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

Chapter 1

Metric Spaces

1.1 Introduction

1.1.1 Basic Definitions and Notation

Definition 1.1A metric space is a pair (X, dX ) of a set X and a function dX : X ×X −→ R, which has the following properties:(a) d

X(x, y) � 0 for all x, y ∈ X and d

X(x, y) = 0 if and only if

x = y;(b) d

X(x, y) = d

X(y, x) for all x, y ∈ X;

(c) dX(x, y) � d

X(x, u) + d

X(u, y) for all x, y, u ∈ X (triangle in-

equality).

The function dX (·, ·) is called a metric or distance.

Remark 1.2If (a) in the above definition is replaced by a weaker requirement:(a)′ dX (x, y) � 0 for all x, y ∈ X and if x = y, then dX (x, y) = 0,then dX is said to be a pseudometric or ecart or semimetric and(X, d

X) is a pseudometric space (or semimetric space).

Example 1.3

(a) Suppose that{(Xk, dXk

)}N

k=1are metric spaces and set

X =

N∏

k=1

Xk

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

1

2 Chapter 1. Metric Spaces

and

dp(x, y) =( N∑

k=1

dXk

(xk, yk)p) 1

p ∀ x, y ∈ X,

d∞(x, y) = max(dXk

(xk, yk) : 1 � k � N) ∀ x, y ∈ X,

where 1 � p < +∞, x = (xk)Nk=1, y = (yk)

Nk=1 ∈ X. Then (X, dp)

(1 � p < +∞) and (X, d∞) are metric spaces. In particular, if Xk = Rand

dR(x, y) = |x− y| ∀ x, y ∈ R,

then dRis a metric on R. For N � 1, d2 is the Euclidean metric on

RN .

(b) Suppose now that we have an infinite family{(Xk, dXk

)}k�1

ofmetric spaces and assume that

sup{dXk

(x, y) : x, y ∈ X, k � 1}

< +∞.

We setX =

∏

k�1

Xk

and

dX(x, y) =

∑

k�1

1

2kdXk

(xk, yk) ∀ x = {xk}k�1 , y = {yk}k�1 ∈ X.

Then (X, dX) is a metric space.

(c) Let X be an arbitrary nonempty set and let

ddX(x, y) =

{0 if x = y,1 if x �= y,

∀ x, y ∈ X.

Then (X, ddX) is a metric space and dd

Xis called the discrete metric

on X.

(d) Let X = C[a, b] (the space of continuous functions on [a, b]). Weset

d1X(f, g) =

∫ b

a

∣∣f(t) − g(t)∣∣ dt and d∞

X(f, g) = max

a�t�b

∣∣f(t) − g(t)∣∣ ∀ f, g ∈ X.

1.1. Introduction 3

Then (X, d1X) and (X, d∞

X) are metric spaces. However, as we will

see later (see Remark 1.82) these metric spaces are fundamentally dif-ferent. We call d1

Xthe L1-metric and d∞

Xthe uniform metric or

supremum metric on X = C[a, b].

(e) Let X = R ∪ {±∞} (extended real line) and set

dX(x, y) =

∣∣ tan−1 x− tan−1 y∣∣ ∀ x, y ∈ X

(recall that tan−1(±∞) = ±π2 and tan−1 is injection). Then (X, d

X)

is a metric space.

Proposition 1.4If (X, d

X) is a metric space and

dX(x, y) =

dX(x, y)

1 + dX(x, y)

∀ x, y ∈ X,

then (X, dX ) is a metric space.

Remark 1.5Note that

dX (x, y) < 1 ∀ x, y ∈ X.

Moreover, using this proposition, we can see that the space X of allreal sequences equipped with the distance

dX(x, y) =

∑

k�1

1

2k|xk − yk|

1 + |xk − yk|

is a metric space (see Example 1.3(b)).

Definition 1.6Let (X, d

X) be a metric space.

(a) The open ball centred at x0 ∈ X of radius r > 0 is defined by

Br(x0) ={x ∈ X : d

X(x, x0) < r

}.

(b) A set C ⊆ X is said to be bounded if it is contained in someopen ball. The set C is unbounded if this is not the case.


(c) The diameter of a set C ⊆ X is given by

diamC = sup{dX(x, y) : x, y ∈ C

}

(if C = ∅, then diamC = 0).(d) For any x ∈ X and A,B ⊆ X, we define:

dist(x,A) = inf{dX(x, a) : a ∈ A

}

and

dist(A,B) = inf{dX(a, b) : a ∈ A, b ∈ B

}.

1.1.2 Sequences and Complete Metric Spaces


X) be a metric space and let {xn}n�1 ⊆ X be a sequence.

(a) We say that the sequence {xn}n�1 ⊆ X converges to x ∈ X if andonly if

dX (xn, x) −→ 0 as n → +∞,

i.e., for any r > 0, we can find an integer n0 = n0(r) � 1 such that

xn ∈ Br(x) ∀ n � n0.

The point x ∈ X is said to be the limit of the convergent sequence{xn}n�1.(b) Let (X, d

X) be a metric space. A sequence {xn}n�1 ⊆ X is said to

be a Cauchy sequence if for any given ε > 0, we can find an integern0 = n0(ε) � 1 such that

dX(xn, xk) � ε ∀ n, k � n0.

(c) A metric space (X, dX) is said to be complete if every Cauchy

sequence in X is convergent in X.

1.1.3 Topology of Metric Spaces

To be able to proceed further with the study of metric spaces, we needto introduce some topological material associated with them. We willreturn to these concepts in a more general setting in Chap. 2.

1.1. Introduction 5



(a) A set U ⊆ X is said to be open if for every x ∈ U we can findr = r(x) > 0 such that Br(x) ⊆ U .(b) A set C ⊆ X is said to be closed if the set X \ C is open.(c) The family of open sets U ⊆ X is called the topology determinedby the metric d

X(metric topology).

(d) A set A ⊆ X is called a neighbourhood of a point x ∈ X if thereexists r > 0 such that Br(x) ⊆ A.

Proposition 1.9If (X, dX ) is a metric space, then(a) ∅ and X are open sets;

(b) if {Ui}i∈I is any family of open sets in X, then⋃

i∈IUi is an open

set too;

(c) if {Uk}Nk=1 is any finite family of open sets in X, then

N⋂

k=1

Uk is

an open set too.

Using de Morgan laws and Definition 1.8(b), we also have


X) is a metric space, then

(a) ∅ and X are closed sets;

(b) if {Ci}i∈I is any family of closed sets in X, then⋂

i∈ICi is a closed

set too;

(c) if {Ck}Nk=1 is any finite family of closed sets in X, thenN⋃

k=1

Ck is

a closed set too.

We can characterize sets in terms of convergent sequences.

Proposition 1.11If (X, dX ) is a metric space,then C ⊆ X is closed if and only if every convergent sequence{xn}n�1 ⊆ C has limit in C.



X) be a metric space and E ⊆ X. The subspace (metric)

topology on E induced by the metric dX

is the family

{E ∩ U : U ⊆ X open

}.


X) be a metric space and let E ⊆ X.

(a) We say that x ∈ E is an interior point of E if there exists r > 0such that Br(x) ⊆ E. The set of all interior points of E is called theinterior of E and is denoted by intE.(b) We say that x ∈ E is a limit (or cluster or accumulation)point of E if Br(x) ∩

(E \ {x}) �= ∅ for every r > 0. The union of E

with the set of all its accumulation points is called the closure of Eand is denoted by E.(c) The boundary of E is defined by ∂E = E ∩ Ec.(d) We say that x ∈ E is an isolated point of E, if there is an r > 0such that Br(x)∩

(E \ {x}) = ∅. Hence {x} is a relatively open subset

of E.(e) We say that E ⊆ X is a perfect set if every point of E is anaccumulation point.

Theorem 1.14Let (X, d

X) be a metric space and let E ⊆ X.

(a) x ∈ intE if and only if there exists r > 0 such that Br(x) ⊆ E.(b) x ∈ E if and only if for every r > 0, we have Br(x) ∩ E �= ∅.(c) x ∈ ∂E if and only if for every r > 0, we have Br(x)∩E �= ∅ andBr(x) \E �= ∅.(d) x is an accumulation point of E if and only if we can find asequence {xn}n�1 ⊆ E \ {x} such that xn −→ x.

Proposition 1.15Let (X, d

X) be a metric space and let E,F ⊆ X. If E ⊆ F , then

intE ⊆ intF and E ⊆ F .

Definition 1.16For every r > 0 and for every x ∈ X, we define a closed ball

Br(x) ={u ∈ X : dX (u, x) � r

}.

1.1. Introduction 7

Remark 1.17Any closed ball is a closed set (in the sense of Definition 1.8(b)) andwe always have

Br(x) ⊆ Br(x) ∀ x ∈ X, r > 0.

The above inclusion in general cannot be replaced by equality. Tosee this let the set X has at least two elements and let (X, dd

X) be a

discrete metric space (see Example 1.3(c)). If x ∈ X, then

B1(x) = {x} � X = B1(x).


X) is a metric space and E ⊆ X, then

(a) intE is open and is the union of all open sets contained in E, i.e.,intE is the largest open set contained in E (see Proposition 1.9(b)).(b) E is closed and is the intersection of all closed sets which con-tain E, i.e., E is the smallest closed set containing E (see Proposi-tion 1.10(b)).

Corollary 1.19If (X, dX ) is a metric space and E ⊆ X, then(a) E is open if and only if E = intE.(b) E is closed if and only if E = E.


X) be a metric space. A set D ⊆ X is dense in E ⊆ X if

E ⊆ D. We say that D is dense (or everywhere dense) if D = X.


X) be a metric space and x ∈ X.

(a) A family of open sets {Ui}i∈I all containing x is said to be a localbase at x if for every open set U containing x, we can find i ∈ I suchthat Ui ⊆ U .(b) A family {Ui}i∈I of nonempty sets in X is a base for the metrictopology of X if every open set in X is the union of a subfamily of{Ui}i∈I .(c) We say that X is second countable if it has a countable base forthe metric topology.


(d) A family F of open sets in X is an open cover of X if⋃

U∈FU = X.

An open subcover of F is a subfamily of F which is an open coverof X.(e) We say that X is Lindelof if every open cover of X contains acountable subcover.(f) We say that X is separable if it has a countable dense subset.

Proposition 1.22Every metric space has a countable local base at every point.

Remark 1.23The above proposition explains the name “second countable” in Def-inition 1.21(c). As we shall see in Chap. 2, for a topological spacethe property of having a countable local base at every point is calledfirst countability . The above proposition says that every metricspace is first countable. This is the reason why sequences suffice todescribe the various topological notions of a metric space. Howevernote that not every metric space is second countable.

Proposition 1.24Let (X, d

X) be a metric space. The following three properties are equiv-

alent:(a) X is second countable.(b) X is Lindelof.(c) X is separable.

1.1.4 Baire Theorem

One of the most important properties of a metric space is complete-ness and many fundamental results of analysis depend critically onthis property. A main device through which completeness becomes apowerful tool is the so called Baire category theorem .


X) be a metric space. A set E ⊆ X is said to be nowhere

dense if intE = ∅. A set E ⊆ X is said to be meager or of firstcategory if E can be written as a countable union of nowhere densesets. If E ⊆ X is not of the first category, then we say that it is ofsecond category.

1.1. Introduction 9

Theorem 1.26 (Baire Category Theorem)If (X, d

X) be a complete metric space, then

(a) the intersection of countably many open dense sets is dense in X;(b) if X =

⋃

n�1Cn with Cn ⊆ X closed, then for at least one n0 � 1,

we have intCn0 �= ∅.

Corollary 1.27

(a) A complete metric space is of second category.(b) In a complete metric space a meager set has empty interior, i.e.,its complement is dense in X.

Theorem 1.28 (Cantor Intersection Theorem)If (X, dX ) be a metric space,then the following two statements are equivalent:(a) X is complete;(a) every decreasing sequence {Cn}n�1 (i.e., Cn+1 ⊆ Cn for all n � 1)of nonempty closed sets in X such that diamCn ↘ 0 as n → +∞ hassingleton intersection.

1.1.5 Continuous and Uniformly Continuous Functions


X) and (Y, d

Y) be two metric spaces. A function f : X −→ Y

is said to be continuous at x ∈ X if for every ε > 0, there existsδ = δ(x, ε) > 0 such that

dX (x, y) < δ =⇒ dY

(f(x), f(y)

)< ε.

We say that f : X −→ Y is continuous, if it is continuous at everyx ∈ X.


X) and (Y, d

Y) are two metric spaces and f : X −→ Y is a

function,then f is continuous at x ∈ X if and only if for any sequence xn −→ xin X, we have that f(xn) −→ f(x) in Y .


Corollary 1.31If X,Y, V are three metric spaces, f : X −→ Y is continuous at x ∈ Xand g : Y −→ V is continuous at f(x) ∈ Y ,then the composition g ◦ f : X −→ V is continuous at x ∈ X.


X) and (Y, d

Y) are two metric spaces and f : X −→ Y is a

function,then the following statements are equivalent:(a) f is continuous;(b) for every open U ⊆ Y , the set f−1(U) ⊆ X is open;(c) for every closed C ⊆ Y , the set f−1(C) ⊆ X is closed;(d) f(E) ⊆ f(E) for all E ⊆ X;(e) f−1(A) ⊆ f−1(A) for all A ⊆ Y .

The last proposition leads to the following definition.

Definition 1.33Let X,Y be two sets and let E be a proper subset of X. If f : E −→ Yis a function, then f : X −→ Y is said to an extension of f , iff |

E= f . The function f is called the restriction of f on E.

Theorem 1.34If X and Y are two metric spaces, D ⊆ X is dense and f : D −→ Yis a function,then f admits a continuous extension f : X −→ Y if and only if forevery x ∈ X the limit lim

u→xf(u) exists in Y and is equal to f(x) if

x ∈ D.When the extension exists, it is unique.

Proposition 1.35If X and Y are two metric spaces, X is a nonempty collection ofmutually disjoint nonempty subsets of X, f :

⋃X −→ Y and for eachE ∈ X , we have E ∩ (⋃

(X \ {E})) = ∅ and f |E is continuous,then f :

⋃X −→ Y is continuous.


X) is a metric space,

then the metric dX: X ×X −→ R is continuous, where in X ×X we

consider any product metric dp with 1 < p < +∞ (see Example 1.3).

1.1. Introduction 11

Definition 1.37Suppose that (X, d

X) and (Y, d

Y) are two metric spaces and

f : X −→ Y is a function. For each x ∈ X, we define

ωf(x)def= inf

r>0diam f

(Br(x)

).

The function ωf is called the oscillation function for f .

Remark 1.38

(a) Suppose that (X, dX) and (Y, d


f : X −→ Y is a function. The function f is continuous at x ∈ Xif and only if ωf (x) = 0 (see Problem 1.46).(b) If (X, d

X) is a metric space and f : X −→ R is a function, then

for every x ∈ X, we have ωf (x) = lim supu→x

f(u)− lim infu→x

f(u).

Definition 1.39Let X and Y be two metric spaces and let f : X −→ Y be a func-tion.(a) We say that f is an open function if for every open set U ⊆ X,the set f(U) is open in Y .(b) We say that f is a closed function if for every closed set C ⊆ X,the set f(C) is closed in Y .(c) We say that f is a homeomorphism if f is continuous, bijectiveand f−1 : Y −→ X is continuous (i.e., f is bicontinuous bijection).The metric spaces X and Y are said to be homeomorphic spaces ifthere exists a homeomorphism f : X −→ Y .

Theorem 1.40If X and Y are two metric spaces and f : X −→ Y is bijective,then the following statements are equivalent:(a) f is a homeomorphism;(b) f is continuous and open;(c) f is continuous and closed;(d) f(E) = f(E) for every E ⊆ X.



X) and (Y, d

Y) be two metric spaces. A function f : X −→ Y

is an isometry if dY

(f(x1), f(x2)

)= d

X(x1, x2) for all x1, x2 ∈ X.

The metric spaces X and Y are said to be isometric.

Evidently an isometry is a continuous function.

Remark 1.42Using the notions of homeomorphism and isometry, we can speakabout the topological properties and the metric properties of a met-ric space X. Topological are those properties which are preserved byhomeomorphisms, and metric are those properties which are preservedby isometrics. For example openness, closedness or being a convergentsequence in X are topological properties of X, while Cauchy sequencesand completeness are metric properties. Of course the topologicalstructure is more flexible than the metric structure.

The next theorem is a particular case of the so called Urysohnlemma (see Theorem 2.136).

Theorem 1.43If X is a metric space and A,C ⊆ X are disjoint closed sets,then there is a continuous function f : X −→ [0, 1] such that

f |A

≡ 0 and f |C

≡ 1.

Similarly the next theorem is a particular case of the so calledTietze extension theorem (see Theorem 2.138).

Theorem 1.44If X is a metric space, C ⊆ X is a nonempty closed subset andf : C −→ R is a continuous function,then f admits a continuous extension f : X −→ Y such that

inf f(X) = inf f(C) and sup f(X) = sup f(C).

In the definition of continuity (see Definition 1.29), δ depends onx and ε. If δ can be chosen independently of x ∈ X, then we have astronger form of continuity called uniform continuity.



X) and (Y, d

Y) be two metric spaces and f : X −→ Y . We

say that f is uniformly continuous if for every ε > 0, there existsδ = δ(ε) > 0 such that

dX(x1, x2) < δ =⇒ d

Y

(f(x1), f(x2)

)< ε.

Proposition 1.46Let X,Y, V be three metric spaces.(a) If f : X −→ Y is uniformly continuous and E ⊆ X, then f |

Eis

uniformly continuous too.(b) If f : X −→ Y is uniformly continuous, then maps Cauchy se-quences to Cauchy sequences.(c) If f : X −→ Y and g : Y −→ V are both uniformly continuousfunctions, then so is g ◦ f : X −→ Y ;

Combining Theorem 1.34 with Proposition 1.46(b), we get at oncethe following fundamental extension theorem.

Theorem 1.47If X and Y are metric spaces, Y is complete, D ⊆ X is dense andf : D −→ Y is a uniformly continuous function,then f admits a uniformly continuous extension f : X −→ Y .

An important subclass of uniformly continuous functions are theso called Lipschitz functions.


X) and (Y, d

Y) be two metric spaces, k � 0 and let

f : X −→ Y be a function.(a) We say that f is Lipschitz continuous with Lipschitz constantk (or k-Lipschitz), if

dY

(f(x1), f(x2)

)� kd

X(x1, x2) ∀ x1, x2 ∈ X.

If k ∈ [0, 1), then we say that f is a contraction or k-contraction.If k = 1, then we say that f is nonexpansive.(b) We say that f is a locally Lipschitz function if for all x ∈ X,we can find an open set Ux such that x ∈ Ux and a constant kx > 0such that

dY

(f(x1), f(x2)

)� kxdX

(x1, x2) ∀ x1, x2 ∈ Ux.


Theorem 1.49 (Banach Fixed Point Theorem)If (X, d

X) is a complete metric space and f : X −→ X is a contraction,

then there exists unique x0 ∈ X such that f(x0) = x0 (x0 is called fixedpoint of f).

1.1.6 Completion of Metric Spaces: Equivalenceof Metrics


X) be a metric space. The metric space (Y, d

Y) is said to be

a completion of (X, dX) if (Y, d

Y) is complete and X is isometric to

a dense subset of Y .

Theorem 1.51Every metric space (X, d

X) has a completion and any two completions

of X are isometric, i.e., if (Y1, dY1) and (Y2, dY2

) are completions of(X, d

X), then there exists an isometry ϕ : Y1 −→ Y2.

As we already mentioned in Remark 1.42, homeomorphic but non-isometric spaces need not have the same Cauchy sequences. To guar-antee this, we need for the metrics to satisfy the following.

Definition 1.52Let d1

Xand d2

Xbe two metrics on X. We say that d1

Xand d2

Xare

(topologically) equivalent if for every sequence {xn}n�1 ⊆ X andx ∈ X, we have

d1X(xn, x) −→ 0 ⇐⇒ d2

X(xn, x) −→ 0.

Then the metric spaces (X, d1X) and (X, d2

X) are said to be (topologi-

cally) equivalent metric spaces.

Remark 1.53By Proposition 1.30, two metrics d1

Xand d2

Xon X are (topologically)

equivalent if and only if the identity function

i : (X, d1X) −→ (X, d2

X)

is a homeomorphism. This is equivalent to saying that the metrictopologies corresponding to d1

Xand d2

X(see Definition 1.8(c)) are iden-

tical. This justifies the term “topologically”. Continuing in this path,


we say that metrics d1X

and d2X

on X are uniformly equivalent ifand only if the identity functions

i : (X, d1X) −→ (X, d2

X) and i′ : (X, d2

X) −→ (X, d1

X)

are both uniformly continuous and we say that metrics d1X

and d2X

onX are Lipschitz equivalent if and only if both the above identityfunctions i and i′ are Lipschitz continuous.

Proposition 1.54Two metrics d1

Xand d2

Xon X are Lipschitz equivalent if and only if

there exist ϑ > β > 0 such that

βd1X(x1, x2) � d2

X(x1, x2) � ϑd1

X(x1, x2) ∀ x1, x2 ∈ X.

Evidently the isometry satisfies all three equivalences (topological,uniform and Lipschitz). So, it is the strongest form of equivalence be-tween metric spaces. Definition 1.52 and Remark 1.53 prompt us tointroduce some weaker forms of equivalence of different metric spaces.In fact the first one (homeomorphic) has already been introduced inDefinition 1.39(c), but for the sake of completeness we recall its defi-nition here.


X) and (Y, d

Y) be two metric spaces.

(a) X and Y are said to be homeomorphic (or topologicallyhomeomorphic) if there exists a homeomorphism (i.e., a bicontin-uous bijection) f : X −→ Y .(b) X and Y are said to be uniformly equivalent if there exists abijective function f : X −→ Y such that both f and f−1 are uniformlycontinuous.(c) X and Y are said to be Lipschitz equivalent if there exists abijective function f : X −→ Y such that both f and f−1 are Lipschitzcontinuous.

Definition 1.56A metric space (X, dX ) is said to be topologically complete if thereis a metric d

Xwhich is (topologically) equivalent to d

Xsuch that the

space (X, dX) is complete.


Definition 1.57Let X be a metric space.(a) A set E ⊆ X is said to be a Gδ-set if E =

⋂

n�1Un with Un ⊆ X

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

⋃

n�1Cn with Cn ⊆ X

closed for all n � 1.

The next theorem characterizes topologically complete metricspaces.

Theorem 1.58 (Alexandrov Theorem)A metric space (X, d

X) is topologically complete if and only if X is a

Gδ-set in its completion.

1.1.7 Pointwise and Uniform Convergence of Maps

Next we introduce two modes of convergence for sequences of functions.

Definition 1.59Let X be a set and (Y, d

Y) a metric space. Consider functions

fn, : X −→ Y for n � 1 and f : X −→ Y .(a) We say that the sequence {fn}n�1 converges pointwise to f if

dY

(fn(x), f(x)

) −→ 0 as n → +∞ ∀ x ∈ X.

We denote it by

fn −→ f.

(b) We say that the sequence {fn}n�1 converges uniformly to f if

supx∈X

dY

(fn(x), f(x)

) −→ 0 as n → +∞.

We denote it by

fn ⇒ f.

Remark 1.60It is clear from the above definition that uniform convergence impliespointwise convergence. The converse is not true.


Example 1.61Continuity is not preserved by pointwise convergence. Let X = Y =[0, 1] with the metric induced from R and consider the sequence offunctions fn : [0, 1] −→ [0, 1], defined by

fn(x) = xn ∀ x ∈ [0, 1], n � 1.

Thenfn −→ f,

where f is the discontinuous (at x = 1) function

f(x) =

{0 if x ∈ [0, 1),1 if x = 1.

The sequence {fn}n�1 does not converge uniformly.


X) and (Y, d

Y) are two metric spaces and {fn : X −→ Y }n�1

is a sequence of continuous functions such that

fn ⇒ f,

then f : X −→ Y is continuous (i.e., the uniform limit of continuousfunctions is a continuous function).

1.1.8 Compact Metric Spaces

In Definition 1.21(d), we introduced the notion of open cover. Usingit we can introduce the important notion of compactness.

Definition 1.63We say that a metric space (X, dX ) is compact if every open cover Fof X has a finite subcover, i.e., there is a finite subfamily {Uk}Nk=1 ⊆ Fsuch that X =

⋃Nk=1 Uk. A set E ⊆ X is said to be compact if it is

compact for the subspace metric topology (see Definition 1.12).

Remark 1.64Evidently E ⊆ X is compact if and only if every family of opensets in X whose union contains E has a finite subfamily whose unioncontains E.

Directly from De Morgan law, we obtain the following result.


Proposition 1.65A metric space X is compact if and only if every family of closed setswith empty intersection has a finite subfamily with empty intersection.

Definition 1.66Let X be a metric space. A family F of subsets of X is said to havethe finite intersection property if every finite subfamily of F has anonempty intersection.

Proposition 1.67A metric space X is compact if and only if every family of closedsubsets of X with the finite intersection property has nonempty inter-section.

A direct consequence of Proposition 1.24 is the following result.

Proposition 1.68Every compact metric space is separable.

Proposition 1.69Let X be a metric space.(a) Every compact subset of X is closed and bounded.(b) Every closed subset of a compact set is compact.



(a) We say that a set E ⊆ X is sequentially compact if everysequence {xn}n�1 ⊆ E has a subsequence {xnk

}k�1 which converges toa point of E.(b) We say that a set E ⊆ X is totally bounded if for any ε > 0there is a finite number of open balls Bε(xk), k = 1, . . . , N , with centresxk ∈ X such that

E ⊆N⋃

k=1

Bε(xk).

The set {xk}Nk=1 of the centres of the balls is called an ε-net of E.

Theorem 1.71Let X be a metric space. The following statements are equivalent:(a) X is compact;


(b) X is sequentially compact;(c) X is complete and totally bounded.

Definition 1.72Suppose that X and Y are two metric spaces and f : X −→ Y is afunction. We say that f is proper if the inverse image of any compactset in Y is compact in X.

Remark 1.73Even if f is continuous, f need not be proper. For example a constantR-valued function on a noncompact metric space is continuous but notproper.

The next few results relate compactness with continuous functions.

Proposition 1.74If X and Y are two metric spaces, f : X −→ Y is a continuousfunction and E ⊆ X is a compact set,then f(E) ⊆ Y is compact (i.e., continuous functions send compactsets to compact sets).

Theorem 1.75 (Weierstrass Theorem)If X is a compact metric space and f : X −→ R is a continuousfunction,then f attains its supremum and infimum, i.e., there exist x, x ∈ Xsuch that

f(x) = inf f(X) and f(x) = sup f(X).

Proposition 1.76If X and Y are two metric spaces, X is compact and f : X −→ Y isa continuous bijection,then f is a homeomorphism.

Proposition 1.77If X and Y are two metric spaces, X is compact and f : X −→ Y iscontinuous,then f is uniformly continuous.


Many important spaces in analysis (such as RN ) are not compact,but behave locally as compact spaces.

Definition 1.78A metric space (X, d

X) is said to be locally compact if each point

x ∈ X has a closed ball Br(x) which is compact. Equivalently, we cansay that a metric space (X, d

X) is locally compact, if every point x ∈ X

has an open neighbourhood U ⊆ X (see Definition 1.8(d)) such that Uis compact (cf. Proposition 1.69).

Theorem 1.79If X is a locally compact metric space,then the following statements are equivalent:(a) X =

⋃

n�1Kn with each Kn ⊆ X compact (i.e., X is σ-compact);

(b) X is separable;(c) there exists an increasing sequence {Un}n�1 of open sets in X

such that Un is compact and Un ⊆ Un+1 for all n � 1 and we haveX =

⋃n�1 Un.

Proposition 1.80Any open or closed subset of a locally compact metric space is itselflocally compact. Also a locally compact metric space is open in itscompletion.

Definition 1.81We say that a metric space X is a Baire metric space if everyintersection of a nonempty countable collection of open dense subsetsof X is dense in X.

For a given metric space X, let

C(X) ={f : X −→ R : f continuous

}.

We furnish C(X) with the supremum metric d∞, defined by

d∞(f, g) = maxx∈X

∣∣f(x)− g(x)∣∣

(see Example 1.3(d), whereX = [a, b] ⊆ R). Then by Proposition 1.62,(C(X), d∞

)is a complete metric space.


Remark 1.82If X = [a, b] and instead of d∞ we use metric

d1(f, g) =

∫ b

a

∣∣f(t)− g(t)

∣∣ dt

(see Example 1.3(d)), then(C(X), d1

)is not a complete metric space.


X) be a metric space and let Y be a family of functions

f : X −→ R.(a) We say that Y is equicontinuous at x ∈ X if for any givenε > 0, we can find δ = δ(ε, x) > 0 such that

dX(x, u) < δ =⇒ ∣∣f(x)− f(u)

∣∣ < ε ∀ f ∈ Y.

(b) We say that Y is pointwise bounded if for every x ∈ X we canfind Mx > 0 such that

∣∣f(x)

∣∣ � Mx ∀ f ∈ Y.

(c) We say that Y is uniformly equicontinuous on X, if it isequicontinuous at every x ∈ X and the δ > 0 in the definition (a)above can be chosen independently of x ∈ X.(d) We say that Y is uniformly bounded if there exists M > 0 suchthat

∣∣f(x)∣∣ � M ∀ f ∈ Y, x ∈ X.

Theorem 1.84 (Arzela–Ascoli Theorem)If X is a compact metric space and K ⊆ C(X) is equicontinuous ateach x ∈ X and pointwise bounded, then(a) K is uniformly equicontinuous and uniformly bounded on X;

(b) K is relatively compact in(C(X), d∞

)(i.e., K

d∞is compact in

C(X));

In fact (a) and (b) are equivalent.


1.1.9 Connectedness

After compactness, we introduce connectedness which is the other fun-damental topological notion in the theory of metric spaces.

Definition 1.85Let X be a metric space. A separation of X is a pair of nonempty,disjoint open sets U, V ⊆ X such that X = U ∪ V . We say thatX is disconnected if there exists a separation of X and connectedotherwise.

Remark 1.86Connectedness is a property of a space, not a property of subsets asare, for example, openness and closedness. Of course, we can talkabout connected subsets of X, by which we mean connected in thesubspace metric topology. In this context we can always consider aseparation of E ⊆ X to be a pair of open sets U, V ⊆ X such thatU ∩ E �= ∅, V ∩ E �= ∅ and E ⊆ U ∪ V . Note that if the pair {U, V }is a separation for the disconnected metric space X, then U and Vare also closed (such sets which are both open and closed are calledclopen). Finally, if A ⊆ Y ⊆ X, then A is connected in X if and onlyif it is connected in Y .

Proposition 1.87A metric space X is connected if and only if the only subsets of Xwhich are both open and closed (i.e., are clopen) are ∅ and X itself.

Proposition 1.88A nonempty subset of R is connected if and only if it is an interval.

Proposition 1.89If X is a metric space and E ⊆ X is a connected set,

then so is every set A such that E ⊆ A ⊆ E.In particular the closure of a connected set is connected.

The well known Bolzano Theorem (or Intermediate ValueTheorem) from introductory calculus is a particular case of the fol-lowing more general theorem.

Theorem 1.90The continuous image of a connected metric space is connected.


Simple examples show that the intersection of two connected metricspaces need not be connected. Under certain conditions, a union ofconnected subsets is connected.

Theorem 1.91If X is a metric space, {Ei}i∈I is a family of connected subsets of Xand at least one of the following conditions is satisfied:(a) Ei ∩ Ej �= ∅ for all i, j ∈ I;(b) there exists i0 such that Ei ∩Ei0 �= ∅ for all i ∈ I,then the set E =

⋃

i∈IEi is connected.

Definition 1.92Let X be a metric space and x ∈ X. The union C(x) of all con-nected subsets of X containing x is called the connected componentof x. Evidently C(x) is a maximal connected subset of X (see Theo-rem 1.91).

Proposition 1.93If X is a metric space, then(a) the distinct connected components of X form a partition of X;(b) every connected component C ⊆ X is closed;(c) every nonempty connected subset of X which is both open andclosed is a connected component of X;(d) if y ∈ C(x), then C(x) = C(y).

Definition 1.94A metric space X in which the connected components are all singletonsets is said to be totally disconnected.

As with compactness (see Definition 1.78), we can have a localversion of the notion of connectedness.

Definition 1.95A metric space X is said to be locally connected at x if it has alocal basis at x (see Definition 1.21(a)) consisting of connected sets.We say that X is locally connected, if it is locally connected at everypoint x ∈ X, i.e., X has a basis for the metric topology (see Defini-tion 1.21(b)) consisting of open connected sets.


Proposition 1.96A metric space X is locally connected if and only if for every open setU ⊆ X, the connected components of U are open too.

This is another notion of connectedness of metric spaces.

Definition 1.97A metric space X is said to be path-connected if for every x1, x2 ∈ Xthere is a continuous function (path) f : [0, 1] −→ X such that f(0) =x1 and f(1) = x2.

Proposition 1.98If X is a metric space and x0 ∈ X,then X is path-connected if and only if every x ∈ X can be joined tox0 by a path.

Proposition 1.99A path-connected metric space is connected, but the converse fails ingeneral.

Example 1.100Consider E ⊆ R2, defined by E = G ∪ T0, where

G ={(x1, x2) ∈ R2 : 0 < x1 � 2

π , x2 = sin 1x

}, T0 = {0} × [−1, 1].

Then E is connected but not path-connected.

1

−1

1

p2p

As for connected metric spaces (see Theorems 1.90 and 1.91), wehave corresponding properties for path-connected spaces.


Proposition 1.101(a) The continuous image of a path-connected metric space is path-connected.(b) If {Ei}i∈I is a family of path-connected subsets of X such that⋂

i∈IEi �= ∅, then the set

⋃

i∈IEi is path-connected too.

By Proposition 1.101(b) one can make the following definition.

Definition 1.102Let X be a metric space and let x ∈ X. The path-connected com-ponent of x in X is the maximal subset of X containing x which ispath-connected.

Definition 1.103A metric space X is said to be locally path-connected at x if it hasa local basis at x consisting of path-connected sets. We say that X islocally path-connected, if it is locally path-connected at every pointx ∈ X, i.e., X has a basis for the metric topology consisting of openpath-connected sets.

Proposition 1.104If X is a metric space,then the following two statements are equivalent:(a) X is locally path-connected;(b) the path-connected components of X are open (hence closed too).

Proposition 1.105If X is a locally path-connected metric space,then the connected components and the path-connected components ofX coincide.Moreover, X is connected if and only if it is path-connected.

Remark 1.106Another way to view the connected and the path-connected compo-nents of a metric space X is the following. We define an equiva-lence relation on X, called the connectivity relation (respectively,


path-connectivity relation), by saying that x1 ∼ x2 if there existsa connected subset of X containing both x1 and x2 (respectively, ifthere is a path in X from x1 to x2). Then X/∼ (the set of equivalenceclasses) are the connected (respectively, path-connected) componentsof X.

Theorem 1.107If X is a connected, locally compact metric space,then X is σ-compact (see Theorem 1.79).


