Introduction to Topological Spaces and Set-Valued Maps ... · Introduction to Topological Spaces...

Introduction to Topological Spaces and Set-Valued Maps

(Lecture Notes)

Abebe Geletu (Dr.)Institute of Mathematics

Department of Operations Research & StochasticsIlmenau University of Technology

August 25, 2006

Contents

1 Preface 1

2 Introduction to Metric Spaces 32.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 Subspaces of Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Sequences, Convergence and Complete Metric Spaces . . . . . . . . . . . . . . . . . . . 7

2.4.1 Complete Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Baire Category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.6 Compact Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.6.1 Bounded Sets and Totally Bounded Metric Spaces . . . . . . . . . . . . . . . . . 132.7 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.7.1 Continuity of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.7.2 Real Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.7.3 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.7.4 Convergence Properties of Sequences of Functions . . . . . . . . . . . . . . . . 202.7.5 Equicontinuiuty and the Ascoli-Arzela Theorem . . . . . . . . . . . . . . . . . . 222.7.6 Homeomorphisms and Isometries in Metric Spaces . . . . . . . . . . . . . . . . 242.7.7 Contractive Maps and Fixed Point Properties . . . . . . . . . . . . . . . . . . . . 25

3 Topological Spaces 293.1 Neighborhood and Neighborhood Systems . . . . . . . . . . . . . . . . . . . . . . . . . 313.2 Bases and Subbases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.3 Sequences, Continuity and Homeomorphism . . . . . . . . . . . . . . . . . . . . . . . 363.4 Classification of Topological Space: Separation Axioms . . . . . . . . . . . . . . . . . . 373.5 Uryson’s Lemma, Tietze’s Extension Theorem and Metrizability . . . . . . . . . . . . . 39

3.5.1 Tietze’s Extension Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.5.2 Urysohn’s Metrizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.6 Compact Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.6.2 The Finite Intersection Property . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.6.3 Compact Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.7 Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.8 Sigma-Compact Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.9 Paracompact Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.10 Partition of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4 The Hausdorff Metric and Convergence of Sequences of Sets 714.1 The Hausdorff Metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.2 Convergence of Sequences of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

4.2.1 Calculus of Limits of Sequences of Sets . . . . . . . . . . . . . . . . . . . . . . . 794.2.2 Convergence w.r.t. the Hausdorff Metric . . . . . . . . . . . . . . . . . . . . . . 82

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5 Set-Valued Maps 895.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5.1.1 Some Examples of Set-Valued Maps . . . . . . . . . . . . . . . . . . . . . . . . . 895.2 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.2.1 Elementary Mathematical Operations with Set-Valued Maps . . . . . . . . . . . 915.3 Semi-Continuity of Set-Valued Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

5.3.1 Properties of Semi-Continuous Set-Valued Maps . . . . . . . . . . . . . . . . . . 945.3.2 Local Uniform Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.3.3 Hausdorff Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5.4 Set-Valued Maps with Given Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6 Measurability of Set-Valued Maps 1116.1 Definitions and Properties of Measurable Set-Valued Maps . . . . . . . . . . . . . . . . 111

6.1.1 Operations with Measurable Set-Valued Maps . . . . . . . . . . . . . . . . . . . 1146.2 Measurable Selections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.3 Measurability of Set-Valued Maps with given Structure . . . . . . . . . . . . . . . . . . 118

7 Comments on Literature 123

Bibliography 125

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1 Preface

These notes are a result of a two semester course that I held at the technical university of Ilmenauduring winter semester 2005 and summer semester 2006. These are simply lecture notes organizedto serve as introductory course for advanced postgraduate and pre-doctoral students. The main objec-tive is to give an introduction to topological spaces and set-valued maps for those who are aspiring towork for their Ph. D. in mathematics. It is assumed that measure theory and metric spaces are alreadyknown to the reader. Hence, only a review has been made of metric spaces. At the same time the top-ics on topological spaces are taken up as long as they are necessary for the discussions on set-valuedmaps. Here are to be found only basic issues on continuity and measurability of set-valued maps.Issues on selection functions, fixed point theory, etc. have not be dealt with due to time constraints.

The is not an original work of the writer. In many cases, I have attempted to mention the sourcesof theorems and statements. I have tried to supply my own versions of simplified proofs, whenever Ifelt necessary. These is not by far an all-inclusive introductory note. In fact, I leave it at the mercy ofthe criticisms, suggestions and comments of the reader. However, it is my belief that the material canserve as spring board to dive into the ocean of set-valued maps.

Abebe Geletu, August 2006.

1

2 Introduction to Metric Spaces

2.1 Introduction

Definition 2.1.1 (metric spaces). Let X be a non-empty set. A function ρ : X × X → R+ is called ametric on X if the following are satisfied

M1: ρ(x, y) ≥ 0, for each x, y ∈ X;

M2: ρ(x, y) = 0 if and only if x = y;

M3: ρ(x, y) = ρ(y, x);

M4: for any z ∈ X : ρ(x, y) ≤ ρ(x, z) + ρ(z, y) (triangle inequality).

The set X together with the metric ρ is called a metric space and this usually depicted by < X, ρ >.

Example 2.1.2. Some standard examples of metric spaces

(a) < Rn, ρ > with ρ(x, y) = (∑n

i=1 (xi − yi)p)

1p , where p > 0. Here, if p = 2 we obtain the Euclidean

or the l2 metric; if p = 1 we have the l1 metric on Rn, and so on.

(b) < Rn, σ > with σ(x, y) = maxi=1...n |xi − yi|.(c) < C[a, b], ρ∞ > with ρ∞(f, g) = maxt∈[a,b] |f(t)− g(t)|; where C[a, b] represents the space of contin-

uous functions on [a, b].

(d) if X is any normed space with a norm ‖ · ‖, then < X, ρ > will be a metric space if we defineρ(x, y) := ‖x− y‖. In this case, ρ is called an induced metric - induced by this particular norm onX.

From Exa. 2.1.2, it is obvious that there might be more than one metric on a given set.

In Def. 2.1.1 M1, M3 and M4 hold true, but M2 fails, then we call the metric ρ a pseudometricand the pair < X, ρ > is called a pseudometric space. For instance the space Lp of functions is apseudometric space; with metric induced by the Lp− norm. If all but M3, then < X, ρ > is called aquasi-metric space. 1

Furthermore, if < X, ρ > is a metric space and x, y ∈ X, then the non-negative real number ρ(x, y)can be interpreted as the distance between the elements x and y in the metric space < X, ρ >. Ingeneral, the distance between x and y can be different for a different metric. Note also that, in ametric space X, x = y iff and only if the distance between x and y is 0.

Definition 2.1.3. Let < X, ρ > be a metric space and S any subset of X. Then the diameter of S w.r.t.the metric ρ is defined as:

diamS := supρ(x, y) | x, y ∈ S.A set S is said to be bounded if diamS < ∞.

1See Chap. 4 for an example of a quasi-metric.

3

Remark 2.1.4. For a subset A of a metric space X, the following are easy to verify

(i) if A is unbounded, then diam(A) = ∞.

(ii) if cardA > 1, then diam(A) > 0.

Definition 2.1.5 (distance between sets). Let < X, ρ > be a metric space and A,B ⊂ X and A,B 6= ∅.Then the distance between the sets A and B with respect to the metric ρ is given by

dist(A,B) := infρ(x, z) | x ∈ A, y ∈ B= inf

z∈A,x∈Bρ(x, z).

If A = x, then we writedist(A,B) = dist(x,B).

Proposition 2.1.6. If A ⊂ B, then dist(A, B) = 0.

Definition 2.1.7 (product metric). Let < Xi, ρi >, i = 1, . . . , n, n ≥ 2 be metric spaces. Then the productmetric space is the Cartesian product

n∏

i=1

Xi := X1 × . . .×Xn

along with the metric given by

ρ(x, y) =

[n∑

i=1

ρi(xi, yi)2] 1

2

,

where x = (x1, . . . , xn) ∈ ∏ni Xi and y = (y1, . . . , yn) ∈ ∏n

i Xi.

2.2 Open and Closed Sets

Definition 2.2.1 (open set). Let < X, ρ > be a metric space. A set O is called open in X if

∀x ∈ O,∃r > 0 : y ∈ X | ρ(x, y) < r ⊂ O.

The setBr(x) := y ∈ X | ρ(x, y) < r

is called the open ball of radius r and center at x. Hence, a set O ⊂ X is open if for each x ∈ X thereis an open ball Br(x) such that Br(x) ⊂ O.• The sets X and ∅ are open. For x ∈ X and r > 0, the open ball Br(x) is an open set.

Definition 2.2.2 (neighborhood). We say that a set U ⊂ X is a neighborhood of a point x in X iffthere is an open set O such that

x ∈ O ⊂ U.

Definition 2.2.3 (interior of a set). Let < X, ρ > be a metric space and A ⊂ X.

(i) A point x ∈ A is called an interior point of A iff

∃r > 0 : Br(x) ⊂ A;

(ii) The collection of all interior points of a set A is known as the interior of A and is denoted by intA.

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Remark 2.2.4. For any set A:

(i) we have intA ⊂ A and intA is an open set.

(ii) A is an open set iff A = intA; i.e. for an open set, all of its elements are in its interior.

Definition 2.2.5 (closed set). Let < X, ρ > be a metric space and F ⊂ X. Then F is a closed set iffX \ F is an open set.

Hence, the complement of an open set is closed and the complement of a closed set is open.

Proposition 2.2.6.

(i) The intersection of any finite number of open sets is an open set;

(ii) the union of any collection of open sets is open.

(iii) The union of a finite collection of closed sets is closed;

(iv) the intersection of an arbitrary collection of closed sets is closed.

Definition 2.2.7 (accumulation point, closure of a set). Let < X, ρ > be a metric space and A ⊂ X.

(i)A point x ∈ X is an accumulation point of A iff

∀r > 0 : Br(x) ∩A 6= ∅.

We denote set of all accumulation points of a set A by A′, and A′ is sometimes called the derivedset of A.

(ii) The closure of a set A, denoted by clA, is defined as

clA := A ∪A′

Proposition 2.2.8. Let A be a subset of a metric space. Then

(i) A ⊂ clA;

(ii) clA is a closed set;

(iii) A is a closed set iff A = clA.

Excercises 2.2.9. Verify the following properties for the interior and closure of sets A and B.

(i) int(A ∪B) ⊃ intA ∪ intB;

(ii) cl(A ∪B) = clA ∪ clB;

(iii) int(A ∩B) = intA ∩ intB;

(iv) cl(A ∩B) ⊂ clA ∩ clB;

(v) diam(A) = diam(clA).

(vi) dist(x,A) = 0 ⇔ x ∈ cl A. Consequently, for any set A, we have dist(A, clA) = 0.

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Definition 2.2.10 (boundary). Let A ⊂ X be non-empty. Then boundary ∂A of A is defined as

∂A := clA \ intA.

Note that, if A is a closed set, then ∂A ⊂ A.

Definition 2.2.11 (dense set). A subset D of a metric space < X, ρ > is dense in X iff

clD = X.

That is, a set is dense if its closure is the whole space.

Proposition 2.2.12. The following are easy to verify:

(i) a set D is a dense set iff int(X \D) = ∅;(ii) if D is a dense in set and D ⊂ B, then B is also a dense set.

Remark 2.2.13 (On the importance of dense sets). Note that the density of a set D in a set X (w.r.t. ametric ρ) implicity contains the possibility of the approximatability of the elements of X by the elementsof D. In fact, if x0 is any element of X, we can find an element d0 of D which is arbitrarily close to xw.r.t. ρ. Specifically, for any ε > 0, the density of D in X w.r.t. ρ implies

Bε(x0) ∩D 6= ∅ ⇒ ∃d0 ∈ D : d0 ∈ Bε(x0) ⇒ ∃d0 ∈ D : d0 : ρ(x0, d0) < ε.

In this respect, one of the well known results of Karl Weierstrass guarantees that: the set of all polynomialsis dense in C[a, b] w.r.t. the metric ρ∞ (see. Example 2.1.2). Implying that, every continuous functionon [a, b] can be approximated by a polynomial on [a, b] (see for instance Meinardus[18]).

Definition 2.2.14 (separable metric space). A metric space X is called separable if it has a countabledense subset; i.e. if there is D ⊂ X such that D is countable and clD = X.

The Euclidean space Rn is a standard example of a separable metric space, since Qn is a countabledense subset of Rn, where Q stands for the set of rational numbers.

Proposition 2.2.15. A metric space X is separable iff for each open set O ⊂ X there is a countable familyOk of open sets such that

O =⋃

Ok⊂O

Ok.

2.3 Subspaces of Metric Spaces

Recall that, a metric ρ : X ×X → R+ = [0,+∞) is a mapping. Hence,

Definition 2.3.1 (subspace of a metric space). Let < X, ρ > be a metric space and S ⊂ X. Therestriction ρS of ρ to S × S is a metric on S and the pair < ρS , S > is called a subspace of < X, ρ >.

A set which is closed relative to S may not be closed in X. For instance, consider the space X = [0, 1]with the absolute valued metric ρ(x, y) = |x− y| and S = (0, 1]. The set A = (0, 1

2 ] is closded w.r.t. to< S, ρS >, but not in X.

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Note that, a set A ⊂ S is open relative to S iff, for each x ∈ A, there is r > 0 such that

x ∈ BSr (x) := y ∈ S | ρ(x, y) < r ⊂ A.

This is the same as∃r > 0 : x ∈ Br(x) ∩ S ⊂ A,

for each x ∈ A. With this observation, the following can be easily verified:

Proposition 2.3.2. Let < X, ρ > be a metric space, S is a subspace of X and A ⊂ S. Then

(i) if A is open relative to S, then there exists an open set O in X such that A = O ∩ S;

(ii) if A is closed relative to S, then there exists a closed set F in X such that A = F ∩ S.

Proof. (i) Let A ⊂ S be open relative to S implies that, for each x ∈ A, there is r(x) > 0 such thatx ∈ Br(x)(x) ∩ S ⊂ A. Hence

A =⋃

x∈A

(Br(x)(x) ∩ S

)=

⋃

x∈A

(Br(x)(x)

) ∩ S

Set O :=⋃

x∈A

(Br(x)(x)

). Then O is an open set in X and

A = O ∩ S.

(ii) Exercise!

Proposition 2.3.3. Every subspace of a separable metric space is separable.

2.4 Sequences, Convergence and Complete Metric Spaces

Given a metric space X, a function f : N → X is called a sequence such that for each n ∈ N, f(n) =xn ∈ X. Traditionally, a sequence is represented by the image set xnn∈N (note that f(N) =xnn∈N). In fact, we can drop ’n ∈ N’ (when not required) and simply write xn for the sequencexnn∈N.

Definition 2.4.1 (convergence). We say that a sequence xn ⊂ X converges to a point x ∈ X iff

∀ε > 0, ∃N : ρ(xn, x) < ε,∀n ≥ N.

In this case we writexn → x

and we call x the limit of the sequence xn.Thus, a sequence xn converges to x iff

∀ε > 0, ∃N : xn ∈ Bε(x), ∀n ≥ N.

Equivalently, a sequence xn converges to x iff

ρ(xn, x) → 0;

i.e. the distance between x and xn goes to zero as n goes to infinity.

It is easy to verify that, in a metric space, a convergent sequence has a unique limit point.

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Proposition 2.4.2. Let X be a metric space and B ⊂ X. Then

(i) a point x is in the closure of B iff there is a sequence xn ⊂ B such that xn → x;

(ii) the set B closed iff for every sequence xn ⊂ B and xn → x implies x ∈ B.

Definition 2.4.3 (subsequence). Let xn be a sequence in a metric space X. A subset xnkk∈N of xn

is called subsequence of xn;Corollary 2.4.4. A point x ∈ X is an accumulation of a sequence xn, then there is a subsequencexnk

k∈N that converges to x.

Obviously, the limit of a sequence is an accumulation point.

2.4.1 Complete Metric Spaces

Definition 2.4.5 (Cauchy sequence). Let < X, ρ > be a metric space. A sequence xn ⊂ X is a Cauchysequence iff

∀ε > 0, ∃N ∈ N : ρ(xn, xm) < ε,∀n, m ≥ N.

It is easy to verify that, in every metric space a convergent sequence is always a Cauchy sequence.However, the converse of this statement is not always true. That is, there are metric spaces in whichCauchy sequences may not converge.

Definition 2.4.6 (complete metric space). A metric space < X, ρ > is said to be a complete metricspace iff every Cauchy sequence xn ⊂ X converges in X.

Example 2.4.7. (Examples of complete metric spaces)

• The euclidean space Rn, n ∈ N, is a complete metric space.

• The space of real-valued continuous functions C[a, b] = f | f : [a, b] → R and f is continuous iscomplete w.r.t. the metric ρ∞.

Proposition 2.4.8. Let < X, ρ > be a complete metric space and S ⊂ X. Then

(i) if S is a closed set in X, then < S, ρS > is a complete metric subspace of X, conversely;

(ii) if < S, ρS > is a complete metric subspace, then S is a closed set in X.

Lemma 2.4.9. Let xn be a sequence. If, for any ε > 0, there is N ∈ N such that

ρ(xn, xn+1) < ε,∀n ≥ N,

then xn is a Cauchy sequence.

Proposition 2.4.10 (Cantor’s Theorem). Let < X, ρ > be a complete metric space and Fn a sequenceof subsets of X. If, for each n, Fn is a non-empty closed set,

Fn ⊃ Fn+1 and limn→∞ diam(Fn) = 0,

then⋂∞

n=1 Fn is non-empty and contains only one element.

Proposition 2.4.11. Let < X, ρ > be a metric space. If every sequence of closed balls Brn(xn), withthe property that Brn(xn) ⊃ Brn+1(xn+1) and rn → 0, has a non-empty intersection, then < X, ρ > is acomplete metric space.

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Excercises 2.4.12. Prove the following

1. If a Cauchy sequence xn has a accumulation point x, then xn converges to x.

2. If xn and yn are two Cauchy sequences in a metric space < X, ρ >, then ρ(xn, yn) is a conver-gent sequence of real numbers.

3. Given a sequence xn. Then xn is a Cauchy sequence iff the sequence dimAn is a decreasingsequence of real number and dimAn ↓ 0; where

An := xk : k ≥ n.

4. The product∏n

i=1 Xi of metric spaces is complete iff each of the metric spaces Xi, i = 1, . . . , n, iscomplete.

2.5 Baire Category

Baire had categorized sets in to two groups as: sets of first category and second category.

Definition 2.5.1 (nowhere dense). Let < X, ρ > be a metric space. A set E ⊂ X is said to be nowheredense if X \ clE is a dense set in X.

If E is nowhere dense and A ⊂ E, then A is also nowhere dense.

Proposition 2.5.2. A subset E of a metric space is nowhere dense if and only if int(clE) = ∅; i.e. clEcontains no open ball.

Proof.

E is nowhere dense ⇔ X \ clE is dense

⇔ int [X \ (X \ clE)] = ∅ (Rem. 2.2.12)

⇔ int(clE) = ∅.

Consequently, given a set A, the boundary ∂A is a nowhere dense set.

Theorem 2.5.3 (Baire). Let X be a complete metric space and Onn∈N a countable collection of denseopen subsets of X. Then ∩On is dense in X.

Proof. See for instance P. 158 of Royden[21].

Definition 2.5.4 (sets of first and second category). A set A is of first category if A is a union of acountable collection of nowhere dense sets. If A is not of first category, then it is of second category.2

A nowhere dense set is of first category.

2sometimes sets of first category are called meager while those of first category are called nonmeager

9

Corollary 2.5.5 (Baire Category Theorem). Any complete metric space is of second category.

Proof. Follows from Thm. 2.5.3 and Def. 2.5.4. ( See also P. 89 of Shirali & Vasudeva[23] for a directproof).

Proposition 2.5.6. Every subset of a set of first category is of first category.

Proposition 2.5.7. The union of a countable collection of sets of first category is a gain of first category.

Corollary 2.5.8. If X is a complete metric space, then every non-empty open subset of X is of secondcategory; i.e. non-empty open subsets of X cannot be given by a union of a countable number of nowheredense sets.

Proof. Let ∅ 6= U ⊂ X be an open set. Assume that there is a countable collection Enn∈N of nowheredense subsets of X such that U =

⋃n∈NEn. Then, for each n ∈ N, On := X \ clEn is a dense open

subset of X. By Thm. 2.5.3, it follows that ⋂

n∈NOn

dense in X. Since U is an open set, there is x ∈ U ∩⋂n∈NOn 6= ∅. This implies x /∈ X \⋂

n∈NOn =⋃n∈NEn. But this contradicts that U =

⋃n∈NEn. Therefore, U is not of first category.

Remark 2.5.9. Let X be a complete metric space and En be a countable collection of nowhere densesets. Then, for any non-empty open set U in X, there is an element x0 ∈ U which does not belong to⋃

n∈NEn.

Theorem 2.5.10 (Baire-Hausdorff). If X is a complete metric space and A ⊂ X a set of first category,then X \A is dense in X.

Proof. LetA =

⋃

n∈NEn, where En is a nowhere dense, for each n ∈ N.

ThenX \A =

⋂

n∈N(X \ En) ⊃

⋂

n∈N(X \ clEn) .

But the sets X \ clEn are dense open subsets. Then Thm. 2.5.3 implies that⋂

n∈N(X \ clEn) is a denseset. Therefore, X \A is a dense set.

Proposition 2.5.11. Let X be a metric space.

(i) If O is an open and F is a closed sets in X, then the sets clO \O and F \ intF are nowhere dense;

(ii) Let X be a complete metric space. If a set F ⊂ X is closed and of first category, then F is nowheredense.

Corollary 2.5.12. The boundary ∂A of any set A is a nowhere dense set.

Excercises 2.5.13. Prove the following statements:

1. A closed set F is nowhere dense iff it contains no open set;

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2. A set E is nowhere dense iff for any nonempty open set O there is a ball contained in O \ E;

3. Rem. 2.5.9;

4. Cor. 2.5.12;

5. If A and B are sets of second category, then what can you say about A ∩B, A ∪B and A \B?

6. If a set E is of first category, then any subset A ⊂ E is also of first category;

7. If En is a sequence of set of first category, then ∪n∈NEn is also of first category;

8. Let E be a subset of a complete metric space. Then if X \E is dense and F ⊂ E is a closed set, thenF is a nowhere dense set.

2.6 Compact Metric Spaces

Definition 2.6.1 (compact set). A subset K of a metric space is called compact if every sequence in Khas a convergent subsequence in K; i.e. every sequence in K has an accumulation point in K. A metricspace < X, ρ > is said to be a compact metric space if X is a compact set.3

Proposition 2.6.2 (a compact set is closed). If K compact subset of a metric space, then K is a closedset.

Proof. Use Prop. 2.4.2 and Def. 2.6.1.

Let < X, ρ > be a metric space and S ⊂ X. A family Aα | α ∈ Ω of subsets of X is a covering of S if

S ⊂⋃

α∈Ω

Aα

for some index set Ω. If each Aα, α ∈ Ω, is an open set, then Aα | α ∈ Ω is called an open covering.If there is Ω′ ⊂ Ω such that

S ⊂⋃

α∈Ω′Aα,

then Aα | α ∈ Ω′ is a subcovering of Aα | α ∈ Ω. If, in this case, the index set Ω′ is finite, then wehave a finite subcovering.

Lemma 2.6.3 (Lebesgue covering Lemma). If K is a compact set and Oα | α ∈ Ω is an open coveringof K, then there is r > 0§ such that, for each x ∈ K the open ball Br(x) is contained in an element ofOα | α ∈ Ω; i.e. if K is compact, then, given x ∈ K,

∃r > 0, ∃α ∈ Ω : Br(x) ⊂ Oα.

Proof. Let Oα | α ∈ Ω is an open covering of K. Assume that, given x ∈ K, for all r > 0,Br(x) isnot a subset of an element of Oα | α ∈ Ω. Consequently, for xn ∈ K

B 1n(xn) \Oα 6= ∅,∀α ∈ Ω (*)

3The property that every sequence has a convergent subsequence is known as the Bolzano-Weierstrass propery.§The number r is usually called the Lebesgue number of the set K.

11

Now, consider the sequence xn ⊂ K. By Def. 2.6.1, xn has a convergent subsequence, sayxnk

⊂ xn and xnk→ x ∈ K. But since

K ⊂⋃

α∈Ω

Oα,

it follows that x ∈ ⋃α∈Ω Oα. This implies that, for some α0 ∈ Ω, x ∈ Oα0 . But, since Oα0 is an open

set, there is r > 0 such thatx ∈ Br(x) ⊂ Oα0 .

Thus, by the convergence of xnkto x, there is a sufficiently large nk such that

1nk

<r

2and xnk

∈ B r2(x).

Hence, for any z ∈ B 1nk

(xnk), we have

ρ(z, x) ≤ ρ(z, xnk) + ρ(xnk

, x) ≤ 1nk

+r

2≤ r

2+

r

2= r

From this follows thatB 1

nk

(xnk) ⊂ Br(x) ⊂ Oα0 .

But this is a contradiction to (*). Hence, the assumption is false and the claim of the lemma is justified.

Theorem 2.6.4. If K is a compact subset of a metric space, then, for every real number r > 0, there is afinite number of elements x1, . . . , xp of K such that the system of balls

Br(xk) | k = 1, . . . , p

is an open covering of K.

Proof. Assume that there is r > 0 such that, for any finite number of elements x1, . . . , xn of K, thesystem Br(xp) | k = 1, . . . , p does not cover K. This implies, given x1 ∈ K, then Br(x1) doesnot cover K. Hence, ∃x2 ∈ K \ Br(x1). Again Br(x1), Br(x2) does not cover K. This implies,∃x3 ∈ K \ (Br(x1) ∪Br(x2)). Proceeding in this way, given n ∈ N,

∃xn ∈ K \n−1⋃

k=1

Br(xk)

Hence, we have constructed a sequence xn ⊂ K with the property that ρ(xn, xm) ≥ r wheneverm 6= n. Since, K is compact, xn has a convergent subsequence xnk

. Hence, there is nk0 such that

ρ(xnk, xnl

) < r,∀k, l ≥ nk0 .

But this contradicts the fact that ρ(xnk, xnl

) ≥ r for nk 6= nl. Hence, the assumption is false and theclaim of the theorem holds true.

The theorem next gives an alternative definition for compactness of a set in a metric space. In fact,the following characterization is used to define compactness of a set in general topological spaces interms of coverings.

Theorem 2.6.5. Let K be a subset of a metric space. Then the following statements are equivalent:

12

(i) The set K is compact;

(ii) Every open covering of K has a finite subcovering.

Proof. ”(i) ⇒ (ii)”: Let K be a compact set and Oα | α ∈ Ω is an open covering of K. Let r > 0be the Lebesgue number of K (as given by Lem. 2.6.3). Hence, by Thm. 2.6.4, there is a finitenumber of elements x1, . . . , xn ∈ K such that the system

Br(xk) | k = 1, . . . , nis a covering of K. Again, by Lem. 2.6.3, each of the balls Br(xk) is contained in some elementof Oα | α ∈ Ω; say Br(xk) ⊂ Oαk

for some αk ∈ Ω, k = 1, . . . , n. Hence

K ⊂n⋃

k=1

Br(xk) ⊂n⋃

k=1

Oαk

Consequently, the covering Oα | α ∈ Ω has a finite subcovering of K.

”(ii) ⇒ (i)”: Let xn be any sequence in K. Define the following family of open subsets of K

O := O ⊂ K | O open and O contains a finite number of elements of the sequence xn.Then O cannot be a covering of K.( Otherwise, there will be a finite number of open setsO1, . . . , On such that K ⊂ ⋃n

k=1 Ok. From this follows that xn ⊂⋃n

k=1 Ok. This implies that,there is at least one Ok0 , 1 ≤ k0 ≤ n, that contains infinitely many element of the sequence xn,but this contradicts the definition of O). Hence,

∃x ∈ K \⋃

O∈OO.

Now, for each k ∈ N, the open ball B 1k(x) does not belong to O or cannot be a subset of any

of the elements of O. Consequently, for each k ∈ N, B 1k(x) contains infinitely many elements of

xn. Hence, there is a subsequence of xn that converges to x. Therefore, K is a compact set.

2.6.1 Bounded Sets and Totally Bounded Metric Spaces

Recall that, by Def. 2.1.3, a set B is bounded if diam B is a finite real number. Equivalently, B is abounded set if there exists a real number M > 0 such that

ρ(x, y) ≤ M,∀x, y ∈ B.

Trivially,

Lemma 2.6.6. A set B is bounded in a metric space X iff there is an open (or closed) set of finite diameterthat contains B.

Proposition 2.6.7. A closed subset of a compact metric space is compact. A compact subset of a metricspace is both closed and bounded.

Proof. (a) Let F ⊂ K be a closed set and Oα | α ∈ Ω be an open covering of F . Then the family

X \ F, Oα, α ∈ Ωis an open covering of K. This implies that, there is a finite subcovering X \ F,O1, . . . , On ofK. Consequently, O1, . . . , On must cover F . Hence, F is a compact set.

13

(b) Let K be a compact subset of a metric space. The closedness of K has been given by Prop. 2.6.2.Hence, it remains to show the boundedness. Then, by Thm. 2.6.4, there exists a real numberr > 0 and finite elements x1, . . . , xN such that

K ⊂N⋃

k=1

Br(xk).

Now define

O :=N⋃

k=1

Br(xk) and r0 := r + . . . + r︸︷︷︸N times

+ max1≤i,j≤n

ρ(xi, xj).

Then diam O ≤ r0 and K ⊂ O. Consequently, K is a bounded set.

Definition 2.6.8 (total boundedness). A metric space X is said to be totally bounded iff, for each ε > 0,there is a finite collection x1, . . . , xn of elements of X such that

∀x ∈ X, ∃xk ∈ x1, . . . , xn : ρ(x, xk) < ε.

Totally bounded metric spaces are also alternatively known as pre-compact metric spaces.

Proposition 2.6.9. If X is a totally bounded metric space, then every sequence in X contains a Cauchysubsequence.

Proposition 2.6.10. A metric space X is compact if and only if it is both complete and totally bounded.

Proof. If K is compact and r > 0, then Br(x) | x ∈ K has a finite subcover.

Obviously, a compact metric space is pre-compact (totally bounded). But, for a pre-compact metric tobe compact, it needs to be complete.

Excercises 2.6.11. Prove that

1. The intersection of any collection of compact sets is again compact; and the union of a finite numberof compact sets is compact.

2. If K is a compact subset of a metric space X, then there is a countable family An | n ∈ N ofsubsets of X, such that

cl

( ⋃

n∈NAn

)= K.

3. Every compact metric space is separable.

4. If < Xi, ρi >, i = 1, . . . , n, are compact metric spaces, then the product∏n

i=1 is also a compactmetric space. Moreover, if < Xk, ρk >k∈N is a countable collection of compact metric spaces, then∏∞

i=1 is also a compact metric space.

5. Let < X, ρ > be a metric space. A collection F of subsets of X is said to have the finite intersectionproperty iff every finite subset of F has a non-empty intersection. Then prove that: if X be acompact, then every collection F of closed subsets of X with the finite intersection property has anonempty intersection.

14

6. A metric space X is compact iff every countable collection Fn of non-empty closed sets in X,with the property Fn ⊃ Fn+1 (i.e. Fn is a nested sequence), has a non-empty intersection; i.e.∩Fn 6= ∅.

7. Show that a totally bounded metric space is second countable (has a countable basis).

8. The product metric space∏n

i=1 Xi is totally bounded iff each Xi, i = 1, . . . , n, is totally bounded.

2.7 Functions

2.7.1 Continuity of Functions

Let < X, ρ > and < Y, σ > be two metric spaces. We consider a function f : X → Y , which associates,to each x ∈ X, a unique element y ∈ Y such that f(x) = y. Sometimes, the terms ”function” and”mapping” are used interchangeably.

Definition 2.7.1 (continuous function). Let f be a function from a metric space < X, ρ > to a metricspace < Y, σ >, written f : X → Y . Then

(i) f is said to be continuous at a point x0 ∈ X, if for every ε > 0, there is a δ > 0¶ such that

σ(f(x), f(x0)) < ε, for each x with ρ(x, x0) < δ;

(ii) f is said to be a continuous function (on X) if it is continuous at every point x in X

Equivalently, f is a continuous function at x0 iff

∀ε > 0,∃δ > 0 : f(x) ∈ Bε(f(x0)), ∀x ∈ Bδ(x0).

This is the same as∀ε > 0, ∃δ > 0 : Bδ(x0) ⊂ f−1

(Bε(f(x0))

).

Let f : X → Y be a function, S ⊂ X and M ⊂ Y . Then

• the image of the set S under the mapping f is give by

f(S) := f(x) ∈ Y | x ∈ S;

• the inverse image of M under f is given by

f−1(M) = x ∈ X | f(x) ∈ M.

Thus we can easily verify the following:

Proposition 2.7.2 (inverse image of an open set). Let X and Y be metric spaces and f : X → Y be afunction. Then f is a continuous function iff, for every open set O ⊂ Y , f−1(O) is an open set in X.

Note, that if f : X → Y is continuous, then, for any closed set F ⊂ Y , f−1(F ) is a closed set in X.

Proposition 2.7.3. The image of a compact set under a continuous mapping is again a compact set.¶Properly speaking, δ depends on x0 and ε; i.e. for a different x0 we may have a different δ and to show this dependence

it is usually written δ(x0, ε).

15

Proof. Let X and Y be arbitrary metric spaces, f : X → Y and K ⊂ X be a compact set. LetOα | α ∈ Ω be an open covering of f(K) in Y . That is,

f(K) ⊂⋃

α∈Ω

Oα.

⇒K ⊂

⋃

α∈Ω

f−1(Oα).

Hence, by the continuity of f the collection f−1(Oα) | α ∈ Ω is an open covering of K (see. Prop.2.7.2). But, since K is compact, there is a finite collection O1, . . . , Op such that

K ⊂p⋃

k=1

f−1(Ok).

Consequently,

f(K) ⊂p⋃

k=1

Ok.

From this we conclude that the image set f(K) is compact.

Continuity of functions can also be characterized in terms of sequences, as indicated next.

Proposition 2.7.4. Let X and Y be metric spaces, f : X → Y and x0 ∈ X. Then f is continuous at x0

iff and only for each sequence xn that converges to x0 in X the sequence f(xn) converges to f(x0) inY .

Proposition 2.7.5 (continuity of compositions). Let < X, ρ >, < Y, σ > and < Z, % > be metric spacesand f : X → Y and g : Y → Z be functions. If f is continuous in X and g is continuous on Y , then thecomposition g f is a continuous function on X.

Proposition 2.7.6 (continuity on a product space). Let < Xi, ρi >, i = 1, . . . , n, be metric spaces and

X =n∏

i

Xi,

with the product metric ρ on X and f : X → Y be a function and < Y, σ > is a metric space. The functionf is continuous at x0 = (x0

1, . . . , x0n) iff each of the functions

fi(·) := f(x01, . . . , x

0i−1, ·, x0

i+1, . . . , x0n),

are continuous on Xi at x0i , i = 1, . . . , n.

