Ring of Sets is Semiring of Sets

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Let $\RR$ be a ring of sets.

Then $\RR$ is also a semiring of sets.


Let $A \in \RR$.

Suppose $A_1 \subseteq A$.

Let $A_2 = A \setminus A_1$, where $A \setminus A_1$ denotes set difference.

By definition, $A_2$ is then the relative complement of $A_1$ with respect to $A$.

From Union with Relative Complement it then follows that $A_1 \cup A_2 = A$.

By Ring of Sets Closed under Various Operations:

$A \setminus A_1 = A_2 \in \RR$

But by Set Difference Intersection with Second Set is Empty Set:

$\paren {A \setminus A_1} \cap A_1 = \O$

Hence for any given $A$ for which we have $A_1 \subseteq A$, we can represent $A$ as the finite expansion $A_1 \cup A_2$ such that $A_1 \cap A_2 = \O$ and $A_1, A_2 \in \RR$

Thus, by definition, $\RR$ is a semiring of sets.