1549901369-Elements_of_Real_Analysis__Denlinger_

1.6 The Completeness Property 37

are not minimum or maximum elements of (a, b), there is something special

about a and b. In technical language, we say that a is the greatest lower bound

of (a, b) and bis the least upper bound of (a, b). We now define these technical

terms. Examples and exercises will clarify them further.

SUPREMA AND INFIMA

Definition 1.6.3 Suppose that F is an ordered field and A ~ F. We say that

an element u E F is

(1) a least upper bound ("supremum") of A if u is an upper bound for

A and\:/ upper bounds v for A , u :::; v. The notation we use is u =sup A.

(2) a greatest lower bound ( "infimum") of A if u is a lower bound for

A·and \:/lower bounds v for A, u 2: v. The notation we use is u =inf A.

We now justify the assertion that a is t he greatest lower bound of (a, b)

and b is the least upper bound of (a, b).

Theorem 1.6.4 Let a < b in an ordered field F. Then

a = inf(a, b), and b =sup( a, b).

Proof. Part 1. First we prove that a= inf(a, b).
(i) By definition of (a, b), a is a lower bound of (a, b).
(ii) Suppose that v is a lower bound of (a, b). We must prove that a 2: v.
For contradiction, suppose a :l v. That is, a < v. Since v is a lower bound for
(a, b) and a~b E (a, b), we have
a+b
a < v :::; -
2

• < b.
  Let c = a!v. Then a < c < v < b. Thus, c E (a, b) and c < v. But v is a
  lower bound for (a, b). Contradiction. Therefore, a 2: v.

(

a

a+v

2

I

c v

a+b

2

I

Figure 1.1

)

b

Part 2. Next, prove that b =sup( a, b). Exercise 6. •

It may be helpful to think of inf A and sup A as "substitutes" for min A and
max A in instances when A is a bounded set that does not have a maximum
or minimum. It should also be observed that although min A and max A must