Download Introduction to Topological Spaces and Set-Valued Maps

Introduction to Topological Spaces and Set-Valued Maps
(Lecture Notes)
Abebe Geletu (Dr.)
Institute of Mathematics
Department of Operations Research & Stochastics
Ilmenau University of Technology
August 25, 2006
Contents
1 Preface
1
2 Introduction to Metric Spaces
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Open and Closed Sets . . . . . . . . . . . . . . . . . . . .
2.3 Subspaces of Metric Spaces . . . . . . . . . . . . . . . . .
2.4 Sequences, Convergence and Complete Metric Spaces . . .
2.4.1 Complete Metric Spaces . . . . . . . . . . . . . . .
2.5 Baire Category . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Compact Metric Spaces . . . . . . . . . . . . . . . . . . . .
2.6.1 Bounded Sets and Totally Bounded Metric Spaces .
2.7 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.1 Continuity of Functions . . . . . . . . . . . . . . .
2.7.2 Real Valued Functions . . . . . . . . . . . . . . . .
2.7.3 Uniform Continuity . . . . . . . . . . . . . . . . . .
2.7.4 Convergence Properties of Sequences of Functions
2.7.5 Equicontinuiuty and the Ascoli-Arzelá Theorem . .
2.7.6 Homeomorphisms and Isometries in Metric Spaces
2.7.7 Contractive Maps and Fixed Point Properties . . . .
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3
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3 Topological Spaces
3.1 Neighborhood and Neighborhood Systems . . . . . . . . . . . .
3.2 Bases and Subbases . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Sequences, Continuity and Homeomorphism . . . . . . . . . .
3.4 Classification of Topological Space: Separation Axioms . . . . .
3.5 Uryson’s Lemma, Tietze’s Extension Theorem and Metrizability
3.5.1 Tietze’s Extension Theorem . . . . . . . . . . . . . . . .
3.5.2 Urysohn’s Metrizability . . . . . . . . . . . . . . . . . . .
3.6 Compact Topological Spaces . . . . . . . . . . . . . . . . . . . .
3.6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.2 The Finite Intersection Property . . . . . . . . . . . . . .
3.6.3 Compact Hausdorff Spaces . . . . . . . . . . . . . . . .
3.7 Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . . .
3.8 Sigma-Compact Topological Spaces . . . . . . . . . . . . . . . .
3.9 Paracompact Topological Spaces . . . . . . . . . . . . . . . . . .
3.10 Partition of Unity . . . . . . . . . . . . . . . . . . . . . . . . . .
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29
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4 The Hausdorff Metric and Convergence of Sequences of
4.1 The Hausdorff Metrics . . . . . . . . . . . . . .
4.2 Convergence of Sequences of Sets . . . . . . . .
4.2.1 Calculus of Limits of Sequences of Sets .
4.2.2 Convergence w.r.t. the Hausdorff Metric
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71
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Sets
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iii
5 Set-Valued Maps
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Some Examples of Set-Valued Maps . . . . . . . . . . . . . .
5.2 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Elementary Mathematical Operations with Set-Valued Maps
5.3 Semi-Continuity of Set-Valued Maps . . . . . . . . . . . . . . . . .
5.3.1 Properties of Semi-Continuous Set-Valued Maps . . . . . . .
5.3.2 Local Uniform Boundedness . . . . . . . . . . . . . . . . . .
5.3.3 Hausdorff Continuity . . . . . . . . . . . . . . . . . . . . . .
5.4 Set-Valued Maps with Given Structures . . . . . . . . . . . . . . . .
6 Measurability of Set-Valued Maps
6.1 Definitions and Properties of Measurable Set-Valued Maps
6.1.1 Operations with Measurable Set-Valued Maps . . .
6.2 Measurable Selections . . . . . . . . . . . . . . . . . . . .
6.3 Measurability of Set-Valued Maps with given Structure . .
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89
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. 101
. 102
. 105
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7 Comments on Literature
123
Bibliography
125
iv
1 Preface
These notes are a result of a two semester course that I held at the technical university of Ilmenau
during winter semester 2005 and summer semester 2006. These are simply lecture notes organized
to serve as introductory course for advanced postgraduate and pre-doctoral students. The main objective is to give an introduction to topological spaces and set-valued maps for those who are aspiring to
work for their Ph. D. in mathematics. It is assumed that measure theory and metric spaces are already
known to the reader. Hence, only a review has been made of metric spaces. At the same time the topics on topological spaces are taken up as long as they are necessary for the discussions on set-valued
maps. 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 sources
of theorems and statements. I have tried to supply my own versions of simplified proofs, whenever I
felt necessary. These is not by far an all-inclusive introductory note. In fact, I leave it at the mercy of
the criticisms, suggestions and comments of the reader. However, it is my belief that the material can
serve as spring board to dive into the ocean of set-valued maps.
Abebe Geletu, August 2006.
1
2
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 a
metric 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
1
P
(a) < Rn , ρ > with ρ(x, y) = ( ni=1 (xi − yi )p ) p , 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 continuous functions on [a, b].
(d) if X is any normed space with a norm k · k, then < X, ρ > will be a metric space if we define
ρ(x, y) := kx − yk. In this case, ρ is called an induced metric - induced by this particular norm on
X.
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 pseudometric
and the pair < X, ρ > is called a pseudometric space. For instance the space Lp of functions is a
pseudometric space; with metric induced by the Lp − norm. If all but M3, then < X, ρ > is called a
quasi-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 >. In
general, the distance between x and y can be different for a different metric. Note also that, in a
metric 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 := sup{ρ(x, y) | x, y ∈ S}.
A set S is said to be bounded if diamS < ∞.
1
See 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)
:= inf{ρ(x, z) | x ∈ A, y ∈ B}
=
inf
z∈A,x∈B
ρ(x, z).
If A = {x}, then we write
dist(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 product
metric space is the Cartesian product
n
Y
Xi := X1 × . . . × Xn
i=1
along with the metric given by
ρ(x, y) =
where x = (x1 , . . . , xn ) ∈
Qn
i
" n
X
#1
2
ρi (xi , yi )2
i=1
Xi and y = (y1 , . . . , yn ) ∈
Qn
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 set
Br (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 there
is 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 iff
there 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.
4
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 iff
X \ 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 A0 , and A0 is sometimes called the derived
set of A.
(ii) The closure of a set A, denoted by clA, is defined as
clA := A ∪ A0
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.
5
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. a
metric ρ) implicity contains the possibility of the approximatability of the elements of X by the elements
of D. In fact, if x0 is any element of X, we can find an element d0 of D which is arbitrarily close to x
w.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 polynomials
is dense in C[a, b] w.r.t. the metric ρ∞ (see. Example 2.1.2). Implying that, every continuous function
on [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 countable
dense 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 countable
dense 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 family
{Ok } of open sets such that
[
O=
Ok .
Ok ⊂O
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. The
restriction 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, 21 ] is closded w.r.t. to
< S, ρS >, but not in X.
6
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 that
x ∈ Br(x) (x) ∩ S ⊂ A. Hence
[¡
[¡
¢
¢
A=
Br(x) (x) ∩ S =
Br(x) (x) ∩ S
x∈A
Set O :=
x∈A
¡
¢
x∈A Br(x) (x) . Then O is an open set in X and
S
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 {xn }n∈N (note that f (N) =
{xn }n∈N ). In fact, we can drop ’n ∈ N’ (when not required) and simply write {xn } for the sequence
{xn }n∈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 write
xn → 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.
7
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 {xnk }k∈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 subsequence
{xnk }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 Cauchy
sequence 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 which
Cauchy sequences may not converge.
Definition 2.4.6 (complete metric space). A metric space < X, ρ > is said to be a complete metric
space 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} is
complete 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 sequence
of subsets of X. If, for each n, Fn is a non-empty closed set,
Fn ⊃ Fn+1 and
then
T∞
n=1 Fn
lim diam(Fn ) = 0,
n→∞
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 )}, with
the property that Brn (xn ) ⊃ Brn+1 (xn+1 ) and rn → 0, has a non-empty intersection, then < X, ρ > is a
complete metric space.
8
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 convergent sequence of real numbers.
3. Given a sequence {xn }. Then {xn } is a Cauchy sequence iff the sequence {dimAn } is a decreasing
sequence of real number and dimAn ↓ 0; where
An := {xk : k ≥ n}.
4. The product
complete.
Qn
i=1 Xi
of metric spaces is complete iff each of the metric spaces Xi , i = 1, . . . , n, is
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 nowhere
dense 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. clE
contains 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 {On }n∈N a countable collection of dense
open 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 a
countable 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.
2
sometimes 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 direct
proof).
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 second
category; i.e. non-empty open subsets of X cannot be given by a union of a countable number of nowhere
dense sets.
Proof. Let ∅ 6= U ⊂ X be an open set.
S Assume that there is a countable collection {En }n∈N of nowhere
dense subsets of X such that U = n∈N En . Then, for each n ∈ N, On := X \ clEn is a dense open
subset of X. By Thm. 2.5.3, it follows that
\
On
n∈N
T
T
O
6
=
∅.
This
implies
x
∈
/
X
\
dense
in
X.
Since
U
is
an
open
set,
there
is
x
∈
U
∩
n
n∈N On =
n∈N
S
S
E
.
But
this
contradicts
that
U
=
E
.
Therefore,
U
is
not
of
first
category.
n
n
n∈N
n∈N
Remark 2.5.9. Let X be a complete metric space and {En } be a countable collection of nowhere dense
sets.
Then, for any non-empty open set U in X, there is an element x0 ∈ U which does not belong to
S
n∈N En .
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. Let
A=
[
En , where En is a nowhere dense, for each n ∈ N.
n∈N
Then
X \A=
\
(X \ En ) ⊃
n∈N
\
(X \ clEn ) .
n∈N
But the sets X \ clEn are dense open subsets. Then Thm. 2.5.3 implies that
set. Therefore, X \ A is a dense set.
T
n∈N (X
\ clEn ) is a dense
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 nowhere
dense.
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;
10
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∈N En 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, then
F 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 K
has a convergent subsequence in K; i.e. every sequence in K has an accumulation point in K. A metric
space < 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 closed
set.
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⊂
Aα
α∈Ω
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 Ω0 ⊂ Ω such that
[
S⊂
Aα ,
α∈Ω0
then {Aα | α ∈ Ω0 } is a subcovering of {Aα | α ∈ Ω}. If, in this case, the index set Ω0 is finite, then we
have a finite subcovering.
Lemma 2.6.3 (Lebesgue covering Lemma). If K is a compact set and {Oα | α ∈ Ω} is an open covering
of K, then there is r > 0§ such that, for each x ∈ K the open ball Br (x) is contained in an element of
{Oα | 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) is
not a subset of an element of {Oα | α ∈ Ω}. Consequently, for xn ∈ K
B 1 (xn ) \ Oα 6= ∅, ∀α ∈ Ω
n
3
§
(*)
The 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, say
{xnk } ⊂ {xn } and xnk → x ∈ K. But since
[
K⊂
Oα ,
α∈Ω
S
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 that
x ∈ Br (x) ⊂ Oα0 .
Thus, by the convergence of xnk to x, there is a sufficiently large nk such that
1
r
<
and xnk ∈ B r2 (x).
nk
2
Hence, for any z ∈ B
1
nk
(xnk ), we have
ρ(z, x) ≤ ρ(z, xnk ) + ρ(xnk , x) ≤
1
r
r r
+ ≤ + =r
nk
2
2 2
From this follows that
B
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 a
finite 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, the
system {Br (xp ) | k = 1, . . . , p} does not cover K. This implies, given x1 ∈ K, then Br (x1 ) does
not 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
[
Br (xk )
k=1
Hence, we have constructed a sequence {xn } ⊂ K with the property that ρ(xn , xm ) ≥ r whenever
m 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 the
claim 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 in
terms 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 > 0
be the Lebesgue number of K (as given by Lem. 2.6.3). Hence, by Thm. 2.6.4, there is a finite
number of elements x1 , . . . , xn ∈ K such that the system
{Br (xk ) | k = 1, . . . , n}
is a covering of K. Again, by Lem. 2.6.3, each of the balls Br (xk ) is contained in some element
of {Oα | α ∈ Ω}; say Br (xk ) ⊂ Oαk for some αk ∈ Ω, k = 1, . . . , n. Hence
K⊂
n
[
Br (xk ) ⊂
k=1
n
[
Oαk
k=1
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
number of open sets
S of K.( Otherwise, there will be a finite
S
O1 , . . . , On such that K ⊂ nk=1 Ok . From this follows that {xn } ⊂ nk=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.
O∈O
Now, for each k ∈ N, the open ball B 1 (x) does not belong to O or cannot be a subset of any
k
of the elements of O. Consequently, for each k ∈ N, B 1 (x) contains infinitely many elements of
k
{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 a
bounded 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 diameter
that contains B.
Proposition 2.6.7. A closed subset of a compact metric space is compact. A compact subset of a metric
space 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 ∈ Ω}
is an open covering of K. This implies that, there is a finite subcovering {X \ F, O1 , . . . , On } of
K. 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 number
r > 0 and finite elements x1 , . . . , xN such that
K⊂
N
[
Br (xk ).
k=1
Now define
O :=
N
[
k=1
Br (xk ) and r0 := r| + .{z
. . + r} + max ρ(xi , xj ).
1≤i,j≤n
N times
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 Cauchy
subsequence.
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 to
be 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 number
of 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} of
subsets of X, such that
Ã
!
[
An = K.
cl
n∈N
3. Every compact metric space is separable.
Q
4. If < Xi , ρi >, i = 1, . . . , n, are compact metric spaces, then the product ni=1 is also a compact
metric
Q∞ 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 intersection
property iff every finite subset of F has a non-empty intersection. Then prove that: if X be a
compact, then every collection F of closed subsets of X with the finite intersection property has a
nonempty 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).
Q
8. The product metric space ni=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 metric
space < 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 a
function. 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. Let
{Oα | α ∈ Ω} be an open covering of f (K) in Y . That is,
[
f (K) ⊂
Oα .
α∈Ω
⇒
K⊂
[
f −1 (Oα ).
α∈Ω
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
[
f −1 (Ok ).
k=1
Consequently,
f (K) ⊂
p
[
Ok .
k=1
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 ) in
Y.
Proposition 2.7.5 (continuity of compositions). Let < X, ρ >, < Y, σ > and < Z, % > be metric spaces
and f : X → Y and g : Y → Z be functions. If f is continuous in X and g is continuous on Y , then the
composition 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
Y
Xi ,
i
with the product metric ρ on X and f : X → Y be a function and < Y, σ > is a metric space. The function
f is continuous at x0 = (x01 , . . . , x0n ) iff each of the functions
fi (·) := f (x01 , . . . , x0i−1 , ·, x0i+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 each
fixed 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 continuous
function.
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 is
closed, 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 (πy−1 (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 valued
functions 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}
and
{x ∈ | 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-continuous
at 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) > α}
is 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) < α}
is 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) ≤ α}
is a closed set; and
(ii) if f is lower semi-continuous, then the set
{x ∈ X | f (x) ≥ α}
is 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 , then
lim sup f (xn ) ≤ f (x0 );
n
(ii) if f is l.s.c. at x0 , then
lim inf f (xn ) ≥ f (x0 ).
n
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
max f (x) = sup
x∈K
has a solution; equivalently, there is
x∗
∈ K such that
f (x∗ ) = max f (x) = sup f (x)
x∈K
x∈K
(ii) if f is lower semi-continuous on X, then f assumes its minimum on K; i.e. the problem
min f (x)
x∈K
has a solution; equivalently, there is x∗ ∈ K such that
f (x∗ ) = min f (x) = inf f (x).
x∈K
x∈K
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 the
image f (S) is a bounded subset of Y . In general, f is called a bounded function if f is bounded on
X.
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. For
instance, consider the real valued function f (x) = x1 .
Proposition 2.7.20. Let < X, ρ > and < Y, σ > be metric spaces and f : X → Y be uniformly
continuous. 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, there
is 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 that
σ(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. Then
the 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 uniformly
continuous 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-empty
subset A ⊂ X such that fT(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 . Consequently, for each fixed x ∈ X, we have a sequence {fn (x)}n∈N of elements of Y . We also refer to the
