660 Appendix C • Answers & Hints for Selected Exercises- f has a removable discontinuity at x 0 ::::} lim f(x) exists, but either
x-+xo
(a) f(x 0 ) does not exist, in which case f is not continuous from the left or
from the right at x 0 , or
(b) lim f(x) = L =f. f(xo). Then lim f(x) = lim f(x) = L =f. f(xo), in
x--+xo x --+xQ x--+xci
which case f is not continuous from the left or from the right at xo. - f is monotone increasing and bounded on (a, b), so
a < x < c::::} f(x)::; f(c); sup{f(x): a < x < c}::; f(c); lim f(x) :S f(c);
x-+c-
c < x < b::::} f(x) ::=: f(c); inf{j(x) : c < x < b} ::=: f(c); lim f(x) ::=: f(c).
x--+c+ - Modify appropriately the proofs given in Thm. 5.2.17 and Exercises 8-10.
- Modify the proof given for the monotone increasing case, using Thm. 5.2.18
instead of Thm. 5.2.17. - By Thm. 5.2.17, lim f(x ) and lim f(x) exist. Define f(a) = lim f(x)
x-+a+ x--+b- x--+a+
and f(b) = lim f(x). Then f: [a, b] ->JR is continuous on (a, b) and, by Thm.
x->b-
5.2.17, f is monotone increasing on [a,b]. Since lim f(x) = f(a), 'tic: > 0,
x-+a+
36 > 0 3 't/x E D(f), a < x < a+ 6::::} lf(x) - f(a)I < c:. Since D(f) = [a, b],
this is equivalent to 'tic:> 0, 36 > 0 3 't/x E D(f), Ix-al< 6::::} lf(x)-f(a)I < c:.
. ·. f is continuous at a.
Continuity at b is proved similarly. - (a) Suppose f is monotone increasing and bounded on I = (a, b), and
c E (a,b]. By Thm. 5.2.17(a), lim f(x) exists and equals sup A, where A=
x--+c-
{f ( x) : a< x < c}. Let B = {f(r) : r E Q, a< r < c}.
Since B ~ A , sup B ::; sup A. On the other hand,
a < x < c ::::} 3r E Q 3 x < r < c ::::} 3r E Q 3 f ( x) ::; f ( r) ::::} f ( x) ::; sup B.
. ·. sup A ::; sup B. We have proved that sup A = sup B.
The other parts are proved similarly. - As shown in Example 4.2.21, the graph off "oscillates" between the lines
y = x and y = -x infinitely often in any nbd of 0, and yet lim f(x) = 0. Since
X->0
x--+lim 0 f(x) exits but is unequal to f(O), f has a removable discontinuity at 0.
EXERCISE SET 5.3- We must prove that Statement #2 ::::} Statement #1 <==? A is open.
(a) Assume A open and Stmnt. #2 true; i.e., f: A-> JR is continuous. Let
c: > 0 and xo EA. Then 361 > 0 3 't/x EA, Ix - xol < 61 ::::} lf(x) - f(xo)I < c:.
Since A open, 362 > 0 3 N 02 ( xo) ~ A. Choose 6 = min { 61, 62}. Then 't/x E
D(f), Ix - xol < 6 ::::} x E A and lf(x) - f(xo)I < c: .. ·. f is continuous on A.
(Stmnt. #1)