Equivalence of Definitions of Compact Topological Space

Theorem
Let $$X$$ be a topological space. The following are equivalent:
 * 1) $$X$$ is compact, i.e. every open cover of $$X$$ has a finite subcover
 * 2) In every set $$\mathcal{A}$$ of closed subsets of $$X$$ satisfying $$\bigcap \mathcal{A} = \varnothing$$ exists a finite subset $$\tilde{\mathcal{A}}$$ such that $$\bigcap \tilde{\mathcal{A}} = \varnothing$$
 * 3) Each filter on $$X$$ has an limit point in $$X$$
 * 4) Each ultrafilter on $$X$$ converges

(1) $$\implies$$ (2)
Let $$\mathcal{A}$$ be any set of closed subsets of $$X$$ satisfying $$\bigcap \mathcal{A} = \varnothing$$.

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

which is clearly an open cover of $$X$$.

From De Morgan:


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

and therefore $$X = \bigcup \mathcal{V}$$.

By (1) there exists a finite subcover $$\tilde{\mathcal{V}} \subseteq \mathcal{V}$$.

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

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

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


 * $$\bigcap \tilde{\mathcal{A}} = \bigcap \{ X \setminus V : V \in \tilde{\mathcal{V}} \} = X \setminus \bigcup \tilde{\mathcal{V}} = \varnothing$$

(2) $$\implies$$ (1)
This part works exactly as the previous, but with the roles of the open cover and $$\mathcal{A}$$ reversed.

(3) $$\implies$$ (4)
Let $$\mathcal{F}$$ be an ultrafilter on $$X$$.

By (3) $$\mathcal{F}$$ has a limit point $$x \in X$$.

Thus there exists a filter $$\mathcal{F}'$$ on $$X$$ which converges to $$x$$ satisfying $$\mathcal{F} \subseteq \mathcal{F}'$$.

Because $$\mathcal{F}$$ is an ultrafilter, $$\mathcal{F} = \mathcal{F}'$$.

Thus $$\mathcal{F}$$ converges to $$x$$.

(4) $$\implies$$ (3)
Let $$\mathcal{F}$$ be a filter on $$X$$.

Then there exists an ultrafilter $$\mathcal{F}'$$ such that $$\mathcal{F} \subseteq \mathcal{F}'$$.

By (4) we know that $$\mathcal{F}'$$ converges to a certain $$x \in X$$.

This implies that $$x$$ is a limit point of $$\mathcal{F}$$.

(2) $$\implies$$ (3)
Let $$\mathcal{F}$$ be a filter on $$X$$.

Assume that $$\mathcal{F}$$ has no limit point.

This would imply that $$\bigcap \{ \overline{F} : F \in \mathcal{F} \} = \varnothing$$.

By (2) there are therefore sets $$F_1, \ldots, F_n \in \mathcal{F}$$ such that $$\overline{F}_1 \cap \ldots \cap \overline{F}_n = \varnothing$$.

Because for any set $$M$$ we have $$M \subseteq \overline{M}$$, we know that $$\overline{F}_1, \ldots, \overline{F}_n \in \mathcal{F}$$.

This contradicts the fact that $$\mathcal{F}$$ is a filter, because filters are closed under finite intersections and must not contain the empty set.

Thus $$\mathcal{F}$$ has a limit point.

(3) $$\implies$$ (2)
Let $$\mathcal{A} \subset \mathcal{P}(X)$$ be a set of closed subsets of $$X$$.

Assume that $$\bigcap \tilde{\mathcal{A}} \ne \varnothing$$ for all finite subsets $$\tilde{\mathcal{A}}$$ of $$\mathcal{A}$$.

We show that this implies $$\bigcap \mathcal{A} \ne \varnothing$$.

Because of our assumption, $$\mathcal{B} := \left\{{\bigcap \tilde{\mathcal{A}} : \tilde{\mathcal{A}} \subseteq \mathcal{A} \text{ finite}}\right\}$$ is a filter basis.

Let $$\mathcal{F}$$ be the corresponding generated filter.

Then $$\mathcal{F}$$ has a limit point by (3) and thus $$\varnothing \ne \bigcup \left\{{\overline{F} : F \in \mathcal{F}}\right\} \subseteq \bigcap \mathcal{B} \subseteq \bigcap \mathcal{A}$$.

Thus $$\bigcap \mathcal{A} \ne \varnothing$$.

Therefore (2) follows.