# Hint on simple problem regarding countably generated sigma algebras

Suppose that $\mathcal{F}$ is the countable/co-countable sigma algebra on ground set $\Omega$. An exercise that I'm working on is:

Show that $\mathcal{F}$ is countably generated if and only if $\Omega$ is itself a countable set.

We may assume that $\Omega$ is not finite (in other words, we do not need to deal with the question as to whether countable sets are taken to be at most countable, etc.), otherwise the claim is trivial.

If $\Omega$ is countable, then $\Omega = (\omega_n)_{n\in \mathbf{N}}$, and now I set $A_n = \{\omega_n\}$, and then we have that $\mathcal{F} = \sigma(A_n, n \in \mathbf{N})$. To see this, one direction is immediate: $A_n$ are singletons, thus co-countable, and hence $A_n \in \mathcal{F}$ for all $n$, so also $\sigma(A_n) \subset \mathcal{F}$. Conversely, if $A \in \mathcal{F}$, then there is $J \subset \mathbf{N}$ so that $A = \cup_{j \in J} A_j \in \sigma(A_n)$.

For the other direction, I've been stuck. We begin by supposing the hypothesis, that $\mathcal{F} = \sigma(A_n)$, where $A_n \subset \Omega$ are a countable family of (possibly uncountable) subsets. Here's what I know: it is quite important that $\mathcal{F}$ is the countable/co-countable sigma algebra because there do exist countably generated sigma algebras for uncountable ground sets (like the Borel sigma algebra on $\mathbf{R}$). I also know that

$$\mathcal{F} = \sigma(A_n) = \cap\{\mathcal{G} : \mathcal{G} \subset 2^\Omega \text{ is a sigma-algebra }, A_n \in \mathcal{G} \text{ for all n}\},$$ which implies in particular that $\mathcal{F} \subset \mathcal{G}$ for all such $\mathcal{G}$. Another thing that seems useful is that if $S \in \mathcal{G}$ for all such $\mathcal{G}$, then $S$ is either countable or co-countable, which tells us for example that $A_n$ must be countable or co-countable.

What I'd like is just a hint. I think I'm close, but I'm missing putting some of these details together.

## 1 Answer

Hint: Try explicitly describing a $\sigma$-algebra which contains every $A_n$ but is not all of $\mathcal{F}$. To find this $\sigma$-algebra, first suppose for simplicity that every $A_n$ is countable. Suppose $B$ is some other countable set that you can build out of the $A_n$ using the $\sigma$-algebra operations. Can you make a guess about some relationship $B$ must have with the $A_n$'s?

Bigger hint:

What relationship would you expect there to be between $B$ and $\bigcup_n A_n$? Can you describe a $\sigma$-subalgebra of $\mathcal{F}$ whose countable sets are exactly the sets $B$ satisfying this relationship?

• Thanks. I appreciate you taking the time to think about what would help me, and especially hiding the bigger hint. I'm anxious to know if I'm going to have to reveal it or not :)! – Drew Brady Jul 14 '18 at 20:43
• For what it's worth, the bigger hint still does not give away the whole answer, it just points a little more directly towards what you might try to prove. – Eric Wofsey Jul 14 '18 at 20:44