Corollary 2.7.7. A function f := X × Y → Z is continuous iff, both functions fy(·) : X → Z, for eachfixed y ∈ Y ; and fx(·) : Y → Z, for each fixed x ∈ X, are continuous; where

fy(·) := f(·, y) and fx(·) := f(x, ·).

Corollary 2.7.8. Let < X, ρ > be a metric space. Then the metric ρ : X × X → R is a continuousfunction.

16

Excercises 2.7.9. Prove the following statements.

(a) Let πx : X × Y → X be the projection mapping and Y be a compact metric space. If F ⊂ X × Y isclosed, then πx(F ) is a closed set in X.

(b) Let f : X → Y be a function and Y is a compact metric space. If the graph of f

Graph(f) := (x, y) ∈ X × Y | y = f(x)

is a closed set in X × Y , then f is a continuous function. (Hint: for B ⊂ Y , show that f−1(B) =πx(π−1

y (B) ∩Graph(f)). Use this for a closed set B and apply excercise (a) )

2.7.2 Real Valued Functions

Given a metric space X, a function f : X → R is a real valued function. Continuous real valuedfunctions posses some very important properties.

Proposition 2.7.10. Let X be a metric space and f : X → R. Then the following hold true:

(i) if f is a continuous function, then −f is also a continuous function;

(ii) if f is a continuous function and α ∈ R, then the set

x ∈ X | f(x) < α

is an open set.

From Prop. 2.7.10, we observe that the set

x ∈ X | f(x) > α

is also an open set.

Corollary 2.7.11. If f is a continuous real valued function on a metric space X and α ∈ R, then the sets

x ∈ X | f(x) ≤ α, x ∈ X | f(x) ≥ α

andx ∈ | f(x) = α

are closed sets.

Definition 2.7.12 (upper and lower semi-continuous functions). Let f : X → R and x0 ∈ X. Then

(i) f is said to be upper semi-continuous at x0 if

∀ε > 0, ∃δ > 0 : f(x)− f(x0) < ε,∀x ∈ Bδ(x0).

Moreover, f is said to be an upper semi-continuous function on X if f is upper semi-continuousat every x ∈ X;

(ii) f is said to be lower semi-continuous(l.s.c.) if −f is upper semi-continuous.

Proposition 2.7.13. Let f : X → R. Then

17

(i) if f is an upper semi-continuous(u.s.c) function, then, for every real number α ∈ R, the set

x ∈ X | f(x) > αis an open set set.

(ii) if f is an lower semi-continuous(l.s.c) function, then, for every real number α ∈ R, the set

x ∈ X | f(x) < αis an open set set.

Corollary 2.7.14. A real valued continuous function is both lower and upper semi-continuous.

Corollary 2.7.15. Let f : X → R and let α ∈ R be any. Then

(i) if f is upper semi-continuous, then the set

x ∈ X | f(x) ≤ αis a closed set; and

(ii) if f is lower semi-continuous, then the set

x ∈ X | f(x) ≥ αis a closed set.

Proposition 2.7.16. Let f : X → R and let xn be a sequence that converges to x0 ∈ X. Then

(i) if f is u.s.c. at x0, thenlim sup

nf(xn) ≤ f(x0);

(ii) if f is l.s.c. at x0, thenlim inf

nf(xn) ≥ f(x0).

Proof. See for instance pp. 42-43 of Aliprantis & Border [1].

Proposition 2.7.17 (Weirstras’ Theorem). Let f : X → R and K a compact subset of X. Then

(i) if f is upper semi-continuous on X, then f assumes its maximum on K; i.e. the problem

maxx∈K

f(x) = sup

has a solution; equivalently, there is x∗ ∈ K such that

f(x∗) = maxx∈K

f(x) = supx∈K

f(x)

(ii) if f is lower semi-continuous on X, then f assumes its minimum on K; i.e. the problem

minx∈K

f(x)

has a solution; equivalently, there is x∗ ∈ K such that

f(x∗) = minx∈K

f(x) = infx∈K

f(x).

Proof. Cf. pp. 55-56 of Kosmol[17].

For metric spaces X and Y , a function f : X → Y is said to be bounded on a subset S ⊂ X if theimage f(S) is a bounded subset of Y . In general, f is called a bounded function if f is bounded onX.

Corollary 2.7.18. Let f : X → R and K is a compact subset of X. If f is a continuous function on X,then f assumes both its maximum and minimum values on K; hence, f is a bounded function on K.

18

2.7.3 Uniform Continuity

Definition 2.7.19 (uniform continuity). Let < X, ρ > and < Y, σ > be metric spaces and f : X → Y .Then f is said to be uniformly continuous (on X) if given ε > 0, there exists δ > 0 such that

σ(f(x), f(z)) < ε whenever ρ(x, z) < δ and x, z ∈ X.

Trivially, a uniformly continuous function, is continuous. But, the converse is not always true. Forinstance, consider the real valued function f(x) = 1

x .

Proposition 2.7.20. Let < X, ρ > and < Y, σ > be metric spaces and f : X → Y be uniformlycontinuous. Then if xn is a Cauchy sequence in X, then f(xn) is a Cauchy sequence in Y .

Proof. Let xn be a Cauchy Sequence. Suppose an ε > 0 be given. Then, by unform continuity, thereis δ > 0 such that

σ(f(x), f(z)) < ε,∀x, z : ρ(x, z) < δ.

Since, xn is a Cauchy Sequence, there is N ∈ N:

ρ(xn, xm) < δ,∀n,m ≥ N.

From this follows thatσ(f(xn), f(xm)) < ε,∀n,m ≥ N.

Proposition 2.7.21. Let < X, ρ > and < Y, σ > be metric spaces and f : X → Y be a function. Thenthe following statements are equivalent

(i) f is uniformly continuous;

(ii) for any pair of sequences xn and yn, if ρ(xn, yn) → 0, then σ(f(xn), f(yn)) → 0.

Proof. (i)” ⇒ ” (ii) Triviall!! (similar to the proof of Prop. 2.7.20).

(ii) ⇐(i) Prove by contradiction.

Proposition 2.7.22. If f : X → Y is a continuous function and X is a compact set, then f is uniformlycontinuous on X.

Proof. Assume that f is not uniformly continuous and arrive at a contradiction.

Excercises 2.7.23. Prove that:

(a) If f : X → Y is uniformly continuous and X is totally bounded, then f(X) is totally bounded.

(b) Let f : X → X be a continuous function. If X is a compact metric space, then there is a non-emptysubset A ⊂ X such that f(A) = A. (Hint: use X1 = f(X), X2 = f(X1), and so on. In general,Xn+1 = f(Xn) and A :=

⋂∞k=1 Xk).

19

2.7.4 Convergence Properties of Sequences of Functions

Let X and Y be metric spaces. Then, for each n ∈ N, we consider a function fn : X → Y . Conse-quently, for each fixed x ∈ X, we have a sequence fn(x)n∈N of elements of Y . We also refer to thesequence fn as a sequence of functions.

Definition 2.7.24 (pointwise convergence). Let X and Y be metric spaces and S ⊂ Y . A sequence offunctions fn is said to be pointwise convergent to a function f : X → Y on S if, for each fixed x ∈ S,we have

limn→∞ fn(x) = f(x).

We write fn → f pointwise on S.

Even if the functions fn are continuous, for all n ∈ N, the pointwise limit function f may not becontinuous.

Example 2.7.25. Let fn(x) = xn, x ∈ [0, 1]. Hence,

limn

fn(x) = f(x),

where

f(x) =

0, x ∈ [0, 1)1, x = 1.

Hence, for f with limn→ fn(x) = f(x) to be continuous we need a strong convergence property.

Definition 2.7.26 (uniform convergence). Let < X, ρ > and < Y, σ > be metric spaces and S ⊂ Y . Asequence of functions fn is said to be uniformly convergent to a function f : X → Y on S if, for everyε > 0, there is N ∈ N such that

σ(fn(x), f(x)) < ε,∀n ≥ N,∀x ∈ S.

In this case, we write fn → f uniformly on S.

Obviously,

• unform convergence is stronger than pointwise convergence; i.e. uniform convergence impliespointwise convergence.

• if D ⊂ S and fn is uniformly convergent on S, then it is also uniformly convergent on D.

Hence, the following is a direct consequence of Def. 2.7.26.

Proposition 2.7.27. Let S ⊂ X, fn be a sequence such that fn → f pointwise on S and, for eachn ∈ N,Mn := supx∈S σ(fn, f(x)). Then fn → f uniformly on S iff Mn → 0. In short

fn → f uniformly on S ⇔ limn→∞

[supx∈S

σ(fn(x), f(x))]

= 0.

Theorem 2.7.28. Let X and Y be metric spaces, S ⊂ X and, for each n ∈ N, fn : X → Y be a continuousfunction. If fn → f uniformly on S, then f is a continuous function on S.

20

Proof. Suppose we are given an arbitrary point x0 ∈ S. If ε > 0, then for any x ∈ S

σ(f(x), f(x0)) ≤ σ(f(x), fn(x)) + σ(fn(x), f(x0))

This implies

σ(f(x), f(x0)) ≤ σ(f(x), fn(x)) + σ(fn(x), fn(x0)) + σ(fn(x0), f(x0))

By the continuity of fn, there is a δ > 0 such that

σ(fn(x), fn(x0)) <ε

3, ∀x ∈ Bδ(x0).

Thus, by the uniform convergence of fn to f on S, we see that fn converges uniformly to f onS ∩Bδ(x0) =: BS

δ (x0) . This implies, there is N > 0 such that

σ(f(x), fn(x)) <ε

3,∀n ≥ N,∀x ∈ BS

δ (x0).

Consequently,σ(f(x), f(x0)) < ε,∀x ∈ BS

δ (x0).

Hence, f is continuous at x0 relative to S. Since, x0 ∈ S is arbitrary, we conclude that f is a continuousfunction relative to S; therefore, f is continuous on S.


(a) Prop. 2.7.27.

(b) Which of the following is uniformly convergent

(i) fn(x) = xe−nx2,

(ii) fn(x) = n2x(1− x)n,

(iii) fn(x) = x(1+nx) ,

(iv) fn(x) = xn

(1+xn) .

(c) Let fn be a sequence of real valued functions. If fn → f uniformly on S and, for each n ∈ N, fn isbounded on D, then

(i) f is bounded on S;

(ii) there is a uniform bound for fn; i.e. there is M ∈ R such that

|fn(x)| ≤ M,∀n ∈ N,∀x ∈ S.

(d) Let fn → f uniformly on S. If, for each n ∈ N, fn is uniformly continuous on X, then f is alsouniformly continuous on X.

21

2.7.5 Equicontinuiuty and the Ascoli-Arzela Theorem

In many situations we may need to know if a sequence of functions fn has a convergent subsequence.

Definition 2.7.30 (equicontinuiuty). Let F be a family of functions from a metric space < X, ρ > to ametric space < Y, σ >. The family F is said to be equicontinuous at x0 ∈ X if, for every ε > 0, there isδ > 0 such that

σ(f(x), f(x0)) < ε,∀x ∈ Bδ(x0),∀f ∈ F .

The family F is called equicontinuous (on X) if it is equicontinuous at each point x in X.

Lemma 2.7.31. Let X and Y be metric spaces and D ⊂ X, and fn be a sequence of functions fromX to Y . If D is a countable set and, for each x ∈ S, the set fn(x) | n ∈ N is compact, then there is asubsequence fnk

of fn such that fnk(x) converges for each x ∈ D.

Proof. Let D = x1, x2, . . .. For x1 ∈ D, there is a convergent subsequence f1n(x1) of fn(x1)(Observethat f1n(x1) ⊂ clf1n(x1) and f1n(x1) is compact)‖. For x2 ∈ D, there is a convergent subse-quence f2n(x2) of f1n(x2), and so on. Proceeding in this manner, for xk ∈ D, we obtain aconvergent subsequence fkn(xk) of f(k−1)n(xk−1). Hence

f11(x1), f12(x1), . . . , f11(x1), . . .f21(x2), f22(x2), . . . , f2n(x2), . . .f31(x3), f32(x3), . . . , f3n(x3), . . ....

......

fk1(xk), fk2(xk), . . . , fkn(xk), . . ....

......

Now, consider the (diagonal) sequence fnn. The sequence fnn is a subsequence of fkn for eachk ≥ n. Hence, for each xk ∈ S, fnn(xk) is convergent.

Remark 2.7.32. In Lem. 2.7.31, it is enough to have clfn(x) | n ∈ N compact, for each x ∈ D.

Lemma 2.7.33. Let fn be an equicontinuous sequence of functions from a metric space X to a completemetric space Y . If the sequence fn(x) converges for each point x in a dense subset D of X, then

(i) fn converges at each point x in X, and

(ii) the limit function is continuous.

Proof. (i) Let x ∈ X and ε > 0 be arbitrary. Then there is δ > 0 such that

σ(fn(x), fn(y)) < ε,∀y ∈ Bδ(x), ∀n ∈ N. (2.1)

Since D is a dense set, there is y ∈ D ∩ Bδ(x). Thus, by assumption, fn(y) is a convergentsequence. This implies that fn(y) is a Cauchy sequence. Hence, there is a sufficiently large Nsuch that

σ(fn(y), fm(y)) <ε

3, ∀n,m ≥ N. (2.2)

‖A discrete subset of a compact set is compact.

22

From (2.1) and (2.2), it follows that

σ(fn(x), fm(x)) ≤ σ(fn(x), fn(y)) + σ(fn(y), fm(y)) + σ(fm(x), fm(y)) < ε,∀n,m ≥ N.

This implies that fn(x) is a Cauchy sequence in Y . Since Y is a complete metric space, weconclude that fn(x) is convergent.

(ii) For x ∈ X, let f(x) = limn→∞ fn(x). To show f is continuous at x, let ε > 0 be given. Since,fn is equicontinuous at x, there is δ > 0 such that

σ(fn(x), fn(y)) < ε,∀x ∈ Bδ(x),∀n ∈ N.

This impliesσ(f(x), f(y)) = lim

n→∞σ(fn(x), fn(y)) ≤ ε, ∀x ∈ Bδ(x).

Since σ : Y × Y → R+ is continuous(see Cor.2.7.8). Therefore, f is continuous at x. Since,x ∈ X is arbitrary, we conclude that f is a continuous function.

Lemma 2.7.34. Let X and Y be metric spaces, K ⊂ X and fn be an equicontinuous sequence. If K iscompact and fn(x) converges to f(x) at each x ∈ K, then fn converges uniformly to f on K.

Proof. Let ε > 0. By the equicontinuiuty of fn, for each x ∈ K, there is an open ball Bδ(x)(x) suchthat

σ(fn(x), fn(y)) <ε

3, ∀y ∈ Bδ(x)(x), ∀n.

This implies thatσ(f(x), f(y)) <

ε

3, ∀y ∈ Bδ(x)(x),

Hence, by the compactness of K, there is a finite collection x1, . . . , xm such that

K ⊂m⋃

i=1

Bδ(xi)(xi).

Since, fn(xi) → f(xi), for each i = 1, . . . , m, choose N sufficiently large so that

σ(fn(xi), f(xi)) <ε

3, ∀i ∈ 1, . . . , m, ∀n ≥ N.

Then, for any y ∈ K, there is i0 ∈ 1, . . . , m such that y ∈ Bδ(xi0)(xi0). Hence

σ(fn(y), f(y)) ≤ σ(fn(y), fn(xi0)) + σ(fn(xi0), f(xi0)) + σ(fn(xi0), f(y)) < ε,∀n ≥ N.

Consequently, fn converges uniformly to f on K.

Using the above three lemmas and the fact that a compact subset of a metric space is compete, onecan verify the validity of the following well known theorem:

Theorem 2.7.35 (Ascoli-Arzela Theorem, Thm. 40, p. 169, Royden[21]). Let F be a family of equicon-tinuous functions from a separable metric space X to a metric space Y . Let fn be a sequence in F suchthat for each x ∈ X the closure of the set fn(x) | n ∈ N is compact. Then there is a subsequence fnk

that converges pointwise to a continuous function f , and the convergence is uniform on each compactsubset of X.

Proof. (see also pp. 189-191 of Shirali & Vasudeva [23])

23

(a) Since X is separable, there is D ⊂ X such that D is dense and countable. Then clfn(x) | n ∈ Nis a compact set, for each x ∈ D. Then, by Lem. 2.7.31, there is a subsequence fnk

of fnsuch that fnk

(x) converges for each x ∈ D.

(b) Since F is equicontinuous, then fnk is equicontinuous, too. Consequently, by Lem. 2.7.33,

fnk→ f pointwise on X, where f is a continuous function.

(c) Since fnk is equicontinuous, Lem. 2.7.34 concludes that fnk

→ f uniformly on X.

2.7.6 Homeomorphisms and Isometries in Metric Spaces

Definition 2.7.36 (a homeomorphism). Let X and Y be metric spaces. A function f : X → Y is anhomeomorphism between X and Y if

(i) f is a one-to-one and onto function; and

(ii) both f and f−1 are continuous functions.

When there is a homeomorphism between two metric spaces X and Y , we say that X and Y arehomeomorphic metric spaces.

Theorem 2.7.37. Let f : X → Y be both one-to-one and onto. Then the following statements areequivalent:

(i) f is a homeomorphism from X to Y ;

(ii) for each subset A ⊂ X, f(clA) = cl(f(A));

(iii) for each closed set F ⊂ X, f(F ) is closed in X; and for each closed set in E ⊂ Y, f−1(E) is closedin X;

(iv) for every open set O ⊂ X, f(O) is open in Y ; and for every open set U ⊂ Y , f−1(U) is open in X.

Consequently, when two metric spaces X and Y are homeomorphic, it follows that: if X is complete,then Y will be complete; if X is separable, then Y will be separable; if X is compact, then Y will becompact, and so on. In general, properties of the metric space X, that could be characterized by opensets, also hold true in Y ; vice versa. Such properties are usually known as topological properties∗∗.Hence, homeomorphic metric spaces have identical topological properties.

However, note that distance is not a topological property; i.e., even if f : X → Y is a homeomorphism,the distance ρ(x, y), for x, y ∈ X, may not be the same as the distance σ(f(x), f(y)).

Example 2.7.38. Let < X, ρ > and < Y, σ > with X = Y = R2 and, for x = (x1, x2), y = (y1, y2) ∈ R2,we have

ρ(x, y) =√

(x1 − y1)2 + (x2 − y2)2 and σ(x, y) = max|x1 − y1|, |x2 − y2|.The identity map ι : X → Y is a homeomorphism between and X and Y , but ρ(x, y) 6= σ(ι(x), ι(y)).

Definition 2.7.39 (isometry). Let < X, ρ > and < Y, σ > be two metric spaces. A mapping f : X → Yis an isometric mapping (or simply an isometry) if, for each x, y ∈ X,

ρ(x, y) = σ(f(x), f(y)).

Two metric spaces X and Y are called isometric if there is an isometry between them.∗∗In other words, a property that remains true under homeomorphic maps is said to be a topological property.

24

Accordingly, an isometry preserves distance.

Proposition 2.7.40. Let X and Y be metric spaces and f : X → Y .

(i) If f : X → Y is an isometry, then f is is a continuous function.

(ii) If f : X → Y is an isometry, then f is one-to-one.

(iii) If f : X → Y is an isometry, then f is one-to-one (injective).

Hence, an isometry f : X → Y will be a homeomorphism between X and Y if it maps X onto Y ; i.e.if it is surjective. In fact, an isometric map is a homeomorphism between X and f(X). But, not everyhomeomorphism is an isometric mapping. Hence, a homeomorphism may not preserve distance.

Excercises 2.7.41. (i) Give some examples of properties which are not topological.

(ii) Let X and Y be a metric spaces, f : X → X be a continuous function and h : Y → X is ahomeomorphism, then f h−1 is a continuous function on Y .

2.7.7 Contractive Maps and Fixed Point Properties

Definition 2.7.42 (Lipschitz Continuity). Let < X, ρ > and < Y, σ > be two metric spaces. A functionf : X → Y is said to be Lipschitz continuous, if there is a constant L > 0 such that, for each x, z ∈ X

σ(f(x), f(z)) ≤ L ρ(x, z). (2.3)

The smallest number L for which (2.3) holds is called the Lipschitz constant of the function f .

It is easy to verify that: an isometry is Lipschitz continuous - with Lipschitz constant L = 1.

Example 2.7.43. Let < X, ρ > and < Y, σ > be as given in Example 2.7.38 with the identity mapι : X = R2 → Y = R2. Then ι is a Lipschitz continuous function. (But, recall that, ι is not an isometry).

Corollary 2.7.44. Every Lipschitz continuous function is uniformly continuous.

Definition 2.7.45 (contraction and non-expansive maps). Let < X, ρ > be a metric space and f : X →X. If there is a constant γ ∈ [0, 1] such that

ρ(f(x), f(y)) ≤ γ ρ(x, y), ∀x, y ∈ X,

then

(i) f is called contractive if 0 ≤ γ < 1;

(ii) f is called non-expansive if 0 ≤ γ ≤ 1.

Trivially, a non-expansive map is Lipschitz continuous. Hence, f : X → X is a contractive map if thereis γ ∈ [0, 1) such that

ρ(f(x), f(y)) ≤ ρ(x, y).

25

Definition 2.7.46. Let < X, ρ > be a metric space and f : X → X a map. Then we say that x ∈ X is afixed point of f if

f(x) = x.

The relation f(x) = x is a fixed point equation.

Theorem 2.7.47 (Banach Fixed point Theorem). Let < X, ρ > and f : X → X. If X is a completemetric space and f is a contractive map (with γ ∈ [0, 1)), then f has a unique fixed point in X; i.e. thereis x ∈ X such that

f(x) = x.

Proof. Existence: Let x0 ∈ X be arbitrary and set x1 = f(x0), x2 = f(x1), and so on, so thatxn+1 = f(xn), n = 0, 1, 2, . . . We show that the sequence xn is convergent. Thus, it is enoughto show that xn is a Cauchy sequence. Let n > m. Note that

ρ(xk+1, xk) = ρ(f(xk), f(xk−1)) ≤ γρ(xk, xk−1) = ρ(f(xk−1), f(xk−2)) ≤ γ2ρ(xk−1, xk−2)≤ . . . ≤ γkρ(x1, x0).

Hence,ρ(xn, xm) ≤ ρ(xn, xn−1) + ρ(xn−1, xn−2) + . . . + ρ(xm+1, xm).

From this it follows that

ρ(xn, xm) ≤ (γn−1 + . . . + γm

)ρ(x1, x0) =

ρ(x1, x0)1− γ

[γm − γn] .

The sequence γn is a Cauchy sequence. Consequently, xn is a Cauchy sequence. Since, X iscomplete, there is x ∈ X such that xn → x. And f is contractive

ρ(f(xn), f(x)) ≤ γρ(xn, x).

This impliesf(x) = lim

n→∞ f(xn) = limn→∞xn+1 = x.

Therefore, f(x) = x.

Uniqueness: Let x, y ∈ X such that f(x) = x and f(y) = y. Then

ρ(x, y) = ρ(f(x), f(y)) ≤ γρ(x, y).

⇒(1− γ)ρ(x, y) ≤ 0 ⇒ ρ(x, y) ≤ 0 ⇒ x = y.

Corollary 2.7.48. Let X be a complete metric space, f : X → X and f is a contractive mapping. Ifx ∈ X is the fixed point of f , then

ρ(x, x) ≤ 11− γ

ρ(f(x), x)

for any x ∈ X.

Proof. In the the proof of Thm. 2.7.47 take x0 = x.

Cor. 2.7.48 gives an estimate of how far the chosen initial iterate x0 lies from the fixed point x of f .

26


(i) Let S ⊂ X and define function

d(x) := infz∈S

ρ(x, z) =: dist(x, S),

for each x ∈ X - known as the distance function. Then d is a Lipschitz continuous function fromX to R (R with the usual metric).

(ii) Let < Rn, d∞ >, with d∞(x, y) = max1≤i≤n |xi − yi|, b ∈ Rn and T : Rn → Rn is a mapping givenby Tx = Ax + b for an n× n matrix A = (aij)1≤1,i,j,≤n. Show that if there is α such that

n∑

j=1

|aij | ≤ α < 1, i = 1, . . . , n,

then the system of equations

xi =n∑

j=1

aijxj + bi, i = 1, . . . , n

has a unique solution. (Hint: Show that T is contractive).

(iii) Suppose X is a complete metric space and T : X → X such that, form some integer n, T (n) is acontraction, where

T (n) = T T . . . T︸︷︷︸r times

,

then T has a unique fixed point x in X and, for any x ∈ X, the sequence Tn(x) converges to x.

27

3 Topological Spaces

Definition 3.0.50 (topological space). Let X be a non-empty set. Then a family τ of subsets of X iscalled a topology on X if the following statements (axioms) hold true

A1: X and ∅ are elements of τ ;

A2: A,B ∈ τ ⇒ A ∩B ∈ τ ;

A3: for any family Aα : | α ∈ Ω ⊂ τ ,⋃

α∈Ω Aα ∈ τ .

The set X with a topology τ is called a topological spaces, denoted by < X, τ >.

Note that if τ is a topology, then the intersection of any finite number of elements of τ is again anelement of τ . In a topological space < X, τ >, the elements of τ are called open sets.

Example 3.0.51. Examples of topological spaces.

(i) Let τ1 = X, ∅, then < X, τ1 > is a topological space. The topology τ1 in known as the trivialtopology. The only open sets are X and ∅.

(ii) For X 6= ∅, let τ2 = 2X , then < X, τ2 > is a topological space. This topology in known as thediscrete topology. Here, every subset of X is an open set.

(iii) If < X, ρ > is a metric spaces and τ3 is a family of open sets of < X, ρ >, then < X, τ3 > is atopological space associated with the metric ρ. Hence, different metrics may give rise to differenttopologies on a given set.∗. A topological space which could associated with a metric space is calledmetrizable.

Observe that, there might be several topologies being defined on a given set. (see Example 3.0.51(i)& (ii)).

Proposition 3.0.52. Let τα | α ∈ Ω be any collection of topologies of a set X. Then the intersection⋂

α∈Ω

τα

is again a topology of X.

Suppose that τ and σ be two topologies on a set X. If τ ⊂ σ, then we say that σ is a stronger (finer)topology than τ ; equivalently, τ is a weaker (coarse) topology than σ. Consequently, topologies arepartially ordered with respect to ”⊂”.

On a given set X, the trivial topology is the weakest topology and the discrete topology is the strongesttopology.

Proposition 3.0.53. Let X be a non-empty set and C be any collection of subsets of X. Then there is aweakest topology W that contains C.∗If two metrics are equivalent, then they give rise to the same topology

29

Proof. Follows from Prop. 3.0.52.

Proposition 3.0.54 (relative topology). Let < X, τ > be a topological space and S ⊂ X. Then thefamily of sets

τS := S ∩O | O ∈ τis a topology on S. The topology τS is called the relative topology or the induced topology on S.

Definition 3.0.55 (closed set). A set F ⊂ X is said to be closed if X \ F ∈ τ .

Proposition 3.0.56. In a topological space < X, τ >,

(i) the sets ∅ and X are both closed and open;

(ii) if F1, . . . , Fn is any finite collection of closed subsets of X, then⋃n

i=1 Fi is closed;

(iii) if Fα | α ∈ Ω is any collection of closed subsets of X, then⋂

α∈Ω Fα is closed.

Definition 3.0.57 (closure,interior). Let < X, τ > be a topological space and A ⊂ X. Then

(i) the intersection of all closed sets containing A is called the closure of A, denoted by clA; i.e.

clA :=⋂F | A ⊂ F and X \ F ∈ τ.

(ii) the union of all open sets contained in A is called the interior of A, denoted by intA; i.e.

intA :=⋃O | O ⊂ A and O ∈ τ.

Corollary 3.0.58. For a set A, clA is the smallest closed set containing A and intA is the largest open setcontained in A.

Proposition 3.0.59. Let < X, τ > be a topological space and A,B ⊂ X. Then

(i) cl(A ∪B) = clA ∪ clB;

(ii) cl(A ∩B) ⊂ clA ∩ clB;

(iii) int(A ∩B) = intA ∩ intB;

(iv) int(A ∪B) ⊃ intA ∪ intB;

(v) int(X \A) = X \ clA.

Definition 3.0.60 (exterior, boundary). Let < X, τ > be a topological space and A ⊂ X. Then

(i) the exterior of the set A, denoted extA, is defined as

extA := int(X \A).

(ii) the boundary of the set A, denoted ∂A, is defined as

∂A := X \ (extA ∪ intA).

Definition 3.0.61 (accumulation point). Let < X, τ > be a topological space and A ⊂ X.

30

(i) A point x ∈ X is an accumulation point of A iff

O ∈ τ, x ∈ O ⇒ O ∩ (A \ x) 6= ∅.

Denote by A′ the set of all accumulation points of A.

(i) A point x ∈ A is an interior point of A iff

∃O ∈ τ, x ∈ O ⊂ A.

Denote by0A the set of all interior points of A.

Proposition 3.0.62. Let < X, τ > be a topological space. Then

(i) for any set A, it follows that A′ ⊂ clA;

(ii) F is a closed set if and only if F = clF = F ∪ F ′; i.e. if and only if F ′ ⊂ F ;

(iii) for any set A, we have0A = intA;

(iv) if O is an open set, then O = intO.

Proof. (ii),(iii) and (iv) follow trivially. Thus it remains to show (i). Let x ∈ X, but x /∈ clA. Thenthere is a closed set F such that A ⊂ F such that x /∈ F .⇒

x ∈ X \ F =: O.

⇒x ∈ O, but O ∩ (A \ x) = ∅ ⇒ x /∈ A′.

Consequently, A′ ⊂ clA.

Definition 3.0.63 (dense set). Let < X, τ > be a topological space. A subset D of X is dense in X iffclD = X.

Thus if a set D is dense in X, then any element x ∈ X is an accumulation point of D. Hence, we have

Proposition 3.0.64. If D is dense in X, then

∀O ∈ τ : O ∩D 6= ∅.

Definition 3.0.65 (separable topological space). A topological space X is separable if it has a countabledense subset D.

3.1 Neighborhood and Neighborhood Systems

Definition 3.1.1 (neighborhood). Let < X, τ > be a topological space and x ∈ X. A set U,U ⊂ X, iscalled a neighborhood of x if there is an open set O ∈ τ such that

x ∈ O ⊂ U.

If a neighborhood U is an open set, then it is called an open neighborhood.

31

Definition 3.1.2. [neighborhood system] Let x ∈ X andNx ⊂ 2X . ThenNx is said to be a neighborhoodsystem of x if the following (neighborhood axioms) are satisfied:

N1: N ∈ Nx ⇒ x ∈ N ;

N2: N ∈ Nx and N ⊂ A ⇒ N ∈ Nx;

N3: N1, N2 ∈ Nx ⇒ N1 ∩N2 ∈ Nx;

N4: N1 ∈ Nx ⇒ ∃N2 ∈ Nx : N2 ⊂ N1 and N2 ∈ Ny,∀y ∈ N2.

According to Def. 3.1.2, the set of all neighborhoods of a point x, satisfies the axioms N1 - N4.

Proposition 3.1.3 (relative neighborhood). Let < X, τ > be a topological space, S ⊂ X with therelative topology τS and x ∈ S. If V ⊂ S is a neighborhood of x w.r.t. the relative topology τA, there is aneighborhood U of x w.r.t. τ such that

V = S ∩ U.

The neighborhood V is called the relative neighborhood of x; conversely, if U is a neighborhood of x inX, then U ∩ S is a neighborhood of x in S.

Excercises 3.1.4. Suppose that < X, τ > is a topological space and prove the following

1. for any set A, ∂A = clA ∪ cl(X \A);

2. if U ∈ τ , then, for each x ∈ U , U is a neighborhood of x;

3. if x is an accumulation point of A, then x is also an accumulation point of A \ x;

4. let S ⊂ X and G ⊂ S, then G is a closed set in S if there is closed set F in X such that G = F ∩ S;

5. given A ⊂ X, what condition should be satisfied, so that a point x ∈ X is not an accumulationpoint of A;

6. if τ is the discrete topology and A ⊂ X, then the set of accumulation points A′ = ∅;

7. if A ⊂ B, then clA ⊂ clB and intA ⊂ intB;

8. if D is dense in X and D ⊂ D2 ⊂ X, then D2 is also dense in X;

9. for any set A, ∂A = extA ∪ intA is cannot be a dense set in X;

10. for any set A, clA = intA ∪ ∂A; i.e. ∂A ⊂ clA;

11. if Nx is a neighborhood system of x, then ∩N | N ∈ Nx ∈ Nx.

32

3.2 Bases and Subbases

Definition 3.2.1 (a base). Let B be any collection of open sets in a topological space < X, τ >; i.e.B ⊂ τ . Then B be is said to be a base for the topology τ iff for each open set O and each x ∈ O, there is aset B ∈ B such that x ∈ B ⊂ O.

Proposition 3.2.2. (base) Let < X, τ > be a topological space. A collection B ⊂ τ is a base for τ iff everyO ∈ τ is a union of sets from B.

Example 3.2.3.

(i) If B := (a, b) | a, b ∈ R is a collection of all open intervals, then B is a base for the usual topologyon R (i.e. for the topology generated by the absolute value metric).

(ii) Let < X, τ > be the discrete topological space. Then the collection B = x | x ∈ X is a base forτ .

Suppose we have some collection B of sets, is there a topology for which B is a base? In other words,given a collection B can we generate a topology for which B is a base? The following statement givesa necessary and sufficient condition that a collection B in order to be a base for some topology τ .

Proposition 3.2.4. Let B be a collection of subsets of X, X 6= ∅. Then B is a base for some topology onX if and only

(i) X =⋃B | B ∈ B;

(ii) for any two sets B1, B2 ∈ B, if x ∈ B1 ∩B2, then there is B3 ∈ B such that x ∈ B3 ⊂ B1 ∩B2.

Proof. ⇒: Suppose B is a base for some topology τ on X. Since X is open and, for B1, B2 ∈ B,B1 ∩B2 is open. Hence, the claim follows by Def. 3.2.1.

⇐: Suppose given B that (i) and (ii) are satisfied. New define the collection

τ := U | x ∈ U ⇒ ∃B ∈ B : x ∈ B ⊂ U.

Claim: (a) τ is a topology and (b) B is a base for τ .

(a) Obviously, B ⊂ τ , X ∈ τ , ∅ ∈ τ . Moreover, for any family Uα | α ∈ Ω, we have∪α∈ΩUα ∈ τ .

Now, let U1, U2 ∈ τ and x ∈ U1 ∩ U2. Since x ∈ U1 and x ∈ U2, there is B1, B2 ∈ B suchthat x ∈ B1 ⊂ U1 and x ∈ B2 ⊂ U2. By (ii), there is B3 such that x ∈ B3 ⊂ B1 ∩ B2.Consequently, x ∈ B3 ⊂ U1 ∩ U2. This implies U1 ∩ U2 ∈ τ . Hence, τ is a topology.