X) be a metric space. For x, u ∈ X and ε > 0 an ε-chain

connecting x and u is a finite set {c0, . . . , cn} ⊆ X such that c0 = u,cn = u and d

X(ck, ck+1) � ε for all k ∈ {0, . . . , n−1}. We write x

ε∼ u,

if there exists an ε-chain connecting x and u (evidently for every ε > 0this is an equivalence relation). We say that X is well-chained if forevery ε > 0 and all x, u ∈ X, we have x

ε∼ u.

1.1.10 Partitions of Unity

Definition 1.109Let X be a cover of a metric space X (i.e., X is a family of sets suchthat

⋃

A∈XA ⊇ X). We say that X is locally finite if for every x ∈ X,

there is an open set U � x (a neighbourhood of x) which intersectsonly a finite number of sets in X .

Definition 1.110Let X be a metric space and let f : X −→ R be a function. Thesupport of f is defined as

supp f ={x ∈ X : f(x) �= 0

}.

Definition 1.111A family {ϕi}i∈I of continuous functions from a metric space X iscalled a partition of unity if(a) the family

{ {x ∈ X : ϕi(x) �= 0}}i∈I is a locally finite open cover

of X;(b) for all x ∈ X, we have

∑

i∈Iϕi(x) = 1 (from (a) we see that for

each x ∈ X, the sum has only finitely many nonzero terms).


If X is an open cover of X and for each i ∈ I, there is some U ∈ Xsuch that suppϕi ⊆ U , then we say that the partition of unity {ϕi}i∈Iis subordinate to X .

Since an open cover may not be locally finite, we need the notionof a refinement.

Definition 1.112Let X be a cover of the metric space X. A cover X0 is called a re-finement of X if for every U0 ∈ X0, we can find a set U ∈ X suchthat U0 ⊆ U .

Theorem 1.113If X is a locally compact, σ-compact metric space and X is an opencover of X,then X has a countable locally finite open refinement X0.Moreover, if U0 ∈ X0, then U0 is compact.

Proposition 1.114If X is a σ-compact metric space and X is a locally finite open coverof X,then X is countable.

The next result provides a crucial tool to construct a partition ofunity for locally compact, σ-compact metric spaces.

Theorem 1.115 (Shrinking Lemma)If X is a locally compact, σ-compact metric spaces and X is a locallyfinite open cover of X,then for each U ∈ X , we can find an open set VU such that V U ⊆ U

and X0def=

{V

U: U ∈ X}

is a locally finite open cover of X.

Theorem 1.116If X is a locally compact, σ-compact metric spaces and X is an opencover of X,then there is a partition of unity subordinate to X .


1.1.11 Products of Metric Spaces

Definition 1.117Let (Xk, dXk

), for k = 1, . . . , N , be metric spaces and for x =

(x1, . . . , xN ), y = (y1, . . . , yN ) ∈ X =N∏

k=1

Xk, we define

d∞X(x, y) = max

{dXk

(xk, yk) : k = 1, . . . , N}

(see also Example 1.3(a)).

Proposition 1.118(X, d∞

X) is a metric space and it is called the product metric space of

X1, . . . ,Xk.

Remark 1.119

Other equivalent metrics on X =N∏

k=1

Xk, are

dpX(x, y) =

( N∑

k=1

(dXk

(xk, yk))p) 1

p

with 1 � p < +∞. Note that

d∞X(x, y) � dp

X(x, y) � d1

X(x, y) � nd∞

X(x, y) ∀ x, y ∈ X =

N∏

k=1

Xk.

Thus all the above metrics are Lipschitz equivalent (see Remark 1.53and Proposition 1.54).

Proposition 1.120

For a given x ∈ X =N∏

k=1

Xk and r > 0, we have

Br(x) =

N∏

k=1

BXkr (xk),

where BXkr (xk) =

{u ∈ Xk : d

Xk(u, xk) < r

}.


Corollary 1.121The family

{ N∏

k=1

BXkr (xk) : xk ∈ Xk, 1 � k � N, r > 0

}

is a basis for the product metric topology on X =N∏

k=1

Xk.

Proposition 1.122If Uk ⊆ Xk, for k = 1, . . . , N , are open subsets of Xk,

then U =N∏

k=1

Uk is an open subset of X =N∏

k=1

Xk.

Definition 1.123

For the given product set X =N∏

k=1

Xk, we can always define the pro-

jection on the nth factor

pn : X =

N∏

k=1

Xk −→ Xn

(n = 1, . . . , N) by

pn(x) = xn ∀ x = (x1, . . . , xN ) ∈ X.

Proposition 1.124

If (Xk, dXk), for k = 1, . . . , N , are metric spaces and X =

N∏

k=1

Xk is

furnished with the d∞X-metric,

then the projection map pn : : X =N∏

k=1

Xk −→ Xn is continuous and

open.

Remark 1.125The projection function pn is in general not closed.

Proposition 1.126If (Xk, dXk

), for k = 1, . . . , N and (Y, dY ) are all metric spaces,then f : Y −→ (X, d∞

X) is continuous (respectively, uniformly contin-

uous) if and only if for every k = 1, . . . , N , the composite functionpk ◦ f : Y −→ Xk is continuous (respectively, uniformly continuous).


Proposition 1.127If (Xk, dXk

), for k = 1, . . . , N are metric spaces and Ek ⊆ Xk, fork = 1, . . . , N ,then

N∏

k=1

Ek

d∞X

=N∏

k=1

EdXk

k .

We can also consider infinite metric products. So, suppose that wehave a sequence

{(Xn, dXn

)}n�1

of metric space such that

sup{dXn

(x, y) : x, y ∈ X, n � 1}

� M

for some M > 0. Then we define

dX(x, y) =

∑

n�1

1

2ndXn

(xn, yn) ∀ x, y ∈ X =∏

n�1

Xn

(see Example 1.3(b)).

Proposition 1.128(X, dX ) is a metric space.

We consider the family

B ={Bm

r (x) =

m∏

n=1

BXnr (xn)×

∏

n�m+1

Xn : x ∈ X =∏

n�1

Xn,m ∈ N, r > 0}.

Proposition 1.129B is a basis for the d

X-metric topology on X =

∏

n�1Xn.

Proposition 1.130(X =

∏

n�1Xn, dX

)is a complete (respectively, compact) metric space

if and only if each (Xn, dXn), for n � 1, is a complete (respectively,

compact) metric space.

We conclude with a theorem, which is useful in many theoreticaland applied settings.


Theorem 1.131(a) If X is any set, (Y, d) is a complete metric space and

Bb(X ;Y ) ={f : X −→ Y : f is bounded (i.e., diam f(X) < +∞)

}

furnished with the supremum distance

d∞(u, v) = supx∈X

d(u(x), v(x)

),

then(Bb(X;Y ), d∞


(b) If (X, dX) and (Y, d

Y) are two metric spaces with Y being complete

andCb(X;Y ) =

{f ∈ C(X;Y ) : f is bounded

},

then(Cb(X;Y ), d∞

)is complete.

1.1.12 Auxiliary Notions

Let us introduce some notions that will be needed in the forthcomingproblems.

Definition 1.132

(a) For any function f : X −→ Y , the graph of f is the set Gr f ⊆X × Y , defined by

Gr f ={(x, y) ∈ X × Y : y = f(x)

}.

(b) For any function f : X −→ R, the epigraph of f is the set epi f ⊆X ×R, defined by

epi f ={(x, λ) ∈ X × R : f(x) � λ

}.


X) is a metric space and {En}n�1 is a sequence

of subsets (possibly empty) of X. We define the lower Kuratowskilimit of the sets {En}n�1, by

lim infn→+∞ En =

{x ∈ X : for every r > 0 there is an integer nr � 1,

such that Br(x) ∩ En �= ∅ for all n � nr

}


and the upper Kuratowski limit of sets {En}n�1, by

lim supn→+∞

En ={x ∈ X : for every r > 0, we have Br(x) ∩ En �= ∅,

for an infinite number of the sets En

}.


X) is a bounded metric space and Pf

(X)denotes the

collection of nonempty closed subsets of X. For every A,B ∈ Pf

(X),

we set

h(A,B)def= sup

{∣∣dist(x,A)− dist(x,B)∣∣ : x ∈ X

}.

The metric space(Pf

(X), h

)is called the Hausdorff metric space

on Pf

(X)(see Problem 1.176).

We write Anh−→ A for the convergence in the Hausdorff metric

space(Pf

(X), h

)(i.e., An

h−→ A, if h(An, A) −→ 0).

Definition 1.135Suppose that X is a metric space and f : X −→ R is a function.(a) We say that f has a local minimum at x ∈ X if there exists anopen set Ux containing x such that f(x) � f(u) for all u ∈ Ux.(b) We say that f has a strict local minimum at x ∈ X if thereexists an open set Ux containing x such that f(x) < f(u) for all u ∈Ux \ {x}.(c) We say that f has a local maximum at x ∈ X if there exists anopen set Ux containing x such that f(x) � f(u) for all u ∈ Ux.(d) We say that f has a strict local maximum at x ∈ X if thereexists an open set Ux containing x such that f(x) > f(u) for all u ∈Ux \ {x}.

1.2. Problems 33

1.2 Problems

Problem 1.1 �

Suppose that (X, dX) is a metric space. Show that a Cauchy sequence

in X converges if and only if it has a convergent subsequence.

Problem 1.2 �

Suppose that (X, dX) is a metric space and {xn}n�1 is a sequence in

X. Assume that subsequences {x2n}n�1, {x2n+1}n�1 and {x3n}n�1

converge. Show that {xn}n�1 is a convergent sequence.

Problem 1.3 �

Suppose that (X, dX) is a metric space, x ∈ X and {xn}n�1 is a

sequence in X such that for any subsequence {xnk}k�1 ⊆ {xn}n�1,

we can find a further subsequence{xnkl

}l�1

⊆ {xnk}k�1 such that

liml→+∞

xnkl= x, then lim

n→+∞xn = x (we call this property Urysohn

criterion for convergence).

Problem 1.4 �

Suppose that (X, dX ) is a metric space and {xn}n�1 ⊆ X is a Cauchysequence. Show that we can find a subsequence {xnk

}k�1 of {xn}n�1

such that

dX(xnk

, xnm) � 12k

∀ k,m � 1, k � m.

Problem 1.5 �

Suppose that (X, dX) is a metric space and {xn}n�1 is a Cauchy se-

quence in X.(a) Show that we can find a subsequence

{xnk

}k�1

of {xn}n�1 suchthat ∑

k�1

dX(xnk

, xnk+1) < +∞.

(b) Show that, if every sequence {yn}n�1 in X, which satisfies

∑

n�1

dX(yn, yn+1) < +∞,

is convergent, then X is complete.


Problem 1.6 �

Let l1def=

{u = {un}n�1 : un ∈ R, ∑

n�1 |un| < +∞}be equipped

with the metric

d1(u, x) =∑

n�1

|un − xn| ∀ u, x ∈ l1.

Also let

c0def=

{u ∈ l1 : there exists n0 � 1 such that un = 0 for all n � n0

}.

Show that:(a) (l1, d1) is a complete metric space;(b) c0 is dense in l1.

Problem 1.7 ��

Show that there is a function f : R −→ R such that

{x ∈ R : f is discontinuous at x

}= Q.

Problem 1.8 �

Let (X, dX) be a metric space. Show that X is a singleton if and only

if every bounded sequence is convergent.

Problem 1.9 �

Show that every complete subspace C of a metric space X is closed.

Problem 1.10 ��

Let M ⊆ C[a, b] be the set of all functions f ∈ C([a, b]

)which are

monotone on [c, d] ⊆ [a, b] with [c, d] nonempty and not a singleton.Show that intM = ∅. As usual on C

([a, b]

)we consider the supremum

metric d∞ (see Example 1.3(d)).

Problem 1.11 �

Suppose that X is a metric space and E ⊆ X. Show that x is anaccumulation point of E if and only if x ∈ E \ {x}.

Problem 1.12 �

Suppose that X is a metric space and D is a dense subset of X. Forevery nonempty open set U ⊆ X, show that U ⊆ U ∩D. In particularU ∩D is dense in U with the subspace metric topology.

1.2. Problems 35

Problem 1.13 ��

Let D ⊆ R be a nonempty set and let

Edef=

{x ∈ D : we can find ε > 0 such that (x, x+ ε) ∩D = ∅}.

Show that E is at most countable.

Problem 1.14 ��

Suppose that (X, dX) is a metric space and Y is a countable family of

separable metric subspaces of X. Show that X0 =⋃

Y ∈YY is separable.

Problem 1.15 �

(a) Give an example of a metric space (X, dX) and of two balls

BR(x) � Br(y), where R > r and x, y ∈ X.(b) Show that, if BR(x) � Br(y) for some R, r > 0 and x, y ∈ X,then R < 2r.

Problem 1.16 ��

Let X be a separable metric space. Show that X = U ∪ C, where Uis open and countable, C is a closed set containing only accumulationpoints (i.e., C is perfect) and U∩C = ∅. This is the so called Cantor–Bendixson theorem .

Problem 1.17 ��

Let {qn}n�1 be an enumeration of the rational numbers in R. For eachx ∈ R, we set

Lxdef=

{n ∈ N : qn � x

}.

We introduce the function f : R −→ R by

f(x) =∑

n∈Lx

1

2n.

Clearly f is strictly increasing, it has a jump discontinuity at everyrational number and

f(x) −→ 0 as x → −∞ and f(x) −→ 1 as x → +∞.

Show that f is right continuous and f |R\Q is continuous.


Problem 1.18 ��

Let l∞ be the space of all bounded sequences in R furnished with thesupremum metric d∞, defined by

d∞(x, y)= sup{|xn−yn| : n � 1

} ∀ x= {xn}n�1 , y = {yn}n�1 ∈ l∞.

Show that l∞ is not separable.

Problem 1.19 ��

Show that if X is a separable metric space, then cardX � c (where c =card 2N; assuming the continuum hypothesis c is the cardinal numberof R).

Problem 1.20 ��

Suppose that (X, dX ) is an unbounded metric space. Show that Xadmits a sequence with no convergent subsequence.

Problem 1.21 ��

Suppose that (X, dX) is a metric space and ϕ : [0,+∞) −→ [0,+∞)

is a nontrivial, increasing concave function such that ϕ(0) = 0. Showthat the function, defined by

dX (x, y) = ϕ(dX (x, y)

) ∀ x, y ∈ X

is a metric on X.

Problem 1.22 ��

(a) Suppose that X and Y are two metric spaces and E,C ⊆ X aretwo nonempty open (or closed) sets such that X = E ∪ C. Assumethat f1 : E −→ Y and f2 : C −→ Y are both continuous and f1

∣∣E∩C

=

f2∣∣E∩C

. Let g : X −→ Y be defined by

g(x)def=

{f1(x) if x ∈ E,f2(x) if x ∈ C.

Show that g is continuous.(b) Is the conclusion true, if we drop the hypothesis that E,C areboth open (or close) subsets of X?

Problem 1.23 ��

Suppose that X is a separable metric space and

Yodef=

{U ⊆ X : U is open

}, Yc

def=

{C ⊆ X : C is closed

}.

1.2. Problems 37

Show that

cardYo = cardYc � c

(c = card 2N; we assume the continuum hypothesis).

Problem 1.24 �

(a) Suppose that (X, dX ) is a metric space and {xn}n�1 ⊆ X is asequence. Suppose that un −→ u in X where for all n � 1, un isan accumulation point of the sequence {xn}n�1. Show that u is anaccumulation point of the sequence {xn}n�1.(b) Show that the set of accumulation points of any sequence in ametric space is closed.

Problem 1.25 ��

Suppose that E is a set and B(E)def=

{f : E −→ R : f is bounded

}.

We furnish B(E) with the supremum metric

d∞(f, g)def= sup

s∈E

∣∣f(s)− g(s)

∣∣.

Show that(B(E), d∞


Problem 1.26 ��

Suppose that X is a separable metric space and f : X −→ R is afunction. Let L be the set of all strict local minimizers of f . Showthat L is at most countable.

Problem 1.27 ��

(a) Let (X, dX) and (Y, d

Y) be two metric spaces, let f : X −→ Y be

a continuous function and let {Cn}n�1 be a sequence of subsets of Xsuch that

⋂

n�1Cn �= ∅ and diamCn −→ 0. Show that diam f(Cn) −→ 0

as n → +∞.(b) Is it possible to drop the hypothesis that

⋂

n�1Cn �= ∅? Justify your

answer.

Problem 1.28 ��

Suppose that X is a metric space and D ⊆ X is a nonempty and closedor open set. Show that ∂D (the boundary of D) is nowhere dense. Isthis true if D is an arbitrary nonempty subset of X?


Problem 1.29 �

Show that in a metric space X every nowhere dense set has a densecomplement. Is the converse true? Justify your answer.

Problem 1.30 �

Suppose that (X, dX) is a metric space and let C be a nonempty, closed

subset of X. Show that C in nowhere dense if and only if for everynonempty open set U , we can find a ball in U \ C.

Problem 1.31 ��

Suppose that (X, dX) is a complete metric space and D is a nonempty

perfect subset of X. Show that cardD � c (c being the cardinality ofthe continuum).

Problem 1.32 �

Use the Baire category theorem, to show that I = [0, 1] is uncountable.

Problem 1.33 �

Show that a countable complete metric space has an isolated point.

Problem 1.34 ��

Suppose that X is a complete metric space and {Cn}n�1 is a sequenceof nonempty closed subsets of X such that X =

⋃

n�1Cn. Show that

the set⋃

n�1intCn is dense in X.

Problem 1.35 ��

Suppose that (l∞, d∞) is as in Problem 1.18 and C ⊆ l∞ is the subsetof all convergent real sequences. Show that C is nowhere dense in l∞.

Problem 1.36 ��

Suppose that X and Y are metric spaces and f : X −→ Y is a function.Show that the following two statements are equivalent:(a) f is uniformly continuous;(b) for all sequences {un}n�1 , {xn}n�1 ⊆ X such that d

X(un, xn)

−→ 0, we have dY

(f(un), f(xn)

) −→ 0.

1.2. Problems 39

Problem 1.37 ��

(a) Let X = Cb(R) (the space of bounded continuous functionsf : R −→ R) be furnished with the supremum metric

d∞(f, g) = supt∈R

∣∣f(t)− g(t)∣∣ ∀ f, g ∈ X.

For f ∈ X and r ∈ R, we set fr(t)def= f(t + r). Then fr ∈ X. Show

that, if f ∈ X is uniformly continuous, then d∞(fr, f) −→ 0 as r → 0+.(b) Does the above remain true if we replace the uniform continuityof f by continuity?

Problem 1.38 �

Suppose that X is a separable metric space, Y is a metric space andf : X −→ Y is a continuous function. Show that f(X) is separable.

Problem 1.39 ��

Suppose that X and Y are two metric spaces, X is complete andf : X −→ Y is homeomorphism which is uniformly continuous. MustY be complete? Justify your answer.

Problem 1.40 ��

Give examples that illustrate each of the following statements:(a) A continuous and open function need not be closed (see Defini-tions 1.29 and 1.39).(b) A continuous and closed function need not be open.(c) An open and closed function need not be continuous.(d) A continuous function need not be open nor closed.

Problem 1.41 �

Suppose that X and Y are two metric spaces and f : X −→ Y is aninjection. Show that f is open if and only if f is closed.

Problem 1.42 ��

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

function such that f (k) is a contraction for some k � 1 (recall thatf (k) = f ◦ . . . ◦ f︸︷︷︸

k-times

). Show that f has a unique fixed point.


Problem 1.43 ��

Let (X, dX) and (Y, d

Y) be two metric spaces and let f : X −→ Y be

a function. We say that f is asymptotically nonexpansive if thereexists a sequence {kn}n�1 ⊆ [1,+∞), with kn −→ 1 as n → +∞ suchthat

dY

(f (n)(u), f (n)(x)

)� kndX (u, x) ∀ u, x ∈ X, n � 1.

Give an example of an asymptotically nonexpansive function which isnot nonexpansive.

Problem 1.44 �

Suppose that M is a metric space, X is a complete metric space,f : M ×X −→ X is a function such that(i) for every x ∈ X, the function r �−→ f(r, x) is continuous on M ;and(ii) for every r ∈ M , the function x �−→ f(r, x) is a contraction withcontraction constant k < 1 independent of r ∈ M .

Show that there exists a unique continuous function u : M −→ Xsuch that u(r) = f

(r, u(r)

)for all r ∈ M .

Problem 1.45 ��

Let us consider X = (0,+∞) with the usual metric. Let f ∈ C1(X;X)be such that x

∣∣f ′(x)

∣∣ � kf(x) for all x ∈ X and some k ∈ [0, 1). Show

that f has a unique fixed point.

Problem 1.46 ��

Show that f is continuous at x ∈ X if and only if ωf (x) = 0 (whereωf is the oscillation function for f).

Problem 1.47 ��

Suppose that (X, dX ) and (Y, dY ) are two metric spaces andf : X −→ Y is a function. Consider the oscillation function ωf . Showthat for every λ ∈ R, the set

{x ∈ X : ωf (x) < λ

}is open (i.e., the

function x �−→ ωf (x) is upper semicontinuous; see Definition 2.46).

Problem 1.48 ��

(a) Let X and Y be two metric spaces with Y complete. Supposef : X −→ Y is bijective and uniformly continuous and that f−1 iscontinuous. Show that X is complete.(b) Is the result still true if f is only continuous (instead of uniformlycontinuous)?

1.2. Problems 41

Problem 1.49 ��

Suppose that X is a complete metric space, {Cn}n�1 is a decreasingsequence of nonempty closed sets in X such that diamCn −→ 0 (seeDefinition 1.6(b)) and f : X −→ X is a continuous function. Showthat f

( ⋂

n�1Cn

)=

⋂

n�1f(Cn).

Problem 1.50 ��

Show that every separable metric space is homeomorphic to a subsetof the Hilbert cube H = [0, 1]N.

Problem 1.51 ��

Suppose that A and B are two dense subsets of R and f : A −→ B isan increasing bijection. Show that f is a homeomorphism.

Problem 1.52 ��


Y) be two metric spaces with Y being complete.

LetD ⊆ X be a dense subset and let f : D −→ Y be a k-Lipschitz func-tion. Show that there exists a unique k-Lipschitz function f : X −→ Ysuch that f |D = f .

Problem 1.53 ��

Show that every metric space is isometrically embedded into the spaceof bounded, uniformly continuous functions.

Problem 1.54 ��

Let X be a normed space and let r > 0. Show that X and Br ={x ∈

X : ‖x‖ < r}are homeomorphic.

Problem 1.55 ��

Suppose thatX is a metric space, Y is a complete metric space, D ⊆ Xis a nonempty set and f : D −→ Y is a continuous function. Show thatf can be extended continuously to a Gδ-subset of X containing D.

Problem 1.56 ��

Suppose that X and Y are complete metric spaces, A ⊆ X, B ⊆ Y arenonempty sets and f : A −→ B is a homeomorphism and surjection.Show that f can be extended to a homeomorphism between Gδ-setscontaining A and B respectively.


Problem 1.57 ��

Let X be a metric space and suppose that A ⊆ X is homeomorphic toa complete metric space. Show that A is a Gδ-set in X.

Problem 1.58 ��

Show that Q is not a Gδ-set in R or equivalently the set of irrationalnumbers R \Q is not an Fσ-set.

Problem 1.59 ��

Show that in a complete metric space which has no isolated points, acountable dense subset cannot be Gδ.

Problem 1.60 ��

Let X be a complete metric space. Show the following:(a) A ⊆ X has a meager complement if and only if A contains a denseGδ-set.(b) A ⊆ X is meager if and only if A is contained in an Fσ-set whosecomplement is dense.

Problem 1.61 ��

(a) Suppose that (X, dX) is a metric space such that every continuous

function f : X −→ R is in fact uniformly continuous. Show that X iscomplete.(b) Show that the inverse theorem is not true, i.e., there exists acomplete metric space X and a continuous function f : X −→ R whichis not uniformly continuous.

Problem 1.62 ��

Suppose that (X, dX) and (Y, d

Y) are two metric spaces and f : X −→

Y is a function. Show that the set

Cdef=

{x ∈ X : f is continuous at x

}

is a Gδ-subset of X.

Problem 1.63 ��

Show that there is no function f : R −→ R for which

{x ∈ R : f is continuous at x

}= Q.

1.2. Problems 43

Problem 1.64 ��

Let X be a metric space. A function f : X −→ R is said to be lowersemicontinuous if for every λ ∈ R, the set

{x ∈ X : f(x) � λ

}is

closed (cf. Definition 2.46). Suppose that X is complete and F is afamily of lower semicontinuous functions f : X −→ R. Assume thatfor every x ∈ X, there exists Mx > 0 such that

f(x) � Mx ∀ f ∈ F .

Show that there exist a nonempty open set U ⊆ X and m0 � 1 suchthat

f(x) � m0 ∀ f ∈ F , x ∈ U.

Problem 1.65 ��

Let χQbe the characteristic function of the set of rational numbers,

i.e.,

χQ(x)

def=

{1 if x ∈ Q,0 if x ∈ R \Q.

Show that χQis not the pointwise limit of a sequence of continuous

functions on R.

Problem 1.66 ��

Let (X, dX) be a complete metric space and consider a sequence{

fn : X −→ R}n�1

of continuous functions such that fn −→ f (point-

wise). Show that the set

Edef=

{x ∈ X : f is continuous at x ∈ X

}

is a dense Gδ-set in X.

Problem 1.67 ��

Let f : R2 −→ R be a function which has partial derivatives at everypoint in R2. Show that f is differentiable on a dense Gδ subset of R2.

Problem 1.68 ��

Find two metrics which are topologically but not uniformly equivalent.

Problem 1.69 ��

Find two metrics which are topologically equivalent but only one de-fines a complete metric space.


Problem 1.70 ��

Find two metrics which are uniformly equivalent but not Lipschitzequivalent.

Problem 1.71 ��

Let X = C1([0, 1]

)and consider the following two metrics on X:

d∞1(u, v) = max

t∈[0,1]∣∣u(t)− v(t)

∣∣+ max

t∈[0,1]∣∣u′(t)− v′(t)

∣∣,

d∞(u, v) = maxt∈[0,1]

∣∣u(t)− v(t)

∣∣.

Show that the two metrics d∞1

and d∞ are not topologically equivalent.

Problem 1.72 ��


Y) be two metric spaces and let f : X −→ Y be a

continuous function. Show that there exists a topologically equivalentmetric d

Xon X such that the function f : (X, d

X) −→ Y is Lipschitz

continuous.

Problem 1.73 ��


Y) be two metric spaces and suppose that

{fn : X −→ Y }n�1 is a sequence of continuous functions. Show thatthere exists a topologically equivalent metric d

Xon X and a topologi-

cally equivalent metric dYon Y such that for each n � 1, the function

fn : (X, dX) −→ (Y, d

Y) is Lipschitz continuous.

Problem 1.74 �

Show that the completion of a separable metric space is separable too.

Problem 1.75 �

Suppose that D ⊆ [a, b] and

Y def=

{f ∈ C[a, b] : f(t) = 0 for all t ∈ D

}.

Show that Y is a closed subset of C[a, b] with the uniform metric d∞C[a,b]

(see Example 1.3(d)).

Problem 1.76 ��

Let X = C([0, 1]; [0, 1]

)be furnished with the supremum metric

d∞(f, g) = maxs∈[0,1]

∣∣f(s)− g(s)

∣∣

1.2. Problems 45

(see Example 1.3(d)). Suppose that

E1def=

{f ∈ X : f is injective

},

E2def=

{f ∈ X : f is surjective

},

Edef= E1 ∩E2 =

{f ∈ X : f is bijective

}.

Check whether E1, E2 and E are closed in X.

Problem 1.77 ��

Let X = C([0, 1]

)be furnished with the supremum metric

d∞(f, g) = maxt∈[0,1]

∣∣f(t)− g(t)∣∣ ∀ f, g ∈ X.

Let

Edef=

{f ∈ X : f is differentiable at at least one point in [0, 1]

}.

Show that E is of first category in X (see Definition 1.25).

Problem 1.78 �

Let (X, dX ) be a metric space and let A ⊆ X be any set. Show thatthe function f : X � x �−→ dist(x,A) ∈ R is Lipschitz continuous withLipschitz constant 1 (so also continuous).

Problem 1.79 �

Show that in a metric space (X, dX) every closed set is Gδ and every

open set is Fσ.

Problem 1.80 ��

Let D ⊆ R be an uncountable set. Show that D has uncountablymany accumulation points (see Definition 1.13(b)).

Problem 1.81 ��

Let X be a metric space and let f : X −→ R be a lower semicontinuousfunction. Show that for every open set U ⊆ R, the set f−1(U) is Fσ.

Problem 1.82 ��

Let X be a complete metric space and let f : X −→ R be a lowersemicontinuous function. Show that f is continuous on a dense Gδ-subset of X.


Problem 1.83 ��

Suppose that X is a nonempty set and (Y, dY) is a metric space.

(a) Show that if {fn : X −→ Y }n�1 is a sequence of bounded functionsand fn ⇒ f , then f is bounded.(b) Show that a sequence of unbounded functions {fn : X −→ Y }n�1

cannot converge uniformly to a bounded function f : X −→ Y .

Problem 1.84 �

Suppose that X and Y are two metric spaces, {fn : X −→ Y }n�1 is asequence of functions continuous at x0 and f : X −→ Y is a functionsuch that fn ⇒ f . Show that f is continuous at x0.

Remark. Applying this problem to a sequence of continuous func-tions, we get the completeness of the space C(X;Y ) (see Proposi-tion 1.62).

Problem 1.85 ��

Let X be a metric space and{fn : X −→ R

}n�1

,{gn : X −→ R

}n�1

two uniformly convergent sequences of functions. Is it true that thesequence {fngn}n�1 is uniformly convergent too?

Problem 1.86 �


Y) are two metric spaces and{

fn : X −→ Y}n�1

is a sequence of Lipschitz continuous functions,

all with the Lipschitz constant k > 0 (k-Lipschitz functions). Supposethat fn ⇒ f . Show that f : X −→ Y is k-Lipschitz too.

Problem 1.87 �

Suppose that (X, dX ) and (Y, dY ) are two metric spaces and{fn : X −→ Y

}n�1

is a sequence of uniformly continuous functionssuch that fn ⇒ f . Show that f : X −→ Y is uniformly continuoustoo.

Problem 1.88 ��

Let X and Y be two metric spaces.(a) Is it possible for a sequence {fn : X −→ Y }n�1 of discontinuousfunctions to converge uniformly to a continuous function f?(b) Let f : X −→ X be a function such that f ◦f is continuous. Doesit imply that f is continuous?

1.2. Problems 47

Problem 1.89 ��

Suppose that X is a metric space, Y is a complete metric space, A ⊆ Xis a set and ϕ : A −→ Y is a continuous function. Show that thereexists a set A0 ⊆ X and a function ϕ0 : A0 −→ Y such that(a) A ⊆ A0

(b) A0 is a Gδ-subset of X and(c) ϕ0 is a continuous extension of ϕ.

Problem 1.90 �

Let X, Y and V be three metric spaces with Y being compact. Letf : X × Y −→ V be a continuous function and let v0 ∈ V . Assumethat for every fixed x ∈ X, the equation f(x, y) = v0 (in y ∈ Y ) hasa unique solution s(x) ∈ Y . Show that the function s : X −→ Y iscontinuous.

Problem 1.91 ��

Suppose that (X, dX) is a compact metric space and f : X −→ X is a

continuous function such that

dX

(f(x), f(u)

)� dX (x, u) ∀ x, u ∈ X.

Show that f is an isometry.

Problem 1.92 �

Show that any uniformly continuous function f : (a, b) −→ R definedon a bounded interval (a, b) is bounded.

Problem 1.93 ��

Suppose that (X, dX) is a compact metric space and f : X −→ X is

surjective and such that

dX

(f(x), f(u)

)� d

X(x, u) ∀ x, u ∈ X

(nonexpansive). Show that f is an isometry (see Definition 1.41).

Problem 1.94 ��

(a) Suppose that (X, dX) is a compact metric space and {Kn}n�1 is

a decreasing sequence of closed subsets of X. We set Kdef=

⋂

n�1Kn.

Show that diamKn −→ diamK.


(b) Show that any of two assumptions of compactness of X and theclosedness of sets Kn cannot be dropped.

Problem 1.95 ��

Suppose that X and Y are two compact metric spaces and con-sider the space C(X;Y ) furnished with the supremum metric d∞. Is(C(X;Y ), d∞

)a compact metric space?

Problem 1.96 ��

Suppose that X is a compact metric space and {fn}n�1 ⊆ C(X) is asequence of functions. Show that(a) if fn ⇒ f and xn −→ x in X, then fn(xn) −→ f(x);(b) if fn(xn) −→ f(x) for all sequences {xn}n�1 ⊆ X such thatxn −→ x, then f ∈ C(X) and fn ⇒ f .

Problem 1.97 ��

(a) Suppose that U is a nonempty open set in RN and {Cn}n�1 is a

decreasing sequence of closed, bounded sets with⋂

n�1Cn ⊆ U . Show

that Cn0 ⊆ U for some n0 � 1.(b) Does the result remain true if we drop the assumption of theboundedness of the sets Cn for n � 1? Justify your answer?

Problem 1.98 ��

Let f : R −→ R be a continuous function such that limz→±∞ f(x) = 0.

Show that f is uniformly continuous.

Problem 1.99 ��

Let f : R −→ R be a continuous function and suppose that thereexists a uniformly continuous function h : R −→ R such thatlim|t|→+∞

(f(t) − h(t)

)= 0. Show that f is uniformly continuous

too.

Problem 1.100 ��

(a) Let X be a metric space and let {xn}n�1 ⊆ X be a sequence suchthat xn −→ x ∈ X as n → +∞. Let C = {x} ∪ {xn : n � 1}. Showthat the set C is compact.(b) Suppose that X and Y are two metric spaces and f : X −→ Y isa function. Show that f is continuous if and only if for every compactset C ⊆ X, the function f |

Cis continuous.

1.2. Problems 49


Show that, if f : X −→ Y is a continuous and proper function, then fis closed.


Suppose that X and Y are two metric spaces and f : X −→ Y isa continuous function. Show that the following two statements areequivalent:(a) Every sequence {xn}n�1 ⊆ X for which f(xn) −→ y in Y has asubsequence {xnk

}k�1 such that xnk−→ x in X.

(b) The function f is closed and for all y ∈ Y , the set f−1(y) iscompact in X.


Suppose that X and Y are two metric spaces and f : X −→ Y is a con-tinuous function which satisfies any of the two equivalent statementsin Problem 1.102. Show that f is proper (see Definition 1.72).


Suppose that X is a compact metric space and{fn : X −→ R

}n�1

isa sequence of continuous functions such that

fn ↘ f or fn ↗ f as n → +∞,

for some continuous function f : X −→ R (by fn ↘ f we denote thepointwise convergence fn −→ f (see Definition 1.59) of a decreasingsequence of functions, i.e., for all n � 1 and all x ∈ X, we havefn+1(x) � fn(x); analogously by fn ↗ f we denote the pointwiseconvergence fn −→ f of an increasing sequence of functions, i.e., forall n � 1 and all x ∈ X, we have fn+1(x) � fn(x)). Show that

fn ⇒ f

(see Definition 1.59). This result is known as Dini theorem .


