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Given a $\sigma$-algebra $\Sigma$ on a set $\Omega$, let's borrow language from topology, and call $B\subseteq\Sigma$ a basis for $\Sigma$ if $\Sigma$ is generated solely from countable unions of $B$, and call $S\subseteq\Sigma$ a subbasis if it generates $\Sigma$ in the usual sense (countable unions and complements).

It is easy to show that (i) $|\Sigma|\ne\beth_0$, (ii) $\Sigma$ is finite iff $B$ is finite iff $S$ is finite, (iii) $|\Sigma|=\beth_n$ (for $n>1$) iff $|B|=\beth_n$ iff $|S|=\beth_n$, (iv) $B$ can be a partition iff $|\Sigma|=\beth_1$ and $B$ is countable (and in fact, there is a unique such partition in that case), and (v) if $|B|=\beth_0$ then there is a (countable) partition basis as well. Obviously, any basis can serve as a subbasis as well, and $\Sigma$ itself can serve as both.

The case $|\Sigma|=\beth_1$ can, from a pure cardinality standpoint, admit bases and subbases of size $\beth_0$ or $\beth_1$. This leaves 3 possibilities for a given $\Sigma$ with $|\Sigma|=\beth_1$:

  1. There exists a countable basis (and therefore there also is a countable subbasis).
  2. There only are uncountable bases but there exists a countable subbasis.
  3. There only are uncountable bases and subbases.

Is it possible to prove that every $\beth_1$ $\sigma$-algebra has a countable basis (or furnish a counterexample)? If not, is it possible to prove it has a countable subbasis (or furnish a counterexample)?

Put another way, is it possible to exclude both (2) and (3), or at least exclude just (3)? If not, does anybody have simple counterexample(s)?

Thanks in advance!

Cheers, Ken

Kensmosis
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    The cardinality of a $\sigma$-algebra generated by $\aleph_0$ distinct sets is always at least $2^{\aleph_0}$. – Chad K Jan 31 '24 at 18:41
  • True, but this is the same as the cardinality of a $\sigma$-algebra generated by $\aleph_1$ distinct sets. The question is whether the latter always can be generated by some set of $\aleph_0$ distinct sets (and, if not, a counter-example). – Kensmosis Jan 31 '24 at 19:07
  • To clarify, by "generate" I'm referring to two distinct concepts here. In terms of subbasis, yes. Every set of $\aleph_0$ sets is a subbasis for a $\sigma$-algebra of cardinality $\aleph_1$ (i.e. generates it in the usual sense). However, not every set of $\aleph_0$ sets is the basis for a $\sigma$-algebra (as I defined it). For example, consider the set ${(1\dots n); n\in N}$ of subsets of $N$. The set of unions of its elements is just itself. Or, on $(0,3)\subset R$, consider $B={(0,2),(1,3)}$. Obviously, the set of its unions is not a $\sigma$-algebra. – Kensmosis Jan 31 '24 at 19:16
  • The core problem is that the set of denumerable subsets of $X$ is $\aleph_1$ for both $|X|=\aleph_0$ and $|X|=\aleph_1$. This makes $|\Sigma|=\aleph_1$ troublesome. – Kensmosis Jan 31 '24 at 19:24
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    There is no such thing as a sigma algebra of cardinality $\aleph_1$ unless the continuum hypothesis holds. However, issues with cardinal arithmetic aside, I think this is the counterexample you’re looking for https://math.stackexchange.com/a/2073254/397125 – spaceisdarkgreen Feb 01 '24 at 00:23
  • @spaceisdarkgreen Fair enough --- I'm being sloppy with $\beth$'s vs $\aleph$'s and am indeed assuming the (general) continuum hypothesis. – Kensmosis Feb 01 '24 at 01:37
  • @Kensmosis You may want to edit that into the question. Does that counterexample suffice as an answer? – spaceisdarkgreen Feb 01 '24 at 01:55
  • Good point. I edited it. – Kensmosis Feb 01 '24 at 02:39
  • @spaceisdarkgreen Thanks so much! I went through that reference and the argument by bof is spot on. For anyone who comes across this, bof's answer doesn't happen to be the accepted answer to the linked question --- but it is a good answer and also happens to address my precise question. It's a clever example and is exactly what I was looking for: a counterexample that shows that (3) is quite possible. – Kensmosis Feb 01 '24 at 02:46
  • @spaceisdarkgreen If you are inclined to upgrade your comment with the link to an answer, I'll be happy to accept it. – Kensmosis Feb 01 '24 at 02:47
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    @Kensmosis I'll just put in a vote to mark it as already having an answer (though the question isn't technically a duplicate), no point making a link-only answer – spaceisdarkgreen Feb 01 '24 at 03:03
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    The $\sigma$-algebra ${X\subseteq\mathbb R:\text{either } $X$\text{ or }\mathbb R\setminus X\text{ is countable}}$ has cardinality $\beth_1$ and so does every subbasis. – bof Feb 01 '24 at 12:23
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    @bof Thanks so much for the two simpler examples! I believe that in the first one you may have meant to swap "basis" and "subbasis". BTW, I like your second example a lot! – Kensmosis Feb 02 '24 at 03:06
  • @Kensmosis You're right of course. Too late to edit the comment, so I'll delete it and add a corrected version. – bof Feb 02 '24 at 04:26
  • The $\sigma$-algebra of Borel subsets of $\mathbb R$ has cardinality $\beth_1$ and it has a countable subbasis, the open intervals with rational endpoints. A basis must have cardinality $\beth_1$ since it must contain all the one-point sets. – bof Feb 02 '24 at 04:28

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