sequence {fn } as a sequence of functions.
Definition 2.7.24 (pointwise convergence). Let X and Y be metric spaces and S ⊂ Y . A sequence of
functions {fn } is said to be pointwise convergent to a function f : X → Y on S if, for each fixed x ∈ S,
we have
lim fn (x) = f (x).
n→∞
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 be
continuous.
Example 2.7.25. Let fn (x) = xn , x ∈ [0, 1]. Hence,
lim fn (x) = f (x),
n
where
½
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 . A
sequence of functions {fn } is said to be uniformly convergent to a function f : X → Y on S if, for every
ε > 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 implies
pointwise 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 each
n ∈ N, Mn := supx∈S σ(fn , f (x)). Then fn → f uniformly on S iff Mn → 0. In short
·
¸
fn → f uniformly on S ⇔ lim sup σ(fn (x), f (x)) = 0.
n→∞ x∈S
Theorem 2.7.28. Let X and Y be metric spaces, S ⊂ X and, for each n ∈ N, fn : X → Y be a continuous
function. 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
ε
σ(fn (x), fn (x0 )) < , ∀x ∈ Bδ (x0 ).
3
Thus, by the uniform convergence of fn to f on S, we see that fn converges uniformly to f on
S ∩ Bδ (x0 ) =: BδS (x0 ) . This implies, there is N > 0 such that
ε
σ(f (x), fn (x)) < , ∀n ≥ N, ∀x ∈ BδS (x0 ).
3
Consequently,
σ(f (x), f (x0 )) < ε, ∀x ∈ BδS (x0 ).
Hence, f is continuous at x0 relative to S. Since, x0 ∈ S is arbitrary, we conclude that f is a continuous
function relative to S; therefore, f is continuous on S.
Excercises 2.7.29. Prove the following statements:
(a) Prop. 2.7.27.
(b) Which of the following is uniformly convergent
2
(i) fn (x) = xe−nx ,
(ii) fn (x) = n2 x(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 is
bounded 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 also
uniformly continuous on X.
21
2.7.5 Equicontinuiuty and the Ascoli-Arzelá 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 a
metric space < Y, σ >. The family F is said to be equicontinuous at x0 ∈ X if, for every ε > 0, there is
δ > 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 from
X to Y . If D is a countable set and, for each x ∈ S, the set {fn (x) | n ∈ N} is compact, then there is a
subsequence {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 )}(Observe
that {f1n (x1 )} ⊂ cl{f1n (x1 )} and {f1n (x1 )} is compact)k . For x2 ∈ D, there is a convergent subsequence {f2n (x2 )} of {f1n (x2 )}, and so on. Proceeding in this manner, for xk ∈ D, we obtain a
convergent 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 each
k ≥ n. Hence, for each xk ∈ S, {fnn (xk )} is convergent.
Remark 2.7.32. In Lem. 2.7.31, it is enough to have cl{fn (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 complete
metric 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 convergent
sequence. This implies that {fn (y)} is a Cauchy sequence. Hence, there is a sufficiently large N
such that
ε
σ(fn (y), fm (y)) < , ∀n, m ≥ N.
3
k
A discrete subset of a compact set is compact.
22
(2.2)
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, we
conclude 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 implies
σ(f (x), f (y)) = lim σ(fn (x), fn (y)) ≤ ε, ∀x ∈ Bδ (x).
n→∞
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 is
compact 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) such
that
ε
σ(fn (x), fn (y)) < , ∀y ∈ Bδ(x) (x), ∀n.
3
This implies that
ε
σ(f (x), f (y)) < , ∀y ∈ Bδ(x) (x),
3
Hence, by the compactness of K, there is a finite collection {x1 , . . . , xm } such that
K⊂
m
[
Bδ(xi ) (xi ).
i=1
Since, fn (xi ) → f (xi ), for each i = 1, . . . , m, choose N sufficiently large so that
ε
σ(fn (xi ), f (xi )) < , ∀i ∈ {1, . . . , m}, ∀n ≥ N.
3
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, one
can verify the validity of the following well known theorem:
Theorem 2.7.35 (Ascoli-Arzelá Theorem, Thm. 40, p. 169, Royden[21]). Let F be a family of equicontinuous functions from a separable metric space X to a metric space Y . Let {fn } be a sequence in F such
that 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 compact
subset 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 cl{fn (x) | n ∈ N}
is a compact set, for each x ∈ D. Then, by Lem. 2.7.31, there is a subsequence {fnk } of {fn }
such 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 an
homeomorphism 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 are
homeomorphic metric spaces.
Theorem 2.7.37. Let f : X → Y be both one-to-one and onto. Then the following statements are
equivalent:
(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 closed
in 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 be
compact, and so on. In general, properties of the metric space X, that could be characterized by open
sets, 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
p
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 → Y
is 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 every
homeomorphism 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 a
homeomorphism, 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 function
f : 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 map
ι : 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 there
is 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 a
fixed 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 complete
metric space and f is a contractive map (with γ ∈ [0, 1)), then f has a unique fixed point in X; i.e. there
is 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 that
xn+1 = f (xn ), n = 0, 1, 2, . . . We show that the sequence {xn } is convergent. Thus, it is enough
to 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 )) ≤ γ 2 ρ(xk−1 , xk−2 )
≤ . . . ≤ γ k ρ(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(x1 , x0 ) m
ρ(xn , xm ) ≤ γ n−1 + . . . + γ m ρ(x1 , x0 ) =
[γ − γ n ] .
1−γ
The sequence {γ n } is a Cauchy sequence. Consequently, {xn } is a Cauchy sequence. Since, X is
complete, there is x ∈ X such that xn → x. And f is contractive
ρ(f (xn ), f (x)) ≤ γρ(xn , x).
This implies
f (x) = lim f (xn ) = lim xn+1 = x.
n→∞
n→∞
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. If
x ∈ X is the fixed point of f , then
1
ρ(f (x), x)
ρ(x, x) ≤
1−γ
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
Excercises 2.7.49. Prove the following statements:
(i) Let S ⊂ X and define function
d(x) := inf ρ(x, z) =: dist(x, S),
z∈S
for each x ∈ X - known as the distance function. Then d is a Lipschitz continuous function from
X 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 given
by T x = Ax + b for an n × n matrix A = (aij )1≤1,i,j,≤n . Show that if there is α such that
n
X
|aij | ≤ α < 1, i = 1, . . . , n,
j=1
then the system of equations
xi =
n
X
aij xj + bi , i = 1, . . . , n
j=1
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 a
contraction, where
T (n) = |T ◦ T ◦{z. . . ◦ T},
r times
then T has a unique fixed point x in X and, for any x ∈ X, the sequence {T n (x)} converges to x.
27
28
3 Topological Spaces
Definition 3.0.50 (topological space). Let X be a non-empty set. Then a family τ of subsets of X is
called 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 ,
S
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 an
element 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 trivial
topology. 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 the
discrete 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 a
topological space associated with the metric ρ. Hence, different metrics may give rise to different
topologies on a given set.∗ . A topological space which could associated with a metric space is called
metrizable.
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
\
τα
α∈Ω
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 are
partially ordered with respect to ”⊂”.
On a given set X, the trivial topology is the weakest topology and the discrete topology is the strongest
topology.
Proposition 3.0.53. Let X be a non-empty set and C be any collection of subsets of X. Then there is a
weakest 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 the
family of sets
τS := {S ∩ O | O ∈ τ }
is 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;
S
(ii) if {F1 , . . . , Fn } is any finite collection of closed subsets of X, then ni=1 Fi is closed;
T
(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 set
contained 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 A0 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.
0
Denote by A 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 A0 ⊂ clA;
(ii) F is a closed set if and only if F = clF = F ∪ F 0 ; i.e. if and only if F 0 ⊂ F ;
0
(iii) for any set A, we have A = 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. Then
there is a closed set F such that A ⊂ F such that x ∈
/ F.
⇒
x ∈ X \ F =: O.
⇒
x ∈ O, but O ∩ (A \ {x}) = ∅ ⇒ x ∈
/ A0 .
Consequently, A0 ⊂ clA.
Definition 3.0.63 (dense set). Let < X, τ > be a topological space. A subset D of X is dense in X iff
clD = 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 countable
dense 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, is
called 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 and Nx ⊂ 2X . Then Nx is said to be a neighborhood
system 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 the
relative topology τS and x ∈ S. If V ⊂ S is a neighborhood of x w.r.t. the relative topology τA , there is a
neighborhood 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 in
X, 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 accumulation
point of A;
6. if τ is the discrete topology and A ⊂ X, then the set of accumulation points A0 = ∅;
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 a
set 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 every
O ∈ 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 topology
on 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 for
τ.
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 gives
a 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 on
X if and only
(i) X =
S
{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∈Ω Uα ∈ τ .
Now, let U1 , U2 ∈ τ and x ∈ U1 ∩ U2 . Since x ∈ U1 and x ∈ U2 , there is B1 , B2 ∈ B such
that 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 .
S
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
S
Proof.
We verify (i) and (ii) of Prop. 3.2.4. In fact, (i) is obvious, since X = {S | S ∈ S} =
S
{B | B ∈ B}. Then, let B1 , B2 ∈ B. Then there are S11 , . . . , Sn11 ∈ S and S12 , . . . , Sn22 ∈ S such that
B1 =
n1
\
Sk1
and B2 =
k=1
Hence, if x ∈ B1 ∩ B2 , then
x ∈ B3 :=
n2
\
Sk2 .
k=1
n1
\
Sk1 ∩
k=1
n2
\
Sk2 .
k=1
Consequently, ∃B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2 . Therefore, by Prop. 3.2.4, B is a base for some
topology on X.
Definition 3.2.6. (subbase) A family S of sets is a subbase
if the collection B of all finite intersections of
S
elements 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 accumulation
point 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 countability (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, say
Nx = {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 second
axiom 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. For
instance, 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. Let
U 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 is
arbitrary, we conclude that x is an accumulation point of D. Hence, clD = X and D is a countable
dense 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} is
said to be a covering of X if
[
X⊂
Uα .
α∈Ω
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 (Lindelöf). Let < X, τ > be second countable and A be any subset of X. Then every
open 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⊂
Oα .
α∈Ω
This implies, for each x ∈ A, there is α ∈ Ω such that x ∈ Oα . Since B is a base for τ , there is Bn ∈ B
such that x ∈ Bn ⊂ Oα . If we now let Oαn = Oα whenever Bn ⊂ Oα , then there is N ⊂ N such that
A⊂
[
Bn ⊂
[
Oαn .
n∈N
Consequently, {Oαn | n ∈ N } is a countable subcovering of A.
Definition 3.2.20 (a Lindelöf topological space). A topological space X is said to be Lindelöf if every
open covering of X has a countable subcover.
Hence, the space Rn is Lindelöf.
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 Lindelöf;
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 is
said to converge to an element x0 ∈ X if for every neighborhood U of x0 (i.e. ∀U ∈ Nx0 ) there is N ∈ N
such 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 sequence
{xn } is called a convergent sequence.
Definition 3.3.2 (continuity). Let < X, τ > and < Y, σ > be topological spaces. Then f : X → Y is a
continuous 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 set
is 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 a
closed set. Since, f −1 (Y \ U ) = X \ f −1 (U ), we have that f −1 (U ) is an open set in X. Hence, f
is a continuous function.
Proposition 3.3.4. Let < X, τ > and < Y, σ > be topological spaces, f : X → Y be a function and {xn }
be 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 for
any 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 every
convergent 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 in
X;
(ii) f is said to be a closed function (a closed map) if, for every closed set F ⊂ X, f (F ) is a closed set
in X.
36
Example 3.3.12. The function π : R2 → R given by π(x, y) = x. Then π(·) is an open mapping, but not
a closed one.
Definition 3.3.13 (homeomorphism). Two topological spaces X and Y are said to be homeomorphic if
there 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 continuous
function 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 and
only if f is a non-to-one open or closed map.
Excercises 3.3.16. Prove the following statements:
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 two
separate open sets. But, this is not always possible in a general topological space(unless the topological
space is metrizable). Hence, we have the following major classification of topological spaces based on
separation axioms.
(i) A topological space X is a T1 -space (Fréchet-Riesz) if whenever x, z ∈ X and x 6= z, there exist
open (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 exist
disjoint 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 X
with 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 singleton
is 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 that
z ∈ Oz but x ∈
/ Oz .
⇒
[
X \ {x} =
Oz .
z∈X\{x}
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} and
O2 := X \ {z}, then O1 and O2 are open sets, z ∈ O1 , x ∈
/ O1 and x ∈ O2 , z ∈
/ O2 . Therefore, X
is 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 a
Hausdorff 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 is
Hausdorff, there are disjoint open sets O1 and O2 such that
x ∈ O1 and z ∈ O2 .
Since xn → x, there is N such that
xn ∈ O1 , ∀n ≥ N.
From this follows that, there are only a finite number of elements of {xn } that can belong to O2 . This
implies that xn cannot converge to z. But this is a contradiction. Hence, the assumption is false and
x = z.
Proposition 3.4.6. Suppose X is a topological space that is first countable. Then the following statements
are 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 that
every 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 neighborhoods
of 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 is
Hausdorff.
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 and
F ∩ F2 = ∅. Since X is a normal space, there exist disjoint opens set U and U1 such that
F ⊂ U and F2 ⊂ U1
Hence
U ∩ 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
1
1 1 3
1 1 3 1 5 3 7
D0 = {0, 1}, D1 = {0, , 1}, D2 = {0, , , , 1}, D3 = D2 = {0, , , , , , , , 1}, . . .
2
4 2 4
8 4 8 2 8 4 8
Define D to be the union of all these set as
D :=
∞
[
Dn .
n=0
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 { 21n } ⊂ D. Moreover, for any k with 1 ≤ k ≤ n,
2k
2n
=
1
2n−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 21n → 0, for the given δ > 0, there is n0 ∈ N such that
0<
Now set q := 2n0 . Then it follows that 0 <
1
q
1
< δ.
2n0
< δ and
¸
n0 · k
[
2 − 1 2k
x ∈ [0, 1] =
,
.
2n0 2n0
k=0
Hence, there is m, 1 ≤ m ≤ q = 2n0 (i.e., m ∈ {2k | k = 0, . . . , n0 }), such that
·
¸
m−1 m
x∈
,
q
q
⇒
m−1
q
≤x≤
m
q .
Since 0 <
1
q
< δ, we have
x−δ <x−
Consequently,
1
m 1
m−1
≤
− =
≤x<x+δ
q
q
q
q
m−1
∈ (x − δ, x + δ).
q
Note that { i−1
q | 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 are
equivalent:
(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
½
f (x) =
0,
1,
if x ∈ F1
if x ∈ F2 .
Proof. (ii) ⇒ (i): Let F1 and F2 be any two closed sets in X. By assumption there is a continuous
function f : X → [0, 1] such that f (F1 ) = {0} and f (F2 ) = {1}. The sets [0, 13 ) and ( 13 , 1] are
open sets in [0, 1]. Consequently, by the continuity of f , it follows that
1
1
U1 := f −1 ([0, )) and U2 := f −1 (( , 1])
3
3
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 such that
2
F1 ⊂ U 1 ⊂ clU 1 ⊂ O2
2
2
Using Thm. again we obtain open sets U 1 and U 3 such that
4
4
F1 ⊂ U 1 ⊂ clU 1 ⊂ U 1 ⊂ clU 1 ⊂ U 3 ⊂ clU 3 ⊂ O2
4
4
2
2
4
4
Proceeding in this manner we construct a sequence of sets {Ut | t ∈ D}, corresponding to the set
D 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
1
2n
, ∀n ∈ N.
Now define the function f : X → [0, 1] such that for x ∈ X
f (x) := inf {t | x ∈ Ut }
Then f is a well defined function.
(a) Now let x ∈ F1 be any. Since x ∈ F1 ⊂ U
0 ≤ f (x) ≤
1
2n
, ∀n ∈ N, it follows that
1
, ∀n ∈ N ⇒ f (x) = 0.
2n
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
n
x∈
/ clUt , ∀t ∈ D, t ≤
n+1
⇒
n
f (x) ≥
.
n+1
But this holds true for any given n ∈ N. Hence,
n
1 ≥ f (x) ≥
, ∀n ∈ N.
n+1
⇒ 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 that
f −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 , there
is 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 ∈ D
such that α < t1 < t2 < f (x). This implies that x ∈
/ Ut2 . Furthermore, from t1 < t2 , we
have 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) = inf {t | 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 any
two 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
½
f (x) =
a,
b,
if x ∈ F1
if x ∈ F2 .
Definition 3.5.5 (Tychonoff or completely regular spaces). A topological space X is completely regular
or Tychonoff iff for any closed set F of X and p ∈ X, with p ∈
/ F , there exists a continuous function
f : 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 1 or Tychonoff space). A T1 topological space which is also completely regular is
2
called T3 1 or a Tychonoff space.
2
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 of
X. If X is normal and f is a continuous real valued function such that f : F → [0, 1]. Then there is a
continuous 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
1
A1 := {x ∈ F | f (x) ≤ }
3
2
B1 := {x ∈ F | f (x) ≥ }.
3
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 → [ 31 , 23 ] such that f (A1 ) = 31 and f (B1 ) = 23 .
Hence, for x ∈ F ,