(b) Suppose O ∈ τ and x ∈ O be any. Then by definition of τ , ∃B ∈ B such that x ∈ B ⊂ O.Thus, B is a base for τ .

Corollary 3.2.5. Let S be an arbitrary collection of set. If X =⋃S | S ∈ S and B is the collection of

all finite intersections of elements of S, then B is a base for some topology on X.

33

Proof. We verify (i) and (ii) of Prop. 3.2.4. In fact, (i) is obvious, since X =⋃S | S ∈ S =⋃B | B ∈ B. Then, let B1, B2 ∈ B. Then there are S1

1 , . . . , S1n1∈ S and S2

1 , . . . , S2n2∈ S such that

B1 =n1⋂

k=1

S1k and B2 =

n2⋂

k=1

S2k .

Hence, if x ∈ B1 ∩B2, then

x ∈ B3 :=n1⋂

k=1

S1k ∩

n2⋂

k=1

S2k .

Consequently, ∃B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2. Therefore, by Prop. 3.2.4, B is a base for sometopology on X.

Definition 3.2.6. (subbase) A family S of sets is a subbase if the collection B of all finite intersections ofelements of S is a base for some topology τ on X =

⋃S | S ∈ S.

Definition 3.2.7 (neighborhood base). Let X be a topological space and x ∈ X. Then the collection Nx

of open set that contain x is said to be a neighborhood (local) base at x if for any set U with x ∈ U ,there is Nx ∈ Nx such that x ∈ Nx ⊂ U .

Proposition 3.2.8. If B is a base for some topology τ on X and x ∈ X, then

Nx := B ∈ B | x ∈ B

forms a neighborhood base at x.

Proposition 3.2.9. Let < X, τ > be a topological space and A ⊂ X. Then x ∈ X is an accumulationpoint of A; i.e. x ∈ clA, if and only if,

∀N ∈ Nx,∃y ∈ N ∩A : y 6= x.

Definition 3.2.10 (first countability). A topological space X is said to satisfy the first axiom of count-ability (or X is first countable) if there exists a countable neighbourhood base Nx, for each x ∈ X.

Proposition 3.2.11. Every metric space is first countable.

Remark 3.2.12. If X is first countable, then every x ∈ X has a countable neighborhood base, sayNx = Bn | n ∈ N, we can also assume, w.l.o.g, that

B1 ⊃ B2 ⊃ B3 ⊃ . . .

Definition 3.2.13 (second countability). A topological space < X, τ > is said to satisfy the secondaxiom of countability (or < X, τ > is second countable) if there is a countable base for τ .

Example 3.2.14. For each, n ∈ N, the space X = Rn with the usual topology is second countable. Forinstance, the collection B := (a, b) | a, b ∈ Q is a countable base for the space X = R.

Proposition 3.2.15. Every second countable space is first countable.

Proposition 3.2.16. If a topological space is second countable, then it is separable.

34

Proof. Let B = Bn | n ∈ N. Define the set

D := xn ∈ Bn.

To show D is a dense subset of X, for x ∈ X \D, we show that x is an accumulation point of D. LetU be any open set such that x ∈ U . Since B is a base, there is some Bn0 ∈ B such that

x ⊂ Bn0 ⊂ U,

xn0 ∈ Bn0 ⊂ U and x 6= xn0 . Consequently, U contains an element of D other that x. Since U isarbitrary, we conclude that x is an accumulation point of D. Hence, clD = X and D is a countabledense subset of X. Therefore, X is separable.

Corollary 3.2.17. For each n ∈ N, then space X = Rn, with the usual topology, is separable.

Definition 3.2.18 (a covering). Let < X, τ > be a topological space. A family of sets Uα | α ∈ Ω issaid to be a covering of X if

X ⊂⋃

α∈Ω

Uα.

A covering Uα | α ∈ Ω is an open covering if each of the sets Uα is an open set; i.e. Uα ∈ τ, ∀α ∈ Ω.

Theorem 3.2.19 (Lindelof). Let < X, τ > be second countable and A be any subset of X. Then everyopen covering of A has a countable subcover.

Proof. Since X is second countable, τ has a countable basis B = Bn | n ∈ N. Suppose that Oα | α ∈Ω is an open covering of A; i.e.

A ⊂⋃

α∈Ω

Oα.

This implies, for each x ∈ A, there is α ∈ Ω such that x ∈ Oα. Since B is a base for τ , there is Bn ∈ Bsuch that x ∈ Bn ⊂ Oα. If we now let Oαn = Oα whenever Bn ⊂ Oα, then there is N ⊂ N such that

A ⊂⋃

n∈N

Bn ⊂⋃

Oαn .

Consequently, Oαn | n ∈ N is a countable subcovering of A.

Definition 3.2.20 (a Lindelof topological space). A topological space X is said to be Lindelof if everyopen covering of X has a countable subcover.

Hence, the space Rn is Lindelof.

Theorem 3.2.21. Let < X, ρ > be a metric space. Then the following statements are equivalent.

(i) < X, ρ > is separable;

(ii) < X, ρ > satisfies the second axiom of coutability;

(i) < X, ρ > is Lindelof;

Proof. Exercise!

35

3.3 Sequences, Continuity and Homeomorphism

Definition 3.3.1 (convergent sequence). Let < X, τ > be a topological space. A sequence xn ⊂ X issaid to converge to an element x0 ∈ X if for every neighborhood U of x0 (i.e. ∀U ∈ Nx0) there is N ∈ Nsuch that

xn ∈ U,∀n ≥ N.

In this case we write xn → x0 and we call x0 the limit of xn . When such an x0 exists the sequencexn is called a convergent sequence.

Definition 3.3.2 (continuity). Let < X, τ > and < Y, σ > be topological spaces. Then f : X → Y is acontinuous function if, for every open set U ⊂ Y , f−1(U) is open in X.

Proposition 3.3.3. A function f : X → Y is continuous if and only if the inverse image of any closed setis closed.

Proof. ”⇒”: Suppose f : X → Y be continuous and G ⊂ Y is a closed set. Then

Y \G open ⇒ f−1(Y \G) is an open set in X .

Since f−1(Y \G) = X \ f−1(G). Hence, f−1(G) is a closed set.

”⇐”: Let U ⊂ Y be any open set. Then Y \ U is a closed set. Then, by assumption, f−1(Y \ U) is aclosed set. Since, f−1(Y \ U) = X \ f−1(U), we have that f−1(U) is an open set in X. Hence, fis a continuous function.

Proposition 3.3.4. Let < X, τ > and < Y, σ > be topological spaces, f : X → Y be a function and xnbe a sequence in X. If xn → x, then f(xn) → f(x).

Definition 3.3.5 (continuity at a point). A function f : X → Y is continuous at a point x ∈ X if forany open neighborhood U of f(x) in Y , there is an open neighborhood O of x such that

∀x ∈ O : f(x) ∈ U ; i.e. O ⊂ f−1(U).

Proposition 3.3.6. A function f : X → Y is continuous iff f is continuous at each point in X.

Definition 3.3.7 (sequential continuity). A function f : X → Y is sequentially continuous if for everyconvergent sequence xn with xn → x0, f(xn) is a convergent sequence and f(xn) → f(x0).

Proposition 3.3.8. If f : X → Y is a continuous function, then f is sequentially continuous.

Proof. Follows directly from Prop. 3.3.4.

Remark 3.3.9. The converse of Prop. 3.3.8 is not always true.

Proposition 3.3.10. If f : X → Y is a sequentially continuous, then, for every A ⊂ X,

f(clA) ⊂ clf(A).

Definition 3.3.11 (open, closed functions). Let f : X → Y be a function. Then

(i) f is said to be an open function (an open map) if, for every open set O ⊂ X, f(O) is an open set inX;

(ii) f is said to be a closed function (a closed map) if, for every closed set F ⊂ X, f(F ) is a closed setin X.

36

Example 3.3.12. The function π : R2 → R given by π(x, y) = x. Then π(·) is an open mapping, but nota closed one.

Definition 3.3.13 (homeomorphism). Two topological spaces X and Y are said to be homeomorphic ifthere is a one-to-one and onto function f : X → Y with both f and f−1 are continuous.

Proposition 3.3.14. Let X and Y be topological spaces and f : X → Y . Then f−1 is a continuousfunction if and only if f is an open map; equivalently, if and only if f is a closed map.

Corollary 3.3.15. Let X and Y be topological spaces. A function f : X → Y is a homeomorphism if andonly if f is a non-to-one open or closed map.


1. Prop. 3.3.4.

2. Prop. 3.3.10.

3.4 Classification of Topological Space: Separation Axioms

If two elements x, y of a metric space X are not equal, then is it possible to put these elements in twoseparate open sets. But, this is not always possible in a general topological space(unless the topologicalspace is metrizable). Hence, we have the following major classification of topological spaces based onseparation axioms.

(i) A topological space X is a T1-space (Frechet-Riesz) if whenever x, z ∈ X and x 6= z, there existopen (not necessarily disjoint) sets O1 and O2 of X such that

x ∈ O1, z /∈ O1 and z ∈ O2, x /∈ O2;

(ii) A topological space X is a T2-space (Hausdorff) if whenever x, z ∈ X and x 6= z, there existdisjoint open sets O1 and O2 of X such that

x ∈ O1 and z ∈ O2;

(iii) A topological space X is called regular (Vietoris) if for each x ∈ X and each closed set F of Xwith x /∈ F , there exist disjoint open sets O1 and O2 such that

x ∈ O1 and F ⊂ O2;

(iv) A topological space is a T3-space if it is regular and T1;

(v) A topological space X is called normal (Tietze) if whenever F1, F2 ⊂ X are disjoint closed sets,there exist disjoint open sets O1 and O2 such that

F1 ⊂ O1 and F2 ⊂ O2;

37

(vi) A topological space is called T4 if it is normal and T1;

Proposition 3.4.1. Let < X, τ > be a topological space. Then

X is T4 ⇒ T3 ⇒ T2 ⇒ T1.

Proposition 3.4.2. A topological space X is T1 iff, for every x ∈ X, x is a closed set; i.e. every singletonis closed.

Proof. ”⇒” : Suppose a topological space X is T1. For x ∈ X, we show that x is a closed set.Consider the set X \ x and let z ∈ X \ x. Then x 6= z. Since X is T1, there is an open set Oz

such thatz ∈ Oz but x /∈ Oz.

⇒X \ x =

⋃

z∈X\xOz.

Consequently, X \ x is open. Hence, x is closed.

”⇐” : Suppose for each x ∈ X, x is closed. Let x, z ∈ X and x 6= y. If O1 := X \ x andO2 := X \ z, then O1 and O2 are open sets, z ∈ O1, x /∈ O1 and x ∈ O2, z /∈ O2. Therefore, Xis T1.

Corollary 3.4.3. Every finite subset of a T1 space is closed.

Proposition 3.4.4. If < X, τ > is a Hausdorff topological space and S ⊂ X, then < S, τS > is also aHausdorff topological space.

Proposition 3.4.5. In a Hausdorff topological space, every convergent sequence has a unique limit.

Proof. Let xn be a sequence such that xn → x and xn → z. Assume that x 6= z. Since, X isHausdorff, there are disjoint open sets O1 and O2 such that

x ∈ O1 and z ∈ O2.

Since xn → x, there is N such thatxn ∈ O1,∀n ≥ N.

From this follows that, there are only a finite number of elements of xn that can belong to O2. Thisimplies that xn cannot converge to z. But this is a contradiction. Hence, the assumption is false andx = z.

Proposition 3.4.6. Suppose X is a topological space that is first countable. Then the following statementsare equivalent:

(i) X is Hausdorff;

(ii) every convergent sequence in X has a unique limit.

Proof. ”(i)⇒ (ii)”: See Prop. 3.4.5.

38

”(ii)⇒ (i)”: Assume that X is not Hausdorff. Hence, there are elements x, z ∈ X, x 6= z, such thatevery open neighborhood Ox of x and Uz of z have a non-empty intersection; i.e. Ox ∩ Uz 6= ∅.

By first countability of X, let On and Un be a decreasing sequence of open neighborhoodsof x and z (see Rem.3.2.12), respectively. Then, there is a sequence xn such that

xn ∈ On ∩ Un 6= ∅.Then xn → x and xn → z. By assumption x = z. But this is a contradiction. Therefore, X isHausdorff.

Theorem 3.4.7. Let X be a topological space. Then the the following statements are equivalent:

(i) X is normal;

(ii) whenever F is a closed and O an open sets, with F ⊂ O, there is an open set U such that

F ⊂ U ⊂ clU ⊂ O.

Proof. (a) Let F be a closed and O be an open sets with F ⊂ U . Then F2 := X \O is a closed set andF ∩ F2 = ∅. Since X is a normal space, there exist disjoint opens set U and U1 such that

F ⊂ U and F2 ⊂ U1

HenceU ∩ U1 = ∅ ⇒ U ⊂ X \ U1 and F2 = X \O ⊂ U1 ⇒ X \ U1 ⊂ O

⇒F ⊂ U ⊂ X \ U1 ⊂ O.

Since X \ U1 is a closed set, we further have clU ⊂ X \ U1. Consequently,

F ⊂ U ⊂ clU ⊂ O.

(b) Suppose F1 and F2 are disjoint closed sets. Then F1 ⊂ X \F2 =: O and O is open. By assumption,there is an open set U such that

F1 ⊂ U ⊂ clU ⊂ X \ F2.

It follows that F1 ⊂ U and F2 ⊂ X \ clU . Setting U1 := X \ clU , we see that U ∩ U1 = ∅.

3.5 Uryson’s Lemma, Tietze’s Extension Theorem and Metrizability

Consider the interval [0, 1], we construct the following set of real numbers

D0 = 0, 1, D1 = 0,12, 1, D2 = 0,

14,12,34, 1, D3 = D2 = 0,

18,14,38,12,58,34,78, 1, . . .

Define D to be the union of all these set as

D :=∞⋃

n=0

Dn.

This set is known as the set of all dyadic rational numbers obtained by dividing [0, 1].

39

Remark 3.5.1. Note that the sequence 12n ⊂ D. Moreover, for any k with 1 ≤ k ≤ n, 2k

2n = 12n−k ∈ D.

Lemma 3.5.2. The set D of dyadic rational numbers of [0, 1] is dense in [0, 1]; i.e. clD = [0, 1].

Proof. We show that every open interval of [0, 1] contains an element of D. Let x ∈ [0, 1] be any and(x− δ, x + δ), for an arbitrary δ > 0.Since 1

2n → 0, for the given δ > 0, there is n0 ∈ N such that

0 <1

2n0< δ.

Now set q := 2n0 . Then it follows that 0 < 1q < δ and

x ∈ [0, 1] =n0⋃

k=0

[2k − 12n0

,2k

2n0

].

Hence, there is m, 1 ≤ m ≤ q = 2n0 (i.e., m ∈ 2k | k = 0, . . . , n0), such that

x ∈[m− 1

q,m

q

]

⇒ m−1q ≤ x ≤ m

q . Since 0 < 1q < δ, we have

x− δ < x− 1q≤ m

q− 1

q=

m− 1q

≤ x < x + δ

Consequently,m− 1

q∈ (x− δ, x + δ).

Note that i−1q | i = 1, . . . q ⊂ D. Hence, (x − δ, x + δ) ∩ D 6= ∅. Since, x ∈ [0, 1] and δ > 0 are

arbitrary, we conclude that D is dense in [0, 1].

Theorem 3.5.3 (Uryson’s Lemma). Let X be a topological space. Then the following statements areequivalent:

(i) X is a normal space;

(ii) for any two disjoint subsets F1 and F2 there is a continuous function f : X → [0, 1] such that

f(F1) = 0, f(F2) = 1.That is

f(x) =

0, if x ∈ F1

1, if x ∈ F2.

Proof. (ii) ⇒ (i): Let F1 and F2 be any two closed sets in X. By assumption there is a continuousfunction f : X → [0, 1] such that f(F1) = 0 and f(F2) = 1. The sets [0, 1

3) and (13 , 1] are

open sets in [0, 1]. Consequently, by the continuity of f , it follows that

U1 := f−1([0,13)) and U2 := f−1((

13, 1])

are open sets in X and U1 ∩ U2 = ∅; moreover

F1 ⊂ U1 and F2 ⊂ U2.

Therefore, X is a normal space.

40

(i) ⇒ (ii): Let X be a normal space and F1 and F2 are any two disjoint closed sets. Then F1 ⊂X \ F2 =: O2 and O2 is an open set. By Thm. , there is an open sets U 1

2such that

F1 ⊂ U 12⊂ clU 1

2⊂ O2

Using Thm. again we obtain open sets U 14

and U 34

such that

F1 ⊂ U 14⊂ clU 1

4⊂ U 1

2⊂ clU 1

2⊂ U 3

4⊂ clU 3

4⊂ O2

Proceeding in this manner we construct a sequence of sets Ut | t ∈ D, corresponding to the setD of dyadic rational numbers of [0, 1]. Then the following hold true

• for each t ∈ D, clUt ⊂ X \ F2 ⇒ clUt ∩ F2 = ∅,∀t ∈ D;

• for t1, t2 ∈ D and t1 < t2, it follows that F1 ⊂ clUt1 ⊂ Ut2 ⊂ X \ F2.

• for each t ∈ D, F1 ⊂ Ut; in particular

F1 ⊂ U 12n

,∀n ∈ N.

Now define the function f : X → [0, 1] such that for x ∈ X

f(x) := inft | x ∈ UtThen f is a well defined function.

(a) Now let x ∈ F1 be any. Since x ∈ F1 ⊂ U 12n

, ∀n ∈ N, it follows that

0 ≤ f(x) ≤ 12n

, ∀n ∈ N⇒ f(x) = 0.

Since x ∈ F1 is arbitrary, we conclude that f(F1) = 0.(b) If x ∈ F2, then x /∈ Ut, ∀t ∈ D. In particular, given n ∈ N

x /∈ clUt,∀t ∈ D, t ≤ n

n + 1⇒

f(x) ≥ n

n + 1.

But this holds true for any given n ∈ N. Hence,

1 ≥ f(x) ≥ n

n + 1, ∀n ∈ N.

⇒ f(x) = 1. Moreover, since x ∈ F2 is arbitrary, we conclude that

f(x) = 1, ∀x ∈ F2 ⇒ f(F2) = 1.(c) It remains, now to show that f is a continuous function. For any α ∈ [0, 1], if we show that

f−1([0, α)) and f−1((α, 1]) are open sets, then we are done.

(i) First, we claim thatf−1([0, α)) = ∪Ut | t ∈ D, t < α.

To show this, let x ∈ f−1([0, α)) ⇒ 0 ≤ f(x) < α. Hence, by the definition of f , thereis t ∈ D such that f(x) < t < α and x ∈ Ut. This implies that f−1([0, α)) ⊂ ∪Ut | t ∈D, t < α.

Conversely, let x ∈ ∪Ut | t ∈ D, t < α. Then x ∈ Ut0 , for some t0 ∈ D and t0 < α ⇒f(x) ≤ t0 < α ⇒ x ∈ f−1([0, α)). Consequently, ∪Ut | t ∈ D, t < α ⊂ f−1([0, α)).

41

(ii) Next, we claim that

f−1((α, 1]) = ∪X \ clUt | t ∈ D, α < t.

Let x ∈ f−1((α, 1]), then α < f(x) ≤ 1. Since D is dense in [0, 1], there are t1, t2 ∈ Dsuch that α < t1 < t2 < f(x). This implies that x /∈ Ut2 . Furthermore, from t1 < t2, wehave clUt1 ⊂ Ut2 . Consequently, x /∈ clUt1 ⇒ x ∈ X \ clUt1 . Hence,

f−1((α, 1]) ⊂ ∪X \ clUt | t ∈ D, α < t.

Conversely, let x ∈ ∪X \ clUt | t ∈ D, α < t. Then x ∈ X \ clUt0 , for some t0 > α ⇒x /∈ clUt0 . Moreover, for any t ∈ D, t < t0, we have Ut ⊂ Ut0 ⊂ clUt0 . Consequently,x /∈ Ut,∀t < t0. From this follows that

f(x) = inft | x ∈ Ut ≥ t0 > α.

⇒ α < f(x) ≤ 1 ⇒ x ∈ f−1((α, 1]). Hence,

X \ clUt | t ∈ D, α < t ⊂ f−1((α, 1]).

From (a),(b) and (c), the claim of the theorem follows.

Corollary 3.5.4 (a generalization of Uryson’s Lemma). Let X be a topological space and a, b are anytwo real numbers, with a < b. Then the following statements are equivalent:

(i) X is a normal space;

(ii) for any two disjoint subsets F1 and F2 there is a continuous function f : X → [a, b] such that

f(F1) = a, f(F2) = b.

That is

f(x) =

a, if x ∈ F1

b, if x ∈ F2.

Definition 3.5.5 (Tychonoff or completely regular spaces). A topological space X is completely regularor Tychonoff iff for any closed set F of X and p ∈ X, with p /∈ F , there exists a continuous functionf : X → [a, b] such that f(p) = a, f(F ) = b.

Proposition 3.5.6. If a topological space X is completely regular, then X is regular.

Definition 3.5.7 (a T3 12

or Tychonoff space). A T1 topological space which is also completely regular iscalled T3 1

2or a Tychonoff space.

3.5.1 Tietze’s Extension Theorem

Theorem 3.5.8 (Tietze’s Extension Theorem ). Let X be a topological space and F be a closed subset ofX. If X is normal and f is a continuous real valued function such that f : F → [0, 1]. Then there is acontinuous real valued function g with g : X → [0, 1] such that g|F = f ; i.e. g(x) = f(x) for x ∈ F †.

†g|F is the restriction of f to the set F .

42

Proof. Define the following sets

A1 := x ∈ F | f(x) ≤ 13

B1 := x ∈ F | f(x) ≥ 23.

Then both A1 and B1 are closed sets and A1 ∩ B1 = ∅. Then, by Uryson’s Lemma (see Cor. 3.5.4),there is a continuous funciton f1 : X → [13 , 2

3 ] such that f(A1) = 13 and f(B1) = 2

3 .

Hence, for x ∈ F ,

|f(x)− f1(x)| ≤

13 − 0 = 1

3 , if f(x) < 13

1− 23 = 1

3 , if f(x) > 23

23 − 1

3 = 13 , if 1

3 ≤ f(x) ≤ 23 , since 1

3 ≤ f(x) ≤ 23 .

⇒ |f(x)− f1(x)| ≤ 13 , for x ∈ F . Then the function h1 := f − f1 maps F to [0, 1

3 ].

Repeating the above process, let

A2 := x ∈ F | h1(x) ≤ 19 = x ∈ F | h1(x) ≤ 1

32

B2 := x ∈ F | h1(x) ≥ 29 = x ∈ F | h1(x) ≥ 2

32.

Thus A2 and B2 are closed and disjoint sets. Hence, there is a continuous function f2 : X → [19 , 29 ]

such that f(A2) = 19 and f(B2) = 2

9 . Furthermore, we have

|f(x)− (f1(x) + f2(x)) | = | (f(x)− f1(x))− f2(x)| = |h1(x)− f2(x)| ≤ 132

.

Proceeding inductively, we construct a sequence of closed sets An and Bn, with An ∩Bn = ∅ anda sequence of continuous functions

fn : X →[

13n

,23n

]

such that fn(An) = 13n and fn(Bn) = 2

3n and

|f(x)−n∑

k=1

fk(x)| ≤ 13n

, ∀x ∈ F.

Now let sn(x) :=∑n

k=1 fk(x). Then, for each n, sn(·) is a continuous function on X. Moreover,

limn→∞ sup

x∈F|f(x)− sn(x)| ≤ lim

n→∞13n

= 0.

This implies, sn(x) → f(x) uniformly on F . Thus

f(x) =∞∑

k=1

fk(x), x ∈ F. (3.1)

Moreover, for each x ∈ X, we have

∞∑

k=1

|fk(x)| ≤∞∑

k=1

23k

= 2∞∑

k=1

13k

= 2

( ∞∑

k=0

13k− 1

)= 2

(32− 1

)= 1. (3.2)

43

This implies∑∞

k=1 fk(x) is summable for each x ∈ X. Now, define

g(x) :=∞∑

k=1

fk(x), x ∈ X.

Claim:

(i) if x ∈ F , then g(x) = f(x);

(ii) 0 ≤ g(x) ≤ 1, ∀x ∈ X;

(iii) g is a continuous function on X.

Claim (i) follows from (3.1) and by the definition of g. Claim (ii) is has been shown in (3.2). Thus itremains to show (iii).

Let x ∈ X be any fixed element. Then for any z ∈ X we have

|g(x)− g(z)| =∣∣∣∣∣∞∑

k=1

fk(x)−∞∑

k=1

fk(z)

∣∣∣∣∣ ≤∞∑

k=1

|fk(x)− fk(z)| . (3.3)

We know that∑n

k=113n →

∑∞k=1

13k . Hence, for any given ε > 0, there is n0 ∈ N such that∣∣∣∣∣∞∑

k=1

13k−

n∑

k=1

13k

∣∣∣∣∣ <ε

2, ∀n ≥ n0.

⇒ ∞∑

k=n0+1

13k

<ε

2.

Thus, from 3.3, it follows that

|g(x)− g(z)| ≤∞∑

k=1

|fk(x)− fk(z)| =n0∑

k=1

|fk(x)− fk(z)|+∞∑

k=n0+1

|fk(x)− fk(z)|

≤n0∑

k=1

|fk(x)− fk(z)|+∞∑

k=n0+1

13k

≤n0∑

k=1

|fk(x)− fk(z)|+ ε

2.

Since, for each k = 1, 2, . . . , n0, the function fk is continuous at x, there is an open neighborhoodUk(x) such that

|fk(x)− fk(z)| < ε

2n0,∀z ∈ Uk(x).

Set U(x) := ∩n0k=1Uk(x). Then U(x) is an open neighborhood of x and

|g(x)− g(z)| < n0

(ε

2n0

)+

ε

2= ε,∀z ∈ U(x).

Consequently, g is continuous at x. Since x ∈ X is arbitrary, we conclude that g is a continuousfunction.

44

Corollary 3.5.9. Let X be a topological space and F be a closed subset of X. If X is normal and f is acontinuous real valued function such that f : F → [a, b], a, b ∈ R and a < b. Then there is a continuousreal valued function g with g : X → [a, b] such that g|F = f ; i.e. g(x) = f(x) for x ∈ F .

Theorem 3.5.10 (generalized Tietze’s extension theorem). Let X be a topological space and F be aclosed subset of X. If X is normal and f is a continuous real valued function such on F . Then there is acontinuous real valued function g on X such that g|F = f ; i.e. g(x) = f(x) for x ∈ F .

3.5.2 Urysohn’s Metrizability

Definition 3.5.11 (metrizabllity). A topological space is said to be metrizable if it is homeomorphic toa metric space.

A metrizable topological space inherits all the properties of the metric space associated with it. Butnot all topological spaces are metrizable; i.e. there are metric spaces which are not metrizable.

Definition 3.5.12 (Hilbert Cube). The cartesian product H =∏∞

k=1[0, 1] =: [0, 1]ℵ0 is known as theHilbert cube.

Lemma 3.5.13. Let Hs be the set of all sequences of the form

Hs := xn | 0 ≤ xn ≤ 1n

and, for x = xn, y = yn ∈ Hs, let

ρ(x, y) :=

[ ∞∑

k=1

(xk − yk)2] 1

2

.

Then

(i) ρ is a metric on Hs;

(ii) < Hs, ρ > is a metric space; and

(iii) < Hs, ρ > is homeomorphic to a subspace of the Hilbert cube H.

Lemma 3.5.14. Let X be a T4 topological space and B be a basis for X. If U ∈ B, then, for every x ∈ U ,there exists Ux ∈ B such that

x ∈ clUx ⊂ U.

Proof. Let U ∈ B and x ∈ U be any. Since X is a T1 space, the set x is closed. Hence, x ⊂ U .Then, by Thm. 3.4.7, there exists an open set G such that

x ⊂ G ⊂ clG ⊂ U.

⇒ x ∈ G and G is an open set. Consequently, there is Ux ∈ B such that x ∈ Ux ⊂ G. This implies

x ∈ Ux ⊂ clUx ⊂ clG ⊂ U.

⇒x ∈ clUx ⊂ U.

45

Theorem 3.5.15 (Urysohn’s Metrizability). Every second countable T4 space is metrizable.

Proof. Let X be a second countable T4 space. If X is finite, the the claim follows trivially. Hence,assume w.l.o.g. that X is infinite. In this case we show that X is homeomorphic to a subspace of Hs.

Since X is second countable, then X has a countable base Un. Then, by Lem. 3.5.14, for eachUk ∈ B, there is Ui ∈ B such that

clUi ⊂ Uk.

Then the system(Ui, Uk) | clUi ⊂ Uk; Ui, Uk ∈ B

is countable. Consequently, we can use the representation Pn := (Uin , Ukn), where clUin ⊂ Ukn , n ∈ N.

Hence, for each n ∈ N and Pn = (Uin , Ukn), the sets clUin and X \ Ukn are disjoint closed sets. Then,by Urysohn’s Lemma, there is a continuous function fn : X → [0, 1] such that

fn(clUin) = 0 and fn(X \ Ukn) = 1.

Hence, for each x ∈ X and each n ∈ N,∣∣∣∣fn(x)

2n

∣∣∣∣ ≤12n

≤ 1n

Now, for x ∈ X, if we define

f(x) =

fn(x)2n

n∈N,

then f is a function from X to Hs; i.e. f : X → Hs. Furthermore, we claim that

• f is a one-to-one;

•f is continuous;

• f−1 is continuous on a subspace of Hs.

(i) Let x, z ∈ X such that x 6= z. Then there exists Uk ∈ B such that x ∈ Uk and z /∈ Uk. Then, byLem. 3.5.14, there is Ui with x ∈ clUi ⊂ Uk and Pm = (Ui, Uk). Since x ∈ clUi and z ∈ X \ Uk,it follows that

fm(x) = 0 and fm(z) = 1.

⇒ fn(x)

2n

n∈N6=

fn(z)2n

n∈N.

⇒ f(x) 6= f(z). Hence, f is one-to-one.

(ii) Since fn ∈ C[0, 1], for x, z ∈ X, it follows that∣∣∣∣fn(x)− fn(z)

22n

∣∣∣∣ ≤1

22n.

Since the series∑∞

n=11

22n converges, there exists n0 ∈ N such that

∞∑

n=n0+1

122n

<ε2

2.

46

Moreover, f(x), f(z) ∈ Hs yields

ρ(f(x), f(z))2 =∞∑

n=1

|fn(x)− fn(z)|22n

=∞∑

n=n0+1

|fn(x)− fn(z)|22n

+n0∑

n=1

|fn(x)− fn(z)|22n

⇒ρ(f(x), f(z))2 ≤

∞∑

n=n0+1

122n

+n0∑

n=1

|fn(x)− fn(z)|22n

<ε2

2+

n0∑

n=1

|fn(x)− fn(z)|22n

For each l = 1, . . . , n0, there is an open neighborhood Ul of x such that

|fk(x)− fk(z)| < ε2

2n0,∀z ∈ Ul.

Then the set O :=⋂n0

l=1 Ul is open and

ρ(f(x), f(z))2 <ε2

2+ n0

(ε2

2n0

)= ε2, ∀z ∈ O.

Consequently, f is a continuous function.

(iii) Let Y := f(X) ⊂ Hs. Next we show that f−1 : Y → X is continuous. Assume that there isy ∈ Y , f−1 is not continuous at y. This implies , there is a sequence yn

yn → y but xn = f−1(yn)9 f−1(y) = x.

Hence, there is a neighborhood U of x that contains only a finite number of elements of xn.This implies, there is N ∈ N such that xn | n ≥ N ⊂ X \ U .

Then there exists Uk ∈ B such that x ∈ Uk ⊂ U . By Lem. 3.5.14, there Ui ∈ B such that

x ∈ Ui ⊂∈ clUi ⊂ Uk ⊂ U.

Hence, for some fixed m ∈ N, Pm = (Hi,Hk) and it follows that

x ∈ Ui and xn ∈ X \ Uk, ∀n ≥ N.

⇒fm(x) = 0 and fm(xn) = 1, ∀n ≥ N.

⇒ for each n ≥ N, |fm(xn)− fm(x)|2 = 1 and

ρ(f(xn), f(x))2 =∞∑

k=1

|fk(xn)− fk(x)|22n

≥ |fm(xn)− fm(x)|222m

=1

22m

⇒ρ(f(xn), f(x)) ≥ 1

22m,∀n ≥ N.

⇒yn = f(xn)9 f(x) = y.

But this a contradiction. Hence, the assumption is false and f−1 must be continuous.

Consequently, f is a homeomorphism between X and Y = f(X) ⊂ Hs. Therefore, X is metrizable.

47

The converse of the above statement is not always true.


1. Let < x, τ > and < Y, σ > be topological spaces, f : X → Y and S is a subbase for the topology σon Y . Then the function f is continuous iff

∀S ∈ S : f−1(S) ∈ τ.

2. Let f : X → Y and B be a basis for the topological space X. If, for every B ∈ B, f(B) is open in Y ,then f is an open map.

3. Let < X, τ > and < Y, σ > be topological space and f : X → Y be a function. Then

(a) f is a closed function if and only if, for any set A ⊂ X, clf(A) ⊂ f(clA);

(b) f is an open function if and only if, for any set A ⊂ X, f(intA) ⊂ intf(A).

4. The topological spaces X = (0, 1) and Y = R, with the usual topology, are homeomorphic;

5. Let X and Y be topological spaces and f : X → Y . If f is a continuous function and Y is aHausdorff space, then

Graph(f) := (x, y) ∈ X × Y | y = f(x) (graph of f)

is a closed set in X × Y .

6. Let X and Y be topological spaces and f, g : X → Y are continuous functions. If Y is a Hausdorffspace, then the set

x ∈ X | f(x) = g(x)is a closed set in X.

7. If f : X → X is a continuous function and X is Hausdorff space, then the set of fixed points of f ,given by

x ∈ X | x = f(x)is a closed set in X.

8. Let < X, ρ > be a metric space. For A ⊂ X, A 6= ∅ and x ∈ X we set the (distance) function as

fA(x) = dist(x,A) = infz∈A

ρ(x, z).

Then the map fA : X → R is continuous. Moreover, for A and B be disjoint closed sets, g : X → Rand g := fA− fB, we have g−1(0,∞)∩ g−1(−∞, 0) = ∅ and g−1(0,∞) ⊂ A and g−1(−∞, 0) ⊂ B.This implies that every metric space is a normal topological space.

48

3.6 Compact Topological Spaces

3.6.1 Definitions

Definition 3.6.1 (a refinement). A covering Vλ | λ ∈ Λ of X is a refinement of a covering Uα | α ∈ Ωof X if

∀λ ∈ Λ, ∃α ∈ Ω : Vλ ⊂ Uα.

