# 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) 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.