1
3
− 0 = 13 ,
1 − 2 = 1,
|f (x) − f1 (x)| ≤
 2 31 31
3 − 3 = 3,
if f (x) < 13
if f (x) > 23
if 13 ≤ 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, 31 ].
Repeating the above process, let
1
A2 := {x ∈ F | h1 (x) ≤ } = {x ∈ F | h1 (x) ≤
9
2
B2 := {x ∈ F | h1 (x) ≥ } = {x ∈ F | h1 (x) ≥
9
1
}
32
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 ) = 91 and f (B2 ) = 29 . Furthermore, we have
|f (x) − (f1 (x) + f2 (x)) | = | (f (x) − f1 (x)) − f2 (x)| = |h1 (x) − f2 (x)| ≤
1
.
32
Proceeding inductively, we construct a sequence of closed sets {An } and {Bn }, with An ∩ Bn = ∅ and
a sequence of continuous functions
·
¸
1 2
fn : X → n , n
3 3
such that fn (An ) =
1
3n
and fn (Bn ) =
2
3n
and
|f (x) −
n
X
fk (x)| ≤
k=1
Now let sn (x) :=
Pn
k=1 fk (x).
1
, ∀x ∈ F.
3n
Then, for each n, sn (·) is a continuous function on X. Moreover,
1
= 0.
n→∞ 3n
lim sup |f (x) − sn (x)| ≤ lim
n→∞ x∈F
This implies, sn (x) → f (x) uniformly on F . Thus
f (x) =
∞
X
fk (x), x ∈ F.
(3.1)
k=1
Moreover, for each x ∈ X, we have
∞
X
k=1
̰
!
µ
¶
∞
∞
X
X
X 1
2
1
3
|fk (x)| ≤
=2
=2
−1 =2
− 1 = 1.
3k
3k
3k
2
k=1
k=1
(3.2)
k=0
43
This implies
P∞
k=1 fk (x)
is summable for each x ∈ X. Now, define
g(x) :=
∞
X
fk (x), x ∈ X.
k=1
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 it
remains to show (iii).
Let x ∈ X be any fixed element. Then for any z ∈ X we have
¯
¯∞
∞
∞
¯ X
¯X
X
¯
¯
fk (z)¯ ≤
|fk (x) − fk (z)| .
fk (x) −
|g(x) − g(z)| = ¯
¯
¯
k=1
We know that
Pn
1
k=1 3n
→
P∞
k=1
(3.3)
k=1
1
k=1 3k .
Hence, for any given ε > 0, there is n0 ∈ N such that
¯
¯∞
n
¯X 1
X
1 ¯¯ ε
¯
−
¯ < , ∀n ≥ n0 .
¯
¯
3k
3k ¯ 2
k=1
⇒
k=1
∞
X
k=n0 +1
1
ε
< .
3k
2
Thus, from 3.3, it follows that
|g(x) − g(z)| ≤
∞
X
|fk (x) − fk (z)| =
k=1
≤
n0
X
k=1
n0
X
k=1
≤
n0
X
k=1
|fk (x) − fk (z)| +
|fk (x) − fk (z)| +
∞
X
k=n0 +1
∞
X
k=n0 +1
|fk (x) − fk (z)|
1
3k
ε
|fk (x) − fk (z)| + .
2
Since, for each k = 1, 2, . . . , n0 , the function fk is continuous at x, there is an open neighborhood
Uk (x) such that
ε
|fk (x) − fk (z)| <
, ∀z ∈ Uk (x).
2n0
0
Set U (x) := ∩nk=1
Uk (x). Then U (x) is an open neighborhood of x and
µ
¶
ε
ε
|g(x) − g(z)| < n0
+ = ε, ∀z ∈ U (x).
2n0
2
Consequently, g is continuous at x. Since x ∈ X is arbitrary, we conclude that g is a continuous
function.
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 a
continuous real valued function such that f : F → [a, b], a, b ∈ R and a < b. Then there is a continuous
real 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 a
closed subset of X. If X is normal and f is a continuous real valued function such on F . Then there is a
continuous 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 to
a metric space.
A metrizable topological space inherits all the properties of the metric space associated with it. But
not all topological spaces are metrizable; i.e. there are metric spaces which are not metrizable.
Q
ℵ0 is known as the
Definition 3.5.12 (Hilbert Cube). The cartesian product H = ∞
k=1 [0, 1] =: [0, 1]
Hilbert cube.
Lemma 3.5.13. Let Hs be the set of all sequences of the form
Hs := {{xn } | 0 ≤ xn ≤
1
}
n
and, for x = {xn }, y = {yn } ∈ Hs , let
#1
"∞
2
X
2
(xk − yk )
ρ(x, y) :=
.
k=1
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 each
Uk ∈ 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) ¯
1
1
¯
¯
¯ 2n ¯ ≤ 2n ≤ n
Now, for x ∈ X, if we define
½
f (x) =
fn (x)
2n
¾
,
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, by
Lem. 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.
⇒
½
fn (x)
2n
¾
½
6=
n∈N
fn (z)
2n
¾
.
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) ¯
¯
¯≤ 1 .
¯
¯ 22n
2n
2
P
1
Since the series ∞
n=1 22n converges, there exists n0 ∈ N such that
∞
X
n=n0 +1
46
1
ε2
<
.
22n
2
Moreover, f (x), f (z) ∈ Hs yields
2
ρ(f (x), f (z)) =
∞
X
|fn (x) − fn (z)|
22n
n=1
⇒
ρ(f (x), f (z))2 ≤
∞
X
n=n0 +1
n
∞
X
=
n=n0 +1
0
|fn (x) − fn (z)| X
|fn (x) − fn (z)|
+
2n
2
22n
n=1
n
n
n=1
n=1
0
0
X
1
|fn (x) − fn (z)|
ε2 X
|fn (x) − fn (z)|
+
<
+
2n
2n
2
2
2
22n
For each l = 1, . . . , n0 , there is an open neighborhood Ul of x such that
|fk (x) − fk (z)| <
Then the set O :=
Tn0
l=1 Ul
ε2
, ∀z ∈ Ul .
2n0
is open and
ε2
+ n0
ρ(f (x), f (z)) <
2
2
µ
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 is
y ∈ 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 =
∞
X
|fk (xn ) − fk (x)|
k=1
22n
⇒
ρ(f (xn ), f (x)) ≥
≥
1
22m
|fm (xn ) − fm (x)|2
1
= 2m
22m
2
, ∀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.
Excercises 3.5.16. Prove the following statements:
1. Let < x, τ > and < Y, σ > be topological spaces, f : X → Y and S is a subbase for the topology σ
on 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 a
Hausdorff 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 Hausdorff
space, 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) = inf ρ(x, z).
z∈A
Then the map fA : X → R is continuous. Moreover, for A and B be disjoint closed sets, g : X → R
and 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 ∈ Ω}
of 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 a
covering of X. If there is a finite index {α1 , . . . , αn } ⊂ Ω such that
n
[
X⊂
Uαk ,
k=1
then the collection {Uα1 , Uα2 , . . . , Uαn } is called a finite subcovering of X; i.e. {Uα | α ∈ Ω} has a finite
subcovering of X.
Definition 3.6.3 (a compact set). Let < X, τ > be a topological space and K ⊂ X. Then the set K is said
to be a compact set if every open covering {Oα | α ∈ Ω} of K has a finite subcovering {Oα1 , Oα2 , . . . , Oαn };
i.e.
n
[
[
K⊂
Oα ⇒ K ⊂
Oαk .
α∈Ω
k=1
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 continuous
function. 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) ⊂
Oα
α∈Ω
⇒
K⊂
[
f −1 (Oα )
α∈Ω
Since f is a continuous function, the collection {f −1 (Oα ) | α ∈ Ω} is an open covering of the compact
set K. Consequently, there exists {Oα1 , . . . , Oαn } such that
K⊂
n
[
f −1 (Oαk )
k=1
⇒
f (K) ⊂
n
[
Oαk .
k=1
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 ∈ Ω}
is an open covering of F . Then the collection
{Oα , X \ F | α ∈ Ω}
is an open covering of X. Hence, there is {Oα1 , . . . , Oαn } ⊂ {Oα | α ∈ Ω} such that
X⊂
n
[
Oαk ∪ (X \ F ).
k=1
⇒
F ⊂
n
[
Oαk ∪ (X \ F ).
k=1
But, F ∩ (X \ F ) = ∅. Hence,
F ⊂
n
[
Oαk .
k=1
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 a
closed 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 subcovering {Ox1 , . . . , Oxn } such that
n
[
Oxk .
K⊂
k=1
Let {Ux1 , . . . , Uxn } be the corresponding collection of open sets such that z ∈ Uxk , k = 1, . . . , n.
Define
n
\
Ux k .
U :=
k=1
Hence, U is an open set, z ∈ U and U ∩ Oxk = ∅ for each k = 1, . . . , n. This implies
U∩
n
[
Oxk = ∅ ⇒ U ∩ K = ∅ ⇒ z ∈ U ⊂ X \ K.
k=1
Consequently, z is an interior point of X \ K. Since z ∈ X \ K is arbitrary, we conclude that
X \ 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 ∈ Ω}
be a collection in X. Then {Aα | α ∈ Ω} is said to have the finite intersection property if for every finite
collection {Aα1 , . . . , Aαn }, the intersection
n
\
Aαk
k=1
is non-empty.
50
Proposition 3.6.7. A topological space X is compact if and only if every collection {Fα | α ∈ Ω} of closed
sets with the finite intersection property has a non-empty intersection.
Proof. ”⇒” : Assume that
\
Fα = ∅.
α
⇒
X=X\
\
Fα =
α
[
(X \ Fα ) .
α∈Ω
Since X is a compact set, there exist Fα1 , . . . , Fαn such that
X⊂
n
[
(X \ Fαk )
k=1
⇒
n
\
Fαk = ∅.
k=1
But this contradicts the finite intersection property. Hence,
T
α Fα
6= ∅.
”⇐”: Assume that X is not a compact set. Then there is an open covering {Oα | α ∈ Ω} of X with no
finite subcovering. This implies, for every finite subcollection {Oα1 , . . . , Oαn }
X\
n
[
Oαk 6= ∅.
k=1
⇒ the family of closed sets {X \ Oα | α ∈ Ω} satisfies the finite intersection property. Then, by
assumption,
\
(X \ Oα ) 6= ∅.
α∈Ω
⇒
[
X\
Oα 6= ∅.
α∈Ω
This implies the collection {Oα | α ∈ Ω} does not cover X. But this is a contradiction. Consequently, 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 are
equivalent
• X is a compact topological space;
• for every family of closed subsets {Fα | α ∈ Ω} of X with the property that
finite subcollection {Oα1 , . . . , Oαn } such that
m
\
T
α∈Ω Fα
= ∅, there is a
Oαk = ∅.
k=1
51
3.6.3 Compact Hausdorff Spaces
Compact Hausdorff topological spaces exhibit very interesting properties, which are very important
from 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 ∈ X
and 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 compact
sets, 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 such
that
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 finite
open covering {Oz1 , . . . , Ozm }, i.e.
m
[
Ozk =: O1
K1 ⊂
k=1
and there is a corresponding finite collection {Uz1 , . . . , Uzm } with K2 ⊂ Uzk for each k = 1, . . . , m.
Then
m
\
Uzk =: O2 .
K2 ⊂
k=1
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 topological
space.
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
T topological space, U be an open subset of X and {Kn }n∈N be a family of compact
subsets of X. If n∈N Kn ⊂ U , then there exists a finite index I ⊂ N such that
\
n∈I
52
Kn ⊂ U.
3. Let X be a compact topological space and let {Kn }n∈N be a family of non-empty closed subsets of X
with Kn+1 ⊂ Kn for each n ≥ 1. Then
\
Kn 6= ∅.
n∈N
4.
(a) It is necessary and sufficient for a topological space X to be compact that: if {Vα |α ∈ Ω} is any
family of closed subsets of X such that ∩α∈Ω0 Vα 6= ∅, for any subset Ω0 ⊂ Ω, then ∩α∈Ω Vα 6= ∅
(b) Let X be a compact topological space and for each n ∈ N, Vn is a non-empty closed subset of
X such that
Vn ⊃ Vn+1 .
Then
\
Vn 6= ∅.
n
(c) Suppose f : X → X is continuous, where X is a compact metric space. Then there exists a
non-empty subset A ⊂ X such that f (A) = A. (Hint: put X1 = f (X), Xn+1 = f (Xn ) and
A = ∩∞
n=1 Xn 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. If
F ⊂ U , then there exists ε > 0 such that
[
Bε (x) ⊂,
x∈F
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-one
function 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, then
f (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 that
ρ(x, z) = dist(K1 , K2 ).
3.7 Locally Compact Spaces
Locally compact Hausdorff space are among the most important topological spaces, say in non-linear
analysis and abstract measure theory. For instance, measure and integral theory on topological space
usually 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 compact
if
∀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 , there
are disjoint open sets Ux and Vz such that x ∈ Ux and z ∈ Vz . This implies
[
[
F1 ⊂
Ux and F2 ⊂
Vz .
x∈F1
z∈F2
(Note that: each Ux is so that Ux ∩ Vz1 = ∅ and Ux ∩ Vz2 = ∅ for z1 6= z2 and vice-versa). By the
compactness of F1 and F2 , there are x1 , . . . , xn ∈ F1 and z1 , . . . , zm ∈ F2 such that
F1 ⊂
n
[
Uxk =: U and F2 ⊂
k=1
m
[
Vzk =: V.
k=1
⇒ 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 every
x ∈ 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 subspace
of 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 the
claim 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⊂
Ox .
x∈K
⇒ there are x1 , . . . , xn ∈ K such that
K⊂
n
[
k=1
Oxk .
S
If we let O := nk=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. By
the compactness of K, there are x1 , . . . , xn ∈ K such that
A=
n
[
Vxk , A is compact and K ⊂ intA ⊂ A ⊂ O.
k=1
By Lem 3.7.4 A is a T4 subspace of X. Hence, by Uryson’s Lemma( Lem. 3.5.3), there is a continuous
function g : A → [0, 1] such that g(K) = 1 and g(A \ intA) = 0. Now define f : X → [0, 1] as
½
g(x),
if x ∈ A
f (x) =
0,
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 at
x0 ∈ ∂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, cl{x ∈ X | f (x) 6= 0} ⊂ A ⊂ O.‡
Excercises 3.7.6. Prove the following statements:
1. If X is a locally compact T3 topological space, then each point x ∈ X has a neighborhood base of
the form
Nx = {x ∈ O ∈ τ | clO compact }.
2. Let
T X be a locally compact space and {Dn | n ∈ N} be a countable collection of dense set. Then
n∈N Dn 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 closed
compact 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 the
union of a countable number of compact sets.
Spaces which are σ-compact are also known as countably compact spaces. Thus, a compact space is
a σ-compact.
‡
The set cl{x ∈ 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 Lindelöf
⇔ (ii) X S
is σ-compact ⇔ (iii) There is a sequence {On } of open sets with clOn compact, clOn ⊂ On+1
and X ⊂ n∈N On .
Proof. (i)⇒(ii): X is locally compact implies thatS
for each x ∈ X, there is a neighborhood Ux such
that x ∈ Ux and clU is compact. Hence, X ⊂ x∈X Ux . Since X is Lindelöf, there is a countable
subcover {Un | n ∈ N} such that
X=
[
Un ⇒ X =
n∈N
[
clUn .
n∈N
Consequently, X is σ-compact.
S
(ii)⇒ (iii): X is σ-compact ⇒ X = n∈N Kn , where for each n ∈ N, Kn is a compact set. Now, since
K1 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 such
that Kn ∪ clOn−1 ⊂ On and clOn is compact. Consequently, we have
clOn ⊂ On+1 , n = 1, . . . and X =
[
On .
n∈N
(iii)⇒ (i): Let C = {Uα | α ∈ Ω} be an open covering of X.SThen, by S
(iii) there is a sequence
{On | n ∈ N} such that clOn ⊂ On , clOn compact and X = n∈N On = n∈N clOn . This implies,
for each n ∈ N,
clOn ⊂
[
Uα ⇒ ∃{Oα1 , . . . , Oαmn } ⊂ C : clOn ⊂
α∈Ω
Set
F :=
[
m
[n
Uαk .
k=1
{Uα1 , . . . , Uαmn | clOn ⊂
n∈N
It follows that F is countable, F ⊂ C and
S
m
[n
Uαk , n ∈ N}.
k=1
F = X. Consequently, X is Lindelöf.
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)
S
F=
S
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} of
subsets of X is said to be locally finite if for each x ∈ X, there is a neighborhood O of x such that O
intersects 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 is
said to be σ-locally finite if
[
C=
Cn ;
n∈N
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 Ce := {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 family
F := {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 closed
or open sets. Recall also that, for a collection {Uα | α ∈ Ω} we have
Ã
!
[
[
cl
Uα ⊂
clUα
α∈Ω
α∈Ω
but equality does not hold always. However,
Lemma 3.9.6. Let C = {Uα | α ∈ Ω} be a locally finite family, then
Ã
!
[
[
cl
Uα =
clUα
α∈Ω
Proof. (a)
[
α∈Ω
Ã
clUα ⊂ cl
α∈Ω
[
!
Uα
is obvious.
α∈Ω
¢
¡S
(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 ∩ Uαk 6= ∅, k = 1, . . . , n. Hence,


