Prove a that a topological space is compact iff

Question

Prove that the topological space $X$ is compact $\Leftrightarrow$ whenever {$C_j:j\in J$} is a collection of closed sets with $\bigcap_{j\in J}C_j = \varnothing$, there is a finite subcollection {$C_k:k\in K$} such that $\bigcap_{k\in K}C_k=\varnothing$.

My attempt:

First note that for the closed sets {$C_i:i\in I$} such that $\bigcap_{i\in I}C_i=\varnothing$, the complements {$X-C_i:i\in I$} are precisely the open covers of $X$, and for the open covers {$U_i:i\in I$} of $X$, the complements {$X-U_i:i\in I$} are precisely the closed sets that intersect to $\varnothing$. Now $X$ is compact $\Leftrightarrow$ for all open covers {$U_i:i\in I$} (that is, $\bigcup_{i\in I}U_i = X$, since each $U_i\subseteq X$) we have a finite subcover {$U_k:k\in K$} (that is, $\bigcup_{k\in K}U_k = X$) where $K$ is finite. This is true $\Leftrightarrow$ {$X-U_i:i\in I$} is a collection of closed set such that $\bigcap_{i\in I}(X-U_i)=\varnothing$ which implies $\exists$ a finite sub-collection {$X-U_k:k\in K$} such that $\bigcap_{k\in K}(X-U_k)=\varnothing$.

I think the basic idea is right, but something about how I'm phrasing it doesn't sound right to me. Does anyone have any suggestions/critiques?

Answer 1

The basic idea is correct (taking complements and using de Morgan, essentially).

As suggestions for write-up: show the directions, for left to right e.g.:

Suppose $X$ is compact. Let $\{ C_j: j \in J \}$ be a collection of closed sets with empty intersection. Then define, for each $j \in J$, $U_j = X \setminus C_j$, which is open in $X$. Then $$\cup_{j \in J} U_j = \cup_{j \in J} (X \setminus C_j) = X \setminus \cap_{j \in J} C_j = X \setminus \emptyset = X$$

using De Morgan's law. So we have an open cover of $X$, and a finite subset $F \subset J$ exists such that $X = \cup_{j \in F} U_j$. But then, using that also $C_j = X \setminus U_j$: $$\cap_{j \in F} C_j = \cap_{j \in F} (X \setminus U_j) = X \setminus (\cup_{j \in F} U_j) = X \setminus X = \emptyset$$

as required. The other direction is similar, of course.