . Arguing bycontradiction, suppose that we could find a continuous surjectionf : X −→ Y . Then there are points x1, x2 ∈ X such that

f(x1) = y1 and f(x2) = y2.

Because X is path-connected, we can find a path ϑ : [0, 1] −→ X suchthat

ϑ(0) = x1 and ϑ(1) = x2.

Then f ◦ ϑ : [0, 1] −→ Y is a path in Y such that

(f ◦ ϑ)(0) = f(x1) = y1 and (f ◦ ϑ)(u) = f(x2) = y2,

a contradiction.

Solution of Problem 2.92For N = 1, we already know that S1 in path-connected (see Defini-tion 2.122), since it is the image of R under the exponential function

e(x) =(cos(2πx), sin(2πx)

) ∀ x ∈ R.

For the general case (N � 1), we know that RN+1 \ {0} is path-

connected and f : RN+1 \ {0} −→ SN , defined by

f(x) = x‖x‖ ∀ x ∈ R

N+1

is a continuous surjection. Problem 2.91 implies that SN must bepath-connected.


Alternative SolutionLet en be the “north pole” of SN (i.e., en = {δk,N}Nk=1) and es is the“south pole” of SN (i.e., es = {−δk,N}Nk=1). Using the stereographicprojection (cf. the solution of Problem 2.58), we have that SN \ {en}and SN \ {es} are both homeomorphic to R

N . Since RN is path-

connected (being convex), it follows that both SN \{en} and SN \{es}are path-connected (see Definition 2.122). Invoking Proposition 2.129,we conclude that

SN = (SN \ {en}) ∪ (SN \ {es})

is path-connected.

Solution of Problem 2.93We do the solution for the locally connected case (the locally compactcase being similar; Definition 2.92). So, assume that X is locallyconnected (see Definition 2.116). Let y ∈ Y and V ∈ NY (y). Theny = f(x) for some x ∈ X and f−1(V ) ∈ NX(x). Let U ∈ NX(x) besuch that U ⊆ f−1(V ) and since f is an open continuous surjection(see Definition 2.59), we have

y ∈ f(U) ⊆ f(f−1(V )

)= V

and f(U) is an open connected neighbourhood of y (see Proposi-tion 2.108). Since y ∈ Y and V ∈ NY (y) were arbitrary, we concludethat Y is locally connected.

Solution of Problem 2.94Let x ∈ C and let C(x) be the connected component containing x (seeDefinition 2.111). Then C ⊆ C(x). If C(x) ∩D �= ∅, then in a similarfashion C(x) ⊇ D and so C(x) = X, a contradiction to the fact that Xis disconnected (see Definition 2.104). So C(x) ∩D = ∅ and it followsthat C(x) = C. Similarly, C(u) = D for u ∈ D. This proves that Cand D are both the connected components of X.

2.3. Solutions 345

Solution of Problem 2.95Let f : C ∪ D −→ {0, 1} be a continuous function. Since C and Dare connected (see Definition 2.104), the functions f |C and f |D areconstant functions (see Theorem 2.105). Note that the set C∩ (C∪D)is the closure of C in C ∪D and so by continuity the value of f |C isalso the value of f |

C∩(C∪D). This is also the value of f |

C∩Dand so we

conclude thatf∣∣C

= f∣∣D

which by Theorem 2.105 establishes the connectedness of C ∪D.

Solution of Problem 2.96We proceed by contradiction. So, suppose that A is disconnected (seeDefinition 2.104). Then

A = C1 ∪C2,

with C1, C2 ⊆ X being nonempty, disjoint, closed sets.Since C1 ∪C2 is the smallest closed set containing A (see Proposi-

tion 2.11(d)), we have that

A ∩ C1 �= ∅ and A ∩ C2 �= ∅.We also have

A = (A ∩ C1) ∪ (A ∩ C2).

Note that∂(A ∩ C1) �= ∅ and ∂(A ∩ C2) �= ∅.

Indeed, if this is not the case, say ∂(A ∩ C1) = 0, then

A ∩ C1 = A ∩ C1 = int (A ∩ C1)

and so, we have a nontrivial clopen set, contradicting the connected-ness of X. Since

∂A = ∂(A ∩ C1) ∪ ∂(A ∩ C2)

and by hypothesis ∂A is connected, we have a contradiction.


Solution of Problem 2.97We know that D = γ

([0, 1]

)is a closed, connected set (see Defini-

tion 2.104 and Proposition 2.108) and by hypothesis

D ∩A �= ∅ and D ∩ (X \ A) �= ∅.

If D ∩ ∂A = ∅, then

D = (D ∩A) ∪ (D ∩X \ A)

and the sets D ∩ A and D ∩X \ A are closed in D and

(D ∩ A) ∩ (D ∩X \ A) = D ∩ (A ∩X \ A) = D ∩ ∂A = ∅.

So, we infer that the set D is disconnected, a contradiction.


(a) This set is disconnected (see Definition 2.104), since its projectionon the x-axis is disconnected.

(b) This set is connected, since any two points in it can be connectedby a piecewise linear path (with at most two segments).

(c) Let

C =⋃

ϑ∈Q\{0}

{(x, y) ∈ R

2 : y = ϑx}.

Clearly the coordinates of an element in C are either both rational orboth irrational. The set C is connected (in fact it is path-connected;see Definition 2.122). Note that our set contains C and is containedin C = R

2 and so it is connected (see Corollary 2.107).

Solution of Problem 2.99If X is not a singleton, then we can find x, u ∈ X with x �= u. Since{x} and {u} are two disjoint closed sets, invoking the Urysohn lemma(see Theorem 2.136), we can find a continuous function f : X −→ [0, 1]such that f(x) = 0 and f(u) = 1. Since X is connected, F (X) is an

2.3. Solutions 347

interval in [0, 1], hence f(X) = [0, 1]. This means that cardX � c (thecardinality of the continuum).

Solution of Problem 2.100For every x ∈ X choose Ux ∈ Y such that x ∈ Ux. Since X is normal,we can apply the Urysohn lemma (see Theorem 2.136) and produce acontinuous function gx : X −→ [0, 1] such that

gx|X\Ux= 0 and gx(x) = 1.

Let us setVx =

{u ∈ X : gx(u) > 0

}.

Evidently Vx ⊆ Ux and so {Vx}x∈X is a locally finite open cover of X.Therefore

g(x) =∑

U∈YgU (x)

is a well defined continuous function on X and

g(x) > 0 ∀ x ∈ X.

Replacing gU by fU = gUg , we obtain fU : X −→ [0, 1], continuous

functions, such that

∑

U∈YfU (x) = 1 ∀ x ∈ X.

Solution of Problem 2.101Because of the normality of X (see Definition 2.4), we can find twoopen sets U, V1 ⊆ X such that A ⊆ U , C ⊆ V1 and U ∩ V1 = ∅. Weclaim that U ∩ V1 = ∅. Proceeding by contradiction, suppose thatx ∈ U ∩ V1. Since x ∈ U and V1 ∈ N (x), we must have U ∩ V1 �= ∅, acontradiction. This proves the claim.

We have U ∩C = ∅ and so the normality of X implies that we canfind two open sets U1, V ⊆ X such that

U ⊆ U1, C ⊆ V and U1 ∩ V = ∅.


As before, we show that U1 ∩ V = ∅. Evidently U and V are thedesired open sets.

Alternative SolutionBecause X is normal, the Urysohn lemma (see Theorem 2.136) impliesthat there exists a continuous function f : X −→ [0, 1] such that

f |A

= 0 and f |C

= 1.

LetU = f−1

([0, 14)

)and V = f−1

((34 , 1

)].

Both are open sets (due to the continuity of f) and they satisfy

A ⊆ U, C ⊆ V and U ∩ V = ∅.


(a) “=⇒”: We have

A = f−1({0}) = f−1

( ⋂

n�1

[0, 1n))

=⋂

n�1

f−1([0, 1n))

and each set f−1([0, 1n)), for n � 1, is open by the continuity of f and

of the fact that the set [0, 1n) is open in [0, 1]. So A is a Gδ-set (seeDefinition 2.18).

“⇐=”: Since A is a Gδ-set in X, A =⋂n�1

Un with Un ⊆ X being

open for n � 1. Note that A and X \ Un are disjoint closed sets.Since X is normal (see Definition 2.4), by the Urysohn lemma (seeTheorem 2.136), for every n � 1, we can find a continuous functionfn : X −→ [0, 1] such that

fn|A = 0 and fn|X\Un= 1.

Let f : X −→ [0, 1] be defined by

f(x) =∑

n�1

12n fn(x) ∀ x ∈ X.

2.3. Solutions 349

Since12n

∣∣fn(x)∣∣ � 1

2n ∀ n � 1, x ∈ X,

by the Weierstrass M -test we infer that

∑

n�1

12n fn(x) converges uniformly on X

and so f is continuous. We claim that f−1({0}) = A. Clearly f |A = 0.

On the other hand, if f(x) = 0, then fn(x) = 0 for all n � 1 (recall thatfn � 0) and so x ∈ Un for all n � 1 (since fn|X\Un

= 1 for all n � 1).

Therefore x ∈ ⋂n�1

Un = A and we have proved that A = f−1({0}).

(b) Since X is normal, we can find two open sets U, V ⊆ X such that

A ⊆ U, C ⊆ V and U ∩ V = ∅.

Since A is a Gδ-set, we have

A =⋂

n�1

Un,

with Un being open sets and we may assume that Un ⊆ U for alln � 1. As in part (a) we obtain continuous functions fn : X −→ [0, 1]for n � 1 such that

fn|A = 0 and fn|X\Un= 1.

Evidently fn|C = 1 and we set

f(x) =∑

n�1

12n fn(x) ∀ x ∈ X,

as in part (a), we show that the function f : X −→ [0, 1] is continuous,A = f−1

({0}) and f |C= 1.

Solution of Problem 2.103Since X is locally compact (see Definition 2.92), for every x ∈ K, wecan find Vx ∈ N (x) such that V x is compact in X. Then {Vx}x∈K is anopen cover of K. We can also assume that Vx ⊆ U for all x ∈ K. The


compactness of K implies that we can find a finite subcover {Vxk}nk=1

of {Vx}x∈K . We have

K ⊆ V =

n⋃

k=1

Vxk and V =

n⋃

k=1

V xk is compact in X.

The set V ⊆ X furnished with the subspace topology is compact (seeProposition 2.79) and so it is normal (see Definition 2.4 and Proposi-tion 2.83(d)). Then the Urysohn lemma (see Theorem 2.136) impliesthat there exists a continuous function g : V −→ [0, 1] such that

g|K

= 0 and g|∂V

= 1.

Let f : X −→ [0, 1] be defined by

f(x) =

{g(x) for x ∈ V ,

1 for x ∈ X \ V .Clearly f is continuous and

f |K

= 0 and f |X\U = 1

(since X \ U ⊆ X \ V ).

Solution of Problem 2.104“=⇒”: Since by hypothesis K is a Gδ-set (see Definition 2.18), we canfind a sequence {Un}n�1 of open sets in X such that

K =⋂

n�1

Un.