n
[
[
x ∈ N, N ∩
Uαk 6= ∅, and N ∩ 
Uα  = ∅.
k=1
α∈Ω\{α1 ,...,αn }
S
S
S
This implies that x ∈ cl( nk=1 Uαk ) = nk=1 clUαk ⊂ α∈Ω clUα § . Consequently,
Ã
!
[
[
cl
Uα ⊂
clUα .
α∈Ω
§
α∈Ω
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 subsets
of 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 only
a finite number of elements of C. Hence,
[
K⊂
Ox .
x∈K
Since K is compact, there are x1 , . . . , xn ∈ K, for some n ∈ N, such that
K⊂
n
[
Oxk .
k=1
But, for each k ∈ {1, . . . , n}, Oxk can intersect
P only a finite number of elements of C, say about mk of
them. Hence, K can only intersect about nk=1 mk elements of C.
Definition 3.9.8 (paracompact topological space). A Hausdorff topological space < X, τ > is said to be
paracompact 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 Lindelöf. Hence, C has a countable sub-cover, say {Un | n ∈ N}; and
• Thm. 3.8.2(iii) ⇒ there is a sequence
S of sets {On | n ∈ N} such that, for each n ∈ N, clOn is
compact, clOn ⊂ On+1 and X = n∈N On .
Now, for each n ∈ N, define the set
en := Un ∩ (On \ clOn−2 ) ,
U
where, w.l.o.g, we assume that O−1 = O0 = ∅.
en | n ∈ N} is an open refinement of C.
(i) The collection {U
en ⊂ Un ∈ C, and
For each n ∈ N, U
[
[
[
[
[
en =
U
Un ∩ (On \ clOn−2 ) = X ∩ (On \ clOn−2 ) =
(On \ clOn−2 ) .
n
n
n
n
n
S
We show next that n (On \ clOn−2 ) = X. Since X =
n(x) ∈ N such that x ∈ On(x) .
(a) If n(x) = 1, then
x ∈ O1 \ clO−1 ⇒ x ∈
[
S
n On ,
for x ∈ X, there is a smallest
(On \ clOn−2 ) .
n
(b) If n(x) ≥ 2, then
x∈
/ On(x)−1 ⇒ x ∈
/ clOn(x)−2 ⇒ x ∈ On(x) \ clOn(x)−2 ⇒ x ∈
[
n
From (a) and (b), we conclude that X =
58
S
(3.4)
n (On
\ clOn−2 ).
(On \ clOn−2 ) .
Consequently, from eqn. (3.4), we conclude that
refinement of C.
S e
e
n Un = X. That is, {Un | n ∈ N} is an open
en | n ∈ N} is locally finite. To see this, let x ∈ X be any. Since X =
(ii) {U
there is a smallest n0 ∈ N such that
x ∈ On0 .
S
n∈N On
=
S
n∈N clOn ,
This implies that
ek = ∅, ∀k ≥ n0 + 2.
On0 ∩ [Uk ∩ (Ok \ clOk−2 )] = ∅, ∀k ≥ n0 + 2 ⇒ On0 ∩ U
en | n ∈ N}. Consequently,
Hence, On0 can only intersect a finite number of elements of {U
e
{Un | n ∈ N} is locally finite.
en | n ∈ N} is a locally finite open refinement of C. Therefore, X is paracomHence, form (i) & (ii) {U
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 finite
refinement, then C has a locally finite refinement.
S
Proof. Let C = {Uα | α ∈ Ω} be an open covering of X. Then by assumption C = n∈N Cn , where for
each 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 .
S
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 family
Bm := {Uα ∈ C | n(α) = m} ⊂ Cm .
is locally finite. Hence, there exists a neighborhood of Om of x such that Om intersects only
a 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 of
{Wα | α ∈ Ω}.
Proposition 3.9.13. Let < X, τ > be a T3 topological space. If every open cover of C of X has a locally
finite 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 that
x ∈ Uα(x) . Since X is T3 , there is an open set Ox such that
x ∈ clOx ⊂ Uα(x) .
S
Then the collection O =: {Ox | x ∈ X} is an open refinement of C (note that X = x∈X Ox ) such that
e := {clOx | x ∈ X} is a closed refinement of C. By assumption, there is a locally finite refinement F
O
of O. Then, the family
Fe := {clV | V ∈ F}
e Consequently, Fe is a closed locally finite refinement of C.
is a closed locally finite refinement of O.
(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 a
closed 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 a
neighborhoods Uz such that
x∈
/ clUz .
Define
C := {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
Ã
!
[
[
cl
V =
clV ;
V ∈A
(b) F ⊂
¶
S
V ∈A
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
Ã
!
[
[
V ⇒x∈
/ clO.
x∈
/
clV ⇒ x ∈
/ cl
V ∈A
V ∈A
⇒
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, then
there 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) = min{1, dist(x, X \ Un )}.
Now let
• V1 := U1 ; and
• for n = 2, . . .
Vn := Un ∩
¯
¾
¯
1
.
x ∈ X ¯¯ ϕk (x) <
n
n−1
\½
k=1
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 for
some n ∈ N. If x ∈ U1 , then we are done. Otherwise, let n0 ∈ N be the smallest index such
that 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 ∩
k=1
Since x ∈ X is arbitrary, we conclude that
[
X=
¯
¾
¯
1
¯
x ∈ X ¯ ϕk (x) <
= Vn0 .
n
n\
0 −1 ½
Vn ;
n∈N
(iii) the family {Vn | n ∈ N} is locally finite. By part (ii), {Vn |
x0 ∈ X there is n0 ∈ N such that x ∈ Vn0 . But
¯
¾
n\
n\
0 −1 ½
0 −1 ½
¯
1
¯
Vn0 = Un0 ∩
x ∈ X ¯ ϕk (x) <
= Un 0 ∩
x∈X
n0
k=1
k=1
n ∈ N} covers X implies for
¯
¾
¯
¯ min{1, dist(x, X \ Uk )} < 1 .
¯
n0
But note that
n\
0 −1
X \ Uk ⊂
k=1
k=1
⇒
Un0 ∩
¯
¾
¯
1
x ∈ X ¯¯ min{1, dist(x, X \ Uk )} <
n0
n\
0 −1 ½
n\
0 −1
X \ Uk ⊂ Vn0 ⇒ Un0 \
k=1
n[
0 −1
Uk ⊂ Vn0 .
k=1
Moreover,
Un0 +1 \
n0
[
Uk ⊂ Vn0 +1 .
k=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 locally
finite 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 ∈ Ω
being the first element. Now for each (n, α) ∈ N × Ω define
¯
¾
¯
1
¯
Hn,α1
¯ dist(x, X \ Uα1 ) ≥ n

