Countably Compact Space satisfies Countable Finite Intersection Axiom

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Then $T$ is countably compact iff it satisfies the Countable Finite Intersection Axiom.

Proof
Let every countable open cover of $S$ have a finite subcover.

Let $\mathcal A$ be any set of closed subsets of $S$ satisfying $\bigcap \mathcal A = \varnothing$.

We define the set:
 * $\mathcal V := \left\{{S \setminus A : A \in \mathcal A}\right\}$

which is clearly an open cover of $S$.

From De Morgan's Laws: Difference with Union:


 * $\displaystyle S \setminus \bigcup \mathcal V = \bigcap \left\{{S \setminus V : V \in \mathcal V}\right\} = \bigcap \left\{{A : A \in \mathcal A}\right\} = \varnothing$

and therefore $S = \bigcup \mathcal V$.

By definition, there exists a finite subcover $\tilde{\mathcal V} \subseteq \mathcal V$.

We define:
 * $\tilde{\mathcal A} := \left\{{S \setminus V : V \in \tilde{\mathcal V}}\right\}$

then $\tilde{\mathcal A} \subseteq \mathcal A$ by definition of $\mathcal V$.

Because $\tilde{\mathcal V}$ covers $S$, it follows directly that:


 * $\displaystyle \bigcap \tilde{\mathcal A} = \bigcap \left\{{S \setminus V : V \in \tilde{\mathcal V}}\right\} = S \setminus \bigcup \tilde{\mathcal V} = \varnothing$

Thus, in every countable set $\mathcal A$ of closed subsets of $S$ satisfying $\displaystyle \bigcap \mathcal A = \varnothing$ exists a finite subset $\tilde{\mathcal A}$ such that $\displaystyle \bigcap \tilde{\mathcal A} = \varnothing$.

That is, $S$ satisfies the Countable Finite Intersection Axiom.

The converse works exactly as the previous, but with the roles of the open cover and $\mathcal A$ reversed.