By Problem 2.103, for every n � 1, we can find a continuous functionfn : X −→ [0, 1] such that

fn|K = 0 and fn|X\Un= 1.

Then, if we set

f(x) =∑

n�1

12n fn(x) ∀ x ∈ X,

2.3. Solutions 351

we have that the function f : X −→ [0, 1] is continuous (see the proofof Problem 2.102) and K = f−1

({0}).

“⇐=”: This part is identical as the first part of the solution of Prob-lem 2.102(a).

Solution of Problem 2.105“=⇒”: LetX be a normal topological space Let C ⊆ X be a nonempty,proper, closed set. If u ∈ X \C, then by the perfect normality of X, wecan find a continuous function f : X −→ [0, 1] such that C = f−1

({0}),{u} = f−1({1}). Then, from Problem 2.102(a), we infer that C is aGδ-set.

“⇐=”: Suppose that every closed subset of X is a Gδ-set. LetC,D ⊆ X be two nonempty disjoint closed sets. Then Prob-lem 2.102(b) implies the existence of two continuous functionsg, h : X −→ [0, 1] such that

g−1({0}) = C and g|D = 1

and

h−1({0}) = D and g|

C= 1.

We set

f = 12g +

12(1− h).

Evidently f : X −→ [0, 1] is continuous and

f−1({0}) = C and f−1

({1}) = D,

which shows that X is perfectly normal.

Solution of Problem 2.106By the Urysohn metrization theorem (see Theorem 2.139), the spaceX is metrizable. Then Problems 1.79 and 2.105 imply that X is


perfectly normal (see Definition 2.137). Thus, if C = X \ U , we canfind a continuous function f : X −→ [0, 1] such that f−1

({0}) = C.So,

f |U> 0 and f

X\U = 0.

Solution of Problem 2.107“=⇒”: Since X is second countable (see Definition 2.24), it is a Lin-delof separable space (see Definition 2.26 and Theorem 2.27). Also,because X is locally compact (see Definition 2.92), it is the union ofrelatively compact open sets. The Lindelof property implies that wecan find a sequence {Un}n�1 of relatively compact sets such that

X =⋃

n�1

Un.

Let {Vk}k�1 be a basis for the topology of X. If by X∗ we denote theone-point compactification of X (see Remark 2.97 and Theorem 2.98),then

{Vk}k�1 ∪{X∗ \ Un

}n�1

is a countable basis for X∗.

So X∗ is compact and second countable. But a compact space is nor-mal, hence regular too. Invoking the Urysohn metrization theorem (seeTheorem 2.139), we infer that X∗ is metrizable, hence X is metrizabletoo.

“⇐=”: Recall that in a metric space, separability is equivalent tosecond countability (see Proposition 1.24).

Solution of Problem 2.108Since X is locally compact (see Definition 2.92), for every x ∈ Uα ∈ Y,we can find a neighbourhood Vα,x of x such that V α,x is compact.Then D′ = {Vα,x}x∈X is an open cover of the space X. Since by

2.3. Solutions 353

hypothesis X is paracompact (see Definition 2.142), it has a locallyfinite refinement D (see Definition 2.140). This is the desired cover.

Solution of Problem 2.109Let {Dn}n�1 be a sequence of open dense subsets of X. Then X \Dn

for n � 1 are closed nowhere dense sets. Then by hypothesis, the set⋃n�1

(X \Dn) is a nowhere dense set. Hence the set int⋃n�1

(X \Dn) is

empty and so the set

X \⋃

n�1

(X \Dn) =⋂

n�1

Dn

is dense. This proves that X is Baire (see Definition 2.148).

The converse is not true. Let X = R. From the Baire categorytheorem (see Theorem 1.26), we know that X is a Baire space. Let{gn}n�1 be an enumeration of the rationals and let Dn = {gn} forevery n � 1. Then each Dn is closed and nowhere dense, but

⋃

n�1

Dn = X.

Solution of Problem 2.110Suppose that

⋃n�1

An = X. Then⋃n�1

An = X and so⋂n�1

(X \An) = ∅.But for every n � 1, the set X \ An is open and dense (see Defini-tion 1.25 and Problem 1.29). Because X is a Baire space (see Defini-tion 2.148), the set

⋂n�1

(X \ An) is dense in X, a contradiction. This

proves that ⋃

n�1

An �= X.


Solution of Problem 2.111No. Consider X = R, which is a Baire space by Theorem 1.26 andlet D1 = Q and D2 = R \ Q. Both are dense subsets of R butD1 ∩D2 = ∅.

Solution of Problem 2.112Let {Un}n�1 be a sequence of open and dense subsets of X. We needto show that ⋂

n�1

Un is dense.

Let x ∈ X and let V ∈ N (x) such that V is a Baire space (seeDefinition 2.148). Then every W ∈ N (x), W ⊆ V is open in V(with the subspace topology) and so W is a Baire space itself. ThenUn ∩W ⊆W is open and dense in W , hence

⋂

n�1

(Un ∩W ) =( ⋂

n�1

Un) ∩W

is dense in W and so

x is a limit point of⋂

n�1

Un,

which proves that⋂n�1

Un is dense and this proves that X is a Baire

space.

Solution of Problem 2.113Let X be a Baire topological space (see Definition 2.148). Since byhypothesis f is lower semicontinuous (see Definition 2.46), for everyn � 1 the set

Cn =(x ∈ X : f(x) � n

)is closed.

Then ⋃

n�1

Cn ={x ∈ X : f(x) < +∞}

2.3. Solutions 355

and by hypothesis this set is not of first category (i.e., is not includedin the union of a sequence of closed nowhere dense sets; see Defini-tion 1.25). So, for some integer n � 1, we have intCn �= ∅ (becauseX is a Baire space). This means that we can find x ∈ intCn andU ∈ N (x) such that

f(x) � n ∀ x ∈ U.

Solution of Problem 2.114Let U ⊆ X be a nonempty and open set. Then U is a Baire space too.We have

U = U ∩X =⋃

n�1

(Cn ∩ U)

and for each n � 1, the set Cn ∩ U is closed in U (with the subspacetopology). Hence for some n0 � 1, we have

intU (Cn0 ∩ U) �= ∅(here intU denotes the interior in U with the subspace topology). Butbecause U is open, we have

intU (Cn0 ∩ U) = (intCn0) ∩ U �= ∅.It follows that ( ⋃

n�1

intCn) ∩ U �= ∅

and as U was an arbitrary open set in X, we conclude that⋃

n�1

intCn is dense in X.

Solution of Problem 2.115Since Y is a separable metric space, for every n � 1, we can find asequence of open sets {Unk }k�1 with diamUnk <

1n and which cover Y .

Then by hypothesis

f−1(Unk ) =⋃

m�1

Cnk,m,


where for every m � 1, the set Cnk,m ⊆ X is closed. Clearly

X =⋃

k,m�1

Cnk,m

and because X is a Baire space, we have that

Vn =⋃

k,m�1

intCnk,m is open, dense in X

(see Problem 2.114). Let ωf be the oscillation function for f (seeDefinition 2.42). Then

ωf (x) < 1n ∀ x ∈ Vn.

Therefore

f is continuous on V =⋂

n�1

Vn

(see Problem 2.23), which is a dense Gδ-set.

Solution of Problem 2.116Let f : X −→ R be a lower semicontinuous function (see Defini-tion 2.46). For every a, c ∈ R, a < c, we have

f−1((a, c)

)=⋃

n�1

(f−1(−∞, c− 1

n

] ∩ f−1(a,+∞)).

Due to the lower semicontinuity of f , we have that the setf−1((−∞, c− 1

n

])is closed and the set f−1

((a,+∞)

)is open in X.

By hypothesis f−1(a,+∞) is an Fσ-set (see Definition 2.18). There-fore f−1

((a, c)

)is an Fσ-set and since the open intervals form a basis

for the usual topology on R which is a separable metric space, weconclude that for every open set V ⊆ R, the set f−1(V ) is an Fσ-set.Because X is also a Baire topological space (see Definition 2.148), wecan apply Problem 2.115 and conclude that f is continuous at everypoint of a dense Gδ-set.

2.3. Solutions 357

When f is an upper semicontinuous function, we apply the aboveresult to −f .

Solution of Problem 2.117Let C be a closed subset of a paracompact space X (see Defini-tion 2.142) and let {Vi}i∈I be an open cover of C. Then for everyi ∈ I, there is an open set Ui ⊆ X such that Vi = Ui ∩C. Let

Y = {Ui}i∈I ∪ {X \ C}.

Then Y is an open cover of X and because X is paracompact, Yhas a locally finite refinement Y ′ (see Definition 2.140). The family{U ′∩C}U ′∈Y ′ is a locally finite refinement of {Vi}i∈I . This proves thatC is paracompact.


(a) Let U ⊆ X be a nonempty open set and let {Un}n�1 be a sequence

of open dense subsets of U . Then the set Vn = Un ∪ (U)c is open in Xand

V n ⊇ Un ∪ (U )c = U ∪ (U)c = X.

So the set Vn is also dense in X. Then, since X is Baire (see Defini-tion 2.148), we have ⋂

n�1

(Vn ∩ U) �= ∅

and so ⋂

n�1

(Un ∩ U) �= ∅,

hence the set⋂n�1

Un is dense in U , which proves that U is a Baire

space.

(b) Suppose that p : X −→ Y = X/∼ is the quotient map and V ⊆ Yis an open dense set. Then the set p−1(V ) is open (see Definition 2.76).Also it is dense in X. Indeed, let V be any open set in X. Then, since


p is an open map, the set p(V ) ⊆ Y is open and so V ∩p(V ) �= ∅ (sinceV is dense). Hence

p−1(V ) ∩ V �= ∅,which shows that the set p−1(V ) is open dense in X. Now, let

{Un}n�1

be a sequence of open dense sets in X. Then from the previous partof the solution, the sequence

{p−1(Un)

}n�1

consists of open dense sets

in X and by hypothesis X is a Baire space. So, the set⋂n�1

p−1(Un) is

dense in X. So, we have

p−1( ⋂

n�1

Un ∩ U)

=⋂

n�1

p−1(Un) ∩ p−1(U) �= ∅,

so ⋂

n�1

Un ∩ U �= ∅.

Thus the set⋂n�1

Un is dense in U and so U is a Baire space.

Solution of Problem 2.119From Problem 2.118(a), we know that U is a Baire space too. Let{un}n�1 be a countable dense subset of X. By Problem 2.116, forevery integer n � 1, there exists a dense Gδ-set Dn of U such that thefunction x �−→ f(x, u) is continuous on Dn. Let

D =⋂

n�1

Dn.

Since U is a Baire space, we have that D ⊆ U is dense (see Def-inition 2.148) and for every n � 1, the function x �−→ f(x, un) iscontinuous on D. The density of {un}n�1 in X and the continuity ofthe function u �−→ f(x, u) imply that for every u ∈ X, the functionx �−→ f(x, u) is continuous on D.


Y = {X \ C} ∪ {Uα}α∈I .

2.3. Solutions 359

This is an open cover of X. Because of the paracompactness of X (seeDefinition 2.142), we can find a locally finite refinement {Vβ}β∈J of Y(see Definition 2.140). Let

K ={β ∈ J : Vβ ∩ C �= ∅}

and thenD =

{Vβ}β∈K

is an open cover of C. Let

W =⋃

β∈KVβ.

