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.
Best 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?