Am I allowed to use distributive law for infinitely many sets? 
Let $X$ be a nonempty set. A collection $S$ of subsets of $X$ is 
  called a semiring if it satisfies the following properties:
  
  
*
  
*The empty set belongs to S; that is $\emptyset \in S$.
  
*If $A,B \in S$; then $A \cap B \in S$; that is, $S$ is closed under finite intersections. 
  
*The set difference of any two sets of $S$ can be written as a finite union of 
  pair-wise disjoint members of $S$. That is, for every $A, B \in S$; there exist
  $C_1, ...,C_n \in S$ (depending on $A$ and $B$) such that $A\setminus B = \cup _{i=1}^n C_i$ and $C_i \cap C_j = \emptyset$ if $i\ne j$. 
  

Now, let $S$ be a semiring of subsets of $X$. A subset $A$ of $X$ is called a $\sigma$-set 
with respect to $S$ (or simply a $\sigma$-set) if there exists a disjoint sequence $\{A_n\}$ of 
$S$ such that $A = \cup_{n=1}^\infty A_n$. One can show easily that for every sequence $\{A_n\}$ of $S$, the set $A = \cup_{n=1}^\infty A_n$ is a $\sigma$-set.
I would like to prove that finite intersections of $\sigma$-sets is a $\sigma$-set. For this purpose, suppose $A ,B$ are $\sigma$-sets then $A = \cup_{i=1}^\infty C_i$ , $B = \cup_{j=1}^\infty D_j$. 
$$A\cap B=(\cup_{i=1}^\infty C_i)\cap (\cup_{j=1}^\infty D_j)$$
In this step I don't know am I allowed to use distributive law for infinitely many sets? Or the law holds only for finitely many sets?
If it holds only for finitely many sets how do I prove that finite intersections of $\sigma$-sets is a $\sigma$-set?
Thanks.
 A: Two sets $X$ and $Y$ are equal exactly when, for all $x$, $x \in X \iff x \in Y$.
Now, take $x \in (\cup_i C_i) \cap (\cup_j C_j)$.
This means precisely that $x$ is in $\cup_i C_i$ and also is in $\cup_j D_j$.
A statement of the form $x \in \cup_k S_k$ means precisely "there exists some $k$ such that $x \in S_k$". So we have from $x \in \cup_i C_i$ and $x \in \cup_j D_j$ that there exist $i$ and $j$ such that $x \in C_i$ and $x \in D_j$.
Or in other words, there exist $i$ and $j$ such that $x \in C_i \cap D_j$.
Or in other words, $x \in \cup_i \cup_j (C_i \cap D_j)$.
So, yes, the distributive law works for infinite, even uncountable families of sets.
A: Yes, both distributive laws generalize. (You need only one of the two, but it’s useful to know both.) The first step in verifying the generalization that you want is to check that 
$$\left(\bigcup_{i\in I}A_i\right)\cap D=\bigcup_{i\in I}(A_i\cap D)\;,\tag{1}$$
and to verify its mate you’ll want to check that
$$\left(\bigcap_{i\in I}A_i\right)\cup D=\bigcap_{i\in I}(A_i\cup D)\;.\tag{2}$$
Both are easily verified by element-chasing. For $(1)$, if $x\in\left(\bigcup_{i\in I}A_k\right)\cap D$, then $x\in\bigcup_{i\in I}A_i$ and $x\in D$. Since $x\in\bigcup_{i\in I}A_i$, there is an $i_0\in I$ such that $x\in A_{i_0}$, and therefore $x\in A_{i_0}\cap D\subseteq\bigcup_{i\in I}(A_i\cap D)$. Conversely, if $x\in\bigcup_{i\in I}(A_i\cap D)$, then there is an $i_0\in I$ such that $x\in A_{i_0}\cap D$. Then $x\in A_{i_0}\subseteq\bigcup_{i\in I}A_i$, and $x\in D$, so $x\in\left(\bigcup_{i\in I}A_i\right)\cap D$.
For $(2)$, if $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$, then $x\in\bigcap_{i\in I}A_i$ or $x\in D$. Let $i_0\in I$ be arbitrary. Then $A_{i_0}\supseteq\bigcap_{i\in I}A_i$, so $x\in A_{i_0}$ or $x\in D$, and therefoer $x\in A_{i_0}\cup D$. Since this holds for each $i_0\in I$, $x\in\bigcap_{i\in I}(A_i\cup D)$. Conversely, if $x\in\bigcap_{i\in I}(A_i\cup D)$, then $x\in A_i\cup D$ for each $i\in I$. If $x\in D$, then certainly $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$. If $x\notin D$, then we must have $x\in A_i$ for each $i\in I$, in which case $x\in\bigcap_{i\in I}A_i\subseteq\left(\bigcap_{i\in I}A_i\right)\cup D$.
Two applications of $(1)$ will give you the distributive law that you want. Suppose that $C=\bigcup_{i\in I}A_i$ and $D=\bigcup_{j\in J}B_j$. Then
$$\begin{align*}
C\cap D&=\left(\bigcup_{i\in I}A_i\right)\cap D\\
&=\bigcup_{i\in I}(A_i\cap D)\\
&=\bigcup_{i\in I}\left(A_i\cap\bigcup_{j\in J}B_j\right)\\
&=\bigcup_{i\in I}\left(\bigcup_{j\in J}(A_i\cap B_j)\right)\\
&=\bigcup_{\langle i,j\rangle\in I\times J}(A_i\cap B_j)\;.
\end{align*}$$
In other words, $C\cap D$ is the union of all possible intersections of the form $A_i\cap B_j$. In particular, if $I$ and $J$ are countable index sets, $I\times J$ is also countable.
Similarly, two applications of $(2)$ will give you the other general distributive law of this kind. This time suppose that $C=\bigcap_{i\in I}A_i$ and $D=\bigcap_{j\in J}B_j$. Then
$$\begin{align*}
C\cup D&=\left(\bigcap_{i\in I}A_i\right)\cup D\\
&=\bigcap_{i\in I}(A_i\cup D)\\
&=\bigcap_{i\in I}\left(A_i\cup\bigcap_{j\in J}B_j\right)\\
&=\bigcap_{i\in I}\left(\bigcap_{j\in J}(A_i\cup B_j)\right)\\
&=\bigcap_{\langle i,j\rangle\in I\times J}(A_i\cup B_j)\;
\end{align*}$$
the intersection of all possible unions of the form $A_i\cup B_j$ as $i$ and $j$ run over their respective index sets.
