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I am trying to prove the following theorem:

A topological space $X$ is compact iff for every collection $\mathscr{C}$, of closed set in $X$ having the Finite Intersection Property (FIP), $\cap C$ 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!

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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.

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