# 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

## 1 Answer

It is not true in ZF alone.

It is consistent with ZF that there exist infinite, Dedekind-finite sets - these are sets which admit no non-surjective self-injection but are nonetheless infinite. It's a good exercise to show that Dedekind-finiteness is equivalent to not containing a countable infinite subset; so on such a set, the co-countable topology is just the cofinite topology. And the cofinite topology on any set is compact.

In fact, the statement "The cocountable topology on an infinite set is never compact" is exactly equivalent to the statement "There are no infinite Dedekind-finite sets."

• 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