1549901369-Elements_of_Real_Analysis__Denlinger_

268 Chapter 5 • Continuous Functions

Since f is monotone increasing on I, and f(xi) < f(xo) < f(x2), we must have
x1 < xo < x2 (by Lemma 5.5.1). Then Vx EI,

x 1 < x < x2 ::::} x E I (since I is an interval)

and f(x1) :::; f(x) :::; f(x2) (by monotonicity off)

::::} f(xo) -c/2:::; f(x):::; f(xo) + c/2

::::} lf(x) - f(xo)I < c. (11)

Choose o = min{x2 - xo,Xo - xi}. Then o > 0 and

Ix - xal < o ::::} -o < x - xo < o

::::} X1 - Xo < x -Xo < X2 - Xo

::::} X1 < X < X2

::::} lf(x) - f(xo)I < c by (11) above.

Therefore, f is continuous at x 0.

Case 2: f(xo) is the left endpoint of J. Then, \Ix EI, f(x) ~ f(xo), which
implies x ~ x 0 by Lemma 5.5.1. So x 0 is the left endpoint of I. Since J is an
interval containing more than one point, :lea> 0 3 (f(xo), f(xo) +co)~ J.
Let c > 0. Without loss of generality, assume c <co. Then (f(xo), f(xo) + c)
~ J = f(I), so 3 X1 EI 3

f(x1) = f (xo) + c/2.

Since f is monotone increasing on I and f(xo) < f(x 1 ), we must have xo < x 1.
Thus, \Ix EI,

xo :::; x < X1 ::::} f(xo) :::; f(x) :::; f(x1) (by monotonicity off on I)

::::} 0:::; f(x) - f(xo):::; f(x1) - f(xo) = c/2

::::} lf(x) - f(xo)I < c. (12)

Choose o = x 1 - xa. Then o > 0 and

Ix - Xo I < o and x E I ::::} -o < x -xa < o and x ~ xo

::::} 0 < x - Xo < X1 - Xo

::::} XQ :S X < X1

::::} lf(x) - f(xo)I < c by (12) above.

Therefore, f is continuous at xa.

Case 3: f (xo) is the right endpoint of J. Exercise 6. •

Corollary 5.5.3 (Inverse Function Theorem for Continuous Mono-
tone Functions) Suppose I is a nonempty interval, and f : I ---+ IR is contin-
uous and strictly monotone. Then 1-^1 : f (I) ---+ I is continuous and strictly
monotone in the same sense. (!(I) is an interval, by Theorem 5.3.8.)