The Finite Intersection Property in Compact Spaces and Countably Compact Spaces

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 X is compact if and only if every collection of close subsets of X 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 \mathcal{C} of subsets of X has the finite interection property if for every finite F \subset \mathcal{C}, \bigcap F \neq \phi.

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

Proof. \Rightarrow Let X be compact. Let \mathcal{C} be a collection of closed subsets of X. We show that if \mathcal{C} has the finite intersection property, then it has non-empty intersection. Suppose that \bigcap \mathcal{C}=\phi. Then \mathcal{U}=\lbrace{X-C:C \in \mathcal{C}}\rbrace is an open cover of X. By the compactness of X, \mathcal{U} has a finite subcover \lbrace{X-C_0,X-C_1,...,X-C_n}\rbrace. it follows that \bigcap_{i \leq n} C_i = \phi.

\Leftarrow Let \mathcal{U} be an open cover of X such that it has no finite subcover. Note that \mathcal{C}=\lbrace{X-U:U \in \mathcal{U}}\rbrace has the finite intersection property and \bigcap \mathcal{C}=\phi.

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

  1. X is countably compact.
  2. Every countable collection of closed subsets of X satisfying the finite intersection property has non-empty intersection.
  3. For every sequence \lbrace{C_0,C_1,C_2,...}\rbrace of non-empty closed subsets of X where C_0 \supset C_1 \supset C_2 \supset ..., we have \bigcap_{n<\omega} C_n \neq \phi.

Proof. The direction of 1 \rightarrow 2 is analogous to the direction of \Rightarrow in Theorem 1. The direction of 2 \rightarrow 3 is obvious.

3 \rightarrow 1. Let \mathcal{U}=\lbrace{U_0,U_1,U_2,...}\rbrace be an open cover of X. For each n, define V_n=\bigcup_{i \leq n} U_i and C_n=X-V_n. Note that C_0 \supset C_1 \supset C_2 \supset .... If X=V_n for some n, then we are done. If not, each C_{n} is non-empty and we have \bigcap_{n<\omega} C_n \neq \phi. Then we have a point that does not belong to all V_n, which is a contradiction. Thus we must have X=V_n for some n.

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

  1. X is countably compact.
  2. Every locally finite collection of subsets of X is finite.
  3. Every infinite subset of X has an accumulation point.

Proof. 1 \rightarrow 2. Suppose we have an infinite locally finite collection \lbrace{A_n \subset X:n \in \omega}\rbrace. For each n, let C_n=\bigcup_{i \ge n} \overline{A_i}, which is a closed set. Note that C_0 \supset C_1 \supset ... and \bigcap_{n<\omega} C_n = \phi. By theorem 2, X is not countably compact.

2 \rightarrow 3. Suppose we have A \subset X such that A has no accumulation point. Then for each x \in X, there is an open set O_x such that x \in O_x and O_x contains no point of A other than x. This means that the singletons \lbrace{x}\rbrace, where x \in A, form a discrete collection of subsets of X (thus a locally finite collection). By (2), A must be finite.

3 \rightarrow 1. Suppose we have a sequence \lbrace{C_0,C_1,C_2,...}\rbrace of closed subsets of X where C_0 \supset C_1 \supset C_2 \supset .... We want to show that \bigcap_{n<\omega} C_n \neq \phi. Then by Theorem 2, X is countably compact. If there is some n such that C_n=C_m for all m \ge n, then we are done. So assume that the sequence of C_n is strictly decreasing. So there is an increasing sequence of integers m(0)<m(1)<m(2)<... such that we can choose x_i \in C_{m(i)}-C_{m(i)+1}. Then A=\lbrace{x_0,x_1,x_2,...}\rbrace is an infinite set. Let x be an accumulation point of A. It follows that x \in \bigcap_{n<\omega} C_n.

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

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