Since D is locally finite, the set E =⋃β∈D

V β is closed. Then W and

X \E are the two disjoint neighbourhoods of C and D respectively.

Solution of Problem 2.121Let Y = {Ui}i∈I . By Problem 2.27, we can find a locally finite opencover {Vi}i∈I of X such that

V i ⊆ Ui ∀ i ∈ I

(see Proposition 2.141). Invoking the Urysohn lemma (see Theo-rem 2.136), we can find a continuous function ψi : X −→ [0, 1] suchthat

supp ψi = X \ Ui and ψ−1i (1) ⊇ V i.

Letψ(x) =

∑

i∈Iψi(x).

This is well defined since Y is locally finite. Let

ψi(x) = ψi(x)

ψ(x)∀ i ∈ I.

Then clearly {ψi}i∈I is the desired partition of unity (see Defini-tion 2.146).


Solution of Problem 2.122Let X be a Lusin space (see Definition 2.167 and Remark 2.168). Thenby definition there is a Polish space Y (see Definition 2.150) and acontinuous bijection f : Y −→ X. Suppose that D ⊆ X is open (orclosed). Then f−1(D) is open (or closed) in Y (due to the continuityof f). Hence by Proposition 2.152, the set f−1(D) is a Polish space.The function f restricted on f−1(D) is a continuous bijection fromf−1(D) to D. Hence by definition D is a Lusin space.

Solution of Problem 2.123From Definition 2.167, for every n � 1, we can find a Polish space Yn(see Definition 2.150) and a continuous bijection

fn : Yn −→ Xn.

Let

Y =∏

n�1

Yn

and let f : Y −→ X be a function, defined by f = (fn)n�1. FromProposition 2.153, we know that Y is a Polish space. Also note thatf is a continuous bijection. Hence by definition X is a Lusin space.

Solution of Problem 2.124Let I = [0, 1] and let Q ⊆ R be a set of rational numbers. Then theset I \ (I ∩Q) is a Gδ-subset of I. So, by Theorem 2.154(a), it is also aPolish space. Evidently I \ (I ∩ Q) is dispersed (see Definition 2.161)and so is I ∩ Q. Therefore by Proposition 2.162(c), the set I is dis-persible and then so is IN (see Proposition 2.162(a)). If Y is a Polishspace, then Y is homeomorphic to a Gδ-subset of I

N and so Proposi-tion 2.162(c) implies that Y is dispersible. Therefore, if X is a Souslinspace (see Definition 2.156), then X is dispersible.

2.3. Solutions 361

Solution of Problem 2.125For every q ∈ Q and i ∈ I, let

Ui,q ={x ∈ X : fi(x) > q

}.

Then each set Ui,q ⊆ X is open (see Definition 2.46). Since X is astrongly Lindelof topological space (see Definition 2.163), there exista countable subset Jq of I such that

⋃

i∈IUi,q =

⋃

j∈JqUj,q.

Let us set

J =⋃

q∈QJq.

Then J is countable. Also, if

f = supi∈I

fi and f(x) > q,

then x ∈ ⋃i∈I

Ui,q and so x ∈ ⋃j∈Jq

Uj,q, hence supj∈J

fj(x) > q, from which

we infer that f = supj∈J

fj.

Solution of Problem 2.126It is enough to show the result for open sets, since the one for closedsets can be obtained by complementation. Let U ⊆ X be an open set.Due to regularity, U is the union of open sets V such that V ⊆ V ⊆ U .Then the strong Lindelof property (see Definition 2.163) implies thatthere is a countable subfamily of such closed sets V whose union is U .Therefore U is a Fσ-set (see Definition 2.18).

Solution of Problem 2.127Let ΔX be the diagonal of X ×X and ΔY the diagonal of Y × Y . As{fi : X −→ Y }i∈I is a separating family (see Definition 2.64), for every(x, x′) ∈ (X×X)\ΔX , we can find i ∈ I such that

(fi(x), fi(x

′)) �∈ ΔY .


This means that the open sets (fi, fi)−1((Y × Y ) \ΔY

)form an open

cover of (X × X) \ ΔX . Because by hypothesis X × X is stronglyLindelof (see Definition 2.163), we can find a countable subset J ⊆ Isuch that the family {(fj , fj)−1

((Y ×Y )\ΔY

)}j∈J is still an open coverof (X ×X) \ΔX . Then the countable family {fj}j∈J is separating.

Solution of Problem 2.128In what follows let proj

Xand proj

Ybe the canonical projections of

X × Y onto X and Y respectively. Let A be a Souslin subspace ofY (see Definition 2.156). Then proj−1

Y(A) = X × A and since by

hypothesis X is Souslin, X × A is a Souslin subspace of X × Y (seeProposition 2.159(b)). Since Gr f is a Souslin subspace of X×Y , then

Gr f ∩ proj−1Y

(A) = Gr f ∩ (X ×A) = D

is Souslin (see Proposition 2.159(c)). So, there exists a Polish space Vand a continuous surjection h : V −→ D. Then proj

X◦h is a continuous

surjection from V onto f−1(A). This proves that f−1(A) ⊆ X is aSouslin set.

Solution of Problem 2.129Let {(xi, yi)}i∈I ⊆ GrF be a net such that

(xi, yi) −→ (x, y) in X × Y.

Suppose that y �∈ F (x). By the regularity of Y we can find U ∈ N (y)and an open set V ⊆ Y such that V ⊃ F (x) and U∩V = ∅. Because Fis upper semicontinuous (see Definition 2.169), we can find W ∈ N (x)such that F (W ) ⊆ V . Since xi −→ x in X and yi −→ y in Y , we canfind i0 ∈ I such that

xi ∈ W and yi ∈ U ∀ i � i0,

soyi ∈ F (xi) ⊆ V and yi ∈ U ∀ i � i0.

But U ∩V = ∅. So, we reach a contradiction. Therefore y ∈ F (x) andso we conclude that the set Gr f ⊆ X × Y is closed.

2.3. Solutions 363

Note that when the multifunction is actually single-valued (i.e., ausual function), then the notion of upper semicontinuity coincides withthat of continuity. For usual functions (i.e., single-valued), closednessof the graph does not imply continuity. For example, consider a func-tion f : R −→ R, defined by

f(x) =

{0 if x = 0,1x if x �= 0.

Then Gr f ⊆ R× R is closed, but f is not continuous.

Solution of Problem 2.130“=⇒”: We argue by contradiction. So, suppose that {(xi, yi)}i∈I ⊆GrF is a net such that xi −→ x in X and {yi}i∈I does not have alimit point in F (x). So, for every y ∈ F (x), we can find i0(y) ∈ Iand V (y) ∈ N (y) such that yi �∈ V (y) for all i � i0(y). The family{V (y)

}y∈F (x)

is an open cover of F (x) and because F (x) is compact,

we can find a finite subcover{V (yk)

}nk=1

. Let

V =

n⋃

k=1

V (yk) ⊃ F (x).

Evidently we can find i0 ∈ I such that yi �∈ V for all i � i1. Onthe other hand the upper semicontinuity of F (see Definition 2.169)implies that there is U ∈ N (x) such that F (U) ⊆ V . Then we can findi2 � i1 such that xi ∈ U for all i � i2. So, yi ∈ F (xi) ⊆ F (U) ⊆ V forall i � i2, a contradiction. Therefore {yi}i∈I has a limit point in F (x).

“⇐=”: By hypothesis every net {yi}i∈I ⊆ F (x) has a limit pointy ∈ F (x), hence a subnet converging to y. This means that F (x) ⊆ Yis compact (see Theorem 2.81). Suppose that F is not upper semicon-tinuous at x. Then we can find a net {xi}i∈I and an open set V suchthat xi −→ x in X, F (x) ⊆ V and F (xi) ∩ (X \ V ) �= ∅ for all i ∈ I.Let yi ∈ F (xi)∩ (X \ V ). Because xi −→ x, by hypothesis {yi}i∈I hasa limit point y ∈ F (x). So, we can find a subnet {yj}j∈J of {yi}i∈I


such that yj −→ y in Y and y ∈ F (x) ∩ (X \ V ) since the latter set isclosed. But recall that F (x) ⊆ V , a contradiction.


(a) Let K be a compact set and let {yi}i∈I be a net in F (K). We

need to show that it has a convergent subnet to some element in F (K).We have yi ∈ F (xi) with xi ∈ K for all i ∈ I. Because K is compact,we can find a subnet {xj}j∈J of {xi}i∈I such that xj −→ x ∈ K. Thenby Problem 2.130, the net {yj}j∈J has a limit point y ∈ F (x) and sowe can find a subnet {yλ}λ∈Λ of {yj}j∈J such that

yλ −→ y ∈ F (x) ⊆ F (K).

This proves the compactness of F (K) (see Theorem 2.81).

(b) Let{(xi, yi)i

}i∈I ⊆ GrF be a net and assume that

(xi, yi) −→ (x, y) in X × Y.

Then by Problem 2.130, we have y ∈ F (x), hence GrF ⊆ X × Y isclosed.

Solution of Problem 2.132Let F : X −→ 2R \ {∅} be the multifunction, defined by

F (x) =[f(x), g(x)

] ∀ x ∈ X.

Clearly F has closed, convex values and

intF (x) �= ∅ ∀ x ∈ X.We claim that F is a lower semicontinuous multifunction. Let us fixx ∈ X. By Definition 2.169(b), it suffices to show that for everya, c ∈ R, a < c such that F (x)∩ (a, c) �= ∅, we can find U ∈ N (x) suchthat

F (u) ∩ (a, c) �= ∅ ∀ u ∈ U.

Since∅ �= F (x) ∩ (a, c) =

[f(x), g(x)

] ∩ (a, c),

2.3. Solutions 365

we havef(x) < c and g(x) > a.

Because f is an upper semicontinuous function and g is a lower semi-continuous function (see Definition 2.46), we can find U ∈ N (x) suchthat

f(u) < c and g(u) > a ∀ u ∈ U,

soF (u) ∩ (a, c) =

[f(u), g(u)

] ∩ (a, c) �= ∅ ∀ u ∈ U

and thus F is lower semicontinuous.Observing that

[intF (x) ∩ (a, c) �= ∅] ⇐⇒ [

F (x) ∩ (a, c) = intF (x) ∩ (a, c) �= ∅]

we infer that the function x �−→ intF (x) is lower semicontinuous.So, we can apply Theorem 2.172 and obtain a continuous functionh : X −→ R such that

h(x) ∈ intF (x) =(f(x), g(x)

) ∀ x ∈ X

and sof(x) < h(x) < g(x) ∀ x ∈ X.

Solution of Problem 2.133Let G : X −→ 2Y \ {∅} be the multifunction, defined by

G(x) =

{F (x) if x �= x,{y} if x = x.

Clearly G is a lower semicontinuous multifunction (see Defini-tion 2.169) with values which are nonempty, closed and convex setsin Y . So, we can apply the Michael selection theorem (see Theo-rem 2.170) and obtain a continuous function g : X −→ Y such that

g(x) ∈ G(x) ∀ x ∈ X.

