1549901369-Elements_of_Real_Analysis__Denlinger_

158 Chapter 3 11 Topology of the Real Number System

Corollary 3.3.9

(a) JR

(b) (a, b)

( c) (a, b]

(d) [a, b)

(e) (-oo,a)

(f) (-oo, a]

None of the following sets is compact (assuming a< b):

(g) (a, +oo)

(h) [a, +oo)

(i) U: n EN}

(j) N

(k) z

(1) Q

Proof. Exercise 3. •

Theorem 3.3.10 (Heine-Borel) Every closed, bounded interval of real num-
bers is compact.

Proof. Let I = [a, b], where a < b. Let U denote any open cover of I. We.
want to show that U has a finite subcover of I. We define the set

S = { x E I : some finite subcollection of U covers [a, x]}.

Then Sis nonempty since a ES. Moreover, Sis bounded since S ~ [a, b].
By completeness of JR, 3 u =sup S.

Claim #1: u ES.
Proof: Since a E S, a ::; u; and since bis an upper bound for S, u ::; b. Thus,
a ::; u ::; b. Thus, u is a member of some set V in U ; i.e., 3 V E U 3 u E V.
But V is open, so 3 c > 0 3

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

By the €-criterion for supremum (Theorem i.6.6) 3 x E S 3 u - € < x ::; u.
Case 1 (x = u): Then u ES as desired.
Case 2 (x < u): Then (x, u] ~ (u - c , u + c) ~ V. Since x E S, the
definition of S says that some finite subcollection of U covers [a, x]. Adding V
to this subcollection gives us a finite subcollection of U that covers [a, u]. Thus,
u ES. (See Figure 3.13.)

In either case, u E S.

x

(! ) I

U-£ U U+t: b

Figure 3.13