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### MA4266 Topology

Wayne Lawton

Department of Mathematics

S17-08-17, 65162749 [email protected]

http://www.math.nus.edu.sg/~matwml/

http://arxiv.org/find/math/1/au:+Lawton_W/0/1/0/all/0/1

Theorem 6.11: A subset of

is compact iff it is

closed and bounded.

Definition: A topological space is countably compact

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 is

compact iff it is countably compact.

Theorem 6.13: The Lindelöf Theorem Every

second countable space is Lindelöf.

Proof see page 175

Definition: A topological space X has the BW-property

if 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 compact

space 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

such that

Therefore

is an open cover of B.

Furthermore

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.

Example 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

in the Hilbert space

does not have the BWP (why?).

Lebesgue Number of an Open Cover

Definition: Let

be a metric space and

an open cover of

A Lebesgue number for

such that every subset of

is a positive number

having diameter less than

is contained in some

element in

Theorem 6.16 If

is a compact metric space

has a Lebesgue number.

then every open cover of

Proof follows from the following Lemma 1 since each

subset having diameter less than

is a subset of

an open ball of radius

BWExistence of Lebesgue Number

Lemma 1: Let

be a metric space that satisfies

the Bolzano-Weierstrass property. Then every open

cover

of

has a Lebesgue number.

and assume to the

Proof Let

be an open cover of

contrary that

does not have a Lebesgue number.

in

such that

Then there exists a sequence

for every

and for every

Then

is infinite (why?) so the BW property

implies that it has a limit point

so there exists

contains

and

with

Then

infinitely many members of

BWExistence of Lebesgue Number

Hence

contains some

with

Then for

so

This contradicting the initial assumption that for all

and completes the proof of Lemma 1.

Definition: Let

be a metric space and

An

net for

is a finite subset

such that

The metric space

is totally bounded if it has an

net for every

Lemma 2: Let

be a metric space that satisfies

the Bolzano-Weierstrass property. Then

is TB.

Proof Assume to the contrary that there exists

such that

does not have an

net.

Choose

and construct a sequence

with

that has no limit point.

Theorem 6.15: For metric spaces compactness = BWP.

Proof Theorem 4.14 implies that compactness BWP.

For the converse let

be an open cover of a metric

space

having the Bolzano-Weierstrass property.

such that for

Lemma 1 implies that there exists

every

the open ball

is contained in some

member of

Lemma 2 implies that there exists a finite

subset

such that

an open cover of

Choose

and observe that

covers

the following

Theorem 6.17: For a subset

conditions are equivalent:

(a)

is compact.

(b)

has the BWP.

(c)

is countably compact.

(d)

is closed and bounded.

Question Are these conditions equivalent for

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