1549901369-Elements_of_Real_Analysis__Denlinger_

3.3 *Compact Sets 159

Claim #2: u = b.

Proof: For contradiction, suppose u "I-b. Then u < b, since a ~ u ~ b.

Since u E S, some finite subcollection V of U covers [a, u]. Thus, some open

set V E V contains u. Since V is open, 3 c > 0 3

(u-c,u+c)~V.

Thus, the finite subcollection V of U covers [a, u+c). Let c be any real number

satisfying u < c < min{ u + c, b }. Then

[u,c] ~ (u-c,u+c) ~ V ,

c

( t )

u-e u u+e

Figure 3.14

so t he finite subcollection V of U covers [a, c]. But then

c ES, and c > u =sup S.

This is a contradiction. Therefore, u = b.

By Claim #1 and Claim #2, b E S. That is, U has a finite subcover of
[a, b]. Therefore, [a, b] is compact. •

Theorem 3.3.11 A closed subset of a compact set is compact.

Proof. Suppose A i.s a closed subset of a compact set C. Let U be any
open cover of A. Since A is closed, Ac is open. Then U U {Ac} is an open cover
of IR, hence of C. But C is compact. Hence U U {Ac} has a finite subcover of
C. But A~ C. Hence Uhas a finite subcover of A. That is , A is compact. •

Corollary 3.3.12 A set of real numbers is compact if and only if it is ••• p
closed and bounded. ---

Proof. A compact set is closed and bounded, by Theorems 3.3.6 and 3.3.8.
To prove the converse, suppose A is closed and bounded. Then A is a closed
subset of a closed interval. By the Heine-Borel Theorem (3.3.10) and Theorem
3.3.11, it follows that A is compact. •

In the remainder of this section, we shall show that compactness is closely
related to several important concepts previously studied. We begin by showing
a connection with the Bolzano-Weierstrass Theorems.