Algebra of Sets is Closed under Intersection

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Let $S$ be a set.

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


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

Proof 1

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:      \(\ds \forall A, B \in \RR:\) \(\ds A \cap B \in \RR \)      


Proof 2

By definition $1$ of algebra of sets, we have that:

\((\text {AS} 2)\)   $:$   Closure under Union:      \(\ds \forall A, B \in \RR:\) \(\ds A \cup B \in \RR \)      
\((\text {AS} 3)\)   $:$   Closure under Complement Relative to $S$:      \(\ds \forall A \in \RR:\) \(\ds \relcomp S A \in \RR \)      


\(\ds A, B\) \(\in\) \(\ds \RR\)
\(\ds \leadsto \ \ \) \(\ds \relcomp S A \cup \relcomp S B\) \(\in\) \(\ds \RR\) Definition of Algebra of Sets: $\text {AS} 2$ and $\text {AS} 3$
\(\ds \leadsto \ \ \) \(\ds \relcomp S {A \cap B}\) \(\in\) \(\ds \RR\) De Morgan's Laws: Complement of Intersection
\(\ds \leadsto \ \ \) \(\ds A \cap B\) \(\in\) \(\ds \RR\) Definition of Algebra of Sets: $\text {AS} 3$