Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods
Theorem
Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $V_1$ and $V_2$ be compact sets in $T$.
Then $V_1$ and $V_2$ have disjoint neighborhoods.
Lemma
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 $\tuple {Z, W}$ such that:
- $Z, W \in \tau$
- $V_1 \subseteq Z$
- $Z \cap W = \O$
By the lemma, $\Img \FF$ covers $V_2$.
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By the definition of compact space, there exists a finite subset $K$ of $\Img \FF$ which also covers $V_2$.
By the definition of topology, $\ds \bigcup K$ is open.
By the Principle of Finite Choice, there exists a bijection $\GG \subseteq \FF$ such that $\Img \GG = K$.
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Then $\GG$, and hence its preimage, will be finite.
Let $\ds J = \bigcap \Preimg \GG$.
By Intersection is Largest Subset, $V_1 \subseteq J$.
By the definition of a topology, $J$ is open.
Then $\ds \bigcup K$ and $J$ are disjoint open sets such that $\ds V_2 \subseteq \bigcup K$ and $V_1 \subseteq J$.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms