Constructively generating a sigma algebra We have a collection $\mathcal{C}$ of sets (includes $\Omega)$ and would like to constructively generate the sigma algebra $\sigma(\mathcal{C})$. Would the following process work?
Let $\mathcal{S}=\mathcal{C}$
1.Take the complement of each set in $\mathcal{S}$ and add it to $\mathcal{S}$
2.Take all possible finite and countably infinite unions of sets in $\mathcal{S}$ and add them to $\mathcal{S}$.
3. goto step 1.
I have seen it claimed that taking countable unions and intersections of open intervals can only give you a proper subset of the Borel sets. This would seem to imply that the above process would not work because any resulting set could only have arisen from countable unions and intersections. 
 A: The process works, but in general you have to repeat it many times through a transfinite process. 
Call $\mathcal C_0=\mathcal C$, and $\mathcal C_\alpha$, for $\alpha>0$ an ordinal, the result of taking complements of sets in $\bigcup_{\beta<\alpha}\mathcal C_\beta$, and all possible finite and countable unions of sets in $\bigcup_{\beta<\alpha}\mathcal C_\beta$. 
We have that $\mathcal C_{\omega_1}=\bigcup_{\alpha<\omega_1}\mathcal C_\alpha=\sigma(\mathcal C)$ is the $\sigma$-algebra generated by $\mathcal C$, where $\omega_1$ is the first uncountable ordinal. In general, this is best possible, for instance, if $\mathcal C$ consists of the open intervals in $\mathbb R$. 
This representation is quite useful, and it appears frequently in descriptive set theory. 
The reason why we can stop at stage $\omega_1$ is because $\omega_1$ is regular, meaning that a countable union of countable ordinals is still a countable ordinal, and therefore strictly below $\omega_1$. This uses a modicum amount of choice. In the absence of choice, we can have the process stop much earlier, or much later, depending on the specific $\mathcal C$ we begin with. Arnold Miller has written extensively on this (and I may have posted a similar answer, with references, a few months ago).
