# Co-countable topology, anticompact and axiom of choice

It's a known result that in an uncountable set $X$ with the topology of countable complementaries (i.e. $U$ is open if $X\backslash U$ is countable), a subset $A\subset X$ is compact if and only if $A$ is finite.

I thought of a proof of this fact. If $A$ is infinite, then let $B\subset A$ a countable set, and for each $x \in B$ let be $U_x = (X\backslash B) \cup \{x\}$, which is open because $B$ is countable. Then $A\subset\cup _{x\in B} U_x$ and it does not admit a finite subcovering. Thus, $A$ is not compact.

The thing is that for proving these I used the axiom of choice (only in $\mathtt{ZF}$ it's not sure that every set will have a countable subset).

My question is, ¿is this result true just on $\mathtt{ZF}$ (without $\mathtt{AC}$)?

• Note that $X$ doesn't need to be uncountable - in ZFC, the cocountable topology on an infinite set is never compact (if the set is countable then the topology is discrete). – Noah Schweber Apr 23 '17 at 11:16

• Thank you! What happens if $X=\mathbb{R}$? In that case, is it consistent with ZF the existence of infinite Dedekind-finite subsets of $\mathbb{R}$? – G. Gallego Apr 23 '17 at 13:41
• @G.Gallego Yes - Cohen's original model of ZF + $\neg$AC contains an infinite Dedekind-finite set of reals. – Noah Schweber Apr 23 '17 at 13:56