Let \(X\) be a compact space and let \(\{C_\alpha\}_{\alpha\in A}\) be a collection of closed sets in \(X\). Let \(C=\bigcap_{\alpha\in A}C_\alpha\) and let \(U\) be an open set containing \(C\). Prove there is a finite set \(\alpha_1,\ldots,\alpha_n\) in \(A\) with \(C_{\alpha_1}\cap\cdots\cap C_{\alpha_n}\subseteq U\).
Solution
1. The family\[ \{U\}\cup\{X\setminus C_\alpha:\alpha\in A\} \]is an open cover of \(X\).
Each \(C_\alpha\) is closed, so every \(X\setminus C_\alpha\) is open. Let \(x\in X\). If \(x\in U\), then \(x\) is covered by \(U\). If \(x\notin U\), then \(x\notin C\) because \(C\subseteq U\). Since \[ C=\bigcap_{\alpha\in A}C_\alpha, \] there is some \(\alpha\) with \(x\notin C_\alpha\), so \(x\in X\setminus C_\alpha\). Thus the displayed family covers \(X\).
2. Compactness of \(X\) gives indices \(\alpha_1,\dots,\alpha_n\in A\) such that\[ X=U\cup(X\setminus C_{\alpha_1})\cup\cdots\cup(X\setminus C_{\alpha_n}). \]
1.If a finite subcover omits \(U\), adjoin \(U\) without affecting finiteness.
3. One has\[ C_{\alpha_1}\cap\cdots\cap C_{\alpha_n}\subseteq U. \]
Let \(x\) lie in the finite intersection on the left. Then \(x\) lies in none of the complements \(X\setminus C_{\alpha_i}\). The covering equality in 2 therefore forces \(x\in U\).
4. This is the required finite family.
The indices \(\alpha_1,\dots,\alpha_n\) from 2 satisfy exactly the inclusion proved in 3.