5.3 Continuity on Compact Sets and Intervals 249Caution: Theorem 5.3.6 says that the continuous image of a closed, bounded
set is closed and bounded. It does not say that the continuous image of a
bounded set must b e bounded,^12 or that the continuous image of a closed set
must be closed. Indeed, neither of these is true! (See Exercise 6.)Corollary 5.3.7 (Extreme Value Theorem): If A is a nonempty compact
set and f : A -+ IR is continuous, then f has the extreme value property on
A:(a) 3u = minf(A) = min{f(x): x EA}, and
(b) 3v = maxf(A) = max{f(x): x EA}.
That is, a continuous function assumes a maximum and a minimum value on
any nonempty compact set.Proof. Exercise 5. •CONTINUITY ON INTERVALS
(INTERMEDIATE VALUE THEOREM)Before discussing continuous functions on intervals, we remind ourselves of
the definition of an "interval,'' given in Section 1.6. There, we first defined a
closed interval [a, b] and then defined an interval^13 in general as any set I that
satisfies the condition
Vx < y in I, [x, y] ~ I.
The following theorem looks innocent enough, but it has far-reaching con-
sequences in calculus and analysis. It leads to the "intermediate value theorem"
and other important consequences.
Theorem 5.3.8 Suppose I is an interval and f : I -+ IR is continuous. Then
f (I) is an interval.^14
Proof. Suppose f : I-+ IR is continuous on an interval I. To prove that f (I)
is an interval, let u,v E f(I), where u < v. We must show that every w E (u,v)
is also in f(I). So, suppose u < w < v. Since u, v E f(I), 3 a, b EI 3 u = f(a)
and v = f(b). Then either a< b orb< a.
But, see Theorem 5.4.6.
See Definition 1.2.16.
In the general setting of topology, this theorem would state that the continuous image of
a "connect ed " set is "connect ed." See references such as [4], [45], [94], or [122].