# Disjoint compact sets in a Hausdorff space can be separated

I want to show that any two disjoint compact sets in a Hausdorff space $X$ can be separated by disjoint open sets. Can you please let me know if the following is correct? Not for homework, just studying for a midterm. I'm trying to improve my writing too.

My work:

Let $C$,$D$ be disjoint compact sets in a Hausdorff space $X$. Now fix $y \in D$ and for each $x \in C$ we can find (using Hausdorffness) disjoint open sets $U_{x}(y)$ and $V_{x}(y)$ such that $x \in U_{x}(y)$ and $y \in V_{x}(y)$. Now the collection $\{U_{x}: x \in C\}$ covers $C$ so by compactness we can find some natural k such that

$C \subseteq \bigcup_{i=1}^{k} U_{x_{i}}(y)$

Now for simplicity let $U = \bigcup_{i=1}^{k} U_{x_{i}}(y)$, then $C \subseteq U$ and let $W(y) = \bigcap_{i=1}^{k} V_{x_{i}}(y)$. Then $W(y)$ is a neighborhood of $y$ and disjoint from $U$.

Now consider the collection $\{W(y): y \in D\}$, this covers D so by compactness we can find some natural q such that $D \subseteq \bigcup_{j=1}^{q} W_{y_{j}}$.

Finally set $V = \bigcup_{j=1}^{q} W_{y_{j}}$, then $U$ and $V$ are disjoint open sets containing $C$ and $D$ respectively.

What do you think?

• This looks good but there's a slight problem in that $U$ depends on $y$ but that's quite easily fixed. I'd like to suggest to do it in two steps: First show that given a compact set $C$ and a point $p \notin C$ you can find disjoint open sets $U_{p} \supset C$ and $V_{p} \ni p$. Now let $p$ run through $D$, and find $p_{1},\ldots,p_{n}$ by compactness and put $U = U_{p_{1}} \cap \cdots \cap U_{p_{n}}$ and $V = V_{p_{1}} \cup \cdots \cup V_{p_{n}}$.
– t.b.
Commented Jan 19, 2011 at 19:51
• Can you prove that a compact set and a point can be separated? Commented Jan 19, 2011 at 20:08

This is a very good start, but there is a slight problem with your argument: as you change $y$, your $U$ changes as well (since $U$ is constructed in terms of $y$); you should really call it $U(y)$.

Your construction gives you an open neighborhood $W(y)$ of $y$ for each $y$; $W(y)$ is disjoint from $U(y)$. But for all you know, $W(y)$ may fail to be disjoint from $U(y')$ with $y'\neq y$.

So you really still have a bit more to go before you are done.

• Thank you, yeah I see now. Can we fix the above construction taking as $U$ the set given by $U = \bigcap_{i=1}^{q} U(y_{i})$ ? Commented Jan 19, 2011 at 20:03
• @student: Exactly; that is, you are using the same "trick" (procedure) for the open set containing $C$ as you used to find the open set $W(y)$ containing $y$. Commented Jan 19, 2011 at 20:11

First prove the lemma:

Let $$X$$ be Hausdorff and $$A$$ be compact and $$p \notin A$$. Then there exist open sets $$U$$ and $$V$$ such that $$A \subseteq U$$, $$p \in V$$ and $$U \cap V = \emptyset$$.

The proof follows your idea: for each $$a \in A$$ we pick $$U(a)$$ open and $$V(a)$$ open and disjoint such that $$a \in U(a), p \in V(a)$$ by Hausdorffness. The $$\{U(a): a \in A\}$$ cover $$A$$ by construction so by compactness of $$A$$, we have finitely many $$a_1, \ldots a_n$$ such that

$$A \subseteq U:=\bigcup_{i=1}^n U(a_i)$$

and as we have finitely many corresponding $$V(a_i)$$ as well, $$p \in V:= \bigcap_{i=1}^n V(a_i)$$ which is then open. And no point is in $$U \cap V$$ or it would be in some $$U(a_i)$$ (from $$U$$ being a union) and also in the same $$V(a_i)$$ ($$V$$ being the intersection), contradicting how they were chosen. So $$U$$ and $$V$$ are as required. QED for the lemma.

Now apply the lemma repeatedly for $$c \in C$$ and $$D$$ when these are disjoint compact. We get $$U(c)$$ and $$V(c)$$ disjoint neighbourhoods of $$c$$ and $$D$$ (!) now, compactness lets us take a finite union of $$U(c)$$ as $$U$$ and the corresponding intersection of $$V(c)$$'s will work again, same argument essentially.

So in two steps is cleanest. Make the inbetween step visible.

$$U=⋃k_i=Ux_i(y)$$ might meet $$V$$, so to avoid this problem use compactness of the second set. I think if your prove is completely correct, we don't need to suppose compactness of the two. and we consider simply $$V=⋃w(y)$$; $$y$$ in $$D$$