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diff --git a/pages/general-topology/compactness/index.md b/pages/general-topology/compactness/index.md
index 60c29a0..37e9b4d 100644
--- a/pages/general-topology/compactness/index.md
+++ b/pages/general-topology/compactness/index.md
@@ -1,9 +1,46 @@
 ---
 title: Compactness
 parent: General Topology
-nav_order: 1
+nav_order: 5
 has_children: true
-# cspell:words
+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 %}
+