# Countably generated sigma-algebras

This question asks whether every family $\mathcal A\subseteq\mathcal P(X)$ is contained in a countably generated
$\sigma$-algebra. (The OP stipulates that $\mathcal A$ is itself a $\sigma$-algebra, but that clearly doesn't matter.) The answers provide counterexamples with either $|\mathcal A|\gt2^{\aleph_0}$ or $|X|=2^{2^{\aleph_0}}.$ I would like to see a counterexample with $|\mathcal A|\le2^{\aleph_0}$ and $|X|\le2^{\aleph_0}.$

Question. Is there a family $\mathcal A\subset\mathcal P(\mathbb R)$ such that $|\mathcal A|\le2^{\aleph_0}$ and $\mathcal A$ is not contained in any countably generated $\sigma$-algebra?

• For what it's worth, the free concrete $\sigma$-algebra on $2^{\aleph_0}$ generators (in fact, even the free concrete $\sigma$-algebra on $2^{2^{\aleph_0}}$ generators) embeds in $\mathcal{P}(\mathbb{R})$, by an argument similar to the argument that the free Boolean algebra on $2^{\kappa}$ generators embeds in $\mathcal{P}(\kappa)$. It seems plausible such an algebra could not be contained in a countably generated $\sigma$-algebra, though I can't see a way to prove it... – Eric Wofsey Jan 3 '17 at 6:54
• The tag [descriptive-set-theory] is relevant here. Agree? – Pedro Sánchez Terraf Jan 3 '17 at 14:23

Proof: Recall that $\mathcal{P}(\omega_1) \otimes \mathcal{P}(\omega_1) = \mathcal{P}(\omega_1 \times \omega_1)$ (a result of B. V. Rao, On discrete Borel spaces and projective sets, Bull Amer. Math. Soc. 75 (1969), 614–617). Given $\{A_i : i < \omega_1\} \in [\omega_1]^{\omega_1}$, choose $\{(X_n, Y_n): n < \omega\}$, $X_n, Y_n \subseteq \omega_1$, such that the sigma algebra generated by $\{X_n \times Y_n : n < \omega \}$ contains $\{(i, x): i < \omega_1, x \in A_i\}$. It follows that the sigma algebra generated by $\{Y_n : n < \omega\}$ contains each $A_i$.
In the other direction, Arnold W. Miller, Generic Souslin sets, Pacific J. Math. 97 (1981), 171–181 showed that it is consistent that there is no countable family $\cal{F} \subseteq \mathcal{P}(\mathbb{R})$ such that the sigma algebra generated by $\mathcal{F}$ contains all analytic sets. So the existence of such a family is independent of ZFC.
• Thanks. I had thought of the analytic sets but guessed wrongly that they would be a real rather than a consistent counterexample. Rao's result seemed amazingly counterintuitive at first, but I must have seen it before in the dim past, because I managed to reconstruct the proof. The key ideas seem to be that (1) in $\omega_1\times\omega_1$ every "curve" $y=f(x)$ or $x=f(y)$ is a "Boolean $\sigma$-combination" of "rectangles" $X\times Y$ and (2) $\omega_1\times\omega_1$ is the union of countably many "curves". Right? – bof Jan 4 '17 at 19:11
• Yes, that's the argument. An interesting followup question is if a negative answer to your question is equivalent to "every subset of plane belongs to the sigma algebra (call it $\Sigma$) generated by abstract rectangles". A natural try would be to take $A \subseteq \mathbb{R}^2$, $A \notin \Sigma$ and show that the sigma algebra generated by $\{A_x : x \in \mathbb{R}\}$ is not countably generated. But I don't see how to do it. – hot_queen Jan 4 '17 at 19:59
• An equivalent formulation of this would be: Suppose for every $\cal{A} \in [\mathcal{P}(\mathbb{R})]^{\mathfrak{c}}$, there is a countable family $\cal{F}$ such that the sigma algebra generated by $\cal{F}$ contains every member of $\cal{A}$. Must there exist, for every such $\mathcal{A}$, a countable family $\cal{E}$ such that for some $\alpha < \omega_1$, every member of $\cal{A}$ is $\Sigma^0_{\alpha}(\cal{E})$ (i.e., their Borel ranks are bounded below $\omega_1$)? – hot_queen Jan 4 '17 at 20:08