# 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:

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

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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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][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

1. 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.

# OLD PAGE

## Theorem statement

If [ilmath]\mathcal{A} [/ilmath] is a class of subsets of [ilmath]\Omega[/ilmath] such that[1]

• $\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}]$ - 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][Note 1]

Proof:

TODO: Page 3 in[1]

## Notes

1. 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. Probability Theory - A comprehensive course - Second Edition - Achim Klenke