point and a compact set in a Hausdorff space have disjoint open neighborhoods.

Theorem.

Let $X$ be a Hausdorff space, let $A$ be a compactnon-empty set in $X$, and let $y$ a point in the complement of $A$.Then there exist disjoint open sets $U$ and $V$ in $X$ such that$A\mathrm{\subset }U$ and $y\mathrm{\in }V$.

Proof.

First we use the fact that $X$ is a Hausdorff space.Thus, for all $x\in A$ there existdisjoint open sets $U_x$ and $V_x$ such that $x\in U_x$ and $y\in V_x$.Then ${\{U_x\}}_{x\in A}$ is an open cover for $A$.Using this characterization of compactness (http://planetmath.org/YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover),it follows that thereexist a finite set $A_0\subset A$ such that ${\{U_x\}}_{x\in A_0}$is a finite open cover for $A$.Let us define

$U={\displaystyle \bigcup _{x\in A_0}}{U}_x,$$\mathrm{\hspace{0.17em}}V={\displaystyle \bigcap _{x\in A_0}}{V}_x.$

Next we show that these sets satisfy the given conditionsfor $U$ and $V$. First, it is clearthat $U$ and $V$ are open. We also have that $A\subset U$ and $y\in V$.To see that $U$ and $V$ are disjoint, suppose $z\in U$. Then$z\in U_x$ for some $x\in A_0$.Since $U_x$ and $V_x$ are disjoint, $z$ can not be in $V_x$, andconsequently $z$ can not be in $V$.∎

The above result and proof follows [1] (Chapter 5, Theorem 7) or [2] (page 27).