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?

**Answer**

Let’s say the σ-algebra on X is generated by the sets Ai⊆X. For each subset I of the natural numbers, consider the set BI=⋂i∈IAi∩⋂i∉I(X∖Ai). For distinct sets I and J, the corresponding sets BI and BJ are disjoint. Now take cases: either only finitely many of the BI are nonempty, or infinitely many are. This will show that the σ-algebra is either finite or has cardinality at least that of the continuum.

To show that the σ-algebra cannot have cardinality strictly above that of the continuum is a bit more involved. I can’t come up with an approach avoiding transfinite induction up the Borel hierarchy. Here’s a sketch of what I have in mind:

We build an increasing family Sα of subsets of the power set of X, as α ranges over the countable ordinals. In the end, ⋃α<ω1Sα will be a σ-algebra of size at most continuum containing our countably many generators (in fact, it will be the σ-algebra they generate, but that's just an added bonus). We start by setting S0 to equal the (countable) set of generators. Given Sα, we let Sα+1 be the collection of subsets which can be written as countable unions of the form ⋃iAi∪⋃j(X∖Bj), where Ai and Bj are chosen from Sα. Note that if |Sα|≤2ℵ0, then |Sα+1|≤2ℵ0 as well (since there are only continuum many choices of ways to write the union: this is essentially the cardinal equality (2ℵ0)ℵ0=2ℵ0). For limit ordinals λ, let Sλ=⋃α<λSα. This will again satisfy |Sλ|≤2ℵ0 provided each Sα in the union does.

Finally, we see ⋃α<ω1Sα has cardinality at most that of the continuum, since ℵ1⋅2ℵ0=2ℵ0. Moreover, it is closed under the σ-algebra operations since any countable sequence of elements is accounted for in some Sα (with α<ω1).

**Attribution***Source : Link , Question Author : Syang Chen , Answer Author : user83827*