Properties of classes of sets closed under setsubtraction
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Contents
Statement
Suppose [ilmath]\mathcal{A} [/ilmath] is an arbitrary class of sets with the property that:
 [ilmath]\forall A,B\in\mathcal{A}[AB\in\mathcal{A}][/ilmath] where [ilmath]AB[/ilmath] denotes "set subtraction" (AKA: relative complement)
Then we claim the following are true of [ilmath]\mathcal{A} [/ilmath]:
 [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]closed. That is: [ilmath]\forall A,B\in\mathcal{A}[A\cap B\in\mathcal{A}][/ilmath]
Proof of claims
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Outlines follow for each claim:
 Prove [ilmath]A\cap B=A(AB)[/ilmath] or indeed [ilmath]B(BA)[/ilmath]. Include picture of Venn diagram
This proof has been marked as an page requiring an easy proof
Notes for further claims
I know from a reference (see the old page below) that:
 If [ilmath]\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath][ilmath]\cup[/ilmath]closed then it is [ilmath]\sigma[/ilmath][ilmath]\cap[/ilmath]closed
 Any countable union of sets in [ilmath]\mathcal{A} [/ilmath] can be expressed as a countable disjoint union of sets in [ilmath]\mathcal{A} [/ilmath]^{[Note 1]}
These feel very "measure theory specific" though and might be true in the case of being closed under arbitrary union or for an arbitrary collection of sets whose union is [ilmath]\mathcal{A} [/ilmath]
Notes
 ↑ Note that this doesn't require [ilmath]\mathcal{A} [/ilmath] to be closed under union, we can still talk about unions we just cannot know that the result of a union is in [ilmath]\mathcal{A} [/ilmath]. This simply says IF there is a sequence of elements of [ilmath]\mathcal{A} [/ilmath], say [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{A} [/ilmath], where [ilmath]\bigcup_{n=1}^\infty A_n[/ilmath] just happens to be in [ilmath]\mathcal{A} [/ilmath] then there exists a sequence of pairwise disjoint sets whose union is the same.
References
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OLD PAGE
Theorem statement
If [ilmath]\mathcal{A} [/ilmath] is a class of subsets of [ilmath]\Omega[/ilmath] such that^{[1]}
 [math]\forall A,B\in\mathcal{A}[AB\in\mathcal{A}][/math]  that is closed under setsubtraction (or [ilmath]\backslash[/ilmath]closed)
Then we have:
 [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]closed
 [ilmath]\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath][ilmath]\cup[/ilmath]closed[ilmath]\implies[/ilmath] [ilmath]\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath][ilmath]\cap[/ilmath]closed
 Any countable union of sets in [ilmath]\mathcal{A} [/ilmath] can be expressed as a countable disjoint union of sets in [ilmath]\mathcal{A} [/ilmath]^{[Note 1]}
Proof:
TODO: Page 3 in^{[1]}
Notes
 ↑ Note that this doesn't require [ilmath]\mathcal{A} [/ilmath] to be closed under union, we can still talk about unions we just cannot know that the result of a union is in [ilmath]\mathcal{A} [/ilmath]
References
 ↑ ^{1.0} ^{1.1} Probability Theory  A comprehensive course  Second Edition  Achim Klenke
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