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:
- $\O \times \O = \O$
$\Box$
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:
\(\ds \paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}\) | \(=\) | \(\ds \paren {S_1 \times \paren {T_1 \setminus T_2} } \cup \paren {\paren {S_1 \setminus S_2} \times T_1}\) | Set Difference of Cartesian Products | |||||||||||
\(\ds \) | \(\) | \(\ds \) | ||||||||||||
\(\ds S_1\) | \(=\) | \(\ds \paren {S_1 \setminus S_2} \sqcup \paren {S_1 \cap S_2}\) | Set Difference Union Intersection | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds S_1 \times \paren {T_1 \setminus T_2}\) | \(=\) | \(\ds \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \cap S_2} \times \paren {T_1 \setminus T_2}\) | Cartesian Product Distributes over Union | ||||||||||
\(\ds \) | \(\) | \(\ds \) | ||||||||||||
\(\ds T_1\) | \(=\) | \(\ds \paren {T_1 \cap T_2} \sqcup \paren {T_1 \setminus T_2}\) | Set Difference Union Intersection | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {S_1 \setminus S_2} \times T_2\) | \(=\) | \(\ds \paren {S_1 \setminus S_2} \times \paren {T_1 \cap T_2} \ \sqcup \ \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2}\) | Cartesian Product Distributes over Union | ||||||||||
\(\ds \) | \(\) | \(\ds \) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}\) | \(=\) | \(\ds \paren {S_1 \setminus S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \cap S_2} \times \paren {T_1 \setminus T_2} \ \sqcup \ \paren {S_1 \setminus S_2} \times \paren {T_1 \cap T_2}\) |
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.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $13.1$