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ANALYSIS II
Metric Spaces: Compactness

Defn   A collection of open sets is said to be an open cover for a set A if the union of the collection contains A. A subset of an open cover whose union also contains the set A is called a subcover of the original cover. A cover is called finite if it has finitely many members.

Defn   A set K in a metric space (X,d) is said to be  compact  if each open cover of K has a finite subcover.

Theorem  Each compact set K in a metric space is closed and bounded.

Proposition  Each closed subset of a compact set is also compact.

Theorem (Heine-Borel Theorem from last term)   Each closed and bounded interval [a,b] is a compact subset of the real numbers.

Pf:  Let C be an open cover for [a,b] and consider the set A := {x | [a,x] has an open cover from C }. Note that A is not empty since a belongs to A.  Let c:= lub (A). It is enough to show that  c > b , since if x₁ belongs to A and a  x  x₁ , then x belongs to A.  Suppose instead that c  b , then there must be some O₀  in C such that c belongs to O₀ .  But O₀  is open, so there exists   > 0 so that N(c) is contained in O₀.  Since c is the least upper bound for A , then there is an x in A such that c- <x  c. But x belongs to A so there are members O₁  , ... , O_n  of  C whose union covers [a,x] . The collection O₁  , ... , O_n covers [a , c+/2] .  Contradiction, since c is the least upper bound for the set A .

Corollary  Each closed and bounded set of real numbers is compact.

Theorem  If a set A is compact in a metric space X and f: X  Y is continuous, then f[A] is compact in Y. 

Corollary  If f : X  Y is continuous and X is compact, then f is a bounded function.

Corollary  If f : X  R is continuous and X is compact, then f attains its extremal values.

Theorem Suppose that f: [a,b]® K is one-to-one, onto and continuous, then f^-1 is continuous.

Pf.  Let O Í [a,b] be relatively open, then (f^-1)^-1(O) = f(O). Let C be the complement in [a,b] of O, then C is closed and hence compact. Therefore f(C) is compact in K and conseqently it is closed. Its complement in K must then be relatively open. That complement however is f(O).   ^[¯]

Compactness Characterization Theorem  Suppose that K is a subset of a metric space X, then the following are equivalent:

1. K is compact.
2. each infinite subset of K has a limit point in K.
3. each sequence from K has a subsequence that converges in K.
   
   (Click here for the details of the proofs.)

Corollary  Each closed and bounded set K in R^k (or C^k) is compact.

Pf: Use the sequential convergence criterium and consider projections into each coordinate. Recall that convergence in R^k is equivalent to convergence in each coordinate.

A set K in a metric space X is said to be totally bounded, if for each  > 0 there are a finite number of open balls with radius  which cover K. Here the centers of the balls and the total number will depend in general on .

Theorem  A set K in a metric space is compact if and only if it is complete and totally bounded.
[Homework.]