Is the power set of $\mathbb R$ a countably generated $\sigma$-algebra? Is the power set of the real line, $\mathcal P(\mathbb R)$, countably generated, i.e., is there a countable subclass $P'\subseteq \mathcal P(\mathbb R)$ such that $\mathcal P(\mathbb R) = \Sigma(P')$?
 A: Given a collection $\Sigma_0$ of subsets of a set $X$, define recursively a transfinite sequence of length $\omega_1$ (the first uncountable ordinal) by setting (for $\alpha<\omega_1$)
 $$ \Sigma_{\alpha+1}=\Sigma_\alpha\cup\{X\setminus A\mid A\in\Sigma_\alpha\}\cup B_\alpha, $$
where $B_\alpha$ consists of all those subsets of $X$ that are a countable union of sets in $\Sigma_\alpha$. Also, set
 $$ \Sigma_\lambda=\bigcup_{\alpha<\lambda}\Sigma_\alpha $$
for $\lambda\le\omega_1$ a limit ordinal.
Then $\Sigma_{\omega_1}$ is the $\sigma$-algebra $\Psi$ generated by $\Sigma_0$. This is easily proved: 


*

*By induction, each $\Sigma_\alpha$ is a subset of $\Psi$. 

*On the other hand, $\Sigma_{\omega_1}$ is closed under countable sequences, because any countable collection of sets in $\Sigma_{\omega_1}$ actually belong to $\Sigma_\beta$ for some $\beta<\omega_1$; this is simply the fact that a countable union of countable sets is countable, which translates into the fact that, given a countable collection of sets in $\Sigma_{\omega_1}$, they all appear at a countable stage, and the supremum of these stages is still countable.


This means that $\Sigma_{\omega_1}$ is a $\sigma$-algebra, so it coincides with $\Psi$.
(It doesn't matter here but, on the other hand, you may really need to go all the way up to $\omega_1$ to get this. For example, if $\Sigma_0$ is the collection of open subsets of ${\mathbb R}$, then you do not reach the $\sigma$-algebra of Borel sets at any countable stage.)
Ok. By induction, if $\Sigma_0$ is countable, then each $\Sigma_\alpha$ for $\alpha$ countable has size at most $|{\mathbb N}^{\mathbb N}|=|{\mathbb R}|$, and $\Sigma_{\omega_1}$ therefore has size at most $|{\mathbb R}|\times\omega_1=|{\mathbb R}|$. This means you cannot reach the whole power set of the reals this way. 
