Theorem 1   Let $X$ be a Hausdorff space, let $A$ be a compact non-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 $A\subset U$ and $y\in V$ .

Proof. First we use the fact that $X$ is a Hausdorff space. Thus, for all $x\in A$ there exist disjoint 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, it follows that there exist 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 \begin{eqnarray*} U= \bigcup_{x\in A_0} U_x,\,\,\,\,\,&\,&\,\,\,\,\, V= \bigcap_{x\in A_0} V_x. \end{eqnarray*}Next we show that these sets satisfy the given conditions for $U$ and $V$ . First, it is clear that $U$ 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 $V_x$ , and consequently $z$ can not be in $V$ .