---
title: Compactness
parent: General Topology
nav_order: 5
has_children: true
has_toc: false
---

# {{ page.title }}

## Compactness in Terms of Closed Sets

{% theorem %}
A topological space $X$ is compact
if and only if it has the following property:
- Given any collection $\mathcal{C}$ of closed subsets of $X$,
  if every finite subcollection of $\mathcal{C}$ has nonempty intersection,
  then $\mathcal{C}$ has nonempty intersection.
{% endtheorem %}

{% proof %}
By definition, a topological space $X$ is compact
if and only if it has the following property:
- Given any collection $\mathcal{O}$ of open subsets of $X$,
  if $\mathcal{O}$ covers $X$,
  then there exists a finite subcollection of $\mathcal{O}$ that covers $X$.

If $\mathcal{A}$ is a collection of subsets of $X$,
let $\mathcal{A}^c = \braces{ X \setminus A : A \in \mathcal{A}}$ denote
the collection of the complements of its members.
Clearly, $\mathcal{B}$ is a subcollection of $\mathcal{A}$
if and only if $\mathcal{B}^c$ is a subcollection of $\mathcal{A}^c$.
Moreover, note that $\mathcal{B}$ covers $X$ if and only if
$\mathcal{B}^c$ has empty intersection.
Taking the contrapositive, we reformulate above property:
- Given any collection $\mathcal{O}$ of open subsets of $X$,
  if every finite subcollection of $\mathcal{O}^c$ has nonempty intersection,
  then $\mathcal{O}^c$ has nonempty intersection.

To complete the proof, observe that a collection $\mathcal{A}$ consists of open subsets of $X$
if and only if $\mathcal{A}^c$ consists of closed subsets of $X$.
{% endproof %}

{% definition Finite Intersection Property%}
TODO
{% enddefinition %}