Characterize sigma algebra generated by a set Let $E$ be a set. Let $C$ be a collection of subsets of E. Denote by $\sigma(C)$ the $\sigma$-algebra generated by $C$ i.e. the intersection of all $\sigma$-algebra containing $C$.
Is it true that any element $A \in \sigma(C)$ can be written as a countable union/intersection/complement of elements of $C$?
I would say that it is true but was told that it is only true if $E$ is a separable set. Why is that so?
 A: When I was studying measure theory my teacher show us a, let's say, explicit way to construct a $\sigma$-algebra. He said when $E=\mathbb{R}$ and $C=\{\text{open sets}\}$ the ordered set $\{\mathcal{B}_\lambda\}_\lambda$ is strictly increasing, but this is very difficult to prove. This solves your problem partially. I wrote the whole construction because I think it will be useful to you and also because I didn't find it explicitly in internet.
For $\mathcal{B}\subseteq 2^E$, define
$$\mathcal{B}_c = \{\cup_{k\in\mathbb{N}}A_k:A_k\in\mathcal{B}\}\cup\{A^c:A\in\mathcal{B}\}$$
and define inductively
$$\mathcal{B}_0 = C,\quad \mathcal{B}_\lambda=\left(\mathcal{B}_{\lambda-1}\right)_c,\quad\mathcal{B}_\eta=\bigcup_{\alpha<\eta}\mathcal{B}_\alpha$$
for a limit ordinal $\eta$ and a non limit ordinal $\lambda$.
Let $\Omega$ be the first uncountable ordinal. Remember that this ordinal has two important properties: it is a limit ordinal and every countable subset of $[0,\Omega[$ has a upper bound there. I claim that $\mathcal{B}_\Omega=\sigma(C)$. To see this it's enough to show that $\mathcal{B}_\Omega$ is a $\sigma$-algebra. 


*

*If $B\in\mathcal{B}_\lambda$, with $\lambda<\Omega$, then $B^c\in\mathcal{B}_{\lambda+1}$.

*It is easy to see that $E\in\mathcal{B}_2$.

*For every $k\in\mathbb{N}$ let $\lambda_k<\Omega$ and $B_{\lambda_k}\in\mathcal{B}_{\lambda_k}$. Then there exists an ordinal $\lambda<\Omega$ such that $\lambda_k\leq\lambda$, for every $k$. Then, $\cup_k B_k\in\mathcal{B}_{\lambda+1}\subseteq\mathcal{B}_\Omega$

