Neighborhood in Compact Hausdorff Space Contains Compact Neighborhood

Theorem
Let $X$ be a compact Hausdorff topological space.

Let $x\in X$.

Let $U$ be a neighborhood of $x$.

Then $U$ contains a compact neighborhood of $x$.

Proof
By definition of neighborhood, there exists an open set $V$ with $x\in V\subset U$.

Then $X \setminus V$ is closed.

By Compact Hausdorff Space is T4, there exist disjoint open sets $A,B$ such that:
 * $X\setminus V\subset A$
 * $x\in B$

Then:
 * $X \setminus A$ is compact by Closed Subspace of Compact Space is Compact
 * $B \subset X \setminus A$ by Empty Intersection iff Subset of Relative Complement
 * $X \setminus A \subset X \setminus \paren{X \setminus V}$ by Relative Complement inverts Subsets
 * $X \setminus \paren{X \setminus V} = V$ by Relative Complement of Relative Complement

Thus $x\in B \subset X\setminus A \subset V$, so $X\setminus A$ is a compact neighborhood of $x$ contained in $V$.

Also see

 * Definition:Locally Compact Hausdorff Space