# compactness of cofinite topology and non-compactness of cocountable topology

Let $$(X, F_1)$$ be a cofinite topology and $$(X, F_2)$$ be a cocountable topology for an uncountable set $$X$$.

The question says that a cofinite topology is compact while cocountable topology is not.

Although the following is not the proof, my idea is as follows: Pick any $$E \in F_1$$, then $$X \setminus E$$ is finite and let $$X \setminus E = \{x_1, ..., x_n\}$$ for some $$n \in \mathbb{N}$$. Then, $$X$$ has a finite subcover (i.e., $$E \cup (\bigcup_{i=1}^n V_{x_i}) = X$$, where $$V_{x_i}$$ is a neighbourhood of $$x_i$$). However, this is not the case for a cocountable topology because for any $$F \in F_2$$, $$X \setminus F$$ is at most countable.

However, I do not know how to prove compactness in general for both cases. I appreciate if you give some help.

## 1 Answer

You have the basic idea for the proof that $$(X,F_1)$$ is compact, but to get an actual proof, you need to show that if $$\mathscr{U}$$ is an arbitrary $$F_1$$-open cover of $$X$$, then some finite subset of $$\mathscr{U}$$ covers $$X$$. Let $$U$$ be any member of the cover $$\mathscr{U}$$; then $$X\setminus U$$ is finite; say $$X\setminus U=\{x_1,\ldots,x_n\}$$. Can you see how to finish it from here?

For the second part of the problem, assume that $$X$$ is uncountable, and let $$A$$ be a countably infinite subset of $$X$$. For each $$x\in A$$ let $$U_x=(X\setminus A)\cup\{x\}$$, and let $$\mathscr{U}=\{U_x:x\in A\}$$. $$\mathscr{U}$$ is an open cover of $$X$$ (why?); does it have a finite subcover?