Definition 3.6.2 (finite subcovering). Let < X, τ > be a topological space and Uα | α ∈ Ω be acovering of X. If there is a finite index α1, . . . , αn ⊂ Ω such that

X ⊂n⋃

k=1

Uαk,

then the collection Uα1 , Uα2 , . . . , Uαn is called a finite subcovering of X; i.e. Uα | α ∈ Ω has a finitesubcovering of X.

Definition 3.6.3 (a compact set). Let < X, τ > be a topological space and K ⊂ X. Then the set K is saidto be a compact set if every open covering Oα | α ∈ Ω of K has a finite subcovering Oα1 , Oα2 , . . . , Oαn;i.e.

K ⊂⋃

α∈Ω

Oα ⇒ K ⊂n⋃

k=1

Oαk.

If X itself is a compact set, then < X, τ > is called a compact topological space.

Proposition 3.6.4. Let < X, τ > and < Y, σ > be topological spaces and f : X → Y be a continuousfunction. If K is a compact set in X, then f(K) is a compact set in Y .

Proof. Let Oα | α ∈ Ω be an open covering of f(K);i.e

f(K) ⊂⋃

α∈Ω

Oα

⇒K ⊂

⋃

α∈Ω

f−1 (Oα)

Since f is a continuous function, the collection f−1 (Oα) | α ∈ Ω is an open covering of the compactset K. Consequently, there exists Oα1 , . . . , Oαn such that

K ⊂n⋃

k=1

f−1 (Oαk)

⇒f(K) ⊂

n⋃

k=1

Oαk.

Hence, Oα1 , . . . , Oαn is a finite subcovering of f(K). Consequently, f(K) is a compact set.

Proposition 3.6.5.

(a) A closed subset of a compact topological space is compact;

(b) A compact subset of a Hausdorff topological space is closed.

49

Proof. (a) Let < X, τ > be a compact topological space and F be a closed set. Suppose Oα | α ∈ Ωis an open covering of F . Then the collection

Oα, X \ F | α ∈ Ωis an open covering of X. Hence, there is Oα1 , . . . , Oαn ⊂ Oα | α ∈ Ω such that

X ⊂n⋃

k=1

Oαk∪ (X \ F ).

⇒F ⊂

n⋃

k=1

Oαk∪ (X \ F ).

But, F ∩ (X \ F ) = ∅. Hence,

F ⊂n⋃

k=1

Oαk.

Consequently, F is a compact set.

(b) Let < X, τ > be a Hausdorff topological space and K be a compact subset of X. To show K is aclosed set, we show X \K is an open set.

Let z ∈ X \K be any fixed element, then z /∈ K. Since X is a Hausdorff space,

∀x ∈ K, ∃Ux ∈ τ,∃Ox ∈ τ : z ∈ Ux, x ∈ Ox and Ux ∩Ox = ∅.Hence, the family Ox | x ∈ K is an open covering of K. This implies there is a finite subcov-ering Ox1 , . . . , Oxn such that

K ⊂n⋃

k=1

Oxk.

Let Ux1 , . . . , Uxn be the corresponding collection of open sets such that z ∈ Uxk, k = 1, . . . , n.

Define

U :=n⋂

k=1

Uxk.

Hence, U is an open set, z ∈ U and U ∩Oxk= ∅ for each k = 1, . . . , n. This implies

U ∩n⋃

k=1

Oxk= ∅ ⇒ U ∩K = ∅ ⇒ z ∈ U ⊂ X \K.

Consequently, z is an interior point of X \ K. Since z ∈ X \ K is arbitrary, we conclude thatX \K is an open set. Therefore, K is a closed set.

3.6.2 The Finite Intersection Property

Definition 3.6.6 (the finite intersection property). Let < X, τ > be a topological space and Aα | α ∈ Ωbe a collection in X. Then Aα | α ∈ Ω is said to have the finite intersection property if for every finitecollection Aα1 , . . . , Aαn, the intersection

n⋂

k=1

Aαk

is non-empty.

50

Proposition 3.6.7. A topological space X is compact if and only if every collection Fα | α ∈ Ω of closedsets with the finite intersection property has a non-empty intersection.

Proof. ”⇒” : Assume that ⋂α

Fα = ∅.

⇒X = X \

⋂α

Fα =⋃

α∈Ω

(X \ Fα) .

Since X is a compact set, there exist Fα1 , . . . , Fαn such that

X ⊂n⋃

k=1

(X \ Fαk)

⇒n⋂

k=1

Fαk= ∅.

But this contradicts the finite intersection property. Hence,⋂

α Fα 6= ∅.

”⇐”: Assume that X is not a compact set. Then there is an open covering Oα | α ∈ Ω of X with nofinite subcovering. This implies, for every finite subcollection Oα1 , . . . , Oαn

X \n⋃

k=1

Oαk6= ∅.

⇒ the family of closed sets X \ Oα | α ∈ Ω satisfies the finite intersection property. Then, byassumption, ⋂

α∈Ω

(X \Oα) 6= ∅.

⇒X \

⋃

α∈Ω

Oα 6= ∅.

This implies the collection Oα | α ∈ Ω does not cover X. But this is a contradiction. Conse-quently, X should be a compact set.

Remark 3.6.8. From the proof of Prop. 3.6.7 we can easily verify that the following two statements areequivalent

• X is a compact topological space;

• for every family of closed subsets Fα | α ∈ Ω of X with the property that⋂

α∈Ω Fα = ∅, there is afinite subcollection Oα1 , . . . , Oαn such that

m⋂

k=1

Oαk= ∅.

51

3.6.3 Compact Hausdorff Spaces

Compact Hausdorff topological spaces exhibit very interesting properties, which are very importantfrom practical point of view.

Corollary 3.6.9. Let < X, τ > be a Hausdorff topological space. If K is a compact subset of X, z ∈ Xand z /∈ K, then there exist disjoint open sets O and U in X such that

K ⊂ U and z ∈ O.

Proof. See the proof of part (b) of Prop. 3.6.5.

Proposition 3.6.10. Let < X, τ > be a Hausdorff topological space. If K1 and K2 are disjoint compactsets, then there exists disjoint open sets O1 and O2 of X such that

K1 ⊂ O1,K2 ⊂ O2.

Proof. Each z ∈ K1 is such that z /∈ K3. Using Cor. 3.6.9, there are disjoint open sets Oz and Uz suchthat

z ∈ Oz and K2 ⊂ Uz.

Hence, the family Oz | z ∈ K1 is an open covering of K1. By the compactness of K1, there is a finiteopen covering Oz1 , . . . , Ozm, i.e.

K1 ⊂m⋃

k=1

Ozk=: O1

and there is a corresponding finite collection Uz1 , . . . , Uzm with K2 ⊂ Uzkfor each k = 1, . . . , m.

Then

K2 ⊂m⋂

k=1

Uzk=: O2.

Then O1 and O2 are the required open sets.

Corollary 3.6.11. If < X, τ > is a compact Hausdorff topological space, then X is a normal topologicalspace.

Proof. Follows from Prop. 3.6.5(a) and Prop. 3.6.10.

Corollary 3.6.12. A compact Hausdorff second countable topological space is metrizable.

Proof. Follows from Cor. 3.6.11 and Thm. 3.5.15 .

Excercises 3.6.13. Show that

1. The finite union of compact sets is again compact.

2. Let X be a compact topological space, U be an open subset of X and Knn∈N be a family of compactsubsets of X. If

⋂n∈NKn ⊂ U , then there exists a finite index I ⊂ N such that

⋂

n∈I

Kn ⊂ U.

52

3. Let X be a compact topological space and let Knn∈N be a family of non-empty closed subsets of Xwith Kn+1 ⊂ Kn for each n ≥ 1. Then

⋂

n∈NKn 6= ∅.

4. (a) It is necessary and sufficient for a topological space X to be compact that: if Vα|α ∈ Ω is anyfamily of closed subsets of X such that ∩α∈Ω′Vα 6= ∅, for any subset Ω′ ⊂ Ω, then ∩α∈ΩVα 6= ∅

(b) Let X be a compact topological space and for each n ∈ N, Vn is a non-empty closed subset ofX such that

Vn ⊃ Vn+1.

Then ⋂n

Vn 6= ∅.

(c) Suppose f : X → X is continuous, where X is a compact metric space. Then there exists anon-empty subset A ⊂ X such that f(A) = A. (Hint: put X1 = f(X), Xn+1 = f(Xn) andA = ∩∞n=1Xn and use (b)),

5. Let X be a compact metric space, F be a closed subset of X and U be an open subset of X. IfF ⊂ U , then there exists ε > 0 such that

⋃

x∈F

Bε(x) ⊂,

where Bε(x) represents the open ball centered at x and with radius ε.

6. Let X be Hausdorff and Y be a compact topological spaces. If f : X → Y is a continuous one-to-onefunction from X on to Y , then f is a Homeomorphism.

7. Let X and Y be metric spaces. If X is compact and f : X → Y be a continuous function, thenf(X) = f(x) | x ∈ X is a bounded subset of Y .

8. If K1 and K2 be compact subsets of a metric space < X, ρ >, then there exist x ∈ K1 and z ∈ K2

such thatρ(x, z) = dist(K1,K2).

3.7 Locally Compact Spaces

Locally compact Hausdorff space are among the most important topological spaces, say in non-linearanalysis and abstract measure theory. For instance, measure and integral theory on topological spaceusually assume the underlying space to be locally compact Hausdorff topological space (eg. Borel,Radon measure on topological spaces, etc).

Definition 3.7.1 (locally compact spaces). A topological space < X, τ > is said to be locally compactif

∀x ∈ X, ∃O ∈ τ : x ∈ O and clO is a compact set.

Proposition 3.7.2. Every compact topological space is locally compact.

53

The converse of Prop. 3.7.2 is not always true. For instance, the Euclidean space Rn is locally compact,but it is not a compact space.

Lemma 3.7.3. If X is a compact T2 space, then X is T4.

Proof. Let F1 and F2 be two closed sets in X such that F1 ∩ F2 = ∅. For each x ∈ F1 and z ∈ F2, thereare disjoint open sets Ux and Vz such that x ∈ Ux and z ∈ Vz. This implies

F1 ⊂⋃

x∈F1

Ux and F2 ⊂⋃

z∈F2

Vz.

(Note that: each Ux is so that Ux ∩ Vz1 = ∅ and Ux ∩ Vz2 = ∅ for z1 6= z2 and vice-versa). By thecompactness of F1 and F2, there are x1, . . . , xn ∈ F1 and z1, . . . , zm ∈ F2 such that

F1 ⊂n⋃

k=1

Uxk=: U and F2 ⊂

m⋃

k=1

Vzk=: V.

⇒ F1 ⊂ U,F2 ⊂ V and U ∩ V = ∅. Therefore, X is T4.

Lemma 3.7.4. Let X be a T2 space. If U ⊂ X is an open set such that clU is compact, then, for everyx ∈ U , there is a compact neighborhood Vx such that

x ∈ Vx ⊂ U.

Proof. A closed subset of a T2 space is T2 (cf. Prop. 3.4.4). Hence, clU is a T2 and compact subspaceof X. By, Lem. 3.7.3, clU is a T4 subspace of X. Since

x ⊂ U ⊂ clU and U is again an open set w.r.t. the topology of clU,

by Thm. 3.4.7, there is an open set O in X such that

x ⊂ O ∩ clU ⊂ cl (O ∩ clU) = cl(O ∩ U) ⊂ U ⊂ clU.

(Note that O∩ clU is an open set w.r.t. the relative topology on clU). Now set, Vx = cl(O∩U) and theclaim follows.

Proposition 3.7.5. Let X be a locally compact Hausdorff space. If K is a compact subset of X, then

(i) there is an open set O such that K ⊂ O, clO is compact; and

(ii) given such a set O there is non-negative function f : X → [0, 1]; i.e. 0 ≤ f(x) ≤ 1 such that

f(X \O) = 0 and f(K) = 1.

Hence, x ∈ X | f(x) 6= 0 ⊂ O.

Proof. (i) For each x ∈ K, there is an open sets Ox such that x ∈ Ox and clOx is compact. Then

K ⊂⋃

x∈K

Ox.

⇒ there are x1, . . . , xn ∈ K such that

K ⊂n⋃

k=1

Oxk.

If we let O :=⋃n

k=1 Oxk, it follows that K ⊂ O and clO is a compact set (note that clO is a finite

union of compact sets, according to Ex. 3.6.13(1) ).

54

(ii) Now for each x ∈ K, (by Lem. 3.7.4) choose a compact neighborhood Vx such that Vx ⊂ O. Bythe compactness of K, there are x1, . . . , xn ∈ K such that

A =n⋃

k=1

Vxk, A is compact and K ⊂ intA ⊂ A ⊂ O.

By Lem 3.7.4 A is a T4 subspace of X. Hence, by Uryson’s Lemma( Lem. 3.5.3), there is a continuousfunction g : A → [0, 1] such that g(K) = 1 and g(A \ intA) = 0. Now define f : X → [0, 1] as

f(x) =

g(x), if x ∈ A0, if x ∈ X \A.

It remains now to show that f is continuous on X. For this it is enough to show that f continuous atx0 ∈ ∂A ⇒ x0 /∈ intA ⇒ x0 ∈ A \K. Thus,

(a) there is a neighborhood Vx0 such that Vx0 ∩A ⊂ (A \K) ⇒ f(x) = g(x) = 0, ∀x ∈ Vx0 ∩A; and

(b) by the definition of f , f(x) = 0,∀x ∈ Vx0 ∩ (X \A).(Note that A \K is an open set relative to A.)

From (a) and (b) it follows that f(x) = 0,∀x ∈ Vx0 . Hence, f is continuous at x0. Moreover,X \O ⊂ X \A implies that f(X \O) = 0. Hence, clx ∈ X | f(x) 6= 0 ⊂ A ⊂ O.‡


1. If X is a locally compact T3 topological space, then each point x ∈ X has a neighborhood base ofthe form

Nx = x ∈ O ∈ τ | clO compact .

2. Let X be a locally compact space and Dn | n ∈ N be a countable collection of dense set. Then⋂n∈NDn is dense in X.

3. Let X be a locally compact space. A subset F of X is closed if and only if F ∩K is closed for closedcompact set K.

4. A closed subset of a locally compact space is locally compact.

5. An open subset of a locally compact space is locally compact.

3.8 Sigma-Compact Topological Spaces

Definition 3.8.1 (σ-compact spaces). A topological space < X, τ > is said to be σ-compact if it is theunion of a countable number of compact sets.

Spaces which are σ-compact are also known as countably compact spaces. Thus, a compact space isa σ-compact.

‡The set clx ∈ X | f(x) 6= 0 is known as the support of the function f on the set X, denoted by sup f ,

55

Theorem 3.8.2. Let < X, τ > be a locally compact Hausdorff topological space. Then (i) X is Lindelof⇔ (ii) X is σ-compact ⇔ (iii) There is a sequence On of open sets with clOn compact, clOn ⊂ On+1

and X ⊂ ⋃n∈NOn.

Proof. (i)⇒(ii): X is locally compact implies that for each x ∈ X, there is a neighborhood Ux suchthat x ∈ Ux and clU is compact. Hence, X ⊂ ⋃

x∈X Ux. Since X is Lindelof, there is a countablesubcover Un | n ∈ N such that

X =⋃

n∈NUn ⇒ X =

⋃

n∈NclUn.

Consequently, X is σ-compact.

(ii)⇒ (iii): X is σ-compact ⇒ X =⋃

n∈NKn, where for each n ∈ N, Kn is a compact set. Now, sinceK1 is compact, by Prop. 3.7.5(i), there is an open set O1 such that K1 ⊂ O1 and clO1 is compact.Proceeding inductively, for n = 2, . . . , the set Kn ∪ clOn−1 is compact. Hence, there is On suchthat Kn ∪ clOn−1 ⊂ On and clOn is compact. Consequently, we have

clOn ⊂ On+1, n = 1, . . . and X =⋃

n∈NOn.

(iii)⇒ (i): Let C = Uα | α ∈ Ω be an open covering of X. Then, by (iii) there is a sequenceOn | n ∈ N such that clOn ⊂ On, clOn compact and X =

⋃n∈NOn =

⋃n∈N clOn. This implies,

for each n ∈ N,

clOn ⊂⋃

α∈Ω

Uα ⇒ ∃Oα1 , . . . , Oαmn ⊂ C : clOn ⊂

mn⋃

k=1

Uαk.

Set

F :=⋃

n∈NUα1 , . . . , Uαmn

| clOn ⊂mn⋃

k=1

Uαk, n ∈ N.

It follows that F is countable, F ⊂ C and⋃F = X. Consequently, X is Lindelof.

3.9 Paracompact Topological Spaces

Definition 3.9.1 (refinement). Let < X, τ > be a topological space and C be a collection of subsets of X.A collection F of subsets of X is said to be a refinement of C if

(i)⋃F =

⋃ C; and

(ii) for each V ∈ F , there is U ∈ C such that V ⊂ U .

Accordingly, if F is a refinement of C and C is a covering of X, then F should also be a covering of X.

Definition 3.9.2 (locally finite). Let < X, τ > be a topological space. A family C = Uα | α ∈ Ω ofsubsets of X is said to be locally finite if for each x ∈ X, there is a neighborhood O of x such that Ointersects only a finite number of elements of C.

56

Definition 3.9.3 (σ-locally finite). Let < X, τ > be a topological space. A family C of subsets of X issaid to be σ-locally finite if

C =⋃

n∈NCn;

where, for each n ∈ N, Cn locally finite. That, is C is a countable union of locally finite families.

Remark 3.9.4. Observe that,

• any finite collection F of subsets of X is locally finite;

• if C = Uα | α ∈ Ω is locally finite, then C := clUα | α ∈ Ω is also locally finite;

• any refinement of a locally finite collection is again locally finite;

• every locally finite collection is also σ-locally finite.

Lemma 3.9.5. Let C = Uα | α ∈ Ω a locally finite family. If, for each α ∈ Ω, Vα ⊂ Uα, then the familyF := Vα | α ∈ Ω is also locally finite.

Proof. Trivial!

In Lem 3.9.5, the elements of F can be any type of sets; i.e. it is not necessarily that they are closedor open sets. Recall also that, for a collection Uα | α ∈ Ω we have

cl

( ⋃

α∈Ω

Uα

)⊂

⋃

α∈Ω

clUα

but equality does not hold always. However,

Lemma 3.9.6. Let C = Uα | α ∈ Ω be a locally finite family, then

cl

( ⋃

α∈Ω

Uα

)=

⋃

α∈Ω

clUα

Proof. (a)⋃

α∈Ω

clUα ⊂ cl

( ⋃

α∈Ω

Uα

)is obvious.

(b) Now, let x ∈ cl(⋃

α∈Ω Uα

). Then there is a neighborhood N of x such that N intersects only a

finite number of elements of C. Let Uα1 , . . . , Uαn ∈ C such that N ∩Uαk6= ∅, k = 1, . . . , n. Hence,

x ∈ N, N ∩n⋃

k=1

Uαk6= ∅, and N ∩

⋃

α∈Ω\α1,...,αnUα

= ∅.

This implies that x ∈ cl(⋃n

k=1 Uαk) =

⋃nk=1 clUαk

⊂ ⋃α∈Ω clUα

§. Consequently,

cl

( ⋃

α∈Ω

Uα

)⊂

⋃

α∈Ω

clUα.

§Observe that: if x ∈ cl(A ∪B) and x /∈ clB, then x ∈ clA.

57

Lemma 3.9.7. Let < X, τ > be a topological space. If C = Uα | α ∈ Ω is locally finite family of subsetsof X and K is a compact subset of X, then K intersects only a finite number of elements of C.

Proof. Take any element x ∈ K. Then there is an open neighborhood Ox such that Ox intersects onlya finite number of elements of C. Hence,

K ⊂⋃

x∈K

Ox.

Since K is compact, there are x1, . . . , xn ∈ K, for some n ∈ N, such that

K ⊂n⋃

k=1

Oxk.

But, for each k ∈ 1, . . . , n, Oxkcan intersect only a finite number of elements of C, say about mk of

them. Hence, K can only intersect about∑n

k=1 mk elements of C.

Definition 3.9.8 (paracompact topological space). A Hausdorff topological space < X, τ > is said to beparacompact if every open cover of X has a locally open refinement.

Proposition 3.9.9. A closed subset of a paracompact space is paracompact.

Proposition 3.9.10. If < X, τ > is a σ-compact locally compact Hausdorff topological space, then <X, τ > is paracompact.

Proof. Let C = Uα | α ∈ Ω be an open covering of X. Then

• Thm. 3.8.2(i) ⇒ X is Lindelof. Hence, C has a countable sub-cover, say Un | n ∈ N; and

• Thm. 3.8.2(iii) ⇒ there is a sequence of sets On | n ∈ N such that, for each n ∈ N, clOn iscompact, clOn ⊂ On+1 and X =

⋃n∈NOn.

Now, for each n ∈ N, define the set

Un := Un ∩ (On \ clOn−2) ,

where, w.l.o.g, we assume that O−1 = O0 = ∅.(i) The collection Un | n ∈ N is an open refinement of C.

For each n ∈ N, Un ⊂ Un ∈ C, and⋃n

Un =⋃n

Un ∩⋃n

(On \ clOn−2) = X ∩⋃n

(On \ clOn−2) =⋃n

(On \ clOn−2) . (3.4)

We show next that⋃

n (On \ clOn−2) = X. Since X =⋃

n On, for x ∈ X, there is a smallestn(x) ∈ N such that x ∈ On(x).

(a) If n(x) = 1, thenx ∈ O1 \ clO−1 ⇒ x ∈

⋃n

(On \ clOn−2) .

(b) If n(x) ≥ 2, then

x /∈ On(x)−1 ⇒ x /∈ clOn(x)−2 ⇒ x ∈ On(x) \ clOn(x)−2 ⇒ x ∈⋃n

(On \ clOn−2) .

From (a) and (b), we conclude that X =⋃

n (On \ clOn−2).

58

Consequently, from eqn. (3.4), we conclude that⋃

n Un = X. That is, Un | n ∈ N is an openrefinement of C.

(ii) Un | n ∈ N is locally finite. To see this, let x ∈ X be any. Since X =⋃

n∈NOn =⋃

n∈N clOn,there is a smallest n0 ∈ N such that

x ∈ On0 .

This implies that

On0 ∩ [Uk ∩ (Ok \ clOk−2)] = ∅, ∀k ≥ n0 + 2 ⇒ On0 ∩ Uk = ∅,∀k ≥ n0 + 2.

Hence, On0 can only intersect a finite number of elements of Un | n ∈ N. Consequently,Un | n ∈ N is locally finite.

Hence, form (i) & (ii) Un | n ∈ N is a locally finite open refinement of C. Therefore, X is paracom-pact.

Corollary 3.9.11. Every compact Hausdorff space is paracompact.

Proposition 3.9.12. Let < X, τ > be a topological space. If an open cover C of X has a σ-locally finiterefinement, then C has a locally finite refinement.

Proof. Let C = Uα | α ∈ Ω be an open covering of X. Then by assumption C =⋃

n∈N Cn, where foreach n ∈ N, Cn is a locally finite family.

Thus, for each α ∈ Ω,∃n ∈ N : Uα ∈ Cn.

Let n(α) be the smallest natural number for which Uα ∈ Cn(α). Define now

Wα := Uα ∩⋂X \ Uβ | Uβ ∈ Cm for some m < n(α) .

It follows that:

(i) for each α ∈ Ω : Wα ⊂ Uα;

(ii) the family Wα | α ∈ Ω is a covering of X. To see this let x ∈ X be any. This implies

∃α0 ∈ Ω : x ∈ Uα0 .

For the corresponding smallest natural number n(α0) it follows that

m < n(α0) and Uβ ∈ Cm ⇒ x /∈ Uβ ⇒ x ∈ X \ Uβ.

Hence, x ∈ Wα0; i.e. X =⋃

α∈Ω Wα.

(iii) The family Wα | α ∈ Ω is locally finite. Let x ∈ X, α0 and n(α0) be as in (ii).

Case a: If α ∈ Ω such that n(α) > n(α0), then x ∈ Uα0 and Uα0 ∩Wα = ∅.Case b: If α ∈ Ω such that n(α) ≤ n(α0); i.e. n(α) ∈ 1, . . . , n(α0); say n(α) = m ≤ n(α0).

Then the familyBm := Uα ∈ C | n(α) = m ⊂ Cm.

is locally finite. Hence, there exists a neighborhood of Om of x such that Om intersects onlya finite number of elements of Bm. Since Wα ⊂ Uα, we also have that

Wm = Wα ∈ C | n(α) = m

59

is also locally finite; i.e. that Om intersects only a finite number of elements of Wm. Now,define the set

Ox := Uα0 ∩O1 ∩O2 ∩ . . . ∩On(α0).

Then Ox is a neighborhood of x which intersects only a finite number of elements ofWα | α ∈ Ω.

Proposition 3.9.13. Let < X, τ > be a T3 topological space. If every open cover of C of X has a locallyfinite refinement, then C has a closed locally finite refinement.

Proof. Let C = Uα | α ∈ Ω be an open cover of X. Then for each x ∈ X, there is Uα(x) ∈ C such thatx ∈ Uα(x). Since X is T3, there is an open set Ox such that

x ∈ clOx ⊂ Uα(x).

Then the collection O =: Ox | x ∈ X is an open refinement of C (note that X =⋃

x∈X Ox) such thatO := clOx | x ∈ X is a closed refinement of C. By assumption, there is a locally finite refinement Fof O. Then, the family

F := clV | V ∈ Fis a closed locally finite refinement of O. Consequently, F is a closed locally finite refinement of C.(Observe that: V ∈ F ⇒ V ⊂ Ox, for some x ∈ X ⇒ clV ⊂ clOx ⊂ Uα(x) ∈ C).

Proposition 3.9.14. Let < X, τ > be a Hausdorff topological space. If every open cover C of X has aclosed locally finite refinement, then X is paracompact.

Theorem 3.9.15. Every paracompact space is normal.

Proof. Let X be a paracompact topological space. First we show that X is regular.

(i) Let x ∈ X, F ⊂ X closed and x /∈ F . Since X is Hausdorff, for each z ∈ F there exists aneighborhoods Uz such that

x /∈ clUz.

DefineC := Uz | z ∈ F ∪ X \ F

Then C is an open covering of X. Hence,

X is paracompact ⇒ ∃F which is a locally finite open refinement of C. (Hence, X =⋃F).

Define now A := V ∈ F | V ⊂ Uz, for some Uz ∈ C¶.

Then

(a) Lem. 3.9.5 implies that A is locally finite. And, from Lem. 3.9.6, it follows that

cl

( ⋃

V ∈AV

)=

⋃

V ∈AclV ;

(b) F ⊂ ⋃A =: O and O is an open set;

¶Indeed A 6= ∅. If z ∈ F , then x ∈ V for some V ∈ F . But then V * X \ F . Hence, there is Uz such that V ⊂ Uz; i.e.z ∈ V ⊂ Uz and V ∈ A. So, A 6= ∅.

60

(c) since x /∈ clUz for each z ∈ F , x /∈ clV for each V ∈ A. This implies that

x /∈⋃

V ∈AclV ⇒ x /∈ cl

( ⋃

V ∈AV

)⇒ x /∈ clO.

⇒x ∈ X \ clO and F ⊂ O.

Consequently, X is regular; i.e. T3.

(ii) Let now F be a closed and U and an open subsets of X such that

F ⊂ U.

Since X is regular, for each x ∈ F there is an open neighborhood Ux such that

x ∈ Ux ⊂ clUx ⊂ U. (?)

Define the collection C := Ux | x ∈ F ∪ X \ F and let F ,A and O be as in part (i).Then, under the condition (?), it is easy to see that

F ⊂ O ⊂ clO ⊂ U.

Which implies that X is normal; hence, T4.

Corollary 3.9.16. A second countable paracompact space is metrizable.

Proof. A direct consequence of Thm. 3.9.15 and Uryson’s metrizability (Thm. 3.5.15).

Lemma 3.9.17. Let < X, ρ > be a metric space. If Un | n ∈ N is a countable open cover of X, thenthere is a locally finite open refinement Vn | n ∈ N such that Vn ⊂ Un, for each n ∈ N.

Proof. For each n ∈ N, define the function

ϕn : X → [0, 1] such that ϕn(x) = min1, dist(x, X \ Un).

Now let

• V1 := U1; and

• for n = 2, . . .

Vn := Un ∩n−1⋂

k=1

x ∈ X

∣∣∣∣ ϕk(x) <1n

.

Then

(i) for each n ∈ N, Vn ⊂ Un and Vn is an open set;

(ii) the family Vn covers X. To see this, let x ∈ X. Since Un covers X, we have x ∈ Un forsome n ∈ N. If x ∈ U1, then we are done. Otherwise, let n0 ∈ N be the smallest index suchthat x ∈ Un0 . This implies

x /∈ Uk, k = 1, . . . , n0−1 ⇒ x ∈ X\Uk, k = 1, . . . , n0−1 ⇒ dist(x,X\Uk) = 0, k = 1, . . . , n0−1.

61

From this follows that

ϕk(x) = 0, k = 1, . . . , n0 − 1 ⇒ x ∈ Un0 ∩n0−1⋂

k=1

x ∈ X

∣∣∣∣ ϕk(x) <1n

= Vn0 .

Since x ∈ X is arbitrary, we conclude that

X =⋃

n∈NVn;

(iii) the family Vn | n ∈ N is locally finite. By part (ii), Vn | n ∈ N covers X implies forx0 ∈ X there is n0 ∈ N such that x ∈ Vn0 . But

Vn0 = Un0∩n0−1⋂

k=1

x ∈ X

∣∣∣∣ ϕk(x) <1n0

= Un0∩

n0−1⋂

k=1

x ∈ X

∣∣∣∣ min1, dist(x, X \ Uk) <1n0

.

But note thatn0−1⋂

k=1

X \ Uk ⊂n0−1⋂

k=1

x ∈ X

∣∣∣∣ min1, dist(x,X \ Uk) <1n0

⇒Un0 ∩

n0−1⋂

k=1

X \ Uk ⊂ Vn0 ⇒ Un0 \n0−1⋃

k=1

Uk ⊂ Vn0 .

Moreover,

Un0+1 \n0⋃

k=1

Uk ⊂ Vn0+1.

Since, Vn0 ⊂ Un0 , we observe that Vn0 ∩ Vn0+1 = ∅. In fact, for n > n0, Vn0 ∩ Vn = ∅.Consequently, x0 ∈ Vn0 and Vn0 can intersect only a finite number of elements of Vn | n ∈N. In particular, Vn0 can only intersect V1, . . . , Vn0−1. Therefore, Vn | n ∈ N is a locallyfinite family.

In the following, for ε > 0, we use Bε(A) := x ∈ X | dist(x,A) < ε. If A = z , then Bε(z) =Bε(z).

Theorem 3.9.18. Every metric space is paracompact.

Proof. Let C = Uα | α ∈ Ω be an open covering of X and let ”4” be a well ordering on Ω with α1 ∈ Ωbeing the first element. Now for each (n, α) ∈ N× Ω define

Hn,α1 :=

x ∈ X

∣∣∣∣ dist(x,X \ Uα1) ≥1n

Hn,α :=

x ∈ X

∣∣∣∣ dist(x,X \ Uα) ≥ 1n

∩

x ∈ X

∣∣∣∣∣∣dist

x,

⋃

λα

Hn,λ

≥ 1

n

Obviously, we have that

x ∈ X

∣∣∣∣ dist(x,X \ Uα) ≥ 1n

\

⋃

λα

Hn,λ ⊂ Hn,α.

Moreover, the following hold true:

62

(a) for each (n, α) ∈ N× Ω, B 12n

(Hn,α) ⊂ Uα. To see this, let z ∈ Hn,α and x ∈ B 12n

(z) and w ∈ X

be any. Thenρ(w, z) ≤ ρ(w, x) + ρ(x, z) ⇒ ρ(w, z)− ρ(x, z) ≤ ρ(w, x).

In particular, for any w ∈ X \ Uα we have

1n− 1

2n< ρ(w, x) ⇒ ρ(w, x) >

12n

, ∀w ∈ X \ Uα ⇒ dist(x,X \ Uα) ≥ 12n

.

⇒ x /∈ X \ Uα ⇒ x ∈ Uα. Consequently, B 12n

(Hn,α) ⊂ Uα.

(b) If λ α, then dist(Hn,λ,Hn,α) ≥ 12n . Let z ∈ Hn,α be arbitrary and consider B 1

n(z) = x ∈

X | ρ(z, x) < 1n. By definition of Hn,α we have that

dist

z,

⋃

λα

Hn,λ

≥ 1

n⇒ B 1

n(z) ∩

⋃

λα

Hn,λ = ∅ ⇒ B 1n(z) ∩Hn,λ = ∅ for each λ with λ α.

Since, B 12n

(z) ⊂ B 1n(z) it follows that

B 12n

(z) ∩Hn,λ = ∅ for each λ with λ α.

Note that z ∈ Hn,α was chose arbitrarily. Consequently,

dist(Hn,λ,Hn,α) ≥ 12n

, for each λ with λ α.

(c)⋃

n∈N⋃

α∈Ω Hn,α = X. To verify this, let x ∈ X. Then

(i) if x ∈ Uα1 , then we are done. Since x /∈ X \ Uα1 , it follows that

dist(x,X \ Uα1) > 0 ⇒ ∃m ∈ N : dist(x,X \ Uα1) ≥1m⇒ x ∈ Hm,α1 .

(ii) Otherwise, by the well ordering principle, the set Ωx = α ∈ Ω | x ∈ Uα has a leastelement, say αx. Hence, x ∈ Uαx and x /∈ Uλ for λ αx. The set Uαx is open implies

∃nx ∈ N : B 1nx

(x) ⊂ Uαx ⇒ dist(x,X \ Uαx) ≥ 1nx

. (**)

Furthermore,

B 1nx

(x) ∩ ⋃

λαx

Hnx,λ

= ∅.

⇒ dist

x,

⋃

λαx

Hnx,λ

≥ 1

nx.

(***)

If we assume that (***) does not hold, then there is z ∈ B 1nx

(x) such that z ∈ Hnx,λ, for

some λ αx. But then

ρ(x, z) <1nx

⇒ dist(x,X \ z) ≥ 1nx

.

and z ∈ Hnx,λ, λ αx, implies z ∈ Uλ. From this follows that

X\Uλ ⊂ X\z ⇒ dist(x,X\Uλ) ≥ dist(x,X\z) ≥ 1nx

.( That is, x /∈ X \ Uλ, so that x ∈ Uλ. )

63

Hence, x ∈ Uλ. But this is a contradiction to the definition of αx. Consequently, (***)holds.From (**) and (***) we conclude now that x ∈ Hnx,α. Therefore,

X =⋃

n∈N

⋃

λ∈Ω

Hn,α.

