The problem: Let $X = \mathbb{R}$ Define $S = \{ G \subseteq \mathbb{R} : G$ is at most countable $\}$ $\cup$ $ \{\emptyset \}$. Show $S$ is a semi-ring.
Here is what I have so far. Well the empty set is in $S$. Also, if we let $A,B \in S$, then $A$ and $B$ is either finite or countable. Without loss of generality, suppose $A,B$ are countable. Then, $A \cap B \subseteq A$. Since $A$ is countable $A \cap B$ is countable. Thus $S$ is closed under finite intersections.
For the third condition, $A = \{a_1, a_2, a_3,... \}$, $B = \{ b_1,b_2,b_3,... \}$. I am stuck with getting a disjoint union from $A \setminus B$. I labeled the intersection elements $A \cap B = \{ c_1,c_2,c_3,... \}$. Then I tried writing the disjoint sets as follows:
For each $n \in \mathbb{N}$, define the sets
$C_n = \{ a_1,a_2,...,a_n \} \setminus\{ c_1,c_2,c_3,... \} $.
Do you think I am on the right track? Thank you very much!
$A \setminus B$ is itself a member of $S$ and so there is nothing to prove. It is the union of a single member of $S$. If you wish you can think of it as its union with the empty set. $A \setminus B=(A \setminus B) \cup \emptyset$. The empty set is at most countable set so it belongs to $S$