Theorem. A topological space is compact if and only if any collection of its closed sets having the finite intersection property has non-empty intersection.

The above theorem is essentially the definition of a compact space rewritten using de Morgan's laws. The usual definition of a compact space is based on open sets and unions. The above characterization, on the other hand, is written using closed sets and intersections.

Proof. Suppose $X$ is compact, i.e., any collection of open subsets that cover $X$ has a finite collection that also cover $X$ . Further, suppose $\{F_i\}_{i\in I}$ is an arbitrary collection of closed subsets with the finite intersection property. We claim that $\cap_{i\in I} F_i$ is non-empty. Suppose otherwise, i.e., suppose $\cap_{i\in I} F_i=\emptyset$ . Then, \begin{eqnarray*} X&=&\left(\bigcap_{i\in I} F_i\right)^c\\ &=&\bigcup_{i\in I} F_i^c. \end{eqnarray*}(Here, the complement of a set $A$ in $X$ is written as $A^c$ .) Since each $F_i$ is closed, the collection $\{F_i^c\}_{i\in I}$ is an open cover for $X$ . By compactness, there is a finite subset $J\subset I$ such that $X=\cup_{i\in J} F_i^c$ . But then $X=(\cap_{i\in J} F_i)^c$ , so $\cap_{i\in J} F_i=\emptyset$ , which contradicts the finite intersection property of $\{F_i\}_{i\in I}$ .

The proof in the other direction is analogous. Suppose $X$ has the finite intersection property. To prove that $X$ is compact, let $\{F_i\}_{i\in I}$ be a collection of open sets in $X$ that cover $X$ . We claim that this collection contains a finite subcollection of sets that also cover $X$ . The proof is by contradiction. Suppose that $X\neq \cup_{i\in J} F_i$ holds for all finite $J\subset I$ . Let us first show that the collection of closed subsets $\{F_i^c\}_{i\in I}$ has the finite intersection property. If $J$ is a finite subset of $I$ , then \begin{eqnarray*} \bigcap_{i\in J} F^c_i &=& \Big(\bigcup_{i\in J} F_i\Big)^c \neq \emptyset, \end{eqnarray*}where the last assertion follows since $J$ was finite. Then, since $X$ has the finite intersection property, \begin{eqnarray*} \emptyset &\neq& \bigcap_{i\in I} F_i^c = \Big(\bigcup_{i\in I} F_i\Big)^c. \end{eqnarray*}This contradicts the assumption that $\{F_i\}_{i\in I}$ is a cover for $X$ .

Bibliography

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R.E. Edwards, Functional Analysis: Theory and Applications, Dover Publications, 1995.