a sigma-algebra on a countable set is a topology I am trying to prove the statement in the title, i.e. that 

a sigma-algebra $\Sigma$ on a countably infinite set $X$ is a topology on $X$. 

I feel like I have an intuition of why this is true, but I can't articulate it. I'd like to request some hints to get me thinking in the right direction. (Also, if the proof involves choice, could you kindly elaborate on that part? I have not studied that before!)
Thanks in advance!
Edit: this fact is stated (without proof) in an answer to a related question: https://math.stackexchange.com/a/51229/39117.
 A: Edit: To get this result, we must (I believe) assume the Axiom of Countable Choice (AC$_\omega$), which says, roughly, that if I have a countably infinite collection of non-empty sets $A_n$, then there is a sequence $\{a_n\}_n$ of elements of $\bigcup_nA_n$ such that $a_n\in A_n$ for each $n$.
The only sticking point of showing that a sigma algebra $\Sigma$ is a topology is in whether or not arbitrary unions of infinitely-many (not necessarily countably-many) $\Sigma$-sets are again $\Sigma$-sets. By AC$_\omega$, we may rewrite any such union as a countable union as follows.
Take any infinite $\mathcal A\subseteq\Sigma$ and let $A:=\bigcup\mathcal A$. We must show that $A\in\Sigma$. Now, if $A$ can be obtained as a finite union of $\Sigma$-sets, then we're done, so suppose not. It follows that $A$ must be infinite, so countably infinite since $A$ is a subset of the countable set $X$. Let $\{x_n\}$ be any enumeration of $A$, and for each $n$, let $$\mathcal A_n:=\{B\in\mathcal A:x_n\in B\}.$$ By definition of $A$ and the $x_n$, each $\mathcal A_n$ is non-empty, and $\mathcal A=\bigcup_n\mathcal A_n$.
By AC$_\omega$, there is a sequence $A_n$ of elements of $\mathcal A,$ such that $A_n\in\mathcal A_n$ for all $n$, meaning in particular that each $A_n$ is a subset of $A$, and is a $\Sigma$-set containing $x_n.$ Thus, $A=\bigcup_n A_n,$ so $A$ is a countable union of $\Sigma$-sets, and so $A\in\Sigma,$ as desired.

Without AC$_\omega$, we certainly can't use this argument, and we may not be able to get there at all (though I'm not sure about that). A lot of things can go very wrong if AC$_\omega$ fails--for example, a countable union of pairwise disjoint, $2$-element sets may be uncountable. This sort of thing is rather upsetting to people, so most people will take AC$_\omega$, at least, for granted, even if they don't assume stronger choice principles than that.
