Disjoint compact subsets of a Hausdorff space are separated by disjoint open neighborhoods

Question

Let $X$ be a Hausdorff space and let $A,B\subseteq X$ two compact subspaces which don't intersect. Show exist $U,V\subseteq X$ open which don't intersect s.t $A\subseteq U,B\subseteq V$.

I thought taking $A\ni a\neq b\in B$ and since $X$ is Hausdorff, $$\exists S_{a,i}\in N(a),S_{b,j}\in N(b):S_{a,i}\cap S_{b,j}=\emptyset$$ and taking $S_a=\bigcap_{a\in A,i\in I}S_{a,i},S_b=\bigcap_{b\in B,j\in J}S_b$ as my $U$ and $V$.

The only problem is I don't use the compactness. What am I missing? Is my solution correct?

Answer 1

Let $b \in B$.

For each $a \in A$, we have $U_a \ni a, \ V_a \ni b$ open such that $U_a \cap V_a=\emptyset$. The $U_a$ form an open cover of $A$ so have a finite sub cover $U_{a_1}, \ldots, U_{a_k}$.

Let $$U_b = U_{a_1}\cup\cdots\cup U_{a_k}$$ which is open as it is a union of a finite number of open sets. Let $$V_b = V_{a_1}\cap\cdots\cap V_{a_k}$$

The $V_b$ form an open cover of $B$, and for each $b \in B$, $V_b \cap U_b = \emptyset$. Also note that for every $b$, $A \subset U_b$.

Can you use the compactness of $B$ to finish the proof?

Answer 2

Hint: you need compactedness to choose a finite subcover of an open cover (which you choose s.t. it has properties that you like).

How is this going to help you? Well, finite intersections of open sets are open...