5.5 *Monotonicity, Continuity, and Inverses 267(b) Prove that the conjecture is true if A and B are bounded and sup A <
inf B.
( c) Prove that the conjecture is true if A and B are compact sets.5.5 *Monotonicity, Continuity, and Inverses
Monotone functions were defined in Definition 5.2.15. Since strictly monotone
functions are 1-1, they must have inverses.^17 We shall now establish some
connections between continuity, monotonicity, and invertibility of functions.
First, we make note of a useful lemma.
Lemma 5.5.1 (a) If f : V(f) ---+ ~ is monotone increasing on a set A, then
Vx1, x2 EA, f(xi) < f(x2) =?xi < x2.
(b) If f : V(f) ---+ ~ is monotone decreasing on a set A, then Vx 1 , x 2 E A,
f(x1) < f(x2) =?xi > x2.
Proof. Exercise 1. •The next theorem is a partial converse of the intermediate value theorem.
Its proof is a good example of the usefulness of the concept of restricted domain.
Theorem 5.5.2 If f :I ---+ ~ is monotone on an interval I, and f(I) is an
interval, then f is continuous on I. [That is, a monotone function with the
intermediate value property on an interval must be continuous on that interval.]
Proof. We shall prove the monotone increasing case and leave the mono-
tone decreasing case to Exercise 7. Suppose f:I---+ ~is monotone increasing on
a nonempty interval I , and f (I) is an interval with nonempty interior (the case
in which f (1)^0 = 0 is trivial; explain). For notational simplicity, let J = f (I).
Let x 0 E I. We shall prove that f is continuous at xo.
Case 1: f(x 0 ) is an interior point of J. Then :lc:o > 0 3 N 00 (J(xo)) ~ J.
Let c; > 0. Without loss of generality, assume c; < c:o. Then N 0 (J(x 0 )) ~
J = f(I), so 3 X1,X2 EI 3
f(xi) = f(xo) - c:/2;
f (x2) = f (xo) + c:/2.- See Appendix B .3, especiall y Definition B.3.11 and Theorem B .3.12.