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 $$\bigcap \mathcal{A} = \emptyset$$ exists a finite subset $$\tilde{\mathcal{A}}$$ such that $$\bigcap \tilde{\mathcal{A}} = \emptyset$$
 * 3) Each filter on $$X$$ has an limit point in $$X$$
 * 4) Each ultrafilter on $$X$$ converges

(1) $$\Rightarrow$$ (2)
$$\mathcal{V} := \{ X \setminus A | A \in \mathcal{A} \}$$ is an open cover of $$X$$ since all sets in $$\mathcal{A}$$ are closed in $$X$$ and $$\bigcap \mathcal{A} = \emptyset$$, which implies $$X \setminus \bigcup \mathcal{V} = \bigcap \{ X \setminus V | V \in \mathcal{V} \} = \bigcap \{ A | A \in \mathcal{A} \} = \emptyset$$ and therefore $$X = \bigcup \mathcal{V}$$. By (1) there exists a finite subcover $$\tilde{\mathcal{V}} \subseteq \mathcal{V}$$. Define $$\tilde{\mathcal{A}} := \{ X \setminus V | V \in \tilde{\mathcal{V}} \}$$, then $$\tilde{\mathcal{A}} \subseteq \mathcal{A}$$ by definition of $$\mathcal{V}$$. Obviously $$\bigcap \tilde{\mathcal{A}} = \bigcap \{ X \setminus V | V \in \tilde{\mathcal{V}} \} = X \setminus \bigcup \tilde{\mathcal{V}} = \emptyset$$ because $$\tilde{\mathcal{V}}$$ covers $$X$$.

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

(3) $$\Rightarrow$$ (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) $$\Rightarrow$$ (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) $$\Rightarrow$$ (3)
Let $$\mathcal{F}$$ be a filter on $$X$$. Assume that $$\mathcal{F}$$ has no limit point. This would imply that $$\bigcap \{ \overline{F} | F \in \mathcal{F} \} = \emptyset$$. By (2) there are therefore sets $$F_1, \ldots, F_n \in \mathcal{F}$$ such that $$\overline{F}_1 \cap \ldots \cap \overline{F}_n = \emptyset$$. 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) $$\Rightarrow$$ (2)
Let $$\mathcal{A} \subset \mathcal{P}(X)$$ be a set of closed subsets of $$X$$. Assume that $$\bigcap \tilde{\mathcal{A}} \ne \emptyset$$ for all finite subsets $$\tilde{\mathcal{A}}$$ of $$\mathcal{A}$$. We show that this implies $$\bigcap \mathcal{A} \ne \emptyset$$. Because of our assumption, $$\mathcal{B} := \{ \bigcap \tilde{\mathcal{A}} | \tilde{\mathcal{A}} \subseteq \mathcal{A} \text{ finite} \}$$ is a filter basis. Let $$\mathcal{F}$$ be the corresponding generated filter. Then $$\mathcal{F}$$ has a limit point by (3) and thus $$\emptyset \ne \bigcup \{ \overline{F} | F \in \mathcal{F} \} \subseteq \bigcap \mathcal{B} \subseteq \bigcap \mathcal{A}$$. Thus $$\bigcap \mathcal{A} \ne \emptyset$$. Therefore (2) follows.

$$\square$$