Prove: If $A,B \in S_Y$ then there exists disjoint sets $C_1, \dots, C_n \in S_Y $ s.t. $A\setminus B = \bigcup \limits_{i=1}^nC_i$.

Question

Let $S$ be a semiring of subsets of a set $X$ and let $Y \subseteq X$.

Let $$S_Y=\{ Y\cap A:A\in S \}$$

I am trying to prove that $S_Y$ is a semiring of $Y$.

I already managed to prove that $\emptyset \in S_Y$ and if $A,B \in S_Y$ then $A\cap B \in S_Y$.

The only thing that remains for me is to prove the following:

If $A,B \in S_Y$ then there exists a finite number of disjoint sets $C_1, \dots, C_n \in S_Y $ such that $A\setminus B = \bigcup \limits_{i=1}^nC_i$.

I wrote down the following so far:

Let $A,B \in S_Y$ then $A = Y\cap K_1$ and $B = Y\cap K_2$ for some $K_1, K_2 \in S$.

Now $A\setminus B = (Y \cap K_1)\setminus(Y\cap K_2)$

And this is where I am stuck. Can anyone please point me in the right direction?

Answer 1

It suffices to observe that $(K_1\cap Y)\setminus (K_2\cap Y) = (K_1 \setminus K_2) \cap Y$. Indeed, denoting by $X^c$ the complement of $X$, one gets \begin{align} (K_1\cap Y)\setminus (K_2\cap Y) &= (K_1\cap Y) \cap (K_2\cap Y_{})^c = (K_1\cap Y_{}) \cap (K_2^c \cup Y_{}^c) \\ &= \bigl((K_1\cap Y_{}) \cap K_2^c\bigr) \cup \bigl((K_1\cap Y_{}) \cap Y_{}^c\bigr) \\ &= (K_1\cap K_2^c \cap Y_{}) \cup \emptyset = (K_1 \setminus K_2) \cap Y_{} \end{align} Now, since $K_1, K_2 \in S$, there exists a finite number of disjoint sets $C_1, \dots, C_n \in S_Y$ such that $K_1 \setminus K_2 = \bigcup \limits_{i=1}^nC_i$. It follows that $(K_1 \setminus K_2) \cap Y_{} = \bigcup \limits_{i=1}^n(C_i \cap S)$.

Answer 2

Let it be that $K_{1}\setminus K_{2}=\bigcup_{i=1}^{n}C_{i}$ where the $C_{i}$ are disjoint elements of the semiring $S$.

Then:

$$A\setminus B=\left(Y\cap K_{1}\right)\setminus\left(Y\cap K_{2}\right)=Y\cap\left(K_{1}\setminus K_{2}\right)=Y\cap\left(\bigcup_{i=1}^{n}C_{i}\right)=\bigcup_{i=1}^{n}\left(Y\cap C_{i}\right)$$

In the $Y\cap C_{i}$ we recognize disjoint elements of $S_{Y}$.