1549901369-Elements_of_Real_Analysis__Denlinger_

1.6 The Completeness Property 43

Definition 1.6.14 (-oo and +oo as infimum and supremum): Let JR denote

the complete ordered field,

(a) If a set A s;;; JR has no lower bound, we say that inf A= -oo;

(b) If a set A s;;; JR has no upper bound, we say that sup A = +oo;

~(c) Since every real number is a lower bound of 0, the empty set has ...----no
greatest lower bound, so we defiEe inf 0 = +oo ..

(d) Since every real number is an upper bound of 0, the empty set has no

least upper bound, so we define sup 0 = -oo.

It is important to remember that -oo and +oo are not real numbers, but are

merely symbols used to conveniently convey certain carefully defined situations

(here, to denote that a set is unbounded below or above). In addition, we adopt

the universal convention that all real numbers x obey the inequalities

-oo < x < +oo.

In analysis, the symbols -oo and +oo are frequently used in inequalities

and equations, whenever it is convenient. But these symbols are always used

with the understanding that they do not represent real numbers.

EXERCISE SET 1.6-B

In Exercises 1- 5 below, assume that A, B s;;; F, a complete ordered field.

l. Prove Theorem 1.6.12. (See Exercise 1.6-A.12.)

2. Suppose A s;;; B and A -j. 0. Prove that

(a) if Bis bounded below, then so is A , and inf A;:::: inf B.

(b) if B is bounded above, then so is A, and sup A ::::; sup B.

3. Suppose A is a nonempty set of F that is bounded above in F , and let
   B be the set of all upper bounds of A. Prove that B has a minimum
   element, and sup A = min B.


4. Suppose A is a nonempty set of F that is bounded below in F, and let B
   be the set of all lower bounds of A. Prove that B has a maximum element,
   and sup A= maxB.


5. Suppose A and Bare nonempty subsets of F bounded above in F. Prove
   that AU Bis bounded above, and sup( AU B) =max{ sup A, sup B}.


6. Modify the proof of Theorem 1.6.10 to prove that if a> 0 in a complete
   ordered field F, then :3 x E F 3 x^2 = a.