10
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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... $\endgroup$ – Eric Wofsey Jan 3 '17 at 6:54
  • 3
    $\begingroup$ The tag [descriptive-set-theory] is relevant here. Agree? $\endgroup$ – Pedro Sánchez Terraf Jan 3 '17 at 14:23
6
$\begingroup$

Claim: Under CH, there is no such family.

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.

$\endgroup$
  • $\begingroup$ Yes. The reference is B.V.Rao, On discrete Borel spaces and projective sets, Bulletin of the AMS, 75 (1969), 614-617 $\endgroup$ – hot_queen Jan 4 '17 at 13:56
  • $\begingroup$ 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? $\endgroup$ – bof Jan 4 '17 at 19:11
  • 2
    $\begingroup$ 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. $\endgroup$ – hot_queen Jan 4 '17 at 19:59
  • 2
    $\begingroup$ 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 $\omega_1$)? $\endgroup$ – hot_queen Jan 4 '17 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.