1549901369-Elements_of_Real_Analysis__Denlinger_

5.5 *Monotonicity, Continuity, and Inverses 269

Proof. We prove the strictly increasing case, and leave the strictly decreas-

ing case as Exercise 8.

Suppose I is a nonempty interval, and f : I ---+ JR is continuous and strictly

increasing. Then f: I---+ f(I) is a 1-1 correspondence, and so has an inverse,^18

1-^1 : f (I) ---+ I, which is also 1-1 and onto.

By Lemma 5.5.1, 1-^1 is strictly increasing on f(I). Thus, by Theorem

5.5.2, 1-^1 : f (I) ---+ I is continuous on f (I). •

The following theorem is intuitively plausible, but its proof is a little tricky.

Theorem 5.5.4 If f : I---+ JR is 1-1 and continuous on an interval I , then f

is strictly monotone on I.

Proof. Suppose I is an interval and f : I---+ JR is 1-1 and continuous.

Claim #1: Va< b < c in I, f(b) is between f(a) and f(c).

Proof: Let a< b < c in I. For contradiction, suppose f(b) is not between

f(a) and f(c). Then

either f(b) > max{f(a), f(c)}
or f(b) < min{f(a), f(c)}
We consider the former case, and leave the latter to Exercise 9. In the
former case, 3 real number y such that

f(b) > y > max{f(a), f(c)}.

Then f(b) > y > f(a), so by the intermediate value theorem, 3 x 1 E (a, b) 3
f(xi) = y. Similarly, f(b) > y > f(c), so 3 x2 E (b, c) 3 f(x2) = y. But then
we have x 1 =f x2 and f(x 1 ) = f(x2), contradicting the hypothesis that f is 1-1
on I. Therefore, in this case, f(b) is between f(a) and f(c).

Claim #2: f is either strictly increasing or strictly decreasing on I.
Proof: It suffices to prove^19 that if f is not strictly increasing on I, then
f is strictly decreasing on I. Suppose f is not strictly increasing on I. Then 3
c < din I 3 f(c) ;:::: f(d). Since f is 1-1, we must have f(c) > f(d). We shall
prove that f must be strictly decreasing on I.
First, we note that Vx EI, one of the following must hold: x < c, c < x < d,
or x > d. Applying Claim #1, we see that

x<c=?x<c<d

::::} f(c) lies between f(x) and f(d)

::::} f(d) < f(c) < f(x)

18. See Theorem B.3.12 in Appendix B.


19. To prove P or Q, it suffices to assume ,..., P and prove Q. (See Proof Strategy "PS-4" in
    Appendix A .3.)