Suppose \(A,B\) are disjoint, compact subspaces of the Hausdorff topological space \(X\). Prove there are open subsets \(U,V\) of \(X\) such that \(A\subseteq U\), \(B\subseteq V\), and \(U\cap V=\varnothing\).
Solution
If \(A=\varnothing\), take \(U=\varnothing\) and \(V=X\); if \(B=\varnothing\), take \(U=X\) and \(V=\varnothing\). Assume from now on that \(A\) and \(B\) are nonempty. For every \((a,b)\in A\times B\) we have \(a\neq b\), so since \(X\) is Hausdorff we may choose disjoint open sets \(U_{a,b},V_{a,b}\subseteq X\) with \(a\in U_{a,b}\) and \(b\in V_{a,b}\).
1. For each \(a\in A\) there are disjoint open sets \(U_a\ni a\) and \(V_a\supseteq B\).
The family \(\{V_{a,b}:b\in B\}\) is an open cover of the compact space \(B\), so there are \(b_1,\ldots,b_n\in B\) with \(B\subseteq V_{a,b_1}\cup\cdots\cup V_{a,b_n}\). Put \(U_a=\bigcap_{i=1}^n U_{a,b_i}\) and \(V_a=\bigcup_{i=1}^n V_{a,b_i}\). These are open as a finite intersection and a finite union of open sets, \(a\in U_a\), and \(B\subseteq V_a\). If \(x\in U_a\cap V_a\), then \(x\in V_{a,b_i}\) for some \(i\) while \(x\in U_a\subseteq U_{a,b_i}\), contradicting \(U_{a,b_i}\cap V_{a,b_i}=\varnothing\).
2. There exist \(a_1,\ldots,a_m\in A\) with \(A\subseteq U_{a_1}\cup\cdots\cup U_{a_m}\).
By step 1, \(\{U_a:a\in A\}\) is an open cover of the compact space \(A\), so it has a finite subcover.
3. Q.E.D.
1.Hence \(U\) and \(V\) are disjoint.