Theng(x) ∈ F (x) ∀ x ∈ X

and g(x) = y.


Solution of Problem 2.134Consider the multifunction G : X −→ 2Y \ {∅}, defined by

G(x) =

{f(x) if x ∈ C,F (x) if x ∈ X \ C.

Evidently G is lower semicontinuous (see Definition 2.169) and hasclosed and convex values. So, we can apply the Michael selection theo-rem (see Theorem 2.170) and obtain a continuous function f : X −→ Ysuch that

f(x) ∈ G(x) ∀ x ∈ X.Then

f(x) ∈ F (x) ∀ x ∈ X

and f |C= f .

Solution of Problem 2.135The family

{F−({y})}

y∈Y is an open cover of X. Because X is com-

pact, we can find a finite subcover{F−({yk})

}nk=1

of{F−({y})}

y∈Y .Recalling that a compact space is paracompact (see Definition 2.142and Theorem 2.144) and invoking Theorem 2.147, we can find a par-tition of unity {pk}nk=1 subordinated to

{F−({yk})

}nk=1

. We set

f(x) =n∑

k=1

pk(x)yk ∀ x ∈ X.

Evidently f : X −→ Y is continuous. Moreover, if pk(x) �= 0, then

x ∈ F−({yk})and so

yk ∈ F (x).

Therefore f(x) is a convex combination of elements of F (x) and so theconvexity of F (x) implies that

f(x) ∈ F (x) ∀ x ∈ X.

2.3. Solutions 367

Solution of Problem 2.136First we do the lower semicontinuous case. In what follows, for everyC ⊆ Y , we set

F−(C) ={x ∈ X : F (x) ∩ C �= ∅}.

Let A ⊆ X be a connected set (see Definition 2.104) and suppose thatV1, V2 ⊆ Y are open sets such that

F (A) ⊆ V1 ∪ V2, F (A) ∩ V1 �= ∅, F (A) ∩ V2 �= ∅.

We need to show that

F (A) ∩ V1 ∩ V2 �= ∅.

To this end, suppose that

A ∩ F−(V1) ∩ F−(V2) = ∅.

Because F is lower semicontinuous, both sets F−(V1) and F−(V2) areopen in X and

A ⊆ F−(V1) ∪ F−(V2), A ∩ F−(V1) �= ∅, A ∩ F−(V2) �= ∅.

These facts contradict the connectedness of A. Hence

A ∩ F−(V1) ∩ F−(V2) �= ∅

and let x ∈ A ∩ F−(V1) ∩ F−(V2). Then

F (x) ∩ V1 �= ∅, F (x) ∩ V2 �= ∅ and F (x) ⊆ V1 ∪ V2.

Because the set F (x) is connected in Y , it follows that

F (x) ∩ V1 ∩ V2 �= ∅,

hence

F (A) ∩ V1 ∩ V2 �= ∅and this proves the connectedness of F (C).

For the upper semicontinuous case, the proof is similar, check-ing this time a disconnection consisting of closed sets. Recall from


Definition 2.169(a) that F is upper semicontinuous if and only if forevery closed set D ⊆ Y , the set

F−(D) ={x ∈ X : F (x) ∩D �= ∅}

is closed.

Solution of Problem 2.137(a) “=⇒”: Let V ⊆ R be an open set. We need to show that the set

L+f (V ) =

{x ∈ X : Lf (x) ⊆ V

}

is open (see Definition 2.169). Let x ∈ L+f (V ). Then Lf (x) ⊆ V and

this means that (μ,+∞) ⊆ V with some μ < f(x). Because f is lowersemicontinuous (see Definition 2.46), we can find U ∈ N (x) such that

μ < f(u) ∀ u ∈ U.

ThenLf (u) ⊆ (μ,+∞) ⊆ V ∀ u ∈ U

and so we have proved that the multifunction x �−→ Lf (x) is uppersemicontinuous.

“⇐=”: Since Lf is upper semicontinuous, for every μ ∈ R, the set

L+f

((μ,+∞)

)={x ∈ X : Lf (x) ⊆ (μ,+∞)

}={x ∈ X : μ < f(x)

}

is open, which is equivalent to saying that f is lower semicontinuous(see Proposition 2.53).

(b) “=⇒”: Let V ⊆ R be an open set. We need to show that theset

L−f (V ) =

{x ∈ X : Lf (x) ∩ V �= ∅}

is open. Let x ∈ L−f (V ). Then we can find λ ∈ V such that f(x) � λ.

Because V is open we can assume that f(x) < λ. The upper semicon-tinuity of f implies that there exists U ∈ N (x) such that

f(u) < λ ∀ u ∈ U.

2.3. Solutions 369

Then

Lf (u) ∩ V �= ∅ ∀ u ∈ U

and so we have proved that the multifunction x �−→ Lf (x) is lowersemicontinuous.

“⇐=”: Since Lf is lower semicontinuous, for every λ ∈ R, the set

L−f

((−∞, λ)

)={x ∈ X : f(x) < λ

}

is open, which is equivalent to saying that f is upper semicontinuous(see Proposition 2.53).

Solution of Problem 2.138We need to show that for every λ ∈ R, the set

Lλ ={y ∈ Y : m(y) � λ

}

is closed (see Definition 2.46 and Proposition 2.53). To this end, let{yi}i∈I be a net in Lλ and assume that yi −→ y. Let x ∈ F (y), take any

V ∈ N (x) and consider the set F−(V )def={y′ ∈ Y : F (y′) ∩ V �= ∅}.

We see that y ∈ F−(V ) and since F is lower semicontinuous (seeDefinition 2.169), the set F−(V ) is open. Because yi −→ y in Y , wecan find i

V∈ I such that

yi ∈ F−(V ) ∀ i � iV.

Hence we can find

xi ∈ F (yi) ∩ V ∀ i � iV,

so

xi ∈ F (yi) and xi ∈ V ∀ i � iV.

Because V ∈ N (x) was arbitrary, we conclude that xi −→ x in X. Wehave

f(xi, yi) � m(yi) � λ ∀ i ∈ I,

so f(x, y) � λ (since f is lower semicontinuous on X × Y and(xi, yi) −→ (x, y) in X × Y ).


Because x ∈ F (y) is arbitrary, we conclude that m(y) � λ andso y ∈ Lλ. This proves that Lλ is closed, hence m is lower semi-continuous.

Solution of Problem 2.139We need to show that for any ε > 0 and y ∈ Y , we can find Uy ∈ N (y)such that for all y′ ∈ Uy, we have

m(y′) � m(y) + ε

(see Definition 2.46). Let ε > 0 and let y ∈ Y . Because f is uppersemicontinuous, for any x ∈ X, we can find two sets Vx ∈ N (x) andU ′x(y) ∈ N (y) such that

f(x′, y′) � f(x, y) + ε ∀ x′ ∈ Vx, y′ ∈ U ′x(y).

Note that{Vx}x∈F (y)

is an open cover of the set F (y) and by hypoth-

esis F (y) is compact. So, we can find a finite subcover{Vxk}nk=1

of{Vx}x∈F (y)

. Then

F (y) ⊆n⋃

k=1

Vxk = V

and because F is upper semicontinuous (see Definition 2.169), we canfind Wy ∈ N (y) such that

F (y′) ⊆ V ∀ y′ ∈Wy.

Let

Uy = Wy ∩( n⋂

k=1

U ′xk(y)) ∈ N (y).

If y′ ∈ Uy and x′ ∈ F (y′), we have x′ ∈ V and so x′ ∈ Vxk for somek ∈ {1, . . . , n}. Since y′ ∈ U ′

xk, we have

f(x′, y′) � f(xk, y) + ε � m(y) + ε,

hencem(y′) � m(y) + ε,

which proves the upper semicontinuity of m.

2.3. Solutions 371

Solution of Problem 2.140Consider the sequence {Kn}n�1, with

Kn = F (n)(X) ∀ n � 1

(here F (n) denotes the composition of F with itself n times which iseasily seen to be upper semicontinuous; see Definition 2.169). FromProblem 2.131, we know that this is a decreasing sequence of compactsubsets of X. Let us set

C =⋂

n�1

Kn.

Since the sequence {Kn}n�1 has the finite intersection property (seeDefinition 2.80), from Theorem 2.81, we infer that

C �= ∅ and F (C) ⊆ C.

We claim that equality holds. Arguing indirectly, suppose that wecan find x ∈ C \ F (C). Recall that a compact space is paracompactand a paracompact space is normal (see Definitions 2.4, 2.142 andTheorem 2.144). So, we can find open sets U1 ⊆ N (x) and U2 ⊇ F (C)such that U1 ∩ U2 = ∅. Since F is upper semicontinuous, the setF+(U2) =

{x ∈ X : F (x) ⊆ U2

}is open and contains C. So, we have

C =⋂

n�1

Kn ⊆ F+(U2),

henceX \ F+(U2) ⊆ X \ C =

⋃

n�1

(X \Kn).

Note that{X \Kn

}n�1

is an open cover of the set X \F+(U2) which is

compact. So, we can find a finite subcover{X \Knk

}Nk=1

of the cover{X \Kn

}n�1

, i.e.,

X \ F+(U2) ⊆N⋃

k=1

(X \Knk).

For every n > max {n1, . . . , nN}, we have

Kn ⊆N⋂

k=1

Knk⊆ F+(U2).


So, the sets {Kn}n�1 eventually reside in U2. This implies that C ⊆ U2,hence x ∈ U2, a contradiction. This proves that F (C) = C.

Solution of Problem 2.141From Theorem 2.183, we know that the c-topology on C(X;Y ) isinduced by the supremum metric

d∞(f, g) = supXdY

(f(x), g(x)

) ∀ f, g ∈ C(X;Y ).

Since the uniform limit of continuous functions is a continuous func-tion, it follows that

(C(X;Y ), d∞

)is a compete metric space. So, we

only need to check that(C(X;Y ), d∞

)is separable. Let k,m, n � 1

be integers. The compactness of X implies that it has an 1m -net

Xm = {x1, . . . , xl} ⊆ X. Since Y is separable, there is a countableopen cover Dk = {Ui}i�1 such that

diamUi <1k ∀ i � 1.

Let

Am,n ={f ∈ C(X;Y ) : ∀x, u ∈ X : dX (x, u) < 1

m =⇒ dY

(f(x), f(y)

)< 1

n

}.

For each l-tupe s = (i1, . . . , il), let us choose fs ∈ Am,n (wheneverpossible) such that fs(xj) ∈ Uj for all 1 � j � l. Let Bm,n,k be thecollection of all these fs and let

Bm,n =⋃

k�1

Bm,n,k.

Claim: For every f ∈ Am,n and every ε > 0, we can find g ∈ Bm,nsuch that

dY

(f(u), g(u)

)< ε ∀ u ∈ Xm.

To this end, take k > 1ε and choose i1, i2, . . . , ik such that f(xj) ∈ Uij

for all 1 � j � k. Hence for s = (i1, . . . , ik), fs exists and we can takeg = fs. This proves the Claim.

LetB =

⋃

m,n�1

Bm,n.