¯
¾ 
½
¯
1
∩ x∈X
Hn,α := x ∈ X ¯¯ dist(x, X \ Uα ) ≥

n
½
:= x ∈ X
Obviously, we have that
½
x∈X
¯
¾ [
¯
¯ dist(x, X \ Uα ) ≥ 1 \
Hn,λ ⊂ Hn,α .
¯
n
λα
Moreover, the following hold true:
62
¯



¯
[
¯
1
¯ dist x,

≥
H
n,λ
¯
n
¯
λα
(a) for each (n, α) ∈ N × Ω, B
1
2n
(Hn,α ) ⊂ Uα . To see this, let z ∈ Hn,α and x ∈ B
1
2n
(z) and w ∈ X
be any. Then
ρ(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
1
1
1
1
−
< ρ(w, x) ⇒ ρ(w, x) >
, ∀w ∈ X \ Uα ⇒ dist(x, X \ Uα ) ≥
.
n 2n
2n
2n
⇒x∈
/ X \ Uα ⇒ x ∈ Uα . Consequently, B
(b) If λ α, then dist(Hn,λ , Hn,α ) ≥
1
2n .
1
2n
(Hn,α ) ⊂ Uα .
Let z ∈ Hn,α be arbitrary and consider B 1 (z) = {x ∈
n
1
n }.
X | ρ(z, x) <
By definition of Hn,α we have that


[
[
1
dist z,
Hn,λ  ≥ ⇒ B 1 (z) ∩
Hn,λ = ∅ ⇒ B 1 (z) ∩ Hn,λ = ∅ for each λ with λ α.
n
n
n
λα
Since, B
1
2n
λα
(z) ⊂ B 1 (z) it follows that
n
B
1
2n
(z) ∩ Hn,λ = ∅ for each λ with λ α.
Note that z ∈ Hn,α was chose arbitrarily. Consequently,
dist(Hn,λ , Hn,α ) ≥
(c)
S
S
n∈N
α∈Ω Hn,α
1
, for each λ with λ α.
2n
= 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 ) ≥
1
⇒ x ∈ Hm,α1 .
m
(ii) Otherwise, by the well ordering principle, the set Ωx = {α ∈ Ω | x ∈ Uα } has a least
element, say αx . Hence, x ∈ Uαx and x ∈
/ Uλ for λ αx . The set Uαx is open implies
∃nx ∈ N : B
1
nx
(x) ⊂ Uαx ⇒ dist(x, X \ Uαx ) ≥
1
.
nx
(**)
Furthermore,


B
1
nx
(x) ∩ 
[



Hnx ,λ  = ∅. ⇒ dist x,
[
λαx
λαx
If we assume that (***) does not hold, then there is z ∈ B
some λ αx . But then
ρ(x, z) <

1
Hnx ,λ  ≥
.
nx
1
nx
(***)
(x) such that z ∈ Hnx ,λ , for
1
1
⇒ dist(x, X \ {z}) ≥
.
nx
nx
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}) ≥
1
.( That is, x ∈
/ X \ Uλ , so that x ∈ Uλ . )
nx
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=
[ [
Hn,α .
n∈N λ∈Ω
Now define
Vn,α = B
1
6n
(Hn,α ). Then it follows that X =
[ [
Vn,α ; since Hn,α ⊂ Vn,α .
(****)
n∈N α∈Ω
That is, the collection {Vn,α | (n, α) ∈ N × Ω} is an open cover of X. Moreover, if
An :=
[
Vn,α ,
α∈Ω
then A := {An | n ∈ N} is a countable open cover for X. S
By Lem. 3.9.17, there is a locally finite
S open
refinement F = {A0n | n ∈ N} of A such that A0n ⊂ An = α∈Ω Vn,α for each n ∈ N and X = n∈N A0n .
Let next
On,α := A0n ∩ Vn,α , n ∈ N, α ∈ Ω.
Then we claim
(d) O := {On,α | n ∈ N, α ∈ Ω} is a cover for X.
S
Let x ∈ X. Since X = n∈N A0n , there is n ∈ N such that
x ∈ An0 ⊂ An =
[
Vn,α ⇒ ∃n ∈ N, α ∈ Ω : x ∈ A0n and x ∈ Vn,α ⇒ ∃n ∈ N, α ∈ Ω : x ∈ On,α .
α∈Ω
Hence,
X=
[ [
On,α .
n∈N α∈Ω
(e) O := {On,α | n ∈ N, α ∈ Ω} is a locally finite open refinement of C.
Note that
On,α ⊂ Vn,α = B 1 (Hn,α ) ⊂ B 1 (Hn,α ) |{z}
⊂ Uα ⇒ On,α ⊂ Uα .
6n
2n
part (a)
Furthermore, for x ∈ X, since F is locally finite, there is a neighborhood N (x) of x such that
N (x) intersects only a finite number of elements of F ⇒
∃n0 ∈ N : N (x) ∩ A0n 6= ∅, 1 ≤ n ≤ n0 and N (x) ∩ A0n = ∅, n ≥ n0 + 1.
Moreover, for λ α it follows that dist(Vn,α , Vn,λ ) ≥
Vn,α , y ∈ Hn,λ , z ∈ Vn,λ . We estimate ρ(x, z).
Repeatdly, using the triangle inequality, we obtain
1
6n .
To verify this, let w ∈ Hn,α , x ∈
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) ⇒
64
1
1
1
−
−
≤ ρ(w, y) − ρ(w, x) − ρ(z, y) ≤ ρ(x, z)
2n 6n 6n
⇒
ρ(x, z) ≥
1
.
6n
1
Hence, dist(Vn,α , Vn,λ ) ≥ 6n
. This implies the open ball B 1 (x) can intersect only one Vn,α
8n
whenever λ α. Consequently, the neighborhood N (x) ∩ B 1 (x) intersects only at most n0
8n
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 is
a real valued function. Then the support of the function f is defined as
supp f := cl{x ∈ 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. Then
a 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 that
K ⊂ O and clO is compact.
Proposition 3.10.4. Let < X, τ > be a locally compact Hausdorff space and K be a compact subset of
X. If C = {Uα | α ∈ Ω} is an open covering of K, then there is a collection {ϕ1 , . . . , ϕn } of continuous
real-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∩
Uα .
α∈Ω
(i) If x0 ∈ K, then
¡ there is α(x
¢ 0 ) ∈ 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 that
fx0 (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 ) is
compact (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] such
that
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
½
Vx0 :=
{x ∈ O ∩ Uα(x0 ) | fx0 (x) > 0},
{x ∈ Ox0 ∩ (X \ K) | gx0 (x) > 0},
if x0 ∈ K;
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 fx10 , . . . , fxn0 , gxn+1 , . . . , gxn+m .
0
0
Set now
f=
n
X
fxk and g =
k=1
0
m
X
gxk .
0
k=n+1
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 , . . . , Vxn0 , Vxn+1 , . . . , Vxn+m }, such that
0
0
n
[
K ⊂ clO ⊂
k=1
By part (a), we have K ∩
Hence, f > 0 on K.
From this follows that
³S
m
m
[
Vxk ∪
0
= ∅ ⇒ K ⊂
0
0
k=n+1
´
k=n+1 Vxk
Vxk
Sn
k=1 Vxk0
f
≡ 1 on K.( Note that g ≡ 0 on K).
f +g
Now define
ϕk :=
fxk
0
f +g
, k = 1, . . . , n.
It follows that ϕ1 + . . . + ϕn = 1, for each k ∈ {1, . . . , n}
supp ϕk ⊂ Uα(xk ) and ϕk ≥ 0, on X.
0
66
⇒ fxk (x) > 0, ∀x ∈ K.
Definition 3.10.5 (partition of unity). Let < X, τ > be a topological space. A family of functions
{ϕλ | 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
X
ϕλ (x) = 1.
λ∈Λ
If ϕλ is continuous (or Lipschitz continuous or differentiable), for each λ ∈ Λ, then the partition is said to
be 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 unity
subordinate 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 a
locally 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 set
Wx 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 ∈ Ω
Aα := {O ∈ O | O ⊂ Uα }.
(v) Next define
½ S
Vα :=
O,
Any open set with ∅ 6= Vα ⊂ clVα ⊂ Uα ,
O∈Aα
if Aα =
6 ∅
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 locally
finite 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 X
has a partition of unity subordinate to it.
67
Proof. Let C = {Uα | α ∈ Ω} be an open cover of X. Then, by Lem. 3.10.6, there is a locally finite
open 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
fα (x)
.
λ∈Ω fα (x)
ϕα (x) := P
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 a
finite number of elements of {clOα | α ∈ Ω}. That is there is {α1 , . . . , αn(x0 ) } such that
U (x0 ) ∩ Oα = ∅, ∀α ∈ Ω \ {α1 , . . . , αn(x0 ) }
k
This implies
U (x0 ) ⊂ X \ Oα , ∀α ∈ Ω \ {α1 , . . . , αn(x0 ) } ⇒ fα (U (x0 )) = 0, ∀α ∈ Ω \ {α1 , . . . , αn(x0 ) } (3.5)
and
U (x0 ) ∩ supp fα = ∅, ∀α ∈ Ω \ {α1 , . . . , αn(x0 ) }.
(3.6)
Thus
X
λ∈Ω
n(x0 )
fα (x) =
X
k=1
X
fαk (x) +
fα (x) ⇒
λ∈Ω\{α1 ,...,αn(x0 ) }
Consequently, for x ∈ U (x0 ):
X
n(x0 )
fα (x) =
λ∈Ω
X
fαk (x), for x ∈ U (x0 ).
k=1
fα (x)
ϕα (x) = Pn(x )
.
0
f
(x)
α
k
k=1
Pn(x )
This implies, ϕα is a continuous function at x0 . Note also that k=10 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.
P
(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 locally
finite open covering C := {Uα | α ∈ Ω} of X, there is locally Lipschitzean partition of unity subordinate
to it.
k
In fact we have here U (x0 ) ∩ clOα = ∅, ∀α ∈ Ω \ {α1 , . . . , αn(x0 ) }.
68
Proof. Let {Uα | α ∈ Ω} be a locally finite open covering of X. Then, by Lem. 3.10.6, there is a locally
finite open covering {Oα | α ∈ Ω} such that clOα ⊂ Uα . Now, define
fα (x) := dist(x, X \ Oα ) and ϕα (x) := P
fα (x)
.
λ∈Ω fλ (x)
Then the following hold true :
(i) fα : X → R+ Lipschitz continuous and
supp fα = supp ϕα = cl{x ∈ 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.
Excercises 3.10.10. Prove the following statements:
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 Lindlöf space, then < X, τ > is paracompact.
4. If < X, τ > is paracompact and sparable, then < X, τ > is Lindelöf.
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
70
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 a
normed 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 the
Hausdorff distance between A and B is defined as
½
¾
h(A, B) := max sup dist(a, B), sup dist(b, A)
a∈A
b∈B
Observe that: dist(A, B) ≤ h(A, B).
Remark 4.1.6. The quantity
h∗ (A, B) := sup dist(a, B).
a∈A
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 {ρ(a, c) + dist(c, B)} ≤ inf ρ(a, c) + sup dist(c, B).
c∈C
c∈C
c∈C
⇒
dist(a, B) ≤ dist(a, C) + h∗ (C, B).
This last inequality holds true for every a ∈ A. Hence
sup dist(a, B) ≤ sup dist(a, C) + h∗ (C, B)
a∈A
a∈A
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 quasi
metric (see sec. 1.1 of chap. 1 ).
Lemma 4.1.8. For any four real numbers a, b, c, d, the following holds
max{a + b, d + e} ≤ max{a, d} + max{b, 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 a
metric 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) = sup dist(a, B) = 0 ⇒ ∀a ∈ A : dist(a, B) = 0, ( since dist(a, B) ≥ 0 for any point a and set B ).
a∈A
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 the
Hausdorff 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},
Ueε (A) := {x ∈ X | dist(x, A) ≤ ε} and clBε (b) := {x ∈ X | ρ(x, b) ≤ ε}
Lemma 4.1.13. For any two subsets A and B of a metric space X we have
h∗ (A, B) = inf{ε > 0 | A ⊂ Uε (B)}
∗
h (B, A) = inf{ε > 0 | B ⊂ Uε (A)}.
(4.1)
(4.2)
Proof. We need only to show that h∗ (A, B) = inf{ε > 0 | A ⊂ Uε (B)}.
• If h∗ (A, B) = 0, then the equality follows by using Lem. 4.1.11.
(a) Let ε0 := inf{ε > 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) ≤
inf
c∈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 dist(a, B) ≤ ε0
a∈A
Implying that
h∗ (A, B) ≤ inf{ε > 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⊂
[
clBr (b).
(4.3)
b∈B
Now, for any b ∈ B, let z ∈ Br (b). Then
dist(z, B) ≤ ρ(z, b) + dist(b, B) ⇒ dist(z, B) ≤ r ⇒ z ∈ Uer (B).
Since, z ∈ Br (b) is arbitrary, we conclude that clBr (b) ⊂ Uer (B). It then follows that,
[
clBr (b) ⊂ Uer (B).
b∈B
Hence, along with (4.3) we find that A ⊂ Uer (B). Furthermore, if we take an arbitrary n ∈ N,
we obtain
A ⊂ Uer (B) ⊂ Ur+ 1 (B).
n
⇒
r+
1
≥ inf{ε > 0 | A ⊂ Uε (B)}, ∀n ∈ N
n
Consequently,
h∗ (A, B) = r ≥ inf{ε > 0 | A ⊂ Uε (B)}.
Therefore, from (a) and (b), we conclude that
h∗ (A, B) = inf{ε > 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 sets
γA = {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 a
normed linear space takes the form
h(A, B) = inf{ε > 0 | A ⊂ B + εB and B ⊂ A + εB}.
74
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} = sup |dist(x, A) − dist(x, B)|.
x∈X
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) ≤ inf ρ(x, b) + dist(b, A) ≤ inf ρ(x, b) + sup dist(b, A) ≤ dist(x, B) + h∗ (B, A).
b∈B
b∈B
b∈B
⇒
dist(x, A) − dist(x, B) ≤ h∗ (B, A).
Similarly, we have
dist(x, B) − dist(x, A) ≤ h∗ (A, B)
Hence,
max{dist(x, A) − dist(x, B), dist(x, B) − dist(x, A)} ≤ max{h∗ (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) = sup dist(b, A)
b∈B
= sup [dist(b, A) − dist(b, B)]
b∈B
≤ sup [dist(x, A) − dist(x, B)]
x∈X
≤ sup |dist(x, A) − dist(x, B)|.
x∈X
Analogously
h∗ (A, B) ≤ sup |dist(x, A) − dist(x, B)|
x∈X
Consequently, we have
max{h∗ (B, A), h∗ (A, B)} ≤ sup |dist(x, A) − dist(x, B)|
x∈X
⇒
h(A, B) ≤ D∞ (A, B).
Which concludes the proof.
75
Excercises 4.1.18. Prove the following statements:
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) = inf{ε > 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 with
Lipschitz 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 {An }n∈N be a sequence
of subsets of X. Then
(i) the upper limit or outer limit of the sequence {An }n∈N is a subset of X given by
lim sup An = {x ∈ X | lim inf dist(x, An ) = 0};
n→∞
n→∞
(ii) the lower limit∗ or inner limit† of the sequence {An }n∈N is a subset of X given by
lim inf An = {x ∈ X | lim dist(x, An ) = 0}.
n→∞
n→∞
If lim supn→∞ An = lim inf n→∞ An , then we say that the limit of {An }n∈N exists and
lim An = lim sup An = lim inf An .
n→∞
∗
n→∞
n→∞
lim supn→∞ An and lim inf n→∞ An are sometimes called Kuratowski limit inferior and limit superior, respectively.(see
Papagorgeu)
†
The notions outer limit and inner limit are introduced by Rockafellar and Wets (see )
76
Remark 4.2.2. For a fixed set A ⊂ X, the distance function dist(·, A) : X → R is a Lipschitz continuous
function. 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 {xn }n∈N is sequence in X such that xn → x, then
lim dist(xn , A) = dist(x, A).
n→∞
With this remark the following proposition follows trivially.
Proposition 4.2.3. Let {An }n be a sequence of subsets of a metric space X. Then
(i) lim inf n→∞ An ⊂ lim supn→∞ An ;
(ii) the sets lim supn→∞ An and lim inf n→∞ An are closed in X .
Proof. (i) is trivial. To show (ii), let x ∈ cl (lim supn→∞ An ). This implies
∃{xk }k∈N ⊂ lim sup An such that xk → x.
n→∞
Consequently, for each k, we have
dist(x, An ) ≤ ρ(x, xk ) + dist(xk , An ).
⇒
lim inf dist(x, An ) ≤ ρ(x, xk ) + lim inf dist(xk , An ) ⇒ lim inf dist(x, An ) ≤ ρ(x, xk ), ∀k.
n→∞
n→∞
n→∞
⇒
lim inf dist(x, An ) ≤ inf ρ(x, xk ) = 0.
n→∞
k
Hence, lim inf n→∞ 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 {An }n∈N is a sequence in a metric space X, then
(i) lim supn→∞ An = {x ∈ X | xnk ∈ Ank : xnk → x}.
(ii) lim inf n→∞ An = {x ∈ X | xn ∈ An : xn → x}.
That is, lim inf n→∞ An is a collection of limits of sequences {xn }n∈N , where xn ∈ An ; whereas
lim supn→∞ An is a collection of cluster points of sequences {xn }n∈N , where xn ∈ An .
Proof. (i) Obviously, lim supn→∞ An ⊃ {x ∈ X | xnk ∈ Ank : xnk → x}. Thus, let x ∈ lim supn→∞ An .
Then
lim inf dist(x, An ) = 0 ⇒ sup inf dist(x, Ai ) = 0.
n→∞
n i≥n
Let k ∈ N be arbitrary, we have
sup inf dist(x, Ai ) ≤
n i≥n
1
k
77
Hence,
∃nk ∈ N : inf dist(x, Ai ) ≤
i≥nk
1
.
k
⇒
∃ik ≥ nk , ∃xik ∈ Aik : ρ(x, xik ) ≤
1
.
k
But this is true for each k ∈ N. Hence, there is {xik }ik ∈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 inf n→∞ 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 dist(x, An ) = sup inf dist(x, Ak ) 6= 0.
n→∞
n k≥n
⇒x∈
/ lim supn→∞ An . But this is a contradiction. Consequently,
x ∈ {x ∈ X | ∀ε > 0, ∀N ∈ N, ∃n ≥ N : Bε (x) ∩ An 6= ∅}
That is
lim sup An ⊂ {x ∈ X | ∀ε > 0, ∀N ∈ N, ∃n ≥ N : Bε (x) ∩ An 6= ∅}
n→∞
To show the reverse inclusion. By assumption, we have
∀k ∈ N, ∃nk ≥ k : B 1 (x) ∩ Ank 6= ∅.
k
⇒ ∃xnk ∈ Ank : ρ(x, xnk ) ≤ k1 . Consequently, the sequence {xnk }k∈N , with xnk ∈ Ank converges
to x. Hence, by Prop. 4.2.4,
x ∈ lim sup An
n→∞
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 of
inferior and superior limits of a sequence of sets as defined above. In particular, from Prop. 4.2.5, it
follows that
78
(i)
lim sup An =
n→∞
(ii)
cl 
n≥1
(iii)
Uε (An )
(4.4)
ε>0 N ≥1 n≥N

