Cartesian Product of Semirings of Sets

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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 $\displaystyle A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.


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:

$\O \times \O = \O$

$\Box$


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

$\Box$


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:

\(\displaystyle \paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}\) \(=\) \(\displaystyle \paren {S_1 \times \paren {T_1 \setminus T_2} } \cup \paren {\paren {S_1 \setminus S_2} \times T_1}\) Set Difference of Cartesian Products
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle S_1\) \(=\) \(\displaystyle \paren {S_1 \setminus S_2} \sqcup \paren {S_1 \cap S_2}\) Set Difference Union Intersection
\(\displaystyle \leadsto \ \ \) \(\displaystyle S_1 \times \paren {T_1 \setminus T_2}\) \(=\) \(\displaystyle \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \cap S_2} \times \paren {T_1 \setminus T_2}\) Cartesian Product Distributes over Union
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle T_1\) \(=\) \(\displaystyle \paren {T_1 \cap T_2} \sqcup \paren {T_1 \setminus T_2}\) Set Difference Union Intersection
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {S_1 \setminus S_2} \times T_2\) \(=\) \(\displaystyle \paren {S_1 \setminus S_2} \times \paren {T_1 \cap T_2} \ \sqcup \ \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2}\) Cartesian Product Distributes over Union
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}\) \(=\) \(\displaystyle \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \cap S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \setminus S_2} \times \paren {T_1 \cap T_2}\)


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.

$\blacksquare$


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