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 $\mathcal{A}\_{\alpha}\neq\mathcal{M}$?



How long can it take to generate a $\sigma$-algebra?
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](http://projecteuclid.org/euclid.lnl/1235423343), 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].


