Equivalence of Definitions of Compact Topological Space

$(2) \implies (4)$
Let $T = \struct {S, \tau}$ satisfy the Finite Intersection Axiom.

Let $\FF$ be a filter on $X$.

that $\FF$ has no limit point.

Thus:
 * $\ds \bigcap_{F \mathop \in \FF} \overline F = \O$

By hypothesis there are therefore sets $F_1, \ldots, F_n \in \FF$ such that:
 * $\overline F_1 \cap \ldots \cap \overline F_n = \O$

Because for any set $M$ we have $M \subseteq \overline M$:
 * $\overline F_1, \ldots, \overline F_n \in \FF$

But by definition of a filter, $\FF$ must not contain the empty set.

Thus $\FF$ has a limit point.

$(4) \implies (2)$
Let $\AA \subset \powerset S$ be a set of closed subsets of $S$.

Let:
 * $\bigcap \tilde \AA \ne \O$

for all finite subsets $\tilde \AA$ of $\AA$.

We show that this implies $\bigcap \AA \ne \O$.

From our assumption, $\BB := \set {\bigcap \tilde \AA : \tilde \AA \subseteq \AA \text{ finite} }$ is a filter basis.

Let $\FF$ be the corresponding generated filter.

By hypothesis $\FF$ has a limit point.

Thus:
 * $\ds \O \ne \bigcap_{F \mathop \in \FF} \overline F \subseteq \bigcap \BB \subseteq \bigcap \AA$.

Thus $\bigcap \AA \ne \O$.

Hence $T = \struct {S, \tau}$ satisfies the Finite Intersection Axiom.