662 Appendix C • Answers & Hints for Selected Exercises
- Suppose 1 : I ~ JR is continuous, strictly monotone, and bounded on
I= (a, b). Let c =inf f(I) and d =sup f(I).
Case 1: (! strictly increasing on I) In this case,
(a) 'r/x E (a, b), l(x) E l(I) soc< f(x) < d. Thus, f(I) ~ (c, d).
(b) Vy E (c, d), c < y < d, so by definition of inf and sup, 3 Y1, Y2 E l(J) 3
c < Y1 < y < Y2 < d. Then, 3 x1,X2 EI 3 l(xi) = Y1 and l(x2) = Y2· Since 1
is strictly increasing on I, x 1 < x 2. Then, by the intermediate value theorem,
3x EI 3 l(x) = y; i.e., (c,d) ~ l(I).
(c) Therefore, l(I) = (c, d).
Extension to [a,b]: Define l(a) = c and l(b) = d. By Thm. 5.2.17,
lim l(x) = inf l(I) = c = l(a) and lim l(x) = sup l(I) = d = l(b).
x-.a+ x~b-
Then 1 is continuous on [a, b] and J[a, b] = [c, d].
To see that 1 is strictly increasing on [a, b], let a < x < b. Then 3 x1, x2 E
I 3 a< x1 < x < x2 < b. So, l(a) = c = infl(J):::; l(x1) < l(x) < l(x2):::;
sup l(I) = d = l(b); i.e., l(a) < l(x) < l(b).
Case 2: (! strictly decreasing on I) Modify the proof of Case 1.
- Let Yo E l(A). We shall prove 1-^1 continuous at Yo· Let {Yn} be a sequence
in l(A) 3 Yn ~ YO· Then 3 Xn, Xo E A 3 l(xn) = Yn and l(xo) = Yo Since
A compact, { xn} is bounded. Let { Xnk} be any convergent subsequence of
{xn}, say Xnk ~ L. Since A is closed, L E A. Since 1 continuous on A,
l(xnk) ~ l(L); i.e., Ynk ~ l(L). But Ynk ~Yo, so by the uniqueness oflimits,
l(L) =Yo= l(xo). Since 1 is 1-1 on A , L = xo. That is , Xnk ~ xo. Thus, all
convergent subsequences of {xn} have the same limit, xo. By Exercise 2.6.21,
this means Xn ~ xo; i.e., 1-^1 (yn) ~ 1-^1 (yo). :. 1-^1 is continuous at Yo· - (a) Define g:[a, a!b] ~JR by g(x) = l(x) -l(x+ b2a). If g(a) = 0,
take x =a, y = a!b· Suppose g(a) =fa 0. Then g (a!b) = -g(a), so by the
intermediate value theorem, 3 c E [a, a!b] 3 g(c) = O; i.e., l(c) = 1 (c+ b;a).
Take x = c and y = c + b2a.
(b) By (a), 3 x1, Y1 E [a, b] 3 Y1 - X1 = ~(b - a) and l(x1) = l(yi).
Continuing by induction, 'r/n EN, 3 Xn, Yn E [a, b] 3 Yn -Xn = ~(Yn-1 -Xn-1) =
.}n (b - a) and 1 (xn) = 1 (Yn)· Since 2 ~ (b -a) ~ 0, 3no E N 3 2 ~ 0 (b - a) < c.
- ( ~) Suppose every continuous 1 : A ~ JR has a max and a min on A.
Since the function i : A~ JR given by i(x) = x is continuous, A must have a
max and a min, so A is bounded. We must prove A closed. Let x 0 E Ac. Define
l(x) = tx!xat · Then 1 is continuous on A since xo t/-A, so by hypothesis 1
has a maximum value on A. That is, 3M > 0 3 'r/x E A, l(x) :::; M; i.e.,
tx!xol :::; M. Then 'r/x EA, Ix - xol 2: '}j. Thus, N-k (xo) contains no point of
A, so N -1.. M (xo) ~Ac. :. Ac is open, so A is closed.