Algebra of Sets is Closed under Intersection/Proof 1

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Theorem

Let $S$ be a set.

Let $\RR$ be an algebra of sets on $S$.

Then:

$\forall A, B \in S: A \cap B \in \RR$


Proof

By definition $2$ of Algebra of Sets:

$\RR$ is a ring of sets with a unit.

By definition $1$ of Ring of Sets:

\((\text {RS} 2_1)\)   $:$   Closure under Intersection:      \(\ds \forall A, B \in \RR:\) \(\ds A \cap B \in \RR \)      

$\blacksquare$