122 9 Number Systems- bounded if E is both bounded above and below.
Example 9.1.5 A = {p G Q : p >- 0,p^2 •< 2} is- bounded above, b — 3/2,2,... are upper bounds.
- bounded below, a = 0, —1/2,... are lower bounds.
Definition 9.1.6 Let S be an ordered set and 0 ^ E C S be bounded above.
Suppose 3 b G S 9 :
i. b is an upper bound of E.- if b' & S and b' -< b then b' is not an upper bound of E. Equivalently, if
b" is any upper bound of E if b" >- b.
Then, b is called least upper bound (lub) or supremum (sup) of E and denoted
by
b = supE = lubE.
Greatest lower bound (gib) or infimum (ml) of E is defined analogously.
Example 9.1.7 S = Q, E = {p G Q : p >• 0, p^2 -< 2} inf £ = 0, but E has
no supremum in S = Q. Suppose po = sup E exists in Q. Then, either po G E
or po <£ E.
If po 6 E, 3q € E 3 po -< Q because E has no largest element; therefore, p is
not an upper bound of E.
IfPo & E, thenpo >- 0 because it is an upper bound andp\ >: 2 because po £ E.
Then, either pg = 2 (not true because po 6 Q) or p^ > 2 (true), then po G
B = {p G Q : p y 0, p^2 >- 2}. Then, 3q 0 e B 3 q 0 -< p 0 (*) because B
has no smallest element. Vp G E, p^2 -< 2 -< q^ =? qo is an upper bound of E.
Moreover, po < qo because lub Contradiction to (*).Definition 9.1.8 Let S be an ordered set. We say that S has the least upper
bound property if every nonempty subset of S which is bounded above has lub
in S.Example 9.1.9 S — Q does not have lub-property.Theorem 9.1.10 Let S be an ordered set with lub-property. Then, every
nonempty subset of S which is bounded below has inf in S.Proof. Let B ^ 0, B c S be bounded below, L be the set of all lower bounds
of B. Then, L ^ 0 (because B is bounded below), y G B be arbitrary, then
for any lEiwe have x <y. So, y is an upper bound of L; i.e. all elements of
B are upper bounds of L => L is bounded above, a = sup L, a G S (because
S has lub property).
Claim (i): a = inf B
Proof (i): Show a is lower bound of B; i.e. show V:r G B, a <x. Assume that
it is not true; i.e. 3a;o G B 9 a >- XQ. Then, XQ is not an upper bound of a
(because a = sup L) => XQ <£ B (because all elements of B are upper bounds
of L). Contradiction! (XQ G B). Therefore, a is a lower bound of B.