248 Chapter 5 • Continuous TunctionsPart 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 \:/n E N, 3an E A 3 lanl > n. By our
hypothesis, the sequence {an} has a convergent subsequence, {ank}. Now, \:/k,
I ank I > nk 2:: 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 closed and bounded, and hence is compact. •
Compactness is a very special property in relation to continuity. The fol-
lowing theorem says, in brief, that continuous functions "preserve" compact-
ness. This leads to the corollary that continuous functions have the "extreme
value property" on compact sets. In less formal words, a continuous function
on a nonempty compact set must have both a maximum and a minimum value
there-a fact of great importance in calculus. See also Exercise 27.
Theorem 5.3.6 The Continuous Image of a Compact Set Is Compact.
That is, if A is a compact set and f : A ---+ IR is continuous,^11 then f(A) is
compact.
Proof. Suppose A is a compact set and f : A---+ IR is continuous. Let {Yn}
be any sequence of points of f(A). Then \:/n EN, Yn E f(A), so 3an EA 3
f(an) = Yn· Consider the sequence {an}· Since A is compact, Theorem 5.3.5
guarantees that {an} has a convergent subsequence { ank} whose limit is in A;
i.e.,
Since f is continuous on A , f is continuous at L, so
f(ank)---+ f(L)
i.e., Ynk ---+ f(L).Thus, {Yn} has a convergent subsequence whose limit is inf (A). Therefore,
f (A) is compact, by Theorem 5.3.5. •
- Notice that our theorem is strengthened by using the weaker statement #2 in the hy-
potheses. The following theorem is also true, but weaker: If f : V(f) -+ IR is continuous on a
compact set A, then f (A) is compact.