Sigma-Ring is Closed under Countable Intersections

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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

From De Morgan's laws: Difference with Intersection:

$\ds \bigcup_{n \mathop = 2}^\infty \paren {A_1 \setminus A_n} = A_1 \setminus \paren {\bigcap_{n \mathop = 2}^\infty A_n}$

From Set Difference with Set Difference:

\(\ds A_1 \setminus \paren {A_1 \setminus \paren {\ds \bigcap_{n \mathop = 2}^\infty A_n} }\) \(=\) \(\ds A_1 \cap \paren {\ds \bigcap_{n \mathop = 2}^\infty A_n}\)
\(\ds \) \(=\) \(\ds {\bigcap_{n \mathop = 1}^\infty A_n}\)

Combining the previous equalities, it follows that:

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

$\blacksquare$


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