Complexity of generated $\sigma$-ring Let $E$ be a family of subsets of some set $X$, and $\varnothing \in E$. Define recursivelly $$E_0 =E,$$ $$E_\alpha= \biggl\{ \bigcup_{n=1}^\infty (A_n \setminus B_n): A_n,B_n \in \bigcup_{\beta < \alpha} E_\beta \biggr\}$$
for every ordinal $\alpha$. 
How can I show $\sigma(E) = \bigcup_{\alpha < \omega_1} E_{\alpha}$, where $\sigma(E)$ is the $\sigma$-ring generated by $E$, and $\omega_1$ is the first uncountable ordinal?
 A: There are two parts in showing that $\sigma(E) = E_{\omega_1}$ (which is the same as the union you describe): 


*

*Showing that $E_{\omega_1} \subseteq \sigma(E)$; and 

*Showing that $E_{\omega_1}$ is actually a $\sigma$-ring (i.e. $\sigma(E) \subseteq E_{\omega_1}$).


The first ought to be obvious; if you want to be formal, use induction over $\alpha$.
The second requires some playing around with set identities; the union case is trivial, for the set difference:
\begin{align*}
& \left(\bigcup_n (A_n \setminus B_n)\right) \setminus \left(\bigcup_m A_m' \setminus B_m'\right) \\
=& \bigcup_n \left( (A_n \setminus B_n) \setminus \left(\bigcup_m A_m'\setminus B_m'\right) \right) \\
=& \bigcup_n A_n \setminus \left(B_n \cup \bigcup_m A_m'\setminus B_m'\right)
\end{align*}
and because the term in brackets is in $E_\alpha$ for any $\alpha$ with $\alpha > \alpha_{A_m'}, \alpha_{B_m'}, B_n$ for all $m,n$, the whole expression is in $E_{\omega_1}$ (for the definition of $\omega_1$ ensures $\alpha < \omega_1$).
Note how $\omega_1$ is really the smallest ordinal that suffices, because in order for this argument to work, the ordinal ought to be inaccessible by countable suprema.
