Math 5615H: Honors: Introduction to Analysis

Compact Sets

Sasha Voronov

University of Minnesota

September 30, 2020

Properties of Compact Sets in Metric Spaces

Theorem

If K is a compact subset of a metric space X, then K is closed

and bounded.

Remark. The converse is true for X = Rn (the Heine-Borel thm)

Examples

2 / 7

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Proof of Theorem

Bounded: Cover X (and thereby K ⇢ X ) with open balls Bn(x0)of radius n 2 N, centered at x0 2 X .

Closed: Prove K c := X ⇢ K is open. Given a 2 K c , consider

Ok = {x 2 R | |x � a| > 1/k}, k 2 N. ThenS1

k=1Ok = X \ {a},

so K ⇢S1

k=1Ok .

3 / 7

Suppose k compact EX .

**¥.¥. x dlx,xoKh⇒Bn¥o

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Kkr c#x,a) b/c * CX#a)

¥;§ Pick a finite sub cover

Oki = Opp where p issuedthat kp=maxkk , -fed

9%9..EE#o..iE*?.omi

Closed Subsets of Compact Sets

Theorem

Let K be a closed subset of a compact metric space X. Then K

is compact. In particular, a closed subset of a compact set in

any metric space is compact.

Proof. Let {O↵} be an open cover of K . Then

{Kc} [ {O↵}

is an open cover of X .

4 / 7

Xcx

*axclosed compact

+open ,

b/c k closedU Op K

, (V Od U ke o X .

L L

choose a finite sub cover 102 if it, in}U kkI→with or without thisDrop Lk

') anyway . pai U a X ⇒ Kati Oj

The Bolzano-Weierstrass Property

Definition

Let X be a metric space. A subset A ⇢ X has the

Bolzano-Weierstrass property if every infinite subset of A has a

limit point (cluster point) that belongs to A.

Theorem

Let A be a subset of a metric space (X , d). Then A is compact

iff A has the Bolzano-Weierstrass property.

Proof.

5 / 7

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here no pout of Ais a hut point of EM! ⇒ ta e-A 3- open

ball

B•

catered at a t.BE#EE--L&afoor. This gives anopen cover {B al a e- Af of A . A compact ⇒ F finitesubcover . It covers A arsed Etherdog E . ⇒ E finite . DNext time ⇐

Nested Intervals

Theorem (Nested interval property)

Proof.

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.. .

SIP Ike, 0. . .Then in

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Then EE,Iq is a single point .

Prodi. next tiene .

The Heine-Borel Theorem

Theorem

A subset K of Rn is compact if and only if it is closed and

bounded.

Proof.

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Next time