# Ring of Sets is Semiring of Sets

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## Theorem

Let $\mathcal R$ be a ring of sets.

Then $\mathcal R$ is also a semiring of sets.

## Proof

Let $A \in \mathcal R$.

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:

- $A \setminus A_1 = A_2 \in \mathcal R$

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

- $\left({A \setminus A_1}\right) \cap A_1 = \varnothing$

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 $A_1 \cap A_2 = \varnothing$ and $A_1, A_2 \in \mathcal R$

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

$\blacksquare$