# Prove that the countable complement topology is not meta compact?

I have seen the proof of (countable complement topology is not meta compact) , which says that the countable intersection of open sets is open and thus uncountable, so this topology cannot be meta compact. It is easy to see that the countable intersection is open and uncountable, but why this implies that this topology is not meta compact?

## 1 Answer

Note that this property shows that there are no infinite point-finite open covers of $X$: any countably infinite subfamily of an infinite open cover has nonempty intersection, and so the cover is not point-finite with respect to the points in this intersection.

It now suffices to construct an open cover with no finite open refinement. Without loss of generality assume that $\mathbb{N} \subseteq X$, and pick a countable family $\mathcal{A}$ of infinite subsets of $\mathbb{N}$ with the finite intersection property, but with empty intersection. Then $\mathcal{O} = \{ X \setminus A : A \in \mathcal{A} \}$ is an open cover of $X$. If $\{ U_1 , \ldots , U_n \}$ was a finite open refinement of $\mathcal{O}$, then for $i \leq n$ let $A_i \in \mathcal{A}$ be such that $U_i \subseteq X \setminus A_i$. But then $\bigcup_{i=1}^n U_i \subseteq \bigcup_{i=1}^n ( X \setminus A_i ) = X \setminus \bigcap_{i=1}^n A_i$, and the right-hand-side is not $X$, contradicting that we had a finite open refinement!

Added:

The second part of the above is a bit of a mess. I am constructing an open cover with no finite open refinement. But any open cover with a finite open refinement actually has a finite subcover, and so all I needed to do was construct an open cover with no finite subcover. Which I do above, but phrase it in a terrible way.

In the end, what is important is this:

• $X$ is not compact (and so has an open cover with no finite subcover); and
• the only point-finite open covers are the finite ones.

From these two facts it follows that $X$ is not metacompact.