Properties of classes of sets closed under set-subtraction

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Statement

Suppose [ilmath]\mathcal{A} [/ilmath] is an arbitrary class of sets with the property that:

• [ilmath]\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}][/ilmath] where [ilmath]A-B[/ilmath] denotes "set subtraction" (AKA: relative complement)

Then we claim the following are true of [ilmath]\mathcal{A} [/ilmath]:

1. [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

Grade: C

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Outlines follow for each claim:

1. Prove [ilmath]A\cap B=A-(A-B)[/ilmath] or indeed [ilmath]B-(B-A)[/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:

1. If [ilmath]\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-[ilmath]\cup[/ilmath]-closed then it is [ilmath]\sigma[/ilmath]-[ilmath]\cap[/ilmath]-closed
2. 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]

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]

OLD PAGE

Theorem statement

If [ilmath]\mathcal{A} [/ilmath] is a class of subsets of [ilmath]\Omega[/ilmath] such that

• [math]\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}][/math] - that is closed under set-subtraction (or [ilmath]\backslash[/ilmath]-closed)

Then we have:

1. [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]-closed
2. [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
3. 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]

Proof:

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TODO: Page 3 in

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