Now define

Vn,α = B 16n

(Hn,α). Then it follows that X =⋃

n∈N

⋃

α∈Ω

Vn,α; since Hn,α ⊂ Vn,α. (****)

That is, the collection Vn,α | (n, α) ∈ N× Ω is an open cover of X. Moreover, if

An :=⋃

α∈Ω

Vn,α,

then A := An | n ∈ N is a countable open cover for X. By Lem. 3.9.17, there is a locally finite openrefinement F = A′n | n ∈ N of A such that A′n ⊂ An =

⋃α∈Ω Vn,α for each n ∈ N and X =

⋃n∈NA′n.

Let nextOn,α := A′n ∩ Vn,α, n ∈ N, α ∈ Ω.

Then we claim

(d) O := On,α | n ∈ N, α ∈ Ω is a cover for X.Let x ∈ X. Since X =

⋃n∈NA′n, there is n ∈ N such that

x ∈ A′n ⊂ An =⋃

α∈Ω

Vn,α ⇒ ∃n ∈ N, α ∈ Ω : x ∈ A′n and x ∈ Vn,α ⇒ ∃n ∈ N, α ∈ Ω : x ∈ On,α.

Hence,X =

⋃

n∈N

⋃

α∈Ω

On,α.

(e) O := On,α | n ∈ N, α ∈ Ω is a locally finite open refinement of C.Note that

On,α ⊂ Vn,α = B 16n

(Hn,α) ⊂ B 12n

(Hn,α) ⊂︸︷︷︸part (a)

Uα ⇒ On,α ⊂ Uα.

Furthermore, for x ∈ X, since F is locally finite, there is a neighborhood N(x) of x such thatN(x) intersects only a finite number of elements of F ⇒

∃n0 ∈ N : N(x) ∩A′n 6= ∅, 1 ≤ n ≤ n0 and N(x) ∩A′n = ∅, n ≥ n0 + 1.

Moreover, for λ α it follows that dist(Vn,α, Vn,λ) ≥ 16n . To verify this, let w ∈ Hn,α, x ∈

Vn,α, y ∈ Hn,λ, z ∈ Vn,λ. We estimate ρ(x, z).Repeatdly, using the triangle inequality, we obtain

ρ(w, y) ≤ ρ(w, x) + ρ(x, y) ≤ ρ(w, x) + ρ(x, z) + ρ(z, y)

⇒ (using part (b), and the definitions of Vn,α and Vn,λ)

ρ(w, y)− ρ(w, x)− ρ(z, y) ≤ ρ(x, z) ⇒ 12n

− 16n

− 16n

≤ ρ(w, y)− ρ(w, x)− ρ(z, y) ≤ ρ(x, z)

64

⇒ρ(x, z) ≥ 1

6n.

Hence, dist(Vn,α, Vn,λ) ≥ 16n . This implies the open ball B 1

8n(x) can intersect only one Vn,α

whenever λ α. Consequently, the neighborhood N(x) ∩ B 18n

(x) intersects only at most n0

elements of O; i.e. O is locally finite.

Remark 3.9.19. Thms. 3.9.18 and 3.9.15 imply that every metric space is normal; hence, Hausdorff.

Excercises 3.9.20. Verify the statements of Rem. 3.9.4.

3.10 Partition of Unity

Definition 3.10.1 (support of a function). Let < X, τ > be a topological space and f : X → R, i.e. f isa real valued function. Then the support of the function f is defined as

supp f := clx ∈ X | f(x) 6= 0.

Definition 3.10.2. Let < X, τ > be a topological space and C = Uα | α ∈ Ω is a covering of X. Thena collection of real-valued functions ϕλ | λ ∈ Λ is said to be subordinate to the covering C on X if

∀λ ∈ Λ, ∃α ∈ Ω : supp ϕλ ⊂ Uα.

Excercises 3.10.3. If K is a compact subset of Hausdorff space, then there is an open set O such thatK ⊂ O and clO is compact.

Proposition 3.10.4. Let < X, τ > be a locally compact Hausdorff space and K be a compact subset ofX. If C = Uα | α ∈ Ω is an open covering of K, then there is a collection ϕ1, . . . , ϕn of continuousreal-valued non-negative functions subordinate to the collection C on K.

Proof. By Prop. 3.7.5(i), there is O open such that K ⊂ O and clO is compact. Hence,

K ⊂ O ∩⋃

α∈Ω

Uα.

(i) If x0 ∈ K, then there is α(x0) ∈ Ω such that x0 ∈ O ∩ Uα(x0). Thus x0 is compact, O ∩ Uα(x0)

open and cl(O ∩ Uα(x0)

)is compact. Using Prop. 3.7.5(ii), there is a continuous function fx0 :

X → [0, 1] such thatfx0(x0) = 1 and f(X \ cl(O ∩ Uα(x0))) = 0.

This implies,

• suppfx0 ⊂ O ∩ Uα(x0); i.e. suppfx0 ⊂ Uα(x0); and

• fx0(x) > 0, ∀x ∈ cl(O ∩ Uα(x0)).

65

(ii) If x0 ∈ clO \K, there is an open neighborhood N(x0) of x0 such that x0 ∈ N(x0) and clN(x0) iscompact (since X is locally compact). Hence, N(x0)∩ (X \K) is open and cl (N(x0) ∩ (X \K))is compact. Once more by Prop. 3.7.5(ii), there is a continuous function gx0 : X → [0, 1] suchthat

gx0(x0) = 1, and gx0(X \ cl (N(x0) ∩ (X \K))) = 0.

This implies

supp gx0 ⊂ cl (N(x0) ∩ (X \K)) = clN(x0) ∩X \K ⇒ supp gx0 ⊂ X \K.

Consequently,supp gx0 ∩K = ∅ ⇒ gx0 = 0,∀x ∈ K ⇒ gx0 ≡ 0 on K.

(iii) Now define the sets

Vx0 := x ∈ O ∩ Uα(x0) | fx0(x) > 0, if x0 ∈ K;x ∈ Ox0 ∩ (X \K) | gx0(x) > 0, if x0 ∈ clO \K.

Thus, for each x0 ∈ clO, the sets Vx0 are open and x0 ∈ Vx0 . Consequently,

clO ⊂⋃Vx0 | x0 ∈ clO.

Since, clO is compact, there is a finite covering of clO from Vx0 | x0 ∈ clO. Correspondingly,there are functions fx1

0, . . . , fxn

0, gxn+1

0, . . . , gxn+m

0.

Set now

f =n∑

k=1

fxk0

and g =m∑

k=n+1

gxk0.

Hence,

(a) Note that for each k ∈ n + 1, . . . , m, supp gxk ⊂ X \K ⇒ K ∩ supp gxk = ∅ ⇒ gxk(x) =0, ∀x ∈ K. Hence, g ≡ 0 on K.

(b) Let the corresponding finite covering be Vx10, . . . , Vxn

0, Vxn+1

0, . . . , Vxn+m

0, such that

K ⊂ clO ⊂n⋃

k=1

Vxk0∪

m⋃

k=n+1

Vxk0

By part (a), we have K ∩(⋃m

k=n+1 Vxk0

)= ∅ ⇒ K ⊂ ⋃n

k=1 Vxk0⇒ fxk(x) > 0, ∀x ∈ K.

Hence, f > 0 on K.

From this follows thatf

f + g≡ 1 on K.( Note that g ≡ 0 on K).

Now define

ϕk :=fxk

0

f + g, k = 1, . . . , n.

It follows that ϕ1 + . . . + ϕn = 1, for each k ∈ 1, . . . , n

supp ϕk ⊂ Uα(xk0) and ϕk ≥ 0, on X.

66

Definition 3.10.5 (partition of unity). Let < X, τ > be a topological space. A family of functionsϕλ | λ ∈ Λ is called a partition of unity on X iff

(i) supp ϕλ | λ ∈ Λ is a closed locally finite covering of X;

(ii) for each λ ∈ Λ, ϕλ ≥ 0 on X;

(iii) for each x ∈ X ∑

λ∈Λ

ϕλ(x) = 1.

If ϕλ is continuous (or Lipschitz continuous or differentiable), for each λ ∈ Λ, then the partition is said tobe a continuous (or Lipschitz continuous or differentiable) partition of unity.

If a partition ϕλ | λ ∈ Λ is subordinate to a family Uα | α ∈ Ω, then it is called a partition of unitysubordinate to Uα | α ∈ Ω.

Lemma 3.10.6. If X is a paracompact space, then every open covering C = Uα | α ∈ Ω of X has alocally finite open refinement Vα | α ∈ Ω such that ∅ 6= clVα ⊂ Uα.

Proof. (i) For each x ∈ X, there is α ∈ Ω such that x ∈ Uα.

(ii) Since X is paracompact, X is normal by Thm. 3.9.15. Then, by Thm.3.4.7, there is an open setWx such that

x ∈ Wx ⊂ clWx ⊂ Uα.

Hence, W := Wx | x ∈ X is an open cover of X which refines C.

(iii) By paracompactness of X, there is a locally finite open refinement O of W.

(iv) Now define, for each α ∈ ΩAα := O ∈ O | O ⊂ Uα.

(v) Next define

Vα := ⋃

O∈AαO, if Aα 6= ∅

Any open set with ∅ 6= Vα ⊂ clVα ⊂ Uα, if Aα = ∅.

(vi) Let x ∈ X, then x ∈ O, for some O ∈ O. Thus, there is Wx ∈ W such that x ∈ O ⊂ Wx.Correspondingly, there is α ∈ Ω such that Wx ⊂ clWx ⊂ Uα. Hence,

∃α ∈ Ω : x ∈ O ⊂ Uα ⇒ x ∈ Vα.

From this follows that Vα | α ∈ Ω is a covering of X. Furthermore, Vα | α ∈ Ω is locallyfinite and, for each α ∈ Ω, Vα ⊂ clVα ⊂ Uα (cf. Lem. 3.9.6).

Theorem 3.10.7. Let < X, τ > be a topological space. If X is paracompact, then every open cover of Xhas a partition of unity subordinate to it.

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Proof. Let C = Uα | α ∈ Ω be an open cover of X. Then, by Lem. 3.10.6, there is a locally finiteopen refinement of C such that clOα ⊂ Uα for each α ∈ Ω. Then, by Thm. 3.5.3 (since X is normal) ,there is a continuous function fα : X → [0, 1] such that

fα(clOα) = 1 and fα(X \ Uα) = 0, for each α ∈ Ω. (That is, supp fα ⊂ Uα.)

Now, for x ∈ X, define

ϕα(x) :=fα(x)∑

λ∈Ω fα(x).

Claim ϕα | α ∈ Ω is a partition of unity subordinate to C.

(i) Obviously, for each α ∈ Ω, ϕ ≥ 0 on X;

(ii) To show continuity, let x0 ∈ X, there is a neighborhood U(x0) such that U(x0) intersects only afinite number of elements of clOα | α ∈ Ω. That is there is α1, . . . , αn(x0) such that

U(x0) ∩Oα = ∅, ∀α ∈ Ω \ α1, . . . , αn(x0)‖ This implies

U(x0) ⊂ X \Oα,∀α ∈ Ω \ α1, . . . , αn(x0) ⇒ fα(U(x0)) = 0,∀α ∈ Ω \ α1, . . . , αn(x0) (3.5)

andU(x0) ∩ supp fα = ∅, ∀α ∈ Ω \ α1, . . . , αn(x0). (3.6)

Thus

∑

λ∈Ω

fα(x) =n(x0)∑

k=1

fαk(x) +

∑

λ∈Ω\α1,...,αn(x0)fα(x) ⇒

∑

λ∈Ω

fα(x) =n(x0)∑

k=1

fαk(x), for x ∈ U(x0).

Consequently, for x ∈ U(x0):

ϕα(x) =fα(x)

∑n(x0)k=1 fαk

(x).

This implies, ϕα is a continuous function at x0. Note also that∑n(x0)

k=1 fαk(x0) 6= 0, since x0 ∈

clOαk0, for some k0 ∈ 1, . . . , n(x0). Hence, ϕα is a continuous on X.

(iii) For each α ∈ Ω, clOα ⊂ supp ϕα ⊂ Uα and , from eqn. (3.6), supp ϕα | α ∈ Ω is locally finite.Hence, supp ϕα | α ∈ Ω is a closed locally finite covering of X.

(iv) Moreover,∑

α∈Ω ϕα(x) = 1 for each x ∈ X.

Consequently, by Def. 3.10.5, ϕα | α ∈ Ω is a partition of unity of subordinate to C.

Remark 3.10.8. The converse of Thm. 3.10.7 also holds true. (Proof is left as an exercise!).

Corollary 3.10.9 (Thm. 2, p. 10, Aubin/Cellina). Let < X, ρ > is a metric space. Then to any locallyfinite open covering C := Uα | α ∈ Ω of X, there is locally Lipschitzean partition of unity subordinateto it.‖In fact we have here U(x0) ∩ clOα = ∅,∀α ∈ Ω \ α1, . . . , αn(x0).

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Proof. Let Uα | α ∈ Ω be a locally finite open covering of X. Then, by Lem. 3.10.6, there is a locallyfinite open covering Oα | α ∈ Ω such that clOα ⊂ Uα. Now, define

fα(x) := dist(x,X \Oα) and ϕα(x) :=fα(x)∑

λ∈Ω fλ(x).

Then the following hold true :

(i) fα : X → R+ Lipschitz continuous and

supp fα = supp ϕα = clx ∈ X | ϕα(x) 6= 0 = clOα;

(ii) ϕα | α ∈ Ω is a partition of unity subordinate to C (as in the proof of Thm. 3.10.7);

(iii) for each α ∈ Ω, ϕα is locally Lipschitz continuous on X.


1. A closed subspace of a paracompact topological space is normal.

2. A closed subspace of a paracompact topological space is paracompact.

3. If < X, τ > is a regular Lindlof space, then < X, τ > is paracompact.

4. If < X, τ > is paracompact and sparable, then < X, τ > is Lindelof.

5. The following statements are equivalent for a T1 space

(i) X is normal;

(ii) every locally finite open cover C = Uα | α ∈ Ω of X has an open refinement O = Oα | α ∈Ω with the property that ∅ 6= clOα ⊂ Uα.

69

4 The Hausdorff Metric and Convergence of Sequences of Sets

4.1 The Hausdorff Metrics

In this this section we consider solely metric spaces and normed linear spaces. Given a metric or anormed space X, we also let that P(X) := S | S ⊂ X to represent the power set of X.

Definition 4.1.1 (distance from a point to a set). Let < X, ρ > be a metric space and A ⊂ X, A 6= ∅and y ∈ X. Then the distance between y to A with respect to ρ is given by

dist(b, A) := inf ρ(b, a) | a ∈ A .

Proposition 4.1.2. Let X be a metric space, b ∈ X and A ⊂ X. Then

(i) dist(b, A) ≥ 0;

(ii) if b ∈ A, then dist(b, A) = 0;

(iii) if dist(b, A) = 0, then b ∈ clA.

Remark 4.1.3. In Def. 4.1.1, we can in fact allow A = ∅. But in this case we define

dist(b, ∅) = ∞,

for any b ∈ X.

Proposition 4.1.4. Given A ⊂ X and b ∈ X it follows that

dist(b, A) ≤ ρ(b, c) + dist(c, A)

for any c ∈ X.

Recall also that the metric distance between two sets A and B is given by

dist(A,B) = inf ρ(x, y) | x ∈ A, y ∈ B .

Definition 4.1.5 (The Hausdorff distance). Let A and B be subsets of a metric space X. Then theHausdorff distance between A and B is defined as

h(A, B) := max

supa∈A

dist(a,B), supb∈B

dist(b, A)

Observe that: dist(A,B) ≤ h(A,B).

Remark 4.1.6. The quantityh∗(A,B) := sup

a∈Adist(a,B).

is sometimes termed as the semi-(Hausdorff)distance from the set A to the set B. Thus, obviously,

• h∗(A,B) 6= h∗(B, A) and• h(A,B) = max h∗(A,B), h∗(B,A) .

71

Lemma 4.1.7. For any three sets A,B and C we have

h∗(A, B) ≤ h∗(A,C) + h∗(C, B).

Proof. Take a ∈ A arbitrarily. Then, by Prop. 4.1.4, for any c ∈ C, we have that

dist(a,B) ≤ ρ(a, c) + dist(c,B).

⇒dist(a, B) ≤ inf

c∈Cρ(a, c) + dist(c, B) ≤ inf

c∈Cρ(a, c) + sup

c∈Cdist(c,B).

⇒dist(a,B) ≤ dist(a,C) + h∗(C,B).

This last inequality holds true for every a ∈ A. Hence

supa∈A

dist(a, B) ≤ supa∈A

dist(a,C) + h∗(C, B)

Therefore,h∗(A, B) ≤ h∗(A,C) + h∗(C, B).

That fact that h∗(A,B) 6= h∗(B, A) and the satisfaction of the triangle inequality make h∗ a quasimetric (see sec. 1.1 of chap. 1 ).

Lemma 4.1.8. For any four real numbers a, b, c, d, the following holds

maxa + b, d + e ≤ maxa, d+ maxb, e

The Hausdorff distance defines some sort of metric on the set P(X) as given by

Theorem 4.1.9. The following statements hold true for the Hausdorff distance h:

(i) for each A,B ∈ P(X), h(A,B) ≥ 0;

(ii) for each A ∈ P(X), h(A,A) = 0;

(iii) for A, B ∈ P(X), h(A,B) = h(B,A);

(iv) for A,B, C ∈ P(X), h(A,B) ≤ h(A,C) + h(C, B) (triangle inequality).

Thus (i)-(iii) imply that the Hausdorff distance h defines a pseudo-metric on P(X) (sec. 1.1. of chap.1).

Proof. (i) - (iii) follow directly from Defs. 4.1.1 and 4.1.5.

(iv) Using Lem. 4.1.7 twice, we find that

h∗(A,B) ≤ h∗(A,C) + h∗(C, B)h∗(B, A) ≤ h∗(B, C) + h∗(C,A).

⇒

h(A,B) = max h∗(A,B), h∗(B,A) ≤ max h∗(A,C) + h∗(C,B), h∗(B,C) + h∗(C, A)= max h∗(A,C) + h∗(C,B), h∗(C,A) + h∗(B, C) .

72

Next, applying Lem. 4.1.8, we obtain that

h(A,B) ≤ max h∗(A,C), h∗(C, A)+ max h∗(C, B), h∗(B,C) = h(A,C) + h(C, B).

Hence,h(A,B) ≤ h(A,C) + h(C,B).

Remark 4.1.10. In general, for A,B ⊂ X, h(A,B) = 0 does not imply that A = B; i.e. h is not ametric on P(X).

Lemma 4.1.11. Let A and B be subsets of a metric space. If h∗(A, B) = 0, then A ⊂ clB.

Proof. By definition

h∗(A, B) = supa∈A

dist(a,B) = 0 ⇒ ∀a ∈ A : dist(a,B) = 0, ( since dist(a,B) ≥ 0 for any point a and set B ).

Hence, for each a ∈ A, a ∈ clB (cf. Prop. 4.1.2); i.e. A ⊂ clB.

Theorem 4.1.12. Let X be a metric space and Pcl(X) is the set of all closed subsets of X. Then theHausdorff distance h defines a metric on Pcl(X); < Pcl(X), h > is a metric space.

Proof. According to Thm.4.1.9, it remains to show that for A,B ∈ Pcl(X), h(A,B) = 0 implies A = B.But then

h(A,B) = 0 ⇒ h∗(A,B) = 0 and h∗(B,A) = 0.

Then, using Lem. 4.1.11, we conclude

A ⊂ clB and B ⊂ clA ⇒ clA = clB ⇒ A = B. ( Since both A and B are closed sets).

Suppose, for a set A ⊂ X, b ∈ X and ε > 0, we have

Uε(A) := x ∈ X | dist(x,A) < ε, Bε(b) := x ∈ X | ρ(x, b) < ε,Uε(A) := x ∈ X | dist(x,A) ≤ ε and clBε(b) := x ∈ X | ρ(x, b) ≤ εLemma 4.1.13. For any two subsets A and B of a metric space X we have

h∗(A,B) = infε > 0 | A ⊂ Uε(B) (4.1)

h∗(B, A) = infε > 0 | B ⊂ Uε(A). (4.2)

Proof. We need only to show that h∗(A,B) = infε > 0 | A ⊂ Uε(B).• If h∗(A,B) = 0, then the equality follows by using Lem. 4.1.11.

(a) Let ε0 := infε > 0 | A ⊂ Uε(B). Then, using Prop. 4.1.4, for a ∈ A and c ∈ Uε0(B) follows

dist(a,B) ≤ ρ(a, c) + dist(c, B) ≤ ρ(a, c) + ε0.

⇒ dist(a, B) ≤ ρ(a, c) + ε0. Since c ∈ Uε0(B) is arbitrary, we obtain

dist(a,B) ≤ infc∈Uε(B)

ρ(a, c) + ε0 = dist(a,Uε(B)) + ε0 = 0 + ε0 = ε0. (Note that A ⊂ Uε0(B)).

Consequently,dist(a,B) ≤ ε0, for each a ∈ A. ⇒ sup

a∈Adist(a, B) ≤ ε0

Implying thath∗(A,B) ≤ infε > 0 | A ⊂ Uε(B).

73

(b) Conversely, suppose h∗(A,B) = r. Then supa∈A dist(a,B) = r. Hence, for a ∈ A, we obtain

r = h∗(A, B) ≥ ρ(a,B).

⇒∃b ∈ B : ρ(a, b) ≤ r ⇒ a ∈ clBr(b) := x ∈ X | ρ(x, b) ≤ r.

⇒A ⊂

⋃

b∈B

clBr(b). (4.3)

Now, for any b ∈ B, let z ∈ Br(b). Then

dist(z, B) ≤ ρ(z, b) + dist(b,B) ⇒ dist(z, B) ≤ r ⇒ z ∈ Ur(B).

Since, z ∈ Br(b) is arbitrary, we conclude that clBr(b) ⊂ Ur(B). It then follows that,⋃

b∈B

clBr(b) ⊂ Ur(B).

Hence, along with (4.3) we find that A ⊂ Ur(B). Furthermore, if we take an arbitrary n ∈ N,we obtain

A ⊂ Ur(B) ⊂ Ur+ 1n(B).

⇒r +

1n≥ infε > 0 | A ⊂ Uε(B),∀n ∈ N

Consequently,h∗(A,B) = r ≥ infε > 0 | A ⊂ Uε(B).

Therefore, from (a) and (b), we conclude that

h∗(A,B) = infε > 0 | A ⊂ Uε(B).

Proposition 4.1.14. Let X be a metric space and A,B ⊂ X. Then

h(A,B) = inf ε > 0 | A ⊂ Uε(B) and B ⊂ Uε(A)Proof. Use Lem. 4.1.13

Remark 4.1.15. Given a normed linear space X and A,B ⊂ X and γ ∈ R we recall that

A + B = a + b | a ∈ A, b ∈ B - the sum of two setsγA = γa | a ∈ A - scalar product.

The sum of sets A + B is commonly known as Kuratowski sum. Thus, for B = b, we write

B + A = b + A = b + a | a ∈ A.Consequently, for a set A ⊂ X and ε > 0 we can now write

Uε(A) = A + εB,

where B represents the unit ball centered at the zero element of X. Thus, the Hausdorff metric in anormed linear space takes the form

h(A,B) = infε > 0 | A ⊂ B + εB and B ⊂ A + εB.

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Definition 4.1.16 (Pompeiu-Hausdorff distance, see [22]). Let < X, ρ > be a metric space, A,B ∈P(X) \ ∅. Then the Pompeiu-Hausdorff distance between A and B is given by

D∞(A,B) = sup|dist(x, A)− dist(x,B)| : x ∈ X = supx∈X

|dist(x,A)− dist(x,B)|.

Note that, if either A = ∅ or B = ∅, then D∞ is undefined.

Proposition 4.1.17. Let < X, ρ > be a metric space and A,B ∈ P(X) \ ∅. Then

h(A,B) = D∞(A,B).

Proof. (a) Using Prop. 4.1.4, for any x ∈ X and b ∈ B, we have

dist(x,A) ≤ ρ(x, b) + dist(b, A).

Since b ∈ B is arbitrary, this implies that

dist(x,A) ≤ infb∈B

ρ(x, b) + dist(b, A) ≤ infb∈B

ρ(x, b) + supb∈B

dist(b, A) ≤ dist(x,B) + h∗(B, A).

⇒dist(x,A)− dist(x,B) ≤ h∗(B,A).

Similarly, we havedist(x,B)− dist(x,A) ≤ h∗(A,B)

Hence,

maxdist(x,A)− dist(x,B), dist(x,B)− dist(x,A) ≤ maxh∗(B,A), h∗(A,B).

⇒|dist(x,A)− dist(x, B)| ≤ h(A,B); i.e. D∞ ≤ h(A,B).

(b) Conversely

h∗(B,A) = supb∈B

dist(b, A) = supb∈B

[dist(b, A)− dist(b,B)]

≤ supx∈X

[dist(x,A)− dist(x, B)]

≤ supx∈X

|dist(x,A)− dist(x, B)|.

Analogouslyh∗(A,B) ≤ sup

x∈X|dist(x,A)− dist(x,B)|

Consequently, we have

maxh∗(B,A), h∗(A,B) ≤ supx∈X

|dist(x,A)− dist(x, B)|

⇒h(A,B) ≤ D∞(A,B).

Which concludes the proof.

75


1. Show that, if dist(b, A) = 0, then b ∈ clA.

2. Show that if B1 ⊂ B2, then dist(A,B1) ≥ dist(A,B2).

3. Prove Prop. 4.1.4.

4. Verify the validity of Lem. 4.1.8.

5. Show that if h∗(A, B) = 0, then h∗(A,B) = infε > 0 | A ⊂ Uε(B).

6. Let X be a normed linear space, A, B,C, D ∈ P(X) and λ ∈ [0, 1]. Then prove the following

(i) h(rA, rB) ≤ |r|h(A,B), ∀r ∈ R;

(ii) h(A + B, C + D) ≤ h(A,C) + h(B,D)

(iii) h(λA + (1− λ)B, λC + (1− λ)D) ≤ λh(A,C) + (1− λ)h(B, D),∀λ ∈ [0, 1].

(iv) h(cl convA, cl convB) ≤ h(A,B);

(v) h(Uε(A), Uε(A)) ≤ h(A,B),∀ε > 0;

where convA represents the convex Hull of A.

7. Considering h : P(X) × P(X) → R+, show that h(·, ·) is a Lipschitz continuous function withLipschitz constant 1; i.e. h(·, ·) is a non-expansive map.

4.2 Convergence of Sequences of Sets

Definition 4.2.1 (Kuratowski Convergence). Let < X, ρ > be a metric space and Ann∈N be a sequenceof subsets of X. Then

(i) the upper limit or outer limit of the sequence Ann∈N is a subset of X given by

lim supn→∞

An = x ∈ X | lim infn→∞ dist(x,An) = 0;

(ii) the lower limit∗ or inner limit† of the sequence Ann∈N is a subset of X given by

lim infn→∞ An = x ∈ X | lim

n→∞ dist(x,An) = 0.

If lim supn→∞An = lim infn→∞An, then we say that the limit of Ann∈N exists and

limn→∞An = lim sup

n→∞An = lim inf

n→∞ An.

∗lim supn→∞An and lim infn→∞An are sometimes called Kuratowski limit inferior and limit superior, respectively.(seePapagorgeu)

†The notions outer limit and inner limit are introduced by Rockafellar and Wets (see )

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Remark 4.2.2. For a fixed set A ⊂ X, the distance function dist(·, A) : X → R is a Lipschitz continuousfunction. This follows from the fact that

dist(x,A) ≤ ρ(x, x) + dist(x,A) ⇒ dist(x,A)− dist(x,A) ≤ ρ(x, x).

for any x, x ∈ X. Hence,|dist(x,A)− dist(x, A)| ≤ ρ(x, x).

Consequently, if xnn∈N is sequence in X such that xn → x, then

limn→∞ dist(xn, A) = dist(x,A).

With this remark the following proposition follows trivially.

Proposition 4.2.3. Let Ann be a sequence of subsets of a metric space X. Then

(i) lim infn→∞An ⊂ lim supn→∞An;

(ii) the sets lim supn→∞An and lim infn→∞An are closed in X .

Proof. (i) is trivial. To show (ii), let x ∈ cl (lim supn→∞An). This implies

∃xkk∈N ⊂ lim supn→∞

An such that xk → x.

Consequently, for each k, we have

dist(x, An) ≤ ρ(x, xk) + dist(xk, An).

⇒lim infn→∞ dist(x,An) ≤ ρ(x, xk) + lim inf

n→∞ dist(xk, An) ⇒ lim infn→∞ dist(x,An) ≤ ρ(x, xk), ∀k.

⇒lim infn→∞ dist(x, An) ≤ inf

kρ(x, xk) = 0.

Hence, lim infn→∞ dist(x,An) = 0 ⇒ x ∈ lim supn→∞An . It follows that cl (lim supn→∞An) ⊂lim supn→∞An. Thus, lim supn→∞An is a closed set. The rest is left as an exercise.

The closure of lim supn→∞An also follows similarly.

Proposition 4.2.4. If Ann∈N is a sequence in a metric space X, then

(i) lim supn→∞An = x ∈ X | xnk∈ Ank

: xnk→ x.

(ii) lim infn→∞An = x ∈ X | xn ∈ An : xn → x.

That is, lim infn→∞An is a collection of limits of sequences xnn∈N, where xn ∈ An; whereaslim supn→∞An is a collection of cluster points of sequences xnn∈N, where xn ∈ An.

Proof. (i) Obviously, lim supn→∞An ⊃ x ∈ X | xnk∈ Ank

: xnk→ x. Thus, let x ∈ lim supn→∞An.

Thenlim infn→∞ dist(x,An) = 0 ⇒ sup

ninfi≥n

dist(x, Ai) = 0.

Let k ∈ N be arbitrary, we have

supn

infi≥n

dist(x, Ai) ≤ 1k

77

Hence,

∃nk ∈ N : infi≥nk

dist(x,Ai) ≤ 1k.

⇒∃ik ≥ nk, ∃xik ∈ Aik : ρ(x, xik) ≤ 1

k.

But this is true for each k ∈ N. Hence, there is xikik∈N such that xik ∈ Aik and xik → x.Consequently,

x ∈ x ∈ X | xnk∈ Ank

: xnk→ x

Therefore, lim supn→∞An ⊂ x ∈ X | xnk∈ Ank

: xnk→ x.

(ii) follows trivially.

Proposition 4.2.5. Let An be a sequence in a metric space < X, ρ >. Then

(i) lim supn→∞An = x ∈ X | ∀ε > 0,∀N ∈ N, ∃n ≥ N : Bε(x) ∩An 6= ∅.

(ii) lim infn→∞An = x ∈ X | ∀ε > 0,∃N(ε) ∈ N : Bε(x) ∩An 6= ∅,∀n ≥ N(ε)

Proof. (i) Suppose x ∈ lim supn→∞An. Assume that

∃ε > 0, ∃N ∈ N : Bε(x) ∩An = ∅,∀n ≥ N.

⇒dist(x,An) ≥ ε, ∀n ≥ N ⇒ lim inf

n→∞ dist(x,An) = supn

infk≥n

dist(x,Ak) 6= 0.

⇒ x /∈ lim supn→∞An. But this is a contradiction. Consequently,

x ∈ x ∈ X | ∀ε > 0, ∀N ∈ N, ∃n ≥ N : Bε(x) ∩An 6= ∅

That islim sup

n→∞An ⊂ x ∈ X | ∀ε > 0,∀N ∈ N, ∃n ≥ N : Bε(x) ∩An 6= ∅

To show the reverse inclusion. By assumption, we have

∀k ∈ N, ∃nk ≥ k : B 1k(x) ∩Ank

6= ∅.

⇒ ∃xnk∈ Ank

: ρ(x, xnk) ≤ 1

k . Consequently, the sequence xnkk∈N, with xnk

∈ Ankconverges

to x. Hence, by Prop. 4.2.4,x ∈ lim sup

n→∞An

Hence, lim supn→∞An = x ∈ X | ∀ε > 0, ∀N ∈ N, ∃n ≥ N : Bε(x) ∩An 6= ∅.

(ii) Use a similar argument as in (i). (Exercise!)

Remark 4.2.6. The statements in Prop. 4.2.4 and Prop. 4.2.5 can be used as alternative definitions ofinferior and superior limits of a sequence of sets as defined above. In particular, from Prop. 4.2.5, itfollows that

78

(i) lim supn→∞

An =⋂

ε>0

⋂

N≥1

⋃

n≥N

Uε(An) (4.4)

(ii)⋂

n≥1

cl

⋃

n≥m

Am

⊂ lim sup

n→∞Am (4.5)

(iii) lim infn→∞ An =

⋂

ε>0

⋃

N≥1

⋂

n≥N

Uε(An). (4.6)

Proposition 4.2.7. If Ann∈N is a sequence such that An ⊃ An+1, n ∈ N (a decreasing sequence),then limn→∞An exists and

limn→∞An =

⋂

n∈N

cl An

Proof. Since dist(x, clAn) = dist(x,An), it follows that,⋂

n∈N

cl An ⊂ lim infn→∞ An. (4.7)

If we now show that lim supn→∞An ⊂⋂

n∈N cl An, then we are done!

• Let x ∈ lim supn→∞An. Hence,sup

ninfk≥n

dist(x,Ak) = 0.

Since, for k ≥ n, An ⊃ Ak, we have dist(x,An) ≤ dist(x,Ak). This implies that

supn

infk≥n

dist(x,Ak) = supn

dist(x,An) = 0 ⇒ dist(x,An) = 0,∀n.

⇒ (by Prop. 4.1.2(iii)) x ∈ clAn, ∀n ∈ N. Hence

lim supn→∞

An ⊂⋂

n∈N

cl An.

• Moreover, by Prop. 4.2.3, we have lim infn→∞An ⊂ lim supn→∞An. Hence

lim infn→∞ An ⊂

⋂

n∈N

cl An.

Finally, using (4.7), limn→∞An exists and

limn→∞An =

⋂

n∈N

cl An.

4.2.1 Calculus of Limits of Sequences of Sets

Theorem 4.2.8. Let Ann∈N and Bnn∈N be two sequences of sets, K ⊂ X be a compact set. If

for every neighborhood U of K, ∃N ∈ N : An ⊂ U,∀n ≥ N,

then

for every neighborhood V of K ∩(

lim supn→∞

Bn

),∃N ∈ N : An ∩Bn ⊂ V,∀n ≥ N.

79

Proof. Let V be any neighborhood of K ∩ (lim supn→∞Bn); i.e. K ∩ (lim supn→∞Bn) ⊂ V .

(i) If K ⊂ V , then taking U = V , we have, by assumption, that

∃N : An ⊂ V, ∀n ≥ N ⇒ An ∩Bn ⊂ An ⊂ V,∀n ≥ N.

