You only need countable many sets each time in a generated $\sigma$-algebra, right? Let $X$ be a non-empty set, then any $\mathcal{E}\subset\mathcal{P}(X)$ can generate a $\sigma$-algebra $\sigma(\mathcal{E})$. Here, $\sigma(\mathcal{E})$ is the intersection of all the $\sigma$-algebras that contains $\mathcal{E}$.
Since the definition of $\sigma$-algebra only involves countable operations, I have the feeling that any element in $\sigma(\mathcal{E})$ only uses countable many elements in $\mathcal{E}$. 
To be precise, I'm saying that if $A\in\sigma(\mathcal{E})$, then we should have that $A\in\sigma(\mathcal{F})$ for some countable $\mathcal{F}\subset\mathcal{E}$.
Is that the case? Can you give a counter-example or a hint on how to prove it?
 A: Ah-hah. I just proved it. My guess is correct. Sometimes literally writing down your question again really helps you to think it more clearly.
The trick is, we can prove that the union of all the $\sigma(\mathcal{F})$ we mentioned forms a $\sigma$-algebra $\bigcup_i{\sigma(\mathcal{F}_i)}$, this is because if $A_j\in\bigcup_i\sigma(\mathcal{F}_i)$ for $j=1,2,...$, then there exists $i_1,i_2,...$, such that $A_j\in\sigma(\mathcal{F}_{i_j})$, but since $\mathcal{F}_{i_j}$ are all countable, so is their union, that is, $$\bigcup_j{A_j}\in\sigma(\bigcup_j\mathcal{F}_{i_j})\subset\bigcup_i\sigma(\mathcal{F}_i).$$ And as we can see $\bigcup_i\sigma(\mathcal{F}_i)$ clearly contains $\mathcal{E}$, therefore it also contains $\sigma(\mathcal{E})$.
But it's not over yet. On the other hand $\bigcup_i{\sigma(\mathcal{F}_i)}$ is also the subset of $\sigma(\mathcal{E})$, because if $A\in\bigcup_i{\sigma(\mathcal{F}_i)}$, then $\exists i$, s.t. $A\in\sigma(\mathcal{F}_i)\subset\sigma(\mathcal{E})$, therefore we have the surprising result that
$$\bigcup_i{\sigma(\mathcal{F}_i)}=\sigma(\mathcal{E}).$$
