1549901369-Elements_of_Real_Analysis__Denlinger_

1.6 The Completeness Property 39

(i) Let x E A. By (a), Ve > 0, x < u + e. Thus, by the "forcing
principle" [Theorem 1.5.9 (b)], x :::; u. Thus, Vx E A, x :::; u. That is, u is an
upper bound for A.
(ii) Suppose that v is an upper bound for A. We need to prove that
u :::; v. For contradiction, suppose u > v. Let e = u - v. Then e > 0, so by
hypothesis (b), =ix EA 3
x>u- e
x>u-(u-v)
x > v.
But vis an upper bound for A , and x EA. Contradiction. Therefore, u:::; v.
By (i) and (ii) together, u = sup A. •

Theorem 1.6.7 (c Criterion for Infimum) Let F be an Archimedean or-
dered field, As;;; F, and u E F. Then u =inf A{::;> Ve> 0,
(a) Vx E A, x > u - e, and
(b) ::3 x E A 3 x < u + e.

3x£A

U- £ u u+c

Figure 1.3

Proof. Exercise 11. •

EXERCISE SET 1.6-A

1. Assume that real numbers exist and behave according to the familiar rules
   of algebra. For each of the following sets of real numbers, tell whether or
   not the given set is bounded above. For those that are, give three different
   upper bounds and find the least upper bound.
   (a) (-1, 3] (b) [-1,3)
   (c) {1,2,3,4} (d) {5}
   (e) (-oo, OJ (f) (0, +oo)

(g) 0 (h) {~:nEN}

(i) {~:x>O} (j) {~:l<x<2}

(k) {2-~:nEN} (1) { 1 +

2 :: n EN}

(m) {x;l :x>2} (n) {sin ( n

2

7r) : n E N}