# How long can it take to generate a $\sigma$-algebra?

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
+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

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].
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