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 $V$.

$\blacksquare$

Also see

• Definition:Locally Compact Hausdorff Space