Principles of Mathematics in Operations Research

148 10 Basic Topology

=> d{p, qi) < d{p, z) + d(z, qi) < d(p, qi).

Contradiction! Therefore, W D V = 0.

Thus, V = Br(p) C Xc C Kc => Kc is open => K is closed. D

Theorem 10.2.9 Closed subsets of compact sets are compact.

Corollary 10.2.10 If F is closed and K is compact, then FC\K is compact.

Theorem 10.2.11 Let {Ki\i G /} be a collection of compact subsets of a

metric space such that the intersection of every finite subcollection of Ki is

nonempty. Then,

Proof. Assume f]ieI Ki — 0.

Fix a member of {Ki, i G 1} and call it K.. Then,

/cn[ fl Ki] = 9^JCc[\J K?].

Since K is compact, 3KX, ...,Kn 3lCc [K[ U • • • U K%\ => K. D Kx n • • • D

Kn = 0, since we intersect a finite subcollection, we have a contraposition

(Contradiction). D

Corollary 10.2.12 If (Kn) is a sequence of nonempty compact sets 3 K\ D

K 2 D---, then OZi Kn? 0-

Theorem 10.2.13 (Nested Intervals) Let(In) be a sequence of non-empty,

closed and bounded intervals in R 3 I\ C I2 C • • •, then

n=l

Proof. Let In = [an,bn] 3 an <bn. Then,

h C h C • • • => ai < a 2 < • • • < a„ < • • • < bn < • • • b 2 < 61.

Moreover, if k < n => Ik C In and a^ < an <bn <bk-
Let E = 01,02,... is bounded above by 61. Let x — sup-E, then Vn, an < x.
Let us show that Vn, x < bn: If not, 3n 3 bn < x => 3a,k £ E 3 bn < a/..

case 1: k < n =>• a^ < an < bn < a^, Contradiction!
case 2: k > n => an < ak < bk < bn < a^, Contradiction!

Thus, Vn, a: < bn =* a; G /„, Vn =>• x e fX°=i A. => lT=i J»? 0- •