The collection of unions of finite disjoint collections of sets in a semiring is closed under finite unions and finite intersections. Define a semiring to be a collection $S$ of subsets of $X$ such that if $A$, $B \in S$, then $A \cap B \in S$, and there is a finite disjoint collection $\{C_k\}$ of sets in $S$ such that $A - B = \bigcup_{k=1}^n C_k$.
Define $S'$ to be the collection of unions of finite disjoint collections of sets in $S$, how to prove that $S'$ is closed with respect to the finite unions and finite intersections.
(please do not assume $S'$ is closed under relative complements, cause later I need to use the above property to prove $S'$ is closed under relative complements.)
Can anyone help me on this please?
 A: Let $A$ and $B$ be sets in $S'$. Then there are finite disjoint collections $\{A_i\}_{i=1}^n$ and $\{B_j\}_{j=1}^m$of sets in $S$, such that $A= \bigcup_{i=1}^n A_i$ and
$B= \bigcup_{j=1}^m B_j$.
It is easy to see that $\{A_i\cap B_j\}_{i,j=1}^n$ is a finite disjoint collection of sets in $S$. So,
$$ A\cap B= \bigcup_{i,j=1}^n (A_i\cap B_j) \in S' \tag{1}$$
Now note that
$$A \cup B_1 = \left (\bigcup_{i=1}^n A_i \right )\cup B_1 = \left (\bigcup_{i=1}^n (A_i \setminus B_1) \right )\cup B_1 $$
For each $i$, $1\leqslant i \leqslant n$ there is a finite disjoint collection $\{D_{i,k}\}_{k=1}^{n_i}$ of set in $S$ such that $A_i \setminus B_1= \bigcup_{k=1}^{n_i}D_{i,k}$. Since $\{A_i\}_{i=1}^n$ is a finite disjoint collection of sets in $S$, we can conclude that the collection $\{ D_{i,k} : 1 \leqslant i \leqslant n, 1 \leqslant k \leqslant n_i \} \cup \{B_1\} $  is a finite disjoint collection of sets in $S$. So, we have
$$ A \cup B_1 = \left (\bigcup_{i=1}^n (A_i \setminus B_1) \right )\cup B_1 = 
 \left (\bigcup_{i=1}^n \bigcup_{k=1}^{n_i} D_{i,k} \right )\cup B_1 \in S'
$$
So, we have that $ A \cup B_1  \in S'$. By finite induction, we have that
$$ A\cup B = A \cup \left (\bigcup_{j=1}^m B_j \right ) =( ( ( (A\cup B_1)\cup B_2) \cdots )\cup B_m) \in S' \tag{2} $$
From $(1)$ and $(2)$, by finite induction, we have that $S'$ is closed with respect to the finite unions and finite intersections.
Remark: Since the question is about semirings, we must be careful about the term "finite intersection". If we take an empty collection of sets in $S'$, the intersection of such empty family is, by definition, $X$ which may not be in $S$ nor in $S'$. So, we understand here "finite intersection" as being the intersection of a non-empty family of sets in $S'$.
A: To show if $A,B \in S'$ then $A \cup B \in S'$.
Let $A$ and $B$ be sets in $S'$. Then there are finite disjoint collections $\{ A_i \}_1^n$ and $\{ B_j \}_1^m$ of sets in $S$ such that $A = \cup_1^n A_i$ and $B = \cup_1^m B_j$.
$A \cup B
= \cup_{ i, j } ( A_i \cap B_j ) \cup
( \cup_i ( A_i - \cup_j B_j ) ) \cup
( \cup_j ( B_j - \cup_i A_i ) )$
$A_i \cap B_j, ( A_i - \cup_j B_j ), ( B_j - \cup_i A_i )$, $1 \le i \le n, 1 \le j \le m$ are disjoint.
Since S is a semiring, $A_i \cap B_j \in S$.
$A_i - \cup_1^m B_j = \cap_{ j = 1 }^m ( A_i - B_j )$
$B_j - \cup_1^m A_i = \cap_{ j = 1 }^m ( B_j - A_i )$
Since $S$ is a semiring, $A_i - B_j$ is a finite union of disjoint sets in $S$.
$A_i - B_j$ is a finite union of disjoint sets in $S$ $\implies$
$\cap_j ( A_i - B_j )$ is a finite union of disjoint sets in $S$ $\implies$
$\cup_i ( A_i - \cup_j B_j )$ is a finite union of disjoint sets in $S$.
Similarly,
$\cup_i ( B_j - \cup_1^m A_i )$ is a finite union of disjoint sets in $S$.
Therefore, $A \cup B$ is a finite union of disjoint sets in $S$.
$A \cup B \in S'$

Note. This question arises from the first sentence in the proof of Proposition 13 in Chapter 17.5 of Royden's Real Analysis. The author says it is clear that [...] and the reader thinks Why is this clear?
