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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.

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For an easier example, try letting $X = \mathbb{R}$ and $M = \{(a, +\infty) : a \in \mathbb{R}\}$ the set of all open intervals that extend to infinity on the right.

Then $M_1 = \{(a, +\infty), (-\infty, a] : a \in \mathbb{R}\}$ consists of open and closed intervals.

Any intersection of intervals is again an interval (easy exercise) so $M_2$ consists only of (open, closed, half-open) intervals.

Thus every set in $M_3$ is a countable union of intervals.

Let $E$ be the irrational numbers. $E$ doesn't contain any nontrivial intervals, so if we are to write $E$ as a union of intervals, each of them must be a single point (like $[a,a] = \{a\}$). But since $E$ is uncountable we cannot write it as a countable union of points.

However, since $M_2$ contains all open intervals, $M_3$ contains all open sets, so $\Sigma(M) = \Sigma(M_3)$ is the Borel $\sigma$-algebra, and $E$ is certainly a Borel set.

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