a space is compact iff any family of closed sets having fip has non-empty intersection

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 essentiallythe definition of a compact space rewritten using de Morgan’s laws.The usual definition of a compact space is based on open sets andunions. The above characterization, on the other hand, is writtenusing closed sets and intersections.

Proof. Suppose $X$ is compact, i.e., any collection of open subsetsthat cover $X$ has a finite collection that also cover $X$. Further, suppose${\{F_i\}}_{i\in I}$ is an arbitrary collection of closed subsetswith the finite intersection property. We claim that ${\cap }_{i\in I}{F}_i$is non-empty.Suppose otherwise, i.e., suppose ${\cap }_{i\in I}{F}_i=\mathrm{\varnothing }$. Then,

$X$$=$${\left({\displaystyle \bigcap _{i\in I}}{F}_i\right)}^c$

$=$${\displaystyle \bigcup _{i\in I}}{F}_i^c.$

(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 afinite subset $J\subset I$ suchthat $X={\cup }_{i\in J}{F}_i^c$. But then$X={({\cap }_{i\in J}{F}_i)}^c$, so ${\cap }_{i\in J}{F}_i=\mathrm{\varnothing }$, whichcontradicts 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 setsin $X$ that cover $X$. We claim that this collection contains a finite subcollectionof sets that also cover $X$.The proof is by contradiction.Supposethat $X\ne {\cup }_{i\in J}{F}_i$ 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

${\displaystyle \bigcap _{i\in J}}{F}_i^c$$=$${\left({\displaystyle \bigcup _{i\in J}}{F}_i\right)}^c\ne \mathrm{\varnothing },$

where the last assertion follows since $J$ was finite.Then, since $X$ has the finite intersection property,

$\mathrm{\varnothing }$$\ne$${\displaystyle \bigcap _{i\in I}}{F}_i^c={\left({\displaystyle \bigcup _{i\in I}}{F}_i\right)}^c.$

This contradicts the assumption that ${\{F_i\}}_{i\in I}$ is a cover for $X$.$\mathrm{\square }$