This is a discussion of a basic result about compact spaces, namely the characterization of compactness in terms of the finite intersection property. All spaces are at least Hausdorff. A space is compact if and only if every collection of close subsets of satisfying the finite intersection property has non-empty intersection. We also present a version of this theorem for countably compactness. We also show that in countably compact space, any locally finite collection of non-empty subsets is finite. As a corollary, in the class of paracompact spaces, countably compactness is equivalent to compactness.

A collection of subsets of has the finite interection property if for every finite , .

** Theorem 1**. A space is compact if and only if every collection of closed subsets of satisfying the finite intersection property has non-empty intersection.

* Proof*. Let be compact. Let be a collection of closed subsets of . We show that if has the finite intersection property, then it has non-empty intersection. Suppose that . Then is an open cover of . By the compactness of , has a finite subcover . it follows that .

Let be an open cover of such that it has no finite subcover. Note that has the finite intersection property and .

* Theorem 2*. Let be a space. The following conditions are equivalent.

- is countably compact.
- Every countable collection of closed subsets of satisfying the finite intersection property has non-empty intersection.
- For every sequence of non-empty closed subsets of where , we have .

* Proof*. The direction of is analogous to the direction of in Theorem 1. The direction of is obvious.

. Let be an open cover of . For each , define and . Note that . If for some , then we are done. If not, each is non-empty and we have . Then we have a point that does not belong to all , which is a contradiction. Thus we must have for some .

* Theorem 3*. Let be a space. Then the following conditions are equivalent.

- is countably compact.
- Every locally finite collection of subsets of is finite.
- Every infinite subset of has an accumulation point.

* Proof*. . Suppose we have an infinite locally finite collection . For each , let , which is a closed set. Note that and . By theorem 2, is not countably compact.

. Suppose we have such that has no accumulation point. Then for each , there is an open set such that and contains no point of other than . This means that the singletons , where , form a discrete collection of subsets of (thus a locally finite collection). By (2), must be finite.

. Suppose we have a sequence of closed subsets of where . We want to show that . Then by Theorem 2, is countably compact. If there is some such that for all , then we are done. So assume that the sequence of is strictly decreasing. So there is an increasing sequence of integers such that we can choose . Then is an infinite set. Let be an accumulation point of . It follows that .

* Corollary*. If is countably compact and paracompact, then is compact.