2.3. Solutions 373

Evidently B is countable. We will show that it is dense in C(X;Y ).So, let f ∈ C(X;Y ) and let ε > 0. Let us take n > 3

ε . The uniformcontinuity of f (recall that the domain X is compact), implies thatf ∈ Am,n for some m. By the Claim, we can find g ∈ Bm,n such that

dY

(f(u), g(u)

)< ε

3 ∀ u ∈ Xm.

Since Xm is an 1m -net, via the triangle inequality, we have

dY

(f(x), g(x)

)< ε ∀ x ∈ X,

so the set B is dense in C(X;Y ) and so(C(X;Y ), d∞

)is separable,

hence Polish.

Solution of Problem 2.142(a) Evidently the sets of the form

[U0;U1, . . . Un] ={C ∈ Pk

(X): C ⊆ U0 and C ∩ Ui �= ∅ for all 1 � i � n

}

form a base for the Vietoris topology on Pk(X). So, let [U0;U1, . . . Un]

be such a nonempty basic open set. Then U0∩Ui �= ∅ for all 1 � i � n.We choose xi ∈ U0 ∩ Ui for 1 � i � n. Then

{x1, . . . , xn} ∈ [U0;U1, . . . Un]

and this proves the density of the finite sets in Pk(X)with the Vietoris

topology.

(b) Let D be a countable dense set in X and let F be the set of allnonempty finite subsets of D. In the proof of (a), choose xi to belongin D (since D is dense in X). Then F is dense in Pk

(X)and F is

countable.

Solution of Problem 2.143First we show that if a set is open in the metric space

(Pk(X), h),

then it is also Vietoris open. Let K ∈ Pk(X)and let ε > 0. The


compactness of K implies that there exists an ε2 -net {x1, . . . , xn} ⊆ K.

LetU0 = Bε(K) =

{x ∈ X : dist(x,K) < ε

}

andUi = Bε(xi) ∀ i ∈ {1, . . . , n}.

It suffices to show that

K ∈ [U0;U1, . . . , Un] ={C ∈ Pk

(X): h(K,C) < ε

}

(see the solution of Problem 2.142). We have that K ⊆ U and

xi ∈ K ∩ Ui ∀1 � i � n.

Therefore K ⊆ [U0;U1, . . . , Un]. Let C ∈ [U0;U1, . . . , Un]. We need toshow that h(K,C) < ε. Since C ⊆ U0 = Bε(K), from the definition ofthe Hausdorff metric, it is sufficient to show that

K ⊆ Bε(C) ={x ∈ X : dist(x,C) < ε

}.

Let x ∈ K and choose xi such that dX(x, xi) <

ε2 . Let y ∈ K ∩Ui �= ∅.

We havedX(x, y) � d

X(x, xi) + d

X(xi, y) < ε,

so x ∈ Bε(C), hence K ⊆ Bε(C).Next we show that every Vietoris open set is also open in the metric

space(Pk(X), h). It suffices to show that every subbasic open set is

open in(Pk(X), h). So, let U be open in X and let K ∈ Pk

(X)be

such that K ⊆ U . We define

ε = min{dist(x,K) : x ∈ X \ U} > 0.

For every C ∈ Pk(X)with h(C,K) < ε, we have C ⊆ Bε(K) ⊆ U ,

which shows that the subbasic open set{C ∈ Pk

(X): C ⊆ U

}is also

open in(Pk(X), h).

Also, let K ∈ Pk(X)and let U ⊆ X be an open set such that

K ∩ U �= ∅. Let x ∈ K ∩ U and let ε > 0 be such that Bε(x) ⊆ U .Suppose that h(K,C) < ε. Since x ∈ K, we have dist(x,C) < ε andso there exists y ∈ C such that y ∈ Bε(x) ⊆ U . Hence C ∩ U �= ∅ andthis proves that the subbasic open set

{C ∈ Pk

(X): C ∩ U �= ∅} is

open in(Pk(X), h).

2.3. Solutions 375

Solution of Problem 2.144Let {xi}i∈I ⊆ A be a net such that xi −→ x. We need to showthat x ∈ A. Let r : X −→ A be the retraction function (see Defini-tion 2.196). Then r(xi) = xi for all i ∈ I. The continuity of r impliesthat r(xi) −→ r(x) ∈ A. Hence r(x) = x ∈ A and so we have provedthat A is closed.

Next, if Y is another topological space and f : A −→ Y is a con-tinuous function, let us set f = f ◦ r : X −→ Y . Then f is continuous(as a composition of two continuous functions; see Proposition 2.41)and f(x) = f

(r(x)

)for all x ∈ A, i.e., f |A = f .

Solution of Problem 2.145Let X be a normal space and let A ⊆ X be a retract of it (see Defini-tion 2.196). Let r : X −→ A be the retraction of X onto A. ConsiderC,D ⊆ A closed (these are closed in X too, since A is closed; seeProblem 2.144) such that C ∩D = ∅. The sets r−1(C) and r−1(D) aredisjoint, closed subsets of X. The normality of X implies that we canfind two open sets U, V ⊆ X such that

r−1(C) ⊆ U, r−1(D) ⊆ V and U ∩ V = ∅.Consider the sets U ∩A and V ∩A. Both are open in A and of coursedisjoint. Since r|

A= id

A, we have

(r|A)−1(C) = C ⊆ U ∩A and (r|A)−1(D) = D ⊆ V ∩A.So, we have two disjoint open sets in A containing C and D respec-tively. This implies the normality of A.

Solution of Problem 2.146Let h : [0, 1] × (RN+1 \ {0}) −→ R

N+1 \ {0} be a function, defined by

h(t, x) = (1− t)x+ t x‖x‖ ∀ (t, x) ∈ [0, 1] × (RN+1 \ {0}).

Then

h(0, ·) = idRN+1\{0} ,

h(1, x) = x‖x‖ ∈ SN ∀ x ∈ R

N+1 \ {0},h(t, ·)|

SN= id

SN.


Therefore h is a strong deformation retraction and so we concludethat SN is a strong deformation retract of R

N+1 \ {0} (see Defini-tion 2.196).


(a) Let X be a compact (respectively, (path-)connected) topologicalspace (see Definitions 2.78, 2.104 and 2.122) and let A ⊆ X be a re-tract (see Definition 2.196). Then there exists a continuous functionr : X −→ A such that r|

A= id

A(a retraction from X to A). Because

of the continuity of the retraction r, we have that r(X) = A is compact(respectively, (path-)connected).

(b) Let X be a simply connected topological space (see Defini-tion 2.208) and let A ⊆ X be a retract. Let r : X −→ A be a re-traction from X to A and let iA : A −→ X be the inclusion function(i.e., i

A(x) = x for all x ∈ A). Then

r ◦ iA = idA

and so

r∗ ◦ (iA)∗ = (r ◦ iA)∗ = (id

A)∗

is the identity function on π1(A). Since (iA)∗ : π1(A) −→ π1(X) andr∗ : π1(X) −→ π1(A), it follows that (iA)∗ is injective and r∗ is surjec-tive. Hence

r∗(π1(X)

)= π1(A).

But since X is simply connected, we have π1(X) = 0. Hence

0 = r∗(π1(X)

)= π1(A)

and this implies that A is simply connected too.(c) Since A is a retract of X, there is a retraction r

A: X −→ A

from X to A. Also, because C is a retract of A, there is a retraction

2.3. Solutions 377

rC : A −→ C from A to C. Let rC = rC ◦ rA . Then rC : X −→ C iscontinuous and it x ∈ C, then

rC(x) = r

C

(rA(x))

= x,

hence

rC|C

= idC,

which means that rC is a retraction from X to C. Therefore C is aretract of X.

Solution of Problem 2.148Let en = (0, . . . , 0, 1) be the “north pole” of SN−1 and let es = −enbe the “south pole”. Consider the open sets

U = SN−1 \ {en} and V = SN−1 \ {es}.

Both are homeomorphic to RN be means of the stereographic projec-

tion (cf. the solution of Problem 2.58). Let f : [0, 1] −→ SN−1 be apath. By Problem 1.137, we can find an integer m � 1 such that forevery k ∈ {0, 1, . . . ,m− 1

}, we have

f([

km ,

k+1m

]) ⊆ U or f([

km ,

k+1m

]) ⊆ V.

Note that V \ {en} is homeomorphic to RN−1 \ {0} which is con-

nected (here the restriction N � 3 plays a crucial role, since R \ {0}is disconnected; Definition 2.104). Then for every path in V , there isanother one also located in V with the same initial and final pointswhich avoids en. Since V is simply connected (being homeomorphic toRN−1; Definition 2.208), the two paths are homotopic in V (see Def-

inition 2.186) and so also in SN−1. Every path located in U alreadyavoids en. Therefore we conclude that f is homotopic to a path in


SN−1 \ {en} and the latter is homeomorphic to RN−1. Therefore f is

nullhomotopic (see Proposition 2.194) and so we conclude that

π1(SN−1) = 0,

i.e., SN−1 is simply connected (see Definition 2.208).The function g : RN \ {0} −→ SN−1 given by

g(x) = x‖x‖

is a strong deformation retraction (see Definition 2.196 and Prob-lem 2.146). Hence

π1(RN \ {0}) = π1(S

N−1) = 0

by the first part of the proof (see Corollary 2.207).

Solution of Problem 2.149First we consider the closed unit interval [0, 1]. Let f : [0, 1] −→ {0} bedefined by f(x) = 0 for all x ∈ [0, 1] and let h : {0} −→ [0, 1] be definedby h(0) = 0. Then f ◦ h : {0} −→ {0} is the identity function and soit is trivially homotopic to the identity function (see Definition 2.186).Conversely, note that h ◦ f : [0, 1] −→ [0, 1] and

(h ◦ f)(x) = 0 ∀ x ∈ [0, 1].

Consider the homotopy G : [0, 1] × [0, 1] −→ [0, 1], defined by

G(t, x) = (1− t)x ∀ (t, x) ∈ [0, 1] × [0, 1].

This shows that h ◦ f ≡ 0 and the identity are homotopic. Therefore[0, 1] and {0} are homotopy equivalent (see Definition 2.191).

For the open unit interval (0, 1), the proof is similar. In this case,we need to change the definition of h : {0} −→ (0, 1), say to h(0) = 1

2 .Then a homotopy G : [0, 1] × (0, 1) −→ (0, 1) which passes from h ◦ fto the identity of (0, 1) is given by

G(t, x) = 12(1− t+ 2tx) ∀ (t, x) ∈ [0, 1] × (0, 1).

2.3. Solutions 379

Evidently

G(0, x) = 12 = h ◦ f and G(1, x) = x.

Therefore (0, 1) and {0} are homotopy equivalent.

Solution of Problem 2.150Let f : S1 −→ A be the canonical inclusion of S1 into A, i.e.,

f(x, y) = (x, y) ∀ (x, y) ∈ A

and let h : A −→ S1 be the radial projection inwards (it moves theouter boundary of A onto S1 which is the inner boundary of A). Wehave

h(x, y) = 1√x2+y2

(x, y) ∀ (x, y) ∈ A.