In the Dini theorem (see Problem 1.104) show that none of the fourconditions can be dropped (namely the compactness of X, the conti-nuity of the limit function f , the continuity of functions fn (at leastfor large n) and the monotonicity of the sequence {fn}n�1).



(a) Suppose that X is a compact metric space and f : X −→ X is anisometry. Show that f(X) = X.(b) Does the result remain true if instead of the compactness of thespace X we assume only that it is bounded? Justify your answer.


Let f : RN −→ RN be an isometry and assume that there exists x ∈

RN such that f(x) = x. Show that f is a homeomorphism.


Suppose that (X, dX) is a totally bounded metric space. Show that

any sequence {xn}n�1 ⊆ X has a Cauchy subsequence.


Suppose that (X, dX), (Y, d

Y), (V, d

V) are three metric spaces, Y is

compact and f : X × Y −→ V is a continuous function. Show that

f(x, ·) ⇒ f(x0, ·) when x −→ x0 in X.


Suppose that X and Y are two metric spaces with Y being compact,f : X × Y −→ R is a continuous function and let us set

m(x)def= inf

{f(x, y) : y ∈ Y

} ∀ x ∈ X.

Show that m : X −→ R is continuous.


Suppose that X and Y are two compact metric spaces, V is a metricspace and f : X ×Y −→ V is a continuous function. For every x ∈ X,let fx be the continuous function Y � y �−→ f(x, y) ∈ V . We furnishC(Y ;V ) with the supremum metric

d∞(g, h) = maxy∈Y

dV

(g(y), h(y)

) ∀ g, h ∈ C(Y ;V ).

Show that the function x �−→ fx is continuous from X into C(Y ;V )and {fx}x∈X is an equicontinuous family in C(Y ;V ).

1.2. Problems 51


(a) Show that the Banach fixed point theorem (see Theorem 1.49) is

no more valid if instead of assuming that f is a contraction, we assumethat

dX

(f(x), f(u)

)< d

X(x, u) ∀ x, u ∈ X, x �= u.

(b) Suppose that (X, dX) is a compact metric space and f : X −→ X

satisfies the inequality

dX

(f(x), f(u)

)< d

X(x, u) ∀ x, u ∈ X, x �= u.

Show that f has a unique fixed point (i.e., there exists x0 ∈ X suchthat f(x0) = x0).


Let X = C([0, b]

)with the supremum metric d∞, defined by

d∞(u, v) = max0�t�b

∣∣u(t)− v(t)

∣∣ ∀ u, v ∈ X.

Let k : [0, b]× [0, b] −→ R be a continuous function, λ ∈ R and f ∈ X.Find a condition on λ so that the integral equation

u(t) = λ

b∫

0

k(t, s)u(s) ds + f(t), t ∈ [0, b],

has a solution u ∈ X.


Show that a locally compact metric space X is a Baire metric space.


Let f : RN −→ R be a continuous function such thatlim

‖x‖→+∞f(x) = +∞. Show that f is bounded below and there exists

x0 ∈ RN such that f(x0) = infRN

f .


Let X be a metric space such that every continuous functionf : X −→ R attains its supremum on X. Show that X is compact.(Compare this problem with the Weierstrass theorem; see Theo-rem 1.75).



(a) Suppose that D ⊆ RN is an Fσ-set and f : RN −→ R is a contin-uous function. Show that f(D) is an Fσ-set too.(b) Is the image of a Gδ-set by a continuous function necessarily aGδ-set?


(a) Suppose that (X, dX) is a metric space and K,C are two

nonempty disjoint subsets of X such that K is compact and C isclosed. Show that dist(K,C) > 0.(b) Is the above conclusion true if K is only closed but not compact?


Suppose that (X, dX) is a metric space and K,C are two nonempty

disjoint subsets of X such that K is compact and C is closed. Showthat there exists an open set U ⊆ X such that K ⊆ U and C ∩U = ∅.Problem 1.120 ��

(a) Suppose that X is a compact metric space, Y is a metric spaceand f : X −→ Y is a continuous function. Show that f(D) = f(D) forall D ⊆ X.(b) Show that the above is not true if X is only complete (but notcompact).


Suppose that (X, dX) is a metric space such that for every continuous

function f : X −→ R and any D ⊆ X, we have f(D) = f(D).(a) Show that X is complete.(b) Is X necessarily compact?


Suppose that (X, dX) is a metric space.

(a) Show that X is compact if and only if for every continuous func-tion f : X −→ R and any D ⊆ X, we have f(D) = f(D).(b) Is the above result true if we replace R by another metric space?


Suppose that (X, dX ) is a locally compact metric space and K is anonempty compact subset of X. Show that there exists r > 0 suchthat Kr is compact, where

Krdef=

{x ∈ X : dist(x,K) < r

}.

1.2. Problems 53




f : X −→ Y is a function.(a) Show that, if f is continuous, then Gr f is closed in X × Y . Givean example showing that the converse is not true.(b) Show that when Y is compact, then f is continuous if and only ifGr f is closed in X × Y .


Show that the interval [0, 1] cannot have a countable partition intononempty closed sets.


Suppose that (X, dX ) is a compact metric space and {Ck}Nk=1 (N > 1)are nonempty closed subsets of X with an empty intersection. Showthat there is a number δ > 0 such that every set which meets all thesets Ck for k = 1, . . . , N has diameter bigger or equal to δ > 0.


Suppose that X is a compact metric space, f : X −→ R is a functionand assume that for every s ∈ R, the set f−1

([s,+∞)

)is closed. Show

that there exists x0 ∈ X such that f(x0) = supR

f < +∞.

Problem 1.128 �

Show that every metric space in which every closed ball is compact iscomplete and a set is complete if and only if it is closed and bounded.


Let f : RN −→ R be a function such that lim inf‖x‖→+∞

f(x)‖x‖ > 0. Show that

for every λ ∈ R, the set{x ∈ RN : f(x) � λ

}is compact.


Let G : [0, 1] × [0, 1] −→ R be a continuous function. Consider thespace C

([0, 1]

)equipped with the supremum metric d∞, i.e.,

d∞(f, g) = maxt∈[0,1]

∣∣f(t)− g(t)

∣∣ ∀ f, g ∈ C

([0, 1]

)


(see Example 1.3). Let L : C([0, 1]

) −→ C([0, 1]

)be the function,

defined by

L(f)(t)def=

1∫

0

G(t, s)f(s) ds.

Show that, if E ⊆ C([0, 1]

)is bounded, then the set L(E) ⊆ C

([0, 1]

)

is compact.


Is the set{sin(nx) : n � 1

} ⊆ C([−π, π]

)relatively compact?


Suppose that d1X

and d2X

are two topologically equivalent metrics onX and E ⊆ X is a nonempty closed set which is locally compact withrespect to d1

X. Show that E is locally compact with respect to d2

Xtoo.


Let X ={(0, 0)

} ∪ {(x, sin 1

x

): x > 0

}be a metric space with the

natural metric induced by the Euclidean metric in R2. Show that Xis not locally compact.


(a) Suppose that X and Y are two metric spaces with X being locallycompact and f : X −→ Y is a continuous and open function. Showthat f(X) is locally compact.(b) Is the above still true if f is only continuous (but not necessarilyopen)?(c) Is the above still true if f is only open (but not necessarily con-tinuous)?


Show that every locally compact metric space X is homeomorphic toa complete metric space.


Show that Q is not homeomorphic to a complete metric space.


Suppose that (X, dX) is a compact metric space and Y is an open cover

of X. Show that there is a number δ > 0 such that for any A ⊆ Xwith diamA < δ (see Definition 1.6), we can find U ∈ Y such thatA ⊆ U .

1.2. Problems 55

Remark. This number δ > 0 is called the Lebesgue number of thecover Y.Problem 1.138 ��

Find a bounded function on R which realizes neither its maximum norminimum on any compact nontrivial interval.

Problem 1.139 �

Show that a totally bounded metric space is separable.


Let (X, dX) be a metric space. Show that the following statements are

equivalent:(a) (X, d

X) is a compact metric space.

(b) If dX is another metric topologically equivalent to dX , then(X, dX ) is a complete metric space.

Problem 1.141 �

Suppose that X and Y are two metric spaces, f : X −→ Y is a con-tinuous function and K ⊆ X is a compact set. Show that for everyε > 0, we can find δ = δ(ε) > 0 such that

∀x ∈ K, u ∈ X :[dX(x, u) � δ =⇒ d

Y

(f(x), f(u)

)� ε

].


Suppose that (X, dX) is a metric space and B is the family of all

bounded sets of X (sets with finite diameter). TheKuratowski mea-sure of noncompactness α : B −→ R+ is defined by

α(B)def= inf

{d > 0 : B can be covered by a finite number

of sets of diameter � d} ∀ B ∈ B.

Show that(a) If B is compact, then α(B) = 0.(b) If α(B) = 0, then B is totally bounded.(c) α is monotone, i.e., if B ⊆ B′, then α(B) � α(B′).(d) α(B) = α(B) for all B ∈ B.(e) If (X, dX ) is a complete metric space and {Cn}n�1 ⊆ B is a de-creasing sequence of nonempty, closed sets and α(Cn) −→ 0, then⋂

n�1

Cn �= ∅ and it is compact.



(a) Suppose that X and Y are two metric spaces and f : X −→ Y isan injection which maps compact sets in X into compact sets in Y .Show that f is continuous.(b) Show that the assumption on the injectivity of f cannot bedropped.


(a) Suppose that (X, dX) is a compact metric space and C1, . . . , Cm

(with m � 2) are nonempty closed sets in X such thatm⋂

k=1

Ck = ∅.Show that there exist open sets U1, . . . , Um in X such that Ck ⊆ Uk

for all k ∈ {1, . . . ,m} andm⋂

k=1

Uk = ∅.(b) Does the above result remains true if X is any metric space (notnecessarily compact)? Justify your answer.


Suppose that X is a normed vector space (for example X = RN ),

f : X −→ X is a locally Lipschitz function (i.e., for all x ∈ X, wecan find an open set Ux such that x ∈ Ux and kx > 0 such that∥∥f(u)− f(u′)

∥∥ � kx‖u− u′‖ for all u, u′ ∈ Ux; see Definition 1.48(b))and C is a compact subset ofX. Show that f |C is Lipschitz continuous,i.e., there exists kC > 0 such that

∥∥f(u)− f(u′)

∥∥ � kC‖u− u′‖ for all

u, u′ ∈ C.

Problem 1.146 �

Let X and Y be two metric spaces with Y being compact and letC ⊆ X × Y be a closed subset. Show that proj

X(C) ⊆ X is a closed

subset (where projX

denotes the projection map on X).

Problem 1.147 �

Let X be a metric space. We say that X has property S if for everyε > 0, X admits an open cover by connected sets of diameter less thanε > 0. Are the following statements true:(a) If X has a dense subset with property S, then X has property S.(b) If X has property S, then any dense subset of X has property S.


Suppose that (X, dX) is a compact metric space, {xn}n�1 ⊆ X is a

sequence such that dX(xn+1, xn) −→ 0 and A is the set of accumulation

points of the sequence {xn}n�1 ⊆ X. Show that A is connected.

1.2. Problems 57


(a) Let f : [0, 1] −→ [0, 1] be a continuous function. Show that f hasa fixed point, i.e., there exists x ∈ [0, 1] such that f(x) = x.(b) Is the above true if we replace the interval [0, 1] by (0, 1)?(c) Is the above true if we replace the interval [0, 1] by any compactmetric space (X, d

X)?


Let A ⊆ R be a nonempty and connected set. Assume that A ⊆ Q.Show that A is a singleton.


Let X be a metric space. Show that the following two statements areequivalent:(a) X is disconnected.(b) There exists a surjective continuous function h : X −→ {0, 1},where {0, 1} is furnished with the discrete metric.


(a) Show that for N > 1, the set RN \ {0} is connected.(b) Show that R and RN (with N > 1) cannot be homeomorphic.


Let ϕ : RN −→ R (with N � 2) be a continuous and surjective func-tion. Show that for all u ∈ R, the set ϕ−1

({u}) is not bounded.


Are the intervals [0, 1] and (0, 1) homeomorphic? Justify your answer.


Let Edef=

{u ∈ RN : u has at least one irrational component

}. Is E

path-connected? Justify your answer.


(a) Let (X, dX) be a compact metric space. Show that the following

statements are equivalent:(i) X is locally connected;(ii) for every ε > 0, X is the finite union of connected sets withdiameter less or equal to ε.


(b) Is the above equivalence still true if we drop the assumption oncompactness of X?


(a) Suppose that X is a metric space and {Ak}mk=1 are connected sub-sets of X such that Ak ∩ Ak+1 �= ∅ for all k � m − 1. Show that the

setm⋃

k=1

Ak is connected.

(b) Is the above result true for a countable family of sets {Ak}∞k=1?Justify your answer.

Problem 1.158 �

Let (X, dX) be a metric space with connected open balls (for example

X = RN for N � 1), let C ⊆ X and for all ε > 0 let

Cεdef=

{x ∈ X : dist(x,C) < ε

}

(the ε-neighbourhood of C). Show that, if C is connected, then Cε isconnected too.


Suppose that f ∈ C(R). Show that epi f is path-connected.


Let (X, dX) be a metric space.

(a) Show that if X is connected, then X is well-chained.(b) Is the opposite true?


Suppose that (X, dX) is a compact metric space. Show that X is

connected if and only if X is well-chained.


Suppose that X and Y are metric spaces and f : X −→ Y is a home-omorphism. Show that f maps connected components of X to con-nected components of Y , i.e., f

(C(x)

)= C

(f(x)

)for all x ∈ X.

Problem 1.163 �

Consider the set Edef=

{(x, sin 1

x) : x ∈ [0, 1]} ∪ ({0} × [−1, 1]

). Is the

set E locally connected? Justify your answer.

1.2. Problems 59


Suppose that T is an interval in R and f : T −→ R is a monotonefunction (i.e., f is either increasing or decreasing). Show that f iscontinuous if and only if f(T ) is an interval.


Suppose that T = [a, b] and f : T −→ R is a differentiable function.Show that f ′ has the Darboux property (i.e., if η, ϑ ∈ f ′(T ), then f ′

takes all values λ ∈ [η, ϑ]).


Let (X, dX) be a connected metric space with at least two elements.

Show that X is uncountable.


Suppose that X is a metric space, {Cn}n�1 is decreasing sequence ofnonempty compact and connected subsets of X. Show that the setC =

⋂

n�1Cn is nonempty, compact and connected too.



Y) are two metric spaces with Y being

complete and g : X × Y −→ Y is a function such that

dY

(g(x, y), g(x, y′)

)� kd

Y(y, y′) ∀ x ∈ X, y, y′ ∈ Y

for some k ∈ (0, 1). Show that there exists a continuous and boundedfunction f : X −→ Y such that g

(x, f(x)

)= f(x) for all x ∈ X.


Suppose that (X, dX) is a metric space and K,C ⊆ X is two nonempty

compact subsets. Show that there exist a ∈ K and c ∈ C such thatdX(a, c) = dist(K,C).

Moreover, if K = C, show that there exist a′, c′ ∈ K such thatdX (a

′, c′) = diamK.


Suppose that (X, dk), k = 1, . . . ,m are metric spaces and X =

m∏

k=1

Xk.

For a sequence {xn}n�1 in X, let{xkn

}n�1

be its projection on Xk,

k = 1, . . . ,m. Show that {xn}n�1 is a Cauchy sequence in X if and

only if{xkn

}n�1

are Cauchy sequences in X for all k = 1, . . . ,m.



Suppose that (X, dX) is a metric space and {En}n�1 is a sequence of

subsets (possibly empty) of X. Show that

lim infn→+∞ En =

{x ∈ X : lim

n→+∞dist(x,En) = 0}

={x ∈ X : there exist xn ∈ En such that

dX(xn, x) −→ 0

};

lim supn→+∞

En ={x ∈ X : lim inf

n→+∞ dist(x,En) = 0}

={x ∈ X : there exist xnk

∈ Enk, with n1 < n2 < . . .,

such that dX(xnk

, x) −→ 0}

=⋂

m�1

⋃

n�m

En;

lim infn→+∞ En ⊆ lim sup

n→+∞En;

lim infn→+∞ En = lim inf

n→+∞ En = lim infn→+∞ En;

lim supn→+∞

En = lim supn→+∞


En.

In particular both sets lim infn→+∞ En and lim sup

n→+∞En are closed.


Suppose that (X, dX) is a metric space, {Cn}n�1 ⊆ 2X \ {∅} is a

sequence of sets and x ∈ X. Show that

lim supn→+∞

dist(x,Cn) � dist(x, lim inf

n→+∞ Cn

).


Suppose that {Cn}n�1 is a sequence of nonempty closed sets in RN

and Cn ⊆ BR for all n � 1. Assume that lim infn→+∞ Cn = lim sup

n→+∞Cn = C.

Show that for every x ∈ RN , we have dist(x,Cn) −→ dist(x,C).


Suppose that (X, dX ) is a metric space, Pf

(X)is the collection of

nonempty closed subsets of X, ϕ : X −→ R is a continuous function,C ∈ Pf

(X), {Cn}n�1 ⊆ Pf

(X)is a sequence such that C ⊆ lim inf

n→+∞ Cn.

Let mn = infCn ϕ and m = infC ϕ. Show that lim supn→+∞

mn � m.

1.2. Problems 61


Suppose that X is a metric space and {xm,n}m,n�1 is a “double”sequence in X such that xm = lim

n→+∞xm,n and x = limm→+∞xm.

Show that there exist sequences {nm}m�1 and {mn}n�1 such thatx = lim

m→+∞xm,nm = limn→+∞xmn,n.


(a) Suppose that (X, dX) is a bounded metric space and Pf

(X)

denotes the collection of nonempty closed subsets of X. For anyA,B ∈ Pf

(X), we set

h(A,B)def= sup

x∈X

∣∣dist(x,A) − dist(x,B)

∣∣.

Show that h is a metric on Pf

(X). It is known as the Hausdorff

metric; see Definition 1.134.(b) For any A,B ∈ Pf

(X), we set

h(A,B)def= max

{supa∈A

dist(a,B), supb∈B

dist(b,A)}.

Show that

h(A,B) = h(A,B) ∀ A,B ∈ Pf

(X).

(c) For any A,B ∈ Pf

(X), we set

h(A,B)def= inf

{ε > 0 : A ⊆ Bε and B ⊆ Aε

},

where

Sεdef=

{x ∈ X : dist(x, S) � ε

} ∀ S ∈ Pf

(X).

Show that

h(A,B) = h(A,B) ∀ A,B ∈ Pf

(X).

So in fact, we have three equivalent definitions of Hausdorff metric onPf

(X).


Let X be a bounded metric space and let(Pf

(X), h

)be the Hausdorff

metric space (see Definition 1.134 and Problem 1.176).


(a) Show that for any D1,D2, E1, E2 ∈ Pf

(X), we have

h(D1 ∪D2, E1 ∪ E2) � max{h(D1, E1), h(D2, E2)

}.

(b) Show that, for any k-contraction η : X −→ X, we have

h(η(D), η(E)

)� kh(D,E) ∀ D,E ∈ Pf

(X).


(a) Show that, if X is a complete bounded metric space, then so is

the Hausdorff metric space(Pf

(X), h

).

(b) Show that, ifX is a compact metric space, then so is the Hausdorffmetric space

(Pf

(X), h

).


Suppose that (X, dX) is a compact metric space and f, g : X −→ X

are two k-contractions with k ∈ [0, 1). Let Pf

(X)be the collection of

all nonempty closed subsets of X and let ξ : Pf

(X) −→ Pf

(X)be the

function, defined by

ξ(C)def= f(C) ∪ g(C).

Show that there exists a unique C0 ∈ Pf

(X)such that ξ(C0) = C0.


Suppose that X is a bounded metric space and h is the Hausdorffmetric on Pf

(X). Let

Pk

(X) def

={C ⊆ X : C is nonempty and compact

}.

(a) Show that Pk

(X)is closed in

(Pf

(X), h

).

(b) Show that, if X is a complete metric space, then(Pk

(X), h

)is a

complete metric space too.


Suppose that (X, dX ) is a bounded metric space and h is the Hausdorffmetric on Pf

(X). Show that the space X is compact if and only if the

space (Pf

(X), h) is compact.

1.2. Problems 63


Suppose that (X, dX) is a bounded metric space,{Cn}n�1 ⊆ Pf

(X)is

a sequence such that Cnh−→ C ∈ Pf

(X)(h denotes the Hausdorff

metric on Pf

(X)), Cn is connected for every n � 1 and C is compact.

Show that C is connected too.


Let X be a separable bounded metric space. Show that(Pk

(X), h

)is

a separable, complete metric space (h is the Hausdorff metric).


Show that the Hausdorff metric h depends on the metric dX

(i.e., it isnot topological; cf. Remark 1.42).


Find a sequence of closed sets {Cn}n�1 in some metric space such that

lim infn→+∞ Cn = lim sup

n→+∞Cn = C

for some closed set C, but Cn �−→ C in the Hausdorff distance.


Suppose that (X, dX) is a bounded metric space, C ∈ Pf

(X),

{Cn}n�1 ⊆ Pf

(X)and Cn

h−→ C (i.e., converges in the sense of Haus-dorff). Show that


n→+∞Cn = C.


Let (X, dX ) be a complete bounded metric space and let {Cn}n�1

be a Cauchy sequence in(Pk

(X), h

). Let C

def=

⋃

n�1Cn. Show that

C ∈ Pk

(X).


1.3 Solutions

Solution of Problem 1.1“=⇒”: If {xn}n�1 is a Cauchy sequence in X (see Definition 1.7) andxn −→ x, then every subsequence of {xn}n�1 also converges to x.

“⇐=”: Suppose that the subsequence {xnk}k�1 of {xn}n�1 converges

to x ∈ X. Then for a given ε > 0, we can find integer n0 = n0(ε) > 0such that

dX (xnk, x) < ε

2 and dX (xn, xm) < ε2 ∀ nk, n,m � n0.

Then

dX(xn, x) � d

X(xn, xnk

) + dX(xnk

, x) < ε ∀ n � n0,

so xn −→ x (X, dX).

Solution of Problem 1.2For a given ε > 0, we can find an integer n0 = n0(ε) � 1 such that

dX (xn, xm) < ε3 ∀ n,m-even, n � n0,

dX(xn, xm) < ε

3 ∀ n,m-odd, n � n0,

dX(xn, xm) < ε

3 ∀ n,m-multiples of 3, n � n0.

Let us choose k, l � n0 such that they are both multiples of 3, k iseven and l is odd. Then for n,m � n0, we need to consider only thecase when n is even and m is odd. We have

dX(xn, xm) � d

X(xn, xk) + d

X(xk, xl) + d

X(xl, xm) < ε,

which proves that the sequence {xn}n�1 is a Cauchy sequence in X.Because it has a convergent subsequence, it converges itself (see Prob-lem 1.1).

1.3. Solutions 65

Solution of Problem 1.3Let us proceed by contradiction. So suppose that x is not the limit ofthe sequence {xn}n�1. Thus, there exists ε > 0 such that for all N ∈ Nthere exists n > N such that d

X(xn, x) � ε. First, for N = 1, let us

find n1 > 1 such that dX(xn1 , x) � ε. Next, for N = n1, let us find

n2 > n1 such that dX(xn2 , x) � ε. So, proceeding by induction, we

can find a subsequence {xnk}k�1 ⊆ {xn}n�1, with the property, that

dX(xnk

, x) � ε ∀ k � 1.

This subsequence has no further subsequence convergent to x. So, weobtained a contradiction with our hypothesis.

Solution of Problem 1.4Since {xn}n�1 ⊆ X is a Cauchy sequence (see Definition 1.7), for agiven ε > 0 we can find N(ε) � 1 such that

dX (xk, xm) < ε ∀ k,m � N(ε).

Let ε = 12 and choose n1 � N(12 ). Then

dX(xn1 , xm) � 1

2 ∀ m � n1.

Suppose that we have produced a finite sequence {xni}ki=1 which isstrictly increasing and satisfies

dX(xnk

, xm) � 12k

∀ m � nk.

Let ε = 12k+1 and choose integer nk+1 > nk, nk+1 > N

(1

2k+1

). Then

dX(xnk+1

, xm) � 12k+1 ∀ m � nk+1

and so by induction, we have a sequence {xnk}k�1 which is a subse-

quence of {xn}n�1, satisfying

dX(xnk

, xnm) � 12k

∀ k � m.


Solution of Problem 1.5

(a) Since {xn}n�1 is a Cauchy sequence, we can find n1 � 1 such that

dX(xm, xk) � 1

2 ∀ m,k � n1.

We set u1 = xn1 . Then choose n2 > n1 such that

dX(xm, xk) � 1

22∀ m,k � n2.

We set u2 = xn2 . Inductively, we produce{um = xnm

}m�1

, a subse-

quence of {xn}n�1 such that

dX (um, um+1) � 12m ∀ m � 1.

Hence ∑

m�1

dX(um, um+1) �

∑

m�1

12m = 1 < +∞.

(b) Next, let {xn}n�1 be a Cauchy sequence in X. From part (a), wecan find a subsequence {um}m�1 of {xn}n�1 such that

∑

m�1

dX(um, um+1) < +∞.

Then by hypothesis {um}m�1 is convergent, hence so is {xn}n�1, whichproves the completeness of X.


(a) Let{uk

}k�1

be a d1-Cauchy sequence. Then, for a given ε > 0,

we can find k0 = k0(ε) � 1 such that

d1

(uk, um

)� ε ∀ k,m � k0.

Hence ∞∑

n=1

|ukn − umn | � ε ∀ k,m � k0 (1.1)

(where uk ={ukn

}n�1

and um = {umn }n�1). It follows that for every

fixed n � 1,{ukn

}k�1

is a Cauchy sequence and so we have

ukn −→ ϑn as k → +∞.

1.3. Solutions 67

Let u = {ϑn}n�1. If in (1.1), we let m → +∞, then

∞∑

n=1

|ukn − ϑn| � ε ∀ k,� k0, (1.2)

so ∞∑

n=1

|ϑn| �∞∑

n=1

|uk0n |+ ε

and thus u = {ϑn}n�1 ∈ l1. Moreover, from (1.2), it follows that

d1(un, u) −→ 0

and so we conclude that (l1, d1) is a complete metric space.

(b) Evidently c0 �= l1 (for example u ={

1nr

}n�1

with r > 1, belongs

in l1 but not in c0). Let u = {un}n�1 ∈ l1. For every k ∈ N, let

uk ={ukn

}n�1

with ukn = un if n � k and ukn = 0 if n > k. Then

uk ∈ c0 and

d1(uk, u) =

∞∑

n=k+1

|un| −→ 0 as k → +∞.

This proves the d1-density of c0 in l1 (see Definition 1.20).

Solution of Problem 1.7Consider the function

f(x) =

⎧⎨

⎩

1q if x = p

q , p, q ∈ Z \ {0}, relatively prime,

1 if x = 0,0 if x ∈ R \Q.

Then f is continuous at every x ∈ R \ Q and discontinuous at everyx ∈ Q. To see this note that, if x ∈ R \ Q and xn −→ x, withxn = pn

qn∈ Q, pn ∈ Z, qn ∈ N, pn, qn relatively prime, then qn −→ +∞

and sof(xn) =

1qn

−→ 0 = f(x).

This proves the continuity of f at every irrational point. Clearly f isdiscontinuous at x = 0. Finally let x ∈ Q\{0}. Then x = p

q with p, q ∈


Z \ {0} relatively prime. Suppose that the sequence {xn}n�1 ⊆ R \Qsatisfies xn −→ x. Then f(xn) = 0 for all n � 1 and so they cannotconverge to f(x) = 1

q . Hence f is discontinuous at every rationalnumber.

Solution of Problem 1.8“=⇒”: If X is a singleton, then every sequence is constant, henceconvergent.“⇐=”: Suppose that X is not a singleton and let x, u ∈ X be suchthat x �= u. Let

xn =

{x if n = 2k,u if n = 2k + 1

∀ k � 1.

then the sequence {xn}n�1 is bounded, but not convergent, a contra-diction.

Solution of Problem 1.9Let x ∈ C. Then we can find a sequence {xn}n�1 ⊆ C such thatxn −→ x. The sequence {xn}n�1 being convergent in X is also aCauchy sequence in C and the latter is complete by hypothesis. Hencex ∈ C and so C is closed.

Solution of Problem 1.10We proceed by contradiction. So, suppose that intM �= ∅. Thus, thereare f ∈ M and ε > 0 such that

B4ε(f) ⊆ M.

We will construct a function h ∈ B4ε(f) \M . Let x0 ∈ (c, d). Becausef is continuous, so we can find δ > 0, δ < 1

2(d− x0) such that

∀x ∈ [c, d] : |x− x0| � 3δ =⇒ ∣∣f(x)− f(x0)∣∣ � ε.

We define the function h ∈ C[a, b] is such a way that

h∣∣[a,x0]

= f∣∣[a,x0]

and h∣∣[x0+2δ,b]

= f∣∣[x0+2δ,b]

,

1.3. Solutions 69

on the interval [x0, x0 + δ], Grh is a segment joining(x0, f(x0)

)with(

x0+δ, f(x0)−2ε)and on the interval [x0+δ, x0+2δ], Grh is a segment

joining(x0 + δ, f(x0)− 2ε

)with

(x0 + 2δ, f(x0 + 2δ)

). Evidently h is

continuous, it is not monotone (as it is strictly decreasing on [x0, x0+δ]and strictly decreasing on [x0 + δ, x0 +2δ]), so h �∈ M and h ∈ B4ε(f),a contradiction.

Solution of Problem 1.11By Theorem 1.14(d), x is an accumulation point of E (see Defini-tion 1.13(b)) if and only if we can find a sequence {xn}n�1 ⊆ E \ {x}such that xn −→ x. Hence x is an accumulation point of E if and onlyif x ∈ E \ {x} (see Proposition 1.11).

Solution of Problem 1.12Suppose that x ∈ U . Since D is dense in X (see Definition 1.20), wehave

B 1n(x) ∩D �= ∅ ∀ n � 1.

Let xn ∈ B 1n(x) ∩D for n � 1. Evidently

xn −→ x in X.

Since U is open we can find n0 � 1 such that

B 1n(x) ⊆ U ∀ n � n0.

Hence

xn ∈ U ∩D ∀ n � n0

and so x ∈ U ∩D. Therefore U ⊆ U ∩D and so U ∩D is dense in Uwith the subspace metric topology (see Definition 1.12).


Solution of Problem 1.13Let x ∈ E and choose ux > x, ux ∈ Q such that (x, ux) ∩D = ∅. Wewill show that, if x, y ∈ E, x �= y, then ux �= uy. Assume withoutany loss of generality that x < y and arguing indirectly suppose thatux = uy. Then y ∈ (x, ux) and so (x, ux) is a neighbourhood of y ∈ D,which means that (x, ux)∩D �= ∅, a contradiction. Therefore we haveestablished an injective function E � x �−→ ux ∈ Q. This implies thatE is at most countable.

Solution of Problem 1.14Let Y ∈ Y and let DY be a countable dense subset of Y (see Defini-tions 1.20 and 1.21). Let

Ddef=

⋃

Y ∈YDY .

Then D is countable (being the countable union of countable sets) andwe claim that D is dense in X0. To this end let x ∈ X0. Then x ∈ Yfor some Y ∈ Y and so dist(x,D

Y) = 0 (see Definition 1.6(d)), hence

dist(x,D) = 0, which proves the claim.


(a) Let X = [0,+∞) with the natural metric induced from R. Notethat

B4(1) = [0, 5) � [0, 6) = B3(3).

(b) As BR(x) � Br(y), so we can find z ∈ Br(y) \BR(x). Then

dX (z, y) < r and dX (z, x) > R.

As x ∈ Br(y), we have dX(x, y) < r. Then

r > dX (x, y) � dX (x, z) − dX (z, y) > R− r,

so R < 2r.

1.3. Solutions 71

Solution of Problem 1.16Let

C ={x ∈ X : for every r > 0, Br(x) is uncountable

}

and let U = X \ C. Suppose that x ∈ U . Then we can find rx > 0such that the ball Brx(x) is at most countable. Note that Brx(x) ⊆ U .Consider the family

{Brx(x)

}x∈U . This is an open cover of U and by

Proposition 1.24, it has a countable subcover{Brxn (xn)

}n�1

. Then

U =⋃

n�1

Brxn (xn) is open and countable

and so C is closed and of course perfect (see Definition 1.13(e)).

Solution of Problem 1.17First we will show that f |

R\Q is continuous. Let us fix an irrationalnumber e ∈ R \Q. Let ε > 0 and choose m = m(ε) ∈ N such that

∑

n�m

1

2n< ε.

We setδ = min

{|qk − e| : k = 1, . . . ,m− 1}.

Let x ∈ R be such that |x − e| < δ. If x < e, then Le \ Lx ⊆ {k}k�m

and so

f(x)− f(e) �∑

n∈Le\Lx

1

2n�

∑

n�m

1

2n< ε.

If x > e, then Lx \ Le ⊆ {k}k�m and so

f(e)− f(x) �∑

n∈Lx\Le

1

2n�

∑

n�m

1

2n< ε.

Thus f is continuous at e. Since e was an arbitrary irrational number,we conclude that f |

R\Q is continuous.Now we will show that f is right continuous at rational numbers.

Let us fix ε > 0 and a rational number u. Let {xk}k�1 ⊆ R be suchthat

xk −→ u+ as k → +∞.


Since f is strictly increasing, we have

f(u) � f(xk) ∀ k � 1.

Then ∣∣f(xk)− f(u)

∣∣ = f(xk)− f(u) =

∑

n∈Sk,u

1

2n,

whereSk,u

def=

{n ∈ N : u < qn � xk

}.

Choose an integer m0 � 1 such that

∑

n�m0

1

2n< ε

and consider the set A ⊆ R, defined by

A =⋂

{n∈N:n<m0, qn>u}(u, qn).

Evidently A = (u, q) for some q ∈ Q. Moreover, every rational numberin A is of the form qn, with n � m0. Because xk −→ u+, we can findan integer k0 � 1 such that

xk ∈ A ∀ k � k0.

Therefore ∣∣f(xk)− f(u)∣∣ < ε ∀ k � k0

and so f is right continuous at u.

Solution of Problem 1.18Let D be any countable subspace of l∞. Let {xn}n�1 be the sequenceof all elements of D. In particular, for all n � 1, we have xn = {xnk}k�1∈ l∞. We define a new real sequence u = {un}n�1 as follows

un =

{1 if xnn < 0,−1 if xnn � 0.

Then u = {un}n�1 ∈ l∞ and dist(u,D) � 1 (see Definition 1.6(d)).So D is not dense in l∞ (see Definition 1.20) and since D was an

1.3. Solutions 73

arbitrary countable subset of l∞, we conclude that l∞ is not separable(see Definition 1.21).

Solution of Problem 1.19Let D = {xn}n�1 be a countable dense subset of X (see Defini-tions 1.20 and 1.21). Consider the following collection of open balls{B 1

k(xn)

}n,k�1

. This collection is countable. Let{Bm

}m�1

be an

enumeration of this countable collection. For every x ∈ X, let

Lx ={m ∈ N : x ∈ Bm

}.

