Compact spaces and closed sets (finite intersection property)

I am trying to prove the following theorem:

A topological space $X$ is compact iff for every collection $\mathscr{C}$ of closed sets in $X$ having the Finite Intersection Property (FIP) the intersection $\bigcap \mathscr{C}$ of all elements of $\mathscr{C}$ is nonempty.

I can easily prove the forward direction but am having trouble with the reverse direction, that is "I cannot prove that $X$ is compact." I know some sort of contraposition should be used but I am not sure how. Any help is appreciated!

Solution 1:

A topological space $X$ is compact if: for every covering of $X$ with open sets there exists a finite subcovering. You want to prove that this property is equivalent to: for every family of closed sets such that every finite subfamily has nonempty intersection then the intersection of the whole family was nonempty.

The equivalence is very simple: to pass from one statement to the other you have just to pass to the complementary of sets.

So: open becomes closed, union becomes intersection, covering becomes empty intersection. With this translation you get the desired equivalence.

Formally: Compactness means that for every family $\mathcal R$ of open sets: $$ \bigcup \mathcal R = X \Longrightarrow \exists \text{finite}\ \mathcal R_0 \subset \mathcal R \colon \bigcup \mathcal R_0 = X, $$ while the other property is that for every family $\mathcal F$ of closed sets: $$ \bigcap \mathcal F \neq \emptyset \Longleftarrow \forall\text{finite}\ \mathcal F_0\subset \mathcal F \colon \bigcap \mathcal F_0 \neq \emptyset. $$

You can reverse the implication by negating both sides. Hence passing to the complementary (as said before) you get the equivalence.