Then h ◦ f : S1 −→ S1 and for all (x, y) ∈ S1, we have

(h ◦ f)(x, y) = h(x, y) = (x, y)

(since√x2 + y2 = 1). Of course h ◦ f = id

S1 is trivially homotopic toid

S1 .On the other hand f ◦ h : A −→ A is defined by

(f ◦ h)(x, y) = 1√x2+y2

(x, y) ∀ (x, y) ∈ A.

Consider the homotopy G : [0, 1] ×A −→ A, defined by

G(t, (x, y)

)=

t√x2+y2+(1−t)√

x2+y2(x, y) ∀ (t, (x, y)) ∈ [0, 1] ×A.

Clearly G is continuous and

G(0, (x, y)

)= (f ◦ h)(x, y), G

(1, (x, y)

)= (x, y) ∀ (x, y) ∈ A.

This proves that A and S1 are homotopy equivalent, i.e., A � S1 (seeDefinition 2.191).


Solution of Problem 2.151The homotopy equivalence f : S1 −→ I × S1 is given by

f(x) = (0, x)

(see Definition 2.191). The homotopy equivalence g : I × S1 −→ S1 isgiven by g = g2 ◦ g1, where

g1(t, x) = (0, x) ∀ (t, x) ∈ I × S1

and g2 = f−1.

Solution of Problem 2.152Since by hypothesis f is not surjective, we have that

f(X) ⊆ SN \ {y0},

for some y0 ∈ SN . But via the stereographic projection (cf. thesolution of Problem 2.58) the set SN \ {y0} is homeomorphic toRN . So, without any loss of generality, we may assume that f

is a continuous function from X into RN . Consider the homotopy

G : [0, 1] ×X −→ SN , defined by

G(t, x) = tf(x) ∀ (t, x) ∈ [0, 1] ×X.

It is continuous and shows that f is nullhomotopic (see Defini-tion 2.186).

Solution of Problem 2.153Let Gr f be a graph of f (see Definition 1.132) and leth : X × Y −→ Gr f be defined by

h(x, y) =(x, f(x)

).

2.3. Solutions 381

Evidently h is continuous and

h|Gr f

= idGr f

.

This means that Gr f is a retract of X × Y (see Definition 2.196).

Solution of Problem 2.154Let X be the union of two tangent circles C1 and C2. Let C1∩C2 = x0,a ∈ C1 \ {x0}, b ∈ C2 \ {x0}. Let

U = X \ {a} and V = X \ {b}.

Both sets U and V are open andX = U∪V . Invoking the Van Kampentheorem (see Theorem 2.229), we infer that π1(X,x0) is a free groupwith two generators {ϑ, η}, which are illustrated by the circles C1 andC2 in the above figure.


(a) Because A is a closed subset of the compact space X, it is itself

compact. Hence due to the continuity of f , the sets fn(A) are compactand connected (see Propositions 2.82 and 2.108). We have

fn+1(A) ⊆ fn(A) ⊆ A ∀ n � 1

and since A is compact, we have that

C =⋂

n�1

fn(A) is nonempty and coampct too

(see Theorem 2.81). We will show that the set C is also connected. Weargue by contradiction. So, suppose that C is not connected. ThenC = D ∪ E, with D,E being nonempty, disjoint, closed subsets ofC, hence of A too. The space A is normal (see Definition 2.4 and


Proposition 2.83(d)). Therefore, we can find U and V disjoint, opensets in A such that D ⊆ U and E ⊆ V . Then C ⊆ U ∪V =W and Wis open in A.

Now we show that there exists n0 � 1 such that

fn0(A) ⊆ W.

If this is not the case, we can find a sequence {xn}n�1 ⊆ A such thatxn ∈ fn(A) ∩ (A \W ) for all n � 1. The set A \W is compact and{xn}n�1 admits a limit point x. Evidently x ∈ C ⊆ W , which is acontradiction. This proves that fn0(A) ⊆W for some n0 � 1.

Thusfn0(A) =

(fn0(A) ∩ U) ∪ (fn0(A) ∩ V ),

which contradicts the fact that fn0(A) is connected.

(b) Let h : [0, 1] × S1 −→ X \ C be a homotopy between f and aconstant function (see Definition 2.186(c)). The set h

([0, 1] × S1

)is

compact inX\C and the latter is open. Hence h([0, 1]×S1

)is compact

in X. Because X is normal (being compact), we can find two opensets U, V ⊆ X such that

h([0, 1] × S1

) ⊆ U and C ⊆ V.

From (a), we know that fn0(A) ⊆ V for some n0 � 1. Therefore

h([0, 1] × S1

) ⊆ U ⊆ X \ fn0(A)

and sog(S1) ⊆ X \ fn0(A).

Solution of Problem 2.156No. Suppose that A is a retract of T (see Definition 2.196) and letr : T −→ A be a retraction (i.e., r is continuous and r|

A= id

A).

Let i : A −→ T be the inclusion function, then r ◦ i = idA. So, by

Proposition 2.206, the induced homomorphism

(idA)∗ = (r ◦ i)∗ = r∗ ◦ i∗ : π1(A) −→ π1(A)

2.3. Solutions 383

is the identity homomorphism. We know that π1(T ) is abelian, whileπ1(A) is nonabelian free with two generators a and b. We have ab �= baand

i∗(ab) = i∗(a)i∗(b) = i∗(b)i∗(a) = i∗(ba).

Therefore

ab = r∗i∗(ab) = r∗(i∗(ba)

)= r∗i∗(b)r∗i∗(a) = ba,

a contradiction. This proves that A is not a retract of T .

Solution of Problem 2.157Since X is path-connected (see Definition 2.122), we can find apath γ : [0, 1] −→ X such that γ(0) = x0 and γ(1) = x1. Thenγ = f ◦ γ : [0, 1] −→ Y is a path joining f(x0) and f(x1). Let[a] ∈ π1

(Y, f(x1)

). Then a is a loop in Y based on f(x1) and γaγ−1

is a loop in Y based on f(x0), hence

[γaγ−1] ∈ π1(Y, f(x0)

).

By assumption, we can find a loop β in X based on x0 such that

f∗([β])

= [γaγ−1].

So f ◦ β and γaγ−1 are homotopic. It follows that f ◦ (γβγ−1)and a

are homotopic and so

pf∗([γβγ−1]

)= [a].

Therefore f∗ : π1(X,x1) −→ π1(Y, f(x1)

)is surjective.

Solution of Problem 2.158From Example 2.219(d), we know that the projection p : SN −→ P

N isa covering function. Since SN is simply connected (see Definition 2.208and Theorem 2.215(b)), we can apply Theorem 2.226 and obtain an


isomorphism η : π1(PN , u) −→ p−1(u). The latter is a two element set,

hence π1(PN , u) is a group of order 2, i.e., π1(P

N , u) = Z2.

Solution of Problem 2.159From Example 2.228(c), we know that

π1(P2 × S2

)= π1

(P2)× π1(S

2).

But from Problem 2.158 (see also Example 2.228(d)) and Theo-rem 2.215(b), we know that

π1(P2) = Z2 and π1(S

2) = 0.

Therefore, we conclude that

π1(P2 × S2

)= Z2.


(a) Note that r ◦ iA= id

Aand so

(r ◦ iA)∗ = r∗ ◦ (iA)∗is the identity homomorphism on π1(A, x) (see Proposition 2.206(b)).From this it follows that (i

A)∗ is injective and r∗ is surjective.

(b) Consider the function r : R2 \ {0} −→ S1, defined by

r(x) = x‖x‖ .

Clearly this is a retraction (see Definition 2.196), i.e., S1 is a retractof R2 \ {0}. By part (a), the homomorphism π1(S

2) −→ π1(R2 \ {0})

induced by the inclusion function is injective. But from Theo-rem 2.215(a), we know that π1(S

1) = Z. Hence π1(R2 \ {0}) has

an infinite cyclic subgroup.Also let A = S1 × {1}. This is homeomorphic to S1. Hence by

Corollary 2.207, we have that

π1(A) = π1(S1) = Z.

2.3. Solutions 385

Moreover, the function r : T −→ A, defined by

r(x, y) = (x, 1)

is a retraction, i.e., A is a retract of T . Hence the homomorphism(i

A)∗ : π1(A) −→ π1(T ) is injective, which by what was observed above,

implies that π1(T ) has an infinite cyclic subgroup.

Solution of Problem 2.161Let p : X −→ X be the universal covering function (see Defini-tion 2.227). Suppose that π1(X,x) is not finite. Then p−1(x) is aninfinite closed subset of X. Because X is compact, the set p−1(x) hasat least one limit point x such that

p(x) = x.

Hence p cannot be a local homeomorphism at x, which contradictsDefinition 2.218.

Solution of Problem 2.162Let ε : R −→ S1 be the exponential covering function, defined by

ε(x) = e2πix

(i.e., ε is the universal covering function of S1; see Definition 2.227).Since X is locally path-connected (see Definition 2.127) and simplyconnected (see Definition 2.208), Theorem 2.221 implies that f has alift f : X −→ R (see Definition 2.220) such that

ε ◦ f = f.

Because R is simply connected, f is nullhomotopic (see Defini-tion 2.186). Let h be the homotopy transforming f continuously to aconstant function c. Then ε ◦ h = b0 is a homotopy transforming fcontinuously to a constant function c. Therefore f is nullhomotopic.


Solution of Problem 2.163Since π1(S

1) = Z (see Theorem 2.215(a)), π1(X) is finite andf∗ : π1(X) −→ π1(S

1) is a homomorphism, we see that f∗(π1(X)

)

is a finite subgroup of π1(S1) = Z, hence

f∗(π1(X)

)= 0.

So, the argument in the solution of Problem 2.162 works and we havethe desired result.

Solution of Problem 2.164Let X and Y be two homotopy equivalent spaces (see Definition 2.191)and let π0(X), π0(Y ) denote the sets of path-connected componentsof X and Y respectively (see Definition 2.130). Let f : X −→ Y andg : Y −→ X be two mutually inverse homotopy equivalences. Since f, gare continuous, they map path-connected sets to path-connected ones.So, f, g induce maps f : π0(X) −→ π0(Y ) and g : π0(Y ) −→ π0(X)respectively. Since g ◦ f � id

X, it follows that each x ∈ X lies in the

same path-connected component as g(f(x)

). Hence g ◦ f = id

π0(X).

Similarly f ◦ g = idπ0(Y )

. Therefore g ◦ f and f ◦ g are mutually inversehomotopy inverses, hence

cardπ0(X) = cardπ0(Y ).

So, X and Y have the same number of path-connected components.

Solution of Problem 2.165Let γ be a path in X connecting x0 and x1 and let v = [p ◦ γ]. Letτγ : π1(X, x0) −→ π1(X, x1) be defined by

τγ([u])

= [γ]−1 ∗ [u] ∗ [γ]

and let h : π1(X, u) −→ π1(X, u) be defined by

h(w)def= v−1 ∗ w ∗ v.

