Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods

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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 $\left({S, \tau}\right)$ 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 $U \cap V = \varnothing$.


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 $V_2$.



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, $\displaystyle \bigcup K$ is open.

By the Principle of Finite Choice, there exists a bijection $\GG \subseteq \FF$ such that $\Img \GG = K$.



Then $\GG$, and hence its preimage, will be finite.

Let $\displaystyle J = \bigcap \Preimg \GG$.

By Intersection is Largest Subset, $V_1 \subseteq J$.

By the definition of a topology, $J$ is open.

Then $\displaystyle \bigcup K$ and $J$ are disjoint open sets such that $\displaystyle V_2 \subseteq \bigcup K$ and $V_1 \subseteq J$.

$\blacksquare$


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