# $(X,T)$ is countably compact iff every countable family of closed sets with the finite intersection property has non-empty intersection

Prove that a topological space $(X,T)$ is countably compact iff every countable family of closed subsets which has the finite intersection property has a non-empty intersection.

My question is what exactly is this asking in the $\Rightarrow$ direction? If I let $\{F\}$ be a countable family of closed subsets that have the finite intersection property, then $(\forall \{F_i\}_{i=1}^{n})(\bigcap_{i=1}^{n}F_i \neq \emptyset)$. Then doesn't it have a non-empty intersection? Or is this trying to have me prove that $\bigcap_{F_i \in \{F\}} F_i \neq \emptyset$?

• You need to prove an equivalence of two statements $A$ and $B$. To prove this, first assume $A$ and under that assumption prove $B$, second assume $B$ and under that assumption prove $A$. Don't let yourself be confused by how the statements look. – Anon Jun 6 '16 at 15:54
• I understand how to prove "if and only if" statements, but what I'm asking is specifically what ( in the $\Rightarrow$ direction) what the "$B$" statement is asking me to prove. – Oliver G Jun 6 '16 at 15:58

For the forward implication you assume that $X$ is countably compact and try to show that whenever $\mathscr{F}$ is a countable family of closed subsets of $X$ with the finite intersection property, then $\bigcap\mathscr{F}\ne\varnothing$. Thus, the most straightforward way to begin your proof is to let $\mathscr{F}$ be an arbitrary countable family of closed sets in $X$ with the f.i.p.; somehow you need to show that $\bigcap\mathscr{F}\ne\varnothing$.
HINT: Assume that $\bigcap\mathscr{F}=\varnothing$, use $\mathscr{F}$ somehow to get a countable open cover $\mathscr{U}$ of $X$, apply the hypothesis of countable compactness to get a finite subcover of $\mathscr{U}$, and somehow use that finite subcover to derive a contradiction. This contradiction shows that $\bigcap\mathscr{F}$ cannot be empty after all.