Let \(X\) be a Hausdorff space, \(p\in X\), and \(A\subseteq X\) a compact subset of \(X\) disjoint from \(p\). Show that there exist disjoint open sets \(U,V\subseteq X\) with \(p\in U\) and \(A\subseteq V\).
Solution
If \(A=\varnothing\), take \[ U=X, \qquad V=\varnothing. \] Thus assume \(A\ne\varnothing\).
1. For every \(a\in A\), there are disjoint open sets \(U_a,V_a\subseteq X\) such that\[ p\in U_a, \qquad a\in V_a. \]
Because \(p\notin A\), we have \(p\ne a\) for every \(a\in A\). Since \(X\) is Hausdorff, any two distinct points admit disjoint open neighborhoods. Apply this to \(p\) and \(a\).
2. There are points \(a_1,\ldots,a_n\in A\) such that\[ A\subseteq V_{a_1}\cup\cdots\cup V_{a_n}. \]
By 1, the family \[ \{V_a:a\in A\} \] is an open cover of \(A\). Since \(A\) is compact, this cover has a finite subcover.
3. Define\[ U=U_{a_1}\cap\cdots\cap U_{a_n}, \qquad V=V_{a_1}\cup\cdots\cup V_{a_n}. \]Then \(U\) and \(V\) are open, \(p\in U\), and \(A\subseteq V\).
A finite intersection of open sets and a finite union of open sets are open. Each \(U_{a_i}\) contains \(p\), so their intersection contains \(p\). The inclusion \(A\subseteq V\) is exactly 2.
4. The open sets \(U\) and \(V\) are disjoint.
1. Hence \(U\cap V=\varnothing\).Together with 3, these are the required neighborhoods.