Countability of generated ring $R(E)$ I am studying Paul R. Halmos Measure theory. In the section 5 of chapter 1, theorem 5 states that : 
If $E$ is a countable class of sets, then $R(E)$ is countable.
The proof uses class of all finite unions of differences of sets of class E.
Can anyone explain this is a simple manner or any other methods of proof?

Let $X = \bigcup E$ be the set of which all members of $E$ are subsets. For good measure, assume that $X\in E$ (if it isn't, use $E'=E\cup {X} in what follows). 
Note that $R(E)$ is closed under intersection: if $A,B\in R(E)$, then $(X\setminus A)\cup (X\setminus B) = X\setminus(A\cap B) \in R(E)$, so $A\cap B = X\setminus (X\setminus (A\cap B)) \in R(E)$. It's easy to show that $R(E)$ is a ring of sets: $\emptyset, X\in R(E)$, and $R(E)$ is closed under union and complement with respect to $X$.
For each $n\ge 1$, consider the set $E^{2n}$ of sequences of length $2n$ of members of $E$. There is a map $f_n\colon E^{2n}\to R(E)$ defined by
$$
f_n(A_0,A_1,\dotsc, A_{2i}, A_{2i+1}, \dotsc A_{2n}, A_{2n+1}) = (A_0\setminus A_1)\cup \dotsc (A_{2i}\setminus A_{2i+1})\dotsc (A_{2n}\setminus A_{2n+1}).
$$
The union of all of these maps is a surjection
$$
f = \bigcup_n f_n\colon \bigcup_n E^{2n}\to R(E).
$$
Every $E^{2n}$ is countable, so the domain of $f$, as the union of countably many of these sets, is itself countable. So $f$ is a surjection of a countable set onto $R(E)$, thus $R(E)$ is countable.
