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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.

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Bachmaninoff
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    I repeat the hint I gave in a comment to Cameron's answer in somewhat more detail: The only problem is the axiom about arbitrary unions. Let $E = \bigcup_{i \in I} A_i$. Enumerate $E$ and pick for each $x \in E$ an $i_x$ such that $x \in A_{i_x}$. Now you can write $E$ as a countable union of measurable sets. – Martin Jun 03 '13 at 00:51
  • Thank you very much, @Martin! How can we show that $E$ is necessarily enumerable? – Bachmaninoff Jun 03 '13 at 01:04
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    Subsets of countable sets are countable. // We also use the axiom of choice (in its countable version if we are a bit clever) by choosing (picking) for each $x \in E$ an $i_x$ such that $x \in A_{i_x}$. I'll leave it to others to elaborate on these subtleties. – Martin Jun 03 '13 at 01:08
  • By the way: if $X$ is a countable set, and ${ X_i : i \in I }$ is a partition of $X$, then $\Sigma = \left{ \bigcup_{i \in J} X_i : J \subset I \right}$ is both a sigma algebra and a topology on $X$. Without too much extra trouble, you can show every sigma algebra on $X$ arises in this way (and so, in particular, all are topologies). – Mike F Jun 11 '13 at 05:45

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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.

Cameron Buie
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  • There are no countably infinite $\sigma$-algebras, so $\Sigma$ is in fact finite. // It helps to enumerate $X$ to reduce an arbitrary union to a countable union. – Martin Jun 03 '13 at 00:38
  • Thanks for the reply, Cameron! Also, I believe that the proof of Martin's statement is here: http://mathproblems123.wordpress.com/2011/09/08/infinitely-countable-sigma-algebra/. – Bachmaninoff Jun 03 '13 at 00:41
  • @Martin: Thanks! Forgot that fact. – Cameron Buie Jun 03 '13 at 00:44
  • @CameronBuie: So, it seems that the best we can say is that the statement is true if $X$ is finite? By the way, I have linked, in an edit to the original post, to an answer on SE which states this as a fact. – Bachmaninoff Jun 03 '13 at 00:51
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    @Bachmaninoff: I can't see any reason to require that $X$ be finite, only that the sigma-algebra be finite.. Partition an infinite set $X$ into finitely many subsets. The set of unions of those subsets (together with $\emptyset$) will be a finite sigma-algebra on $X$. Of course it is sufficient that $X$ be finite. – Cameron Buie Jun 03 '13 at 01:09
  • @Bachmaninoff: Oops! I just figured out why it's true (if you have enough Choice). Will fix my answer when I return from a quick errand – Cameron Buie Jun 03 '13 at 01:21
  • @Bachmaninoff: Answer updated. Let me know if you've any questions. – Cameron Buie Jun 03 '13 at 01:50
  • @CameronBuie: Thank you, this answers my question exactly (and the AC part was very clear)! Accepted! – Bachmaninoff Jun 03 '13 at 05:40
  • @Bachmaninoff: Glad to help! Sorry for the false start. – Cameron Buie Jun 03 '13 at 18:49