Cartesian Product of Semirings of Sets

Theorem
Let $\mathcal S$ and $\mathcal T$ be semirings of sets.

Then $\mathcal S \times \mathcal T$ is also a semiring of sets.

Here, $\times$ denotes Cartesian product.

Proof
Recall the conditions for $\mathcal S \times \mathcal T$ to be a semiring of sets:


 * $(1): \quad \varnothing \in \mathcal S \times \mathcal T$
 * $(2): \quad \mathcal S \times \mathcal T$ is $\cap$-stable
 * $(3'):\quad$ If $A, B \in \mathcal S \times \mathcal T$, then there exists a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \mathcal S \times \mathcal T$ such that $\displaystyle A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.

Proof of $(1)$
From Empty Set is Subset of All Sets:
 * $\varnothing \in \mathcal S$

and:
 * $\varnothing \in \mathcal T$

So:
 * $\varnothing \times \varnothing \in \mathcal S \times \mathcal T$

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

Proof of $(2)$
Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\mathcal S \times \mathcal T$.

Then from Cartesian Product of Intersections:


 * $\left({S_1 \times T_1}\right) \cap \left({S_2 \times T_2}\right) = \left({S_1 \cap S_2}\right) \times \left({T_1 \cap T_2}\right)$

Since $\mathcal S$ and $\mathcal T$ are $\cap$-stable, $S_1 \cap S_2 \in \mathcal S$ and $T_1 \cap T_2 \in \mathcal T$.

Hence $\left({S_1 \times T_1}\right) \cap \left({S_2 \times T_2}\right) \in \mathcal S \times \mathcal T$.

Proof of $(3')$
Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\mathcal S \times \mathcal T$.

Let $\sqcup$ signify union of disjoint sets.

Then:

Now recall that $\mathcal S$ and $\mathcal T$ 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 $\left({S_1 \times T_1}\right) \setminus \left({S_2 \times T_2}\right)$ as well, completing the proof.