Clearly the function

X � x �−→ Lx ∈ 2N

is an injection. Therefore

cardX � card 2N = c.

Solution of Problem 1.20Evidently X is infinite (as it is unbounded; see Definition 1.6(b)). Letu ∈ X and for every n � 1, define

En ={x ∈ X : d

X(x, u) > n

}.

Since X is unbounded, we have En �= ∅ for all n � 1. Let xn ∈ En forevery n � 1. Let y ∈ X and let m � 1 be an integer such that

dX (u, y) + 1 � m.

Then for all n � m, we have

1 � n+ 1−m � dX(xn, u)− d

X(u, y) � d

X(xn, y),

so the sequence {xn}n�1 does not converge to y. Since y ∈ X wasarbitrary, we infer that the sequence {xn}n�1 has no convergent sub-sequence.


Solution of Problem 1.21First we show that ϕ−1(0) = {0}. Indeed, from the concavity of ϕ, wehave that

ϕ(u+ v)− ϕ(v) � ϕ(u) − ϕ(0) = ϕ(u) ∀ u, v > 0.

So, if ϕ(u) = 0 for some u, then by the fact that ϕ is increasing, wehave

ϕ(u+ v) = ϕ(v) ∀ v > 0.

Let v = u. Then ϕ(2u) = 0 and so inductively

ϕ(ku) = 0 ∀ k � 0,

from which we conclude that ϕ ≡ 0, a contradiction (since ϕ is non-trivial). Hence ϕ(u) > 0 for all u > 0 and so ϕ−1(0) = {0}. Thisimplies that

dX(x, y) = 0 ⇐⇒ x = y.

Also, clearly we have

dX(x, y) = d

X(y, x) ∀ x, y ∈ X.

Finally, for x, y, z ∈ X, by the concavity of ϕ, we have

ϕ(dX(x, y)+d

X(y, z)

)−ϕ(dX(y, z)

)�ϕ

(dX(x, y)

)−ϕ(0) = ϕ(dX(x, y)

)

and

ϕ(dX (x, z)

)� ϕ

(dX (x, y) + dX (y, z)

)

(since ϕ is increasing). Therefore

ϕ(dX (x, z)

)� ϕ

(dX (x, y)

)+ ϕ

(dX (y, z)

)

so

dX(x, z) � d

X(x, y) + d

X(y, z),

i.e., dX

satisfies the triangle inequality. We conclude that dX

is ametric on X.

1.3. Solutions 75


(a) We assume that E and C are both open subsets of X (the proof

is similar if we assume that both are closed). Let U ⊆ Y be an openset. Then

g−1(U) ∩ E = f−11 (U) and g−1(U) ∩ C = f−1

2 (U).

Because f1 and f2 are both continuous, the set f−11 (U) is open in E

and the set f−12 (U) is open in C. Since E and C are both open in X,

it follows that the sets g−1(U) ∩ E and g−1(U) ∩ C are both open inX. But

g−1(U) = g−1(U) ∩X = g−1(U) ∩ (E ∪ C)

=(g−1(U) ∩ E

) ∪ (g−1(U) ∩ C

),

so g−1(U) is also open and thus g is continuous.

(b) If we drop the hypothesis that E and C are both open (or closed),the result fails. Let

X = R, E = Q and C = R \Q.We consider the functions f1 : E −→ R and f2 : C −→ R, defined by

f1(x) = 1 ∀ x ∈ E,

f2(x) = 0 ∀ x ∈ C.

Evidently both f1 and f2 are continuous. Then

g(x) =

{1 if x ∈ Q,0 if x ∈ R \Q,

and this function is discontinuous everywhere.

Solution of Problem 1.23By Proposition 1.24, we know that X is second countable. So, letB = {Un}n�1 be a basis for the metric topology on X. Every openset in X is the countable union of elements in B, hence cardYc � c.Finally note that the function Y0 � U �−→ U c ∈ Yc is a bijection.


Solution of Problem 1.24(a) Let A be the set of accumulation points of the sequence {xn}n�1

(see Definition 1.13(b) and Theorem 1.14(d)). We have

A =⋂

n�1

{xk : k � n

}.

The set A is closed and since un ∈ A for all n � 1, we have u ∈ A.

(b) This comes immediately from (a) and Proposition 1.11.

Solution of Problem 1.25Let {fn}n�1 ⊆ B(E) be a d∞-Cauchy sequence. So, for a given ε > 0,we can find n0 = n0(ε) � 1 such that

d∞(fn, fk) � ε ∀ n, k � n0,

so ∣∣fn(s)− fk(s)

∣∣ � ε ∀ n, k � n0, s ∈ E (1.3)

and thus{fn(s)

}n�1

⊆ R is a Cauchy sequence for all s ∈ [0, 1].

So fn(s) −→ f(s) for all s ∈ [0, 1], for some f(s) ∈ R. Thenf : E −→ R is a function and we will show that f ∈ B(E). If in (1.3)we let k → +∞, we have

∣∣fn(s)− f(s)

∣∣ � ε ∀ n � n0, s ∈ E (1.4)

so ∣∣f(s)∣∣ � sup

s∈E

∣∣fn(s)∣∣+ ε ∀ n � n0, s ∈ E

and thus f ∈ B(E).From (1.4), it is clear that d∞(fn, f) −→ 0 and so we conclude

that(B(E), d∞


Solution of Problem 1.26Let x ∈ L (see Definition 1.135(b)) and let Ux be the open set con-taining x such that

f(x) < f(u) ∀ u ∈ Ux \ {x}.

1.3. Solutions 77

Since X is separable (see Definition 1.21), it is second countable (seeProposition 1.24) and so it has a countable basis B. We can find at leastone Bx ∈ B such that x ∈ Bx ⊆ Ux. Let ξ : L −→ B be the functionwhich assigns to x ∈ L this set Bx ∈ B. This function is injective,since if x, y ∈ L, x �= y and ξ(x) = ξ(y), then x, y ∈ Ux ∩ Uy and sof(x) < f(y), f(y) < f(x), a contradiction. This proves the injectivityof ξ, which in turn by the countability of the basis B, implies that Lis at most countable.

Solution of Problem 1.27(a) Since diamCn −→ 0 and

⋂

n�1Cn �= ∅, we have that

⋂

n�1

Cn = {x}.

Since f is continuous, for a given ε > 0, we can find δ > 0 such that

dX(u, x) � δ =⇒ d

Y

(f(u), f(x)

)� ε.

Also, we can find n0 � 1 such that

dX(u, x) � δ ∀ n � n0, u ∈ Cn.

Hence for n � n0, we have that

f(Cn) ⊆ BYε

(f(x)

)=

{y ∈ Y : d

Y

(y, f(x)

)� ε

},

sodiam f(Cn) � 2ε ∀ n � n0.

Since ε > 0 was arbitrary, we conclude that

diam f(Cn) −→ 0 as n → +∞.

(b) If we drop the hypothesis that⋂

n�1Cn �= ∅, then the conclusion of

the problem fails. To see this let X = (0, 1] and Y = R with the usualmetric. Let Cn =

(0, 1

n

]and

f(x) = sin 1x ∀ x ∈ X.


Then diamCn −→ 0, but for each n � 1, f(Cn) = [−1, 1] and so

diam f(Cn) = 2 ∀ n � 1.

Solution of Problem 1.28By definition (see Definition 1.13(c)), we have

∂D = D ∩Dc.

Therefore without any loss of generality we may assume that D isclosed. We need to show that int ∂D = ∅ (see Definition 1.25). Arguingby contradiction, suppose that int ∂D �= ∅ and let x ∈ int ∂D. Thenwe can find r > 0 such that

Br(x) ⊆ ∂D = D ∩Dc ⊆ D.

Hence

Br(x) ∩Dc = ∅,a contradiction to the fact that x ∈ Dc.

For a general nonempty subset D of X, this is no longer true.Consider X = R and D = Q. Then ∂D = R.

Solution of Problem 1.29Let D ⊆ X be a nowhere dense set (see Definition 1.25). We knowthat intD ⊆ intD = ∅ (see Proposition 1.15). Hence (intD)c = Xand so Dc = X, which means that Dc is dense in X and X \ Dc isnowhere dense (see Definition 1.20).

The converse is not true, as the set of rationals Q has a densecomplement in R but is not nowhere dense.

1.3. Solutions 79

Solution of Problem 1.30“=⇒”: Suppose that C is nowhere dense (see Definition 1.25). ThenCc is dense in X (see Definition 1.20 and Problem 1.29). So, for everynonempty open set U , we have U \ C = U ∩ Cc �= ∅. The set U ∩ Cc

is open. So, it contains a ball.

“⇐=”: Every nonempty open set intersects Cc, hence Cc is densein X, which in turn implies that C is nowhere dense (because the setC is closed).

Solution of Problem 1.31To prove the claim of the problem, we will construct an injectionof {0, 1}N into D. Because D is nonempty and perfect (see Def-inition 1.13(e)), D is infinite. Let u0 �= u1 in D and let us setε1 = min

{12 ,

13dX

(u0, u1)}. We define

D(0) ={u ∈ D : d

X(u, u0) � ε1

},

D(1) ={u ∈ D : dX (u, u1) � ε1

}.

These are disjoint infinite and closed subsets of D such thatdiamD0 � 1 and diamD1 � 1. Now, let n � 1 and for each n-tuple (a1, . . . an) ∈ {0, 1}n, suppose that we have infinitely manyclosed subsets D(a1, . . . , an) of D with diameter less or equal to 1

nand that these sets are pairwise disjoint. We choose u(a1, . . . , an, 0) �=u(a1, . . . , an, 1), both in D(a1, . . . , an) and we set

εn+1 = min{

12(n+1) ,

13dX

(u(a1, . . . , an, 0), u(a1, . . . , an, 1)

)}.

For k = 0, 1, we define

D(a1, . . . , an, k)={u ∈ D(a1, . . . , an) : dX

(u(a1, . . . , an, k), u

)� εn+1

}.

Then the family{D(a1, . . . , an+1) : (a1, . . . , an+1) ∈ {0, 1}n+1

}con-

sists of pairwise disjoint infinite closed subsets ofD, each with diameterless or equal to 1

n+1 . Therefore, for every a = {an}n�1 ∈ {0, 1}N, wehave a decreasing sequence

{D(a1, . . . , an)

}n�1

of infinite closed sub-

sets of D whose diameter tend to 0 as n → +∞. Since (X, dX) is


complete, by the Cantor intersection theorem (see Theorem 1.28), wehave ⋂

n�1

D(a1, . . . , an) ={u(a)

}.

If a �= b, a, b ∈ {0, 1}N, then for some n0 ∈ N, we have an0 �= bn0 andso

u(a) ∈ D(a1, . . . , an0) but u(b) �∈ D(a1, . . . , an0).

Therefore u(a) �= u(b) and so the function a −→ u(a) is a bijectionfrom {0, 1}N into A. So, we conclude that cardD � card {0, 1}N = c.

Solution of Problem 1.32Proceeding by contradiction, suppose that [0, 1] is countable and putits elements in a sequence {xn}n�1. Let

Cn = {xn} ∀ n � 1.

Then each Cn is nowhere dense (see Definition 1.25) and

[0, 1] =⋃

n�1

Cn.

But [0, 1] is a complete metric space, a contradiction (see Theo-rem 1.26).

Solution of Problem 1.33Suppose that no point is isolated (see Definition 1.13). Then eachsingleton is a closed nowhere dense set and their union is the wholespace, which being complete is of second category (see Definition 1.25),a contradiction (see Theorem 1.26).

Solution of Problem 1.34Let U ⊆ X be a nonempty open set. We have

U =⋃

n�1

(Cn ∩ U)

1.3. Solutions 81

and{Cn ∩ U

}n�1

is a sequence of sets closed in U . Since U is openand X is a complete metric space, by Theorem 1.26, we can find atleast one n0 � 1 such that

int (Cn0 ∩ U) = intU (Cn0 ∩ U) �= ∅.

Hence ( ⋃

n�1

intCn

) ∩ U �= ∅

and since U was arbitrary open set in X, we conclude that⋃

n�1intCn

is dense in X (see Definition 1.20).

Solution of Problem 1.35First we show that C is a closed subset of l∞. To this end let{xn}n�1 ⊆ C be a sequence such that

xn −→ x ∈ l∞ in (l∞, d∞).

We havexn = {xnm}m�1 and x = {xm}m�1 .

Let ε > 0. As {xn}n�1 is a Cauchy sequence, there existsk0 = k0(ε) � 1 such that

d∞(xm, xk) < ε3 ∀ m,k � k0.

So, in particular

∣∣xmn − xkn

∣∣ < ε

3 ∀ m,k � k0, n � 1.

Next, as xn −→ x, we can find n0 = n0(ε) � 1 such that

∣∣xnm − xm

∣∣ < ε

3 ∀ n � n0, m � 1.

Then for m � k � k0 and for fixed n � n0, we have

∣∣xm − xk

∣∣ �

∣∣xm − xnm

∣∣+

∣∣xnm − xnk

∣∣+

∣∣xnk − xk

∣∣ < ε.

But ε > 0 was arbitrary. Therefore {xm}m�1 is a Cauchy sequenceand so x ∈ C. Thus C is a closed set.


To show that C is nowhere dense in l∞ (see Definition 1.25), itsuffices to show that l∞ \ C is dense in l∞ (see Definition 1.20). So,let x ∈ l∞ and ε > 0. We need to show that

Bε(x) ∩ (l∞ \ C) �= ∅.

If x ∈ l∞ \ C, then since the latter is open, the above intersection isindeed nonempty. Hence the interesting case is when x ∈ C. Thenx = {xn}n�1 and xn −→ b. We can find n0 = n0(ε) � 1 such that

∣∣xn − b

∣∣ � ε

2 ∀ n � n0.

Choose u = {un}n�1 ∈ l∞ as follows

un =

⎧⎨

⎩

xn if n < n0,b+ ε

2 if n � n0 and n is odd,b− ε

2 if n � n0 and n is even.

Then, we have

d∞(x, u) = supn�1

|xn − un| < ε.

Therefore u ∈ Bε(x). Note that

lim supn→+∞

un = b+ ε2 and lim inf

n→+∞ un = b− ε2 ,

so u = {un}n�1 ∈ l∞ \C and thus Bε(x) ∩ (l∞ \C) �= ∅, which provesthat l∞ \ C is dense in l∞.

Solution of Problem 1.36“(a) =⇒ (b)”: Let {un}n�1 , {xn}n�1 ⊆ X be two sequences suchthat d

X(un, xn) −→ 0. Let ε > 0. Since f is uniformly continuous (see

Definition 1.45), we can find δ = δ(ε) > 0 such that

dX (u, x) � δ =⇒ dY

(f(u), f(x)

)� ε.

We can find n0 � 1 such that

dX(un, xn) � δ ∀ n � n0.

1.3. Solutions 83

Hence

dY

(f(un), f(xn)

)� ε ∀ n � n0

and so we conclude that

dY

(f(un), f(xn)

) −→ 0.

“(b) =⇒ (a)”: Suppose that f is not uniformly continuous. Then

∃ε > 0 ∀δ > 0 ∃u, x ∈ X : dX (u, x) � δ and dY

(f(u), f(x)

)> ε.

So, for any n � 1, we can find un, xn ∈ X such that

dX (un, xn) � 1n and dY

(f(un), f(un)

)> ε ∀ n � 1.

Thus we have obtained two sequences {un}n�1 , {xn}n�1 ⊆ X such

that dX(un, xn) −→ 0, but d

Y

(f(un), f(xn)

) �−→ 0, a contradiction.


(a) Let ε > 0. Since f is uniformly continuous (see Definition 1.45),we can find δ = δ(ε) > 0 such that

|t− s| � δ =⇒ ∣∣f(t)− f(s)

∣∣ � ε.

Let |r| � δ. We have

|t+ r − t| = |r| � δ

and so ∣∣fr(t)− f(t)

∣∣ =

∣∣f(t+ r)− f(t)

∣∣ � ε,

so

d∞(fr, f) � ε ∀ |r| � δ.

This proves that d∞(fr, f) −→ 0 as r → 0+.

(b) No. To see this, let

Sn =

n∑

k=1

1k ∀ n � 1.


We have that Sn −→ +∞ (as these are the partial sums of the har-monic series). Let f : R −→ R be the function, defined by

f(t) =

{0 if t < 1,sin(n+ 1)π(t− Sn) if t ∈ [Sn, Sn+1), n � 1.

The function f is continuous, but not uniformly continuous.

s1 s2 s3 s4 s5 s6 s7

1

We have

d∞(f 12n, f) = sup

t∈R

∣∣f 12n(t)− f(t)

∣∣ � f 12n(Sn)− f(Sn)

= sinnπ(Sn + 12n − Sn)− sinnπ(Sn − Sn)

= sin π2 − sin 0 = 1.

So, in particular d∞(fr, f) �−→ 0 as r → 0+.

Solution of Problem 1.38Let D be a countable dense subset of X (see Definitions 1.20 and 1.21).Then by the continuity of f , we have f(X) = f(D) ⊆ f(D) (seeProposition 1.32(d)). Note that f(D) is countable. Therefore f(X) isseparable.

1.3. Solutions 85

Solution of Problem 1.39No. Let X = R, Y = (−1, 1) and let f : X −→ Y be the function,defined by

f(x) =

{ x1−x if x ∈ (−∞, 0),x

1+x if x ∈ [0,+∞).

Then f is a bijective and

f−1(y) = h(x) =

{y

1+y if y ∈ (−1, 0),y

1−y if y ∈ [0, 1).

Note that f and h are both continuous and so f is a homeomorphism.Also for x, u � 0:

∣∣f(x)− f(u)∣∣ =

∣∣ x1+x − u

1+u

∣∣ = |x−u|(1+x)(1+u) � |x− u|,

for x, u � 0:

∣∣f(x)− f(u)∣∣ =

∣∣ x1−x − u

1−u

∣∣ = |x−u|(1−x)(1−u) � |x− u|,

and for u � 0 � x:

∣∣f(x)−f(u)

∣∣ =

∣∣ x1+x− u

1−u

∣∣ = |x−u−2xu|

(1+x)(1−u) � (x−u)(1+x−u)(1+x)(1−u) � |x−u|.

This proves that f is Lipschitz continuous, hence uniformly continuous(see Definition 1.45). However, note that X = R is complete, butf(X) = Y = (−1, 1) is not.


(a) Consider the function f : R2 −→ R, defined by

f(u1, u2) = u1.

Evidently f is continuous and open (as the projection function on thefirst coordinate), but it is not closed. Consider the closed set

C ={u = (u1, u2) ∈ R2 : u1u2 = 1

}

(a hyperbola). Then f(C) = R \ {0} which is not closed.


(b) Consider the function f : R −→ R, defined by f(x) = c, wherec ∈ R (constant function). Then f is continuous and closed, but it isnot open.

(c) Let f : R −→ {0, 1} (where on {0, 1} we have the discrete metric)be the function, defined by

f(x) =

{1 if x ∈ Q,0 if x ∈ R \Q.

Then f is open and closed, but it is not continuous.

(d) Consider the function f : R −→ R, defined by f(x) = e−x. Thenf is continuous. However, f is not closed. Indeed let C = [0,+∞).Then C is closed in R, but f(C) = (0, 1] is not closed in R.

Also, f is not open. Let U = (−1, 1). Then U is open in R, butf(U) =

(1e , 1

]is not open in R.

Solution of Problem 1.41“=⇒”: Let f : X −→ Y be an open injection (see Definition 1.39). Weclaim that f−1 : f(X) −→ X is continuous. Let U ⊆ X be a nonemptyopen set. Since f is open, the set f(U) is open in Y , hence in f(X)too (see Definition 1.12). But

f(U) = (f−1)−1(U).

So, indeed f−1 is continuous (see Proposition 1.32). Then for everynonempty closed set C ⊆ X, we have that the set f(C) = (f−1)−1(C)is closed (due to the continuity of f−1 : f(X) −→ X; again see Propo-sition 1.32) and this proves that f is closed.

“⇐=”: The proof is similar, reversing the roles of open and closedsets.

Solution of Problem 1.42Suppose that

dX

(f (k)(x), f (k)(u)

)� cd

X(x, u) ∀ x, u ∈ X,

1.3. Solutions 87

with 0 � c < 1. Then by the Banach fixed point theorem (see Theo-rem 1.49), we can find a unique x ∈ X such that

f (k)(x) = x.

We have

dX

(f(x), x

)= dX

(f(f (k)(x)

), f (k)(x)

)

= dX

(f (k)

(f(x)

), f (k)(x)

)� cd

X

(f(x), x

),

so0 � (1− c)d

X

(f(x), x

)� 0

and thusdX

(f(x), x

)= 0,

i.e., f(x) = x.So, x ∈ X is a fixed point of f too.Finally, if f(u) = u, then f (k)(u) = u and from the uniqueness of

x, we have u = x.


l1 ={u = {un}n�1 : un ∈ R,

∑

n�1

|un| < +∞}

be equipped with the metric

d1(u, x) =∑

n�1

|un − xn| ∀ u, x ∈ l1

(see Problem 1.6).Let {an}n�2 ⊆ R be a sequence such that an ∈ (0, 1) for all n � 1

and

limn→+∞πn = 1

2 , where πn =

n∏

k=2

ak for n � 1.

For example, one can take an = (n−1)(n+1)n2 for all n � 2 and then, it

is easy to check that

πn =n+ 1

2n∀ n � 2.


Let X = Bl11 (0) and consider the function f : X −→ X, defined by

f(u1, u2, u3, . . .) = (0, u21, a2u2, a3u3, . . .) ∀ u = (u1, u2, u3, . . .)∈X.

Then

dY

(f(u), f(x)

)= |u21 − x21|+

∑

n�2

an|un − xn|

= |u1 − x1| · |u1 + x1|+∑

n�2

an|un − xn|

� 2∑

n�1

an|un − xn| = 2dX(u, x) ∀ u, x ∈ X,

so f is Lipschitz continuous with Lipschitz constant 2. Note that theconstant 2 cannot be improved (decreased), as taking

un =(1− 1

n , 0, 0, 0 . . .)

and xn =(1− 2

n , 0, 0, 0 . . .) ∀ n � 2,

we get

dX

(f(un), f(xn)

)=

(2− 3

n

) · 1n =

(2− 3

n

)dX(un, xn) ∀ n � 1.

Thus f is not nonexpansive (see Definition 1.48).On the other hand, for all n � 2 and all u, x ∈ X, we have

dX

(f (n)(u), f (n)(x)

)=

n∏

k=2

ak∣∣u21 − x21

∣∣+∑

s�2

s+n−1∏

k=2

ak|us − xs|

� 2∑

s�1

n∏

k=2

ak|us − xs| = 2∑

s�1

πn|us − xs|

= 2πn∑

s�1

|us − xs| = 2πndX (u, x)

and also

limn→+∞ 2πn = 1,

so f is asymptotically nonexpansive function.

1.3. Solutions 89

Solution of Problem 1.44By the Banach fixed point theorem (see Theorem 1.49), for every r ∈M , we can find a unique u(r) ∈ X such that

u(r) = f(r, u(r)

).

It remains to show that the function r �−→ u(r) is continuous. So, letrn −→ r. Then

dX

(u(rn), u(r)

)= d

X

(f(rn, u(rn)), f(r, u(r))

)

� dX

(f(rn, u(rn)), f(rn, u(r))

)

+dX

(f(rn, u(r)), f(r, u(r))

)

� kdX

(u(rn), u(r)

)+ dX

(f(rn, u(r)), f(r, u(r))

)

(see Definition 1.48), so

dX

(u(rn), u(r)

)� 1

1−kdX

(f(rn, u(r)), f(r, u(r))

) −→ 0

(from hypothesis (i)) and hence u(rn) −→ u(r). Thus the functionr �−→ u(r) is continuous.

Solution of Problem 1.45Let 0 < x < u. We have

∣∣ ln f(u)− ln f(x)∣∣ =

∣∣u∫

x

f ′(s)f(s)

ds∣∣ �

u∫

x

|f ′(s)|f(s)

ds

�u∫

x

k

sds = k

∣∣ lnu− lnx

∣∣.

On X we consider the metric

dX (u, x) =∣∣ lnu− lnx

∣∣ ∀ u, x ∈ X.

Then (X, dX) is a complete metric space and from the above estimate,

we see that f is k-contraction for the metric dX. So, by the Banach

fixed point theorem (see Theorem 1.49), f has a unique fixed point.


Solution of Problem 1.46“=⇒”: Suppose that f is continuous at x ∈ X. Then for a given ε > 0,we can find δ = δ(ε, x) ∈ (0, ε) such that

dY

(f(x), f(u)

)< ε

2 ∀ u ∈ Bδ(x).

Hence, we have

dY

(f(v), f(u)

)< ε ∀ u, v ∈ Bδ(x)

(triangle inequality), so

diam f(Bδ(x)

)� ε.

Because ε > 0 was arbitrary, letting ε ↘ 0, we conclude that ωf (x) = 0(see Definition 1.37).

“⇐=”: Suppose that ωf (x) = 0. The for a given ε > 0, we canfind δε > 0 such that

diam f(Bδ(x)

)� ε ∀ δ ∈ (0, δε).

Thus, if dX(x, u) < δ, then

dY

(f(x), f(u)

)� diam f

(Bδ(x)

)� ε

and this proves that f is continuous at x ∈ X.

Solution of Problem 1.47It suffices to show that for every λ ∈ R the complement

Dλ ={x ∈ X : ωf (x) � λ

}

is closed (see Definition 1.37). To this end let {xn}n�1 ⊆ Dλ be asequence such that xn −→ x. Then for a given ε > 0, we can findn0 ∈ N such that

dX(x, xn) � δ

2 ∀ n � n0,

soB δ

2(xn) ⊆ Bδ(x) ∀ n � n0.

Hence

diam f(Bδ(x)

)� diam f

(B δ

2(xn)

)� ωf (xn) � λ ∀ n � n0.

1.3. Solutions 91

Since δ > 0 was arbitrary, we get that ωf (x) � λ, so x ∈ Dλ. ThereforeDλ is closed in X (see Proposition 1.11).


(a) Let {xn}n�1 ⊆ X be a Cauchy sequence. Since by hypothesis f is

uniformly continuous (see Definition 1.45), the sequence{f(xn)

}n�1

is

a Cauchy sequence in Y (see Proposition 1.46(b)). By the completenessof Y , we have f(xn) −→ y. Then the continuity of f−1 implies that

xn −→ f−1(y).

Therefore the Cauchy sequence converges in X and so X is complete.

(b) No. Let X = (0,+∞), Y = R and f(x) = lnx. Then f and f−1

are continuous bijections and Y is complete, but X is not complete.

Solution of Problem 1.49We know that

f( ⋂

n�1

Cn

) ⊆⋂

n�1

f(Cn).

So, we need to show that the opposite inclusion also holds. To thisend, let y ∈ ⋂

n�1f(Cn) and set

En = Cn ∩ f−1(y) ∀ n � 1.

As Cn is closed and f−1(y) is closed (because {y} is closed in X anduse Proposition 1.32), then En ⊆ X is closed for every n � 1 and

diamEn � diamCn −→ 0 as n → +∞.

Since X is complete, we can apply Theorem 1.28 and have that

⋂

n�1

En = {x0},


hencef−1(y) ∩ ( ⋂

n�1

Cn

)= {x0}

and so y ∈ f( ⋂

n�1

Cn

). This proves that

f( ⋂

n�1

Cn

) ⊇⋂

n�1

f(Cn)

and so finally

f( ⋂

n�1

Cn

)=

⋂

n�1

f(Cn).

Solution of Problem 1.50Let {xn}n�1 be a countable dense subset of a metric space (X, d

X) (see

Definitions 1.20 and 1.21). Let

hm(x) = min{1, d

X(x, xm)

} ∀ m � 1.

We consider the function h : X −→ H, defined by

h(x) =(hm(x)

)m�1

∀ x ∈ X.

Since each hm is continuous, it follows that h is continuous. Supposethat h(x) = h(y) and let xnk

−→ x (recall that {xn}n�1 is dense inX). Then

dX(y, xnk

) = dX(x, xnk

) −→ 0 as k → +∞and so x = y, i.e., h is injective. Finally, we need to show that h−1 iscontinuous. So, suppose that h(yn) −→ h(y). For a given ε ∈ (0, 1),choose xm such that

dX(y, xm) < ε.

Because h(yn) −→ h(y), we have

dX (yn, xm) −→ dX (y, xm)

and sodX(yn, xm) < ε ∀ n � n0,

1.3. Solutions 93

for some n0 � 1, hence

dX(yn, y) � 2ε ∀ n � n0,

hence yn −→ y in X and this proves the continuity of h−1. Thereforeh is a homeomorphism into H.

Solution of Problem 1.51Since f is an increasing bijection, it is strictly increasing. Let x0 ∈ A.Since A is dense in R (see Definition 1.20), we have

C = A ∩ (−∞, x0) �= ∅ and f |C

< f(x0).

Let

m(x0) = sup{f(x) : x ∈ A, x < x0

}.

Evidently

m(x0) � f(x0).

If m(x0) < f(x0), then exploiting the density of B in R, we can findu ∈ B such that

m(x0) < u < f(x0).

Let x ∈ A be such that u = f(x). Then x < x0 and so

u = f(x) � m(x0),

a contradiction. So, m(x0) = f(x0).Similarly, we show that

f(x0) = inf{f(x) : x ∈ A, x > x0

}.

Now, let ε > 0 be given. We can find x1, x2 ∈ A such that x1 < x0 < x2and

f(x0)− ε < f(x1) � f(x2) � f(x0) + ε,

so

f(x0)− ε � f(x) � f(x0) + ε ∀ x ∈ [x1, x2] ∩A.

Thus we conclude that f is continuous at x0, hence f is continuous.


Also f−1 is an increasing bijection from B to A and so from thefirst part f−1 is also continuous. So, we conclude that f is homeomor-phism.

Solution of Problem 1.52Let x ∈ X and let {xn}n�1 ⊆ D be a sequence such thatdX(xn, x) −→ 0 (recall that D is dense in X; see Definition 1.20). Note

that for all n,m � 1, we have

dY

(f(xn), f(xm)

)� kd

X(xn, xm)

and dX(xn, xm) −→ as n,m −→ +∞. So

{f(xn)

}n�1

⊆ Y is a Cauchy

sequence. Since Y is complete, the sequence{f(xn)

}n�1

is convergent.Let us set

f(x) = limn→+∞ f(xn).

Note that due to the Lipschitz property of f , this definition is in-dependent of the approximating sequence {xn}n�1 ⊆ D. Indeed, if{x′n}n�1 ⊆ D is another sequence such that d

X(x′n, x) −→ 0, then

dY

(f(xn), f(x

′n))

� kdX (xn, x′n) � kdX (xn, x)+ kdX (x, x

′n) −→ 0.

So, we have a function f : X −→ Y . If x, u ∈ X and{xn}n�1 , {un}n�1 ⊆ D are two sequences such that dX (xn, x) −→ 0and dX (un, u) −→ 0, then

dY

(f(xn), f(un)

)� kd

X(xn, un) ∀ n � 1,

sodY

(f(x), f(u)

)� kd

X(x, u),

i.e., f is k-Lipschitz and f |D = f .

Solution of Problem 1.53Let (X, dX ) be a metric space and let us fix x0 ∈ X. For every x ∈ X,we consider the function ξx : X −→ R, defined by

ξx(u) = dX(x, u)− d

X(u, x0).

1.3. Solutions 95

We have

∣∣ξx(u)−ξx(v)∣∣�

∣∣dX(x, u)−d

X(x, v)

∣∣+∣∣d

X(u, x0)−d

X(v, x0)

∣∣�2dX(u, v),

so ξx is uniformly continuous (in fact 2-Lipschitz; see Definition 1.45).Also, from the triangle inequality, we have

∣∣ξx(u)∣∣ � d

X(x, x0) ∀ u ∈ X

and thus ξx is bounded. Finally, note that

∣∣ξx(u)− ξv(u)∣∣ =

∣∣dX(x, u)− d

X(u, x0)− d

X(v, u) + d

X(u, x0)

∣∣=

∣∣dX(x, u)− d

X(v, u)

∣∣ � dX(x, v) ∀ x, v, u ∈ X

and ∣∣ξx(v)− ξv(v)

∣∣ = d

X(x, v) ∀ x, v ∈ X.

Therefore

d∞(ξx, ξv) = supu∈X

∣∣ξx(u)− ξv(u)

∣∣ = d

X(x, v),

so x �−→ ξx is an isometry.

Solution of Problem 1.54Let f : X −→ X be defined by

f(x) = rx1+‖x‖ .

Then ∥∥f(x)

∥∥ < r ∀ x ∈ X.

So, the range of f is in Br. Let u ∈ Br. Then there exists a uniquex ∈ X such that u = f(x); in fact we have x = 1

r−‖u‖u. So, f is a

bijection from X onto Br and g = f−1 is given by

g(u) = ur−‖u‖ .

Evidently f and g = f−1 are both continuous. Therefore f is a home-omorphism.


Solution of Problem 1.55Let d

Ybe the metric on Y . For every x ∈ D, let ωf (x) be the oscillation

of f at x (see Definition 1.37). We set

C ={x ∈ D : ωf (x) = 0

}.

We haveC =

⋂{x ∈ D : ωf (x) <

1n

},

so C is a Gδ-set in X (see Problem 1.47).We define a continuous function g : C −→ Y , which extends f , as

follows. Let x ∈ C and let {xn}n�1 ⊆ D be a sequence such that

xn −→ x. Since ωf (x) = 0, it follows that{f(xn)

}n�1

⊆ Y is a

Cauchy sequence in (Y, dY). The completeness of (Y, d

Y) implies that

the sequence{f(xn)

}n�1

is convergent. Let

g(x) = limn→+∞ f(xn).

Then it is clear that g is well defined, continuous and g∣∣D= f . There-

fore g is the desired extension of f .

Solution of Problem 1.56Let h = f−1. By Problem 1.55, we can find a Gδ-set A ⊆ X containingA and a continuous extension f : A −→ Y of f . Similarly, we can finda Gδ-set B ⊆ Y containing B and a continuous extension h : B −→ Xof h. We introduce the sets

G(f ) = Gr f ={(x, y) ∈ A× Y : y = f(x)

},

G(h) ={(x, y) ∈ X × B : x = h(y)

}.

Let projX

and projYbe the projections on X and Y respectively. We

define

A∗ = projX

(G(f) ∩G(h)

)=

{x ∈ A : (x, f(x)) ∈ G(f)

},

B∗ = projY

(G(f) ∩G(h)

)=

{y ∈ B : (h(y), y) ∈ G(h)

}.

The set G(h) is closed in X × B and B is a Gδ-set in Y . Hence G(h)is a Gδ-set too. Since f is continuous on the set Gδ-set A, it follows

1.3. Solutions 97

that A∗ is a Gδ-set too. Similarly, we show that B∗ is a Gδ-set. Notethat f∗ = f

∣∣A∗ is a homeomorphism from A∗ onto B∗ and f∗∣∣

A= f .

