Compact Subsets of T3 Spaces

Theorem
Let $T = \struct {S, \tau}$ be a $T_3$ space.

Let $A \subseteq S$ be compact in $T$.

Then for each $U \in \tau$ such that $A \subseteq U$:
 * $\exists V \in \tau: A \subseteq V \subseteq V^- \subseteq U$

where $V^-$ denotes the closure of $V$.

Proof
Let $A \subseteq S$ be compact in $T$.

Let $U \in \tau$ such that $A \subseteq U$.

Since $T$ is $T_3$:


 * Each open set contains a closed neighborhood around each of its points:


 * $\forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U$

Note that $\set {V_x: x \in A}$ forms an open cover for $A$.

By compactness of $A$, $\set {V_x: x \in A}$ has a finite subcover.

Therefore there is a finite subset $I \subseteq A$ such that:
 * $\set {V_x: x \in I}$ is an open cover for $A$.

Let $V = \displaystyle \bigcup_{i \mathop \in I} V_x$, $N = \displaystyle \bigcup_{i \mathop \in I} N_x$.

By definition of a topology, the set $\displaystyle \bigcup_{i \mathop \in I} V_x$ is open since it is a union of open sets.

By Finite Union of Closed Sets is Closed in Topological Space, the set $\displaystyle \bigcup_{i \mathop \in I} N_x$ is closed since it is a finite union of closed sets.

By Set Union Preserves Subsets, since $V_x \subseteq N_x \; \forall x \in I \subseteq U$:
 * $V \subseteq N$

By Set Closure is Smallest Closed Set in Topological Space:
 * $V \subseteq V^- \subseteq N$

Thus:

satisfies our claim.