Equivalence of Definitions of Semiring of Sets

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Theorem

The following definitions of the concept of Semiring of Sets are equivalent:

A collection $\SS$ of subsets of a set $X$ is a semiring (of sets) if and only if:

$(1): \quad \O \in \SS$
$(2): \quad A, B \in \SS \implies A \cap B \in \SS$; that is, $\SS$ is $\cap$-stable
$(3): \quad$ If $A, A_1 \in \SS$ such that $A_1 \subseteq A$, then there exists a finite sequence $A_2, A_3, \ldots, A_n \in \SS$ such that:
$(3 \text a): \quad \ds A = \bigcup_{k \mathop = 1}^n A_k$
$(3 \text b): \quad$ The $A_k$ are pairwise disjoint

We prove that criterion $(3)$ can be replaced by:

$(3'):\quad$ If $A, B \in \SS$, then there exist finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS$ such that $\ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.


Proof

$(3)$ implies $(3')$

Let $X$ be a set, and let $\SS$ be a collection of subsets of $X$.


Suppose that for all $A, A_1 \in \SS$ such that $A_1 \subseteq A$, there exists a finite sequence of sets $A_2, A_3, \ldots, A_n \in \SS$ such that:

$A_1, A_2, \ldots, A_n$ are pairwise disjoint
$\ds A = \bigcup_{k \mathop = 1}^n A_k$

Let $B \in \SS$, and let $A_1 = A \cap B$.

It follows that $A_1 \in \SS$, by definition.

Also, $A_1 \subseteq A$ by Intersection is Subset.

Then:

\(\ds A \setminus B\) \(=\) \(\ds A \setminus \paren {A \cap B}\) Set Difference with Intersection is Difference
\(\ds \) \(=\) \(\ds A \setminus A_1\)
\(\ds \) \(=\) \(\ds \paren {\bigcup_{k \mathop = 1}^n A_k} \setminus A_1\)
\(\ds \) \(=\) \(\ds \bigcup_{k \mathop = 1}^n \paren {A_k \setminus A_1}\) Set Difference is Right Distributive over Union
\(\ds \) \(=\) \(\ds \bigcup_{k \mathop = 2}^n \paren {A_k \setminus A_1}\) Set Difference with Self is Empty Set and Union with Empty Set
\(\ds \) \(=\) \(\ds \bigcup_{k \mathop = 2}^n A_k\) Set Difference with Disjoint Set

as required.

$\Box$


$(3')$ implies $(3)$

Now suppose that for all $A, B \in \SS$, there exists a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS$ such that $\ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.

Then $B$ is disjoint with each of the sets $A_k$.

Let $B \subseteq A$. Then:

\(\ds B \cup \bigcup_{k \mathop = 1}^n A_k\) \(=\) \(\ds \paren {A \setminus B} \cup B\)
\(\ds \) \(=\) \(\ds A \cup B\) Set Difference Union Second Set is Union
\(\ds \) \(=\) \(\ds A\) Union with Superset is Superset

as required.

$\blacksquare$