# Kelley's topology: Compact spaces and finite intersection property

Lemma

A topological space is compact iff each family of closed sets which have the finite intersection property has a non-void intersection.

I've proved the same result in another way but I really can't catch his argument in the second statement of the proof.

Proof

If A is a family of subsets of a topological space X, then, according to the De Morgan formulae, $X\setminus\bigcup\{a : a\in A\} = \bigcap\{X\setminus a : a\in A\}$ and hence A is a cover of X iff the intersection of the complements of the members of A is void. The space X is compact iff each family of open sets, such that no finite subfamily covers X, fails to be a cover, and this is true iff each family of closed sets which possesses the finite intersection property has a non-void intersection.

This proof is from John Kelley "General Topology".

• What is unclear for you ? – user171326 Jul 1 '15 at 12:48

"The space $X$ is compact iff each family of open sets, such that no finite subfamily covers $X$, fails to be a cover" $\Rightarrow$ This follow from the definition of compact. Now the last argument is to show equivalence between :
1. Every family $\mathcal A$ of open sets which has no finitely subcover is not a cover itself
2. Each family $\mathcal B$ of closed sets which has the finite intersection property has a non-void intersection
I will show for example $1 \Rightarrow 2$ and the other way is similar. Assume $\mathcal B = (B_i)_{i \in I}$ is a family of closed subset with the finite intersection property. For every $B_1, \dots, B_n \in \mathcal B$, $\cap_{j=1}^n B_j \neq \emptyset$ by definition. Now define for all $i \in I$, $O_i = B_i^c$. Then, there is not $(O_j)_{j=1}^n$ which cover $X$, because if some finite open $O_j$ cover $X$, then the corresponding $B_j$ has empty intersection, contradicting the hypothesis ! So by $1)$ the full family $(O_i)$ fails to cover $X$, i.e $\mathcal B$ has a non-empty intersection.