21. Compact Metric Spaces

21.1 Definition:

Given a Metric Space $\QTR{Large}{(M,d)}$ , and a subset MATH we say $\QTR{Large}{x}$ is a limit point of $\QTR{Large}{A\ }$if

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MATH.

That is $\QTR{Large}{x\ }$is in the closure of MATH

Note: It is not necessarily the case that the set of limit points of $\QTR{Large}{A\ }$is the closure of $\QTR{Large}{A}$.

For example, a singleton set $\QTR{Large}{\{x\}}$ has no limit points but is its own closure.

21.2 Definition:

A Metric Space, $\QTR{Large}{(M,d)}$ , is called compact if every infinite subset MATH has a limit point.

21.3 Theorem:

The closed unit interval $\QTR{Large}{\ [}$0,1$\QTR{Large}{]}$ is compact.

Proof:

Choose an infinite subset MATH and write $\QTR{Large}{[}$0,1$\QTR{Large}{]=}$ $\QTR{Large}{[}$0,MATH MATH,1$\QTR{Large}{]}$. Since $\QTR{Large}{A}$ is infinite we can choose one of these subintervals, written MATH,MATH such that

MATH,MATH is infinite. Call this intersection MATH and set MATH

By induction, assume we have chosen a sequence

MATH in $\QTR{Large}{[}$0,1$\QTR{Large}{]}$, and infinite subsets MATH

such that

  1. MATH,MATH

    and

  2. MATH,MATH

Let MATH . One verifies that $\QTR{Large}{r\ }$is in the closure of MATH hence a limit point.

Assignment: (Due April 20) provide the details.

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Solution:

First note that $\ $0 MATH

moreover, since MATH is infinite, we can find MATH, MATH, and such that

$\QTR{Large}{\ }$0 MATH

Hence MATH and MATH

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21.4 Theorem:

Let $\QTR{Large}{(M,d)}$ be a Metric Space. The following are equivalent:

  1. $\QTR{Large}{(M,d)}$ is compact.

  2. Every sequence in $\QTR{Large}{M}$ has a convergent subsequence.

Proof:

1. MATH2.

Let MATH be the sequence. If it only takes on a finite number of different values select an infinite constant subsequence. If not choose a limit point and a sequence that converges to that limit point.

2. MATH1.

Since $\QTR{Large}{A}$ is infinite we can choose a sequence without repeating values. That is a

map MATH such that MATH for MATH Select a convergent subsequence and a limit point.

21.4.1 Corollary:

Every compact Metric Space is complete:

Proof:

A Cauchy sequence is its own "convergent subseqence."

Boundedness and Separability

21.5 Theorem:

Given a compact Metric Spaces $\QTR{Large}{(M,d)}$ :

  1. For any MATH0 and some nMATH there exists points MATH MATH

    such that MATH.

  2. In particular, M is bounded in the sense that there exists some bMATH0 such that MATHb for any MATH .

Proof:

1. Choose any $\QTR{Large}{x_{1}}$ . By induction, suppose we have found MATH such that

MATH. Select MATH If does not exist then we can find a sequence MATH such that MATH. However since $\QTR{Large}{M}$ is compact we can find a convergent subsequence MATH and some

$\QTR{Large}{y}$ MATHwith MATHChoose n such that MATH for $\QTR{Large}{i>\ }$n. But then for MATHn, MATH contradicting the choice of MATH .

2. Choose b $\QTR{Large}{=\ }$2MATH where MATH is as in 1.

Next choose MATH with MATH and MATH.

MATH

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21.6 Theorem:

Every compact metric space has a countable dense subset.

Proof:

For each mMATH0 choose nMATH and MATH MATH

such that MATH.

Clearly MATH is countable.

Assignment: (Due April 20) show that it is dense in $\QTR{Large}{M}$.

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Solution:

Select MATH ,MATH0$\QTR{Large}{\ }$and $\QTR{Large}{m}$ such that MATH

Since MATH

we know that for some i MATH

and thus MATH

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Properties of Continuous Maps

21.7 Theorem:

Given Metric Spaces $\QTR{Large}{(M,d)}$ and MATH with $\QTR{Large}{M}$ compact, and a continuous map MATH then $\QTR{Large}{f(M)}$ is compact.

Proof:

Let MATH MATH and let MATH be such that MATH. Since M is compact we can select a convergent subsequence MATHand MATH such that MATH and, by continuity MATH. But MATH

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21.8 Theorem:

Given Metric Spaces $\QTR{Large}{(M,d)}$ and MATH with $\QTR{Large}{M}$ compact, and a continuous map MATH then $\QTR{Large}{f}$ is also uniformly continuous.

Proof:

If $\QTR{Large}{f\ }$was not uniformly continuous then there exists some MATH 0 such that for any MATH0 there exists $\QTR{Large}{x}$ and $\QTR{Large}{y}$ with MATH and MATH

Select sequences MATH and MATH such that

MATH and MATH

Since M is compact we can assume, without loss of generality that converges.(why?)

let MATH.

Since f is continuous we can choose MATH0 such that if MATH then MATH

Finally, choose n such that MATH and MATH

Note that this implies that MATH

hence

MATH

and

MATH

which implies that

MATH

contradicting the definition of the sequences MATH and MATH .

A Construction

21.9 Theorem:

Given Metric Spaces $\QTR{Large}{(M,d)}$ and MATH with $\QTR{Large}{M}$ and MATHcompact, then MATH is compact.

Proof:

Let MATH. Since $\QTR{Large}{M}$ is compact, we can select a subsequence MATH such that the sequence MATH

Next, since MATH is compact, we can select a subsequence MATH of MATH such that the sequence MATH

One checks that MATH

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