A topological space \(X\) is said to be regular if disjoint singleton and closed sets can be separated by disjoint open sets. Prove that in a regular space disjoint closed and compact sets can be separated by disjoint open sets.
Solution
Let \(F,K\subseteq X\) be disjoint, with \(F\) closed and \(K\) compact.
1. For each \(x\in K\), there are disjoint open sets \(U_x,V_x\subseteq X\) such that\[ x\in U_x \qquad\text{and}\qquad F\subseteq V_x. \]
For each \(x\in K\), disjointness of \(F\) and \(K\) gives \[ x\notin F. \] Thus the singleton \(\{x\}\) and the closed set \(F\) are disjoint. By regularity, they can be separated by disjoint open sets. Hence there are open \(U_x,V_x\) with \[ x\in U_x, \qquad F\subseteq V_x, \qquad U_x\cap V_x=\varnothing. \]
2. Finitely many of the sets \(U_x\) cover \(K\).
1. Since \(K\) is compact, there exist points\[ x_1,\ldots,x_n\in K \]such that\[ K\subseteq U_{x_1}\cup\cdots\cup U_{x_n}. \]
3. Define\[ U=U_{x_1}\cup\cdots\cup U_{x_n}, \qquad V=V_{x_1}\cap\cdots\cap V_{x_n}. \]Then \(U\) and \(V\) are open, with\[ K\subseteq U \qquad\text{and}\qquad F\subseteq V. \]
2. The set \(V\) is a finite intersection of open sets, hence open.By 1, every \(V_{x_i}\) contains \(F\), so their intersection \(V\) also contains \(F\).
4. The open sets \(U\) and \(V\) are disjoint.
1. Therefore\[ U\cap V=\varnothing. \]
5. Hence the closed set \(F\) and the compact set \(K\) can be separated by disjoint open sets.
By 3, \(U\) and \(V\) are open neighborhoods of \(K\) and \(F\), respectively, and by 4 they are disjoint. This is exactly the required separation.