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:
- $\ds C \subseteq V = \bigcup \Img \GG$
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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 $U \cap V = \O$.
$\blacksquare$