Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods

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Theorem

Let $T = \left({S, \tau}\right)$ 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 $\mathcal F$ be the set of all ordered pairs $\left({Z, W}\right)$ such that:

$Z, W \in \tau$
$V_1 \subseteq Z$
$Z \cap W = \varnothing$

By the lemma, $\operatorname{Im} \left({\mathcal F}\right)$ covers $V_2$.

By the definition of compact space, there exists a finite subset $K$ of $\operatorname{Im} \left({\mathcal F}\right)$ 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 $\mathcal G \subseteq \mathcal F$ such that $\operatorname{Im} \left({\mathcal G}\right) = K$.

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

Let $\displaystyle J = \bigcap \operatorname{Im}^{-1} \left({\mathcal G}\right)$

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 $V_1 \subseteq J$.

$\blacksquare$