# Disjoint Compact Subsets in Hausdorff Space can be Contained in Open Disjoint Sets

To begin with, I know this question has been answered here multiple times. However, I want to gain some intuition about the proof and be lead to the right direction without being given the proof explicitly.

Let $$A,B$$ be disjoint compact subsets in a Hausdorff space.
I want to show $$\exists \; U,V \; : \; A \subseteq U, \; B\subseteq V$$
Where $$U,V \subseteq X$$ are disjoint and open

We know that subsets of Hausdorff spaces are also Hausdorff spaces. So every point in $$A$$ is contained in some open set that is disjoint from B (similarly for $$B$$). From here, we have open covers for $$A$$ and $$B$$, from which we can extract finitely many open sets to cover their respective sets by compactness.

So now we have disjoint finite covers for both sets, my goal now is to create a bigger set for $$A$$ and $$B$$ respectively. However I'm not quite sure if I'm heading in the right direction or if I've made a mistake anywhere.

Hint: It is helpful to first prove the following fact to be used in the proof: for any point $$x \notin B$$, there exist disjoint open sets $$U,V$$ with $$x \in U$$ and $$B \subset V$$.

Proof of the initial fact:

For each $$b \in B$$, there exist disjoint open sets $$U_b,V_b$$ such that $$x \in U_b$$ and $$b \in V_b$$. The set $$\{V_b: b \in B\}$$ is an open cover of $$B$$, so there exists a finite set $$b_1,\dots,b_n$$ such that $$\{V_{b_1},\dots,V_{b_n}\}$$ is a finite open cover of $$B$$.

Take $$U = \bigcap_{j=1}^n U_{b_j}$$ and $$V = \bigcup_{j=1}^n V_{b_j}$$. Verify that $$U,V$$ are disjoint open sets with $$x \in U$$ and $$B \subset V$$.

• I can see how $x \in U$ and that it's disjoint from $B$, but I'm not quite sure how to show the existence of $V$ Commented Apr 2, 2020 at 0:42
• My only point of trouble with the statement is coming up with the $V$. Can we just say that $V = B \cup W$ where $W$ is some open set that's disjoint from $U$ ? Commented Apr 2, 2020 at 1:08
• @HoopsMcCann see my latest edit. The point is that now we can directly apply the sort of argument that you were considering. Commented Apr 2, 2020 at 1:27
• from the proof, you showed that $V = B$, from there is it sufficient to say $B \subseteq V$ ? Also, why do we intersect all the $U_b$ sets? Can't we just have a single set $U$ with $x$ being an element by the Hausdorff property? Commented Apr 2, 2020 at 2:32
• @HoopsMcCann in what way did I show that $V = B$? Commented Apr 2, 2020 at 2:46