a space is compact iff any family of closed sets having fip has non-empty intersection
The above theorem is essentially the definition of a compact space rewritten using de Morgan’s laws. The usual definition of a compact space is based on open sets and unions. The above characterization, on the other hand, is written using closed sets and intersections.
Proof. Suppose is compact, i.e., any collection of open subsets that cover has a finite collection that also cover . Further, suppose is an arbitrary collection of closed subsets with the finite intersection property. We claim that is non-empty. Suppose otherwise, i.e., suppose . Then,
(Here, the complement of a set in is written as .) Since each is closed, the collection is an open cover for . By compactness, there is a finite subset such that . But then , so , which contradicts the finite intersection property of .
The proof in the other direction is analogous. Suppose has the finite intersection property. To prove that is compact, let be a collection of open sets in that cover . We claim that this collection contains a finite subcollection of sets that also cover . The proof is by contradiction. Suppose that holds for all finite . Let us first show that the collection of closed subsets has the finite intersection property. If is a finite subset of , then
where the last assertion follows since was finite. Then, since has the finite intersection property,
This contradicts the assumption that is a cover for .
- 1 R.E. Edwards, Functional Analysis: Theory and Applications, Dover Publications, 1995.
|Title||a space is compact iff any family of closed sets having fip has non-empty intersection|
|Date of creation||2013-03-22 13:34:10|
|Last modified on||2013-03-22 13:34:10|
|Last modified by||CWoo (3771)|