(ii) If K is not a subset of V , then M := K \ V 6= ∅, M is compact and M ∩ (lim supn→∞Bn) = ∅.Hence,

z ∈ M ⇒ z /∈ lim supn→∞

BnProp. 4.2.5⇒ ∃ε(z) > 0, ∃N(z) ∈ N : Bε(z)(z) ∩Bn = ∅,∀n ≥ N.

Hence,M ⊂

⋃

z∈M

Bε(z)(z).

Since M is compact, there are zk ∈ M, k = 1, . . . , m, such that

M ⊂m⋃

k=1

Bε(zk)(zk) =: W.

Now define, N0 := maxN(zk) | k = 1, . . . , m. It follows that

W ∩Bn = ∅,∀n ≥ N0.

Moreover,K \ V = M ⊂ W ⇒ K ⊂ V ∪W.

Thus K ⊂ U := V ∪W . By assumption,

∃N1 : An ⊂ U,∀n ≥ N1 ⇒ An ∩Bn ⊂ An ⊂ U = V ∩W,∀n ≥ N1.

However, Bn ∩W = ∅, ∀n ≥ N0. Thus, if we set N := maxN0, N1, then we obtain

An ∩Bn ⊂ V, ∀n ≥ N.

Theorem 4.2.9. (Thm. 5.2.4, p 221, Aubin[2], Thm. 1.1.4, p. 21, Aubin Frankowska[4], ) Let Ann∈Nbe a sequence in a metric space X and K ⊂ X. If, for every neighborhood U of K

∃N ∈ N : An ⊂ U,∀n ≥ N,

thenlim sup

n→∞An ⊂ cl(K).

Conversely, if X is a compact metric space, then, for every neighborhood U of lim supn→∞An,

∃N ∈ N : An ⊂ U,∀n ≥ N.

80

Proposition 4.2.10. Let Ann∈N and Bnn∈N be two sequences of subsets of a metric space X. Thenthe following hold true

(i) lim supn→∞

(An ∩Bn) ⊂ lim supn→∞

An ∩ lim supn→∞

Bn

(ii) lim infn→∞ (An ∩Bn) ⊂ lim inf

n→∞ An ∩ lim infn→∞ Bn

(iii) lim supn→∞

(An ∪Bn) = lim supn→∞

An ∪ lim supn→∞

Bn

(iv) lim infn→∞ (An ∪Bn) ⊃ lim inf

n→∞ An ∪ lim infn→∞ Bn

(v) lim supn→∞

(An ×Bn) ⊂ lim supn→∞

An × lim supn→∞

Bn

(vi) lim infn→∞ (An ×Bn) = lim inf

n→∞ An × lim infn→∞ Bn.

Proposition 4.2.11. Let X and Y be metric spaces, Ann∈N and Bnn∈N are sequence in X and Y ,respectively. If f : X → Y is a continuous function, then the following hold true

(i) f

(lim sup

n→∞An

)⊂ lim sup

n→∞f (An) ; (4.8)

(ii) f(lim infn→∞

)An ⊂ lim inf

n→∞ f (An) ; (4.9)

(iii) lim supn→∞

f−1 (Bn) ⊂ f−1

(lim sup

n→∞Bn

); (4.10)

(iv) lim infn→∞ f−1 (Bn) ⊂ f−1

(lim infn→∞ Bn

). (4.11)

Proof. We prove (i) and the rest is transparent. We mainly use here Prop. 4.2.4. Let y ∈ f (lim supn→∞An).Then y = f(x) for some x ∈ lim supn→∞An. This implies,

∃xnk, xnk

∈ Ank: xnk

→ x.

Since, f is a continuous function, we have

f(xnk) → f(x) = y.

Moreover, f(xnk) ∈ f(Ank

). Hence,

y = f(x) ∈ lim supn→∞

f (Ank)

Hence,

f

(lim sup

n→∞Ank

)⊂ lim sup

n→∞f (Ank

) .

Excercises 4.2.12. Let An be a sequence of sets in a metric space.

1. Show that the set lim infn→∞An is a closed set.

81

2. Verify that the limit superior can be also written as

lim supn→∞

An =⋂

N≥1

cl

⋃

n≥N

An

.

3. Verify the statements of Rem. 4.2.6.

4. Prove part (ii) of Prop. 4.2.4; i.e. lim infn→∞An = x ∈ X | xn ∈ An : xn → x.5. Prove part (ii) of Prop. 4.2.5; i.e. lim infn→∞An = x ∈ X | ∀ε > 0,∃N(ε) ∈ N : Bε(x) ∩ An 6=∅,∀n ≥ N(ε).

6. Prove statements (ii)-(iv) of Prop. 4.2.11.

7. Let X be a metric space and An ⊂ P(X) and A ∈ P(X). Then prove that

limn→∞An = A ⇔ lim

n→∞ dist(x,An) = 0,∀x ∈ A and lim infn→∞ dist(x,An) > 0, ∀x /∈ A.

4.2.2 Convergence w.r.t. the Hausdorff Metric

Definition 4.2.13 (Convergence in Hausdorff metric). Let An be a sequence of closed subsets of Xand A ⊂ X be also closed. We say that An converges to A w.r.t. the Hausdorff metric iff

limn→∞h(An, A) = 0.

We denote this byAn

h→ A.

Remark 4.2.14. It is easy to verify that: if limn→∞ h(An, A) = 0, then

∀ε > 0 : ∃N ∈ N : An ⊂ Uε(A) = x ∈ X | dist(x,A) < ε,∀n ≥ N. (4.12)

Note that, An ⊂ Uε(A) implies that dist(An, A) < ε. This in turn implies that A ⊂ Uε(An). We can nowwrite (4.12) equivalently as

A ⊂⋂

ε>0

⋃

n≥1

⋂

m≥n

Uε(Am).

Theorem 4.2.15. Let X be a metric space, An ∈ Pcl(X) and A ∈ Pcl(X) . Then

Anh→ A ⇒ An → A

That is, Hausdorff-convergence implies Kuratowski-convergence.

Proof. We show that A ⊂ lim infn→∞An and lim supn→∞An ⊂ A.

(i) By assumption

limn→∞h(An, A) = 0 ⇒ lim

n→∞h∗(A,An) = 0 ⇒ limn→∞ sup

x∈Adist(x,An) = 0.

⇒∀x ∈ A : lim

n→∞ dist(x,An) = 0.

From this it follows thatA ⊂ lim inf

n→∞ An.

82

(ii) Hence, according to Prop. 4.2.3, if we show that lim supn→∞An ⊂ A, then we are done.Assume that there is x ∈ lim supn→∞An, but x /∈ A. Hence, dist(x,A) = γ > 0, for some γ ∈ R.

Now letB := x ∈ X | dist(x,A) ≤ γ

2 = clU γ

2(A)

Thus, x /∈ B and Rem. 4.2.14 implies that

∃N ∈ N : An ⊂ B, ∀n ≥ N.

Consequently, we have

dist(x,An) ≥ dist(x,B) ≥ γ

2> 0, ∀n ≥ N.

⇒lim infn→∞ dist(x,An) ≥ lim inf

n→∞ dist(x,B) ≥ γ

2> 0

⇒ x /∈ lim supn→∞An, which is a contradiction.

Therefore, we havelim sup

n→∞distAn ⊂ A.

Remark 4.2.16. However, the converse of Thm. 4.2.15 does not hold always true. To see this considerthe sequence

An := 0, 1

n, if n is even;0, n, if n is odd;

Note that:

• xn = 1n ∈ A2n and xn → 0 ⇒ 0 ∈ lim supn→∞An. Moreover, for x ∈ R arbitrary, we have

dist(x,An) = infz∈An

ρ(x, z) = infz∈An

|x− z| =

infz∈0, 1n |x− z|, if n is even;

infz∈0,n |x− z|, if n is odd;

⇒dist(x, An) =

min|x|, |x− 1

n |, if n is even;min|x|, |x− n|, if n is odd;

Hence,

lim infn→∞ dist(x,An) = 0 ⇔

lim infn→∞min|x|, |x− 1

n |, if n is even;lim infn→∞min|x|, |x− n|, if n is odd;

= 0.

⇔

lim infn→∞ min|x|, |x− 1

n| = 0, if n is even; and

lim infn→∞ min|x|, |x− n| = 0, if n is odd

But these two limits are zero if and only if x = 0; i.e. lim supn→∞An = 0. Similarly,lim infn→∞An = 0. Consequently,

limn→∞An = 0.

83

• Moreover,h(0, An) = maxh∗(0, An), h∗(An, 0)

Thus,h∗(0, An) = sup

x∈0dist(x,An) = dist(0, An) = 0.

But

h∗(An, 0) = supx∈An

dist(x, 0) = supx∈An

|x− 0| = supx∈An

|x| =

1n , if n is even;n, if n is odd;

⇒lim

n→∞h∗(0, An) 6= 0.

Consequently,lim

n→∞h(An, 0) 6= 0. That is, An 9 0.

When do we have equality between Kuratowski and Hausdorff convergence? The answer is given inthe following statement.

Proposition 4.2.17. Let X be a metric space, An be a sequence of compact subsets of X and A ⊂ Xcompact. Then, if there is a compact subset K of X such that An ⊂ K, ∀n; and limn→∞An = A, thenlimn→∞ h(An, A) = 0.

Proof. By assumption and Prop. 4.2.3, we see that A is a compact set and, by Thm. 4.2.9, A ⊂ K.Assume that h(An, A)9 0; i.e. An

h9 A. This implies

limn→∞maxh∗(An, A), h∗(A,An) 6= 0 ⇒ lim

n→∞maxh∗(An, A), h∗(A,An) > γ, for some γ > 0.

⇒lim

n→∞h∗(A, An) 6= 0 or limn→∞h∗(An, A) 6= 0.

(i) If limn→∞ h∗(A,An) 6= 0, thenlim

n→∞ supx∈A

dist(x, An) > 0

⇒ (by the compactness of A)

∀n,∃xn ∈ A : dist(xn, An) > 0.

But then xn ⊂ K and K is compact implying that there is a subsequence xnk such that

xnk→ x, where x ∈ A. But

∀nk : dist(xnk, Ank

) > 0. (4.13)

Moreover, since A = lim infn→∞An, by Prop. 4.2.4, there is a sequence zn with zn ∈ An suchthat zn → x. Consequently,

dist(xnk, Ank

) ≤ ρ(xnk, znk

) ≤ ρ(xnk, x) + ρ(x, znk

), ∀n ∈ N.

However, from this follows thatlim

k→∞dist(xnk

, Ank) = 0

which is a contradiction to (4.13). Consequently, it must be that limn→∞ h∗(A,An) = 0.

84

(ii) If limn→∞ h∗(An, A) 6= 0, then there is a subsequence All∈L of Ann∈N, where L ⊂ N suchthat

liml→∞

h∗(Al, A) = liml→∞

supy∈Al

dist(y, A) > 0.

⇒∀l ∈ L,∃xl ∈ Al : dist(xl, A) > 0.

But, again xl ⊂ K and K is compact implies there is a convergent subsequence xlk suchthat xlk → x, for some x ∈ X. From this follows that x ∈ lim supn→∞An = A (see Prop. 4.2.4).However,

0 < dist(xlk , A) ≤ ρ(xlk , x), ∀lk ⇒ 0 < dist(x,A) = 0.

Which is a contradiction.

Therefore, from (i) and (ii), we conclude that limn→∞ h(An, A) = 0.

Proposition 4.2.18. If An ⊂ Pcl(X), A ∈ Pcl(X) and Anh→ A, then

A =⋃

n≥1

cl

⋃

m≥n

Am

.

Proof. Since Anh→ A implies An → A, by Rem 4.2.6, we have that

⋂

n≥1

cl

⋃

m≥n

Am

⊂ A.

Moreover, using Rem. 4.2.14, given any n ≥ 1, for an arbitrary ε > 0,

∃m ≥ n : A ⊂ Uε(Am) ⇒ A ⊂ Uε

⋃

m≥n

Am

.

Since, ε > 0 is arbitrary, we have

A ⊂ cl

⋃

m≥n

Am

. This holds true for any n ≥ 1. Consequently, A ⊂

⋂

n≥1

cl

⋃

m≥n

Am

.

Therefore, the claim of the proposition follows.

A sequence of non-empty closed sets An is said to be a Cauchy sequence in < Pcl(X), h > if

∀ε > 0,∃N : h(An, Am) < ε,∀n,m ≥ N.

Theorem 4.2.19. If < X, ρ > is a complete metric space, then < Pcl, h > is also a compete metric space.

Proof. Let An be any Cauchy sequence of non-empty closed sets in < Pcl, h >. Then we show thatAn is convergent. According to Prop. 4.2.18, we need only to verify that An converges to the set

A =⋂

n≥1

cl

⋃

m≥n

Am

.

Thus, we have to show that: (i) A is closed. (Obvious!) (ii) A is non-empty. (iii) Anh→ A.

85

(ii) Since An a Cauchy sequence, given δ > 0 (say δ = ε3 , where ε > 0), for each k ≥ 0, there is Nk

such thath(An, Am) <

δ

2k+1, ∀n,m ≥ Nk.

• Now, for k=0,

∃N0 : h(An, Am) <δ

2,∀n,m ≥ N0.

Then for any n0 ≥ N0

h(An, An0) <δ

2, ∀n ≥ N0 ⇒ sup

x∈An0

dist(x,An) <δ

2, ∀n ≥ N0.

Thus, for any fixed xn0 ∈ An0

dist(xn0 , An) <δ

2, ∀n ≥ N0.

• For k = 1, there exists N1 such that

h(An, Am) <δ

22, ∀n,m ≥ N1.

Then, for any n1 ≥ maxN0, N1 we have

dist(x,An) <δ

22, ∀x ∈ An1 , ∀n ≥ N1.

Choose xn1 ∈ An1 , then

ρ(xn1 , xn0) ≤ ρ(xn1 , x) + ρ(x, xn0), for any x ∈ An and n ≥ N1.

⇒ρ(xn1 , xn0) <

δ

22+

δ

2= 3

δ

22=

ε

22.

Proceeding in this way, for nk ≥ maxN0, N1, . . . , Nk, we can choose xnk+1∈ Ank+1

such that

ρ(xnk+1, xnk

) <ε

2k+1.

Thus the sequence xnk is a Cauchy sequence. Since X is a complete metric space, there is

x ∈ X such that xnk→ x.

• Furthermore, for each n ≥ 1, there is nk0 ≥ n such that

xnk∈

⋃

m≥n

Am, ∀nk ≥ nk0 ⇒ x ∈ cl

⋃

m≥n

Am

.

This is true for all n ≥ 1.

x ∈ cl

⋃

m≥n

Am

, ∀n ≥ 1 ⇒ x ∈

⋂

n≥1

cl

⋃

m≥n

Am

= A.

Consequently, A 6= ∅.

86

(iii) Next, we show that Anh→ A.

• From (i), for each n0 ≥ N0 and any xn0 ∈ An0 we obtain, by the continuity of ρ(·, x0), that

ρ(x, xn0) = limnk→∞

ρ(xnk, xn0) ≤ lim

nk→∞

nk∑

i=1

ρ(xni , xni−1) < limnk→∞

nk∑

i=1

ε

2i= ε.

⇒∀n0 ≥ N0, ∀xn0 ∈ An0 : ρ(x, xn0) < ε ⇒ An0 ⊂ Bε(x) ⊂ Uε(A), ∀n0 6= N0.

⇒

∃N0 : h∗(An, A) < ε,∀n ≥ N0. (4.14)

• Conversely, let x ∈ A be arbitrary, then

x ∈ cl

⋃

m≥N0

Am

⇒ ∃m ≥ N0, ∃z ∈ Am : ρ(x, z) <

ε

2. (4.15)

Moreover, the following holds true (see the proof of Lem. 4.1.7)

dist(x,An) ≤ dist(x,Am) + h(Am, An).

Since An is a Cauchy sequence, there is N ′0 such that h(An, Am) < ε

2 , ∀n ≥ N ′0. This,

along with (4.15), yields that

dist(x,An) ≤ dist(x,Am)+h(An, Am) ≤ ρ(x, z)+h(An, Am) <ε

2+

ε

2, ∀n,m ≥ maxN0, N

′0.

⇒dist(x, An) < ε,∀n ≥ N ′

1 := maxN0, N′0 ⇒ x ∈ Uε(An),∀n ≥ N ′

1.

Since, x ∈ A is arbitrary, it follows that

A ⊂ Uε(An), ∀n ≥ N ′1 ⇒ h∗(A,An) < ε,∀n ≥ N ′

1. (4.16)

Finally, from (4.14) and (4.16) we conclude that

h(A,An) < ε,∀n ≥ N ′1.

Therefore, Anh→ A.

Excercises 4.2.20. Verify the following statements.

1. if limn→∞An = A, thenlim

n→∞h∗(A,An) = 0.

2. Let An be a sequence of closed sets and A ∈ Pcl(X). Then

87

(i) if limn→ h∗(An, A) = 0, then lim supn An ⊂ A;

(ii) if limn→ h∗(A,An) = 0, then lim infn An ⊃ A.

3. Let X be a metric space, An ⊂ Pcl(X) and A ∈ Pcl(X) such that

limn→∞h(An, A) = 0.

If xn ∈ An and xn → x, then x ∈ A.

4. Let X be a metric space, An ⊂ Pcl(X), A ∈ Pcl(X). If Anh→ A, then ∃εk, εk 0, such that,

for each k ∈ N : dist(y,A) < εk,∀y ∈ Ak.

5. If Anh→ A, then

A =⋃

n≥1

cl

⋃

m≥n

Am

=

⋂

ε>0

⋃

n≥1

⋂

m≥n

Uε(Am).

6. (see pp. 108-109 Aliprantis & Border [1]) Define Uε(F ) = A ∈ Pcl(X) | h(A,F ) < ε.Then

(a) the collection Uε(F ) | F ∈ Pcl(X), ε ∈ (0,∞) forms a base for a topology τh on Pcl(X) andthis topology is first countable;

(b) if < X, ρ > is a compact metric space, then < Pcl(X), τh > is a compact topological space;

(c) if < X, ρ > is a separable metric space, then < Pcl(X), τh > is also a separable topologicalspace.

88

5 Set-Valued Maps

5.1 Introduction

In this chapter we, generally, assume X and Y to be at least Hausdorff topological spaces. But prac-tically set-valued maps reveal interesting properties when X and Y are taken to be normed linearspaces.

Definition 5.1.1 (set-valued map). Let X and Y be topological spaces. If for each x ∈ X there is acorresponding set F (x) ⊂ Y , then F (·) is called a set-valued map from X to Y . We denote this by

F : X −→→ Y.

A function f : X → Y can be treated as a special set-valued map if we define F (x) := f(x). For thesake of brevity we write ’SV-map’ for ’set-valued map’∗.

In some books a set-valued map from X to Y is denoted by F : X → 2Y , or F : X Ã Y , etc. Butwe exclusively use here the notation F : X −→→ Y . Furthermore, the terms ’set-valued map’ (Aubin &Frankowska [4]),’point-to-set map’ (Hogan [13]),’correspondences’ (Aliprantis & Boder [1]), ’multi-valued maps’ (Robinson [19, 20]) and (Hu & Papagorgious [14]), ’multifunctions’ (Castaing & Valadier[8]), are usually used interchangeably; while the first being frequently used in current literature.

In many cases we would like to know how a slight change in a parameter(s) of a given mathematicalproblem could affect the solution or solution set (or even the structure) of the problem. Currently,such a study is, in fact, very important as many useful mathematical problems are usually approx-imately solved on the computer. Thus sensitivity analysis could guarantee the acceptability of theobtained approximate solution(s), based on certain allowed error on the parameters of the problem.For instance, the characterization of the variation (due to, say, data perturbation) of solution sets ofoptimization problems, partial differential equations, etc., is done through set-valued maps. Conse-quently, Set-valued maps are indispensable tools in stability and sensitivity analysis of mathematicalproblems. Beside these, there are several other applications for set-valued maps.

5.1.1 Some Examples of Set-Valued Maps

Set-valued maps arise under several instances.

1. The inverse image of a non-bijective functionLet the function f : R→ R+ be given by f(x) = x2. For y ∈ R+, the inverse image of y under fis either x = −√y or x = +

√y. That is,

f−1(y) = −√y,√

y.Hence, for each y ∈ R, f−1(y) represents more than one value. That is, f−1 is not single valuedinstead it is a multivalued.

∗Actually, the short form ’SVM’ would have been quite practical, but it has been widely used for ’Support Vector Machines’.

89

2.The Subdifferential map of a convex Function

Let f : Rn → R be a convex function. Then, for x ∈ Rn, the map

∂f(x) := s ∈ Rn | s>(y − x) ≤ f(y)− f(x), ∀y ∈ dom(f)is called the subdifferential map of f at x.

2. Solution Sets of Metric ProjectionsLet S ⊂ Rn and x ∈ Rn. Define the map

ΠS(x) = y ∈ S | ‖x− y‖ = dist(x, S)known us the metric projection of X onto S. Note that, if S is a closed convex set, then ΠS(x)contains only a single element; otherwise ΠS(x) a multivalued map.

3. The normal and tangent cone maps of a convex set Let C ⊂ Rn be a convex set. Then forx ∈ C

• the normal cone map: N(x) = y ∈ Rn | y>(z − x) ≤ 0, ∀x ∈ C;• the tangent cone map: TC(x) = cly ∈ Rn | y = α(z−x), x ∈ C,α ≥ 0 = cl

(⋃γ>0

1γ (C − x)

).

4. The feasible and solution sets of a Parametric Optimization ProblemsLet f, hi, gj : Rn × Rm → R be functions. Then given the parametric optimization problem

(P (t)) f(x, t) → infhi(x, t) = 0, i = 1, . . . , m;gj(x, t) ≤ 0, j = 1, . . . , p.

we have

• the feasible set map of (P(t)): M(t) := x ∈ Rn | hi(x, t) = 0, i = 1, . . . ,m; gj(x, t) ≤ 0, j =1, . . . , p;

• the optimal solution map of (P(t)): S(t) := x ∈ Rn | w(t) = f(x, t);where

w(t) := infx∈M(t)

f(x, t)

is the the (marginal) value function of (P(t)). Here, both M(·) and S(·) are set valued mapsfrom T to X.

5.2 Basic Definitions

Definition 5.2.1 (Domain, Range and Graph). Let X and Y be metric spaces and F : X −→→ Y . Thedomain of F (·), denoted by Dom(F ), is defined as:

Dom(F ) := x ∈ X | F (x) 6= ∅;the range of F (·) is defined as

Rang(F ) :=⋃

x∈Dom(F )

F (x);

and the graph of F (·), denoted by Graph(F ), is defined as

Graph(F ) := (x, t) | t ∈ F (x), x ∈ Dom(F ).

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Definition 5.2.2. Let X and Y be topological spaces.

(i) A set valued map F : X −→→ Y is said to be closed valued, open valued or compact valued if, for eachx ∈ X, F (x) is a closed, open or compact set, respectively, in Y . Furthermore, if Y is a topologicallinear space and F (x) is a convex set in Y for each x ∈ X, then F (·) is called convex valued.

(ii) A set valued map F : X −→→ Y is said to be a closed, open or compact set-valued map Graph(F ) isa closed, open or compact set w.r.t. the product topology of X × Y . Furthermore, if X and Y aretopological linear spaces; then F (·) called a convex set-valued map if Graph(F ) is a convex set inw.r.t. X × Y .

Remark 5.2.3. In Def. 5.2.2 care must be taken not to confuse closed valued maps and closed maps. Theformer refers to values of the map; while the latter refers to the graph of the map.

5.2.1 Elementary Mathematical Operations with Set-Valued Maps

Definition 5.2.4 (closure, interior, convex-hull sv-maps). Let X and Y be topological spaces andF : X −→→ Y . Then

• the closure sv-map associated with F is the map

cl(F ) : X −→→ Y, where cl(F )(x) = cl(F (x)), for each x ∈ X.

• the interior sv-map associated with F is the map

int(F ) : X −→→ Y, where int(F )(x) = int(F (x)), for each x ∈ X.

Moreover, if Y is a topological linear space, then

• the convex-hull sv-map associated with F is the map

conv(F ) : X −→→ Y, where conv(F )(x) = conv(F (x)), for each x ∈ X.

Example 5.2.5. For instance, let f : R → R+ with f(x) = x2 and F (x) = f−1(x). Then F : R+−→→ R

and F (x) = −√x,√

x. It follows that conv(F (x)) = [−√x,√

x], for each x ∈ R+.

If F : X −→→ Y and A ⊂ X, then the image of the set A under F is given by

F (A) =⋃

x∈A

F (x).

Proposition 5.2.6 (Aubin & Frankowska [4]). Let X and Y be topological spaces and F : X −→→ Y ,A,B ⊂ X. Then

(i) F (A ∪B) = F (A) ∪ F (B);

(ii) F (A ∩B) ⊂ F (A) ∩ F (B);

(iii) F (X \A) ⊃ Range(F ) \ F (A);

(iv) F (X \A) ⊃ Range(F ) \ F (A);

(v) if A ⊂ B, then F (A) ⊂ F (B).

Proof. Trivial!

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Definition 5.2.7 (combination and composition of sv-maps). Let X,Y and Z be topological spaces, F1

and F2 are two sv-maps from X to Y . Then

• the union map of F1 and F2 is the map F1 ∪F2 : X −→→ Y given by (F1 ∪F2)(x) = F1(x)∪F2(x), foreach x ∈ X;

• the intersection map of F1 and F2 is the map F1 ∩ F2 : X −→→ Y is given by (F1 ∩ F2)(x) =F1(x) ∩ F2(x), for each x ∈ X.

• If F1 : X −→→ Y and F2 : X −→→ Z, then the production map of F1 and F2 the map F1×F2 : X −→→ Y ×Zis given by (F1 × F2)(x) = F1(x)× F2(x), for each x ∈ X.

• If Y is a linear space, the sum and difference can be defined likewise. Thus

(F1 ± F2)(x) = F1(x)± F2(x).

Furthermore, if F1 : X −→→ Y and F2 : Y −→→ Z, then composition map of F1 and F2 is the map

F2 F1 : X −→→ Z

such that(F2 F1)(x) =

⋃

y∈F1(x)

F2(y).

Definition 5.2.8 (lower inverse of a SV-map). Let F : X −→→ Y . For any V ⊂ Y the (lower) inverseimage of V under F (·) is denoted by F−(V ) and is defined as:

F−(V ) := x ∈ X | F (x) ∩ V 6= ∅ =⋃

y∈V

F−(y).

Definition 5.2.9 (upper inverse of a SV-map). † Let F : X −→→ Y . Then for any V ⊂ Y , the upper inverseof V under F (·), denoted by F+1(V ), is defined as:

F+(V ) := x ∈ X | F (x) ⊂ V .

In Defs. 5.2.8 and 5.2.9, if V = ∅, then we have

F+(V ) = F−(V ) = ∅.

Thus, in general, we have

Proposition 5.2.10. If F : X −→→ Y and V ⊂ Y any, then

F+(V ) ⊂ F−(V ).

The above two definitions of inverses of a SV-map lead into two types of continuities - upper and lowersemi-continuity.

Remark 5.2.11. For a function f : X → Y and V ⊂ Y , we have

f+(V ) = f−(V ) = f−1(V ).†The terminologies lower- and upper-inverse are from Berge [6]; while in the book of Aubin & Frankowska [4] the former

is simply termed inverse, and F+1(V ) is termed the core of the set V under F (·). However, the naming ’weak inverseimage’ and ’strong inverse image’, from Hu & Papageorgiou [14], could be more appropriate instead of ’upper-’ and’lower inverse’, respectively.

92

Excercises 5.2.12. Let F : X −→→ Y and V, W ⊂ Y . Then verify the validity of the following statements.

1. (i) F−(V ∪W ) = F−(V ) ∪ F−(W );

(ii) F−(V ∩W ) ⊃ F−(V ) ∩ F−(W );

(iii) F+(V ∪W ) ⊂ F+(V ) ∪ F+(W );

(iv) F+(V ∩W ) = F+(V ) ∩ F+(W ).

2. Let x0 ∈ X and S ⊂ Y . Then

(i) iff x1 ∈ F+(F (x0)), then F (x1) ⊂ F (x0);

(ii) F (F+(S)) ⊂ S;

3. Let F1, F2 : X −→→ Y and S ⊂ Y . Then

(i) (F1 ∪ F2)−(S) = F−1 (S) ∪ F−1

2 (S);

(ii) (F1 ∩ F2)−(S) ⊂ F−1 (S) ∩ F−

2 (S);

(iii) (F1 ∪ F2)+(S) ⊂ F+1 (S) ∪ F+

2 (S);

(iv) (F1 ∩ F2)+(S) ⊃ F+1 (S) ∩ F+

2 (S);

5.3 Semi-Continuity of Set-Valued Maps

Assume that X and Y are Hausdorff topological spaces.

Definition 5.3.1 (upper semi-continuous SV-map). Let F : X −→→ Y and Dom(F ) 6= ∅. Then F (·) issaid to be upper semi-continuous (u.s.c) at x0 ∈ X iff for any open set V ⊂ Y , where F (x0) ⊂ V , thereexists a neighborhood U ⊂ X of x0 such that

∀x ∈ U : F (x) ⊂ V, i.e. U ⊂ F+(V ).

The map F (·) is said to be u.s.c. on X if it is u.s.c. at every x ∈ X.

Definition 5.3.2 (lower semi-continuous SV-map). Let F : X −→→ Y . Then F (·) is said to be lowersemi-continuous (l.s.c.) at x0 ∈ X iff for any open set V ⊂ Y such that F (x0) ∩ V 6= ∅, there exists aneighborhood U ⊂ X of x0 such that

∀x ∈ U : F (x) ∩ V 6= ∅; i.e. U ⊂ F−(V ).

The map F (·) is said to be l.s.c. on X if F (·) is l.s.c. at every x ∈ X.

A set-valued map which is both lower and upper semi-continuous is called continuous.

Example 5.3.3. An upper semi-continuous map need not be lower semi-continuous and vice versa.

93

1. The set valued map F : R −→→ R given by

F (x) =

[1, 4], if x = 0[2, 3], if x 6= 0.

is upper semi-continuous at x = 0, but not lower semi-continuous x = 0. To see the upper semi-continuity at x = 0, let V be such that

F (0) ⊂ V ⇒ [1, 4] ⊂ V.

Take any neighborhood U of x = 0, then we have either F (x) = [1, 4] or F (x) = [2, 3] for x ∈ U .This implies

∀x ∈ U : F (x) ⊂ [1, 4] ⊂ U.

Hence, F (·) is u.s.c. at x = 0. However, if r ∈ F (0) = [1, 4] such that 3 < r < 4 and V =(r − ε, r + ε) ⊂ (3, 4), for a sufficiently small ε > 0, then

F (0) ∩ V = ∅.

That implies that F (·) is not lower semi-continuous at x = 0.

2. The set-valued map

F (x) =

[1, 4], if x < 0[2, 3], if x ≥ 0.

is lower semi-continuous but not upper semi-continuous. (Exercise!)

5.3.1 Properties of Semi-Continuous Set-Valued Maps

We will make use of the following lemma repeatedly.

Lemma 5.3.4. Let F : X −→→ Y and Dom(F ) = X. For any subset W ⊂ Y we have

(i) X \ F−(W ) = F+(Y \W ).

and(ii) X \ F+(W ) = F−(Y \W ).

A. Properties of Upper Semi-Continuous Set-Valued Maps

Proposition 5.3.5. Let F : X −→→ Y and Dom(F ) = X. Then the following statements are equivalent

(i) F (·) is u.s.c.;

(ii) for each open set V ⊂ Y , F+(V ) is an open set in X;

(iii) for each closed set W ⊂ Y , F−(W ) is a closed set in X.

Proof. From Def. 5.3.1, (i) ⇒ (ii) .

(ii) ⇒ (iii): For a closed subset W of Y we have Y \W is open in Y . Thus by (ii), F+(Y \W ) is openin X. Thus, by Lem. 5.3.4, we have X \ F−(W ) is an open set in X. Hence, F−(W ) is a closedset in X.

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(iii) ⇒ (i): Let x0 ∈ X be such that F (x0) ⊂ V for some open set V ⊂ Y . Hence, Y \ V is closed inY and F (x0) ∩ (Y \ V ) = ∅. Using (iii) and Lem. 5.3.4, we have X \ F+(V ) is closed in X andx0 /∈ (X \ F+(V )). Hence, x0 ∈ F+(V ) and F+(V ) is an open set implies

∃U(x0) : U(x0) ⊂ F+(V ); i.e., F (·) is u.s.c. at x0.

Proposition 5.3.6. Let X and Y be metric spaces and F : X −→→ Y and x0 ∈ X. Then the following areequivalent

(i) F (·) is u.s.c. at x0;

(i) if x0 ∈ X and xn is any sequence such that xn → x0 and V ⊂ Y an open subset such thatF (x0) ⊂ V , then

∃N ≥ 1 : F (xn) ⊂ V,∀n ≥ N.

Proof. (i) ⇒ (ii): Follows by definition of u.s.c. and properties of convergence of sequences.

(ii) ⇒ (i): Assume that F (·) is not u.s.c. at x0 and arrive at a contradiction.

Corollary 5.3.7. Let X and Y be metric spaces and F : X −→→ Y and x0 ∈ X. If F (·) is u.s.c. and xn isa sequence such that xn → x0, then

lim supn→∞

F (xn) ⊂ clF (x0).

Proof. Follows from Prop. 5.3.6 and Thm. 4.2.9.

Next we find some basic results on upper semi-continuity properties combination and composition ofsv-maps.

Proposition 5.3.8. Let F : X −→→ Y . If F (·) is u.s.c. and Y is a normal topological space, then clF (·) isu.s.c.

Proposition 5.3.9. Let X, Y, Z be topological spaces, F : X −→→ Y and G : Y −→→ Z. If F and G are uppersemi-continuous, then G F is u.s.c.

Proposition 5.3.10. Let F1 : X −→→ Y and F2 : X −→→ Y . Then

(i) if F1(·) and F2(·) are u.s.c., then F1 ∪ F2 is u.s.c.;

(i) Y is a normal topological space and if F1(·) and F2(·) are u.s.c. such that

F1(x) ∩ F2(x) 6= ∅, ∀x ∈ X,

then F1 ∩ F2 is u.s.c.

Proposition 5.3.11. Let F1 : X −→→ Y , F2 : X −→→ Y be compact valued and Y be a topological linearspace. If F1 and F2 are u.s.c., then F1 + F2 is also u.s.c.

Proposition 5.3.12. Let F : X −→→ Y and Y be a complete normed linear space. If F (·) is u.s.c. andcompact valued, then cl(coF (·)) is u.s.c.

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B. Properties of Lower Semi-Continuous of Set-Valued Maps

Proposition 5.3.13. Let F : X −→→ Y . If F (·) is an open map, then F (·) is l.s.c.

Proof. Exercise!

Proposition 5.3.14. Let F : X −→→ Y and Dom(F ) = X. Then the following statements are equivalent

(i) F (·) is l.s.c.;

(ii) for each open set V ⊂ Y , F−(V ) is an open set in X;

(iii) for each closed set W ⊂ Y , F+(W ) is a closed set in X.

