# Algebra of Sets is Closed under Intersection

## 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 1

$\RR$ is a ring of sets with a unit.
 $(\text {RS} 2_1)$ $:$ Closure under Intersection: $\ds \forall A, B \in \RR:$ $\ds A \cap B \in \RR$

$\blacksquare$

## 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 $S$: $\ds \forall A \in \RR:$ $\ds \relcomp S A \in \RR$

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

$\blacksquare$