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I base my question on this. I saw it at Math Overflow and as nobody answer it, although its bounty, I decided to post it here.

Consider the statements $P$ and $Q$:

$P$: Every subset of the plane belongs to the $\sigma$-algebra generated by arbitrary rectangle.

That is, $C\subseteq \Bbb{R}^2 \implies C\in \sigma(\{A\times B : A\subseteq \Bbb{R} \mbox{ and } B\subseteq \Bbb{R}\})$. And

$Q$: Every continuum-sized family of subsets of $\mathbb{R}$ is contained in a countably generated $\sigma$-algebra.

Here is proved that statement $P$ is independent from $ZFC$; and here is (partially) proved that $Q$ is also independent from $ZFC$.

Question 1: Why does $P\implies Q$ ? As the author of the initial question said.

Question 2: Does $Q\implies P$ ?

I think Question 1 must be much simpler, but I'm just curious about this and I'm far from being a specialist at the matter. I'm sorry for the lack of development of a try :(

Thank you for any help

Henno Brandsma
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Marcelo
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    The argument for $P \implies Q$ is essentially in the first paragraph of this anwer, I think. There it is done for $\aleph_1$ instead of $\mathfrak{c}$. – Henno Brandsma Aug 27 '19 at 04:15
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    About 3 yrs ago I asked about P here and cross-posted it to MathOverflow, where I got the A that it was undecidable in ZFC, which surprised me because it once appeared as a Problem in American Mathematical Monthly, circa 1972. (with an asterisk, meaning neither the proposer nor the editors had a solution). – DanielWainfleet Sep 03 '19 at 06:31
  • @DanielWainfleet , interesting! Do you know how to find some pdf of this event (which appears the problem $P$)? I searched for it quickly and found nothing... – Marcelo Sep 04 '19 at 00:05
  • If by "This event" you mean the appearance of the problem in Amer. Math. Monthly, all I can suggest is a search of their archives. I no longer have the paper copy. – DanielWainfleet Sep 04 '19 at 06:35

1 Answers1

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First part of this must help you on $P\implies Q$

Kya Narendra
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