# A question concerning compactness - Topology

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.

• Looks good to me! – Paul VanKoughnett Mar 22 '13 at 0:38
• That's great, +1. I hope you don't mind: I changed the complements to $\setminus$. Maybe you could conclude your argument by "So we have a finite collection bla" with empty intersection. – Julien Mar 22 '13 at 0:42
• Thanks. The only reason I did not use the other backlash is because I dont know how to do it in latex lol – Roger Mar 22 '13 at 0:49
• The $\LaTeX$ set difference symbol is \setminus. – Brian M. Scott Mar 22 '13 at 4:48