Solution of Problem 1.57By hypothesis there exists a complete metric space (Y, d

Y) and a home-

omorphism h : A −→ Y . For x ∈ A, we set

hr(x) = diamh(A ∩Br(x)

).

If hr(x) <1n , then diamh(A ∩B r

2(x)) � 1

n and so the sets

Cn ={x ∈ A : hr(x) � 1

n for all r > 0}

are closed in A, hence closed in X. The set

D = A \⋃

n�1

Cn

is a Gδ set. We will show that A = D. If x ∈ A, then due to thecontinuity of h, we have that x �∈ Cn for all n � 1 and so x ∈ D. Thisshows that A ⊆ D.

Next, let x ∈ D. Let {xn}n�1 ⊆ A be a sequence such thatxn −→ x. For every ε > 0, we can find an integer k � 1 such that1k < ε. Since x �∈ Ck, we can find rk > 0 such that

hrk(x) < 1k < ε.

We can find n0 � 1 such that xn ∈ Brk(x) for all n � n0. Thedefinition of hrk implies that

dY

(h(xn), h(xm)

)< ε ∀ n,m � n0,

so{h(xn)

}n�1

is a Cauchy sequence in Y .

The completeness of Y implies that h(xn) −→ y ∈ Y . Recall thath is a homeomorphism. So, we have xn −→ h−1(y). Since xn −→ x,it follows that

x = h−1(y) ∈ h−1(Y ) = A.

This proves that D ⊆ A and so we conclude that D = A and we seethat A is a dense Gδ-set in X.


Solution of Problem 1.58We argue by contradiction. So, suppose that Q is a Gδ-set in R (seeDefinition 1.57), so

Q =⋂

n�1

Un with Un ⊆ R open.

Then R \ Q =⋃

n�1U cn and U c

n is closed for every n � 1. Because U cn

contains only irrationals, it can not contain a nonempty interval andso

intU cn = ∅ ∀ n � 1,

which means that U cn is nowhere dense (see Definition 1.25). Let

{qn}n�1 be an enumeration of the rationals and set

Cn = U cn ∪ {qn} ∀ n � 1.

Then Cn is closed and

intCn = intU cn ∪ int {qn} = ∅,

i.e., Cn is nowhere dense. Note that R =⋃

n�1Cn, which contradicts

the Baire category theorem (see Theorem 1.26).

Solution of Problem 1.59Let D = {xn}n�1 be a countable dense set (see Definition 1.20) ina complete metric space X. Suppose that D is a Gδ-set (see Defi-nition 1.57). Then we can find open sets Un ⊆ X for n � 1 suchthat

D =⋂

n�1

Un.

Evidently, since D ⊆ Un for all n � 1, each Un is open and dense inX. Let us set

Vn = Un \n⋃

k=1

{xk} ∀ n � 1.

1.3. Solutions 99

Since X has no isolated points (see Definition 1.13(d)), each Vn is openand dense. Note that

⋂

n�1Vn = ∅, which contradicts the Baire category

theorem (see Theorem 1.26).

Solution of Problem 1.60(a) “=⇒”: Since by hypothesis the complement of A is meager (seeDefinition 1.25), then

A =( ⋃

n�1

Dn

)c, with Dn being nowhere dence.

Hence, we have

A =⋂

n�1

Dcn ⊇

⋂

n�1

(Dn

)c= C.

But the sets(Dn

)care open and dense (see Definition 1.20 and Prob-

lem 1.29). So C is a Gδ-set (see Definition 1.57) and because X iscomplete, by Theorem 1.26, it is also dense.

“⇐=”: Suppose that A contains a dense Gδ-set C. Hence

C =⋂

n�1

Un, with Un open.

Evidently for every n � 1, Un is dense in X. Hence

int(U cn

)=

(Un

)c= Xc = ∅,

so for all n � 1, U cn is a nowhere dense closed set. We have

Ac ⊆ Cc =⋃

n�1

U cn

and so Ac is meager.

(b) “=⇒”: Suppose that A is meager. Then by part (a) we canfind a dense Gδ-set C such that C ⊆ Ac. Therefore

A ⊆ Cc


and Cc is a Fσ-set (being the complement of a Gδ-set), whose comple-ment C is dense.

“⇐=”: Suppose that A ⊆ E with E being an Fσ-set with dense com-plement. Then Ec ⊆ Ac, where Ec is a dense Gδ-set. Invoking part(a), we conclude that A = (Ac)c is meager.


(a) Let X∗ be the completion of X (see Theorem 1.51). Reasoning

indirectly, suppose that X∗ �= X and let x∗ ∈ X∗ \X. Consider thecontinuous function f : X −→ R, defined by

f(x) = 1dX∗ (x∗,x) ∀ x ∈ X,

with dX∗ being the metric of the completion X∗.

As x∗ ∈ X∗ \X, we see that f is continuous (see Proposition 1.36).But f is not uniformly continuous. Indeed, assume that f is uni-

formly continuous (see Definition 1.45). So, for ε = 1, there existsδ > 0 such that

∀x, y ∈ X : dX (x, y) < δ =⇒ ∣∣f(x)− f(y)

∣∣ < 1.

Because X is dense in X∗ (see Definitions 1.20 and 1.50), there existsx0 ∈ X such that

dX∗ (x0, x

∗) < δ2 .

Again using the density of X in X∗, we can choose x1 ∈ X such that

dX∗ (x1, x

∗) < min{δ2 ,

11+ 1

dX∗ (x0,x

∗)

}.

Then

dX∗ (x0, x1) � d

X∗ (x0, x∗) + d

X∗ (x∗, x1) < δ

2 + δ2 = δ.

But ∣∣f(x0)− f(x1)

∣∣ =

∣∣ 1dX∗ (x0,x∗) − 1

dX∗ (x1,x∗)

∣∣ > 1,

a contradiction.

1.3. Solutions 101

(b) Let X = R. This a complete metric space, but the function

f : R � x �−→ x2 ∈ Ris continuous and not uniformly continuous.

Solution of Problem 1.62From Problem 1.46, we know that f is continuous at x ∈ X if andonly if ωf (x) = 0. So

C ={x ∈ X : ωf (x) = 0

}=

⋂

n�1

{x ∈ X : ωf (x) <

1n

}.

But by Problem 1.47, for each n � 1, the set

{x ∈ X : ωf (x) <

1n

}

is open. Hence C is a Gδ-set (see Definition 1.57).

Solution of Problem 1.63Suppose that we can find a function f : R −→ R for which

{x ∈ R : f is continuous at x

}= Q.

From Problem 1.62, we would have that Q is a Gδ-set (see Defini-tion 1.57), which contradicts Problem 1.58.

Solution of Problem 1.64For every integer m � 1 and every f ∈ F , we consider the set

Cm,f ={x ∈ X : f(x) � m

}.

Since f is lower semicontinuous, the set Cm,f is closed for all f ∈ Fand all m � 1, hence the set

Cm =⋂

f∈FCm,f


is closed too. By hypothesis

X =⋃

m�1

Cm.

Since X is complete, by the Baire category theorem (see Theo-rem 1.26), we can find an integer m0 � 1 such that intCm0 �= ∅.Hence

f(x) � m0 ∀ f ∈ F , x ∈ U = intCm0 .

Solution of Problem 1.65Arguing by contradiction, suppose that we can find a sequence of con-tinuous functions fn : R −→ R such that

fn(x) −→ χQ(x) ∀ x ∈ R.

Let

Un = f−1n

((12 ,+∞)) ∀ n � 1.

Then Un is open (since fn is continuous) for n � 1 and from thepointwise convergence of the sequence {fn}n�1 to χ

Q, we have

⋂

n�1

Un = Q,

so Q is a Gδ-set (see Definition 1.57), which contradicts Prob-lem 1.58.

Solution of Problem 1.66For a given ε > 0, let

Aεm =

{x ∈ X :

∣∣fm(x)− f(x)

∣∣ � ε

} ∀ m � 1,

Dε =⋃

m�1

intAεm,

E =⋂

n�1

D1n .

1.3. Solutions 103

We claim that E = E. To this end let x ∈ E. Since fn −→ f , wecan find an integer m � 1 such that |fm(x)− f(x)| � ε

3 . Also thecontinuity of fm (by hypothesis) and of f at x ∈ E, implies that thereexists δ = δ(x, ε) > 0 such that

∣∣fm(u)− fm(x)∣∣ � ε

3 and∣∣f(u)− f(x)

∣∣ � ε3 ∀ u ∈ Bδ(x).

Hence by the triangle inequality, we have

∣∣fm(u)− f(u)∣∣ �

∣∣fm(u)− fm(x)∣∣+

∣∣fm(x)− f(x)∣∣+

∣∣f(u)− f(x)∣∣

� ε ∀ u ∈ Bδ(x),

so x ∈ intAεm ⊆ Dε. Since ε > 0 was arbitrary, it follows that x ∈ E.

Therefore E ⊆ E.Next let x ∈ E. For a given ε > 0, let n � 3

ε . We have that x ∈ D1n

and so there is an integer m � 1 such that x ∈ intA1nm. Then we can

find δ > 0 such that Bδ(x) ⊆ A1nm, so

∣∣fm(u)− f(u)∣∣ � 1

n∀ u ∈ Bδ(x)

and ∣∣fm(u)− fm(x)

∣∣ � 1

n∀ u ∈ Bδ(x)

(from the continuity of fm). Then, the triangle inequality implies that

∣∣f(u)− f(x)

∣∣ � 3

n� ε ∀ u ∈ Bδ(x),

so f is continuous at x, i.e., x ∈ E and so E ⊆ E. Therefore finallyE = E.

Next, for all ε > 0 and m � 1, we introduce the sets

Gεm =

{x ∈ X :

∣∣fm(x)− fm+k(x)∣∣ � ε for all integers k � 1

}.

Clearly each Gεm is closed and by the pointwise convergence of the

sequence {fn}n�1 to f , we have

X =⋃

m�1

Gεm ∀ ε > 0,

Gεm ⊆ Aε

m ∀ ε > 0, m � 1,


so ⋃

m�1

intGεm ⊆ Dε ∀ ε > 0.

Hence

X\Dε ⊆ X\⋃

m�1

intGεm =

⋃

m�1

Gεm\

⋃

m�1

intGεm ⊆

⋃

m�1

(Gε

m\intGεm

),

so ⋃

k�1

(X \D 1

k) ⊆

⋃

k�1

⋃

m�1

(G

1km \ intG

1km

)

and thus

X \E = X \ E = X \⋂

k�1

D1k =

⋃

n�1

(X \D 1k )

⊆⋃

k�1

⋃

m�1

(G

1km \ intG

1km

) ⊆⋃

k�1

⋃

m�1

∂G1km.

But ∂G1km is a nowhere dense set (see Definition 1.25). Therefore X \E

is a subset of a meager set, hence E contains a countable intersectionof open dense sets (see Definition 1.20 and Problem 1.60(a)), which isdense too, since X is complete (see Theorem 1.26). So, finally E is adense Gδ-set in X (see Definition 1.57).

Solution of Problem 1.67We have

∂f∂x(x, y) = lim

n→+∞n∣∣f(x+ 1

n , y)− f(x, y)

∣∣,

∂f∂y (x, y) = lim

n→+∞n∣∣f(x, y + 1

n

)− f(x, y)∣∣.

From Problem 1.66, we know that ∂f∂x is continuous an a dense Gδ set

E1 ⊆ R2 (see Definitions 1.57 and 1.20) and ∂f

∂y is continuous an o

dense Gδ set E2 ⊆ R2. We set E = E1 ∩ E2. Then on E both partialderivatives of f exist and are continuous, hence f is differentiable onE. Clearly E is dense and Gδ .

1.3. Solutions 105

Solution of Problem 1.68Let X = (0, 1] and let d1 be the metric on X induced by the Euclideanmetric on R, i.e.,

d1(x, u) = |x− u| ∀ x, u ∈ X.

Also, let d2 : X ×X −→ R+ be defined by

d2(x, u) =∣∣ 1x − 1

u

∣∣ ∀ x, u ∈ X.

Clearly d2 is a metric. Note that

d1(x, u) = |x− u| � |x−u|xu = d2(x, u) ∀ x, u ∈ X.

Therefore the metric d2 is stronger than d1 (i.e., the metric topology onX generated by d2 is richer (has more open sets) than that generatedby d1) and the identity function i : (X, d2) −→ (X, d1) is continuous(in fact Lipschitz continuous with Lipschitz constant 1).

Let V = [1,+∞) and let d be the metric induced by the usualmetric on R. We consider the bijections f : X −→ V and h : V −→ X,defined by

f(x) = 1x and h(v) = 1

v ∀ x ∈ X, v ∈ V.

Both functions are bijections and f : (X, d1) −→ (V, d) is continuous.Also, for all v, y ∈ V , we have

d2(h(v), h(y)

)=

∣∣ 1h(v) − 1

h(y)

∣∣ = |v − y| = d(v, y)

and so h : (V, d) −→ (X, d2) is an isometry.Note that i = h ◦ f : (X, d1) −→ (X, d2) is continuous and so we

have proved that (X, d1) and (X, d2) are indeed topologically equiva-lent.

Next we show that the two metrics d1 and d2 are not uniformlyequivalent. If this is not the case, i.e., the two metrics are uniformlyequivalent, then i : (X, d1) −→ (X, d2) is uniformly continuous (seeDefinition 1.45) and so we can find δ > 0 such that

d2(x, y) < 1 ∀ x, y ∈ X, with d1(x, y) < δ. (1.5)

Let {xn}n�1 ⊆ X be a sequence, given by xn = 1n for n � 1. Then

d1(xn, xn+1) = 1n(n+1) and d2(xn, xn+1) =

∣∣n− (n+ 1)∣∣ = 1.


For our δ > 0, we can find n0 � 1 such that

d1(xn, xn+1) < δ ∀ n � n0,

whiled2(xn, xn+1) = 1 ∀ n � 1,

a contradiction with (1.5). This proves that d1 and d2 are not uni-formly equivalent (see Remark 1.53).

Solution of Problem 1.69Let X = (0, 1] and consider the metrics d1 and d2 by

d1(x, u) = |x− u| and d2(x, u) =∣∣ 1x − 1

u

∣∣ ∀ x, u ∈ X.

From Problem 1.68, we know that d1 and d2 are topologically equiva-lent. Let xn = 1

n for n � 1. Then {xn}n�1 is d1-Cauchy and

xn −→ 0 �∈ X.

So, (X, d1) is not complete.From the solution of Problem 1.68, we know that the function

h : V = [1,+∞) −→ X, defined by

h(v) = 1v ∀ v ∈ V

is an isometry from V with the usual metric it inherits from R onto(X, d2). So, h preserves Cauchy sequences. Because V is complete, itfollows that (X, d2) is complete.

Solution of Problem 1.70Let X = (0,+∞) and let f : X −→ (0, 1) and h : (0, 1) −→ X bedefined by

f(x)def= x

1+x and h(v)def= v

1−v ∀ x ∈ X, v ∈ (0, 1).

Both functions are continuous and f−1 = h. Hence f is a homeomor-phism on R ontoX (both spaces with the Euclidean metric topologies).Note that

f ∈ C1(X), with f ′(x) = 1(1+x)2 > 0.

1.3. Solutions 107

Therefore f is strictly increasing. So, for all r, s � 0, we have

f(r + s) = r+s1+(r+s) � r+s+rs

1+r+s+rs = r+s(1+r)(1+r)(1+s)

� r(1+s)+s(1+r)(1+r)(1+s) � r

1+r +s

1+s = f(r) + f(s).

Let d1 be the usual metric on R and let d2 : X ×X −→ R+ be definedby

d2(x, u)def= f

(|x− u|) ∀ x, u ∈ X.

Using the properties of f established above, we verify that d2 is ametric on R. Note that f(r) � r for all r � 0 and so

d2(x, u) � d1(x, u) ∀ x, u ∈ X.

Then the identity function i : (X, d1) −→ (X, d2) is Lipschitz contin-uous with Lipschitz constant 1, hence uniformly continuous too (seeDefinition 1.45).

We will show that also the identity function i : (X, d2) −→ (X, d1)is uniformly continuous. Let ε > 0 be given. Let us set δ = ε

1+ε . Then,if d2(x, u) � δ, so

|x−u|1+|x−u| � ε

1+ε and thus d1(x, u) = |x− u| � ε.

Hence, we have shown that

∀ε > 0 ∃δ > 0 ∀x, u ∈ X : d2(x, u) < δ =⇒ d1(x, u) < ε,

so the identity function i : (X, d2) −→ (X, d1) is uniformly continuous.Thus the metrics d1 and d2 are uniformly equivalent.

On the other hand, for every u � 0, we have d1(0, u) = u andd2(0, y) = f(y) < 1. So there is no k > 0 such that

d1(0, u) � kd2(0, u) ∀ u ∈ X

(let u = k+1). This implies that d1 and d2 are not Lipschitz equivalent(see Proposition 1.54).


un(t) = tn

n ∀ t ∈ [0, 1], n � 1


and letu∗(t) = 0 ∀ t ∈ [0, 1].

Thend∞(un, u

∗) = 1n −→ 0 as n → +∞.

On the other hand

d∞1(un, u

∗) = 1n + 1 −→ 1 as n → +∞

(see Definition 1.52).


dX (x, u) = dX (x, u) + dY

(f(x), f(u)

) ∀ x, u ∈ X.

Clearly dX

is a metric on X. Moreover, note that

limn→+∞ d

X(xn, x) = 0 ⇐⇒ lim

n→+∞ dX(xn, x) = 0.

Therefore dXand d

Xare equivalent metrics on X (see Definition 1.52).

Note that

dY

(f(x), f(u)

)� d

X(x, u) ∀ x, u ∈ X,

hence the function f : (X, dX ) −→ Y is Lipschitz continuous (see Def-inition 1.48).


dY(y, v) =

dY(y, v)

1 + dY(y, v)

∀ y, v ∈ Y.

This is a bounded metric on Y which is topologically equivalent to dY

(see Definition 1.52). Also on X we introduce the distance function

dX(u, x) = d

X(u, x) +

∑

n�1

12n dY

(fn(u), fn(x)

) ∀ u, x ∈ X.

1.3. Solutions 109

It is easy to check that dX is a metric on X and

dX(un, u) −→ 0 ⇐⇒ d

X(un, u) −→ 0,

so dX

and dX

are topologically equivalent.Moreover, we have

dY

(fn(u), fn(x)

)� 2ndX (u, x) ∀ u, x ∈ X, n � 1,

so for each n � 1, the function fn : (X, dX) −→ (Y, d

Y) is Lipschitz

continuous (see Definition 1.48).

Solution of Problem 1.74Let Y be the completion of the metric space X and let h : X −→h(X) ⊆ Y be the isometry such that h(X) = Y (see Definition 1.50).Since X is separable (see Definition 1.21), it admits a countable denseset D (see Definition 1.20). Then the countable set h(D) ⊆ Y satisfies

h(X) = h(D) ⊆ h(D) ⊆ h(X) = Y

(see Proposition 1.32(d)), so

Y = h(X) = h(D),

i.e., Y is separable too.

Solution of Problem 1.75If D = ∅, then the result is obvious, so let us suppose that D �= ∅. Fixt ∈ D and let

Yt ={f ∈ C[a, b] : f(t) = 0

}.

Then since uniform convergence (i.e., convergence in the d∞C[a,b]

-metric;

see Definition 1.59) implies pointwise convergence, we infer that Yt isd∞C[a,b]

-closed. Note that Y =⋂

t∈DYt. Hence Y is d∞

C[a,b]-closed.


Solution of Problem 1.76The set E1 is not closed in X. To see this, consider the functionsfn ∈ X for n � 1, defined by

fn(s) = sn ∀ n � 1, s ∈ [0, 1]

Then fn ∈ E1 for each n � 1 but fnd∞−→ 0 �∈ E1.

The set E2 is closed in X. To see that, let {fn}n�1 ⊆ E2 be asequence and assume that

fnd∞−→ f

(i.e., fn ⇒ f ; see Definition 1.59). Let t ∈ [0, 1] and n � 1. Wecan find sn ∈ [0, 1] such that fn(sn) = t. Passing to a subsequence ifnecessary, we may assume that sn −→ s. Then f(sn) −→ f(s). Wehave

∣∣t− f(sn)

∣∣ =

∣∣fn(sn)− f(sn)

∣∣ � d∞(fn, f) −→ 0,

so f(s) = t. Since t ∈ [0, 1] was arbitrary, we conclude that f ∈ E2.

Finally E is not closed. To see this, consider functions

fn(s) =

{2n−1

n s if s ∈ [0, 12

],

2ns+

n−2n if s ∈ (

12 , 1

],

∀ n � 2.

Then fn ∈ E and fnd∞−→ f , where

f(s) =

{2s if s ∈ [

0, 12],

1 if s ∈ (12 , 1

].

However, it is clear that f �∈ E (it has a flat part on[12 , 1

]).

Solution of Problem 1.77For integers m,n � 1, we define

Cm,n ={f ∈ X : ∃t ∈ [0, 1] ∀s ∈ [0, 1] :

|s− t| � 1m =⇒ ∣

∣f(s)− f(t)∣∣ � n|s− t|}.

1.3. Solutions 111

We will show that Cm,n is nowhere dense in X (see Definition 1.25).In fact, if f ∈ Cm,n, we let

m(f) = maxt∈[0,1]

∣∣f(t± 1

m

)− f(t)1m

∣∣.

Let a(x) be a continuous, piecewise linear function bounded by 1 andwith each linear piece having slope 3m(f). Then, for every ε > 0, wehave

d∞(f + εa, f) � ε and f + εa �∈ Cm,n.

So d∞-intCm,n = ∅, i.e., Cm,n is nowhere dense.Since (X, d∞) is a complete metric space (see Proposition 1.62), by

the Baire category theorem (see Theorem 1.26), the set⋃

m,n�1Cm,n is

nowhere dense in X. Hence

D = X \⋃

m,n�1

Cm,n is of second category

(see Definition 1.25). But if f ∈ D, then for every t ∈ [0, 1] and everyn,m � 1, we can find points s ∈ [0, 1] such that

|s− t| � 1m and

∣∣f(s)−f(t)

s−t

∣∣ > n

and so f is not differentiable at any t ∈ [0, 1].

Solution of Problem 1.78Let x, y ∈ X and let ε > 0. From the definition of dist (see Defini-tion 1.6(d)), there exists z ∈ A such that

dX(x, z) < f(x) + ε.

As we also have dX(y, z) � f(y), so

f(y)− f(x) � dX(y, z)− d

X(x, z) + ε � d

X(x, y) + ε.

Because ε > 0 was arbitrary, we conclude that

f(y)− f(x) � dX(x, y).


Exchanging the roles of x and y, we also have

f(x)− f(y) � dX(x, y),

so finally ∣∣f(x)− f(y)

∣∣ � dX (x, y).

This proves that f is Lipschitz continuous with Lipschitz constant 1.

Solution of Problem 1.79Let C ⊆ X be a closed set. Then

C ={x ∈ X : dist

X(x,C) = 0

}

(see Definition 1.6(d)) Hence

C =⋂

n�1

{x ∈ X : dist

X(x,C) < 1

n

}.

The continuity of the function x �−→ distX (x,C) (see Problem 1.78),implies that the set

{x ∈ X : dist

X(x,C) < 1

n

}is open

(see Proposition 1.32). Therefore it follows that C is Gδ (see Defini-tion 1.57). Because the complement of a Gδ set is an Fσ-set. Thereforethe second part of the problem follows from the first.

Solution of Problem 1.80For every n ∈ Z let Dn = D ∩ [n, n + 1). Then D =

⋃

n�1Dn. Since D

is uncountable, at least one of the sets Dn, say Dn0 , is uncountable.So, in Dn0 we can find a sequence with distinct terms, which has aconvergent subsequence. The limit of this sequence is an accumulationpoint of D (see Definition 1.13(b) and Theorem 1.14(d)). Now let

S ={x ∈ R : x is an accumulation point in D

}.

1.3. Solutions 113

We have just seen that S �= ∅ and of course S is closed. Hence Sc isopen and so by Problem 1.79, Sc =

⋃

n�1Cn with Cn ⊆ R closed for

every n � 1. If S is countable, then D must have uncountable manyelements in Sc and so in some Cn0 . From the first part of the proof,D ∩Cn0 has an accumulation point x. Then

x ∈ D ∩Cn0 ⊆ D ∩ Cn0 = D ∩ Cn0 ,

so x ∈ Cn0 and x ∈ S. But S ∩ Cn0 = ∅, a contradiction.

Solution of Problem 1.81Evidently it suffices to consider U = (a, b), a, b ∈ R, a < b. We have

f−1((a, b)

)=

⋃

n�1

[f−1

((−∞, b− 1n

]) ∩ f−1((a,+∞)

)].

But from the lower semicontinuity of f , we have that

f−1((−∞, b− 1

n

])is closed and f−1

((a,+∞)

)is open

(see Problem 1.64).Also, from Problem 1.79, we have that f−1

((a,+∞)

)is an Fσ-set

(see Definition 1.57(b)). Hence

f−1((a, b)

)is an Fσ-set.

Solution of Problem 1.82For every n � 1, we cover R with a countable family of open sets{Un

k }k�1 such that

diamUnk < 1

n ∀ n, k � 1

(see Proposition 1.24). Note that all sets f−1(Unk ) are Fσ-sets (see

Problem 1.81). Hence

f−1(Unk ) =

⋃

m�1

Cnk,m with Cn

k,m closed for all m � 1.

We have X =⋃

k,m�1

Cnk,m, so

Dn =⋃

k,m�1

intCnk,m is open dense in X for all n � 1


(see Problem 1.34). We have D =⋂

n�1Dn is a dense Gδ-set in X (seeTheorem 1.26 and Definitions 1.57 and 1.20) and since the oscillationof f on each Dn is less than 1

n , we infer that f |D is continuous.

Solution of Problem 1.83(a) Since by hypothesis fn ⇒ f (see Definition 1.59), we can findan integer n0 � 1 such that sup

x∈XdY

(fn0(x), f(x)

)< 1. So, for all

x1, x2 ∈ X, we have

dY

(f(x1), f(x2)

)

� dY

(f(x1), fn0(x1)

)+ d

Y

(fn0(x1), fn0(x2)

)+ d

Y

(fn0(x2), f(x2)

)

� 2 + diam fn0(X)

(see Definition 1.6(b)), hence diam f(X) < +∞, i.e., f is boundedtoo.

(b) By way of contradiction, suppose that fn ⇒ f and f : X −→ Yis bounded. Then we can find x0 ∈ X and r0 > 0 such that f(X) ⊆Br0(x0). Again we can find an integer n0 � 1 such that

supx∈X

dY

(fn0(x), f(x)

)< 1.

Hence fn0(X) ⊆ Br0+1(x0), a contradiction to the fact that fn0 isunbounded. This shows that the limit function f cannot be bounded.

Solution of Problem 1.84Let ε > 0. Since fn ⇒ f (see Definition 1.59), we can find n0 =n0(ε) � 1 such that

dY

(fn0(x), f(x)

)� ε

3 ∀ x ∈ X.

The continuity of fn0 at x0 implies that there exists δ > 0 such that

dY

(fn0(x), fn0(x0)

)� ε

3 ∀ x ∈ Bδ(x0)

1.3. Solutions 115

(where Bδ(x0) ={x ∈ X : dX (x, x0) � δ

}). Then for every x ∈

Bδ(x0), we have

dY

(f(x), f(x0)

)� d

Y

(f(x), fn0(x)

)+ d

Y

(fn0(x), fn0(x0)

)

+ dY

(fn0(x0), f(x0)

)� ε,

so f is continuous at x0.

Solution of Problem 1.85No. Let X = R and for every n � 1, let

fn(x) = x ∀ n � 1, x ∈ R, gn(x) = 1n ∀ n � 1, x ∈ R.

Evidently fn ⇒ f = idRand gn ⇒ 0 (see Definition 1.59) Then

(fngn)(x) = xn ∀ n � 1, x ∈ R.

Then fngn −→ 0, but the above convergence is not uniform.

Solution of Problem 1.86Since by hypothesis fn ⇒ f (see Definition 1.59), for a given ε > 0,we can find an integer n0 � 1 such that

dY

(fn(x), f(x)

)< ε

2 ∀ x ∈ X, n � n0.

For x, u ∈ X and all n � n0, we have

dY

(f(x), f(u)

)� d

Y

(f(x), fn(x)

)+d

Y

(fn(x), fn(u)

)+d

Y

(fn(u), f(u)

)

� ε+ kdY(x, u)

(see Definition 1.48). Because ε > 0 is arbitrary we let ε ↘ 0 andobtain

dY

(f(x), f(u)

)� kd

Y(x, u) ∀ x, u ∈ X,

hence f is k-Lipschitz.


Solution of Problem 1.87Because fn ⇒ f (see Definition 1.59), for a given ε > 0, we can findn0 � 1 such that

dY

(fn(x), f(x)

)< ε

3 ∀ x ∈ X, n � n0.

By hypothesis fn0 is uniformly continuous (see Definition 1.45). So,we can find δ = δ(ε) > 0 such that

dY

(fn0(x), fn0(u)

)< ε

3 ∀ x, u ∈ X, with dX (x, u) < δ.

Then for all x, u ∈ X with dX(x, u) < δ, we have

dY

(f(x), f(u)

)� d

Y

(f(x), fn0(x)

)+d

Y

(fn0(x), fn0(u)

)

+ dY

(fn0(u), f(u)

)

< ε3 +

ε3 +

ε3 = ε,

so f is uniformly continuous.


(a) Yes. Let X = Y = [0, 1] and let{fn : [0, 1] −→ [0, 1]

}n�1

be thesequence of discontinuous functions, defined by

fn(x) =

{1n if x ∈ [0, 1] ∩Q0 if x ∈ [0, 1] \Q ∀ n � 1.

Note thatfn ⇒ f ≡ 0

(see Definition 1.59) and of course f is continuous (in fact the func-tions fn are discontinuous at every x ∈ X).

(b) No. Consider the function f : R −→ R, defined by

f(x) =

{1 if x ∈ Q,−1 if x ∈ R \Q.

Thenf ◦ f ≡ 1.

1.3. Solutions 117

So, f ◦f is continuous, but f is discontinuous (in fact f is discontinuousat every x ∈ X).

Solution of Problem 1.89As in Definition 1.37, for every x ∈ X, let

ωϕ(x) = infr>0

diamϕ(Br(x) ∩A

).

We know that ϕ is continuous at x if and only if ωϕ(x) = 0 (seeProblem 1.46). We set

A0 ={x ∈ A : ωϕ(x) = 0

}

(note that, if x ∈ A, then Br(x) ∩ A �= ∅). Evidently A ⊆ A0. Letx ∈ A0. Then we can find a sequence {xn}n�1 ⊆ A such that xn −→ x.

Since ωϕ(x) = 0, we see that{ϕ(xn)

}n�1

⊆ Y is a Cauchy sequence.Since Y is a complete metric space, we can find y ∈ Y such that

ϕ(xn) −→ y in Y

and we can easily see that this limit is independent of the choice ofthe approximating sequence {xn}n�1. So, if we set ϕ0(x) = y, thenϕ0 : A0 −→ Y is well defined and continuous because

ωϕ0(x) = 0 ∀ x ∈ A0.

We need to show that A0 is a Gδ-subset of X (see Definition 1.57(a)).Let

Un =(x ∈ X : ωϕ(x) <

1n+1

).

From Problem 1.47, we know that the set Un is open in X. We have

A0 =( ⋂

n�1

Un

) ∩A.

But from Problem 1.79, we know that A is a Gδ-subset of X. So, weconclude that A0 is a Gδ-set.


Solution of Problem 1.90Let xn −→ x in X. Then

f(xn, s(xn)

)= v0 ∀ n � 1.

Since Y is compact, we can find a subsequence{s(xnk

)}k�1

of{s(xk)

}n�1

and y ∈ Y such that s(xnk) −→ y in Y . Exploiting the

continuity of f , we have

f(xnk

, s(xnk)) −→ f(x, y),

hence f(x, y) = v0 and so by hypothesis y = s(x). Since everysubsequence of

{s(xn)

}n�1

has a further subsequence which con-

verges to s(x), we conclude that for the original sequence, we havesn(x) −→ s(x) (see Problem 1.3) and so s is continuous.

Solution of Problem 1.91We need to show that f is a bijection and that f−1 is continuous.Note that if f(x) = f(u) then d(x, u) = 0 and so x = u. This provesthat f is injective. Next we show that f is surjective. Arguing bycontradiction, suppose that there exists x ∈ X such that x �∈ f(X).Then since f(X) is compact, we have

dX

(x, f(X)

)= c > 0.

We have

c � dX

(x, f (k)(x)

)� d

X

(f (n)(x), f (n+k)(x)

) ∀ k, n � 1

(recall that f (m) = f ◦ . . . ◦ f m-times for all m � 1). Then the se-quence

{f (n)(x)

}n�1

has no convergent subsequence in f(X) ⊆ X, acontradiction. This proves the surjectivity of f , hence f is a bijection.

Let x, u ∈ X. Consider two sequences{f (n)(x)

}n�1

and{f (n)(u)

}n�1

. Because X is compact, we can find a subsequence

{nk}k�1 such that both sequences{f (nk)(x)

}k�1

and{f (nk)(u)

}k�1

are convergent and thus they are Cauchy sequences.Let ε > 0. We can find k � 1 such that

dX

(f (nk)(x), f (nl)(x)

)� ε ∀ k, l � k0

1.3. Solutions 119

and

dX

(f (nk)(u), f (nl)(u)

)� ε ∀ k, l � k0.

Let n = nk0+1 − nk0 . Then, using the hypothesis on f , we have

dX

(x, fn(x)

)� d

X

(fnk0 (x), fnk0+1(x)

)� ε

and

dX

(u, fn(u)

)� d

X

(fnk0 (u), fnk0+1(u)

)� ε

Hence

dX

(f(x), f(u)

)� d

X

(f (n)(x), f (n)(u)

)

� dX

(f (n)(x), x

)+ d

X(x, u) + d

X

(u, f (n)(u)

)

� 2ε+ dX(x, u),

so, letting ε ↘ 0, we have

dX

(f(x), f(u)

)� d

X(x, u)

and thus we conclude that f is an isometry (see Definition 1.41).

Solution of Problem 1.92By Theorem 1.47, f admits a uniformly continuous extension (seeDefinition 1.45) f : [a, b] −→ R and f

([a, b]

)is compact. So, f is

bounded (see Definition 1.6(b)).

Solution of Problem 1.93Let x, u ∈ X.Let us define two sequences

x−1 = f(x), x0 = x and xn ∈ f−1(xn−1) ∀ n � 1,u−1 = f(u), u0 = u and un ∈ f−1(un−1) ∀ n � 1.

Because X is compact, we can find a subsequence {nk}k�1 such that

xnk−→ x and unk

−→ u as k → +∞.


Let ε > 0. We can find k0 � 2 such that

dX

(xnk

, x)

� ε2 ∀ k, l � k0,

dX

(unk

, u)

� ε2 ∀ k, l � k0.

Then

dX

(xk0 , xnk0+1

)� d

X

(xk0 , x

)+ d

X

(x, xnk0+1

)� ε,

dX

(uk0 , unk0+1

)� d

X

(uk0 , u

)+ d

X

(u, unk0+1

)� ε.

