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Lecture 11. 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. Basics. Theorem 6.11: A subset of. is compact iff it is. closed and bounded.

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Ma4266 topology

Lecture 11

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


Basics

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


Bolzano-Weierstrass Property

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.


Examples

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


BWExistence 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


BWExistence 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.


Total Boundedness

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.


Compactness and the BWP

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


Compactness for Subsets of

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


Assignment 11

Assignment 11

Read pages 175-180

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


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