1549901369-Elements_of_Real_Analysis__Denlinger_

38 Chapter 1 • The Real Number System

be members of A, inf A and sup A do not necessarily pelongj;o A. In fact, we

must be very careful when we read that a set A "has_:: an inf or a sup not to

conclude that these are elements of A.

Theorem 1.6.5 (a) A set cannot have more than one greatest lower bound.

(b) A set cannot have more than one least upper bound.

~

(c) If a set has a minimum (or maximum) element, then that element is

he greatest lower bound (or least upper bound) of A.

(d) If a set contains a greatest lower bound (or least upper bound) then

that element is the minimum (or maximum) element of A. (11vl/i«5)

Proof. (a) Let u and u' be greatest lower bounds of A. Then

(i) u =inf A and u' is a lower bound of A=? u;::: u'.

(ii) u' =inf A and u is a lower bound of A=? u' ;::: u.

(iii) By (i) and (ii), u = u'. Therefore, A cannot have more than one

greatest lower bound.

(b) Exercise 7.

(c) Exercise 8.

( d) Exercise 9. •

Theorem 1.6.6 (e Criterion for Supremum) Let F be an Archimedean

ordered field, A~ F, and u E F. Then u =sup A{::} Ye> 0,

(a) Yx EA, x < u+c, and

(b) ::J x E A 3 x > u - c.

3xt:A

U-€ u

Figure 1.2

u+t:

Proof. Let F be an ordered field, A ~ F, and u E F.

(1) First, the(=;.) part. Suppose u =sup A. Let c > 0.

(i) Let x EA. Then x::::; supA = u < u+c. Thus, Yx EA, x < u+c.

(ii) u - c < u, which is the least upper bound for A, so u - c cannot

be an upper bound for A. Hence, ::J x E A 3 x > u - c.

(2) To prove the({=) part, suppose u E F satisfies conditions (a) and (b)

of the hypotheses.