1549901369-Elements_of_Real_Analysis__Denlinger_

640 Appendix C • Answers & Hints for Selected Exercises

(b) If v is any lower bound for A, then by Exercise 1.6-A. 12 , - v is an
upper bound for -A, so - v :=:; u. That is, v 2:: - u.

By (a) and (b) together, - u =inf A.

3. Let A be a nonempty subset of ordered F with an upper bound in F , and B
   be the set of all upper bounds of A in F. By completeness, 3u = sup A. Then
   u E B and \lb E B , u '.S b .. ·. u = min B.


4. Let a = sup A, b = supB and c =max{ a, b}. Then
   (a) Let x E AU B. Then either x E A so x :=:; a :=:; c, or x E B so
   x '.S b ::::; c. Thus x '.S c.
   (b) If d is any upper bound for AU B then d is any upper bound for A and
   d is any upper bound for B , so d 2:: a and d 2:: b; thus d 2:: c.
   By (a) and (b) together, c =supAUB.


5. (a) \Ix EX, f(x) + g(x)::::; sup{f(x): x EX}+ sup{g(x): x EX}. Thus,
   sup{f(x):x E X}+ sup{g(x):x EX} is an upper bound for {f(x )+g(x):x EX}.

Chapter 2

EXERCISE SET 2. 1

(^1). (a) (^1) , 4, 1 1 9, 16,^1 25 1 , 36 1 , 49^1 , 8^1 1

2. (a) 23; 101 ; no > j"f (c) 700 ; 14,000; no> f

(e) 1, 197; 23 , 997; n 0 >^102 -4 (^102 will do.)

(g) 64; 1, 266; no > f 910 + ~ G + 1 will do.)

(i) 50 1; 10, 001; no > ~ (k) 101; 2001 ; n 0 > max{2, ~}

(m) 34; 668; no> fi + 1 (o) 304; 6004; n 0 >max{lO,t+1}

EXERCISE SET 2.2

1. Use Def. 2.1.l directly. 3. \In EN, lxn -cl = 0.
   7. Take an = n, bn = ~.


2. When c = 0, {cxn} is a constant sequence.


3. Take an= (-1r, bn = (- 1 r+^1. Thm. 2.2. 13 applies only when {an} and
   {bn} both converge, so this example does not contradict that theorem.


4. By the algebra of limits, (a) 19; (b) - l 0 ; (c) ~; (d) i+ 2 ../5


5. We give only the numerical answers, except for (1), for which we give a
   complete solution. You must supply the reasons.
   (a) O; (c) ~; (e) 25 ; (g) O; (i) O; (k) 3; (m) 0; (o) ~; (q) -1