2.3. Solutions 387

We consider the following diagram

π1(X, x0) π1(X, x1)

π1(X, u) π1(X, u)

τγ

p∗ p∗

h

We claim that the above diagram is commutative. Note that

τγ([u])

=[γ−1 ∗ u ∗ γ],

so

p∗(τγ([u]))

=[p ◦ (γ−1 ∗ u ∗ γ)] =

[(p ◦ γ−1) ∗ (p ◦ u) ∗ (p ◦ γ)]

= v−1 ∗ p∗([u]) ∗ v = h

(p∗([u])).

So, the diagram is commutative. This proves that the two subgroupsof π1(X, u) are conjugate as claimed by the problem.

Solution of Problem 2.166Let r : X −→ A be the retraction (see Definition 2.196) and leti : A −→ X be the inclusion function. Then r ◦ i = id

Aand so

r∗ ◦ i∗ = (r ◦ i)∗ = (idA)∗.

It follows that the homomorphism i∗ : Hn(A) −→ Hn(X) is injectiveand the homomorphism r∗ : Hn(X) −→ Hn(A) is surjective. Considerthe short exact sequence

0 −→ Hn(A)i∗−→ Hn(X)

j∗−→ Hn(X,A) −→ 0,

where j : X = (X, ∅) −→ (X,A) is the inclusion function. Since i∗ isinjective, we see that the above sequence splits as

Hn(X) � im i∗ ⊕ ker r∗ � Hn(A)⊕ ker r∗.

But we know that

ker r∗ = Hn(X)/Hn(A)� Hn(X,A).


Therefore, finally

Hn(X) � Hn(A)⊕Hn(X,A).

Solution of Problem 2.167Since A is deformation retract of X (see Definition 2.196), we can finda homotopy h : [0, 1] ×X −→ X such that

h(0, ·) = idX,

h(1, x) ∈ A ∀ x ∈ X

h(1, ·)|A = idA

(see Definition 2.196(b)). Then the function r : X −→ A, defined by

r(x) = h(1, x) ∀ x ∈ X

is a retraction of X onto A. Therefore A is a retract of X and we canapply the result of Problem 2.166 and obtain

Hn(X) � Hn(A)⊕Hn(X,A) ∀ n � 0.

But X and A are homotopy equivalent (see Definition 2.191 and Theo-rem 2.198). Therefore Theorem 2.283 implies that

Hn(X) � Hn(A) ∀ n � 0,

soHn(X,A) = 0 ∀ n � 0.

Solution of Problem 2.168Note that � is a retract of X (indeed consider the retraction r(x) = �for all x ∈ X; Definition 2.196). Then according to Problem 2.166, wehave

Hn(X) = Hn(X, �) ⊕Hn(�) ∀ n � 0.

2.3. Solutions 389

Solution of Problem 2.169We do the proof for m = 2, the general case following by induction.Then we have

X = A1 ∪A2 and A1 ∩A2 = ∅.

Let e ∈ A1 and let r : X −→ X be defined by

r(x) =

{x if x ∈ A1,e if x ∈ A2.

Evidently r is continuous and r|A1

= idA1. Hence A1 is a retract of X

and so by Problem 2.166, we have

Hn(X) = Hn(A1)⊕Hn(X,A1) ∀ n � 0.

But by the excision property (see Theorem 2.278), we have

Hn(X,A1) � Hn(A2, ∅) = Hn(A2) ∀ n � 0.

Hence

Hn(X) = Hn(A1)⊕Hn(A2) ∀ n � 0.

By induction we have the general case m � 2.

Solution of Problem 2.170Let � ∈ X and let f : X −→ {�} be the constant function

f(x) = � ∈ X ∀ x ∈ X

(hence f is continuous). Also let i : {�} −→ X be the inclusion func-tion, i.e., i(�) = � ∈ X. Then f ◦ i = id{�} . Also, due to the fact thatX is contractible (see Definition 2.192), we have that i ◦ f is homo-topic to id

X(see Definition 2.186). Therefore X and {�} are homotopy

equivalent (see Definition 2.191) and so

Hn(X) = Hn(�) ∀ n � 0.


Invoking Problem 2.168, we conclude that

Hn(X, �) = 0 ∀ n � 0.

Solution of Problem 2.171From Theorem 2.282, we have the following exact sequence of homo-logical groups:

. . . Hn(A, �)i∗−→ Hn(X, �)

j∗−→ Hn(X,A)∂∗−→ Hn−1(A, �) . . . ∀ n � 0,

where i∗ and j∗ are the homomorphisms induced by the correspondinginclusion functions and ∂∗ is the boundary homomorphism. Since Ais contractible, from Problem 2.170, we have

Hn(A, �) = 0 ∀ n � 0

(note that by definition H−1(A, �) = 0). Then from the long exactsequence, we have

{0} = im i∗ = ker j∗ and im j∗ = ker ∂∗ = Hn(X,A).

Therefore j∗ is an isomorphism and so we conclude that

Hn(X,A) � Hn(X, �) ∀n � 0.

Solution of Problem 2.172Let en and es denote the “north” and “south” poles respectively in S1.Let U = S1 \ {en} and V = S1 \ {es}. Then S1 = U ∪ V . Moreover,using the stereographic projection (cf. the solution of Problem 2.58),we see that both U and V are homeomorphic to R and so

Hn(U) = Hn(V ) = 0 ∀ n � 1

(see Proposition 2.270(c)). Note that U ∩ V is the disjoint union oftwo spaces (two half-circles) each homeomorphic to R. Using Propo-sition 2.270(b), the Mayer–Vietoris exact sequence in Theorem 2.281becomes

. . . −→ 0 −→ Hn(S1) −→ . . . −→ 0 −→

−→ H1(S1) −→ Z⊕ Z −→ Z⊕ Z −→ H0(S

1) −→ 0.

2.3. Solutions 391

Since S1 is path-connected, we have

H0(S1) = Z

(see Proposition 2.270(b)). So, the kernel of the last function Z⊕Z −→H0(S

1) is Z. Similarly, the kernel of the function Z⊕Z −→ Z⊕Z mustbe Z. The exactness of the Mayer–Vietoris sequence, implies that

H1(S1) = Z.

For n > 1, Hn(S1) is between two trivial groups and so

Hn(S1) = 0 ∀ n � 2.

Similarly, for the 2-sphere S2. Again we consider

U = S2 \ {en} and V = S2 \ {es}.

We have

S2 = U ∪ V.Again using the stereographic projection, we have that U and V areboth homeomorphic to R

2 and U∩V is homeomorphic to R×S2 whichis homotopy equivalent to S1 (see Definition 2.191). Therefore fromthe first part of the proof, we have

H0(U ∩ V ) = H1(U ∩ V ) = Z and Hn(U ∩ V ) = 0 ∀ n � 2.

Then the Mayer–Vietoris exact sequence (see Theorem 2.281) becomes

. . . −→ 0 −→ Hn(S2) −→ 0 . . . −→ 0 −→ H1(S

2) −→ Z −→−→ 0 −→ H1(S

2) −→ Z −→ Z⊕ Z −→ Z −→ 0.

It follows that

Hn(S2) = 0 ∀ n � 3 and H2(S

2) = Z.

Also as for S1, we have

H0(S2) = Z.


Finally because the function Z −→ Z⊕Z has image Z and is injective,we have

H1(S2) = 0.


(a) No. Let p : R −→ S1 be the covering function, defined by

p(t) = e2πit

and let f : SN −→ S1 be a continuous function. From Theo-rem 2.215(b), we know that

π1(SN ) = 0 ∀ N � 2.

Hence by Theorem 2.221, f admits a lift f : SN −→ R (see Defini-tion 2.220) such that p ◦ f = f . But R is contractible (see Defini-tion 2.192 and Remark 2.193). Hence f is homotopic to a constantfunction c. This implies that f is homotopic to p ◦ c (see Proposi-tion 2.190) which is a constant function. Therefore, we conclude thatthere is no continuous function f : SN −→ S1 (N � 2) which is notnullhomotopic.

(b) Yes. Note that (x, y) ∈ T if and only if x = e2πit1 , y = e2πit2 witht1, t2 ∈ [0, 1]. Let f : T −→ S1 be defined by

f(e2πit1,2πit2

)= e2πit1 ∈ S.

Then the induced homomorphism f∗ : H1(T ) −→ H1(S1) maps one

generator of H1(T ) to the generator of H1(S1) and the other generator

of H1(T ) to zero. So, f∗ is not trivial, which implies that f is notnullhomotopic.

Solution of Problem 2.174Let y1 and y2 be the two identification points. We introduce the sets

U = ΣX \ {y1} and V = ΣX \ {y2}.

2.3. Solutions 393

Then U and V are open subsets of ΣX and ΣX = U ∪ V . Moreover,U and V are contractible spaces and deformation retracts of U ∩ V(see Definitions 2.192 2.196). By the Mayer–Vietoris sequence (seeTheorem 2.281), we have

Hn(ΣX) =

{Z if n = 0,Hn−1(X, �) if n � 1,

with � ∈ X.

Solution of Problem 2.175Let � ∈ U ∩ V and consider the Mayer–Vietoris sequence for (U, V )(see Theorem 2.281)

. . . −→ H2(X, �)∂∗−→ H1(U ∩ V, �) (j1∗ ,−j2∗)−→ H1(U, �) ⊕H1(V, �)

(i1∗,i2∗)−→(i1∗,i2∗)−→ H1(X, �)

∂∗−→ H0(U ∩ V, �),

where i1 : U −→ X, i2 : V −→ X, j1 : U ∩ V −→ U , j2 : U ∩ V −→ Vare the inclusion functions. Since U ∩ V is path-connected, we haveH0(U ∩ V, �) = 0. Also the exactness of the Mayer–Vietoris sequenceimplies that

im (i1∗, i2∗) = ker ∂∗ = H1(X, �) = H1(X).

But by hypothesis (j1∗ ,−j2∗) is surjective and from the exactness, wehave

ker (i1∗, i2∗) = im (j1∗ ,−j2∗) = H1(U, �) ⊕H1(V, �),

so

im (i1∗, i2∗) = H1(X) = 0.

The result fails if throughout H1 is replaced by H2. To see this letX = S2 and let

U ={(x, y, z) ∈ S2 : z > −1

2

}and V =

{(x, y, z) ∈ S2 : z < 1

2

}.

Then X = U ∪ V . Clearly U , V and U ∩ V are path-connected andH2(U) = H2(V ) = 0. So, the homomorphism H2(U ∩ V ) −→ H2(V )


induced by the corresponding inclusion function is surjective. ButH2(S

2) = Z (see Proposition 2.272(b)).

Solution of Problem 2.176We argue by contradiction. So suppose that there is x ∈ B

N1 \ f(BN

1 ).

Then x ∈ BN1 \ SN−1. Consider the function r : B

N1 \ {x} −→ SN−1,

defined as follows. Consider u ∈ BN1 \ {x} and draw the line be-

tween u and x. Extend this line beyond u until it meets SN−1 andlet r(u) be the point where this happens. Clearly r is continuous andr|

SN−1= id

SN−1. Hence r is a retraction (see Definition 2.196) and

so SN−1 is a retract of BN1 \ {x}. Then h = r ◦ f : BN

1 −→ SN−1 isa continuous function and h|

SN−1= f |

SN−1which by hypothesis is a

homeomorphism. Hence if i : SN−1 −→ BN1 is the inclusion function,

then h ◦ i : SN−1 −→ SN−1 is a homeomorphism. This implies that(h ◦ i)∗ : HN−1(S

N−1) −→ HN−1(SN−1) is an isomorphism (see Theo-

rem 2.283). But (h◦ i)∗ = h∗ ◦ i∗ and i∗ : HN−1(SN−1) −→ HN−1(B

N1 )

is the trivial homomorphism since HN−1(BN1 ) = 0 (see Proposi-

tion 2.272(a)). Therefore (h ◦ i)∗ is the trivial homomorphism, a con-tradiction.

