160 Chapter 3 • Topology of the Real Number System
Theorem 3.3.13 (Sequential Criterion for Compactness) A set A of
real numbers is compact if and only if every sequence of points of A has a
subsequence that converges to a point of A.
Proof. Let A be a set of real numbers.
Part 1 ( :::::> ): Suppose A is compact. Let {an} be a sequence of points of
A. Now A is a bounded set, so {an} is a bounded sequence. By the Bolzano-
Weierstrass Theorem for sequences, {an} has a convergent subsequence { ank }.
Let L = lim ank. Now A is closed, since it is compact. So, by the sequential cri-
k->oo
terion for closed sets (3.2.19), LE A. Thus, {an} has a subsequence converging
to a point of A.
Part 2 ( ~): Suppose every sequence of points of A has a subsequence
converging to a point of A. We want to prove that A is compact; i.e., closed
and bounded.
Suppose A is not bounded. Then \:In E N, 3an E A 3 lanl > n. By our
hypothesis, the sequence {an} has a convergent subsequence, {ank}. Now, \:/k,
lank I > nk ?: k. This means { ank} is unbounded. But every convergent sequence
is bounded. Contradiction. Therefore, A is bounded.
We shall prove that A is closed using the sequential criterion for closed
sets. Suppose {bn} is a sequence of points of A that converges; say, bn ---> M.
Then {bn} is bounded. So, by our hypothesis, {bn} must have a convergent
subsequence {bnk} whose limit is in A. By Theorem 2.6.8, this limit must be
M. Therefore, M E A. So, by the sequential criterion for closed sets, A is closed.
Therefore, A is compact. •
Theorem 3.3.14 A set A of real numbers is compact if and only if every
infinite subset of A has a cluster point in A.
Proof. Part 1 ( =?): Suppose A is compact. Let S be an infinite subset of A.
Then S is a bounded, infinite set. By the Bolzano-Weierstrass Theorem, S has
a cluster point, say x. From the definition of cluster point, x is also a cluster
point of A. But A is closed, so x E A (Theorem 3.2.8). Thus, every infinite
subset of A has a cluster point in A.
Part 2 ( ~): Suppose every infinite subset of A has a cluster point in A. We
are going to apply Theorem 3.3.13 to show that A is compact. Suppose {xn}
is a sequence of.points of A.
Case 1: The set { Xn : n E N} is finite. Then, 3 c E A 3 Xn = c for infinitely
many n. So, {xn} has a constant subsequence. This subsequence converges to
c, a point of A.
Case 2: The set { Xn : n E N} is infinite. By our hypothesis, this set must
have a cluster point, a E A. Then, by Theorem 3.2. 11 , every neighborhood of
a contains Xn for infinitely many n. By Theorem 2.6.7, this means that {xn}
has a subsequence converging to a, and we already know that a E A.