Equivalence of Definitions of Compact Topological Space

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


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

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:


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

(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.