Countably Compact Space satisfies Countable Finite Intersection Axiom

Theorem

The following definitions of the concept of Countably Compact Space are equivalent:

Definition by Finite Subcover

A topological space $T = \left({S, \tau}\right)$ is countably compact if and only if:

every countable open cover of $T$ has a finite subcover.

Definition by Countable Finite Intersection Axiom

A topological space $T = \left({S, \tau}\right)$ is countably compact if and only if:

every countable set of closed sets of $T$ whose intersection is empty has a finite subset whose intersection is empty.

That is, $T$ 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 $S$.

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

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.

$\Box$

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

$\blacksquare$