why S' is closed with respect to the formation of finite unions? $Proposition$ : Let S be a semiring of subsets of a set X. Define S' to be the collection of
unions of finite disjoint collections of sets in S. Then S' is closed with respect to the formation
of relative complements.
my question is
why S' is closed with respect to the formation of finite unions ?
 A: First note that $S'$ is clearly closed under finite unions of pairwise disjoint sets: if $S_1,\dots,S_n$ are pairwise disjoint members of $S'$, then each is the union of a pairwise disjoint family of subsets of $S$, and the union of those families is a pairwise disjoint family of subsets of $S$ whose union is $S_1\cup\ldots\cup S_n$.
Now suppose that $S_1=F_1\cup\dots\cup F_m$ and $S_2=G_1\cup\dots\cup G_n$ are in $S'$, where $\{F_1,\dots,F_m\}$ and $\{G_1,\dots,G_n\}$ are collections of pairwise disjoint members of $S$. For $1\le i\le m$ and $1\le k\le n$ let $H(i,k)=F_i\cap G_k$. Then the sets $H(i,k)$ are pairwise disjoint members of $S$, and 
$$S_1\cap S_2=\bigcup_{i=1}^m\bigcup_{k=1}^nH(i,k)\;.$$
An easy induction on the number of sets now shows that $S'$ is closed under finite intersections. 
Now suppose that you’ve already done the stated exercise of showing that $S'$ is closed under the formation of relative complements; then you know that $S_1\setminus S_2$ and $S_2\setminus S_1$ are in $S'$. And $S_1\cup S_2$ is the disjoint union of $S_1\setminus S_2$, $S_1\cap S_2$, and $S_2\setminus S_1$, all three of which are in $S'$, so $S_1\cup S_2\in S'$. Another easy induction on the number of sets now shows that $S'$ is closed under taking finite unions.
