# Sigma-Ring is Closed under Countable Intersections

## Theorem

Let $\RR$ be a $\sigma$-ring.

Let $\sequence {A_n}_{n \mathop \in \N} \in \RR$ be a sequence of sets in $\RR$.

Then:

$\ds \bigcap_{n \mathop = 1}^\infty A_n \in \RR$

## Proof

 $\ds \forall n \in \N_{>0}: \,$ $\ds A_1, A_n \in \RR$ $\leadsto$ $\ds A_1 \setminus A_n \in \RR$ Axiom $(\text {SR} 2)$ for $\sigma$-rings $\ds$ $\leadsto$ $\ds \bigcup_{n \mathop = 2}^\infty \paren {A_1 \setminus A_n} \in \RR$ Axiom $(\text {SR} 3)$ for $\sigma$-rings $\ds$ $\leadsto$ $\ds A_1 \setminus \paren {\bigcup_{n \mathop = 2}^\infty \paren {A_1 \setminus A_n} } \in \RR$ Axiom $(\text {SR} 2)$ for $\sigma$-rings
$\ds \bigcup_{n \mathop = 2}^\infty \paren {A_1 \setminus A_n} = A_1 \setminus \paren {\bigcap_{n \mathop = 2}^\infty A_n}$
 $\ds A_1 \setminus \paren {A_1 \setminus \paren {\ds \bigcap_{n \mathop = 2}^\infty A_n} }$ $=$ $\ds A_1 \cap \paren {\ds \bigcap_{n \mathop = 2}^\infty A_n}$ $\ds$ $=$ $\ds {\bigcap_{n \mathop = 1}^\infty A_n}$

Combining the previous equalities, it follows that:

$\ds \bigcap_{n \mathop = 1}^\infty A_n \in \RR$

$\blacksquare$