Finite intersection property

Statement

Let $X$ be a topological space and let $K \subseteq X$. Then $K$ is compact if and only if it has the property that follows. Let $(F_i)_{i \in I}$ be a family of closed subsets of $X$ with $K \cap F_{i_1} \cap \cdots \cap F_{i_n} \neq \emptyset$ for all finitely many indices $i_1, \ldots, i_n \in I$. Then also $K \cap \bigcap_{i \in I} F_i \neq \emptyset$.

Remarks

A family of sets has the finite intersection property when each finite subfamily has a nonempty intersection. Proof: take complements. The sets $X \setminus F_i$ cover $K$ if and only if $K \cap \bigcap_i F_i = \emptyset$. Then use compactness via open covers. For $K = X$: a space is compact if and only if each family of closed sets with the finite intersection property has nonempty intersection. Cantor's intersection theorem is a special case.

Explanations

L1 human

This is the dual form of compactness, with closed sets instead of open sets. Suppose that each finite subfamily of the closed sets has a common point in $K$. Then the whole family has a common point in $K$.