\
\ \ [

[
Am  ⊂ lim sup Am
n≥m
lim inf An =
n→∞
n→∞
\ [ \
Uε (An ).
(4.5)
(4.6)
ε>0 N ≥1 n≥N
Proposition 4.2.7. If {An }n∈N is a sequence such that An ⊃ An+1 , n ∈ N (a decreasing sequence),
then limn→∞ An exists and
\
lim An =
cl An
n→∞
n∈N
Proof. Since dist(x, clAn ) = dist(x, An ), it follows that,
\
cl An ⊂ lim inf An .
n→∞
n∈N
If we now show that lim supn→∞ An ⊂
T
n∈N cl
(4.7)
An , then we are done!
• Let x ∈ lim supn→∞ An . Hence,
sup inf dist(x, Ak ) = 0.
n k≥n
Since, for k ≥ n, An ⊃ Ak , we have dist(x, An ) ≤ dist(x, Ak ). This implies that
sup inf dist(x, Ak ) = sup dist(x, An ) = 0 ⇒ dist(x, An ) = 0, ∀n.
n k≥n
n
⇒ (by Prop. 4.1.2(iii)) x ∈ clAn , ∀n ∈ N. Hence
lim sup An ⊂
n→∞
\
cl An .
n∈N
• Moreover, by Prop. 4.2.3, we have lim inf n→∞ An ⊂ lim supn→∞ An . Hence
\
lim inf An ⊂
cl An .
n→∞
n∈N
Finally, using (4.7), limn→∞ An exists and
lim An =
n→∞
\
cl An .
n∈N
4.2.1 Calculus of Limits of Sequences of Sets
Theorem 4.2.8. Let {An }n∈N and {Bn }n∈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
µ
¶
for every neighborhood V of K ∩ lim sup Bn , ∃N ∈ N : An ∩ Bn ⊂ V, ∀n ≥ 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 sup Bn
n→∞
Prop. 4.2.5
⇒
∃ε(z) > 0, ∃N (z) ∈ N : Bε(z) (z) ∩ Bn = ∅, ∀n ≥ N.
Hence,
M⊂
[
Bε(z) (z).
z∈M
Since M is compact, there are zk ∈ M, k = 1, . . . , m, such that
M⊂
m
[
Bε(zk ) (zk ) =: W.
k=1
Now define, N0 := max{N (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 := max{N0 , 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 {An }n∈N
be a sequence in a metric space X and K ⊂ X. If, for every neighborhood U of K
∃N ∈ N : An ⊂ U, ∀n ≥ N,
then
lim sup An ⊂ cl(K).
n→∞
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 {An }n∈N and {Bn }n∈N be two sequences of subsets of a metric space X. Then
the following hold true
(i)
lim sup(An ∩ Bn ) ⊂ lim sup An ∩ lim sup Bn
(ii)
lim inf (An ∩ Bn ) ⊂ lim inf An ∩ lim inf Bn
(iii)
lim sup(An ∪ Bn ) = lim sup An ∪ lim sup Bn
(iv)
lim inf (An ∪ Bn ) ⊃ lim inf An ∪ lim inf Bn
(v)
lim sup(An × Bn ) ⊂ lim sup An × lim sup Bn
(vi)
lim inf (An × Bn ) = lim inf An × lim inf Bn .
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
Proposition 4.2.11. Let X and Y be metric spaces, {An }n∈N and {Bn }n∈N are sequence in X and Y ,
respectively. If f : X → Y is a continuous function, then the following hold true
(i)
(ii)
(iii)
(iv)
µ
¶
f lim sup An ⊂ lim sup f (An ) ;
n→∞
n→∞
³
´
f lim inf An ⊂ lim inf f (An ) ;
n→∞
n→∞
µ
¶
lim sup f −1 (Bn ) ⊂ f −1 lim sup Bn ;
n→∞
n→∞
³
´
−1
−1
lim inf f (Bn ) ⊂ f
lim inf Bn .
n→∞
n→∞
(4.8)
(4.9)
(4.10)
(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 sup f (Ank )
n→∞
Hence,
µ
¶
f lim sup Ank ⊂ lim sup f (Ank ) .
n→∞
n→∞
Excercises 4.2.12. Let {An } be a sequence of sets in a metric space.
1. Show that the set lim inf n→∞ An is a closed set.
81
2. Verify that the limit superior can be also written as

\
lim sup An =
n→∞
cl 
N ≥1

[
An  .
n≥N
3. Verify the statements of Rem. 4.2.6.
4. Prove part (ii) of Prop. 4.2.4; i.e. lim inf n→∞ An = {x ∈ X | xn ∈ An : xn → x}.
5. Prove part (ii) of Prop. 4.2.5; i.e. lim inf n→∞ 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
lim An = A
n→∞
⇔
lim dist(x, An ) = 0, ∀x ∈ A and lim inf dist(x, An ) > 0, ∀x ∈
/ A.
n→∞
n→∞
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 X
and A ⊂ X be also closed. We say that An converges to A w.r.t. the Hausdorff metric iff
lim h(An , A) = 0.
n→∞
We denote this by
h
An → 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 now
write (4.12) equivalently as
\ [ \
A⊂
Uε (Am ).
ε>0 n≥1 m≥n
Theorem 4.2.15. Let X be a metric space, {An } ∈ Pcl (X) and A ∈ Pcl (X) . Then
h
An → A
⇒
An → A
That is, Hausdorff-convergence implies Kuratowski-convergence.
Proof. We show that A ⊂ lim inf n→∞ An and lim supn→∞ An ⊂ A.
(i) By assumption
lim h(An , A) = 0 ⇒ lim h∗ (A, An ) = 0 ⇒ lim sup dist(x, An ) = 0.
n→∞
n→∞
n→∞ x∈A
⇒
∀x ∈ A : lim dist(x, An ) = 0.
n→∞
From this it follows that
A ⊂ lim inf An .
n→∞
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 let
B := {x ∈ X | dist(x, A) ≤
γ
} = clU γ (A)
2
2
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
> 0, ∀n ≥ N.
2
⇒
lim inf dist(x, An ) ≥ lim inf dist(x, B) ≥
n→∞
n→∞
γ
>0
2
⇒x∈
/ lim supn→∞ An , which is a contradiction.
Therefore, we have
lim sup distAn ⊂ A.
n→∞
Remark 4.2.16. However, the converse of Thm. 4.2.15 does not hold always true. To see this consider
the sequence
½
if n is even;
{0, n1 },
An :=
{0, n},
if n is odd;
Note that:
• xn =
1
n
∈ A2n and xn → 0 ⇒ 0 ∈ lim supn→∞ An . Moreover, for x ∈ R arbitrary, we have
(
inf z∈{0, 1 } |x − z|,
if n is even;
n
dist(x, An ) = inf ρ(x, z) = inf |x − z| =
z∈An
z∈An
inf z∈{0,n} |x − z|,
if n is odd;
⇒
½
dist(x, An ) =
min{|x|, |x − n1 |,
min{|x|, |x − n|},
if n is even;
if n is odd;
Hence,
½
lim inf dist(x, An ) = 0 ⇔
n→∞
lim inf n→∞ min{|x|, |x − n1 |,
lim inf n→∞ min{|x|, |x − n|},
if n is even;
if n is odd;
¾
= 0.
⇔
1
| = 0, if n is even; and
n→∞
n
lim inf min{|x|, |x − n|} = 0, if n is odd
lim inf min{|x|, |x −
n→∞
But these two limits are zero if and only if x = 0; i.e. lim supn→∞ An = {0}. Similarly,
lim inf n→∞ An = {0}. Consequently,
lim An = {0}.
n→∞
83
• Moreover,
h({0}, An ) = max{h∗ ({0}, An ), h∗ (An , {0})}
Thus,
h∗ ({0}, An ) = sup dist(x, An ) = dist(0, An ) = 0.
x∈{0}
But
½
∗
h (An , {0}) = sup dist(x, {0}) = sup |x − 0| = sup |x| =
x∈An
x∈An
x∈An
1
n,
n,
if n is even;
if n is odd;
⇒
lim h∗ ({0}, An ) 6= 0.
n→∞
Consequently,
lim h(An , {0}) 6= 0. That is, An 9 {0}.
n→∞
When do we have equality between Kuratowski and Hausdorff convergence? The answer is given in
the following statement.
Proposition 4.2.17. Let X be a metric space, {An } be a sequence of compact subsets of X and A ⊂ X
compact. Then, if there is a compact subset K of X such that An ⊂ K, ∀n; and limn→∞ An = A, then
limn→∞ 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.
h
Assume that h(An , A) 9 0; i.e. An 9 A. This implies
lim max{h∗ (An , A), h∗ (A, An )} 6= 0 ⇒ lim max{h∗ (An , A), h∗ (A, An )} > γ, for some γ > 0.
n→∞
n→∞
⇒
lim h∗ (A, An ) 6= 0
n→∞
or
lim h∗ (An , A) 6= 0.
n→∞
(i) If limn→∞ h∗ (A, An ) 6= 0, then
lim sup dist(x, An ) > 0
n→∞ x∈A
⇒ (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 inf n→∞ An , by Prop. 4.2.4, there is a sequence {zn } with zn ∈ An such
that zn → x. Consequently,
dist(xnk , Ank ) ≤ ρ(xnk , znk ) ≤ ρ(xnk , x) + ρ(x, znk ), ∀n ∈ N.
However, from this follows that
lim dist(xnk , Ank ) = 0
k→∞
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 {Al }l∈L of {An }n∈N , where L ⊂ N such
that
lim h∗ (Al , A) = lim sup dist(y, A) > 0.
l→∞
l→∞ y∈Al
⇒
∀l ∈ L, ∃xl ∈ Al : dist(xl , A) > 0.
But, again {xl } ⊂ K and K is compact implies there is a convergent subsequence {xlk } such
that 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.
h
Proposition 4.2.18. If {An } ⊂ Pcl (X), A ∈ Pcl (X) and An → A, then


[
[
A=
cl 
Am  .
n≥1
m≥n
h
Proof. Since An → A implies An → A, by Rem 4.2.6, we have that


\
[
cl 
Am  ⊂ A.
n≥1
m≥n
Moreover, using Rem. 4.2.14, given any n ≥ 1, for an arbitrary ε > 0,


[
∃m ≥ n : A ⊂ Uε (Am ) ⇒ A ⊂ Uε 
Am  .
m≥n
Since, ε > 0 is arbitrary, we have




[
\
[
A ⊂ cl 
Am  . This holds true for any n ≥ 1. Consequently, A ⊂
cl 
Am  .
m≥n
n≥1
m≥n
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 that
{An } is convergent. According to Prop. 4.2.18, we need only to verify that {An } converges to the set


