13.1. THE GENERAL CONCEPT OF A FILTER 5 The generalization of the finite intersection property is a bit trickier. As a candidate, one might consider the following notion: If (P, ) is a p.o. and A is a subset of P, then A is centered if every finite subset of A has a lower bound in P. Unfortunately, it is not always true that every centered subset of a p.o. is contained in a filter. EXERCISE 13.9 (G). Let P = {{n} : n G uo} U {u\{n} : n e LU}. Show that A = {uj\{n} : n G UJ} is a centered subset of (P, C), but A is not contained in any filter in (P,C). The trouble with generalizing the fip is caused by the lack of "intersections" in some p.o.'s. If (P, ) is a p.o. and p, q G P, then an element r of P is called the greatest lower bound or meet of p and q if r is a lower bound of {p, q} and s r for every lower bound s of {p, q}. A p.o. (P, ) is called well-met if every two compatible elements of P have a greatest lower bound in P.1 EXERCISE 13.10 (G). Convince yourself that the p.o. of Exercise 13.9 is not well-met, but the p.o.'s of Examples 13.4—13.8 are. For well-met p.o.'s, the analogue of Exercise 13.2(a) holds. THEOREM 13.5. Let (P, ) be a well-met p.o., and let A be a centered subset o/P. Then there exists a smallest filter F in (P, ) such that A C F. EXERCISE 13.11 (G). Prove Theorem 13.5. Of course, the smallest filter containing A will be called the filter generated by A. By Theorems 13.4 and 13.5, every centered subset of a well-met p.o. (P, ) is contained in an ultrafilter in this p.o. Let us now present two important applications of ultrafilters in topology. THEOREM 13.6. Let X be a topological space. The following are equivalent: (i) X is compact (ii) Every filter of closed subsets of X has nonempty intersection (iii) Every ultrafilter of closed subsets of X is fixed. PROOF. Let X be a fixed topological space. For any filter F of closed subsets of X, let F# = {X\Y : Y G F}. Of course, F* is a family of open sets. EXERCISE 13.12 (G). Show that for every filter in JC(X) the following are equivalent: (1) f| F = 0 (2) F* is an open cover of X without finite subcover. Moreover, show that if F is an ultrafilter in /C(X), then F is free if and only if it satisfies conditions (1) and (2). The implications (i) = (ii) and (ii) = (iii) follow immediately from Exer- cise 13.12. For the implication (iii) = (i), assume that X is not compact, and let U be an open cover of X without finite subcover. Then the set Um — {X\U : U G K\ is a centered subset of /C(X), and since (/C(X),C) is a well-met p.o., by Theo- rem 13.5 there exists an ultrafilter F of closed subsets of X with W C F. Since [\W = 0, also f]¥ = 0, and Exercise 13.12 implies that F is not fixed. • X A well-met p.o. with no incompatible elements is called a lower semilattice.

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 1997 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.