I am having trouble proving this statement (from Conway. Functions of one complex variable)
If $\mathcal{F}$ is a collection of closed subsets of $K\subseteq X$ which satisfies the f.i.p. and $\bigcap\{F:F\in \mathcal{F}\}\neq\emptyset$ then $K$ is compact.
I tried a proof by contradiction with $K$ supposed not compact. Then it exists an open covering $\mathcal{G}$ which does not contain a finite subcover of $K$. From this point, i thought i could build up a collection of closed sets by taking the complements of the open covering $\mathcal{G}$. But are those $X-G$ subsets of $K$? I mean, each $G\in\mathcal{G}$ is a subset of $K$, which made me believe that $X-G$ could not be possibly a subset of $K$. What's wrong?
Best Answer
Not by contradiction but by contrapositive:
Let $\mathcal{U}$ be an open cover for $X$ (which satisfies the FIP property). Then $\mathcal{F} = \{X\setminus U: U \in \mathcal{U}\}$ is a family of closed sets such that $\bigcap \mathcal{F} = X \setminus \bigcup \{U : U \in \mathcal{U}\} =\emptyset$, by de Morgan. This means that $\mathcal{F}$ does not have the FIP. So there finitely many $F_1 =X\setminus U_1, \ldots F_n = X\setminus U_n$ from $\mathcal{F}$ that have empty intersection. The $U_1,\ldots,U_n$ then form a finite subcover for $\mathcal{U}$ as required, again by de Morgan.