MA4266 Topology

Wayne LawtonDepartment of Mathematics

S17-08-17, 65162749 matwml@nus.edu.sg

http://www.math.nus.edu.sg/~matwml/http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1

Lecture 11

Basics

Theorem 6.11: A subset of is compact iff it isnR

Definition: A topological space is countably compact

closed and bounded.

if every countable open cover has a finite subcover.

Definition: A topological space is a Lindelöf space

if every cover has a countable subcover.

Theorem 6.12: If X is a Lindelöf space, then X iscompact iff it is countably compact.

Theorem 6.13: The Lindelöf Theorem Everysecond countable space is Lindelöf.

Proof see page 175

Bolzano-Weierstrass PropertyDefinition: A topological space X has the BW-propertyif every infinite subset of X has a limit point. Theorem 2.14: Every compact space has the BWP.Proof Assume to the contrary that X is a compactspace and that B is an infinite subset of X that has no limit points. Then B is closed (why?) and

B is compact (why?). Since B has no limit points, for every point x in B there exists an open set

XOx such that }.{xBOx Therefore

}:{ BxOx O is an open cover of B. FurthermoreO does not have a finite subcover of B (why?).

Definition p is an isolated point if {p} is open. See Problem 10 on page 186.

ExamplesExample 6.3.1(a) Closed bounded intervals [a,b] have the BWP.

(b) Open intervals do not have the BWP.

(c) Unbounded subsets of R do not have the BWP.

(d) The unit sphere

does not have the BWP (why?).

)(2 N,...}4,3,2,1{N

}||||,:,...),({)(1

2221

2

k kk xxRxxxN}1||||:)({ 2 xNxS

in the Hilbert space S

Definition: Let be a metric space and

such that every subset of

.X A Lebesgue number foran open cover of

),( dX

is a positive number X

Lebesgue Number of an Open Cover

OO

having diameter less than is contained in some

element in .OTheorem 6.16 If ),( dX is a compact metric space

then every open cover of X has a Lebesgue number.

Proof follows from the following Lemma 1 since eachsubset having diameter less than an open ball of radius

is a subset of .

Lemma 1: Let be a metric space that satisfies

of

1}{ nnx

and assume to the

the Bolzano-Weierstrass property. Then every open

),( dX

cover X

BWExistence of Lebesgue Number

O

O

has a Lebesgue number.

OxB nn ),( 1

be an open cover of

contrary that

O

Then there exists a sequence Xdoes not have a Lebesgue number.

in

X

for every

Proof Let

and for everynThen

such that

.OO1}{ nnx is infinite (why?) so the BW property

implies that it has a limit point a so there exists 0and OO with .),( OaB Then ),( 2

aB contains

infinitely many members of .}{ 1nnx

Hence with

Then for

BWExistence of Lebesgue Number

22),(),(),( zxdxadzad nn

so

.21 nnx

This contradicting the initial assumption that for all

),( 1nnxBz

),( 2aB contains some

.),(),( 1 OaBxB nn

O OnOxB nn ,1,),( 1

and completes the proof of Lemma 1.

Definition: Let

An

be a metric space and

Total Boundedness

net for

.0XA

such that

),( dX

is a finite subset

.,),( XxAxd The metric space

XX

is totally bounded if it has an net for every .0Lemma 2: Let ),( dX be a metric space that satisfiesthe Bolzano-Weierstrass property. Then X is TB.

Proof Assume to the contrary that there exists 0such that X does not have an net. Choose

Xa 1 and construct a sequence 1}{ kka with

),(11 k

j jk aBa that has no limit point.

Theorem 6.15: For metric spaces compactness = BWP.

Compactness and the BWP

For the converse let

space

O be an open cover of a metric

Lemma 1 implies that there exists

),( dX having the Bolzano-Weierstrass property.

such that for

Xx is contained in some

Proof Theorem 4.14 implies that compactness BWP.

0the open ball

XxxA n },...,{ 1

every ),( xB

subset

Lemma 2 implies that there exists a finitemember of

an open cover of

.Osuch that }:),({ ,...,1 nkkxB

.X Choose nkkk OxB ,...,1,),( Oand observe that }:{ ,...,1 nkkO covers .X

Theorem 6.17: For a subset

Compactness for Subsets of

conditions are equivalent:

(a)

nR

is compact.

the following

has the BWP.

nRA

A(b) A(c) A is countably compact.

(d) A is closed and bounded.

Question Are these conditions equivalent for ?)(2 NA

Assignment 11

Read pages 175-180

Exercise 6.3 problems 2, 3, 4, 5, 9, 13, 14, 15