Proof. (i) ⇒ (ii): Let V be an open set in Y and x ∈ F−(V ) ⇒ x /∈ X \ F−(V ) = F+(Y \ V ) (Lem.5.3.4(i)). Hence,

x /∈ F+(Y \ V ) ⇒ F (x) * Y \ V ⇒ F (x) ∩ V 6= ∅.Since, by (i), F (·) is l.s.c. at x0,

∃U(x0) : F (x) ∩ V 6= ∅, ∀x ∈ U(x0) ⇒ U(x0) ⊂ F−(V ).

Consequently, F−(V ) is an open set.

(ii) ⇒ (iii): Follows from Lem. 5.3.4(ii).

(iii) ⇒ (i): Let x ∈ X with F (x) ∩ V 6= ∅ and V is an open set in Y .⇒ Y \ V is closed in Y ⇒ F+(Y \ V ) is closed in X and F (x) ∩ V 6= ∅.⇒ x /∈ F+(Y \ V ) = X \ F−(V ) and X \ F−(V ) is a closed set (Lem. 5.3.4(i)).⇒ x ∈ F−(V ) and F−(V ) is an open set.⇒

∃U(x) : U(x) ⊂ F−(V ) ⇒ ∀x ∈ U(x) : F (x) ∩ V 6= ∅.Consequently, F (·) is l.s.c at x0.

Proposition 5.3.15. Let X and Y be metric spaces and F : X −→→ Y and x0 ∈ X. Then the following areequivalent

(i) if xn is any sequence such that xn → x0 and V ⊂ Y an open subset such that F (x0)∩ V 6= ∅, then

∃N ≥ 1 : F (xn) ∩ V 6= ∅,∀n ≥ N ;

(ii) if xn is a sequence such that xn → x0 and y0 ∈ F (x0) arbitrary, then there is a sequence ynwith yn ∈ F (xn) such that yn → y0;

(ii) F (·) is l.s.c. at x0.

Proof.

(i) ⇒ (ii): Let xn → x0 and y0 ∈ F (x0). Hence, given ε > 0, then Bε(y0) ∩ F (x0) 6= ∅. Hence, by (i)

∃N : F (xn) ∩Bε(y0) 6= ∅, ∀n ≥ N ⇒ ∀n ≥ N,∃yn ∈ F (xn) ∩Bε(y0).

Consequently, for n ∈ 1, . . . , N − 1, choosing yn ∈ F (xn), we will have a sequence yn such thatyn ∈ F (xn) and yn → y0.

96

(ii) ⇒ (iii): Assume that there is x0 ∈ X such that F (·) is not l.s.c. at x0. This implies, by definition,

∃V ⊂ Y open : F (x0) ∩ V 6= ∅, but for any neighborhood U(x0) of x0, ∃x ∈ U(x0) : F (x) ∩ V = ∅.

In particular,∀n ∈ N,∃xn ∈ B 1

n(x0) : F (xn) ∩ V = ∅.

This implies, if y0 ∈ F (x0) ∩ V , there is not sequence no sequence yn such that yn ∈ F (xn) andyn → y0. But this contradicts (ii). Consequently, F (·) should be l.s.c. at x0.

Sometimes the statement of Prop. 5.3.15(ii) is given as a definition for a lower semi-continuousset-valued map on metric spaces. As such the terminology ’open set valued map’ refers to lowersemi-continuous maps. (see for instance Shimizu et. al. [24] ).

Corollary 5.3.16. Let X and Y be metric spaces and F : X −→→ Y and x0 ∈ X. If F (·) is l.s.c. at x0, thenfor every sequence xn → x0 we have

F (x0) ⊂ lim infn→∞ F (xn).

Proof. Follows from Prop. 5.3.15.

Similarly, we have lower semi-continuity properties for combination and composition of sv-maps.

Proposition 5.3.17. Let F : X −→→ Y . If F (·) is l.s.c. , then clF (·) is l.s.c.

Proposition 5.3.18. Let F : X −→→ Y and Y be a topological linear space. If F (·) is l.s.c. and compactvalued, then both coF (·) and cl(coF (·)) are l.s.c.

Proposition 5.3.19. Let X,Y, Z be topological spaces, F : X −→→ Y and G : Y −→→ Z. If F and G are lowersemi-continuous, then G F is l.s.c.

Proposition 5.3.20. Let F1 : X −→→ Y and F2 : X −→→ Y . Then

(i) if F1(·) and F2(·) are l.s.c. closed sv-maps, then F1 ∪ F2 is l.s.c.;

(ii) if F1(·) is l.s.c., GraphF2 is open in X × Y and

(F1 ∩ F2) (x) = F1(x) ∩ F2(x) 6= ∅,∀x ∈ X,

then F1 ∩ F2 is l.s.c.

(iii) if Y is linear topological space, F1(·) is l.s.c., F2(·) has open convex values and

(F1 ∩ F2) (x) = F1(x) ∩ F2(x) 6= ∅,∀x ∈ X,

then F1 ∩ F2 is l.s.c.

Proposition 5.3.21. Let F1 : X −→→ Y , F2 : X −→→ Y and Y be a topological linear space. If F1 and F2 arel.s.c., then F1 + F2 is also l.s.c.

97

C. Outer , Inner, Upper and Lower Semi-Continuity

The terms ”inner” and ”outer” semi-continuous have been recently introduced by Rockafellar & Wets[22],but they also known as Kuratowski upper and lower semi-continuity, respectively (see Berger[7] ).

Definition 5.3.22 (outer, inner semi-continuity, Rockafellar & Wets). Let X and Y be metric spaces,F : X −→→ Y , x0 ∈ X and define

lim supx→x0

F (x) = y | ∃xn → x0, ∃yn → y, yn ∈ F (xn) (5.1)

lim infx→x0

F (x) = y | ∀xn → x0,∃yn → y, yn ∈ F (xn). (5.2)

Then , F : X −→→ Y is said to be

(i) outer semi-continuous at x0 if lim supx→x0

F (x) ⊂ F (x0); (5.3)

(ii) inner semi-continuous at x0 if lim infx→x0

F (x) ⊃ F (x0). (5.4)

Next, we would like to find out relations between inner and lower semi-continuity; as well as betweenouter and upper semi-continuity.

Proposition 5.3.23. Let X and Y be metric spaces and F : X −→→ Y . Then F (·) is inner semi-continuousif and only if F (·) is lower semi-continuous.

Proof. Follows from Prop. 5.3.15 and Def. 5.3.22.

Proposition 5.3.24. Let X and Y be metric spaces, F : X −→→ Y . Then F (·) is outer semi-continuous ifand only if F (·) is a closed map; i.e. Graph(F ) is a closed set in X × Y .

Proof. ”⇒”: Suppose F (·) is outer semi-continuous and (xn, yn) ∈ Graph(F ) such that (xn, yn) →(x0, y0). This implies, xn → x0, yn → y0 and yn ∈ F (xn). But, since F (·) is outer semi-continuous at x0, we have y0 ∈ F (x0). Consequently, (x0, y0) ∈ Graph(F ).

”⇐”: Suppose Graph(F ) is a closed set in X × Y . This implies, if there is a sequence (xn, yn) ∈Graph(F ) such that (xn, yn) → (x0, y0), then (x0, y0) ∈ Graph(F ); i.e. y0 ∈ F (x0).

Proposition 5.3.25. Let X and Y be metric spaces and F : X −→→ Y be closed valued. If F (·) is uppersemi-continuous at x0, then F (·) is outer semi-continuous at x0.

Proof. Assume that F (·) is not outer semi-continuous at x0. This implies

∀xn → x0, ∀yn → y0, yn ∈ F (xn), but y0 /∈ F (x0).

According to Cor. 5.3.7, for any xn → x0 and F (·) is u.s.c, we have

lim supn

F (xn) ⊂ clF (x0) = F (x0).

And, by Prop. 4.2.4, we havey0 ∈ lim sup

nF (xn) ⊂ F (x0).

Thus, y0 ∈ F (x0). But this is a contradiction. Therefore, F (·) is outer semi-continuous.

Corollary 5.3.26. (Closed Map, Aubin[2]) Let X and Y be metric spaces. If F : X −→→ Y is closed valuedand u.s.c., then F (·) is a closed map.

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Proof. Prop. 5.3.25 implies that F (·) is outer semi-continuous and, Prop. 5.3.24, yields that F (·) is aclosed map.

Remark 5.3.27. Cor. 5.3.26 implies that, an upper semi-continuous closed valued map has a closedgraph. However, the converse is not always true; i.e. the closedness of F (·) may not imply its uppersemi-continuity, even if F (·) is compact valued. In other words, there is an outer semi-continuous setvalued map which is not upper semi-continuous.

The example below indicates that the converse of Prop. 5.3.25 also may not be true.

Example 5.3.28 (Rem. 2.1, Kisielewicz [15]). Let F : R+−→→ R be given by

F (x) := 0, 1

x, if x > 0;0, if x = 0.

Observe that, F (·) is compact valued.

• F (·) is outer semi-continuous. To see this, let

xn =1n

and yn =1n

, then yn ∈ F (xn).

In additionxn → 0, yn → 0 and 0 ∈ F (0).

Hence, F (·) is outer semi-continuous at x0 = 0.

• F (·) is not upper semi-continuous. Let ε > 0, then Uε(F (0)) = (−ε, ε). For any δ > 0, let [0, δ) be aneighborhood of x0 = 0 in R+. Now take δ ∈ [0, δ) with

0 < δ < min

1ε, δ, 1

.

It follows that

F (δ) =

0,1δ

* (−ε, ε), since

1δ

> ε.

This implies, F (·) is not upper semi-continuous at x0 = 0.

The following statement guarantees the equivalence of outer semi-continuity and upper semi-continuity.

Proposition 5.3.29. Let X be a metric space, Y be a compact metric space and F : X −→→ Y . If F (·) is aclosed valued outer semi-continuous map, then F (·) is u.s.c.

Proof. Suppose F (·) a closed valued closed map. Assume there is x0 ∈ X such that F (·) is not u.s.c.x0. This implies, there is an open set V such that F (x0) ⊂ V and

∀n ∈ N : ∃xn ∈ B 1n(x0),∃yn ∈ F (xn) : yn /∈ V. (5.5)

Hence, xn → x0 and, by the compactness of Y , there is a subsequence ynk such that ynk → y0 ∈ Y .This implies,

(xnk , ynk) → (x0, y0) and ynk ∈ F (xnk).

Since, F (·) is a closed map, we have y0 ∈ F (x0) ⊂ V . Since, V is an open set and ynk → y0, there iskn0 > 0 such that

ynk ∈ V,∀nk ≥ nk0 .

But this is a contradiction to (5.5). Hence, F (·) should be u.s.c.

99

Observe that,in Prop. 5.3.29, F (·) is implicitly assumed to be compact valued.

Proposition 5.3.30 (Kisielewicz[15]). Let X and Y be metric spaces and F : X −→→ Y . If F (·) is u.s.c.and compact valued on X, then, for every compact set K ⊂ X, F (K) is a compact set in Y .

Proof. Let Vα | α ∈ Ω be an open covering of F (K). Let x ∈ K be any, then F (x) is compact andF (x) ⊂ ⋃

α∈Ω Vα. Hence, there is sub-covering Vαk| k = 1, . . . , n(x) such that

F (x) ⊂n(x)⋃

k=1

Vαk:= Vx and Vx is an open set.

By Prop. 5.3.5, F+(Vx) is an open set. Since, x ∈ K is arbitrary, it follows that

K ⊂⋃

x∈K

F+(Vx) and K is a compact set.

This implies there are x1, . . . , xm ∈ K such that

K ⊂m⋃

i=1

F+(Vxi).

Hence, using Prop. 5.2.6 and Exer. 5.2.12(2(ii)), we obtain that

F (K) ⊂ F

(m⋃

i=1

F+(Vxi)

)=

m⋃

i=1

F (F+(Vxi)) ⊂m⋃

i=1

Vxi

Since, each Vxi is a finite union of elements of Vα | α ∈ Ω, then Vα | α ∈ Ω has a finite sub-coverfor F (K). Consequently, F (K) is compact.

Proposition 5.3.31 (Kisielewicz[15]). Let X and Y be metric spaces and F : X −→→ Y be compactvalued. Then F (·) is u.s.c. at each x ∈ X if and only if, for every sequence xn, xn → x and everysequence yn, yn ∈ F (xn), there is a subsequence ynk

such that ynk→ y ∈ F (x).

Proof. ” ⇒ ” : Suppose F (·) is u.s.c. at x ∈ X, xn → x and yn ∈ F (xn).Set

K := x, x1, x2, . . . , xn, . . ..Then K is a compact set in X‡. Thus, by Prop. 5.3.30, F (K) is a compact set in Y and yn ⊂F (K). Consequently, there is a subsequence ynk of yn such that ynk → y for some y ∈ Y .This implies, by Prop. 4.2.4, we have

y ∈ lim supn

F (xn)

Moreover, by Cor. 5.3.7, we have

lim supn

F (xn) ⊂ clF (x) = F (x), by the compactness of F (x).

Consequently, y ∈ F (x).

‡Note that, if the limit point x of the sequence xn is not an element of K, the compactness of K may not be guaranteed(Why?).

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” ⇐ ” : Assume that F (·) is not u.s.c. at x0. This implies there is an open set V , F (x0) ⊂ V suchthat

∀n ∈ N : ∃xn ∈ B 1n(x0) : F (xn) * V ⇒ ∀n ∈ N : ∃xn ∈ B 1

n(x0), ∃yn ∈ F (xn) such that yn /∈ V.

Hence, xn → x0 and yn ∈ F (xn) and yn ∈ Y \V . Consequently, for any convergent subsequenceynk with limit y, y ∈ Y \ V ; i.e. y /∈ F (x). But this contradicts the assumption. Hence, F (·)should be u.s.c. at x0.

Once again, Prop. 5.3.31 indicates the equivalence of outer and upper semi-continuity for compactvalued set-valued maps. But, in general, upper semi-continuity is stronger than outer semi-continuity.

5.3.2 Local Uniform Boundedness

Definition 5.3.32 (local uniform boundedness). Let X and Y be metric spaces and F : X −→→ Y . ThenF (·) is called locally uniformly bounded at x0 ∈ X iff there is a neighborhood U(x0) of x0 such thatthe set ⋃

x∈U(x0)

F (x)

is a bounded set in Y . And F (·) is called locally uniformly bounded iff it is locally uniformly bounded atevery x ∈ X.

If Y is a finite dimensional or compact metric space, then F (·) is locally uniformly bounded impliesthat, for each x ∈ X, there is a neighborhood U(x) of x such that

cl

⋃

z∈U(x)

F (z)

is bounded - thus, compact in Y . Thus, in some literature we find such a term like locally uniformlycompactness being considered, but the local boundedness is more general .

The following statement includes a weaker form of the one given in Prop. 5.3.29.

Proposition 5.3.33. Let X and Y be metric spaces and F : X −→→ Y . If F (·) is u.s.c. and compactvalued, the F (·) locally uniformly bounded.

Proof. Let x0 ∈ X. Since F (x0) is compact. There is a abounded open set V such that F (x0) ⊂ V andF (·) is u.s.c. at x0 imply that

∃U(x0) :⋃

x∈U(x0)

F (x) ⊂ V.

Consequently, F (·) is locally bounded.

Proposition 5.3.34 (see also Hogan [13]). Let X be a metric and Y be a compact metric spaces andF : X −→→ Y be closed valued. If F (·) is closed and locally uniformly bounded, then Then F (·) is u.s.c.

Proof. Given F (·) is a closed and locally uniformly bounded map, assume there is x0 ∈ X such thatF (·) is not u.s.c. x0. This implies, there is an open set V such that F (x0) ⊂ V and

∀n ∈ N : ∃xn ∈ B 1n(x0), ∃yn ∈ F (xn) : yn /∈ V.

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Hence, xn → x0 and yn ∈ F (xn). By the local uniform boundedness,

∃U(x0) :⋃

x∈U(x0)

F (x) is bounded,

and there is N ∈ N such that

xn ∈ U(x0), ∀n ≥ N ⇒ yn ∈ F (xn) ⊂⋃

x∈U(x0)

F (x), ∀n ≥ N.

Consequently, yn | n ≥ N is bounded; hence, there is ynk such that ynk → y0 ∈ Y . . . (The rest ofthe proof is as in Prop. 5.3.29).

Local uniform boundedness property are useful in characterizing upper semi-continuity of set-valuedmaps with given structure.

5.3.3 Hausdorff Continuity

In this section we assume that X and Y are normed spaces and F : X −→→ Y .

Definition 5.3.35 (ε-upper semi-continuity). The map F (·) is said to be u.s.c. at x0 ∈ X in the ε senseif, for any ε > 0, there is δ > 0 such that

∀x ∈ Bδ(x0), F (x) ⊂ F (x0) + Bε

Definition 5.3.36 (ε-lower semi-continuity). The map F (·) is said to be l.s.c. at x0 ∈ X in the ε senseif, for any ε > 0, there is δ > 0 such that

∀x ∈ Bδ(x0), F (x0) ⊂ F (x) + Bε

Remark 5.3.37. Using the Hausdorff metric, in particular h∗ (see Rem. 4.1.6 and Lem. 4.1.13),

• F (·) is l.s.c. at x0 ∈ X in the ε sense is equivalent to

∀ε > 0, ∃δ > 0 : h∗(F (x), F (x0)) < ε,∀x ∈ Bδ(x0);

In this case, F (·) is said to be Hausdorff upper semi-continuous(H-u.s.c.) at x0.

• F (·) is u.s.c. at x0 ∈ X in the ε sense is equivalent to

∀ε > 0, ∃δ > 0 : h∗(F (x0), F (x)) < ε,∀x ∈ Bδ(x0).

Here, F (·) is said to be Hausdorff lower semi-continuous(H-l.s.c.) at x0.

Hence, in the following, instead to ε-u.s.c. and ε-l.s.c. we simply say H-u.s.c. and H-l.s.c., respec-tively.

Definition 5.3.38 (Hausdorff Continuity). Let x0 ∈ X. Then F (·) is said to be Hausdorff continuous(H-continuous) at x0 if F (·) is both Hausdorff upper and lower semi-continuous at x0.

Proposition 5.3.39. If F (·) is upper semi-continuous at x0, then F (·) is H-u.s.c. at x0.

Proof. Exercise!

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The converse of Prop. 5.3.39 is not always true.

Example 5.3.40 (see Aubin & Cellina[3]). Let F : R −→→ R2 given by

F (x) = (x, y) | y = x.

Then F (·) is H-u.s.c., but it is not u.s.c.Let x0 ∈ R and ε > 0 and

F (x) ⊂ F (x0) + Bε ⇒ F (x) ⊂ Bε

((x0, x0)

) ⇒ (x, x) ∈ Bε

((x0, x0)

) ⇒√

2(x− x0)2 < ε.

Thus, if we choose δ = ε2 , then

∀x : |x− x0| < δ, F (x) ⊂ F (x0) + Bε.

That is, F (·) is H-u.s.c. However, if V = (x, y) | |y| < 1x, then an open set in R2, but F+(V ) = 0

which is a closed set in R. Hence, according to Prop. 5.3.5, F (·) is not u.s.c.

Proposition 5.3.41. If F (·) is H-u.s.c. and closed valued, then F (·) is a closed map; hence, F (·) is outersemi-continuous.

Proof. Let (xn, yn) ∈ Graph(F ) such that (xn, yn) → (x0, y0). We want to show that y0 ∈ F (x0); i.e.(x0, y0) ∈ Graph(F ). Then we have xn → x0 and yn → y0. Since, F (·) s H-u.s.c. at x0,

∀ε > 0, ∃δ > 0 : h∗(F (x), F (x0)) < ε,∀x ∈ Bδ(x0).

Hence, there is N such that xn ∈ Bδ(x0), ∀n ≥ N . It follows that

∀n ≥ N : h∗(F (xn), F (x0)) < ε ⇒ limn→∞h∗(F (xn), F (x0)) = 0.

This implieslim

n→∞ dist(yn, F (x0)) = 0.

Since dist(·, F (x0)) is a continuous function and F (x0) is a closed set, we obtain

0 = limn→∞ dist(yn, F (x0)) = dist(y0, F (x0)) ⇒ y0 ∈ clF (x0) = F (x0).

Therefore, F (·) is a closed map. The fact that F (·) is outer semi-continuous follows from Prop. 5.3.24.

Observe that, the closed valuedness of an H-u.s.c. map is not enough to guarantee that is u.s.c. (seeexample 5.3.28). Thus, we have

Proposition 5.3.42. If F (·) is H-u.s.c. and compact valued, then F (·) is u.s.c.

Proof. We can either use here Prop. 5.3.31 or Prop. 5.3.34 for the proof. We use the latter. Letx0 ∈ X be any. Since F (x0) is compact, there is ε > 0 such that Uε(F (x0)) is bounded and F (x0) ⊂Uε(F (x0)) = F (x0) + Bε. By H-u.s.c., there is δ > 0 such that

⋃

x∈Bδ(x0)

F (x) ⊂ F (x0) + Bε.

Consequently, F (·) is locally uniformly bounded and F (·) is compact valued. Therefore, by Prop.5.3.34, F (·) is u.s.c.

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In contrast to H-u.s.c. we have the following result for H-l.s.c.

Proposition 5.3.43. If F (·) is H-l.s.c., then F (·) is l.s.c.

Proof. Exercise!

But the converse requires stronger assumptions.

Proposition 5.3.44. If F (·) l.s.c. and compact valued, then F (·) is H-l.s.c.

Proof. Let ε > 0 be given and x0 ∈ X be any. Since F (x0) is compact, we have

F (x0) ⊂⋃

y∈F (x0)

B ε2(y) ⇒ ∃y1, . . . , ym : F (x0) ⊂

m⋃

k=1

B ε2(yk).

Hence, F (x0) ∩B ε2(yk) 6= ∅. By the lower semi-continuity of F (·), for each k ∈ 1, . . . , m

∃Uk(x0) : F (x) ∩B ε2(yk) 6= ∅, ∀x ∈ Uk(x0).

Thenm⋂

k=1

Uk(x0) is an open set. Thus, ∃δ > 0 : Bδ(x0) ⊂m⋂

k=1

Uk(x0).

⇒

∀x ∈ Bδ(x0) : F (x) ∩B ε2(yk) 6= ∅,∀k ∈ 1, . . . , m. (5.6)

ButF (x) ∩B ε

2(yk) 6= ∅ ⇒ yk ∈ F (x) + B ε

2⇒ B ε

2(yk) ⊂ F (x) + Bε (Verify!)

Hence, from (5.6), we have

∀x ∈ Bδ(x0) : B ε2(yk) ⊂ F (x) + Bε, ∀k ∈ 1, . . . , m ⇒ ∀x ∈ Bδ(x0) :

m⋃

k=1

B ε2(yk) ⊂ F (x) + Bε.

Consequently,∀x ∈ Bδ(x0) : F (x0) ⊂ F (x) + Bε.

Therefore, F (·) is H-l.s.c.

A Hausdorff continuous set-valued map behaves like a continuous single valued map.

Proposition 5.3.45. Let X and Y be normed spaces, F : X −→→ Y . If F (·) is H-continuous, then for everyconvergent sequence xn, xn → x0, we have F (xn) h→ F (x0).

Proof. Exercise!

Excercises 5.3.46. Prove the following statements

1. If F : X −→→ Y is u.s.c., then the set x ∈ X | F (x) = ∅ is open in X.

2. If F : X −→→ Y is l.s.c., then the set x ∈ X | F (x) = ∅ is closed in X.

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3. For Fk : X −→→ Y, k ∈ N. If for each x ∈ X

⋂

k∈NFk(x) 6= ∅,

then, for any set A ⊂ Y , ( ⋂

k∈NFk

)−(A) =

⋂

k∈NF−

k (A).

4. Let X be a metric space and S ⊂ X, S 6= ∅. The metric projection on to S is PS : X −→→ S given by

PS(x) = y ∈ S | dist(x, S) = ρ(x, y).

Then PS(·) is compact valued and u.s.c. If S is also convex, then PS(·) is also convex.

5. Let F : X −→→ Y , where X = [0, 1] and Y = R given by

F (x) =

[0, 1], if 0 ≤ x ≤ 1;[0, 1), if x = 1.

Then F (·) is H-u.s.c., but not u.s.c.

6. Prove Prop. 5.3.42 using Prop. 5.3.31.

7. Let X and Y be normed spaces and F : X −→→ Y .

(i) If F (·) is H-u.s.c., then clF (·) is H-u.s.c.

(ii) If F (·) is H-l.s.c., then clF (·) is H-l.s.c.

8. Let X and Y be normed spaces, F : X −→→ Y and F (·) is compact valued. Then F (·) is continuousif and only if F (·) is H-continuous.

9. Prove or disprove the converse of Prop. 5.3.45.

10. Let X and Y be normed spaces, F1, F2 : X −→→ Y . Then prove that

(i) if F1, F2 are H-u.s.c., then F1 ∪ F2 is H-u.s.c.;

(ii) if F1, F2 are H-l.s.c., then F1 ∪ F2 is H-l.s.c.

11. If F (·) is H-l.s.c., then clF (·) and convF (·) are also H-l.s.c.

5.4 Set-Valued Maps with Given Structures

In this section we assume all space X,Y, T , etc., to be finite dimensional, like Rn,Rm, and so on.

Set-valued maps which are defined using a parametric family of functions play a vital role in paramet-ric optimization, in sensitivity and perturbation analysis of optimization problems. The main issue,behind set-valued maps with such given structures, is to characterize them through the topologicalproperties of their defining functions. As such, one obtains u.s.c property under weaker assumptions,while the l.s.c. requires stronger ones.

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Of interest are set-valued maps F : X −→→ T and M : X −→→ Y with structures:

F (x) := t ∈ T | gj(x, t) = 0, j ∈ J ; hi(x, t) ≤ 0, i ∈ I;

andM(x) := y ∈ Y | fk(x, y) = 0, k ∈ K;G(x, y, t) ≤ 0, t ∈ F (x) , x ∈ X,

under the following general assumptions

• I = 1, . . . , p, J := 1, . . . , q and K := 1, . . . , r are finite index sets;

• X ⊂ Rn, T ⊂ Rm and Y ⊂ Rl;

• hi : Rn × Rm → R, i ∈ I; fk : Rn × Rl → R, k ∈ K; and G : Rn × Rl × Rm → R are continuousfunctions.

Proposition 5.4.1 (Thm. 3.1.1 Bank et al. [5]). Let

F (x) = t ∈ T | gj(x, t) = 0, j ∈ J ;hi(x, t) ≤ 0, i ∈ I.

If the sets X and T are closed and the functions gj , j ∈ J ; hi, i ∈ I are continuous, then F (·) is a closedset valued map.

Proof. Note that for (xn, tn) ∈ Graph(F ), we have tn ∈ F (xn). This in turn implies,

gj(xn, tn) = 0, j ∈ J and hi(xn, tn) ≤ 0, i ∈ I.

Hence, if (xn, tn) → (x0, t0), it follows by the continuity of the g′js and h′is that

gj(x0, t0) = 0, j ∈ J and hi(x0, t0) ≤ 0, i ∈ I.

That is t0 ∈ F (x0). Therefore, F (·) is a closed map.

In fact, in Prop. 5.4.1, the upper semi-continuity of the functions gj , j ∈ J, could have been sufficient.

Corollary 5.4.2. Let X and T be closed sets and F (x) = t ∈ T | hi(x, t) ≤ 0, i ∈ I; gj(x, t) = 0, j ∈ Jand the functions hi, i ∈ I; gj , j ∈ J are continuous. If F (·) is locally uniformly bounded, then F (·) isu.s.c. and compact valued.

Proof. Follows from Prop. 5.4.1 and Prop. 5.3.34.

Remark 5.4.3. In Cor. 5.4.2 if T is assumed to be a compact set, then

F (x) = t ∈ T | gj(x, t) = 0, j ∈ J ; hi(x, t) ≤ 0, i ∈ I

will be locally uniformly bounded.

Thus, the upper semi-continuity of a SV-map with a given structure could be seen to hold true undersomehow weaker assumptions. However, to guarantee the lower semi-continuity we need regularityconditions, like Metric regularity and constraint qualifications, etc.

Definition 5.4.4 (Slater Constraint Qualification(SCQ)). Let, for each i ∈ I = 1, . . . , m, the functionhi : Rn × Rm → R be continuous, x0 ∈ Rn and hi(x0, ·) be convex w.r.t. t ∈ Rm. Then the SlaterConstraint Qualification is said to be satisfied at x0 if there is t∗ ∈ Rm such that

hi(x0, t∗) < 0, ∀i ∈ I.

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Proposition 5.4.5. Let, for each i ∈ I = 1, . . . , m, the function hi : Rn × Rm → R, hi(·, t) : Rn → Rbe continuous, x0 ∈ Rn, hi(x0, ·) be convex w.r.t. t ∈ Rm, and

F (x) = t ∈ Rm | hi(x, t) ≤ 0, i ∈ I.

If the SCQ holds at x0, then F (·) is l.s.c. at x0.

Proof. First not that F (·) is a closed and convex valued map. Now, let V ⊂ Rm be an open set andF (x0) ∩ V 6= ∅; i.e. ∃t ∈ F (x0) ∩ V . Hence, for some r > 0, t ∈ Br(t) ⊂ V . By the SCQ, there is t∗

such thathi(x0, t∗) < 0, ∀i ∈ I.

Since a convex function on Rm is continuous, we can find an open neighborhood V (t∗) such that

hi(x0, t) < 0, ∀i ∈ I, ∀t ∈ V (t∗).

⇒ ∀t ∈ V (t∗) : t ∈ F (x0); i.e. t∗ ∈ int F (x0).Furthermore, we can find λ ∈ (0, 1), sufficiently small, so that tλ := λt∗ + (1− λ)t ∈ Br(t) ⊂ V . But,since F (x0) is a closed convex set, we also have tλ ∈ F (x0). Consequently, tλ ∈ F (x0) ∩ V and

∀i ∈ I : hi(x0, tλ) ≤ λhi(x0, t∗) + (1− λ)hi(x0, t) < λ · 0 + (1− λ) · 0 = 0.

Next, by the continuity of the h′is, there is U(x0) such that

hi(x, tλ) < 0, ∀x ∈ U(x0) ⇒ tλ ∈ F (x),∀x ∈ U(x0),

and tλ ∈ V . Consequently,∀x ∈ U(x0) : tλ ∈ F (x) ∩ V 6= ∅.

Therefore, F (·) is l.s.c. at x0.

If (SCQ) does not hold, then the map F (·) might not be l.s.c.Next, we would like to characterize semi-continuity when convexity is not available.

Definition 5.4.6 (Mangasarian-Fromovitz Constraint Qualification (MFCQ)). Let F (x) = t ∈ Rm | hi(x, t) ≤0, i ∈ I; the functions hi, i ∈ I be continuous in Rn×Rm; for each x ∈ Rn, hi(x, ·) : Rm → R is continu-ously differentiable; and for x0 ∈ Rn, let t0 ∈ B(x0). The Mangasarian-Fromovitz constraint qualification(MFCQ) is said to hold at (x0, t0) iff there exists a vector ξ ∈ Rm such that

ξ>∇thi(x0, t0) < 0, ∀i ∈ I(x0, t0);

where I(x0, t0) = i ∈ I | hi(x0, t0) = 0. The vector ξ with the above property is known as an (MFCQ)vector.

Proposition 5.4.7. If t0 ∈ B(x0) and (MFCQ) holds at (x0, t0), then the map F (·) is lower semi-continuous at x0.

Proof. By the satisfaction of (MFCQ), for each i ∈ I(x0, t0), there is λi0 such that

hi(x0, t0 + λξ) = hi(x0, t0) + λξ>∇thi(x0, t0) + o(λ), ∀λ ∈ (0, λi0),

which is the first order Taylor expansion of hi(x0, ·) at t0. Thus, using t0 ∈ F (x0) and (MFCQ), wehave (w.l.o.g.) that

hi(x0, t0 + λξ) ≤ 0, ∀λ ∈ (0, λi0),

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for each i ∈ I(x0, t0). Since, the h′is are continuous, there is Ui(x0) such that

hi(x, t0 + λξ) ≤ 0, ∀x ∈ Ui(x0), ∀λ ∈ (0, λi0),

i ∈ I(x0, t0). Moreover,hi(x0, t0) < 0, i ∈ I \ I(x0, t0).

Hence, for each i ∈ I \ I(x0, t0),

∃Ui(x0), ∃Vi(t0) : hi(x, t) < 0, ∀x ∈ Ui(x0), ∀t ∈ Vi(t0)

Now, set

U(x0) =⋂

i∈I(x0,t0)

Ui(x0) ∩⋂

i∈I\I(x0,t0)

Ui(x0) and V (t0) =⋂

i∈I\I(x0,t0)

Vi(t0).

Then, for a sufficiently small λ0 (say λ0 ≤ minλi0 | i ∈ I(x0, t0)), we obtain t0 + λξ ∈ V (t0), ∀λ ∈

(0, λ0). It follows that, for each i ∈ I

hi(x, t0 + λξ) ≤ 0,∀x ∈ U(x0), ∀λ ∈ (0, λ0).

Now if xn → x0, then for λn ∈ (0, λ0) and λn → 0, we have tn = t0 + λnξ ∈ F (xn) and tn → t0.Therefore, by Prop. 5.3.15, F (·) is l.s.c.

In Prop. 5.4.7, if (MFCQ) is not satisfied at (x0, t0), then F (·) may fail to be l.s.c. at x0.Let us next characterize the lower semi-continuity of the map

M(x) = y ∈ Y | G(x, y, t) ≤ 0, t ∈ F (x) , x ∈ Rn.

For (x, y) ∈ Rn × Rm, we define the set of active constraints as

E(x, y) := t ∈ F (x) | G(x, y, t) = 0.

Definition 5.4.8 (Extended Mangasarian-Fromowitz Constraint Qualification). Let (x0, y0) ∈ Rn×Rq,the function G : Rn × Rq × Rm → R be continuous, and G(x, ·, t) differentiable w.r.t. y and ∇yG(·, ·, t)is continuous for each t ∈ F (x). Then the extended Mangasarian-Fromowitz constraint qualification(EMFCQ) is said to be satisfied at (x0, y0) if there exists a vector ξ ∈ Rq such that

ξ>∇yG(x0, y0, t) < 0, ∀t ∈ E(x0, y0).

Proposition 5.4.9. Let F (·) be u.s.c. and compact valued and y0 ∈ M(x0). Then if (EMFCQ) is satisfiedat (x0, y0), then M(·) is l.s.c. at x0.

Proof. Let y0 ∈ M(x0) be arbitrary, V (y0) be any neighborhood of y0 and EMFCQ be satisfied w.r.t. yat (x0, y0). Then there exists ξ ∈ Rq such that

∇yG(x0, y0, t)ξ < 0,∀t ∈ E(x0, y0).

