Let $X$ be a Hausdorff topological space.
Prove that if $\{ C_i | i \in I \}$ is an infinite collection of compact subsets of $X$ such that $\cap_{i \in I} C_i = \emptyset$, then some finite sub collection of $\{ C_i | i \in I \}$ also has an empty intersection.
Well my idea is that I want to show that $X$ is compact. Since each $C_i$ is a compact subset of $X$, this means each $C_i$ is closed, which means $(X \setminus C_i)$ is an open set. I want to say that $\cup_{i \in I} (X \setminus C_i)$ is an open cover of X but I don't know if this proves $X$ is a compact set. Do you think I am on the right track? Thank you very much.
Best Answer
$X$ need not be compact but the implication is true as long as $X$ is Hausdorff.
Fix $j \in I$. Then $C_j$ is covered by $X \setminus C_i, i \neq j$. Hence there is a finite subcover, say $C_{i_1},C_{i_2}...,C_{i_n}$. Now the intersection of $C_j,C_{i_1},C_{i_2}...,C_{i_n}$ is empty.