Cartesian Product of Semirings of Sets

Theorem
Let $\SS$ and $\TT$ be semirings of sets.

Then $\SS \times \TT$ is also a semiring of sets.

Here, $\times$ denotes Cartesian product.

Proof
Recall the conditions for $\SS \times \TT$ to be a semiring of sets:


 * $(1): \quad \O \in \SS \times \TT$
 * $(2): \quad \SS \times \TT$ is $\cap$-stable
 * $(3'):\quad$ If $A, B \in \SS \times \TT$, then there exists a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS \times \TT$ such that $\ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.

Proof of $(1)$
From Empty Set is Subset of All Sets:
 * $\O \in \SS$

and:
 * $\O \in \TT$

So:
 * $\O \times \O \in \SS \times \TT$

From Cartesian Product is Empty iff Factor is Empty:
 * $\O \times \O = \O$

Proof of $(2)$
Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\SS \times \TT$.

Then from Cartesian Product of Intersections:


 * $\paren {S_1 \times T_1} \cap \paren {S_2 \times T_2} = \paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2}$

Since $\SS$ and $\TT$ are $\cap$-stable, $S_1 \cap S_2 \in \SS$ and $T_1 \cap T_2 \in \TT$.

Hence $\paren {S_1 \times T_1} \cap \paren {S_2 \times T_2} \in \SS \times \TT$.

Proof of $(3')$
Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\SS \times \TT$.

Let $\sqcup$ signify union of disjoint sets.

Then:

Now recall that $\SS$ and $\TT$ are semirings of sets.

Thence the expressions $S_1 \setminus S_2$ and $T_1 \setminus T_2$ may be written as finite disjoint unions.

Applying Cartesian Product Distributes over Union again, each of the three $\sqcup$-operands in the expression above may thus be written as a finite disjoint union.

This yields the same fact for $\paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}$ as well, completing the proof.