Let n = nk0+1 − nk0 − 1. Then

dX

(xn, x−1

)� dX

(xnk0+1

, xk0)

� ε,

dX

(un, u−1

)� d

X

(unk0+1

, uk0)

� ε

Hence

dX(x, u) = d

X(x0, y0) = d

X

(f(x1), f(u1)

)� d

X(x1, u1) � . . .

� dX(xn, un) � d

X(xn, x−1)+d

X(x−1, u−1)+d

X(u−1, un)

� dX

(f(x), f(u)

)+ 2ε.

Because ε > 0 was arbitrary, we get

dX(x, u) � d

X

(f(x), f(u)

),

thus using the assumption, we deduce that f is isometry.

Solution of Problem 1.94(a) If any of the sets Kn is empty, then the result is obvious. Solet us assume that all sets Kn are nonempty. Clearly the sequence{diamKn

}n�1

⊆ [0,+∞) is decreasing and bounded below by diamK.So

diamK � limn→+∞diamKn.

Because Kn is compact and the distance function dXis continuous (see

Proposition 1.36), we can find xn, un ∈ Kn for n � 1 such that

dX(xn, un) = diamKn ∀ n � 1.

1.3. Solutions 121

By the compactness of X and by passing to a suitable subsequence ifnecessary, we may assume that

xn −→ x and un −→ u in X.

We claim that x, u ∈ K. Note that

xn, un ∈ Km ∀ n � m � 1.

Hencex, u ∈ Km ∀ m � 1

and so x, u ∈ K. We have

diamK � dX(x, u) = lim

n→+∞ dX(xn, un) = lim

n→+∞diamKn,

sodiamK = lim

n→+∞diamKn.

(b) Let us consider X ={1n : n � 1

}with the natural metric induced

from R. Then X is not compact (as the Cauchy sequence{

1n

}n�1

has

no limit in X). Let

Kk = {1} ∪ {1n : n � k

} ∀ k � 1.

Then the sequence {Kn}n�1 is a decreasing sequence of closed subsetsof X with

diamKk = 1 ∀ k � 1.

But K =⋂

k�1

Kk = {1} and so

diamKk �−→ diamK = ∅.Now, letX = [0, 1] (which is compact). Let {qn}n�1 be an enumerationof rationals in X. Let

Kk ={qn : n � k

} ∀ k � 1.

Then {Kk}k�1 is a sequence of decreasing nonempty sets which arenot closed, with

diamKk = 1 ∀ k � 1.

But K =⋂

k�1

Kk = ∅ and so

diamKk �−→ diamK = 0.


Solution of Problem 1.95No. To see this let X = Y = [0, 1] and let {fn}n�1 ⊆ C(X;Y ) be thesequence, defined by

fn(x) =

{nx if 0 � x � 1

n1 if 1

n < x � 1∀ n � 1.

Then fn −→ f (pointwise), where

f(x) =

{0 if x = 0,1 if x = 1.

Since the limit function is discontinuous, we infer that the convergenceis not uniform (see Proposition 1.62), i.e., it is not in the d∞-metric.Therefore {fn}n�1 is a sequence in

(C(X;Y ), d∞

)with no convergent

subsequence. So, we conclude that(C(X;Y ), d∞

)is not compact.


(a) We have

∣∣fn(xn)− f(x)∣∣ �

∣∣fn(xn)− f(xn)∣∣+

∣∣f(xn)− f(x)∣∣

� ‖fn − f‖∞ +∣∣f(xn)− f(x)

∣∣.

Since fn ⇒ f (see Definition 1.59), we have ‖fn − f‖∞ −→ 0. Also,since f ∈ C(X) (see Proposition 1.62), we have that f(xn) −→ f(x).Therefore, finally we have fn(xn) −→ f(x).

(b) First we show that if xn −→ x in X and rn −→ +∞ with{rn}n�1 ⊆ N strictly increasing, then frn(xn) −→ f(x). To this end,let

un =

{xk if n = rk,x if n �∈ {rk : k � 1}.

Evidently un −→ x and so fn(un) −→ f(x). In particular thenfrk(urk) = frk(xk) −→ f(x).

Now suppose that f is not continuous. Then we can find x ∈ X,xk −→ x and ε > 0 such that

∣∣f(xk)− f(x)

∣∣ > ε ∀ k � 1.

1.3. Solutions 123

The hypothesis implies that fn −→ f . So, for every k � 1,we have fn(xk) −→ f(xk) as n → +∞. Then, by inductionwe can find a strictly increasing sequence {rk}k�1 ⊆ N such that|frk(xk)− f(x)| > ε, contradicting the first part of the solution ofstatement (b).

Next suppose that the sequence {fn}n�1 does not converge uni-formly to f . So, by passing to a subsequence if necessary, we mayassume that for some ε > 0, we have ‖fn− f‖∞ � ε for all n � 1. Thecompactness of X implies that we can find a sequence {xn}n�1 ⊆ Xsuch that ∣

∣fn(xn)− f(xn)∣∣ � ε ∀ n � 1.

Passing to a next subsequence if necessary, we may assumethat xn −→ x (recall that X is compact). Then by hypothesis,fn(xn) −→ f(x). So, we have

ε �∣∣fn(xn)− f(xn)

∣∣ �∣∣fn(xn)− f(x)

∣∣+∣∣f(x)− f(xn)

∣∣ −→ 0,

a contradiction. This proves that fn ⇒ f .


(a) Evidently each Cn is compact and so is Cn \ U = Cn ∩ U c. Wehave ⋂

n�1

(Cn ∩ U c) =( ⋂

n�1

Cn

) ∩ U c = U ∩ U c = ∅.

Then by Proposition 1.67 the sequence of closed sets {Cn ∩ U c}n�1

cannot have the finite intersection property (see Definition 1.66).

Hence we can find an integer n0 � 1 such thatn0⋂

n=1Cn ∩ U c = ∅.

Since the sequence {Cn}n�1 is decreasing (i.e., C1 ⊇ C2 ⊇ C3 ⊇ . . .)we must have Cn0 ∩ U c = ∅, hence Cn0 ⊆ U .

(b) If the sets Cn are not bounded, the result does not hold. Tosee this, let U = (0, 1) ⊆ R and let Cn =

{12

} ∪ [n,+∞) forn � 1. Then {Cn}n�1 is a decreasing family of closed sets such that⋂

n�1Cn =

{12

} ⊆ U , but Cn � U for all n � 1.


Solution of Problem 1.98From the hypothesis, we know that for a given ε > 0, we can findM > 0 such that

∣∣f(x)

∣∣ < ε

2 ∀ x �∈ [−M,M ].

By Proposition 1.77, f |[−M,M]

is uniformly continuous (see Defini-tion 1.45). So, we can find δ1 such that

∣∣f(x)− f(y)

∣∣ < ε

2 ∀ x, y ∈ [−M,M ], with |x− y| < δ1.

Since f is continuous at ±M , we can find δ2 = δ2(M,ε) > 0 such that

∣∣f(x) + f(±M)∣∣ < ε, whenever

∣∣x− (±M)∣∣ < δ2.

Let δ = min{δ1, δ2}. Then for x < M < y with |x− y| < δ, we have

∣∣f(x)− f(y)

∣∣ �

∣∣f(x)− f(M)

∣∣+

∣∣f(M)− f(y)

∣∣ < ε,

and for x > −M > y with |x− y| < δ, we have

∣∣f(x)− f(y)∣∣ �

∣∣f(x)− f(−M)∣∣+

∣∣f(−M)− f(y)∣∣ < ε,

which proves the uniform continuity of f .

Solution of Problem 1.99Let ε > 0. By hypothesis, we can find M > 0 such that

∣∣f(t)− h(t)∣∣ � ε

3 ∀ |t| � M.

Let Tε =[ − M − ε,M + ε

]. Then f |Tε is uniformly continuous (see

Definition 1.45 and Proposition 1.77). So, we can find δ = δ(ε) ∈ (0, ε)such that

∣∣f(t)− f(s)∣∣ � ε ∀ t, s ∈ Tε, |t− s| � δ.

Since h is uniformly continuous, we can choose δ = δ(ε) ∈ (0, ε) evensmaller if necessary, so that

∣∣h(t)− h(s)

∣∣ � ε

3 ∀ t, s ∈ R, |t− s| � δ.

1.3. Solutions 125

Now, if |t|, |s| � M and |t− s| � δ, then

∣∣f(t)− f(s)∣∣ �

∣∣f(t)− h(t)∣∣ +

∣∣h(t)− h(s)∣∣+

∣∣h(s)− f(s)∣∣ � ε.

Finally, if |t| � M and |s| � M or conversely |t| � M and |s| � Mand |t− s| � δ, then

|t| � |s|+ ε � M + ε ∀ t, s, |t| − |s| � |t− s| � ε,

so, we infer that t, s ∈ Tε and thus∣∣f(t)− f(s)

∣∣ � ε. This proves theuniform continuity of f .

Solution of Problem 1.100(a) Let F be an open cover of C. Because F covers C, there existsU0 ∈ F such that x ∈ U0. Because the set U0 is open and x ∈ U0,so there exists r > 0 such that Br(x) ⊆ U0 (see Definition 1.8(a)).Because xn −→ x, so there exists n0 � 1 such that xn ∈ Br(x) ⊆ U0

for all n � n0 (see Definition 1.7). Now, for all k ∈ {1, . . . , n0 − 1}, wecan find Uk ∈ F such that xk ∈ Uk. So, for an arbitrary open coverF , we have found a finite subcover {Uk}n0

k=0 of the set C and thus theset C is compact (see Definition 1.63).

(b) “=⇒”: This is obvious.

“⇐=”: Suppose that xn −→ x. Then C = {x} ∪ {xn : n � 1} isa compact subset of X and so by hypothesis f |

Cis continuous. There-

fore f(xn) −→ f(x), which proves the continuity of f (see Proposi-tion 1.30).

Solution of Problem 1.101Let f : X −→ Y be a continuous and proper function (see Defini-tion 1.72). Let C ⊆ X be a nonempty and closed set. We need toshow that f(C) is closed. So, let {yn}n�1 ⊆ f(C) be a sequence suchthat

yn −→ y in Y.


Then yn = f(xn) with xn ∈ C for all n � 1 and the set

K = {y} ∪ {yn : n � 1

} ⊆ Y

is compact (see Problem 1.100(a)). By the properties of f , the setf−1(K) is compact and {xn}n�1 ⊆ f−1(K). So, by passing to a sub-sequence if necessary, we may assume that

xn −→ x in C

(since C is closed; see Proposition 1.11). Then f(xn) −→ f(x) (sincef is continuous; see Proposition 1.30) and so y = f(x) ∈ f(C), whichshows that f(C) is closed (see Proposition 1.11). Therefore f is aclosed function (see Definition 1.39(b)).

Solution of Problem 1.102“(a) =⇒ (b)”: Let C ⊆ X be a nonempty closed set. We need toshow that f(C) is closed. So, let {yn}n�1 ⊆ f(C) be a sequence suchthat yn −→ y in Y . Then yn = f(xn), with xn ∈ C for all n � 1.By hypothesis we can find a subsequence {xnk

}k�1 of {xn}n�1 suchthat xnk

−→ x in X. Then x ∈ C (as C is closed) and because f iscontinuous, we have

f(xnk) −→ f(x) = y ∈ f(C),

which proves that f(C) is closed. Also, if {xn}n�1 ⊆ f−1(y), then

f(xn) = y ∀ n � 1

and so by hypothesis {xn}n�1 admits a subsequence {xnk}k�1 such

that xnk−→ x in X. Note that f(x) = y and so f−1(y) is compact in

X.

“(b) =⇒ (a)”: Let {xn}n�1 ⊆ X be a sequence such that f(xn) −→ yin Y . Let

Cn ={xk : k � n

}and K = f−1(y).

Since by hypothesis f is closed, we have

f(Cn) = f(Cn)

1.3. Solutions 127

(see Proposition 1.32(d)) and because f(xn) −→ y, we have

{y} =⋂

n�1

f(Cn) =⋂

n�1

f(Cn).

So, if x ∈ K, then

f(x) = y ∈⋂

n�1

f(Cn),

which implies that the closed subsetsDn = K∩Cn of K are nonempty.Note that every finite subfamily of {Dn}n�1 has a nonempty intersec-tion. Due to the compactness of K, Proposition 1.67 implies that

⋂

n�1

Dn =⋂

n�1

(K ∩ Cn) = K ∩ ( ⋂

n�1

Cn

) �= ∅.

This means that {xn}n�1 has an accumulation point in K (see Def-inition 1.13(b) and Theorem 1.14(d)), which is equivalent to sayingthat there is a subsequence {xnk

}k�1 of {xn}n�1 and x ∈ K such thatxnk

−→ x in X.

Solution of Problem 1.103Let K ⊆ Y be a nonempty compact set and let {xn}n�1 ⊆ f−1(K) bea sequence. Then

f(xn) ∈ K ∀ n � 1

and since K is compact, we can find a subsequence{f(xnk

)}k�1

of{f(xn)

}n�1

such that

f(xnk) −→ y in Y.

By statement (a) in Problem 1.102, we can find a further subsequence{xnkm

}m�1

such that xnkm−→ x in X. This shows that f−1(K) is

compact.


Solution of Problem 1.104We do the case when fn ↘ f (the proof of the case fn ↗ f beingsimilar). We have

fn − f � 0 ∀ n � 1.

For a given ε > 0, let

Un ={x ∈ X : 0 � (fn − f)(x) < ε

} ∀ n � 1.

For every n � 1, the set Un is open (since the function fn − f iscontinuous; see Proposition 1.32) and

⋃

n�1

Un = X.

By the compactness of X, we can find an integer m � 1 such that

m⋃

n=1

Un = X

(see Definition 1.63). Note that {Un}n�1 is an increasing sequence(i.e., Un ⊆ Un+1 for n � 1). Hence Um = X and so

0 � (fn − f)(x) < ε ∀ n � m, x ∈ X,

which implies that fn ⇒ f (see Definition 1.59).

Solution of Problem 1.105First consider X = [0, 1) and

fn(x) = xn ∀ x ∈ [0, 1), n � 1.

Then fn ↘ 0 but not uniformly (see Definition 1.59). So, the com-pactness of X cannot be dropped.

Next let X = [0, 1] and

fn(x) = xn ∀ x ∈ [0, 1], n � 1.

1.3. Solutions 129

Then fn ↘ χ{1} but not uniformly. So, the continuity of the limitfunction cannot be dropped.

Next let X = [0, 1] and let fn : [0, 1] −→ R, n � 1 be the functions,defined by

fn(x) =

⎧⎨

⎩

0 if x = 0,1 if 0 < x < 1

n ,0 if 1

n � x < 1.

Then fn ↘ 0, but not uniformly. So, the continuity of functions fn(at least for large n’s) cannot be dropped.

Finally, let X = [0, 1] and let fn : [0, 1] −→ R, n � 2 be thefunctions, defined by

fn(x) =

⎧⎨

⎩

nx if 0 � x < 1n ,

2− nx if 1n � x < 2

n ,0 if 2

n � x � 1.

Note that {fn}n�2 is not monotone, fn −→ 0 but we do not haveuniform convergence (see Definition 1.59). Indeed,

βn = supx∈[0,1]

∣∣fn(x)

∣∣ = sup

x∈[0,1]fn(x)

= fn(1n

)= 1 �−→ 0 as n → +∞.

So, the monotonicity condition on {fn}n�1 cannot be dropped.


(a) Proceeding by contradiction, suppose that f(X) �= X. So wecan find u ∈ X such that u �∈ f(X). Since f(X) is compact (seeDefinition 1.63 and Proposition 1.74), we have

dist(u, f(X)) = ε > 0

(see Definition 1.6). For every integer n � 1, let

f (n) = f ◦ . . . ◦ f︸︷︷︸n-times

.

Evidently for every n � 1, f (n) is an isometry (see Definition 1.41) and

f (n)(X) ⊆ f(X) ∀ n � 1.


Then, if n > m � 1, we have

ε = dist(u, f(X)

)� dist

(u, f (n−m)(X)

)� d

X(u, f (n−m)(u)),

so

ε � dX

(f (m)(u), f (n)(u)

) ∀ n,m � 1.

From this it follows that the sequence{f (n)(u)

}n�1

⊆ X has no con-vergent subsequence, a contradiction to the fact that X is compact(see Definition 1.70 and Theorem 1.71).

(b) Let X = N be equipped with the discrete metric ddN(see Exam-

ple 1.3) and let

f : N � x �−→ x+ 1 ∈ N.Then (N, dd

N) is bounded (see Definition 1.6(b)) but not compact, f is

continuous, but f(N) �= N.

Solution of Problem 1.107Let x ∈ R

N be the fixed point of f , i.e., f(x) = x. Let r > 0 bearbitrary and let and u ∈ Br(x) =

{u ∈ X : ‖u − x‖ � r

}. Then,

since f is an isometry, we have

∥∥f(u)− x∥∥ =

∥∥f(u)− f(x)∥∥ = ‖x− u‖ � r,

so f(u) ∈ Br(x). Thus f : Br(x) −→ Br(x) and by Problem 1.106, wehave that

f(Br(x)

)= Br(x).

Then, Proposition 1.76, implies that f : Br(x) −→ Br(x) is a homeo-morphism.

Let u ∈ RN and consider the closed ball Br(x) with r = ‖u−x‖+1.Then u ∈ Br(x) and from the previous part of the solution, we canfind v ∈ Br(x) such that

f(v) = u.

1.3. Solutions 131

Since u ∈ RN was arbitrary, we infer that f is surjective. Therefore,f is a homeomorphism on all of RN .

Solution of Problem 1.108Let {xn}n�1 ⊆ X be a sequence. As X is totally bounded (see Defini-

tion 1.70(b)), so we can find a finite cover of X by balls of radius 12 :

X =

N1⋃

i=1

B 12(a1i )

(for someN1 � 1, a1i ∈ X with i = 1, . . . , N1). Because the above cover

is finite, we can find a subsequence of {xn}n�1 denoted by{x(1)n

}n�1

such that all values of{x(1)n

}n�1

are in one of the balls of the abovecover. So, in fact

diam{x(1)n : n � 1

}� 1

2 .

Next we choose further subsequences proceeding by induction. Sup-

pose that we have already chosen subsequence{x(k)n

}k�1

with the prop-erty, that

diam{x(k)n : n � k

}� 1

k .

Because X is totally bounded, we can find a finite cover of X by ballsof radius 1

2(k+1) :

X =

Nk+1⋃

i=1

B 12(k+1)

(ak+1i )

(for some Nk+1 � 1, ak+1i ∈ X with i = 1, . . . , Nk+1). Because the

above cover is finite, we can find a subsequence of{x(k)n

}n�k

denoted

by{x(k+1)n

}n�k+1

such that all values of{x(k+1)n

}n�k+1

are in one ofthe balls of the above cover. So, in fact

diam{x(k+1)n : n � k + 1

}� 1

k+1 .

Finally, let us take a “diagonal subsequence”{x(n)n

}n�1

⊆ {xn}n�1.

Note that{x(n)n

}n�1

is a Cauchy sequence (see Definition 1.7). To see


this, let ε > 0 be arbitrary and let us choose N > 1ε , N ∈ N. Then for

any m > n � N , we have

dX(x(n)n , x(m)

m ) � diam{x(N)k : k � N

}� 1

N < ε.

Solution of Problem 1.109Let ε > 0. For a given (x0, y) ∈ X × Y , by the continuity of f(·, ·), wecan find δ1(y), δ2(y) > 0 such that

ifdX(x, x0)<δ1(y)and dY

(u, y)<δ2(y), then dV

(f(x, u), f(x0, y)

)< ε

2

(see Definition 1.29). The balls{B

δ2(y)(y)

}y∈Y form an open cover of

Y . The compactness of Y (see Definition 1.63) implies that, we can

find a finite subcover{B

δ2(yn)(yn)

}N

n=1of Y . Let

δ1 = min{δ1(yn) : 1 � n � N

}

and let u ∈ Y . Assume that

dX(x, x0) < δ1

and n ∈ {1, . . . , N} is such that

u ∈ Bδ2(yn)

(yn).

Then

dX(x, x0) < δ1 � δ1(yn) and d

Y(u, yn) < δ2(yn)

and sodV

(f(x, u), f(x0, yn)

)< ε

2 .

It also follows that

dV

(f(x0, u), f(x0, yn)

)< ε

2 .

Then using the triangle inequality, we conclude that

dV

(f(x, u), f(x0, u)

)< ε ∀ u ∈ Y, with d

X(x, x0) < δ1,

1.3. Solutions 133

so

f(x, ·) ⇒ f(x0, ·) when x −→ x0 in X.

Solution of Problem 1.110Let {xn}n�1 ⊆ X be a sequence such that xn −→ x in X. We chooseyn ∈ Y such that

f(xn, yn) � m(xn) +1n ∀ n � 1.

Since Y is compact, we can find a subsequence{ynk

}k�1

of {yn}n�1

such that

ynk−→ y ∈ Y.

The continuity of f implies that

f(xnk, ynk

) −→ f(x, y) as k → +∞

and from the definition of m, we have

m(x) � f(x, y).

Hence

m(x) � f(x, y) = limk→+∞

f(xnk, ynk

) � lim infk→+∞

m(xnk).

On the other hand, from the Weierstrass theorem (see Theorem 1.75),we know that we can find y = y(x) ∈ Y such that

m(x) = f(x, y).

We have

m(xn) � f(xn, y) ∀ n � 1

and so

lim supn→+∞

m(xn) � f(x, y) = m(x).


It follows that m(xnk) −→ m(x). Since every subsequence of{

m(xn)}n�1

has a further subsequence converging to m(x), we con-

clude that m(xn) −→ m(x) (Urysohn criterion for convergence; seeProblem 1.3). This implies that m is continuous.

Solution of Problem 1.111Since X×Y is compact (see Proposition 1.130), f is uniformly contin-uous (see Definition 1.45 and Proposition 1.77). So, for a given ε > 0,we can find δ = δ(ε) > 0 such that

dX (x1, x2) < δ and dY (y1, y2) < δ =⇒ dV

(f(x1, y1), f(x2, y2)

)< ε.

In particular keeping y ∈ Y fixed, we have

dX(x1, x2) < δ =⇒ d

V

(f(x1, y), f(x2, y)

)< ε,

sodX(x1, x2) < δ =⇒ d∞

(fx1 , fx2

)< ε.

This proves that the function x �−→ fx is continuous from X into(C(Y ;V ), d∞

). Since X is compact, the continuity of x �−→ fx im-

plies that the set {fx : x ∈ X} ⊆ C(Y ;V ) is compact (see Proposi-tion 1.74). So, by the Arzela–Ascoli theorem (see Theorem 1.84), weinfer that family {fx}x∈X is equicontinuous (see Definition 1.83).


(a) For any n � 1, let

Sn =

n∑

k=1

1

k.

Let C = {Sk : k � 1} be equipped with the natural metric inducedfrom R. Then the set C is complete (but not compact).

Let f : C � Sn �−→ Sn+1 ∈ C. Then f has no fixed point and forany n > m, we have

dC

(f(Sn), f(Sm)

)= Sn+1 − Sm+1 =

n+1∑

k=m+2

1

k<

n∑

k=m+1

1

k

= Sn − Sm = dC (Sn, Sm).

1.3. Solutions 135

(b) First we show the uniqueness of the fixed point. Indeed, if we canfind x, u ∈ X such that x �= u, f(x) = x and f(u) = u, then

dX (x, u) = dX

(f(x), f(u)

)< dX (x, u),

a contradiction. So, the fixed point (if it exists) is unique.Next we turn our attention to the existence of a fixed point. Con-

sider the function ϕ : X −→ R+, defined by

ϕ(x)def= dX

(f(x), x

).

Then, for all x, u ∈ X, we have

∣∣ϕ(x)− ϕ(u)∣∣ =

∣∣dX

(f(x), x

)− dX

(f(u), u

)∣∣

�∣∣d

X

(f(x), x

)− dX

(x, f(u)

)∣∣

+∣∣d

X

(x, f(u)

) − dX

(f(u), u

)∣∣

� dX

(f(x), f(u)

)+ dX (x, u) < 2dX (x, u),

so ϕ is Lipschitz continuous.

Since X is compact, by Theorem 1.75, we can find x0 ∈ X suchthat

ϕ(x0) = inf{ϕ(x) : x ∈ X

}.

Suppose that ϕ(x0) > 0. Then f(x0) �= x0 and so

ϕ(f(x0)

)= d

X

(f (2)(x0), f(x0)

)< d

X

(f(x0), x0

)= ϕ(x0),

which contradicts the minimality of x0. Therefore ϕ(x0) = 0 and sof(x0) = x0.

Solution of Problem 1.113Consider the (affine) function S, defined by

S(u)(t) = λ

b∫

0

k(t, s)u(s) ds + f(t) ∀ u ∈ X, t ∈ [0, b].


Clearly, the continuity of k implies that S(u) ∈ X for all u ∈ X. Forevery u, v ∈ X, we have

d∞(S(u), S(v)

)= max

0�t�b

∣∣λ

b∫

0

k(t, s)u(s) ds − λ

b∫

0

k(t, s)v(s) ds∣∣

= max0�t�b

|λ|∣∣b∫

0

k(t, s)(u(s)− v(s)

)ds∣∣.

The continuity of k on the compact [0, b] × [0, b] implies that thereexists M > 0 such that

∣∣k(t, s)∣∣ � M ∀ (t, s) ∈ [0, b] × [0, b]

(from the Weierstrass theorem; see Theorem 1.75). Then

d∞(S(u), S(v)

)� |λ|Md∞(u, v)b.

So, if |λ| < 1Mb , then

d∞(S(u), S(v)

)� kd∞(u, v) ∀ u, v ∈ X,

with k ∈ (0, 1).Since (X, d∞) is a complete metric space, invoking the Banach

fixed point theorem (see Theorem 1.49), we can find u ∈ X such that

u(t) = λ

b∫

0

k(t, s)u(s) ds + f(t) ∀ t ∈ [0, 1].

Solution of Problem 1.114Let {Un}n�1 be open dense subsets ofX (see Definitions 1.20 and 1.78).We need to show that

⋂

n�1

Un is dense in X

1.3. Solutions 137

or equivalently that for every nonempty open V ⊆ X, we have

V ∩ ( ⋂

n�1

Un

) �= ∅.

Since U1 is dense in X, we have

V ∩ U1 �= ∅.Because X is locally compact, we can find a nonempty relatively com-pact open set V1 such that

V 1 ⊆ V ∩ U1

(see Definition 1.78 and Proposition 1.80). Then due to the density ofU2, we have

V1 ∩ U2 �= ∅and so we can find a nonempty relatively compact open set V2 suchthat

V 2 ⊆ V1 ∩ U2.

By induction, we generate a whole sequence {Vn}n�1 of nonemptyrelatively compact open sets such that

V n ⊆ Vn−1 ∩ Un ∀ n � 1

(with V0 = V ). Note that the sequence {Vn}n�1 is decreasing and byProposition 1.67, we have

⋂

n�1

V n �= ∅.

Hence, if x ∈ ⋂

n�1V n, then we have x ∈ V ∩ ( ⋂

n�1Un

), which proves

that⋂

n�1Un is dense in X. Therefore X is a Baire metric space (see

Definition 1.81).

Solution of Problem 1.115For any given η > 0, we can find M = M(η) > 0 such that

f(x) � η ∀ x ∈ RN , ‖x‖ � M.


Let u ∈ RN , η = f(u) and M = M(η) > 0 be as above. Since f |BM

is continuous, it is bounded below (see Definition 1.6(b)) and we canfind x0 ∈ BM such that

f(x0) = infBM

f � f(u)

[see the Weierstrass theorem (Theorem 1.75)]. If x ∈ RN with

‖x‖ � M , then

f(x) � η = f(u) � f(x0)

and so we conclude that

f(x0) = infRN

f.

Solution of Problem 1.116Arguing by contradiction, suppose that X is not compact. Then wecan find a sequence {xn}n�1 ⊆ X with no convergent subsequence (see

Theorem 1.71). Then the set C ={xn : n � 1

}is closed in X and

let f : C −→ R be defined by f(xn) = n. Then f is continuous andso by the Tietze extension theorem (see Theorem 1.44), we can finda continuous extension f : X −→ R of f . Evidently f is a continuousfunction on X which does not attain its supremum, a contradiction toour hypothesis.

Solution of Problem 1.117(a) Since by hypothesis D ⊆ R

N is an Fσ-set (see Definition 1.57),we have that

D =⋃

n�1

Cn,

with Cn ⊆ RN being closed for all n � 1. Without any loss of gener-ality, we may assume that

Cn ⊆ Cn+1 ∀ n � 1

1.3. Solutions 139

(indeed we can always replace Cn byn⋃

k=1

Cn). Let

Cn = Cn ∩ [0, n]N ∀ n � 1.

Then, for every n � 1, the set Cn is compact and

D =⋃

n�1

Cn.

Thenf(D) = f

( ⋃

n�1

Cn

)=

⋃

n�1

f(Cn)

and for all n � 1, the set f(Cn) is compact (see Proposition 1.74)hence closed (see Proposition 1.69). Therefore f(D) is an Fσ-set too.

(b) No. To show this, we will construct a continuous functionf : R −→ R such that f(Z) = Q. The set Z is a Gδ-set (since Z isclosed in R) but Q is not a Gδ-set (see Problem 1.58). Let {qn}n�1 ⊆ Qbe an enumeration of Q. Then for every x ∈ R, we can find k ∈ Z suchthat k < x � k + 1 and we define

f(x) = qk+1 +(qk+1 − qk

)(x− k − 1).

Then, the function f : R −→ R is continuous and f(Z) = Q.


(a) We argue by contradiction. So, suppose that dist(K,C) = 0(see Definition 1.6). We can find a sequences {un}n�1 ⊆ K and{xn}n�1 ⊆ C such that

dX(un, xn) −→ 0.

Because of the compactness of K, by passing to a suitable subsequenceif necessary, we may assume that

un −→ u∗ ∈ K.


Then alsoxn −→ u∗

and because C is closed, we have u∗ ∈ C. Therefore u∗ ∈ K ∩ C, acontradiction to the hypothesis that K ∩ C = ∅.(b) No. To see this, let X = R2 and consider two sets

K ={(

x, 1x

): x > 0

}and C =

{(x, 0) : x � 0

}.

Then K and C are nonempty closed disjoint sets, but dist(K,C) = 0.

Solution of Problem 1.119Let r

def= 1

2dist(K,C) (see Definition 1.6). From Problem 1.118(a), weknow that r > 0. Let

Udef=

⋃

x∈KBr(x).

Then the set U is open (see Proposition 1.9(b)), K ⊆ U and C∩U = ∅(from the choice of r > 0).


(a) Note that D ⊆ X is compact. So, due to the continuity of f , theset f(D) is compact (see Proposition 1.74), hence closed (see Proposi-tion 1.69(a)). Since f(D) ⊆ f(D), it follows that

f(D) ⊆ f(D).

On the other hand the continuity of f implies that

f(D) ⊆ f(D)

(see Proposition 1.32(d)). So, we conclude that

f(D) = f(D).

1.3. Solutions 141

(b) The inclusion f(D) ⊆ f(D) is always true (see Proposi-tion 1.32(d)). The opposite inclusion can fail.

Let X = Y = D = R and f(x) = ex. Then

f(D) = (0,+∞) � [0,+∞) = f(D).


(a) Let X∗ be the completion of X (see Definition 1.50 and Theo-rem 1.51). Reasoning indirectly, suppose that X∗ �= X and letx∗ ∈ X∗ \X. Consider the function f : X −→ R, defined by

f(x) = dX∗ (x

∗, x) ∀ x ∈ X,

with dX∗ being the metric of the completion X∗. The function f is

continuous (see Proposition 1.36).Let D = X. Then

D = D and 0 �∈ f(D),

while 0 ∈ f(D) (because X is dense in X∗; Definition 1.20), so

f(D) � f(D),

a contradiction. Therefore X = X∗ and so X is complete.

(b) Yes. Let us proceed by contradiction and suppose that X is notcompact. Thus, there exists a sequence {xn}n�1 ⊆ X with no conver-gent subsequence. Note that the set

D ={xn : n � 1

}

is closed, and the function

f : D � xn �−→ 1

n∈ R

is continuous on D. Using Tietze extension theorem (see Theo-rem 1.44), we can extent f to a function continuous on X (still denotedby f). Note, that

f(D) = f(D) ={

1n : n � 1

}�

{1n : n � 1

}

= {0} ∪ {1n : n � 1

}= f(D),

a contradiction. So X is compact.


Solution of Problem 1.122(a) “=⇒”: This comes from Problem 1.120(a).

“⇐=”: This comes from Problem 1.121(b).

(b) The inclusion “=⇒” is always true as it was stated in Prob-lem 1.120(a).

The inclusion “⇐=” need not be true if we replace R by anothermetric space. Let Y = {0, 1} be with the discrete metric. Then anycontinuous function f : X −→ Y is constant and of course the condi-tion

f(D) = f(D)

is satisfied for any D ⊆ X. But X can be any metric space (notnecessarily compact).

Solution of Problem 1.123Since X is locally compact, for every x ∈ K, we can find r(x) > 0such that the ball Br(x)(x) is compact. The family

{Br(x)(x)

}x∈K is

an open cover of K. Because K is compact, we can find a finite set{xk}Nk=1 such that

K ⊆N⋃

n=1

Br(xn)(xn) = U.

The set U is open and

U =N⋃

n=1

Br(xn)(xn) ⊆N⋃

n=1

Br(xn)(xn) = K

and K is compact. Note that the function h : X −→ R+, defined by

h(x) = dist(x,U c)

is continuous (see Definition 1.6 and Problem 1.78) and h|K

> 0 (seeProblem 1.118). Invoking the Weierstrass theorem (see Theorem 1.75),we can find x0 ∈ K such that

h(x0) = dist(x0, Uc) = inf

{dist(x,U c) : x ∈ K

},

1.3. Solutions 143

so h(x0) = r > 0. We claim that Kr ⊆ U . Indeed, if v �∈ U , then

dist(v,K) = inf{dX(v, y) : y ∈ K

}� dist(v, U c) = h(x0) = r > 0.

So, v �∈ Kr and we have proved the claim. Finally note that the com-pactness of U implies that Kr is compact too (see Proposition 1.69).

Solution of Problem 1.124(a) Let

{(xn, f(xn)

)}n�1

⊆ Gr f (see Definition 1.132) be a sequenceand assume that

(xn, f(xn)

) −→ (x, y) in X × Y.

Due to the continuity of f , we have

f(xn) −→ f(x) in Y.

Hence f(x) = y and so

(x, y) =(x, f(x)

) ∈ Gr f,

which proves that the set Gr f ⊆ X × Y is closed.

To show that the converse is not in general true, let us considerthe function f : R −→ R, defined by

f(x) =

{1x if x �= 0,0 if x = 0.

Then Gr f is closed in R× R = R2 but f is discontinuous at x = 0.

(b) “=⇒”: This comes from part (a).

“⇐=”: Let xn −→ x in X. Then, because of the compactness ofY , we can find a subsequence

{xnk

}k�1

of {xn}n�1 such that

f(xnk) −→ y in Y.

