Showing that each Element in an Uncountably Generated $\sigma$-algebra belongs to a Smaller $\sigma$-algebra Let $\Omega\not=\emptyset$ be a set. Suppose $\Gamma$ is uncountable and $A_i \subset 2^\Omega$ for each $i\in \Gamma$. Then each $B\in \sigma(A_i, i\in \Gamma)$ corresponds to a countable subcollection $\lbrace A_{i_j}, j\in \mathbb{N}\rbrace \subset \lbrace A_i, i\in \Gamma \rbrace$ such that $B \in \sigma (\lbrace A_{i_j}, j\in \mathbb{N} \rbrace)$.
I am not sure how to do show this since there is not much I can say about $B$. I.e., just because $B$ belongs to $\sigma(A_i, i\in \Gamma)$ does not mean there is an effective way to describe $B$. I don't even think it is correct to say that $B$ is generated by the sets $A_i$ by using countable set operations (but I am not exactly sure). 
May I have a hint on how to proceed?
 A: Let $\mathcal{B}$ be the union of all the $\sigma$-algebras generated by countable subcollections of $\Gamma$. If you can show that $\mathcal{B}$ is a $\sigma$-algebra, then this will show that $\mathcal{B}$ is in fact the $\sigma$-algebra generated by $\Gamma$, itself. Meanwhile, clearly every element of $\mathcal{B}$ satisfies the property in the question.
A: Show that the set of $B$ such that $B$ is in the $\sigma$-algebra generated by some countable collection of $A_i$ is a $\sigma$-algebra. This algebra certainly contains every $A_i$, so it contains the $\sigma$-algebra generated by all the $A_i$.
A: While there isn't exactly an effective way to describe $\sigma(A_i, i\in\Gamma)$, there is a construction of it in just-barely-uncountably-many stages which will let you get a handle on how its elements are all built up. By transfinite recursion (which, we'll see, really only goes up to the first uncountable ordinal $\omega_1$), define:
$$
\begin{align}
S_0 &= \{A_i\mid i\in \Gamma\}, \\
S_{\alpha +1} &= \text{set of all complements of $X\in S_{\alpha}$ and countable unions of $(X_n)_{n\ge 0}$ with all $X_n\in S_{\alpha}$,} \\
S_{\lambda} &= \bigcup_{\alpha<\lambda} S_{\alpha} \quad\text{for limit ordinals $\lambda$.}\\
\end{align}
$$
It's easy to show that the construction stops adding new sets at stage $\omega_1$, so $S_{\omega_1}$ is a $\sigma$-algebra containing all the $A_i, i\in \Gamma$. Furthermore, it's contained in every $\sigma$-algebra containing all the $A_i$. Thus,
$$
S_{\omega_1} = \sigma(A_i, i\in\Gamma).
$$
Now, by (transfinite) induction on $\alpha < \omega_1$, it's also easy to see that any $B\in S_{\omega_1}$ involves only countably many of the $A_i$.
The essential fact used in establishing that last claim, as well as the claim that $S_{\omega_1}$ is a $\sigma$-algebra, is: the union of countably many countable sets is countable.
