Disclaimer: sorry for my poor english and edition.
Claim: If $M\subseteq \mathcal{P}(X)$, then $\Sigma(M)=M_3$, where:
- $\Sigma(M)$ is the sigma-algebra generated by $M$
- $M_1=\{A\subseteq X:(A\in M) \lor(A^c\in M)\}$,
- $M_2=\{\cap_{i\in I} A_i:A_i\in M_1,|I|\leq |\mathbb{N}|\}$
- $M_3=\{\cup_{j\in J} A_j:A_j\in M_2,|J|\leq |\mathbb{N}|\}$
- $|I|\leq |\mathbb{N}|$ means $I$ is countable
(Note: Is easy to prove that my claim is equivalent to say that $M_3$ is a sigma-algebra)
I proved that $\mathcal{A}(M)=M'_3$, where:
- $\mathcal{A}(M)$ is the algebra generated by $M$
- $M'_2=\{\cap_{i\in I} A_i:A_i\in M_1,|I|< |\mathbb{N}|\}$
- $M'_3=\{\cup_{j\in J} A_j:A_j\in M'_2,|J|< |\mathbb{N}|\}$
- $|I|< |\mathbb{N}|$ means $I$ is finite
But my proof uses induction (on $|J|$ to iterate "intersections distributes over unions") to guarantee that $M'_3$ is close under complements,* and therefore fails to prove my claim. However, this fact allows me to reformulate my claim like this: $\mathcal{M}(M'_3)=M_3$, where $\mathcal{M}(M'_3)$ is the monotone class generated by $M'_3$.
*More explicitly: $$A^c=\cap_{j\in J}\left(\cup_{i_{j}\in I_{j}}A_{i_{j},j}\right)=\cup{}_{i_{1}\in I_{1}}\left(\dots\left(\cup{}_{i_{n}\in I_{n}}\left(\cap_{j\in J}A_{i_{j},j}\right)\right)\right)$$ In fact, $$A^c=\cup{}_{\left(i_{1}\dots i_{n}\right)\in I_{1}\times\dots\times I_{n}}\left(\cap_{j\in J}A_{i_{j},j}\right)=\cup{}_{i\in I}\left(\cap_{j\in J}A_{i,j}\right)$$ with $I=\prod_i^{|J|}I_i$, where $J$ needs to be finite (I think, because my proof of this is by induction on $|J|$ via "intersections distributes unions"), but $I_i$ can be countable.
(Note that, even if I admit $J$ infinite and then $I=\prod_i^\infty I_i$, then $I$ is uncountable, link, and therefore $A^c$ is not necessarily of $M_3$)
Then I saw a description of "sigma-algebra generated by collection of subsets" here that suggested me that I was wrong, and then exists $X,M$ such that $\Sigma(M)\neq M_3$ (or $\mathcal{M}(M'_3)\neq M_3$).
Because my above cited proof fails when $J$ is infinite, I thought to search "big" $M$ for a counterexample, e.g., open sets (and then $\Sigma(M)$ are the borel sets). I found that exists borel sets that are not in $(G_\delta)_\sigma$, but $(G_\delta)_\sigma\subseteq M_3$ and I'm not sure that the other inclusion holds.
I try and find more things, but the above are (I think) the more relevant. Any help?
P.S. (related questions)
1) In 1 and 2 exists proofs by transfinite induction that $\Sigma(M)=M_{\omega_1}$ (similar to the above cited proof of Asaf Karagila), but I insist: ¿is this the finest construction? 3 examine this question but not solve my problem (if I understood correct: my set theory is also poor).
2) A very close question (without desired answer) is 5. Point of view of Victor in the conversation with Matt is similar to mine (up to myside bias): if $X$ is finite (equivalent, $\mathcal{P}(X)$ is finite), then the process stop somewhere, but otherwise it can escape of $M_3$; in other words, the same idea that (1).
3) A similar but different question is 4.