\
[
A=
cl 
Am  .
n≥1
m≥n
h
Thus, we have to show that: (i) A is closed. (Obvious!) (ii) A is non-empty. (iii) An → A.
85
(ii) Since {An } a Cauchy sequence, given δ > 0 (say δ = 3ε , where ε > 0), for each k ≥ 0, there is Nk
such that
δ
h(An , Am ) < k+1 , ∀n, m ≥ Nk .
2
• Now, for k=0,
δ
∃N0 : h(An , Am ) < , ∀n, m ≥ N0 .
2
Then for any n0 ≥ N0
δ
δ
h(An , An0 ) < , ∀n ≥ N0 ⇒ sup dist(x, An ) < , ∀n ≥ N0 .
2
2
x∈An0
Thus, for any fixed xn0 ∈ An0
δ
dist(xn0 , An ) < , ∀n ≥ N0 .
2
• For k = 1, there exists N1 such that
h(An , Am ) <
δ
, ∀n, m ≥ N1 .
22
Then, for any n1 ≥ max{N0 , N1 } we have
dist(x, An ) <
δ
, ∀x ∈ An1 , ∀n ≥ N1 .
22
Choose xn1 ∈ An1 , then
ρ(xn1 , xn0 ) ≤ ρ(xn1 , x) + ρ(x, xn0 ),
⇒
ρ(xn1 , xn0 ) <
for any x ∈ An and n ≥ N1 .
δ
δ
δ
ε
+ = 3 2 = 2.
2
2
2
2
2
Proceeding in this way, for nk ≥ max{N0 , 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 ∈
Am , ∀nk ≥ nk0 ⇒ x ∈ cl 
m≥n

[
Am  .
m≥n
This is true for all n ≥ 1.

x ∈ cl 

[
m≥n
Consequently, A 6= ∅.
86
Am  , ∀n ≥ 1 ⇒ x ∈

\
n≥1
cl 

[
m≥n
Am  = A.
h
(iii) Next, we show that An → A.
• From (i), for each n0 ≥ N0 and any xn0 ∈ An0 we obtain, by the continuity of ρ(·, x0 ), that
ρ(x, xn0 ) = lim ρ(xnk , xn0 ) ≤ lim
nk →∞
nk →∞
nk
X
i=1
nk
X
ε
ρ(xni , xni−1 ) < lim
= ε.
nk →∞
2i
i=1
⇒
∀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


[
ε
x ∈ cl 
Am  ⇒ ∃m ≥ N0 , ∃z ∈ Am : ρ(x, z) < .
2
(4.15)
m≥N0
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 N00 such that h(An , Am ) <
along with (4.15), yields that
ε
2 , ∀n
≥ N00 . This,
ε ε
dist(x, An ) ≤ dist(x, Am )+h(An , Am ) ≤ ρ(x, z)+h(An , Am ) < + , ∀n, m ≥ max{N0 , N00 }.
2 2
⇒
dist(x, An ) < ε, ∀n ≥ N10 := max{N0 , N00 } ⇒ x ∈ Uε (An ), ∀n ≥ N10 .
Since, x ∈ A is arbitrary, it follows that
A ⊂ Uε (An ), ∀n ≥ N10 ⇒ h∗ (A, An ) < ε, ∀n ≥ N10 .
(4.16)
Finally, from (4.14) and (4.16) we conclude that
h(A, An ) < ε, ∀n ≥ N10 .
h
Therefore, An → A.
Excercises 4.2.20. Verify the following statements.
1. if limn→∞ An = A, then
lim h∗ (A, An ) = 0.
n→∞
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 inf n An ⊃ A.
3. Let X be a metric space, {An } ⊂ Pcl (X) and A ∈ Pcl (X) such that
lim h(An , A) = 0.
n→∞
If xn ∈ An and xn → x, then x ∈ A.
h
4. Let X be a metric space, {An } ⊂ Pcl (X), A ∈ Pcl (X). If An → A, then ∃{εk }, εk & 0, such that,
for each k ∈ N : dist(y, A) < εk , ∀y ∈ Ak .
h
5. If An → A, then


A=
[
n≥1
cl 
[
m≥n
Am  =
\ [ \
Uε (Am ).
ε>0 n≥1 m≥n
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) and
this 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 topological
space.
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 practically set-valued maps reveal interesting properties when X and Y are taken to be normed linear
spaces.
Definition 5.1.1 (set-valued map). Let X and Y be topological spaces. If for each x ∈ X there is a
corresponding set F (x) ⊂ Y , then F (·) is called a set-valued map from X to Y . We denote this by
→ Y.
F :X−
→
A function f : X → Y can be treated as a special set-valued map if we define F (x) := {f (x)}. For the
sake 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. But
→ Y . Furthermore, the terms ’set-valued map’ (Aubin &
we exclusively use here the notation F : X −
→
Frankowska [4]),’point-to-set map’ (Hogan [13]),’correspondences’ (Aliprantis & Boder [1]), ’multivalued 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 mathematical
problem 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 approximately solved on the computer. Thus sensitivity analysis could guarantee the acceptability of the
obtained 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 of
optimization problems, partial differential equations, etc., is done through set-valued maps. Consequently, Set-valued maps are indispensable tools in stability and sensitivity analysis of mathematical
problems. 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 function
Let the function f : R → R+ be given by f (x) = x2 . For y ∈ R+ , the inverse image of y under f
√
√
is 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 valued
instead 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 Projections
Let S ⊂ Rn and x ∈ Rn . Define the map
ΠS (x) = {y ∈ S | kx − yk = 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 for
x∈C
• the normal cone map: N (x) = {y ∈ Rn | y > (z − x) ≤ 0, ∀x ∈ C};
• the tangent cone map: TC (x) = cl{y ∈ Rn | y = α(z−x), x ∈ C, α ≥ 0} = cl
³S
1
γ>0 γ (C
´
− x) .
4. The feasible and solution sets of a Parametric Optimization Problems
Let f, hi , gj : Rn × Rm → R be functions. Then given the parametric optimization problem
(P (t))
f (x, t) → inf
hi (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) :=
inf f (x, t)
x∈M (t)
is the the (marginal) value function of (P(t)). Here, both M (·) and S(·) are set valued maps
from 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 . The
domain 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 ) :=
[
F (x);
x∈Dom(F )
and the graph of F (·), denoted by Graph(F ), is defined as
Graph(F ) := {(x, t) | t ∈ F (x), x ∈ Dom(F )}.
90
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 each
x ∈ X, F (x) is a closed, open or compact set, respectively, in Y . Furthermore, if Y is a topological
linear 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 ) is
a closed, open or compact set w.r.t. the product topology of X × Y . Furthermore, if X and Y are
topological linear spaces; then F (·) called a convex set-valued map if Graph(F ) is a convex set in
w.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. The
former 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 and
→ Y . Then
F : X−
→
• 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
→ Y, where int(F )(x) = int(F (x)), for each x ∈ X.
int(F ) : 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.
→R
Example 5.2.5. For instance, let f : R → R+ with f (x) = x2 and F (x) = f −1 (x). Then F : 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) =
F (x).
x∈A
→ Y,
Proposition 5.2.6 (Aubin & Frankowska [4]). Let X and Y be topological spaces and F : X −
→
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!
91
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
→ Y given by (F ∪ F )(x) = F (x) ∪ F (x), for
• the union map of F1 and F2 is the map F1 ∪ F2 : X −
→
1
2
1
2
each 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.
→ Y and F : X →
−
→ Y ×Z
• If F1 : X −
→
→ Z, then the production map of F1 and F2 the map F1 × F2 : X −
→
2
is 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).
→ Y and F : Y −
→ Z, then composition map of F and F is the map
Furthermore, if F1 : X −
→
→
2
1
2
→Z
F2 ◦ F1 : X −
→
such that
(F2 ◦ F1 )(x) =
[
F2 (y).
y∈F1 (x)
→ Y . For any V ⊂ Y the (lower) inverse
Definition 5.2.8 (lower inverse of a SV-map). Let F : X −
→
−
image of V under F (·) is denoted by F (V ) and is defined as:
[
F − (V ) := {x ∈ X | F (x) ∩ V 6= ∅} =
F − (y).
y∈V
−
Definition 5.2.9 (upper inverse of a SV-map). † Let F : X →
→ Y . Then for any V ⊂ Y , the upper inverse
+1
of V under F (·), denoted by F (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
→ Y and V ⊂ Y any, then
Proposition 5.2.10. If F : X −
→
F + (V ) ⊂ F − (V ).
The above two definitions of inverses of a SV-map lead into two types of continuities - upper and lower
semi-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 inverse
image’ and ’strong inverse image’, from Hu & Papageorgiou [14], could be more appropriate instead of ’upper-’ and
’lower inverse’, respectively.
92
→ Y and V, W ⊂ Y . Then verify the validity of the following statements.
Excercises 5.2.12. Let F : X −
→
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;
→ Y and S ⊂ Y . Then
3. Let F1 , F2 : X −
→
(i) (F1 ∪ F2 )− (S) = F1− (S) ∪ F2−1 (S);
(ii) (F1 ∩ F2 )− (S) ⊂ F1− (S) ∩ F2− (S);
(iii) (F1 ∪ F2 )+ (S) ⊂ F1+ (S) ∪ F2+ (S);
(iv) (F1 ∩ F2 )+ (S) ⊃ F1+ (S) ∩ F2+ (S);
5.3 Semi-Continuity of Set-Valued Maps
Assume that X and Y are Hausdorff topological spaces.
→ Y and Dom(F ) 6= ∅. Then F (·) is
Definition 5.3.1 (upper semi-continuous SV-map). Let F : X −
→
said to be upper semi-continuous (u.s.c) at x0 ∈ X iff for any open set V ⊂ Y , where F (x0 ) ⊂ V , there
exists 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.
→ Y . Then F (·) is said to be lower
Definition 5.3.2 (lower semi-continuous SV-map). Let F : X −
→
0
semi-continuous (l.s.c.) at x ∈ X iff for any open set V ⊂ Y such that F (x0 ) ∩ V 6= ∅, there exists a
neighborhood 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
→ R given by
1. The set valued map F : R −
→
½
F (x) =
[1, 4],
[2, 3],
if x = 0
if x 6= 0.
is upper semi-continuous at x = 0, but not lower semi-continuous x = 0. To see the upper semicontinuity 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
½
F (x) =
[1, 4],
[2, 3],
if x < 0
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.
→ Y and Dom(F ) = X. For any subset W ⊂ Y we have
Lemma 5.3.4. Let F : X −
→
(i)
X \ F − (W ) = F + (Y \ W ).
and
(ii) X \ F + (W ) = F − (Y \ W ).
A. Properties of Upper Semi-Continuous Set-Valued Maps
→ Y and Dom(F ) = X. Then the following statements are equivalent
Proposition 5.3.5. Let F : X −
→
(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 open
in X. Thus, by Lem. 5.3.4, we have X \ F − (W ) is an open set in X. Hence, F − (W ) is a closed
set in X.
94
(iii) ⇒ (i): Let x0 ∈ X be such that F (x0 ) ⊂ V for some open set V ⊂ Y . Hence, Y \ V is closed in
Y and F (x0 ) ∩ (Y \ V ) = ∅. Using (iii) and Lem. 5.3.4, we have X \ F + (V ) is closed in X and
x0 ∈
/ (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 .
→ Y and x ∈ X. Then the following are
Proposition 5.3.6. Let X and Y be metric spaces and F : X −
→
0
equivalent
(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 that
F (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.
→ Y and x ∈ X. If F (·) is u.s.c. and {x } is
Corollary 5.3.7. Let X and Y be metric spaces and F : X −
→
0
n
a sequence such that xn → x0 , then
lim sup F (xn ) ⊂ clF (x0 ).
n→∞
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 of
sv-maps.
→ Y . If F (·) is u.s.c. and Y is a normal topological space, then clF (·) is
Proposition 5.3.8. Let F : X −
→
u.s.c.
→ Y and G : Y −
→ Z. If F and G are upper
Proposition 5.3.9. Let X, Y, Z be topological spaces, F : X −
→
→
semi-continuous, then G ◦ F is u.s.c.
−
→ Y . Then
Proposition 5.3.10. Let F1 : X →
→ Y and F2 : X −
→
(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.
→ Y, F : X −
→ Y be compact valued and Y be a topological linear
Proposition 5.3.11. Let F1 : X −
→
→
2
space. If F1 and F2 are u.s.c., then F1 + F2 is also u.s.c.
→ Y and Y be a complete normed linear space. If F (·) is u.s.c. and
Proposition 5.3.12. Let F : X −
→
compact valued, then cl(coF (·)) is u.s.c.
95
B. Properties of Lower Semi-Continuous of Set-Valued Maps
→ Y . If F (·) is an open map, then F (·) is l.s.c.
Proposition 5.3.13. Let F : X −
→
Proof. Exercise!
→ Y and Dom(F ) = X. Then the following statements are equivalent
Proposition 5.3.14. Let F : X −
→
(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 .
→ Y and x ∈ X. Then the following are
Proposition 5.3.15. Let X and Y be metric spaces and F : X −
→
0
equivalent
(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 {yn }
with 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 that
yn ∈ 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 (x0 ) : F (xn ) ∩ V = ∅.
n
This implies, if y0 ∈ F (x0 ) ∩ V , there is not sequence no sequence {yn } such that yn ∈ F (xn ) and
yn → 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-continuous
set-valued map on metric spaces. As such the terminology ’open set valued map’ refers to lower
semi-continuous maps. (see for instance Shimizu et. al. [24] ).
→ Y and x ∈ X. If F (·) is l.s.c. at x , then
Corollary 5.3.16. Let X and Y be metric spaces and F : X −
→
0
0
for every sequence xn → x0 we have
F (x0 ) ⊂ lim inf F (xn ).
n→∞
Proof. Follows from Prop. 5.3.15.
Similarly, we have lower semi-continuity properties for combination and composition of sv-maps.
→ Y . If F (·) is l.s.c. , then clF (·) is l.s.c.
Proposition 5.3.17. Let F : X −
→
→ Y and Y be a topological linear space. If F (·) is l.s.c. and compact
Proposition 5.3.18. Let F : X −
→
valued, then both coF (·) and cl(coF (·)) are l.s.c.
→ Y and G : Y −
→ Z. If F and G are lower
Proposition 5.3.19. Let X, Y, Z be topological spaces, F : X −
→
→
semi-continuous, then G ◦ F is l.s.c.
−
→ Y . Then
Proposition 5.3.20. Let F1 : X →
→ Y and F2 : X −
→
(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.
→Y,F :X →
−
Proposition 5.3.21. Let F1 : X −
→
→ Y and Y be a topological linear space. If F1 and F2 are
2
l.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,
→ Y , x0 ∈ X and define
F : X−
→
lim sup F (x) = {y | ∃xn → x0 , ∃y n → y, y n ∈ F (xn )}
(5.1)
lim inf F (x) = {y | ∀xn → x0 , ∃y n → y, y n ∈ F (xn )}.
(5.2)
x→x0
x→x0
→ Y is said to be
Then , F : X −
→
(i)
outer semi-continuous at x0 if
lim sup F (x) ⊂ F (x0 );
(5.3)
x→x0
(ii)
inner semi-continuous at x0 if
lim inf F (x) ⊃ F (x0 ).
x→x0
(5.4)
Next, we would like to find out relations between inner and lower semi-continuity; as well as between
outer and upper semi-continuity.
→ Y . Then F (·) is inner semi-continuous
Proposition 5.3.23. Let X and Y be metric spaces and F : X −
→
if 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 if
and 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 , y n ) ∈ Graph(F ) such that (xn , y n ) →
(x0 , y 0 ). This implies, xn → x0 , y n → y 0 and y n ∈ F (xn ). But, since F (·) is outer semicontinuous at x0 , we have y 0 ∈ F (x0 ). Consequently, (x0 , y 0 ) ∈ Graph(F ).
”⇐”: Suppose Graph(F ) is a closed set in X × Y . This implies, if there is a sequence (xn , y n ) ∈
Graph(F ) such that (xn , y n ) → (x0 , y 0 ), then (x0 , y 0 ) ∈ Graph(F ); i.e. y 0 ∈ F (x0 ).
→ Y be closed valued. If F (·) is upper
Proposition 5.3.25. Let X and Y be metric spaces and F : X −
→
semi-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 , ∀y n → y 0 , y n ∈ F (xn ), but y 0 ∈
/ F (x0 ).
According to Cor. 5.3.7, for any xn → x0 and F (·) is u.s.c, we have
lim sup F (xn ) ⊂ clF (x0 ) = F (x0 ).
n
And, by Prop. 4.2.4, we have
y 0 ∈ lim sup F (xn ) ⊂ F (x0 ).
n
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 valued
and u.s.c., then F (·) is a closed map.
98
Proof. Prop. 5.3.25 implies that F (·) is outer semi-continuous and, Prop. 5.3.24, yields that F (·) is a
closed map.
Remark 5.3.27. Cor. 5.3.26 implies that, an upper semi-continuous closed valued map has a closed
graph. However, the converse is not always true; i.e. the closedness of F (·) may not imply its upper
semi-continuity, even if F (·) is compact valued. In other words, there is an outer semi-continuous set
valued 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
½
{0, x1 },
if x > 0;
F (x) :=
{0},
if x = 0.
Observe that, F (·) is compact valued.
• F (·) is outer semi-continuous. To see this, let
xn =
1
1
and y n = , then y n ∈ F (xn ).
n
n
In addition
xn → 0, y n → 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 a
neighborhood of x0 = 0 in R+ . Now take δ ∈ [0, δ) with
½
¾
1
0 < δ < min
, δ, 1 .
ε
It follows that
¾
½
1
1
F (δ) = 0,
* (−ε, ε), since > ε.
δ
δ
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.
→ Y . If F (·) is a
Proposition 5.3.29. Let X be a metric space, Y be a compact metric space and F : X −
→
closed 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 1 (x0 ), ∃y n ∈ F (xn ) : y n ∈
/ V.
n
(5.5)
Hence, xn → x0 and, by the compactness of Y , there is a subsequence y nk such that y nk → y 0 ∈ Y .
This implies,
(xnk , y nk ) → (x0 , y 0 ) and y nk ∈ F (xnk ).
Since, F (·) is a closed map, we have y 0 ∈ F (x0 ) ⊂ V . Since, V is an open set and y nk → y 0 , there is
kn0 > 0 such that
y nk ∈ 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.
→ Y . If F (·) is u.s.c.
Proposition 5.3.30 (Kisielewicz[15]). Let X and Y be metric spaces and F : X −
→
and compact valued on X, then, for every compact set K ⊂ X, F (K) is a compact set in Y .
Proof. Let
S {Vα | α ∈ Ω} be an open covering of F (K). Let x ∈ K be any, then F (x) is compact and
F (x) ⊂ α∈Ω Vα . Hence, there is sub-covering {Vαk | k = 1, . . . , n(x)} such that
n(x)
F (x) ⊂
[
Vαk := Vx and Vx is an open set.
k=1
By Prop. 5.3.5, F + (Vx ) is an open set. Since, x ∈ K is arbitrary, it follows that
[
K⊂
F + (Vx ) and K is a compact set.
x∈K
This implies there are x1 , . . . , xm ∈ K such that
K⊂
m
[
F + (Vxi ).
i=1
Hence, using Prop. 5.2.6 and Exer. 5.2.12(2(ii)), we obtain that
Ãm
!
m
m
[
[
[
F (K) ⊂ F
F + (Vxi ) =
F (F + (Vxi )) ⊂
Vxi
i=1
i=1
i=1
Since, each Vxi is a finite union of elements of {Vα | α ∈ Ω}, then {Vα | α ∈ Ω} has a finite sub-cover
for F (K). Consequently, F (K) is compact.
→ Y be compact
Proposition 5.3.31 (Kisielewicz[15]). Let X and Y be metric spaces and F : X −
→
n
valued. Then F (·) is u.s.c. at each x ∈ X if and only if, for every sequence {xn }, x → x and every
sequence {y n }, y n ∈ 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 y n ∈ 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 {y n } ⊂
F (K). Consequently, there is a subsequence {y nk } of {y n } such that y nk → y for some y ∈ Y .
This implies, by Prop. 4.2.4, we have
y ∈ lim sup F (xn )
n
Moreover, by Cor. 5.3.7, we have
lim sup F (xn ) ⊂ clF (x) = F (x), by the compactness of F (x).
n
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?).
100
” ⇐ ” : Assume that F (·) is not u.s.c. at x0 . This implies there is an open set V , F (x0 ) ⊂ V such
that
∀n ∈ N : ∃xn ∈ B 1 (x0 ) : F (xn ) * V ⇒ ∀n ∈ N : ∃xn ∈ B 1 (x0 ), ∃y n ∈ F (xn ) such that y n ∈
/ V.
n
n
Hence, xn → x0 and y n ∈ F (xn ) and y n ∈ Y \ V . Consequently, for any convergent subsequence
y nk with limit y, y ∈ Y \ V ; i.e. y ∈
/ F (x). But this contradicts the assumption. Hence, F (·)
0
should be u.s.c. at x .
Once again, Prop. 5.3.31 indicates the equivalence of outer and upper semi-continuity for compact
valued 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 . Then
0
0
F (·) is called locally uniformly bounded at x ∈ X iff there is a neighborhood U (x ) of x0 such that
the set
[
F (x)
x∈U (x0 )
is a bounded set in Y . And F (·) is called locally uniformly bounded iff it is locally uniformly bounded at
every x ∈ X.
If Y is a finite dimensional or compact metric space, then F (·) is locally uniformly bounded implies
that, for each x ∈ X, there is a neighborhood U (x) of x such that