Since Gr f is closed, we have y = f(x). So, every subsequence of{f(xn)}n�1 has a further subsequence, which converges to f(x). Thisimplies that f(xn) −→ f(x) (see Problem 1.3) and proves the conti-nuity of f .


Solution of Problem 1.125We proceed by contradiction. So, suppose that

[0, 1] =⋃

n�1

Cn,

with Cn being nonempty closed sets which are pairwise disjoint. Wecan find an open set U2 such that

C2 ⊆ U2 and U2 ∩ C1 = ∅(see Problem 1.119). Then some component D2 of U2 (the componentis of course a closed interval) satisfies

D2 ∩ C2 �= ∅.If D2 ⊆ U2, then we can find an open interval (d, d∗) such that

D2 ⊆ (d, d∗) ⊆ U2

(see Problem 1.119 with K = D2 and C = U c2), which implies that D2

is not a component of D2, a contradiction. Therefore D2 ⊆ U2 cannotoccur and we have D2 \ U2 �= ∅, hence

D2 \ C2 �= ∅.It follows that

∅ �= D2 \ C2 ⊆⋃

n�3

(D2 ∩Cn)

and so for some integer n � 3, we have D2 ∩ Cn �= ∅. We repeatthis argument to the closed interval D2 and this way by induction wegenerate a decreasing sequence of closed intervals {Dm}m�2 such that

Dm �= ∅, Dm+1 ⊆ Dm, and Dm ∩ Cm−1 = ∅ ∀ m � 2.

Therefore⋂

m�2Dm �= ∅ (see Proposition 1.67) and

( ⋂

m�2

Dm

) ∩ ( ⋃

n�1

Cn

)=

⋃

n�1

(Cn ∩ ( ⋂

m�2

Dm

))= ∅,

a contradiction.

1.3. Solutions 145


SinceN⋂

k=1

Ck = ∅, we have

N⋃

k=1

Uk = X, with Uk = Cck (k = 1, . . . , N).

So, if x ∈ X, then we can find k0 ∈{1, . . . , N

}such that x ∈ Uk0 . The

set Uk0 is open, hence there exists rx > 0 such that B2rx(x) ⊆ Uk0 .The family

{Brx(x)

}x∈X is an open cover of X, so there is a finite

subcover{Brxs (xs)

}M

s=1. Let

δ = min{rxs : s = 1, . . . ,M

}.

Let E ⊆ X be a set which intersects all the sets Ck for k = 1, . . . , N .Suppose that diamE < δ and let y ∈ E. Then we can find s0 ∈{1, . . . ,M

}and k0 ∈

{1, . . . , N

}such that

y ∈ Brxs0(xs0) ⊆ B2rxs0

(xs0) ⊆ Uk0 .

For every other point u ∈ E, we have

dX(u, xs0) � d

X(u, y) + d

X(y, xs0) < δ + rxs0

� 2rxs0,

soE ⊆ B2rxs0

(xs0) ⊆ Uk0

and thusE ∩ Ck0 = ∅,

a contradiction. Therefore diamE � δ.

Solution of Problem 1.127First we show that sup

R

f < +∞. If this is not the case, then for

every n � 1, we can find xn ∈ X such that f(xn) > n. Then{f−1

([n,+∞)

)}n�1

is a closed subsets of X with the finite intersection

property (see Definition 1.66). By Proposition 1.67, it has a nonemptyintersection. Let u be an element in that intersection. Then

f(u) � n ∀ n � 1,


a contradiction. Hence

supR

f = M < +∞.

Next, let

Cn ={x ∈ X : f(x) � M − 1

n

}= f−1

([M − 1

n ,+∞)) ∀ n � 1.

By hypothesis, each Cn is a closed set in X and

Cn ⊇ Cn+1 ∀ n � 1.

It follows that ⋂

n�1

Cn �= ∅

(see Proposition 1.67). We choose x0 ∈⋂

n�1Cn. Then

f(x0) = M = supR

f < +∞.

Solution of Problem 1.128Let {xn}n�1 be a Cauchy sequence in X. The sequence {xn}n�1 isbounded, i.e.,

diam{xn : n � 1

}< +∞.

So, we can find r > 0 and y ∈ X such that {xn}n�1 ⊆ Br(y). But byhypothesis the latter is compact. Hence, we can find a subsequence{xnk

}k�1

of {xn}n�1 such that

xnk−→ x ∈ Br(y) as k → +∞.

Then xn −→ x and this proves the completeness of X. Recall thatevery compact set is closed and bounded (see Proposition 1.69(a)).On the other hand every closed and bounded set, is a closed subset ofsome closed ball, hence it is compact (see Proposition 1.69(b)).

1.3. Solutions 147

Solution of Problem 1.129By hypothesis, there exist β > 0 and M = M(β) > 0 such that

f(x) � β‖x‖ ∀ ‖x‖ � M.

For a given λ ∈ R. Let r = max{M, λβ

}. Then

Lλ ={x ∈ RN : f(x) � λ

} ⊆ Br(0),

so Lλ is compact (see Proposition 1.69(b)).

Solution of Problem 1.130Since E ⊆ Br(0) for some r > 0, it suffices to prove the problem forthe case when E = Br(0). We set

‖f‖∞ = maxt∈[0,1]

∣∣f(t)

∣∣ ∀ f ∈ C

([0, 1]

)

and‖G‖∞ = max

{∣∣G(t, s)∣∣ : (t, s) ∈ [0, 1] × [0, 1]

}.

Then for all t ∈ [0, 1], we have

∣∣L(f)(t)

∣∣ �

1∫

0

∣∣G(t, s)

∣∣∣∣f(s)

∣∣ ds � ‖G‖∞‖f‖∞ � r‖G‖∞,

soL(Br(0)

) ⊆ Br‖G‖∞(0).

Also, if f ∈ Br(0) and t, τ ∈ [0, 1], then

∣∣L(f)(t)− L(f)(τ)∣∣ �

1∫

0

∣∣G(t, s)−G(τ, s)∣∣∣∣f(s)

∣∣ dz

� r

1∫

0

∣∣G(t, s) −G(τ, s)

∣∣ ds.

Note that [0, 1] × [0, 1] is compact and by hypothesis G is continuouson [0, 1] × [0, 1]. Hence it is uniformly continuous (see Definition 1.45


and Proposition 1.77). Therefore, for a given ε > 0, we can findδ = δ(ε) > 0 such that

∣∣G(t, s) −G(τ, s)

∣∣ < ε

r ∀ s, t, τ ∈ [0, 1], with |t− τ | < δ.

Then∣∣L(f)(t)−L(f)(τ)

∣∣ < ε ∀ f ∈ Br(0), t, τ ∈ [0, 1], with |t−τ | < δ,

so L(Br(0)

)is equicontinuous. Invoking the Arzela–Ascoli theorem

(see Theorem 1.84), we conclude that

L(Br(0)

)is compact in C

([0, 1]

).

Solution of Problem 1.131No. According to the Arzela–Ascoli theorem (see Theorem 1.84), ifthe set

{sin(nx) : n � 1

}is relatively compact (for the d∞ metric),

it must be equicontinuous (uniformly since [−π, π] is compact) anduniformly bounded (see Definition 1.83). Clearly

{sin(nx) : n � 1

}is

uniformly bounded. However, for a given δ, we can find n0 � 1 largeenough such that π

n0< δ. If

x = − π2n0

and y = π2n0

,

then

|x− y| = πn0

< δ and∣∣ sin(n0x)− sin(n0y)

∣∣ = 2.

Therefore the set{sin(nx) : n � 1

}is not equicontinuous and so it

cannot be relatively compact in C([−π, π]

).

Solution of Problem 1.132Let x ∈ E. Since by hypothesis E is locally compact with respectto the metric d1

X(see Definition 1.78), we can find an open set U in

(E, d1X) such that U

Eis compact in (E, d1

X) (by U

Ewe denote the

closure of U in (E, d1X)). Since

UE

= E ∩ U,

1.3. Solutions 149

it follows that UE

is compact in (X, d1X). Also since U is open in

(E, d1X), we can find a set V open in (X, d1

X) such that

U = E ∩ V.

Since spaces (X, d1X) and (X, d2

X) are topologically equivalent (see Def-

inition 1.52), we have that UE

is compact in (X, d2X) and V is open

in (X, d2X). Hence U = E ∩ V is open in (E, d2

X) and U

Eis compact

in (E, d2X). Because x ∈ E was arbitrary, we conclude that (E, d2

X) is

locally compact.

Solution of Problem 1.133Let r > 0 be arbitrary. We will show that the closed ball Br

((0, 0)

)

is not compact. Let s = min{r2 , 1

}. Let us fix x0 ∈ (0, s) such that

sin 1x0

= s. Let

xk =1

1x0

+ 2kπ∀ k = 1, 2, . . . .

Then

sin 1xk

= sin(

1x0

+ 2kπ)

= sin 1x0

= s,

so limk→+∞

(xk, sin

1xk

)= (0, s) �∈ X and the sequence

{(xk, sin

1xk

)}k�1

has no limit in Br

((0, 0)

). Thus the closed ball Br

((0, 0)

)is not com-

pact. This proves that X is not locally compact (see Definition 1.78).

Solution of Problem 1.134(a) Let y ∈ f(X). Then y = f(x) for some x ∈ X. Since X islocally compact, we can find r > 0 such that Br(x) is compact. Sinceby hypothesis f is open, the set f

(Br(x)

)is open and y ∈ f

(Br(x)

).

Also f(Br(x)

)is compact (see Proposition 1.74), hence closed in Y .

By the continuity of f , we have

f(Br(x)

) ⊆ f(Br(x)

)


(see Proposition 1.32(d)) and since f(Br(x)

)is compact, we have

f(Br(x)

)= f

(Br(x)

)

(see Problem 1.120). Therefore, f(Br(x)

)is a relatively compact

neighbourhood of y ∈ f(X). Since y ∈ f(X) was arbitrary, we con-clude that f(X) is relatively compact.

(b) No. Let

X ={(−1, 0)

} ∪ ((0,+∞) × R),

Y ={(0, 0)

} ∪ ((0,+∞)× R),

both with the natural metric induced by the Euclidean metric in R2

and let f : X −→ Y be defined by

f(x) =

{x if x ∈ (0,+∞) ×R,(0, 0) if x = (−1, 0).

Then f is a continuous bijection (but not open), X is locally compactand Y = f(X) is not locally compact.

(c) No. Let X = [0,+∞) (with the natural metric induced by theEuclidean metric in R) and let

Y ={(0, 0)

} ∪ {(x, sin 1

x

): x > 0

}

(with the natural metric induced by the Euclidean metric in R2). Letf : X −→ Y be the function, defined by

f(x) =

{ (x, sin 1

x

)if x > 0,

(0, 0) if x = 0.

The f is an open bijection (but not continuous), X is locally compactand Y = f(X) is not locally compact (see Problem 1.133).

Solution of Problem 1.135From Proposition 1.80, we know thatX is an open subset of its comple-tion. Invoking the Alexandrov theorem (see Theorem 1.58), we inferthat X is topologically complete. In particular, it is homeomorphic(see Definition 1.39) to a complete metric space (see Remark 1.53).

1.3. Solutions 151

Solution of Problem 1.136If Q is homeomorphic (see Definition 1.39) to a complete metric space,then Q is a Baire metric space(see Problem 1.114). Let {qn}n�1 be anenumeration of Q. Then for each n � 1, the set Q \ {qn} is open anddense (see Definition 1.20). Hence the set

⋂

n�1

(Q \ {qn}

)is dense in

Q. But ⋂

n�1

(Q \ {qn}

)= ∅,

a contradiction.

Solution of Problem 1.137Let x ∈ X. Then x ∈ U for some U ∈ Y. Since U is open, we can findrx > 0 such that B2rx

(x) ⊆ U . The balls{B2rx

(x)}x∈X form an open

cover of X and so we can find a finite subcover{B2rxk

(xk)}n

k=1. Let

δ = min{rxk

}n

k=1. We will show that this is the desired δ > 0. To this

end, let A ⊆ X be a nonempty set with diamA < δ. Let u ∈ A. Wecan find k ∈ {1, . . . , n} such that u ∈ Brxk

(xk). Let y ∈ A. Using thetriangle inequality, we have

dX (y, xk) � dX (y, u) + dX (u, xk) < δ + rxk� 2rxk

,

so y ∈ B2rxk(xk) ⊆ U and thus A ⊆ U .

Solution of Problem 1.138Let f : R −→ R be the function, defined by

f(x) =

⎧⎪⎨

⎪⎩

0 if x is irrational or x = 0,(−1)q(q−1)

q if x ∈ Q \ {0} and x = pq with (p, q)

irreducible representation of x and q > 0.

For every rational r �= 0, we have

∣∣f(r)

∣∣ � q−1

q < 1.


So, f is bounded above by 1 and bounded below by −1. Let T be acompact interval with nonempty interior. Let

A ={q > 0 : (p, q) is the irreducible representation

with positive denominator of a rational x ∈ T}.

We will show that A is unbounded. Let M > 0 be arbitrary and wewill find k > M such that k ∈ A. Let T = [a, b] with a < b and letx = p

q ∈ (a, b) with (p, q) being irreducible representation of x, withq � 2. Note that the function

ϕ(t) = tptq+p ∀ t ∈ R \ {− p

q

}

is continuous and

limt→±∞ϕ(t) = p

q .

So, we can choose r ∈ N large such that

a < qrpqrq+p < b and qrq + p > M.

But note that the fraction

qrpqrq+p is irreducible.

So k = qrq+p satisfies k ∈ A and k > M . Hence A ⊆ N is unbounded.Since n−1

n −→ 1 as n → +∞, we infer that

−1 ∈ inf f(T ) and 1 = sup f(T ).

But

−1 < f(x) < 1 ∀ x ∈ T.

Solution of Problem 1.139Suppose that (X, d

X) is totally bounded metric space (see Defini-

tions 1.70 and 1.21(d)). Then for every n � 1, we can find a finiteset Fn ⊆ X such that

X =⋃

x∈Fn

B 1n(x),

1.3. Solutions 153

where B 1n(x) =

{u ∈ X : dX (u, x) <

1n

}. We set

D =⋃

n�1

Fn.

Then D is countable and dense in X (see Definition 1.20). ThereforeX is separable.

Solution of Problem 1.140“(a) =⇒ (b)”: Since the identity i : (X, d

X) −→ (X, d

X) is a homeo-

morphism (see Remark 1.53 and Definition 1.52), then (X, dX) is com-

pact (see Proposition 1.74), hence complete too (see Theorem 1.71).“(b) =⇒ (a)”: We argue by contradiction. So, suppose that (X, d

X)

is not compact. We may assume that dX

is bounded and more pre-

cisely dX� 1 (if this is not the case, we replace d

Xby

dX

1+dX; note that

the new metric is topologically equivalent (or even uniformly equiva-lent) to the previous one (see the solution of Problem 1.70) and thenew metric still remains non-compact (see Remark 1.53 and Proposi-tion 1.74)). Since we assume that (X, d

X) is not compact, we can find

a sequence {xn}n�1 ⊆ X with no convergent subsequence.Let

L ={f : X −→ R :

∣∣f(x)− f(u)∣∣ � d

X(x, u) for all x, u ∈ X

}.

Then

dX(x, u) = sup

f∈L

∣∣f(x)− f(u)∣∣.

Let

Ln ={f : X −→ R :

∣∣f(x)−f(u)

∣∣ � 1

ndX(x, u) and f(xk) = 0 ∀k > n

}

and let us set

L =⋃

n�1

Ln.

We define

dX(x, u) = sup

f∈L

∣∣f(x)− f(u)

∣∣.


Claim 1. {xn}n�1 is a Cauchy sequence in (X, dX ).

Let ε > 0, N � 1ε , m,k � N and f ∈ L. If f ∈ Ln and n � N ,

then

∣∣f(xm)− f(xk)

∣∣ � 1

ndX(xm, xk) � 1

n � 1N � ε.

If f ∈ Ln and n < N , then m,k > n and so

f(xm) = f(xk) = 0.

HencedX (xm, xk) � ε ∀ m,k � N.

So, we conclude that {xn}n�1 is a Cauchy sequence in (X, dX). This

proves Claim 1.

Claim 2. dX

is topologically equivalent to dX.

Since dX

� dX, the identity function i : (X, d

X) −→ (X, d

X) is

continuous. We consider the inverse function

i−1 : (X, dX) −→ (X, d

X).

Let u ∈ X and ε > 0. By hypothesis u is not an accumulation point ofthe sequence {xn}n�1 (see Definition 1.13(b) and Theorem 1.14(d)).So, we can find n0 � 1 such that

dX (xn, u) � ε ∀ n � n0.

Letf(x) = 1

n0

(ε− d

X(x, u)

)+.

Evidently f is 1n0-Lipschitz continuous and so f ∈ Ln0 . Let δ = ε

n0> 0.

Suppose thatdX(x, u) < δ.

Then

1n0

∣∣ε− (

ε− dX(x, u)

)+∣∣ =∣∣f(x)− f(u)

∣∣ � d

X(x, u) < δ,

sodX(x, u) < ε.

This proves the continuity of i−1. Therefore dX

and dX

are topologi-cally equivalent. This proves Claim 2.

1.3. Solutions 155

But the sequence {xn}n�1 ⊆ (X, dX ) has no accumulation point(since from Definition 1.52, the sequence {xn}n�1 has the same accu-

mulation points in (X, dX) and in (X, d

X)) and so by Claim 1, the

metric space (X, dX) is not complete, a contradiction.

Solution of Problem 1.141Arguing by contradiction, suppose that we could find two sequences{xn}n�1 ⊆ K and {un}n�1 ⊆ X such that

dX (xn, un) −→ 0

and ε > 0 such that

dY

(f(xn), f(un)

)> ε ∀ n � 1.

Since K is compact and {xn}n�1 ⊆ K, by passing to a subsequence ifnecessary, we may assume that xn −→ x ∈ K. Then

un −→ x in X.

So,

f(un) −→ f(x) in Y and f(xn) −→ f(x) in Y.

Thus, by the triangle inequality, we have

dY

(f(xn), f(un)

) −→ 0,

a contradiction.

Solution of Problem 1.142(a) Recall that if B is compact, then B is totally bounded (see Theo-rem 1.71). So, α(B) = 0 by Definition 1.70(b).

(b) Let B ⊆ X be a set such that α(B) = 0. Let ε > 0. From thedefinition of α, we can find a finite family of sets A1, . . . , An such that

B ⊆n⋃

k=1

Ak with diamAk � ε2 for all k ∈ {1, . . . , n} .


Let us fix xk ∈ Ak for all k ∈ {1, . . . , n}. Because diamAk � ε2 , we

haveAk ⊆ Bε(xk) ∀ k ∈ {1, . . . , n} ,

Thus

B ⊆n⋃

k=1

Ak ⊆n⋃

k=1

Bε(xk)

and so the set B is totally bounded (see Definition 1.70(b)).

(c) If

B′ ⊆n⋃

k=1

Ck with diamCk � d for all k ∈ {1, . . . , n} ,

then

B ⊆n⋃

k=1

Ck

and so α(B) � α(B′).

(d) Since B ⊆ B, from part (c), we have α(B) � α(B). On the other

hand, if B ⊆n⋃

k=1

Ck, then

B ⊆n⋃

k=1

Ck =n⋃

k=1

Ck

and since diamCk = diamCk, it follows that α(B) � α(B), henceα(B) = α(B).

(e) First note that

α( ⋂

n�1

Cn

)� α(Ck) ∀ k � 1

(see part (c)) and so

α( ⋂

n�1

Cn

)= 0,

thus in particular the set⋂

n�1

Cn is totally bounded (see (b)). Hence

⋂

n�1

Cn being also closed is complete, thus compact (see Theorem 1.71).

1.3. Solutions 157

Passing to a subsequence if necessary, we may assume that

α(Ck) � 1k+1 ∀ k � 1

(cf. Problem 1.4). The set C1 can be covered by a finite family ofsets: A1,1, . . . , A1,s1 , all of them of diameter less or equal to 1. Wecan assume that sets A1,1, . . . , A1,s1 are closed (as for any set A, wehave that A ⊆ A and diamA = diamA). Note that there existsi1 ∈ {1, . . . , s1} such that

A1,i1 ∩ Ck �= ∅ ∀ k � 1.

Indeed, if this is not the case, then we can find k0 � 1 such that

A1,i ∩ Ck = ∅ ∀ k � k0, i ∈ {1, . . . , s1} .

But sets A1,1, . . . , A1,s1 cover C1, so they also cover Ck0 (becauseCk0 ⊆ C1), a contradiction.

Let us replace sets C1, C2, . . . by sets A1,i1 ∩ C1, A1,i1 ∩ C2, . . . ob-taining a decreasing sequence of nonempty closed sets (still denotedby {Ck}k�1), with α(Ck) −→ 0 and additionally such that

diam (Ck) � 1 ∀ k � 1.

Now, we proceed by induction. So, suppose that we have changed setsC1, C2, . . . is such a way that they are nonempty, closed, Ck+1 ⊆ Ck

for k � 1,

diamCk � 1k ∀ k ∈ {1, . . . , n} ,

α(Ck) � 1k+1 ∀ k � 1

and α(Ck) −→ 0.Now the set Cn+1 can be covered by a finite family of sets:

An+1,1, . . . , An+1,sn+1 , all of them of diameter less or equal to 1n+1 .

We can assume that sets An+1,1, . . . , An+1,sn+1 are closed. Note thatthere exists in+1 ∈ {1, . . . , sn+1} such that

An+1,in+1 ∩Ck �= ∅ ∀ k � 1.


Let us replace sets Cn+1, Cn+2, . . . by sets

An+1,in+1 ∩ Cn+1, An+1,in+1 ∩ Cn+2, . . .

obtaining a decreasing sequence of nonempty closed sets (still denotedby {Ck}k�1), with α(Ck) −→ 0 and additionally such that

diamCk � 1k ∀ k ∈ {1, . . . , n+ 1} ,

α(Ck) � 1k+1 ∀ k � 1

and α(Ck) −→ 0.So, finally, we obtain a decreasing sequence of nonempty closed

sets {Cn}n�1 such that

diamCn � 1n ∀ n � 1.

Let us choose xn ∈ Cn for all n � 1. By the above procedure, we knowthat

dX(xn, xm) � 1

n ∀ m � n,

so {xn}n�1 ⊆ X is a Cauchy sequence. But by hypothesis, X iscomplete. So,

xn −→ x ∈ X

and x ∈ Cn for all n � 1. Hence

x ∈⋂

n�1

Cn �= ∅.


(a) Let xn −→ x in X. Then by hypothesis the sets

{f(x)

} ∪ {f(xn) : n � k

} ⊆ Y

are compact for all k � 1. From Proposition 1.67, we have that⋂

n�1

({f(xn) : n � k

} ∪ {f(x)

}) �= ∅.

Since f is injective,⋂

n�1

{f(xn) : n � k

}= ∅

1.3. Solutions 159

and so ⋂

n�1

({f(xn) : n � k

} ∪ {f(x)

})=

{f(x)

}.

From Problem 1.94, we have

diam({

f(xn) : n � k}) −→ diam

{f(x)

}= 0 as k → +∞,

thus{f(xn)

}n�1

is a Cauchy sequence. As the set{f(x)

} ∪ {f(xn) : n � 1

}

is complete (being compact), we have that f(xn) −→ y for some y andof course y = f(x), i.e., f is continuous.

(b) Let X = [0, 1] and Y = {0, 1} and

f(x) =

{0 if x = 0,1 if x ∈ (0, 1].

Then f maps compact sets into compact sets, but f is neither injectionnor continuous.

Solution of Problem 1.144(a) For every k ∈ {1, . . . ,m} and n � 1, let

Unk =

{x ∈ X : dist(x,Ck) <

1n

}.

These are open sets and Ck ⊆ Unk . Suppose that for every n � 1, we

havem⋂

k=1

Unk �= ∅. Then we can find xn ∈

m⋂

k=1

Unk and so

dist(xn, Ck) < 1n ∀ k = 1, . . . ,m

(see Definition 1.6). Since X is compact and {xn}n�1 ⊆ X, so passingto a subsequence if necessary, we may assume that xn −→ x. Thendist(x,Ck) = 0 for all k = 1, . . . ,m, so x ∈ Ck for all k = 1, . . . ,m

(since each Ck is closed) and thus x ∈m⋂

k=1

Ck, a contradiction. So, for

some n � 1, we must havem⋂

k=1

Unk = ∅ ∀ k = 1, . . . ,m

and of course Ck ⊆ Uk for all k ∈ {1, . . . ,m}.


(b) Yes. In the above proof of part (a), we have used the compactnessof X. But the result can be proved in another way (and the compact-ness is in fact not needed).

For k ∈ {1, . . . ,m}, let

fk(x) = dist(x,Ck) ∀ x ∈ X.

We know that functions fk are continuous (in fact Lipschitz continuous;see Problem 1.78). Let us define

Uk ={x ∈ X : fk(x) <

1m

m∑

i = 1i = k

fi(x)} ∀ k ∈ {1, . . . ,m} .

The sets Uk are open and Ck ⊆ Uk for all k ∈ {1, . . . ,m}. We willshow that

m⋂

k=1

Uk = ∅.

If in contrast, we assume that there exists x ∈m⋂

k=1

Uk, then

fk(x) < 1m

m∑

i = 1i = k

fi(x) ∀ k ∈ {1, . . . ,m} .

Adding the above inequalities for k = 1, . . . ,m, we get

m∑

k=1

fk(x) <m− 1

m

m∑

i=1

fi(x),

a contradiction. This ends the proof.

Solution of Problem 1.145Arguing by contradiction, suppose that we cannot find such Lipschitzconstant. Then for every n � 1, we can find un, u

′n ∈ X such that

∥∥f(un)− f(u′n)

∥∥ > n‖un − u′n‖. (1.6)

1.3. Solutions 161

LetM = max

u∈C∥∥f(u)

∥∥

(M is finite by the Weierstrass theorem; see Theorem 1.75).From (1.6), we have

‖un − u′n‖ � 2Mn

and so ‖un −u′n‖ −→ 0. On the other hand, by the compactness of Kand by passing to a suitable subsequences if necessary, we may assumethat

un −→ u in K and u′n −→ u′ in K.

We infer that u = u′. Let Uu be an open set containing u and ku > 0such that

∥∥f(x)− f(x′)∥∥ � ku‖x− x′‖ ∀ x, x′ ∈ Uu.

We can find n0 � 1 such that un, u′n ∈ Uu for all n � n0 and so

∥∥f(un)− f(u′n)∥∥ � ku‖un − u′n‖ ∀ n � n0

which contradicts (1.6).

Solution of Problem 1.146Let {xn}n�1 ⊆ proj

X(C) be a sequence and suppose that xn −→ x. For

every n � 1, we can find yn ∈ Y such that (xn, yn) ∈ C for all n � 1.The compactness of Y implies that by passing to a subsequence ifnecessary, we may assume that yn −→ y in Y . Then (xn, yn) −→ (x, y)in X × Y and (x, y) ∈ C (since C is by hypothesis closed). Hencex ∈ proj

X(C) and so we conclude that proj

X(C) is closed in X.


(a) No. Let

D ={

1n : n ∈ N+

}and X = D ∪ {0}

(with the natural metric induced from R). Then D has property Sand it is dense in X (see Definition 1.20), while X has not property S.


(b) No. Let X = R and D = Q. Clearly X has property S but Ddoes not.

Solution of Problem 1.148First note that the set A is closed (see Definition 1.13(b), Theo-rem 1.14(d) and Problem 1.24(b)).To prove that A is connected, weargue by contradiction. So, suppose that A is not connected, thusA = C1 ∪ C2, with C1, C2 ⊆ X being nonempty, closed and disjointsets (see Definition 1.85 and Remark 1.86). Sets C1 and C2 are alsoclosed in X and so they are compact. Thus there exists ε > 0 suchthat

dist(C1, C2) = inf{dX (u, u

′) : u ∈ C1, u′ ∈ C2

}= 3ε > 0

(see Definition 1.6 and Problem 1.118). We set

D1 = (C1)ε ={x ∈ X : dist(x,C1) < ε

}

D2 = (C2)ε ={x ∈ X : dist(x,C2) < ε

}.

Then, the sets D1 and D2 are open and dist(D1,D2) � ε > 0. Induc-tively, we can generate a subsequence {xnk

}k�1 of {xn}n�1 such that{xnk

}k�1 ⊆ X \ (D1 ∪ D2). Because dX(xn+1, xn) −→ 0, then there

exists n � 1 such that

dX (xn+1, xn) < ε ∀ n � n.

First let l,m � n0 be such that xl ∈ D1 and xm ∈ D2. We choosen1 ∈ (l,m) such that xn1 ∈ X \ (D1 ∪ D2) (the existence of such anelement xn1 comes from the fact that dist(D1,D2) � ε). Then, supposethat we have produced xn1 , xn2 , . . . , xnk

. Then, let l,m � nk + 1 besuch that xl ∈ D1 and xm ∈ D2. We choose nk+1 ∈ (l,m) such thatxnk

∈ X \ (D1 ∪D2).Because the set X \ (D1 ∪ D2) is compact (as a closed subset of

a compact metric space; see Proposition 1.69(b)), and{xnk

}k�1

⊆X \(D1∪D2), there exists a convergent subsequence of

{xnk

}k�1

with

1.3. Solutions 163

a limit x ∈ X \(D1∪D2). Thus x is an accumulation point of {xn}n�1

and x �∈ A, a contradiction to the definition of A.

Solution of Problem 1.149(a) Since f is continuous, the set f([0, 1]) is a closed interval [a, b] ⊆[0, 1]. Let

h(x) = x− f(x).

Evidently h is continuous and

h(a) = a− f(a) � 0 � b− f(b) = h(b).

Then the intermediate value theorem (Bolzano theorem; see also Theo-rem 1.90) implies that there exist x ∈ [0, 1] such that h(x) = 0, hencef(x) = x.

(b) No. Let f : (0, 1) −→ (0, 1) be the function, defined by

f(x) = x2.

Then, f is continuous, but it has no fixed point.

(c) No. Let X = [−2,−1] ∪ [1, 2] and let f : X −→ X be defined byf(x) = −x. Then, f is continuous, but it has no fixed point.

Solution of Problem 1.150We argue indirectly. So, suppose that A is not a singleton and letx, u ∈ A with x �= u. We may assume that x < u. Let y ∈ R \ Q besuch that x < y < u. Let

U = (−∞, y) ∩A and V = (y,+∞) ∩A.

Both U and V are nonempty open subsets of A such that A = U ∪ V(as y �∈ A). So, A admits a separation and this contradicts the hy-pothesis that A is connected (see Definition 1.85). This proves that Ais a singleton.


Solution of Problem 1.151“(a) =⇒ (b)”: Since X is disconnected, X = U ∪ V with U and Vbeing nonempty disjoint open sets (see Definition 1.85).

Let h : X −→ {0, 1} be defined by

h(x) =

{0 if x ∈ U,1 if x ∈ V.

Clearly h is surjective. We claim that h is continuous. The open setsof the discrete space {0, 1} are: ∅, {0}, {1} and {0, 1}. We have

h−1(∅) = ∅, h−1({0}) = U, h−1({1}) = V, h−1({0, 1}) = X.

So, all these inverse images are open sets in X and this implies that his continuous (see Proposition 1.32).

“(b) =⇒ (a)”: Let U = h−1({0}) and V = h−1({1}). Both arenonempty (since h is surjective) and open (since h is continuous).Moreover U ∩ V = ∅ and X = U ∪ V . Therefore X is disconnected(see Definition 1.85).

Solution of Problem 1.152(a) Suppose that RN \{0} is disconnected. So, we can find nonemptydisjoint open sets U and V such that RN \ {0} = U ∪ V . Let x ∈ Uand y ∈ V . Consider the line interval joining x and y, i.e.,

[x, y] ={z = (1− t)x+ ty : 0 � t � 1

}.

If 0 �∈ [x, y], then [x, y] ∈ RN \{0} and we contradict the fact that [x, y]is connected (as the continuous image of the interval [0, 1] ⊆ R whichis connected; see Proposition 1.88 and Theorem 1.90). So, 0 ∈ [x, y].Let z ∈ RN be such that it is not on a straight line through x andy. Then z ∈ RN \ {0} and so z ∈ U or z ∈ V . If z ∈ U , then wereplace x by z and we consider [z, y] ⊆ R

N \ {0}, a contradiction tothe fact that [z, y] is connected. Similarly, if z ∈ V , then we consider[x, z] ⊆ RN \ {0}, again a contradiction. This proves that RN \ {0} isconnected (cf. Theorem 1.90).

(b) Suppose that R and RN \ {0} are homeomorphic (see Defini-tion 1.39). Let h be a homeomorphism such that h(0) = 0 (clearly

1.3. Solutions 165

we can always have this). Then R \ {0} and RN \ {0} are homeo-morphic under h, a contradiction since R \ {0} is disconnected, whileRN \ {0} is connected.

Solution of Problem 1.153We argue by contradiction. So, suppose that for some u ∈ R, the setϕ−1

({u}) is bounded. Then we can find R > 0 such that ϕ−1({u}) ⊆

BR(0). We have u �∈ ϕ(RN \BR(0)

). The set RN \BR(0) is connected

and so by the continuity of ϕ, the set ϕ(RN \BR(0)

)is connected (see

Theorem 1.90), hence it is an interval I ⊆ R (see Proposition 1.88).Because u �∈ I, we have I ⊆ (−∞, u) or I ⊆ (u,+∞). To fix thingsassume that I ⊆ (u,+∞) (the argument is similar if I ⊆ (−∞, u)).Note that ϕ

(BR(0)

) ⊆ R is compact (see Proposition 1.74) and so itis bounded above by some λ ∈ R. Then

ϕ(RN ) = ϕ(BR(0)

) ∪ ϕ(RN \BR(0)

) ⊆ (−∞, μ],

with μ = max{λ, u}, which contradicts the hypothesis that ϕ is sur-jective.

Solution of Problem 1.154No. If h : [0, 1] −→ (0, 1) is a homeomorphism, then [0, 1]\{1} = [0, 1)must be homeomorphic to (0, 1) \ {

h(1)}. But [0, 1] \ {1} = [0, 1)

is connected, while (0, 1) \ {h(1)

}is not (it is not an interval; see

Proposition 1.88 and Theorem 1.90). So [0, 1] and (0, 1) cannot behomeomorphic.

Solution of Problem 1.155Yes. Let u0 =

(uk0 = e

)Nk=1

∈ RN . Let u ∈ E and suppose that

ui ∈ R \Q, for some i ∈ {1, . . . , N}. Then we connect u0 and u ∈ RN ,the vector with all components equal to e except ui = ui, by theline segment starting from u0 and ending at u. Evidently this linesegment lies in E. Next connect u and u again with the line segment


determined by these two vectors. This line segment is also in E (sincethe ith component of all its elements is ui ∈ R\Q). Therefore the pathfrom u0 to u which results by concatenating the two line segments isin E and so by Proposition 1.98, E is path-connected.

Solution of Problem 1.156(a) “(i) =⇒ (ii)”: Since X is locally connected (see Definition 1.95),for every x ∈ X there is a connected open set Ux containing x ∈ Xsuch that

Ux ⊆ B ε2(x).

