7.1 Refresher on Suprema, Infima, and the Forcing Principle 359Proof. Suppose A, B ~ JR. such that Va E A , Vb E B , a :::; b. Then Va E A,
a is a lower bound for B , so a:::; inf B. That is , Va EA, a:::; inf B. Thus, inf B
is an upper bound for A, so sup A :::; inf B.
Part 1 [(a)::::} (b)]: Suppose u =sup A= inf B. Let c: > 0. By the c: criteria
for supremum and for infimum (Theorems 1.6.6 and 1.6. 7) 3 a E A 3 a > u - ~,
and 3 b E B 3 b < u + ~. Then -a < -u + ~. Thus,b +(-a) < ( u + ~) + (-u + ~) i.e.,
b-a< c:.Part 2 [(b)::::} (c)]: Trivial. Suppose (b) is true, and take K = 1.
Part 3 [ ( c) ::::} ( d)]: Suppose ( c) is true. Let u = sup A. Then, as we proved
at the beginning, u :::; inf B. Thus, Va E A, Vb E B , a :::; u :::; b. We must now
show that there is not more than one such number. For contradiction, suppose
3 real number v such that v =I-u and
Va E A , Vb E B , a:::; v :::; b.lu-vl
Let c: =~·Then c: > 0 so by (c), :la EA, b EB 3
b-a <Kc:
b-a<lu-vj.But u, v E [a, b], so lu - vi < b - a. Contradiction. Therefore, (d) is true.
Part 4 [(d) ::::} (a)]: Suppose (d) is true. We want to prove that supA =
inf B. For contradiction, suppose sup A =I-inf B. Then sup A < inf B , since we
showed at the beginning that sup A:::; inf B. Then Va EA, Vb EB,
a :::; sup A < inf B :::; b.This contradicts (d). Therefore, sup A= inf B. •
Finally, we shall have occasional use of the following generalized version of
the forcing principle, first encountered in Theorem 1.5.9.Theorem 7.1.6 (Generalized Forcing Principle)^1 Suppose x, a E JR..
(a) lf3K>03Vc:>O,x::;a+Kc:, thenx::;a.
(b) If 3 K > 0 3 Ve: > 0, x 2: a - Kc:, then x 2: a.(c) Jf3K>03Vc:>O, Ix-al ::;Kc:, thenx=a.
- Compare with the forcing principle, Theorem 1.5.9.