Cardinality of Borel sigma algebra It seems it's well known that if a sigma algebra is generated by countably many sets, then the cardinality of it is either finite or $c$ (the cardinality of continuum). But it seems hard to prove it, and actually hard to find a proof of it. Can anyone help me out?

It is easy to prove that the $\sigma$-algebra is either finite or has cardinality at least $2^{\aleph_0}$.
One way to prove that it has cardinality at most $2^{\aleph_0}$, without explicitly using transfinite recursion, is the following.
It is easy to see that it is enough to prove this upper bound for a "generic" $\sigma$-algebra, e.g., for the Borel $\sigma$-algebra of $\{0,1\}^{\omega}$, or for the Borel $\sigma$-algebra of the Baire space $\mathcal{N} = \omega^{\omega}$.
Note that $\mathcal{N}$ is a Polish space, so we can talk about analytic subsets of $\mathcal{N}$.
Every Borel subset is an analytic subset of $\mathcal{N}$ (in fact, $A \subseteq \mathcal{N}$ is Borel if and only if $A$ and $X \setminus A$ are both analytic).
So it is enough to prove that $\mathcal{N}$ has $2^{\aleph_0}$ analytic subsets.
Now use the theorem stating that every analytic subset of $\mathcal{N}$ is the projection of a closed subset of $\mathcal{N} \times \mathcal{N}$.
Since $\mathcal{N} \times \mathcal{N}$ has a countable basis of open subsets, it has $2^{\aleph_0}$ open subsets, so it has $2^{\aleph_0}$ closed subsets.
So $\mathcal{N}$ has $2^{\aleph_0}$ analytic subsets.
The proof using transfinite recursion might be simpler, but I think the analytic subset description gives a slightly different, kind of direct ("less transfinite") view on the Borel sets, that could be useful to know.