Solution of Problem 2.177Evidently X = Px ∪ Py, where Px is the yz-plane and Py is the xz-plane. Let Cx be the unit circle in Px and let Cy be the unit circlein Py. Then Cx ∪Cy is a strong deformation retract of X \ {(0, 0, 0)}(see Definition 2.196). Let

U = Cx ∪ Cy \{(0, 0, 1)

}and V = Cx ∪ Cy \

{(0, 0,−1)

}.

These sets are contractible, U ∪V = Cx∪Cy and U ∩V has four path-connected components which are contractible. The Mayer–Vietorissequence for the pair (U, V ) (see Theorem 2.281), implies that

H1

(X \ {(0, 0, 0)}) = H1(U ∪ V ) = H1(Cx ∪ Cy) = Z⊕ Z⊕ Z.

2.3. Solutions 395

Suppose that X and R2 are homeomorphic and let f : X −→ R

2 be ahomeomorphism. Then f |

X\{(0,0,0)} : X\{(0, 0, 0)} −→ R2\{f(0, 0, 0)}

is a homeomorphism too. Then by Theorem 2.283, the groups

H1

((X \ {(0, 0, 0)})) and H1

((R2 \ {(0, 0)}))

are isomorphic. But

H1

((R2 \ {(0, 0)})) = H1(S

1) = Z

(see Proposition 2.272(b)), a contradiction. This proves that X andR2 are not homeomorphic.Let h : [0, 1] ×X −→ X be define dy

h(t, (x, y, z)

)=

{(x, y, z) if t = 0,(tx, y, z) if t ∈ (0, 1].

Clearly this is a deformation retraction of X onto Py. This proves thatX and R

2 are of the same homotopy type (see Definition 2.191).

Solution of Problem 2.178Let A1 and A2 be two copies of the unit sphere in R

2 together withthe portion of the z-axis inside them, which are touching at the origin.The union of A1 and A2 is called the wedge product of A1 and A2 andis denoted by A1 ∨A2. For example, the figure eight in Problem 2.154is such a wedge product and is a deformation retract of T = S1 × S2

(the torus) minus a point by the function

h(t, (x, y)

)=((1− t)x+ t x|x| , (1− t)y + t y|y|

)

(see Definition 2.196). Let U = A1 ∨ A2 \ {en} (en being the northpole of S2) and V = A1 ∨ A2 \ {es} (es being the south pole of S2).Then U ∪ V = A1 ∨A2. Note that U ∩ V is contractible (hence path-connected). So, by the Van Kampen theorem (see Theorem 2.229),we have that π1(A1 ∨A2) is the free product of the groups π1(U) andπ1(V ). Note that

π1(U) = π1(V ) = π1(A1) = π1(A2) = π1(S1).


Since X is of the same homotopy type as A1 ∨A2, we have

π1(X) = π1(A1 ∨A2)

(see Corollary 2.207) and so π1(X) is a free product generated by twogenerators.

Using the Mayer–Vietoris sequence (see Theorem 2.281) for (U, V ),we obtain

Hk(X) = Hk(U)⊕Hk(V ) = Hk(A1)⊕Hk(A2) ∀ k � 0.

Note thatHk(A1) = Hk(A2) ∀ k � 0

and it is infinite cyclic for k = 0, 1, 2. Hence

Hk(A) =

⎧⎨

⎩

Z if k = 0,Z× Z if k ∈ {1, 2},0 if k � 3.

Solution of Problem 2.179Using the homotopy h : [0, 1] × C −→ C, defined by

h(t, (x, u)

)= (x, tu) ∀ t ∈ [0, 1], (x, u) ∈ C,

we see that C deformation retracts to S1 (see Definition 2.196). Hence

Hk(C) = Hk(S1) =

{Z if k ∈ {0, 1},0 otherwise

(see Proposition 2.272(b)).

Solution of Problem 2.180We argue by contradiction. So, suppose that SN is a retract of

BN+11 (see Definition 2.196) and let r : B

N+11 −→ SN be a retraction.

Also let i : SN −→ BN+11 be the inclusion function. We have that

r ◦ i : SN −→ SN and r ◦ i = idSN

. Note that by Proposition 2.272,we have

i∗ : HN (SN ) = Z −→ HN (B

N+1) = 0.

So (r ◦ i)∗ = r∗ ◦ i∗ = 0. On the other hand, since r ◦ i = idSN

, we havethat (r ◦ i)∗ = r∗ ◦ i∗ is the identity homomorphism, a contradiction.

2.3. Solutions 397

Solution of Problem 2.181Choose r > 0 small such that Br(x) ⊆ U . Using the inclusion func-tions, we have the following commutative diagram

∂Br(x)

U \ {x} RN \ {x}

is

j

These functions induce homology homomorphisms:

HN−1(∂Br(x))

HN−1(U \ {x}) HN−1(RN \ {x})

i∗s∗

j∗

Note that s is a homotopy equivalence (see Definition 2.191).Hence s∗ is an isomorphism (see Corollary 2.207) and so i∗ is injec-tive and j∗ is surjective. But by Proposition 2.272(b), we know thatHN−1

(∂Br(x)

) �= 0. Hence

HN−1

(U \ {x}) �= 0.

Solution of Problem 2.182We argue by contradiction. So, suppose that U and V are homeo-morphic. Let h : U −→ V be the homeomorphism. Then by Prob-lem 2.181, we have that

HN−1

(V \ {h(x)}) �= ∅ ∀ x ∈ U.

On the other hand, the set V \ {h(x)} is homeomorphic to RM \ {0}

which in turn is homotopy equivalent to SM−1 (see Definition 2.191;in fact SM−1 is a strong deformation retract of RM \ {0}; see Defini-tion 2.196). So

HN−1

(V \ {h(x)}) = HN−1(S

M−1) = 0

(since N �=M ; see Proposition 2.272), a contradiction.


Solution of Problem 2.183We write SN = SN+ ∪SN− , where SN+ is the “northern” hemisphere andSN− is the “southern” hemisphere. By the Mayer–Vietoris sequence(see Theorem 2.281), for � ∈ SN+ ∩ SN− we have the exact sequence:

. . . −→ Hk(SN+ , �) ⊕Hk(S

N− , �) −→ Hk(SN+ ∪ SN− , �) −→

−→ Hk−1(SN+ ∪ SN− , �) −→ Hk−1(S

N+ , �) ⊕Hk−1(S

N− , �) −→ . . . .

So, because SN+ and SN− are both contractible (see Definition 2.192),we have

Hk(SN+ , �) = Hk(S

N− , �) = 0

(see Problem 2.170). It follows that

Hk(SN , �) =

⎧⎨

⎩

Hk−1(SN−1, �) = . . . = H0(S

N−k, �) if k < N,H0(S

0, �) if k = N,Hk−N(S0, �) if k > N.

Since

H0(SN−k+ , �)⊕H0(S

N−k− , �)

ξ−→ H0(SN−k+ ∪ SN−k

− , �) −→ 0,

the direct sum is trivial and the chain is exact, we infer that

H0(SN−k+ , �) = 0 ∀ k < N.

Also, we have

Hk−N(S0, �) = Hk−N({x} ∪ {�}, �) = Hk−N(�) ∀ k � N.

Solution of Problem 2.184We consider the triple (X1,X2,X3) and the corresponding long exactsequence

. . . −→ Hk(X3,X1)i∗−→ Hk(X3,X2)

∂∗−→ Hk−1(X2,X1) −→ . . .

(see Theorem 2.277). From the rank theorem, we have

rankHk(X3, X2) = rank ker ∂∗ + rank im ∂∗ = rank im i∗ + rank im ∂∗� rankHk(X3, X1) + rankHk−1(X2, X1) (2.1)

2.3. Solutions 399

(using the exactness of the sequence). Similarly considering the triple(X1,X3,X4), we obtain

rankHk(X3,X1) � rankHk(X4,X1) + rankHk+1(X4,X3). (2.2)

Adding (2.1) and (2.2), we obtain

rankHk(X3,X2)− rankHk(X4,X1) � rankHk−1(X2,X1)

+rankHk+1(X4,X3).

Bibliography

[1] Armostrong, M.A.: Basic Topology. Undergraduate Texts inMathematics. Springer, New York (1983)

[2] Berge, C.: Espaces Topologiques: Fonctions Multivoques. Col-lection Universitaire de Mathematiques, vol. III. Dunod, Paris(1959)

[3] Buskes, G., van Rooij, A.: Topological Spaces. From Distance toNeighborhood. Undergraduate Texts in Mathematics. Springer,New York (1997)

[4] Choquet, G.: In: Marsden, J., Lance, T., Gelbart, S. (eds.) Lec-tures on Analysis. Volume I: Integration and Topological VectorSpaces. W.A. Benjamin, Inc., New York (1969)

[5] Choquet, G.: In: Marsden, J., Lance, T., Gelbart, S. (eds.) Lec-tures on Analysis. Volume II: Representation Theory. W.A. Ben-jamin, Inc., New York (1969)

[6] Dold, A.: Lectures on Algebraic Topology. Grundlehren derMathematischen Wissenschaften, vol. 200. Springer-Verlag, Berlin(1980)

[7] Dugundji, J.: Topology. Allyn and Bacon Series in AdvancedMathematics. Allyn and Bacon, Inc., Boston (1978)

[8] Engelking, R.: General Topology. Sigma Series in Pure Mathe-matics, vol. 6. Heldermann Verlag, Berlin (1989)

[9] Janisch, K.: Algebraic Topology. Springer, New York (1984)

[10] Kelley, J.L.: General Topology. Graduate Texts in Mathematics,vol. 27. Springer, New York (1975)

401

402 Bibliography

[11] Kuratowski, K.: Topology: Volume I. Academic, New York (1966)

[12] Kuratowski, K.: Topology: Volume II. Academic, New York(1968)

[13] Munkres, J.R.: Topology, a First Course. Prentice-Hall, Engle-wood Cliffs (1975)

[14] Maunder, C.R.F.: Algebraic Topology. Dover, Mineola (1996)

[15] Nagata, J.-I.: Modern General Topology. North-Holland Mathe-matical Library, vol. 33. North-Holland Publishing Co., Amster-dam (1985)

[16] Oxtoby, J.C.: Measure and Category. A Survey of the Analo-gies between Topological and Measure Spaces. Graduate Texts inMathematics, vol. 2. Springer, New York (1971)

[17] Spanier, E.H.: Algebraic Topology. Springer, New York (1981)

[18] Steen, L.A., Seebach, J.A.J.: Counterexamples in Topology.Springer, New York (1978)

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