The family {Ux}x∈X is an open cover of X and since X is compact,we can find a finite subcover {Uxk

}mk=1. So,

X =

m⋃

k=1

Uxk, Uxk

is connected and diamUxk

� diamB ε2(xk) = ε.

“(ii) =⇒ (i)”: Let ε > 0 and let {Ek}mk=1 be a finite cover of X witheach Ek connected and

diamEk � ε.

Since Ek is connected too (see Proposition 1.89) and diamEk =diamEk, we may assume that each Ek is closed. Let u ∈ X andlet

Iu ={k ∈ {1, . . . ,m} : u ∈ Ek

},

E =⋃

k∈IuEk and U =

⋂

k∈IcuEc

k.

Evidently U is open and u ∈ U ⊆ E. Also E is connected (see Theo-rem 1.91). Also, if x ∈ E, then x ∈ Ek for some k ∈ Iu and so

dX(x, u) � diamEk � ε.

Hence E ⊆ Bε(u). Therefore u admits a local basis consisting ofconnected sets. Since u is arbitrary, we conclude that X is locallyconnected.

1.3. Solutions 167

(b) The implication “(ii) =⇒ (i)” remains true, as we do not exploitthe compactness of X in the above proof.

The implication “(i) =⇒ (ii)” does not hold. Let X = R. Then Xis locally connected, but X is not a finite union of sets with diameterless or equal to 1.

Solution of Problem 1.157(a) We argue by induction on m � 1. For m = 1 by hypothesis A1 isconnected. Suppose that the result is true for some m � 1. Then weset

Bm =m⋃

k=1

Ak,

which by hypothesis is connected and by hypothesis Bm ∩ Am+1 �= ∅.So, Theorem 1.91 implies that the set Bm ∪ Am+1 =

m+1⋃

k=1

Ak is con-

nected.

(b) Yes. We proceed by contradiction. So let us suppose that the set

A =∞⋃

k=1

Ak

is not connected. Then, there exists a pair of nonempty, disjoint setsU, V ⊆ A both open in A such that A = U ∪ V , both open (in A).First, note that for all k � 1, we have

Ak ⊆ U or Ak ⊆ V

(if this is not the case for some k � 1, then the pair Ak∩U , Ak∩V is aseparation of the set Ak and so Ak is not connected, a contradiction).Let us define

kU = inf{k : Ak ⊆ U

}and kV = inf

{k : Ak ⊆ V

}.

One of the integers kU or kV is equal to 1 and the other is greaterthan 1. Without loss of generality, we can assume that kU > 1. Then


AkU ⊆ U and AkU−1 ⊆ V . But from the assumption, we have thatAkU ∩ AkU−1 �= ∅, so also U ∩ V �= ∅, a contradiction to the fact thatthe pair U , V is a separation for A.

Solution of Problem 1.158Note that

Cε = C ∪⋃

x∈CBε(x)

(see Definition 1.6). By hypothesis C and Bε(x) (with x ∈ C) areconnected sets and

C ∩Bε(x) �= ∅ ∀ x ∈ C.

So, we can apply Theorem 1.91 and conclude that for every ε > 0, theset Cε is connected.

Solution of Problem 1.159Let u1 = (x1, λ1) and u2 = (x2, λ2) be two elements in epi f (seeDefinition 1.132(b)). Let M be the maximum of f on the boundedclosed interval [x1, x2]. We connect u1 and u2 with the following con-tinuous path. First we connect u1 and v1 = (x1,M) with the verticalline segment [u1, v1]. Then we connect v1 and v2 = (x2,M) with thehorizontal line segment [v1, v2]. Finally we connect v2 and u2 withthe vertical line segment [v2, u2]. The result is a piecewise linear pathwhich is continuous and connects u1 and u2. Therefore we concludethat epi f is path-connected (see Definition 1.97).

Solution of Problem 1.160(a) Let x ∈ X and ε > 0 be fixed and define

Cε ={u ∈ X : x

ε∼ u

}.

1.3. Solutions 169

We show that Cε is nonempty, open and closed in X. Since x ∈ Cε,we have that Cε �= ∅.

Let u ∈ Cε and let y ∈ Bε(u). Then ({u, y} is an ε-chain con-necting u and y and so u

ε∼ y, hence by transitivity x

ε∼ y and so

Bε(u) ⊆ Cε which proves that Cε is open.Next, let u ∈ Cε and let y ∈ Bε(u)∩Cε. As before, we have y

ε∼ u,

while uε∼ x. Hence, by transitivity, y

ε∼ x and so y ∈ Cε, which proves

that Cε is also closed.Since by hypothesis X is connected, we have X = Cε (see Propo-

sition 1.87) and this is true for all ε > 0. Therefore X is well-chained(see Definition 1.108).

(b) No. The space Q (with the metric induced from R) is well-chainedbut not connected.

Solution of Problem 1.161“=⇒”: This is Problem 1.160.

“⇐=”: Suppose that X is well-chained (see Definition 1.108). Letξ : X −→ {0, 1} be any continuous function. Since X is compact, ξ isuniformly continuous (see Definition 1.45 and Proposition 1.77) andso we can find ε > 0 such that

dX(x, u) � ε =⇒ ∣∣ξ(x)− ξ(u)

∣∣ < 1.

Therefore ξ(x) = ξ(u). For x, u ∈ X, let (c0, . . . , cn) be the ε-chainconnecting x and u. Then d

X(ck, ck+1) � ε for k = 1, . . . , n−1, implies

that ξ(ck) = ξ(ck+1) and so

ξ(x) = ξ(c1) = . . . = ξ(cn−1) = ξ(cn) = ξ(u),

hence ξ is constant. This, by Problem 1.151 implies that X is con-nected.


Solution of Problem 1.162Let u = f(x). The set f

(C(x)

)(see Definition 1.92) is a connected

subset of Y containing u (see Theorem 1.90 and Definition 1.39). So,we have f

(C(x)

) ⊆ C(u) (see Definition 1.92). Similarly, we showthat f−1

(C(u)

) ⊆ C(x). Acting with f , we obtain C(u) ⊆ f(C(x)

).

Therefore

f(C(x)

)= C(u) = C

(f(x)

) ∀ x ∈ X.

Solution of Problem 1.163Let U be the open set in E, defined by

U = E ∩ (R× (− 1

2 ,12

)).

One of the connected components of U is the interval {0} × (− 12 ,

12

),

which is not open in E. So, by Proposition 1.96, the set E is notlocally connected.

Solution of Problem 1.164“=⇒”: Follows from Theorem 1.90.

“⇐=”: Without any loss of generality, we can assume that f is in-creasing (the proof for f being decreasing is similar). Let t0 ∈ T . If fis not continuous at t0, then

f(t−0 ) = limt → t0t < t0

f(t) < f(t0) or f(t0) < f(t+0 ) = limt → t0t > t0

f(t).

To fix things, we assume that f(t0) < f(t+0 ) (the reasoning is similar ifthe other possibility is true). Since the limit f(t+0 ) exists, we can findt ∈ T ∩ (t0,+∞) such that

f(t0) < f(t+0 ) � f(t).

Let y ∈ (f(t0), f(t

+0 )

) ⊆ (f(t0), f(t)

). Since by hypothesis f(T ) is

an interval, we have y ∈ f(T ) and so, we can find s ∈ T such that

1.3. Solutions 171

y = f(s). If s � t0, then y � f(t0), a contradiction. If s > t0, theny � f(t+0 ) again a contradiction. This prove the continuity of f on T .


C ={(x, u) ∈ T × T : x < u

}

and let g : C −→ R be defined by

g(x, u) = f(u)−f(x)u−x ∀ (x, u) ∈ C.

Let (x, u) ∈ C. Then by the mean value theorem, we can find y ∈ (x, u)such that

g(x, u) = f ′(y),

so

g(C) ⊆ f ′(T ).

On the other hand, from the definition of the derivative, we have

f ′(T ) ⊆ g(C).

So, we have

g(C) ⊆ f ′(T ) ⊆ g(C).

But the set C is connected (in fact it is convex in R2) and g is con-tinuous. So g(C) is connected (see Theorem 1.90). Invoking Proposi-tion 1.89, we infer that f ′(T ) is connected which implies that f ′ hasthe Darboux property.

Solution of Problem 1.166Let x0, x1 ∈ X with x0 �= x1 and let ϑ : X −→ R be defined by

ϑ(x) = dX(x, x0) ∀ x ∈ X.

Then ϑ is continuous (see Proposition 1.36) and so ϑ(X) is connected(see Theorem 1.90), thus an interval (see Proposition 1.88), which is


not a singleton (as 0 ∈ ϑ(X) and 0 �= dX (x0, x1) ∈ ϑ(X)). So the setϑ(X) is uncountable, and hence X is uncountable.

Solution of Problem 1.167The nonemptiness of C follows from Proposition 1.67 and the com-pactness of C follows from the fact that C is a closed subset of thecompact set C1. It remains to show the connectedness of C. Supposethat C = K1 ∪ K2 with K1,K2 being closed subsets of C which aredisjoint. Evidently K1 and K2 are compact and so we can find twodisjoint open sets U1 and U2 such that K1 ⊆ U1 and K2 ⊆ U2. Thenwe can find n0 � 1 such that Cn ⊆ U1∪U2 for all n � n0. The connect-edness of Cn implies that Cn ⊆ U1 (or Cn ⊆ U2) for all n � n0, henceC ⊆ K1 (or C ⊆ K2) and so we conclude that K2 = ∅ (or K1 = ∅),which proves that C is connected.

Solution of Problem 1.168Let V = Cb(X;Y ) be furnished with the supremum metric

d∞(f, h) = supx∈X

dY

(f(x), h(x)

).

From Theorem 1.131(b), we know that (V, d∞) is a complete metricspace. Let ξ : V −→ V be defined by

ξ(h)(x) = g(x, h(x)

) ∀ h ∈ V, x ∈ X.

Let us check that this function is well defined, i.e., for all h ∈ V ,ξ(h) ∈ V . First, note that the function x �−→ ξ(h)(x) = g

(x, h(x)

)is

continuous, being the composition of two continuous functions. Also,we will show that ξ(h)(·) is bounded. To this end, fix x ∈ X andy ∈ Y . Then for all u ∈ V and all x ∈ X, we have

dY

(ξ(u)(x), g(x, y)

)� sup

x∈XdY

(ξ(u)(x), g(x, y)

)

= supx∈X

dY

(g(x, u(x)

), g(x, y)

)

� k supx∈X

dY

(u(x), y

)< +∞

1.3. Solutions 173

(since u ∈ V ). So, indeed ξ : V −→ V . For all u, v ∈ V , we have

d∞(ξ(u), ξ(v)

)= sup

x∈XdY

(g(x, u(x)

), g(x, v(x)

))

� k supx∈X

dY

(u(x), v(x)

)= kd∞(u, v),

so ξ is a contraction on V .Because (V, d∞) is a complete metric space, we can apply the

Banach fixed point theorem (see Theorem 1.49) and obtain f ∈ V =Cb(X;Y ) such that

ξ(f) = f,

sog(x, f(x)

)= f(x) ∀ x ∈ X.

Solution of Problem 1.169The function

X ×X � (x, u) �−→ dX (x, u) ∈ Ris continuous (see Proposition 1.36). Also the set K×C is compact inX ×X (see Proposition 1.130). So, invoking the Weierstrass theorem(see Theorem 1.75), we can find (a, c) ∈ K × C such that

dX(a, c) = inf

{dX(x, u) : x ∈ K, u ∈ C

}= dist(K,C)

(see Definition 1.6).Also, if K = C, then again from the Weierstrass theorem (see

Theorem 1.75), we can find a′, c′ ∈ K such that

dX (a′, c′) = sup

{dX (x, u) : x, u ∈ K

}= diamK

(see Definition 1.6(c)).

Solution of Problem 1.170Recall that all the metrics dp , 1 � p � +∞ are Lipschitz equiva-lent (see Example 1.3(a) and Remark 1.119). So, without any loss ofgenerality, we may assume that the metric on X is the metric

d1(x, u) =

m∑

k=1

dk(xk, uk) ∀ x, u ∈ X.


The projection function projk: X −→ Xk, defined by proj

k(x) = xk for

all x = (x1, . . . , xm) ∈ X is 1-Lipschitz continuous, hence uniformlycontinuous (see Definition 1.45). So, by Proposition 1.46(b), proj

k

maps Cauchy sequences to Cauchy sequences. Therefore, we concludethat

{xkn

}n�1

is a Cauchy sequence in Xk.

Conversely, if each{xkn

}n�1

, for k = 1, . . . ,m is a Cauchy sequence,then from

d1(xn, xl) =m∑

k=1

dk(xkn, x

kl )

we infer that {xn}n�1 is a Cauchy sequence in X.

Solution of Problem 1.171Let x ∈ lim inf

n→+∞ En (see Definition 1.133). Then for every integer m � 1

there exists an index nm � 1 such that

B 1m(x) ∩ En �= ∅ ∀ n � nm. (1.7)

We may assume that nm > nm−1 for m � 2. Let xn ∈ B 1m

∩ En for

nm � n < nm+1. We see that

dX(xn, x) < 1

m ∀ n � nm,

hencedX(xn, x) −→ 0 as n → +∞.

This proves the expressions for lim infn→+∞ En.

Next let x ∈ lim supn→+∞

En. Then by definition for every integers

m,k � 1, we can find an integer nm,k � k such that

B 1m(x) ∩ Enm,k

�= ∅. (1.8)

Therefore the sequence{dist(x,En)

}n�1

of nonnegative numbers has

0 as an accumulation point (see Definitions 1.6, 1.13(b) and Theo-rem 1.14(d)) and so

lim infn→+∞ dist(x,En) = 0.

1.3. Solutions 175

So, we can find a subsequence {Enk}k�1 of {En}n�1 such that

limk→+∞

dist(x,Enk) = lim inf

n→+∞ dist(x,En) = 0.

Let xnk∈ Enk

be such that

dX(x, xnk

) � dist(x,Enk) + 1

k −→ 0 as k → +∞.

Hencexnk

−→ x in X,

with xnk∈ Enk

for all k � 1. Finally directly from the definition oflim supn→+∞

En, we see that for all m � 1, we have x ∈ lim supn→+∞

En if and

only if

x ∈⋃

n�m

En ∀ m � 1.

This proves the different expressions for the set lim supn→+∞

En.

From the above expressions it is clear that we always have

lim infn→+∞ En ⊆ lim sup

n→+∞En.

Moreover, it is clear that lim supn→+∞

En is closed (being the intersection

of the closed sets Cm =⋃

n�1En). Also, note that the condition (1.8) is

equivalent to

∀m,k � 1 ∃nm,k � 1 B 1m(x) ∩ Enm,k

�= ∅.

Therefore, we infer that

lim supn→+∞



En.

Next let x ∈ lim infn→+∞ En. Then for every r > 0, we have

Br(x) ∩ lim infn→+∞ En �= ∅

(see Definition 1.13(b)), hence we can find an integer nr � 1 such that

Br(x) ∩ En �= ∅ ∀ n � nr,


hence x ∈ lim infn→+∞En. In addition, as before, since the equiva-lence of (1.7) to

∀m � 1 ∃nm � 1 ∀n � mn Br(x) ∩ En �= ∅,

we conclude that

lim infn→+∞ En = lim inf

n→+∞ En = lim infn→+∞ En.

Hence it follows that the two sets lim infn→+∞ En and lim sup

n→+∞En are closed

(possibly empty).

Solution of Problem 1.172If lim inf

n→+∞ Cn = ∅, then

dist(x, lim inf

n→+∞ Cn

)= +∞

(see Definition 1.133, Problem 1.171 and Definition 1.6) and sothe inequality is trivially true. Therefore, we may assume thatlim infn→+∞ Cn �= ∅. So, let u ∈ lim inf

n→+∞ Cn. Then, we can find un ∈ Cn

such that

dX(un, u) −→ 0

(see Problem 1.171). We have

dist(x,Cn) � dX(x, un) ∀ n � 1

and so

lim supn→+∞

dist(x,Cn) � dX(x, u).

Since u ∈ lim infn→+∞ Cn is arbitrary, it follows that

lim supn→+∞

dist(x,Cn) � dist(x, lim inf

n→+∞ Cn

).

1.3. Solutions 177

Solution of Problem 1.173Since the sets Cn are compact for n � 1, we can find un ∈ Cn suchthat

dist(x,Cn) = dRN

(x, un)

(see Definition 1.6). From the definition of lim inf (see Definition 1.133and Problem 1.171), there exists a subsequence

{unk

}k�1

of {un}n�1

such that

lim infn→+∞ d

RN(x, un) = lim

k→+∞dRN

(x, unk).

Next, since {un}n�1 ⊆ BR and the latter is compact, there exists a

further subsequence{unkl

}l�1

of{unk

}k�1

such that unkl−→ u in RN .

From the definition of Kuratowski upper limit, we know that u ∈ C.From the continuity of the distance function (see Proposition 1.36),we have

dist(x,Cnkl) = d

RN(x, unkl

) −→ dRN

(x, u) � dist(x,C),

so

lim infn→+∞ dist(x,Cn) = lim

l→+∞dRN

(x, unkl) = d

RN(x, u) � dist(x,C).

Because C = lim infn→+∞ Cn, from Problem 1.172, we also have

lim supn→+∞

dist(x,Cn) � dist(x,C).

Thus, we conclude that dist(x,Cn) −→ dist(x,C).

Solution of Problem 1.174For every ε > 0, we can find x ∈ C such that

m � ϕ(x) � m+ ε.

Let {xn}n�1 ⊆ X be such that xn ∈ Cn for all n � 1 anddX(xn, x) −→ 0 (Problem 1.171). For every n � 1, we have

mn � ϕ(xn). Due to the continuity of ϕ, we have

ϕ(xn) −→ ϕ(x),


so

lim supn→+∞

mn � m+ ε.

Since ε > 0 was arbitrary, we let ε ↘ 0 to conclude that

lim supn→+∞

mn � m.

Solution of Problem 1.175Let us set

En ={xm,n : m � 1

} ∀ n � 1.

Then from Problem 1.171, we have

xm ∈ lim infn→+∞ En ∀ m � 1

(see Definition 1.132). Because lim infn→+∞ En is closed (see Problem 1.171)

and xm −→ x, we have that x ∈ lim infn→+∞ En. So, x = lim

n→+∞ yn, with

yn ∈ En. But from the definition of the sets En, these elements havethe form yn = xmn,n. Therefore

x = limn→+∞xmn,n.

Next let us set

Gm ={xm,n : n � 1

} ∀ m � 1.

Note that

xm ∈ Gm ∀ m � 1.

So

x = limm→+∞xm ∈ lim inf

m→+∞Gm = lim infm→+∞Gm

(see Problem 1.171). Therefore x = limm→+∞um, with um ∈ Gm

for all m � 1. Every um has the form um = xm,nm. Hencex = lim

m→+∞xm,nm .

1.3. Solutions 179

Solution of Problem 1.176(a) Of course h(A,A) = 0. Let A,B ∈ Pf

(X)

be such thath(A,B) = 0. Let a ∈ A. Then

0 = h(A,B) = supx∈X

∣∣dist(x,A) − dist(x,B)∣∣

�∣∣dist(a,A) − dist(a,B)

∣∣ = dist(a,B),

thus a ∈ B (as B is closed). Analogously, we show that for any b ∈ B,we have b ∈ A. Thus A = B.

The symmetry

h(A,B) = h(B,A) ∀ A,B ∈ Pf

(X)

is obvious.To show the triangle inequality, let A,B,C ∈ Pf

(X). Then

h(A,C) = supx∈X

∣∣dist(x,A)− dist(x,C)∣∣

� supx∈X

(∣∣dist(x,A)− dist(x,B)∣∣+

∣∣dist(x,B)− dist(x,C)∣∣)

� supx∈X

∣∣dist(x,A)− dist(x,B)∣∣+ sup

x∈X

∣∣dist(x,B)− dist(x,C)∣∣

= h(A,B) + h(B,C).

(b) Let A,B ∈ Pf

(X). Note that

supa∈A

dist(a,B) = supa∈A

∣∣dist(a,A) − dist(a,B)∣∣

� supx∈X

∣∣dist(a,A) − dist(a,B)∣∣ = h(A,B).

Analogously, we have

supb∈B

dist(b,A) � h(A,B).

Thus

h(A,B) � h(A,B).

On the other hand, let x ∈ X be fixed. Then

dist(x,A)− dist(x,B) = dist(x,A)− infb∈B

dX (x, b)

= supb∈B

(dist(x,A)− dX (x, b)

)= sup

b∈B

(infa∈A

dX (x, a)− dX (x, b))

= supb∈B

infa∈A

(dX (x, a)− dX (x, b)

)� sup

b∈Binfa∈A

dX (a, b) = supb∈B

dist(b, A).


Analogously, we show that

dist(x,B)− dist(x,A) � supa∈A

dist(a,B).

Thus

∣∣dist(x,A)−dist(x,B)∣∣ � max

{supa∈A

dist(a,B), supb∈B

dist(b,A)}

= h(A,B).

As the above inequality holds for every x ∈ X, taking the supremumon the left hand side, we obtain

h(A,B) � h(A,B).

(c) Let A,B ∈ Pf

(X). Let

E ={ε > 0 : A ⊆ Bε and B ⊆ Aε

}.

If ε ∈ E , then A ⊆ Bε and B ⊆ Aε. Thus

{dist(a,B) � ε ∀ a ∈ A,dist(b,A) � ε ∀ b ∈ B.

So

supa∈A

dist(a,B) � ε and supb∈B

dist(b,BA) � ε

and hence

h(A,B) � ε.

As, it holds for any ε ∈ E , thus

h(A,B) � h(A,B).

On the other hand, from the definition of h, we have that

dist(a,B) � h(A,B) ∀ a ∈ A,

so

a ∈ Bh(A,B) ∀ a ∈ A

and thus

A ⊆ Bh(A,B).

1.3. Solutions 181

Analogously, we show that

B ⊆ Ah(A,B).

Thus finally

h(A,B) � h(A,B).


(a) Let D1,D2, E1, E2 ∈ Pf

(X)and let us set

ε = max{h(D1, E1), h(D2, E2)

}and D = D1 ∪D2,

E = E1 ∪ E2,

then

D ⊆ (E1)ε ∪ (E2)ε = (E1 ∪ E2)ε = Eε

E ⊆ (D1)ε ∪ (D2)ε = (D1 ∪D2)ε = Dε,

where for each S ∈ Pf

(X), Sε =

{x ∈ X : dist(x, S) � ε

}(see

Definition 1.6 and Problem 1.176(c)). From these inclusions and thedefinition of the Hausdorff metric (see Problem 1.176(c)), we obtainthat

h(D1 ∪D2, E1 ∪ E2) � ε = max{h(D1, E1), h(D2, E2)

}.

(b) Let η : X −→ X be a k-contraction and let D,E ∈ Pf

(X).

Let ε = h(D,E) and let us set

D = η(D), E = η(E).

If y = η(x) ∈ D, let u ∈ E be such that dX(x, u) � ε. Then, we have

dist(y, E) � dX

(y, η(u)

)= d

X

(η(x), η(u)

)� kd

X(x, u) � kε,

so D ⊆ Ekε and similarly E ⊆ Dkε. Thus

h(η(D), η(E)

)= h(D, E) � kε = kh(D,E).


Solution of Problem 1.178(a) Let (X, dX ) be a complete bounded metric space (see Defini-tion 1.7). To show that the Hausdorff metric space

(Pf

(X), h

)(see

Definition 1.134 and Problem 1.176) is complete, let us take anyCauchy sequence {Dk}k�1 ⊆ Pf

(X)(see Definition 1.7).

Let us start with two observations. First note that

∀i � 1 ∃Ni ∈ N ∀m,n � Ni : h(Dm,Dn) � ε2i

(1.9)

(see Definition 1.7). Next, note that

∀i � 1 ∀k � Ni ∀x ∈ Dk ∀j � k ∃y ∈ Dj : dX(x, y) < ε

2i. (1.10)

To see this, let us fix i � 1, k � Ni, x ∈ Dk and j � k. From (1.9), wehave that

h(Dj ,Dk) < ε2i,

so

Dk ⊆ (Dj) ε

2i

and so, there exists y ∈ Dj such that x ∈ B ε

2i(y). This proves (1.10).

Let us define

F ={x ∈ X : x is an accumulation point of some

sequence {dk}k�1 ⊆ X, such that dk ∈ Dk for k � 1}

(see Definition 1.13(b) and Theorem 1.14(d)). We will show that

Dkh−→ F . To this end, let ε > 0. We will show that

∀k � N2 : h(F,Dk) � ε

(see (1.10)). In other words, we need to show that

∀k � N2 : F ⊆ (Dk)ε and Dk ⊆ Fε.

First, let us fix k � N2 and x ∈ F . From the definition of F , thereexists a sequence {dn}n�1 ⊆ X such that dn ∈ Dn for n � 1 and x isan accumulation point of {dn}n�1. From (1.9), for all m,n � N2, wehave

h(Dm,Dn) <ε

2,

1.3. Solutions 183

so for all n � N2, we have Dn ⊆ (Dk) ε2and thus dn ∈ (Dk) ε

2for all

n � N2. Hence all accumulation points of the sequence {dn}n�1 are in

(Dk) ε2, so x ∈ (Dk)ε. Thus, we have proved that F ⊆ (Dk)ε.

Next, let us fix k � N2 (see (1.9)) and x ∈ Dk. We will show thatx ∈ F2ε. To this end, let i0 be the biggest i � 1 such that k � Ni, sok � Ni0 . We define

fk = x ∈ Dk.

From (1.10), there exists

fNi0+1 ∈ DNi0+1 , with dX

(fNi0+1 , x

)< ε

2i0.

Then proceeding by induction, for any i > i0 we choose

fNi ∈ DNi , with dX

(fNi , fNi−1

)< ε

2i−1 .

For any other l > ni0 (namely for l �= Ni for all i � n0), we choose anyfl ∈ Dl. Note that

dX(x, fNi) � d

X(x, fNi0

) + dX(fNi0

, fNi0+1) + . . .+ dX(fNi−1, fNi)

<ε

2i0+

ε

2i0+1+ . . .+

ε

2s< ε

2 ,

so{fNi

}i�i0

⊆ B ε2(x) and thus for any accumulation point y of

{fNi

}i�i0

(so also for some accumulation point of {fn}n�1), we have

dX(x, y) � ε

2 . As y ∈ F , so x ∈ Fε.

(b) Since X is compact, X is complete (see Theorem 1.71) and so(Pf

(X), h

)is complete (see part (a)). Let us fix b ∈ X and let

g :(Pf

(X), h

) −→ (C(X;R), d∞

)be the function, defined by

g(A)(x) = dist(x,A)− dX (x, b).

First we show that g is an isometry. Note that

d∞(g(A), g(B)

)= sup

x∈X

∣∣dist(x,A)− dist(x,B)∣∣ = h(A,B),

so g is an isometry.Next we show that g

(Pf

(X)) ⊆ C(X;R) is equicontinuous. For all

A ∈ Pf

(X), we have

∣∣g(A)(x) − g(A)(u)∣∣ =

∣∣dist(x,A) − dist(u,B)− (dX(x, b) − d

X(u, b)

)∣∣

� 2dX (x, u),

so g(Pf

(X))

is (uniformly) equicontinuous in C(X;R).


Because g is isometry, the set g(Pf

(X))

is closed in C(X,R). Alsog(Pf

(X))

is equicontinuous and for all x ∈ X, the set{g(A)(x) : A ∈

Pf

(X)}

is bounded (since by hypothesis X is bounded). Therefore wecan apply the Arzela–Ascoli theorem (see Theorem 1.84) and obtainthat g

(Pf

(X))

is compact in C(X;R). Because g is an isometry,(Pf

(X), h

)is a compact metric space.

Solution of Problem 1.179Let h be the Hausdorff metric (see Definition 1.134 and Prob-lem 1.176). From Problem 1.178, we know that

(Pf

(X), h

)is a com-

plete metric space.Using Problem 1.177(a) and (b), we have

h(ξ(D), ξ(E)

)= h

(f(D) ∪ g(D), f(E) ∪ g(E)

)

� max{h(f(D), f(E)

), h

(g(D), g(E)

)}

� kh(D,E).

Since the space(Pf

(X), h

)is complete and ξ is a k-contraction with

k ∈ [0, 1), we can apply the Banach fixed point theorem (see Theo-rem 1.49) and obtain a unique C0 ∈ Pf

(X)such that ξ(C0) = C0.


(a) Let {Cn}n�1 ⊆ Pk

(X)be a sequence such that

Cnh−→ C ∈ Pf

(X).

For a given ε > 0, we can find n0 = n0(ε) � 1 such that

h(Cn, C) � ε2 ∀ n � n0.

The set Cn0 is compact, hence it is also totally bounded (see Theo-rem 1.71). Thus we can find a finite set {xk}mk=1 ⊆ X such that

Cn0 ⊆m⋃

k=1

B ε2(xk).

1.3. Solutions 185

It follows that

C ⊆m⋃

k=1

Bε(xk).

and so C is totally bounded. Since C ∈ Pf

(X), once again Theo-

rem 1.71 implies that C ∈ Pk

(X). Therefore Pk

(X)is a closed subset

of Pf

(X).

(b) Now, if X is complete, then from Problem 1.178(a), we know that(Pf

(X), h

)is complete too. Thus

(Pk

(X), h

)is complete (as a closed

subset of a complete space).

Solution of Problem 1.181“=⇒”: This is exactly Problem 1.178(b).

“⇐=”: Let {xn}n�1 be a sequence in X and let

Cn = {xn} ∈ Pf

(X) ∀ n � 1.

The compactness of(Pf

(X), h

)implies that we can find a subsequence

{Cnk}k�1 of {Cn}n�1 such that

Cnk

h−→ C ∈ Pf

(X)

(see Definition 1.134 and Problem 1.176). So, for a given ε > 0, wecan find k0 = k0(ε) � 1 such that

C ⊆ (Cnk)ε = Bε(xnk

) ∀ k � k0

(see Problem 1.176(c)). Hence diamC � 2ε. Since ε > 0 was arbitrary,we conclude that diamC = 0, i.e., C = {x}. Then, we have

xnk∈ Bε(x) ∀ k � k0

and so xnk−→ x. This proves the compactness of X.


Solution of Problem 1.182We argue by contradiction. So, suppose that C is not connected. Thenwe can find two open nonempty sets U1, U2 ⊆ X such that

C ⊆ U1 ∪ U2, C ∩ U1 �= ∅, C ∩ U2 �= ∅ and U1 ∩ U2 = ∅.

Let x ∈ C ∩ U1. Since by hypothesis Cnh−→ C, we have that

dist(x,Cn) −→ 0

(see Problem 1.176(b)). For every n � 1, we can find xn ∈ Cn suchthat

dX(x, xn) � dist(x,Cn) +

1

n,

hence dX (x, xn) −→ 0. Since U1 is open and x ∈ U1, we can findn1 � 1 such that

xn ∈ U1 ∀ n � n1.

SoCn ∩ U1 �= ∅ ∀ n � n1.

Similarly, we show that

Cn ∩ U2 �= ∅ ∀ n � n2,

for some n2 � 1. The connectedness of Cn implies that

Cn � U1 ∪ U2 ∀ n � n0 = max{n1, n2}.So, for all n � n0, we can find un ∈ Cn, un �∈ U1 ∪ U2. Since the set(U1 ∪U2)

c is closed and the set C is compact, from Problem 1.118(a),we have

0 < dist((U1 ∪ U2)

c, C)

� dist(un, C) � h(Cn, C) −→ 0,

a contradiction. This proves that the set C is connected too.

Solution of Problem 1.183From Problem 1.180, we already have that

(Pk

(X), h

)is complete. So,

it remain to show the separability of(Pk

(X), h

)(see Definition 1.21).

Let {xn}n�1 be dense in X (it exists since X is separable; Defini-tion 1.20) and let F be the family of all finite subsets of {xn}n�1.

1.3. Solutions 187

Evidently F is countable and clearly it is dense in(Pk

(X), h

)(see

Definition 1.20).

Solution of Problem 1.184Let X = N and consider two metrics

d1(n,m) =

{1 if n �= m,0 if n = m,

∀ n,m � 1

d2(n,m) =∣∣ 1n − 1

m

∣∣ ∀ n,m � 1.

The two metrics are topologically equivalent (they both generate thediscrete topology). Let h1 be the Hausdorff metric for d1 and let h2 bethe Hausdorff metric for d2 (see Definition 1.134 and Problem 1.176).Setting Cn = {1, . . . , n}, we have

h1(Cn,N) = 1 ∀ n � 1,

h2(Cn,N) = supk

dist2(k,Cn) = limk→+∞

(1n − 1

k

)= 1

n ,

hence h2(Cn,N) −→ 0. So, the Hausdorff metrics h1 and h2 are nottopologically equivalent.

Solution of Problem 1.185Consider N furnished with the discrete metric. Let Cn = {1, . . . , n}.Then these are closed sets and


n→+∞Cn = N

(see Definition 1.133 and Problem 1.171). On the other hand,

h(Cn,N) = 1 ∀ n � 1,

so

Cn �−→ N in the h-metric.


Solution of Problem 1.186First note, that we always have

lim infn→+∞ Cn ⊆ lim sup

n→+∞Cn

(see Definition 1.133 and Problem 1.171). Let x ∈ C. Since Cnh−→ C

see Definition 1.134 and Problem 1.176, we have

dist(x,Cn) −→ dist(x,C) = 0,

hence x ∈ lim infn→+∞ Cn (see Problem 1.171) and thus

C ⊆ lim infn→+∞ Cn.

Next, let x ∈ lim supn→+∞

Cn. Then we can find a subsequence {nk}k�1 of

{n}n�1 and xnk∈ Cnk

such that xnk−→ x in X (see Problem 1.171).

We havedist(x,Cnk

) � dX(x, xnk

) −→ 0

and since Cnk

h−→ C, we also have

dist(x,Cnk) −→ dist(x,C)

(see Problem 1.176). Therefore dist(x,C) = 0, hence x ∈ C (sinceC ∈ Pf

(X)). This proves that

lim supn→+∞

Cn ⊆ C ⊆ lim infn→+∞ Cn

and so finallylim infn→+∞ Cn = lim inf

n→+∞ Cn = C.


A =⋃

n�1

Cn

It suffices to show that (A, dX) is totally bounded. For a given ε > 0,

we can find N = N(ε) � 1 such that

h(Cn, Cm) < ε2 ∀ n,m � N.

1.3. Solutions 189

The setN⋃

i=1Ci is compact, hence totally bounded. So, we can find an

ε2 -net {x1, . . . , xk} ⊆

N⋃

i=1Ci. We will show that {x1, . . . , xk} is an ε-net

in A. To this end let x ∈ A. If x ∈N⋃

i=1C1, then clearly d

X(x, xi) for

some i ∈ {1, . . . , N}. If x ∈ Ci with i > N , then h(Ci, CN ) < ε2 , hence

dist(x,Cn) <ε2 . Choosing y ∈ CN with d

X(x, y) < ε

2 and j � 1 suchthat d

X(y, xj) <

ε2 , we set d

X(x, xj) < ε.

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