Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Lemma

Theorem
Let $\struct {S, \tau}$ be a Hausdorff space.

Let $C$ be a compact subspace of $S$.

Let $x \in S \setminus C$.

Then there exist open sets $U$ and $V$ such that $x \in U$, $C \subseteq V$, and $U \cap V = \O$.

Proof
Let $\FF$ be the set of all ordered pairs $\tuple {A, B}$ such that $A$ and $B$ are open, $x \in A$, and $A \cap B = \O$.

As a set of ordered pairs, $\FF$ constitutes a relation on $\tau$:
 * $\FF \subseteq \tau \times \tau$

By the definition of Hausdorff space, for each $y \in C$ there exists an element $\tuple {A, B} \in \FF$ such that $y \in B$.

Thus the image of $\FF$ covers $C$.

By the definition of compactness, there exists a finite subset $\GG \subseteq \FF$ such that:
 * $\ds C \subseteq V = \bigcup \Img \GG$

Then $\Preimg \GG$ is also finite, so by the definition of a topology:


 * $\ds U = \bigcap \Preimg \GG$ is open.

Then $x \in U$, $C \subseteq V$, and $U \cap V = \O$.