# Compactness

## Compactness in Terms of Closed Sets

A topological space $X$ is compact if and only if it has the following property:

- Given any collection $C$ of closed subsets of $X$, if every finite subcollection of $C$ has nonempty intersection, then $C$ has nonempty intersection.

Proof By definition, a topological space $X$ is compact if and only if it has the following property:

- Given any collection $O$ of open subsets of $X$, if $O$ covers $X$, then there exists a finite subcollection of $O$ that covers $X$.

If $A$ is a collection of subsets of $X$, let $A_{c}={X∖A:A∈A}$ denote the collection of the complements of its members. Clearly, $B$ is a subcollection of $A$ if and only if $B_{c}$ is a subcollection of $A_{c}$. Moreover, note that $B$ covers $X$ if and only if $B_{c}$ has empty intersection. Taking the contrapositive, we reformulate above property:

- Given any collection $O$ of open subsets of $X$, if every finite subcollection of $O_{c}$ has nonempty intersection, then $O_{c}$ has nonempty intersection.

To complete the proof, observe that a collection $A$ consists of open subsets of $X$ if and only if $A_{c}$ consists of closed subsets of $X$. $□$

Definition (Finite Intersection Property)

TODO