One-to-One Continuous Functions
Fall 2004
Theorem 1 If f : R R is continuous and one-to-one, then f is either increasing or decreasing. Lemma 2 If f : [a, b] R is continuous and one-to-one, f (a) < f (b) and a < c < b, then f (a) < f (c) < f (b). Proof Suppose not. Since f is one-to-one, we must have f (c) < f (a) or f (c) > f (b). Suppose f (c) < f (a). Choose R with f (c) < < f (a) < f (b). By the Intermediate Value Theorem, there are x, y such that c < x < a < y < b and f (x) = f (y) = , contradicting the fact that f is one-to-one. Thus we must have f (c) > f (b). But then we choose such that f (b) < < f (c). Again we nd x, y such that c < x < a < y < b and f (x) = f (y) = , contradicting the fact that f is one-to-one. Thus we must have f (a) f < (c) < f (b). Corollary 3 If f : [a, b] R is continuous, one-to-one, and f (a) < f (b), then f is increasing on [a, b]. Proof Otherwise, there are a c < d b with f (c) < f (d). By Lemma 2 f (a) f (c) < f (b). If b = d we are done. If c < d < b, then we apply Lemma 2 to the restriction of f to [c, b]. Hence, f (c) < f (d), as desired. An analogous argumen shows that if f : [a, b] R is continous, one-toone, and f (a) > f (b), then f is decreasing on [a,...