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I want to know if there is a $\sigma$-algebra such that for every countable ordinal $\alpha$ the $\sigma$-algebra can be generated in more than $\alpha$ steps but less than $\omega_{1}$ steps.

Given an algebra of sets $(X,\mathcal{A})$, let $\mathcal{A}_{0}=\mathcal{A}$, and for all ordinals $0<\alpha\leq\omega_{1}$, let $\mathcal{A}_{\alpha}$ be the algebra of sets generated by countable unions from the collection $\bigcup_{\beta<\alpha}\mathcal{A}_{\beta}$. Clearly $\mathcal{A}_{\omega_{1}}$ is the $\sigma$-algebra generated by $\mathcal{A}$.

  1. For each countable ordinal $\alpha$ does there exist an algebra of sets $(X,\mathcal{A})$ such that $\mathcal{A}_{\alpha}\neq\mathcal{A}_{\omega_{1}}$, but where $\mathcal{A}_{\beta}=\mathcal{A}_{\omega_{1}}$ for some countable ordinal $\beta$?

  2. Does there exist a $\sigma$-algebra $(X,\mathcal{M})$ such that

    1. If $(X,\mathcal{A})$ is an algebra of sets that generates $\mathcal{M}$, then $\mathcal{A}_{\alpha}=\mathcal{M}$ for some countable ordinal $\alpha$, and

    2. for each countable ordinal $\alpha$ there is an algebra of sets $(X,\mathcal{A})$ that generates $\mathcal{M}$ but where $\mathcal{A}_{\alpha}\neq\mathcal{M}$?

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Countable unions are not enough. You will also need either complements or countable intersections. –  Gerald Edgar May 8 '13 at 19:22
    
Gerald, he said "the algebra...generated by countable unions", which would provide the complements. –  Joel David Hamkins May 8 '13 at 19:26
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+1. I like this question very much! –  Joel David Hamkins May 8 '13 at 20:08
    
I have taken a small liberty of using nested lists and improve the second question. I hope you don't mind. –  Asaf Karagila May 8 '13 at 21:40
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1 Answer

up vote 8 down vote accepted

The answer to the first question is in the positive; thanks to a corollary drawn by Ken Kunen from a key theorem (about Boolean algebras) of Arnie Miller. More specifically, as shown in Theorem 9.2 of Arnie Miller's beautiful monograph, we have the following:

Theorem (Miller-Kunen) For every countable ordinal $\alpha$ there is a field $H$ of sets such that $o(H)=\alpha$.

In the above, $H$ is a field of sets means that $H$ is a family of subsets of some set $X$, and $H$ is closed under complements and finite unions; and $o(H)$ measures the length of the $\sigma$-algebra generated by $H$ [see p.10 of the aforementioned reference for the official definition].

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Do you mean Theorem 9.2 here: math.wisc.edu/~miller/res/dstfor.pdf ? Well, it does not say anything about $\sigma$-completeness. –  Tomek Kania May 8 '13 at 21:34
    
@Tomek: thanks, you are right. I will edit my answer. –  Ali Enayat May 8 '13 at 23:54
    
Wait, what is the issue? It seems that Miller's definitions match the OP's, modulo the synonyms 'field of sets' and 'algebra of sets'. –  François G. Dorais May 9 '13 at 1:10
    
@François: there is only issue of clarity, not veracity; the current edit addresses that. –  Ali Enayat May 9 '13 at 1:19
    
OK, I see. (I fixed the link to Miller's book which got mangled after the edit.) –  François G. Dorais May 9 '13 at 2:00
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