Algebra of Sets is Closed under Intersection/Proof 2

Theorem

Let $S$ be a set.

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

Then:

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

$(\text {AS} 2)$	$:$		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 $

Thus:

$\ds A, B$

$\in$

$\ds \RR$

$\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 \ \ $

$\ds \relcomp S {A \cap B}$

$\in$

$\ds \RR$

$\ds \leadsto \ \ $

$\ds A \cap B$

$\in$

$\ds \RR$

Definition of Algebra of Sets: $\text {AS} 3$

$\blacksquare$

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