Answers & Hints for Selected Exercises 639
by Thm. 1.5.3. Suppose x < 0. Then -x > 0 so by Thm. 1.5.3, :Jm E N 3
m - 1 < -x < m. Then -m < x < -m + 1, so we may taken= -m + 1.
Uniqueness follows by the argument given in Thm. 1.5.3.
- Let x <yin an ordered F. Then x < ~ < y (Thm. 1.2.10 (d)).
- Suppose that l::/c > 0, x ~ a+c:. Then l::/c: > 0, x-a ~ c:, so by (a), x-a ~ O;
i.e., x ~a.
- Suppose that l::/c: > 0, la -bl ~ c:. By (a), la -bl ~ 0. But la -bl 2: 0, so
la -bl = O; i.e., a= b.
EXERCISE SET 1.6-A
- (a) Yes; 3, 4,86; 3 (c) Yes; 4,4.01,86; 3 (e) Yes; 0,0.2,86; 0
(g) Yes; -100, 0, 25; none (i) No (k) Yes; 2, 3, 86; 2
(m) Yes; 2, 3, 86; 1.5 [Draw graph of f(x) = 1 + ~ for x > 2.]
- (a) Yes; -1, -2, -100; -1 (c) Yes; 1, 0, -20; 1 (e) No
(g) Yes; -100, 0, 25; none
(i) Yes; 0, -1, -100; 0 [Draw graph of f(x) = ~-]
(k) Yes; 1, 0, -100; 1 [Draw graph.] (m) Yes; 2, 1, -100; 1 [Draw graph.]
- Examples given in Exercises 1 and 2.
- Alter the proof already given that shows S has a maximum element.
- Alter the proof of Part (a) given.
- If u =inf A EA, then u EA and l::/a EA, a 2: u, so by defn., u = minA.
- Let F be Archimedean, A~ F, and u E F.
(::::}) Suppose u = inf A. Let£ > 0. Then l::/x E A, x 2: u > u - c:. Also,
u + c: >inf A, sou+ c: is not a lower bound for A, so :Jx EA 3 x < u + c:.
( <=) Suppose (a) and (b) hold. Then,
(1) l::/x EA, x > u - c:. By Exercise 1.5.12, x 2: u.
(2) Suppose v is a lower bound for A. For contradiction, suppose v > u. Let
c: = v - u. By (b), 3x E A 3 x < u + c: = v. Contradiction. Therefore, all lower
bounds of A are ~ u.
By (1) and (2) together, u =inf A.
EXERCISE SET 1.6-B
- Suppose A is a nonempty set with a lower bound in a complete ordered
F. By Exercise 1.6-A.12, the set -A = {-a : a E A} is bounded above. By
completeness, :Ju= sup(-A). Then
(a) l::/a EA, -a~ u, so a 2: -u.
