312 PROPERTIES OF NUMBER SYSTEMS Chapter 9
DEFINITION 3
Let F be an ordered field and S c F. An element u of F is said to be a least
upper bound for S in F, denoted u = lub, S (or most often u = lub S when there
is no possibility of confusion about the underlying ordered field) if and only if:
(a) x I u for all x E S
(b) For all E > 0 (E an element of F), there exists x E S such that x > u - ECondition (a) asserts that u is an upper bound of S in F, whereas (b) states
that no element of F smaller than u is an upper bound of S. The concept
of greatest lower bound, denoted glb, S or simply glb S, is defined in a
similar manner. You are asked to write out this definition in Exercise 3(a).
Also, it is easy to show that lub's and glb's, as defined in the context of
an ordered field, are unique whenever they exist [see Exercise 3(b)], so that
we may correctly refer to t& lub and the glb of S (also known, incidentally,
as sup S and inf S, respectively).
EXAMPLE 1 Use the definitions of lub and glb to show that the subset
S = {n + (- l)"/n 1 n E NJ of the ordered field R is not bounded above in
R, but is bounded below with 0 = glb S.
Solution For each positive integer n, we have n - 1 5 n + ( - l)"/n. Since
the set {n - 1 In E N} is clearly not bounded above, we conclude the
same for S, by Exercise 4(a). Also, 0 is clearly a lower bound for S,
since n + (- l)"/n equals zero when n = 1 and is clearly greater than 1
if n 2 2. Finally, 0 is the greatest lower bound of S, because for any
E > 0, there exists x E S such that x < 0 + E, namely, take x equal to 0.
0
EXAMPLE 2 Prove that (- a, 21 has least upper bound in R equal to 2.
Solution By definition, S = (-a, 21 = {x E Rlx s 2). Since x s 2 for all
x E S, then 2 is an upper bound for S. To show that 2 is the least upper
bound of S, let c > 0 be given. Then x = 2 - (~12) is clearly an element
of S, and furthermore, x = 2 - (812) > 2 - E.
It is clear that, in an ordered field, any subset having a least upper bound
in the field is bounded above in that field. Thus the set S in Example 1
has no least upper bound in F = R. But what about the converse? That
is, suppose we know that a nonempty subset S of an ordered field F is
bounded above in F. Can we conclude that S has a least upper bound in
F? In Example 2 that conclusion was warranted. Consider, however, the
following example.
EXAMPLE 3 Show that the subset S = {x E Q lx2 c 31 of the ordered field
Q has no least upper bound in Q.