Prove that if $X$ is a compact space and $H = \{h_{\alpha} : \alpha \in A\}$ is any collection of closed subsets with the property that $\cap_{\alpha}h_{\alpha} = \emptyset$, then there is a finite collection $\{h_{i}\}$ of sets from $H$ with the property that $\cap_{i}h_{i} = \emptyset$.
My attempt : By De Morgans Laws, it follows that $\cup(X \setminus h_{\alpha}) = X$. Since each $h_{\alpha}$ is closed then the collection $\{X \setminus h_{\alpha}: \alpha \in A\}$ is an open cover of $X$. By assumption, $X$ is compact, so this cover admits a finite subcover. Therefore, $\exists$ some $K$ such that $X = (X \setminus h_{\alpha_{1}}) \cup ... \cup (X \setminus h_{\alpha_{k}})$. So we have a finite collection $\{h_{\alpha_{1}}...h_{\alpha_{k}}\}$.
If their is a better way to write this proof, or its wrong let me know. I think its the right idea though.
\setminus. $\endgroup$ – Brian M. Scott Mar 22 '13 at 4:48