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:


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

(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 $$\bigcap \left\{{\overline F : F \in \mathcal F}\right\} = \varnothing$$.

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 $$\mathcal A \subset \mathcal P \left({X}\right)$$ 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.