[
F (z)
cl 
z∈U (x)
is bounded - thus, compact in Y . Thus, in some literature we find such a term like locally uniformly
compactness 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.
→ Y . If F (·) is u.s.c. and compact
Proposition 5.3.33. Let X and Y be metric spaces and F : X −
→
valued, 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 and
F (·) is u.s.c. at x0 imply that
[
∃U (x0 ) :
F (x) ⊂ V.
x∈U (x0 )
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 and
→ Y be closed valued. If F (·) is closed and locally uniformly bounded, then Then F (·) is u.s.c.
F : X−
→
Proof. Given F (·) is a closed and locally uniformly bounded 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 1 (x0 ), ∃y n ∈ F (xn ) : y n ∈
/ V.
n
101
Hence, xn → x0 and y n ∈ F (xn ). By the local uniform boundedness,
[
∃U (x0 ) :
F (x) is bounded,
x∈U (x0 )
and there is N ∈ N such that
xn ∈ U (x0 ), ∀n ≥ N ⇒ y n ∈ F (xn ) ⊂
[
F (x), ∀n ≥ N.
x∈U (x0 )
Consequently, {y n | n ≥ N } is bounded; hence, there is {y nk } such that y nk → y 0 ∈ Y . . . (The rest of
the proof is as in Prop. 5.3.29).
Local uniform boundedness property are useful in characterizing upper semi-continuity of set-valued
maps with given structure.
5.3.3 Hausdorff Continuity
→Y.
In this section we assume that X and Y are normed spaces and F : X −
→
Definition 5.3.35 (ε-upper semi-continuity). The map F (·) is said to be u.s.c. at x0 ∈ X in the ε sense
if, 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 sense
if, 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., respectively.
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!
102
The converse of Prop. 5.3.39 is not always true.
→ R2 given by
Example 5.3.40 (see Aubin & Cellina[3]). Let F : R −
→
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
p
¡
¢
¡
¢
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 = 2ε , then
∀x : |x − x0 | < δ, F (x) ⊂ F (x0 ) + Bε .
That is, F (·) is H-u.s.c. However, if V = {(x, y) | |y| < x1 }, 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 outer
semi-continuous.
Proof. Let (xn , y n ) ∈ Graph(F ) such that (xn , y n ) → (x0 , y 0 ). We want to show that y 0 ∈ F (x0 ); i.e.
(x0 , y 0 ) ∈ Graph(F ). Then we have xn → x0 and y n → y 0 . 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 ⇒ lim h∗ (F (xn ), F (x0 )) = 0.
n→∞
This implies
lim dist(y n , F (x0 )) = 0.
n→∞
Since dist(·, F (x0 )) is a continuous function and F (x0 ) is a closed set, we obtain
0 = lim dist(y n , F (x0 )) = dist(y 0 , F (x0 )) ⇒ y 0 ∈ clF (x0 ) = F (x0 ).
n→∞
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. (see
example 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. Let
x0 ∈ 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
[
F (x) ⊂ F (x0 ) + Bε .
x∈Bδ (x0 )
Consequently, F (·) is locally uniformly bounded and F (·) is compact valued. Therefore, by Prop.
5.3.34, F (·) is u.s.c.
103
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 ) ⊂
[
B 2ε (y) ⇒ ∃y1 , . . . , ym : F (x0 ) ⊂
m
[
B 2ε (yk ).
k=1
y∈F (x0 )
Hence, F (x0 ) ∩ B 2ε (yk ) 6= ∅. By the lower semi-continuity of F (·), for each k ∈ {1, . . . , m}
∃Uk (x0 ) : F (x) ∩ B 2ε (yk ) 6= ∅, ∀x ∈ Uk (x0 ).
Then
m
\
Uk (x0 ) is an open set. Thus, ∃δ > 0 : Bδ (x0 ) ⊂
m
\
Uk (x0 ).
k=1
k=1
⇒
∀x ∈ Bδ (x0 ) : F (x) ∩ B 2ε (yk ) 6= ∅, ∀k ∈ {1, . . . , m}.
(5.6)
But
F (x) ∩ B 2ε (yk ) 6= ∅ ⇒ yk ∈ F (x) + B 2ε ⇒ B 2ε (yk ) ⊂ F (x) + Bε (Verify!)
Hence, from (5.6), we have
0
m
[
0
∀x ∈ Bδ (x ) : B (yk ) ⊂ F (x) + Bε , ∀k ∈ {1, . . . , m} ⇒ ∀x ∈ Bδ (x ) :
ε
2
B 2ε (yk ) ⊂ F (x) + Bε .
k=1
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 every
h
convergent sequence {xn }, xn → x0 , we have F (xn ) → F (x0 ).
Proof. Exercise!
Excercises 5.3.46. Prove the following statements
→ Y is u.s.c., then the set {x ∈ X | F (x) = ∅} is open in X.
1. If F : X −
→
→ Y is l.s.c., then the set {x ∈ X | F (x) = ∅} is closed in X.
2. If F : X −
→
104
−
3. For Fk : X →
→ Y, k ∈ N. If for each x ∈ X
\
Fk (x) 6= ∅,
k∈N
then, for any set A ⊂ Y ,
Ã
\
!−
Fk
(A) =
k∈N
\
Fk− (A).
k∈N
→ S given by
4. Let X be a metric space and S ⊂ X, S 6= ∅. The metric projection on to S is PS : X −
→
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.
→ Y , where X = [0, 1] and Y = R given by
5. Let F : X −
→
½
[0, 1],
if 0 ≤ x ≤ 1;
F (x) =
[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.
→Y.
7. Let X and Y be normed spaces and F : X −
→
(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.
→ Y and F (·) is compact valued. Then F (·) is continuous
8. Let X and Y be normed spaces, F : X −
→
if and only if F (·) is H-continuous.
9. Prove or disprove the converse of Prop. 5.3.45.
→ Y . Then prove that
10. Let X and Y be normed spaces, F1 , F2 : X −
→
(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 parametric 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 topological
properties of their defining functions. As such, one obtains u.s.c property under weaker assumptions,
while the l.s.c. requires stronger ones.
105
→ T and M : X −
→ Y with structures:
Of interest are set-valued maps F : X −
→
→
F (x) := {t ∈ T | gj (x, t) = 0, j ∈ J; hi (x, t) ≤ 0, i ∈ I};
and
M (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 continuous
functions.
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 closed
set 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 gj0 s and h0i s 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 ∈ J}
and the functions hi , i ∈ I; gj , j ∈ J are continuous. If F (·) is locally uniformly bounded, then F (·) is
u.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 under
somehow weaker assumptions. However, to guarantee the lower semi-continuity we need regularity
conditions, like Metric regularity and constraint qualifications, etc.
Definition 5.4.4 (Slater Constraint Qualification(SCQ)). Let, for each i ∈ I = {1, . . . , m}, the function
hi : Rn × Rm → R be continuous, x0 ∈ Rn and hi (x0 , ·) be convex w.r.t. t ∈ Rm . Then the Slater
Constraint Qualification is said to be satisfied at x0 if there is t∗ ∈ Rm such that
hi (x0 , t∗ ) < 0, ∀i ∈ I.
106
Proposition 5.4.5. Let, for each i ∈ I = {1, . . . , m}, the function hi : Rn × Rm → R, hi (·, t) : Rn → R
be 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 and
F (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 that
hi (x0 , t∗ ) < 0, ∀i ∈ I.
Since a convex function on Rm is continuous, we can find an open neighborhood Ve (t∗ ) such that
hi (x0 , t) < 0, ∀i ∈ I, ∀t ∈ Ve (t∗ ).
⇒ ∀t ∈ Ve (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 h0i s, 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 continuously 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 > ∇t hi (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 semicontinuous 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 > ∇t hi (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), we
have (w.l.o.g.) that
hi (x0 , t0 + λξ) ≤ 0, ∀λ ∈ (0, λi0 ),
107
ei (x0 ) such that
for each i ∈ I(x0 , t0 ). Since, the h0i s are continuous, there is U
ei (x0 ), ∀λ ∈ (0, λi0 ),
hi (x, t0 + λξ) ≤ 0, ∀x ∈ U
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 )
ei (x0 ) ∩
U
\
Ui (x0 )
and
\
V (t0 ) =
i∈I\I(x0 ,t0 )
Vi (t0 ).
i∈I\I(x0 ,t0 )
Then, for a sufficiently small λ0 (say λ0 ≤ min{λi0 | 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 + λn ξ ∈ 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 , y 0 ) ∈ Rn × Rq ,
the function G : Rn × Rq × Rm → R be continuous, and G(x, ·, t) differentiable w.r.t. y and ∇y G(·, ·, t)
is continuous for each t ∈ F (x). Then the extended Mangasarian-Fromowitz constraint qualification
(EMFCQ) is said to be satisfied at (x0 , y 0 ) if there exists a vector ξ ∈ Rq such that
ξ > ∇y G(x0 , y 0 , t) < 0, ∀t ∈ E(x0 , y 0 ).
Proposition 5.4.9. Let F (·) be u.s.c. and compact valued and y 0 ∈ M (x0 ). Then if (EMFCQ) is satisfied
at (x0 , y 0 ), then M (·) is l.s.c. at x0 .
Proof. Let y 0 ∈ M (x0 ) be arbitrary, V (y 0 ) be any neighborhood of y 0 and EMFCQ be satisfied w.r.t. y
at (x0 , y 0 ). Then there exists ξ ∈ Rq such that
∇y G(x0 , y 0 , t)ξ < 0, ∀t ∈ E(x0 , y 0 ).
For a fixed tl ∈ E(x0 , y 0 ), using Taylor’s theorem, there is λl0 := λ0 (tl ):
G(x0 , y 0 + λξ, tl ) = G(x0 , y 0 , tl ) + λ∇y G(x0 , y 0 , tl ) + o(λ), ∀λ ∈ (0, λl0 ).
⇒ (w.o.l.g.)
G(x0 , y 0 + λξ, tl ) ≤ 0, ∀λ ∈ (0, λl0 ).
108
(i) By the continuity of G(·, ·, ·), there are neighborhoods U1l (x0 ) and W1 (tl ) (with an appropriate λl0
)such that
G(x, y 0 + λξ, t) ≤ 0, ∀λ ∈ (0, λl0 ), ∀x ∈ U1l (x0 ), ∀t ∈ W1 (tl ).
(ii) Moreover, for each t ∈ F (x0 ) \ E(x0 , y 0 )
G(x0 , y 0 , t) < 0.
Thus, (as above) for each fixed t ∈ F (x0 ) \ E(x0 , y 0 ), there are neighborhoods U2t (x0 ), W2 (t) and
V t (y 0 ) such that
G(x, y, t) < 0, ∀x ∈ U2t (x0 ), ∀y ∈ V t (y 0 ), ∀t ∈ W2 (t).
The family {W1 (tl ), W2 (t) | tl ∈ E(x0 , y 0 ), t ∈ B(x0 ) \ E(x0 , y 0 )} forms an open covering of F (x0 ).
l
By assumption, F (x0 ) is a compact set. Hence, there is a finite sub-covering {W (t ) | l = 1, . . . , p} of
F (x0 ). Moreover, corresponding to this finite sub-covering we can find
T
• neighborhoods {U l (x0 ) | l = 1, . . . , p} of x0 , so that U (x0 ) := pl=1 U l (x0 ) is a neighborhood of x0 .
l
• a sufficiently small λ0 (say λ0 := min{λ0 | l = 1, . . . , p}) is such a way that y 0 + λξ ∈ V (y 0 ), ∀λ ∈
(0, λ0 ),
so that
G(x, y 0 + λξ, t) ≤ 0, ∀λ ∈ (0, λ0 ), ∀x ∈ U (x0 ), ∀t ∈ B(x0 ) ⊂
p
[
l
W (t ).
(5.7)
l=1
e (x0 ) such that
The map F (·) is u.s.c. This implies, there is a neighborhood U
F (x) ⊂
p
[
l
e (x0 ).
W (t ), ∀x ∈ U
l=1
Using this with (5.7) we obtain
e (x0 ), ∀t ∈ B(x).
G(x, y 0 + λξ, t) ≤ 0, ∀λ ∈ (0, λ0 ), ∀x ∈ U (x0 ) ∩ U
⇒
e (x0 ), ∀λ ∈ (0, λ0 ) : y 0 + λξ ∈ M (x).
∀x ∈ U (x0 ) ∩ U
Hence,
e (x0 ), M (x) ∩ V (y 0 ) 6= ∅.
∀x ∈ U (x0 ) ∩ U
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 ;
and
F (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.5.3.34
F (·) 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 to
hold 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 be
a 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 a
σ-algebra if
(i) ∅, Ω ∈ F;
(ii) A ∈ F ⇒ Ω \ A ∈ F;
(iii) If {Ak }k∈N ⊂ F is any countable collection, then
[
Ak ∈ F.
k∈N
Definition 6.1.2 (a measurable space). A measurable space (Ω, F) is a non-empty set Ω along with a
σ−algebra F defined on Ω.
Let X be a complete metric space. Then smallest σ-algebra containing all open sets in X is called the
Borel σ-algebra on X denoted by B(X). The measurable space (X, B(X)) is also called the Borel
measurable 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 map
→ Y is said to be measurable (or F-measurable) on X if, for every open set O ⊂ Y , F − (O) is
F : Ω−
→
measurable; i.e. F − (O) ∈ F.
→Y,
Hence, for F : Ω −
→
• 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. This
follows from the fact that, for any open set O ⊂ Y , we have
½
Ω,
if O ∩ B 6= ∅;
−
F (O) =
∅,
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 be
Borel-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).
→ Y is measurable if and only of the distance function
Proposition 6.1.5. A set valued map F : X −
→
ϕ(x) := dist(y, F (x)),
ϕ : X → R+ is measurable, for each fixed y ∈ Rm .
→ Y be a closed valued map and Y be a
Proposition 6.1.6. Let (Ω, F) be a measurable space, F : Ω −
→
separable 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 countable
family of closed sets {Cn | n ∈ N} such that
[
O=
Cn
n∈N
Thus, by a property of F − we have that
Ã
F − (O) = F −
[
!
Cn
n∈N
=
[
F − (Cn ).
n∈N
But, for each n ∈ N, F − (Cn ) is measurable. Consequently, F − (O) is a countable union of measurable
sets 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
¾
½
1
.
On := y ∈ Y | dist(y, K) <
n
Then, it is easy to verify that On is open in Y , clOn is compact and clOn+1 ⊂ On .
Claim:
\
F − (K) =
F − (On ) .
n∈N
F − (K).
Let x ∈
This implies, F (x) ∩ K 6= ∅; i.e. ∃y ∈ F (x) ∩ K. Hence, dist(y, K) = 0 so that
y ∈ On , ∀n ∈ N . From this follows that
\
y ∈ F (x) ∩ On , ∀n ∈ N ⇒ x ∈ F − (On ), ∀n ∈ N ⇒ x ∈
F − (On ) .
n∈N
Consequently, F − (K) ⊂
T
n∈N F
− (O ).
n
Conversely, let z ∈
T
n∈N F
− (O ).
n
This implies
z ∈ F − (On ) , ∀n ∈ N ⇒ F (z) ∩ On 6= ∅, ∀n ∈ N.
Hence,
∀n ∈ N, ∃yn ∈ F (z) ∩ On ⇒ lim dist(yn , K) = 0.
n→∞
112
Since, {yn }n∈N ⊂ clO1 and clO1 is compact, there is a subsequence {ynk } such that ynk → y, for some
y ∈ Y . Then, using the continuity of the distance function, we find that
dist(y, K) = lim = dist(ynk , K) = dist(yn , K) = 0.
k→∞
Which implies y ∈ clK = K. Since, F (z) is closed and {yn } ⊂ F (z), we also have y ∈ F (z). From this
follows that
y ∈ F (z) ∩ K ⇒ z ∈ F − (K).
Hence,
\
F − (On ) ⊂ F − (K).
n∈N
Consequently,
F − (K) =
\
F − (On ) .
n∈N
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⊂
[
Kn ⇒ C =
n∈N
[
(C ∩ Kn ) .
n∈N
But, for each n ∈ N, C ∩ Kn is compact. Hence, by assumption F − (Cn ∩ Kn ) is measurable. Hence,
F − (C) =
[
F − (C ∩ Kn )
n∈N
is measurable.
−
Corollary 6.1.7. Let F : Ω →
→ Y be a closed valued map and Y be a separable metric space. If F (·) is
measurable, 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.
→ Y and Dom(F ) = X.
Proposition 6.1.8. Let X be a complete and Y a separable metric spaces, 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.
113
6.1.1 Operations with Measurable Set-Valued Maps
→ Y is measurable if and only if clF (·) is
Proposition 6.1.9 (measurability of closure). The map F : Ω −
→
measurable.
Proof. Use the fact that for any set B and an open set O we have
B ∩ O 6= ∅ ⇔ clB ∩ O 6= ∅.
→ Y and G : Ω →
−
Corollary 6.1.10. Let F : 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.
→ Y, k ∈ N, are measurable maps, then the union map
Proposition 6.1.11. If Fk : Ω −
→
F (ω) :=
[
Fk (ω), for ω ∈ Ω
k∈N
is measurable.
Proof. Follows from the fact that
Ã
[
!−
Fk
(O) =
[
Fk− (O)
k∈N
k∈N
for an open set O ⊂ Y .
→ Y is measurable
Proposition 6.1.12 (measurability of the intersection). If, for each α ∈ Θ, Fα : Ω −
→
maps, then the intersection map
\
F (ω) :=
Fα (ω), for ω ∈ Ω
α∈Θ
is measurable.
Proposition 6.1.13. Let Y be a normed topological space.
→ Y and F : Ω −
→ Y are measurable, then F + F is measurable.
(i) If F1 : Ω −
→
→
2
1
2
→ Y is measurable and γ ∈ R, then (γF )(·) is measurable; where
(ii) If F1 : Ω −
→
(γF )(x) = γF (x), for x ∈ Ω.
→ Rn . If G(·) is a measurable, then the convex hull map co G(·) is
Proposition 6.1.14. Let G : Ω −
→
measurable.
114
Proof. Define the set
(
Λ=
(λ1 , λ2 , . . . , λn+1 ) ∈
Qn+1
+
¯
)
¯ n+1
¯ X
λk = 1 ,
¯
¯
k=1
where Q+ represents the set of non-negative rational numbers. For λ := (λ1 , λ2 , . . . , λn+1 ) ∈ Λ let
Fλ (ω) :=
n+1
X
λk F (ω).
k=1
Then, by Prop. 6.1.13, for each λ ∈ Λ, Fλ (·) is measurable. Furthermore, the countability of Λ and
Lem. 6.1.10 imply that
[
F (ω) =
Fλ (ω)
λ∈Λ
is measurable. Since
F (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 measurable
set-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) = F1− (πY (O)) ∩ F2− (πZ (O));
where πY : Y × Z → Y and πZ : Y × Z → Z are projections maps onto Y and Z, respectively.
→ Y is
Proposition 6.1.16 (measurability of composition). Let X, Y and Z be metric spaces. If F : X −
→
→ Z is u.s.c., then G ◦ F is also measurable.
closed valued and measurable G : Y −
→
Remark 6.1.17. In Prop. 6.1.16, if the map F2 (·) fails to be u.s.c., then measurability of the composition
is not guaranteed.
6.2 Measurable Selections
Definition 6.2.1 (measurable selection). Let (Ω, F) be a measurable space, X a metric space and
→
F : 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 subset
of X. If A ⊂ X, A 6= ∅, then for each fixed n ∈ N
[
A∩
B 1 (xk ) 6= ∅.
k∈N
n
Proposition 6.2.3 (Kuratowski-Ryll-Nardzewski selection Theorem, see Kisielewicz). If (Ω, F) be a
→ X and Dom(F ) = X is a closed
measurable space, X be a complete separable metric space, F : Ω −
→
set-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) ∩
clB 1 (xk ) 6= ∅
k∈N
n+1
for each fixed n ∈ N. This implies,
∃k ∈ N : F (ω) ∩ B
1
n+1
(xk ) 6= ∅.
Let
kn (ω) := min{k ∈ 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)) ≤
1
→0
n
and Fn (ω) is a closed set. Consequently, using Prop. 2.4.10, we have
\
\
Fn (ω) contains a single element; say {f (ω)} =
Fn (ω) ⊂ F (ω).
n∈N
n∈N
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 a
closed set. Then
{ω ∈ Ω | Fn+1 (ω) ∩ C 6= ∅} = {ω ∈ Ω | Fn (ω) ∩ clBn+1 (xkn (ω) ) ∩ C 6= ∅}
[
({ω | Fn (ω) ∩ clBn+1 (xk ) ∩ C 6= ∅} ∩ {ω ∈ Ω | kn (ω) = k}) .
=
k∈N
By induction assumption {ω | Fn (ω) ∩ clBn+1 (xk ) ∩ C 6= ∅} ∈ F. Moreover
{ω ∈ Ω | kn (ω) = k} =
k−1
\
({ω ∈ Ω | Fn (ω) ∩ clBn+1 (xi ) = ∅} ∩ {ω ∈ Ω | Fn (ω) ∩ clBn+1 (xk ) 6= ∅}) .
i=1
⇒ {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) =
Fn (ω)
n∈N
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
→ Y . If U ⊂ Y and C ⊂ Y are any two
Lemma 6.2.4. Let X and Y be topological spaces and F : X −
→
subsets, 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→ X, Dom(F ) = X and F (·) is closed valued,
able space, X is a separable complete metric space, F : Ω −
→
then there is a sequence of measurable selectors {fn (·)} such that
F (ω) = cl{fn (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
(
F (ω) ∩ B 1 (xn ),
if F (ω) ∩ B 1 (xn ) 6= ∅,
2k
2k
Fn,k (ω) =
F (ω),
otherwise.
Then, by Lem 6.2.4, for any open subset O ⊂ X, we have that
³
´ h
³
F − (O) = F − O ∩ B 1 (xn ) ∪ F − (O) \ F − B
2k
1
2k
By the definition of the Fn0 s we observe that
´ h
³
³
−
Fn,k
(O) = F − O ∩ B 1 (xn ) ∪ F − (O) \ F − B
2k
1
2k
´i
(xn ) .
´i
(xn ) .
But, the measurability of F (·) implies that F − (O ∩ Un ) ∪ [F − (O) \ F − (Un )] is measurable. Consequently, for each n, k ∈ N, Fn,k (·) is measurable and Dom(Fn,k ) = Ω. Using Prop. 6.1.9, we find that
clFn,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) = cl{fn,k (ω) | n, k ∈ N}. From (6.1), the following inclusion is obvious
cl{fn,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 that
1
< ε, and by the density of D, there is n ∈ N such that
2k−1
³
´
−
x ∈ B 1 (xn ) ⇒ x ∈ F (ω) ∩ B 1 (xn ) ⇒ ω ∈ Fn,k
B 1 (xn )
2k
2k
2k
Thus,
fn,k (ω) ∈ clFn,k (ω) = clF (ω) ∩ clB
1
2k
(xn ) ⇒ fn,k (ω) ∈ clB
Consequently,
ρ(fn,k (ω), x) ≤ ρ(fn,k (ω), xn ) + ρ(xn , x) ≤
1
2k
(xn ).
1
1
1
+ k ≤ k−1 < ε.
k
2
2
2
117
⇒ fn,k (ω) ∈ Bε (x). Since ε > 0 is arbitrary, we see that
x ∈ cl{fn,k (ω) | n, k ∈ N}.
Therefore,
F (ω) ⊂ cl{fn,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 Caratheodory
function 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 Caratheodory
function.
Definition 6.3.3 (Epigraphical and Domain maps). Let f : Ω × X → R. Then
→ × R given by
• the epigraphical map associated with f (·, ·) is the set-valued map Ef : Ω−
→X
Ef (ω) = {(x, λ) ∈ X × R | f (ω, x) ≤ λ}
→ given by
• the domain map associated with f (·, ·) is the set-valued map Df : Ω−
→X
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 → R
→ X given by
is a Caratheodory function, then the set-valued map F : Ω −
→
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 countable
dense subset of X and C ⊂ X be a closed set. Then C ∩ D is a countable dense subset of C, say
D ∩ 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 that
f (ω, x) ≤ 0, ∀x ∈ U (x). Hence, by the density of D ∩ C in C, there is xn ∈ U (x) ∩ (D ∩ C) such that
f (ω, xn ) ≤ 0. Consequently, we can write
F − (C) = {ω ∈ Ω | f (ω, xn ) ≤ 0, for some n ∈ N}
[
=
{ω ∈ Ω | f (ω, xn ) ≤ 0}
(6.2)
(6.3)
n∈N
But, for each n ∈ N, {ω ∈ Ω | f (ω, xn ) ≤ 0} = f −1 (·, xn )[(−∞, 0] is measurable, since f (·, xn ) is
measurable. Consequently, F − (C) ∈ F.
Proposition 6.3.5 (measurability of the epigraphical map). Let (Ω, F) is a measurable space, X is
a separable metric space. If f : Ω × X → R is a Caratheodory function, then the epigraphical map
→ X × R of f is closed-valued and measurable.
Ef (·) : Ω −
→
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 on
Ω × (X × R); i.e. g(·, (x, λ)) is measurable for each fixed (x, λ) ∈ X × R and g(ω, ·) is continuous for
each fixed ω ∈ Ω. Therefore,
Ef (ω) = {(x, λ) ∈ X × R | g(ω, (x, λ)) ≤ 0}
is 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 separable
→ X is
metric space. If f : Ω × X → R is a Caratheodory function, then the domain map Df (·) : Ω −
→
measurable.
Proof. Let O ⊂ X be an open set. Then
ω ∈ Df− (O) ⇒ Df (ω) ∩ O 6= ∅.
⇒ ∃x ∈ Df (ω) ∩ O, ∃λ ∈ R such that
f (ω, x) < λ < ∞ ⇒ ω ∈ Ef− (O × (−∞, λ)).
Conversely, if ω ∈ Ef− (O × (−∞, λ)), then for some x ∈ O we have
f (ω, x) < λ < ∞ ⇒ x ∈ Df (ω) ⇒ Df (ω) ∩ O 6= ∅ ⇒ ω ∈ Df− (O).
Hence,
Df− (O) = Ef− (O × (−∞, λ)).
Since Ef− (·) is measurable, we conclude that Df− (·) is also measurable.
Proposition 6.3.7. If f : Ω × X → R is a Caratheodory function and x : Ω → X is a measurable
function, 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).
→ ×R
Since, x(·) : Ω → X is measurable and (−∞, α) is a constant, the set value map G : Ω−
→X
is measurable, by Prop.6.1.15 also Ef (·) is measurable (Prop. 6.3.5). Thus, using Prop. 6.1.12 we
conclude that the intersection map (Ef ∩ R)(·) is measurable; hence, Dom(Ef ∩ R) is a measurable
set 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 separable
metric space. If f : Ω × X → R is a lower semi-continuous function with respect to both Ω and X, then
→ X given by
the set-valued map F : Ω −
→
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 → R
→ X given by
is a Caratheodory function, then the set-valued map F : Ω −
→
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
Set
F1 (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 (·) is
measurable.
Proposition 6.3.10 (measurability of feasible set maps, see also Rockfellar/Wets). Let (Ω, F) be a
measurable space and X be a separable metric space. If gi : Ω × X → R, i ∈ I, and hj : Ω × X → R, j ∈ J
→ X given by
are Caratheodory functions, then the set-valued map F : Ω −
→
F (ω) = {x ∈ X | hj (ω, x) ≤ 0, j ∈ J; gi (ω, x) = 0, i ∈ I}, ω ∈ Ω
is measurable.
120
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 be
→ X is closed
a complete separable metric space. If f : Ω × X → R a Caratheodory function and F : Ω−
→
valued and measurable, then the marginal value function
ϕ(ω) := sup f (ω, x)
x∈F (ω)
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 |
sup f (ω, x) ≤ γ}.
x∈F (ω)
Using Cor. 6.2.5, there is a sequence of measurable functions {xn (·)}, xn : Ω → X such that F (ω) =
cl{xn (ω) | n ∈ N}, ω ∈ Ω. Hence,
{ω ∈ Ω | ϕ(ω) ≤ γ} = {ω ∈ Ω |
sup
f (ω, x) ≤ γ}
x∈cl{xn (ω) | n∈N}
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} =
\
{ω ∈ Ω |f (ω, xn (ω)) ≤ γ}.
n∈N
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 metricTspace. Prove that if, for each
→ X is closed valued and measurable and, for each x ∈ X,
k ∈ N, Fk : Ω −
→
k∈N Fk (x) 6= ∅ , then the set
valued map
Ã
!
\
→X
Fk : Ω −
→
k∈N
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 any
one who wants to know about set-valued maps can begin with the list given below. However, this
list is by no means exhaustive and is biased by my personal preferences and repeated citations in the
literature.
• General set valued maps: Aubin & Cellina [3], Aubin & Frakowsk[4], Berge[6], HU & Papageoriou[14],
Kiesielewichz[15], Rockafellar & Wets[22], Aliprantis & Border[1], Göfert 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], Göpfert[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 as
applied to optimization problems Shimizu et al.[24], Berger[7], etc. In an case, it worth paying
attention 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, Birkhäuser, 1999.
[3] J.-P. Aubin, A. Cellina, Differential inclusions, Springer Verlag, Berlin 1984.
[4] J.-P. Aubin, H. Frankowska, Set valued analysis, Birkhäuser, 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, Beiträge zur Realisierung von Lösungsverfahren der nichtlinearen Optimierung. Ph.D.
Dissertation, Technical University of Ilemanu, Fakulty of Mathematics and Natural Scinces, January, 1978.
[8] C. Castaing, M. Valadier, Convex analysis and measurable multifunctions. Lecture Notes in Mathematics, Vol. 580, Springer-Verlag, 1977.
[9] K. Deimling, Multivalued Differential Equations, Walter de Gruyter & Co., 1992.
[10] A. Göfert, H. Riahi, C. Tammer, C. Zǎlinescu, Variational methods in partially ordered spaces.
Springer, 2003.
[11] K. Goebel , W. A. Kirk, Topics in metric fixed point theory. Cambridge studies in advanced mathematics, V. 28, Cambridge University Press, 1990.
[12] A. Granas, J. Dugundji, Fixed point theory. Springer Monongrphs in Mathematics, SpringerVerlag, 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 Publishers, 1997.
[15] M. Kisielewicz, Differential Inclusions and Optimal Control. Polish Scientific Publishers & Kluwer
Academic Publishers, 1991.
[16] D. Klatte, R. Henrion, Regularity and stability in non-linear semi-infinite optimization. Semiinfinite Programming, R. Reemtsen and J.-J. Rückmann (eds.), pp. 69-102, Kluwer Academic
Pres, 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 Programming. Kluwer Academic Publishers, 1997.
[25] J.-B. H. Urruty, C. Lemaréchal, Convex analysis and minimization algorithms I. Springer Verlag,1993.
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