For a fixed tl ∈ E(x0, y0), using Taylor’s theorem, there is λl0 := λ0(tl):

G(x0, y0 + λξ, tl) = G(x0, y0, tl) + λ∇yG(x0, y0, tl) + o(λ),∀λ ∈ (0, λl0).

⇒ (w.o.l.g.)G(x0, y0 + λξ, tl) ≤ 0,∀λ ∈ (0, λl

0).

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(i) By the continuity of G(·, ·, ·), there are neighborhoods U l1(x

0) and W1(tl) (with an appropriate λl0

)such thatG(x, y0 + λξ, t) ≤ 0, ∀λ ∈ (0, λl

0),∀x ∈ U l1(x

0), ∀t ∈ W1(tl).

(ii) Moreover, for each t ∈ F (x0) \ E(x0, y0)

G(x0, y0, t) < 0.

Thus, (as above) for each fixed t ∈ F (x0)\E(x0, y0), there are neighborhoods U t2(x

0),W2(t) andV t(y0) such that

G(x, y, t) < 0, ∀x ∈ U t2(x

0), ∀y ∈ V t(y0),∀t ∈ W2(t).

The family W1(tl),W2(t) | tl ∈ E(x0, y0), t ∈ B(x0) \ E(x0, y0) forms an open covering of F (x0).By assumption, F (x0) is a compact set. Hence, there is a finite sub-covering W (tl) | l = 1, . . . , p ofF (x0). Moreover, corresponding to this finite sub-covering we can find

• neighborhoods U l(x0) | l = 1, . . . , p of x0, so that U(x0) :=⋂p

l=1 U l(x0) is a neighborhood of x0.

• a sufficiently small λ0 (say λ0 := minλ0l | l = 1, . . . , p) is such a way that y0 + λξ ∈ V (y0),∀λ ∈

(0, λ0),

so that

G(x, y0 + λξ, t) ≤ 0, ∀λ ∈ (0, λ0), ∀x ∈ U(x0), ∀t ∈ B(x0) ⊂p⋃

l=1

W (tl). (5.7)

The map F (·) is u.s.c. This implies, there is a neighborhood U(x0) such that

F (x) ⊂p⋃

l=1

W (tl), ∀x ∈ U(x0).

Using this with (5.7) we obtain

G(x, y0 + λξ, t) ≤ 0, ∀λ ∈ (0, λ0), ∀x ∈ U(x0) ∩ U(x0), ∀t ∈ B(x).

⇒∀x ∈ U(x0) ∩ U(x0), ∀λ ∈ (0, λ0) : y0 + λξ ∈ M(x).

Hence,∀x ∈ U(x0) ∩ U(x0),M(x) ∩ V (y0) 6= ∅.

Therefore, M(·) is a lower semi-continuous map.

Excercises 5.4.10. Prove the following statements.

1. Let F : R −→→ R given by F (x) = [a(x), b(x)], where a, b : R→ R, a(x) ≤ b(x), ∀x ∈ R, a(·) an u.s.c.and b(·) a lower semi-continuous functions. Then F (·) is a lower semi-continuous sv-map.

109

2. Let f : Rn → R and F : R−→→Rn be given by

F (t) = x ∈ Rn | f(x) ≤ t.

If f is a convex function, then F (·) is a convex set-valued map.

3. Let gi : R× R2, i = 1, . . . , 4 given by

h1(x, t) = −t1 − 1;h2(x, t) = t1 − 1;h3(x, t) = −t2 − 1;h4(x, t) = t2 + xt1;

andF (x) = t ∈ R2 | hi(x, t) ≤ 0, i = 1, 2, 3, 4.

Then

(i) show that F (·) is compact valued, closed and locally uniformly bounded (Hence, by Prop.5.3.34F (·) is u.s.c.); but,

(ii) show that F (·) is not l.s.c. at x = 0. (Hint: argue graphically). That is, the (SCQ ) fails tohold at x = 0.

110

6 Measurability of Set-Valued Maps

Unless explicitly specified, we assume here the spaces X, Y and Z to be are metric spaces and Ω to bea subset of a metric space.

6.1 Definitions and Properties of Measurable Set-Valued Maps

First we begin by recalling the definition of a σ-algebra and a measurable space.

Definition 6.1.1 (σ-Algebra). Let Ω be a non-empty set. A collection F of subsets of Ω is said to be aσ-algebra if

(i) ∅,Ω ∈ F ;

(ii) A ∈ F ⇒ Ω \A ∈ F ;

(iii) If Akk∈N ⊂ F is any countable collection, then⋃

k∈NAk ∈ F .

Definition 6.1.2 (a measurable space). A measurable space (Ω,F) is a non-empty set Ω along with aσ−algebra F defined on Ω.

Let X be a complete metric space. Then smallest σ-algebra containing all open sets in X is called theBorel σ-algebra on X denoted by B(X). The measurable space (X,B(X)) is also called the Borelmeasurable space on X. Moreover, if A ∈ B(X), then A is called Borel measurable w.r.t. X.

Definition 6.1.3 (measurable set-valued map). Let (Ω,F) be a measurable space. A set valued mapF : Ω −→→ Y is said to be measurable (or F -measurable) on X if, for every open set O ⊂ Y , F−(O) ismeasurable; i.e. F−(O) ∈ F .

Hence, for F : Ω −→→ Y ,

• if F (·) is measurable, then the sets F−(∅) and Dom(F ) = F−(Y ) are measurable.

• if B ⊂ Y and F (x) = B, ∀x ∈ X (i.e. F (·) is a constant valued map), then F (·) is measurable. Thisfollows from the fact that, for any open set O ⊂ Y , we have

F−(O) =

Ω, if O ∩B 6= ∅;∅, if O ∩B = ∅.

In Def. 6.1.3 if X is a complete metric space, F : X −→→ Y and F = B(X), then F (·) is said to beBorel-measurable.

Proposition 6.1.4. Let X be a complete metric space, F : X −→→ Y and Dom(F ) = X. If F (·) is l.s.c.,then F (·) is Borel-measurable.

111

Proof. For O ⊂ Y is open, Prop. 5.3.14, implies that F−(O) is open. Hence, F−(O) ∈ B(X).

Proposition 6.1.5. A set valued map F : X −→→ Y is measurable if and only of the distance function

ϕ(x) := dist(y, F (x)),

ϕ : X → R+ is measurable, for each fixed y ∈ Rm.

Proposition 6.1.6. Let (Ω,F) be a measurable space, F : Ω −→→ Y be a closed valued map and Y be aseparable metric space. Then the following statements are equivalent:

(i) F−(C) is measurable for all closed sets C ⊂ Y ;

(ii) F−(O) is measurable for all open sets O ⊂ Y ;

(iii) F−(K) is measurable for all compact sets K ⊂ Y ;

Proof.

(i) ⇒ (ii): Let O ⊂ Y be open. Since Y is a metric space, then O is an Fσ set; i.e. there is a countablefamily of closed sets Cn | n ∈ N such that

O =⋃

n∈NCn

Thus, by a property of F− we have that

F−(O) = F−( ⋃

n∈NCn

)=

⋃

n∈NF−(Cn).

But, for each n ∈ N, F−(Cn) is measurable. Consequently, F−(O) is a countable union of measurablesets and F is a σ-algebra imply that F−(O) is measurable.

(ii) ⇒ (iii): Let K ⊂ Y be a compact set. For each n ∈ N define the set

On :=

y ∈ Y | dist(y,K) <1n

.

Then, it is easy to verify that On is open in Y , clOn is compact and clOn+1 ⊂ On.Claim:

F−(K) =⋂

n∈NF− (On) .

Let x ∈ F−(K). This implies, F (x) ∩ K 6= ∅; i.e. ∃y ∈ F (x) ∩ K. Hence, dist(y, K) = 0 so thaty ∈ On, ∀n ∈ N . From this follows that

y ∈ F (x) ∩On,∀n ∈ N⇒ x ∈ F−(On), ∀n ∈ N⇒ x ∈⋂

n∈NF− (On) .

Consequently, F−(K) ⊂ ⋂n∈N F− (On). Conversely, let z ∈ ⋂

n∈N F− (On). This implies

z ∈ F− (On) , ∀n ∈ N⇒ F (z) ∩On 6= ∅, ∀n ∈ N.

Hence,∀n ∈ N, ∃yn ∈ F (z) ∩On ⇒ lim

n→∞ dist(yn,K) = 0.

112

Since, ynn∈N ⊂ clO1 and clO1 is compact, there is a subsequence ynk such that ynk

→ y, for somey ∈ Y . Then, using the continuity of the distance function, we find that

dist(y, K) = limk→∞

= dist(ynk, K) = dist(yn,K) = 0.

Which implies y ∈ clK = K. Since, F (z) is closed and yn ⊂ F (z), we also have y ∈ F (z). From thisfollows that

y ∈ F (z) ∩K ⇒ z ∈ F−(K).

Hence, ⋂

n∈NF− (On) ⊂ F−(K).

Consequently,

F−(K) =⋂

n∈NF− (On) .

By assumption, for each n ∈ N, F− (On) is measurable. Hence, F−(K) is measurable.

(iii) ⇒ (i): Let C be a closed set in Y . Since, Y is separable, there is a countable dense set D =y1, y2, . . .. Given ε > 0, define the sets

Kn := y ∈ Y | ρ(yn, y) ≤ ε.

Then, for each n ∈ N, Kn is a compact set and by density of D, we have

C ⊂⋃

n∈NKn ⇒ C =

⋃

n∈N(C ∩Kn) .

But, for each n ∈ N, C ∩Kn is compact. Hence, by assumption F−(Cn ∩Kn) is measurable. Hence,

F−(C) =⋃

n∈NF−(C ∩Kn)

is measurable.

Corollary 6.1.7. Let F : Ω −→→ Y be a closed valued map and Y be a separable metric space. If F (·) ismeasurable, then, for any closed set C ⊂ Y , F+(Y ) is measurable.

Proof. The set C is closed, implies Y \C is open. Then, by Prop. 6.1.6, F−(Y \C) is measurable. But,by Lem. 5.3.4, we have

F−(Y \ C) = X \ F+(C).

Hence, X \ F+(C) is measurable. Therefore, F+(C) measurable.

Proposition 6.1.8. Let X be a complete and Y a separable metric spaces, F : X −→→ Y and Dom(F ) = X.If F (·) is u.s.c. and closed valued, then F (·) is Borel-measurable.

Proof. Follows from Prop. 5.3.5 and Prop. 6.1.6.

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6.1.1 Operations with Measurable Set-Valued Maps

Proposition 6.1.9 (measurability of closure). The map F : Ω −→→ Y is measurable if and only if clF (·) ismeasurable.

Proof. Use the fact that for any set B and an open set O we have

B ∩O 6= ∅ ⇔ clB ∩O 6= ∅.

Corollary 6.1.10. Let F : Ω −→→ Y and G : Ω −→→ Y . If F is measurable and

F (ω) ⊂ G(ω) ⊂ clF (ω), and for each ω ∈ Ω,

then G(·) is measurable.

Proof. Use the same idea as in Prop. 6.1.9.

Proposition 6.1.11. If Fk : Ω −→→ Y, k ∈ N, are measurable maps, then the union map

F (ω) :=⋃

k∈NFk(ω), for ω ∈ Ω

is measurable.

Proof. Follows from the fact that

( ⋃

k∈NFk

)−(O) =

⋃

k∈NF−

k (O)

for an open set O ⊂ Y .

Proposition 6.1.12 (measurability of the intersection). If, for each α ∈ Θ, Fα : Ω −→→ Y is measurablemaps, then the intersection map

F (ω) :=⋂

α∈Θ

Fα(ω), for ω ∈ Ω

is measurable.

Proposition 6.1.13. Let Y be a normed topological space.

(i) If F1 : Ω −→→ Y and F2 : Ω −→→ Y are measurable, then F1 + F2 is measurable.

(ii) If F1 : Ω −→→ Y is measurable and γ ∈ R, then (γF )(·) is measurable; where

(γF )(x) = γF (x), for x ∈ Ω.

Proposition 6.1.14. Let G : Ω −→→ Rn. If G(·) is a measurable, then the convex hull map co G(·) ismeasurable.

114

Proof. Define the set

Λ =

(λ1, λ2, . . . , λn+1) ∈ Qn+1

+

∣∣∣∣∣n+1∑

k=1

λk = 1

,

where Q+ represents the set of non-negative rational numbers. For λ := (λ1, λ2, . . . , λn+1) ∈ Λ let

Fλ(ω) :=n+1∑

k=1

λkF (ω).

Then, by Prop. 6.1.13, for each λ ∈ Λ, Fλ(·) is measurable. Furthermore, the countability of Λ andLem. 6.1.10 imply that

F (ω) =⋃

λ∈Λ

Fλ(ω)

is measurable. SinceF (ω) ⊂ co G(ω) ⊂ cl F (ω)

for each ω ∈ Ω, we conclude by Prop. 6.1.11 that co G(·) is measurable.

Proposition 6.1.15 (measurability of a product). If F1 : X −→→ Y and F2 : X −→→ Z are measurableset-valued maps, then (F1 × F2)(·) is also a measurable set-valued map.

Proof. The proof follows if we observe for an open set O ⊂ Y × Z that

(F1 × F2)−(O) = F−1 (πY (O)) ∩ F−

2 (πZ(O));

where πY : Y × Z → Y and πZ : Y × Z → Z are projections maps onto Y and Z, respectively.

Proposition 6.1.16 (measurability of composition). Let X,Y and Z be metric spaces. If F : X −→→ Y isclosed valued and measurable G : Y −→→ Z is u.s.c., then G F is also measurable.

Remark 6.1.17. In Prop. 6.1.16, if the map F2(·) fails to be u.s.c., then measurability of the compositionis not guaranteed.

6.2 Measurable Selections

Definition 6.2.1 (measurable selection). Let (Ω,F) be a measurable space, X a metric space andF : Ω −→→ X. A function f : Ω → X is a selector of F (·) if

f(ω) ∈ F (ω), for each ω ∈ Ω.

A selector f is said to be a measurable selector if f(·) is measurable.

Lemma 6.2.2. Let X be a separable metric space with D = x1, x2, . . . being a countable dense subsetof X. If A ⊂ X, A 6= ∅, then for each fixed n ∈ N

A ∩⋃

k∈NB 1

n(xk) 6= ∅.

Proposition 6.2.3 (Kuratowski-Ryll-Nardzewski selection Theorem, see Kisielewicz). If (Ω,F) be ameasurable space, X be a complete separable metric space, F : Ω −→→ X and Dom(F ) = X is a closedset-valued map, then F (·) has a measurable selector.

115

Proof. Let D = x1, x2, . . . be a countable dense subset of X. By Lem.6.2.2, we have

F (ω) ∩⋃

k∈NclB 1

n+1(xk) 6= ∅

for each fixed n ∈ N. This implies,

∃k ∈ N : F (ω) ∩B 1n+1

(xk) 6= ∅.

Letkn(ω) := mink ∈ N | F (ω) ∩ clB 1

n+1(xk) 6= ∅.

Define now, inductively, the set-valued maps

F0(·) = F (·) and Fn+1(ω) = Fn(ω) ∩ clBn+1(xkn(ω)).

Now we have for each ω ∈ Ω,

Fn(ω) ⊃ Fn+1(ω), diam(Fn(ω)) ≤ 1n→ 0

and Fn(ω) is a closed set. Consequently, using Prop. 2.4.10, we have⋂

n∈NFn(ω) contains a single element; say f(ω) =

⋂

n∈NFn(ω) ⊂ F (ω).

Hence, f(·) is a selector for F (·).Claim: f(·) is measurable.

(i) Given F0 = F is measurable. By induction, assume that Fn(·) be measurable and C ⊂ X is aclosed set. Then

ω ∈ Ω | Fn+1(ω) ∩ C 6= ∅ = ω ∈ Ω | Fn(ω) ∩ clBn+1(xkn(ω)) ∩ C 6= ∅=

⋃

k∈N(ω | Fn(ω) ∩ clBn+1(xk) ∩ C 6= ∅ ∩ ω ∈ Ω | kn(ω) = k) .

By induction assumption ω | Fn(ω) ∩ clBn+1(xk) ∩ C 6= ∅ ∈ F . Moreover

ω ∈ Ω | kn(ω) = k =k−1⋂

i=1

(ω ∈ Ω | Fn(ω) ∩ clBn+1(xi) = ∅ ∩ ω ∈ Ω | Fn(ω) ∩ clBn+1(xk) 6= ∅) .

⇒ ω ∈ Ω | kn(ω) = k ∈ F . Consequently,

ω ∈ Ω | Fn+1(ω) ∩ C 6= ∅ ∈ F .

Hence, for each n ∈ N, Fn(·) is measurable.

(ii) Let G(ω) = f(ω). By Prop. 6.1.12, the measurability of Fn(·), n ∈ N, and

G(ω) =⋂

n∈NFn(ω)

we conclude that G(·) is measurable. Now, for a closed set C ⊂ X

G−(C) = ω ∈ Ω | G(ω)∩C 6= ∅ = ω ∈ Ω | f(ω) ∩C 6= ∅ = ω ∈ Ω | f(ω) ∈ C = f−1(C).

This implies, f−1(C) ∈ F . Therefore, f(·) is measurable.

116

Lemma 6.2.4. Let X and Y be topological spaces and F : X −→→ Y . If U ⊂ Y and C ⊂ Y are any twosubsets, then

F−(C) = F−(C ∩ U) ∪ [F−(C) \ F−(U)

].

Proof. Trivial!

Corollary 6.2.5 (Castaing representation, Aliprantis Border, Rockafellar/Wets). If (Ω,F) be a measur-able space, X is a separable complete metric space, F : Ω −→→ X, Dom(F ) = X and F (·) is closed valued,then there is a sequence of measurable selectors fn(·) such that

F (ω) = clfn(ω) | n ∈ N,for each ω ∈ Ω.

Proof. Since X is a separable metric space, there is a countable dense subset D of X such that D =x1, x2, . . . , xn, . . .. Thus, for each n, k ∈ N and ω ∈ Ω, define

Fn,k(ω) =

F (ω) ∩B 1

2k(xn), if F (ω) ∩B 1

2k(xn) 6= ∅,

F (ω), otherwise.

Then, by Lem 6.2.4, for any open subset O ⊂ X, we have that

F−(O) = F−(O ∩B 1

2k(xn)

)∪

[F−(O) \ F−

(B 1

2k(xn)

)].

By the definition of the F ′ns we observe that

F−n,k(O) = F−

(O ∩B 1

2k(xn)

)∪

[F−(O) \ F−

(B 1

2k(xn)

)].

But, the measurability of F (·) implies that F−(O ∩ Un) ∪ [F−(O) \ F−(Un)] is measurable. Conse-quently, for each n, k ∈ N, Fn,k(·) is measurable and Dom(Fn,k) = Ω. Using Prop. 6.1.9, we find thatclFn,k(·) is measurable and the assumptions of Prop. 6.2.3 are also satisfied. Hence, for each n, k ∈ N,there is a measurable selector fn,k : Ω → X such that

fn,k(ω) ∈ clFn,k(ω) and clFn(ω) ⊂ F (ω), ω ∈ Ω. (6.1)

for each fixed ω ∈ Ω.Claim: For each ω ∈ Ω, F (ω) = clfn,k(ω) | n, k ∈ N. From (6.1), the following inclusion is obvious

clfn,k(ω) | n, k ∈ N ⊂ F (ω)

Now, let x ∈ F (ω), ε > 0 and Bε(x) be a neighborhood of x. Then there exists k ∈ N such that1

2k−1 < ε, and by the density of D, there is n ∈ N such that

x ∈ B 1

2k(xn) ⇒ x ∈ F (ω) ∩B 1

2k(xn) ⇒ ω ∈ F−

n,k

(B 1

2k(xn)

)

Thus,fn,k(ω) ∈ clFn,k(ω) = clF (ω) ∩ clB 1

2k(xn) ⇒ fn,k(ω) ∈ clB 1

2k(xn).

Consequently,

ρ(fn,k(ω), x) ≤ ρ(fn,k(ω), xn) + ρ(xn, x) ≤ 12k

+12k≤ 1

2k−1< ε.

117

⇒ fn,k(ω) ∈ Bε(x). Since ε > 0 is arbitrary, we see that

x ∈ clfn,k(ω) | n, k ∈ N.

Therefore,F (ω) ⊂ clfn,k(ω) | n, k ∈ N.

Then the claim of follows by re-indexing the countable set fn,k(ω) | n, k ∈ N.

6.3 Measurability of Set-Valued Maps with given Structure

Let (X,F) and (Y,G) be two measurable spaces. We say that a function f : X → Y is measurable (i.e.(F ,G) measurable) if

∀A ∈ G : f−1(A) ∈ F .

In particular, a real valued function f : X → R is measurable if

f−1(A) ∈ F

whenever A is a Borel set in R.

Definition 6.3.1 (Caratheodory function). Let f : Ω × X → R. Then f is said to be a Caratheodoryfunction if

(i) for each fixed x ∈ X, f(·, x) : Ω → R is measurable; and

(ii) for each fixed ω ∈ Ω, f(ω, ·) : Y → R is continuous.

Remark 6.3.2. Note that if f : Ω×X → R is a Caratheodory function, then −f is also a Caratheodoryfunction.

Definition 6.3.3 (Epigraphical and Domain maps). Let f : Ω×X → R. Then

• the epigraphical map associated with f(·, ·) is the set-valued map Ef : Ω−→→X × R given by

Ef (ω) = (x, λ) ∈ X × R | f(ω, x) ≤ λ

• the domain map associated with f(·, ·) is the set-valued map Df : Ω−→→X given by

Df (ω) = x ∈ X| f(ω, x) < ∞

Proposition 6.3.4. Let (Ω,F) be a measurable space and X be a separable metric space. If f : Ω×X → Ris a Caratheodory function, then the set-valued map F : Ω −→→ X given by

F (ω) = x ∈ X | f(ω, x) ≤ 0, ω ∈ Ω

is measurable.

Proof. Since F (·) is closed valued and X is separable, we use Prop. 6.1.6. Thus, let D a countabledense subset of X and C ⊂ X be a closed set. Then C ∩ D is a countable dense subset of C, sayD ∩ C = x1, x2, . . . and

F−(C) = ω ∈ O | F (ω) ∩ C 6= ∅= ω ∈ Ω | f(ω, x) ≤ 0, for some x ∈ C.

118

Since f(ω, ·) is continuous, for x ∈ C, f(ω, x) ≤ 0 implies there exists a neighborhood U(x) such thatf(ω, x) ≤ 0,∀x ∈ U(x). Hence, by the density of D ∩ C in C, there is xn ∈ U(x) ∩ (D ∩ C) such thatf(ω, xn) ≤ 0. Consequently, we can write

F−(C) = ω ∈ Ω | f(ω, xn) ≤ 0, for some n ∈ N (6.2)

=⋃

n∈Nω ∈ Ω | f(ω, xn) ≤ 0 (6.3)

But, for each n ∈ N, ω ∈ Ω | f(ω, xn) ≤ 0 = f−1(·, xn)[(−∞, 0] is measurable, since f(·, xn) ismeasurable. Consequently, F−(C) ∈ F .

Proposition 6.3.5 (measurability of the epigraphical map). Let (Ω,F) is a measurable space, X isa separable metric space. If f : Ω × X → R is a Caratheodory function, then the epigraphical mapEf (·) : Ω −→→ X × R of f is closed-valued and measurable.

Proof. Since

Ef (ω) = (x, λ) ∈ X × R | f(ω, x) ≤ λ = (x, λ) ∈ X × R | f(ω, x)− λ ≤ 0,defining g(ω, (x, λ)) := f(ω, x) − λ, we have that g : Ω × X × R → R is a Caratheodory function onΩ× (X × R); i.e. g(·, (x, λ)) is measurable for each fixed (x, λ) ∈ X × R and g(ω, ·) is continuous foreach fixed ω ∈ Ω. Therefore,

Ef (ω) = (x, λ) ∈ X × R | g(ω, (x, λ)) ≤ 0is measurable, according to Prop. 6.3.4.

Corollary 6.3.6 (measurability of the domain map). Let (Ω,F) is a measurable space, X is a separablemetric space. If f : Ω × X → R is a Caratheodory function, then the domain map Df (·) : Ω −→→ X ismeasurable.

Proof. Let O ⊂ X be an open set. Then

ω ∈ D−f (O) ⇒ Df (ω) ∩O 6= ∅.

⇒ ∃x ∈ Df (ω) ∩O, ∃λ ∈ R such that

f(ω, x) < λ < ∞⇒ ω ∈ E−f (O × (−∞, λ)).

Conversely, if ω ∈ E−f (O × (−∞, λ)), then for some x ∈ O we have

f(ω, x) < λ < ∞⇒ x ∈ Df (ω) ⇒ Df (ω) ∩O 6= ∅ ⇒ ω ∈ D−f (O).

Hence,D−

f (O) = E−f (O × (−∞, λ)).

Since E−f (·) is measurable, we conclude that D−

f (·) is also measurable.

Proposition 6.3.7. If f : Ω × X → R is a Caratheodory function and x : Ω → X is a measurablefunction, then the function g : Ω → R given by

g(ω) = f(ω, x(ω)), ω ∈ Ω,

is measurable.

119

Proof. Let α ∈ R be any. We consider the set

ω ∈ Ω | g(ω) < α = ω ∈ Ω | f(ω, x(ω)) < α.

Which we can also write as(see also Rockfellar/Wets)

ω ∈ Ω | f(ω, x(ω)) < α = ω ∈ Ω | ∃λ ∈ R : f(ω, x) ≤ λ, x(ω) = x, λ < α (6.4)

= ω ∈ Ω | ∃(x, λ) ∈ Ef (ω), (x, λ) ∈ x(ω) × (−∞, α) (6.5)

= ω ∈ Ω | Ef (ω) ∩R(ω) 6= ∅ , (6.6)

where R(ω) := x(ω) × (−∞, α). Consequently, we obtain that

ω ∈ Ω | f(ω, x(ω)) < α = Dom(Ef ∩R).

Since, x(·) : Ω → X is measurable and (−∞, α) is a constant, the set value map G : Ω−→→X × Ris measurable, by Prop.6.1.15 also Ef (·) is measurable (Prop. 6.3.5). Thus, using Prop. 6.1.12 weconclude that the intersection map (Ef ∩ R)(·) is measurable; hence, Dom(Ef ∩ R) is a measurableset in Ω. Therefore, g(·) is a measurable function.

Proposition 6.3.8. Let Ω be a complete metric space with F being the Borel σ-algebra and X a separablemetric space. If f : Ω ×X → R is a lower semi-continuous function with respect to both Ω and X, thenthe set-valued map F : Ω −→→ X given by

F (ω) = x ∈ X | f(ω, x) ≤ 0, ω ∈ Ω

is Borel-measurable.

Proof. Uses the same ideas as in Prop. 6.3.4.

Proposition 6.3.9. Let (Ω,F) be a measurable space and X be a separable metric space. If g : Ω×X → Ris a Caratheodory function, then the set-valued map F : Ω −→→ X given by

F (ω) = x ∈ X | g(ω, x) = 0, ω ∈ Ω

is measurable.

Proof. We can write

F (ω) = x ∈ X | g(ω, x) ≤ 0 ∩ x ∈ X | − g(ω, x) ≤ 0, ω ∈ Ω

SetF1(ω) = x ∈ X | g(ω, x) ≤ 0 and F2 = x ∈ X | − g(ω, x) ≤ 0

Then, Prop. 6.3.4 yields that both F1(·) and F2(·) are measurable. Therefore, by Prop. ??, F (·) ismeasurable.

Proposition 6.3.10 (measurability of feasible set maps, see also Rockfellar/Wets). Let (Ω,F) be ameasurable space and X be a separable metric space. If gi : Ω×X → R, i ∈ I, and hj : Ω×X → R, j ∈ Jare Caratheodory functions, then the set-valued map F : Ω −→→ X given by

F (ω) = x ∈ X | hj(ω, x) ≤ 0, j ∈ J ; gi(ω, x) = 0, i ∈ I, ω ∈ Ω

is measurable.

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Proof. Define

F1(ω) = x ∈ X | hj(ω, x) ≤ 0, j ∈ J (6.7)

F2(ω) = x ∈ X | gi(ω, x) = 0, i ∈ I. (6.8)

and apply Props. 6.3.4, 6.3.9 and 6.1.12.

Proposition 6.3.11 (measurability of Marginal Functions). Let (Ω,F) be a measurable space and X bea complete separable metric space. If f : Ω ×X → R a Caratheodory function and F : Ω−→→ X is closedvalued and measurable, then the marginal value function

ϕ(ω) := supx∈F (ω)

f(ω, x)

is measurable as a function ϕ : Ω → R.

Proof. Since ϕ(·) is a real valued function, for γ ∈ R, we consider the set

ω ∈ Ω | ϕ(ω) ≤ γ = ω ∈ Ω | supx∈F (ω)

f(ω, x) ≤ γ.

Using Cor. 6.2.5, there is a sequence of measurable functions xn(·), xn : Ω → X such that F (ω) =clxn(ω) | n ∈ N, ω ∈ Ω. Hence,

ω ∈ Ω | ϕ(ω) ≤ γ = ω ∈ Ω | supx∈clxn(ω) | n∈N

f(ω, x) ≤ γ

Since f(ω, ·) is continuous, the above can be written as (why?)

ω ∈ Ω | ϕ(ω) ≤ γ = ω ∈ Ω |f(ω, xn(ω)) ≤ γ, ∀n ∈ N =⋂

n∈Nω ∈ Ω |f(ω, xn(ω)) ≤ γ.

Thus, Prop. 6.3.7 implies that, for each n ∈ N, then function gn : Ω → R given by gn(ω) = f(ω, xn(ω))is measurable. Consequently, ω ∈ Ω | ϕ(ω) ≤ γ is a countable intersection of measurable sets.Therefore, ϕ(·) is measurable.

Excercises 6.3.12. Suppose (Ω,F) be a measruable space and X a metric space. Prove that if, for eachk ∈ N, Fk : Ω −→→ X is closed valued and measurable and, for each x ∈ X,

⋂k∈N Fk(x) 6= ∅ , then the set

valued map ( ⋂

k∈NFk

): Ω −→→ X

is measurable. (From this, it follows that the feasible set-valued map is measurable.)

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7 Comments on Literature

There are several books and literature dealing with set-valued maps and their applications. But anyone who wants to know about set-valued maps can begin with the list given below. However, thislist is by no means exhaustive and is biased by my personal preferences and repeated citations in theliterature.

• General set valued maps: Aubin & Cellina [3], Aubin & Frakowsk[4], Berge[6], HU & Papageoriou[14],Kiesielewichz[15], Rockafellar & Wets[22], Aliprantis & Border[1], Gofert et al. [10] etc.

• Set valued maps defined using parametric systems of functions: Bank et al.[5], Shimizu et al.[24],etc.

• Measurable set-valued maps: Rockafellar & Wets[22], Castaing & Valadier[8], etc.

• Differentiability properties of set-valued maps: Aubin[2], Gopfert[10], etc.

• Fixed Point Properties: Granas & Dugundji[12], Goebe & Kirk[11], etc.

Some of the books cited above still discuss other issues related with set valued maps. For instance,differential inclusions are discussed by Aubin & Cellina[3], Demiling[9], etc. Set valued maps asapplied to optimization problems Shimizu et al.[24], Berger[7], etc. In an case, it worth payingattention to the literature cited in each of the books suggested above.

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Bibliography

[1] C. D. Aliprantis, K. C. Border, Infinite Dimensional Analysis - A Hitchhiker’s Guide (2nd. ed),Springer-Verlag, 1999.

[2] J.-P. Aubin, Mutational and Morphological Analysis, Birkhauser, 1999.

[3] J.-P. Aubin, A. Cellina, Differential inclusions, Springer Verlag, Berlin 1984.

[4] J.-P. Aubin, H. Frankowska, Set valued analysis, Birkhauser, Basel, 1990.

[5] B. Bank, J. Guddat, D. Klatte, B. Kummer, K. Tammer: Non-linear parametric optimization,Akademie-Verlag, Berlin, 1982.

[6] C. Berge, Topological Spaces. Oliver & Boyd, Edinburg, London, 1963.

[7] A. Berger, Beitrage zur Realisierung von Losungsverfahren der nichtlinearen Optimierung. Ph.D.Dissertation, Technical University of Ilemanu, Fakulty of Mathematics and Natural Scinces, Jan-uary, 1978.

[8] C. Castaing, M. Valadier, Convex analysis and measurable multifunctions. Lecture Notes in Math-ematics, Vol. 580, Springer-Verlag, 1977.

[9] K. Deimling, Multivalued Differential Equations, Walter de Gruyter & Co., 1992.

[10] A. Gofert, H. Riahi, C. Tammer, C. Zalinescu, Variational methods in partially ordered spaces.Springer, 2003.

[11] K. Goebel , W. A. Kirk, Topics in metric fixed point theory. Cambridge studies in advanced math-ematics, V. 28, Cambridge University Press, 1990.

[12] A. Granas, J. Dugundji, Fixed point theory. Springer Monongrphs in Mathematics, Springer-Verlag, 2003.

[13] W. W. Hogan, Point-to-set maps in mathematical programming. SIAM Review, Vol. 15, No.3, pp.591 - 603, 1973.

[14] S. Hu, N. S. Papageorgiou, Handbook of multivalued analysis: Volume I. Kluwer Academic Pub-lishers, 1997.

[15] M. Kisielewicz, Differential Inclusions and Optimal Control. Polish Scientific Publishers & KluwerAcademic Publishers, 1991.

[16] D. Klatte, R. Henrion, Regularity and stability in non-linear semi-infinite optimization. Semi-infinite Programming, R. Reemtsen and J.-J. Ruckmann (eds.), pp. 69-102, Kluwer AcademicPres, 1998.

[17] P. Kosmol, Optimierung und Approximation. Walter de Guyter & Co., 1991.

[18] G. Meinardus, Approximation of Functions: Theory and Numerical Methods. Springer-Verlag,1967.

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[19] Robinson, S. M., Some continuity properties of polyhedral multifunctions. Math. Prog. Study, V.14, pp. 206-214, 1991.

[20] Robinson, S. M., Regularity and stability for convex multivalued functions. Math. of OR, V. 1,pp. 130-143, 1976.

[21] H. Royden, Real Analysis (3rd. edition). Macmillan Publishing Company, 1988.

[22] R. T. Rockafellar, R. J.-B. Wets, Varaitional Analysis, Springer Verlag, 1998.

[23] S. Shirali, H. L. Vasudeva, Metric Spaces. Springer-Verlag, 2006.

[24] K. Shimizu, Y. Ishizuka and J. Bard, Nondifferentiable and Two-Level Mathematical Program-ming. Kluwer Academic Publishers, 1997.

[25] J.-B. H. Urruty, C. Lemarechal, Convex analysis and minimization algorithms I. Springer Ver-lag,1993.

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