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 $\struct {S, \tau}$ 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 1

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 $V_1 \subseteq J$.

$\blacksquare$


Proof 2

We first suppose that $V_1 = \set x$ and $x \not \in V_2$.

Since $T$ is Hausdorff:

for each $y \in V_2$ there exists an open neighborhood $O^1_y$ of $x$ and an open neighborhood $O^2_y$ of $y$ such that $O^1_y \cap O^2_y = \O$.

Hence we have:

$\ds V_2 \subseteq \bigcup_{y \mathop \in V_2} O_y^2$

Since $V_2$ is compact, there exists $y_1, \ldots, y_n \in V_2$ such that:

$\ds V_2 \subseteq \bigcup_{i \mathop = 1}^n O_{y_i}^2$

Now let:

$\ds U_2 = \bigcup_{i \mathop = 1}^n O_{y_i}^2$

and:

$\ds U_1 = \bigcap_{i \mathop = 1}^n O_{y_i}^1$

We have that $U_1$ and $U_2$ are open, $x \in U_1$ and $V_2 \subseteq U_2$.

Further, if $x \in U_2$ then $x \in O_{y_i}^2$ for some $i$.

Since $O_{y_i}^2 \cap O_{y_i}^1 = \O$, it follows that $x \not \in O_{y_i}^1$.

Hence $x \not \in U_1$.

So $U_1 \cap U_2 = \O$.

Hence $U_1$ and $U_2$ are disjoint open sets with $V_1 \subseteq U_1$ and $V_2 \subseteq U_2$.

Hence we have proven the claim in the case $V_1 = \set x$.


Now take $V_1$ to be general.

For each $x \in V_1$, there exists open sets $U_{1, x}$ and $U_{2, x}$ such that $U_{1, x} \cap U_{2, x} = \O$, $x \in U_{1, x}$ and $V_2 \subseteq U_{2, x}$.

In particular:

$\ds V_1 \subseteq \bigcup_{x \in V_1} U_{1, x}$

Since $V_1$ is compact, there exists $x_1, \ldots, x_n \in V_1$ such that:

$\ds V_1 \subseteq \bigcup_{i \mathop = 1}^n U_{1, x_i}$

Now set:

$\ds U_1 = \bigcup_{i \mathop = 1}^n U_{1, x_i}$

and:

$\ds U_2 = \bigcap_{i \mathop = 1}^n U_{2, x_i}$

We have $U_1 \cap U_2 = \O$ by an identical argument to above.

Further, $U_1$ and $U_2$ are both open.

Since $V_2 \subseteq U_{2, x_i}$ for each $i$ it follows that $V_2 \subseteq U_2$.

Hence $U_1$ and $U_2$ are the desired open sets.

$\blacksquare$


Sources

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