Notes:Hereditary sigma-ring/Proof of facts
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- An hereditary system is a sigma-ring [ilmath]\iff[/ilmath] it is closed under countable unions.
- Hereditary system is a sigma-ring [ilmath]\implies[/ilmath] closed under countable unions
- It is a [ilmath]\sigma[/ilmath]-ring which means it is closed under countable unions. Done
- A hereditary system closed under countable union [ilmath]\implies[/ilmath] it is a [ilmath]\sigma[/ilmath]-ring
- closed under set-subtraction
- Let [ilmath]A,B\in\mathcal{H} [/ilmath] for some hereditary system [ilmath]\mathcal{H} [/ilmath]. Then:
- [ilmath]A-B\subseteq A[/ilmath], but [ilmath]\mathcal{H} [/ilmath] contains [ilmath]A[/ilmath] and therefore all subsets of [ilmath]A[/ilmath]
- Thus [ilmath]\mathcal{H} [/ilmath] is closed under set subtraction.
- Let [ilmath]A,B\in\mathcal{H} [/ilmath] for some hereditary system [ilmath]\mathcal{H} [/ilmath]. Then:
- Closed under countable union is given.
- closed under set-subtraction
- Hereditary system is a sigma-ring [ilmath]\implies[/ilmath] closed under countable unions
- [ilmath]\mathcal{H}(\mathcal{R})[/ilmath] is a [ilmath]\sigma[/ilmath]-ring (for any [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath])
- It is already shown that a hereditary system is closed under set subtraction, only remains to be shown closed under countable union
- Closed under countable union
- Let [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{H}(\mathcal{R})[/ilmath] (we need to show [ilmath]\implies\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R})[/ilmath])
- This means, for each [ilmath]A_n\in\mathcal{H}(\mathcal{R})[/ilmath] there is a [ilmath]B_n\in\mathcal{R} [/ilmath] with [ilmath]A_n\subseteq B_n[/ilmath] thus:
- [ilmath]\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}(\mathcal{R})\exists(B_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A_i\subseteq B_i][/ilmath]
- However [ilmath]\mathcal{R} [/ilmath] is a [ilmath]\sigma[/ilmath]-ring, thus:
- Define [ilmath]B:=\bigcup_{n=1}^\infty B_n[/ilmath], notice [ilmath]B\in\mathcal{R} [/ilmath]
- But a union of subsets is a subset of the union, thus:
- [ilmath]\bigcup_{n=1}^\infty A_n\subseteq\bigcup_{n=1}^\infty B_n:=B[/ilmath], thus
- [ilmath]\bigcup_{n=1}^\infty A_n\subseteq B[/ilmath]
- BUT [ilmath]\mathcal{H}(\mathcal{R})[/ilmath] contains all subsets of all things in [ilmath]\mathcal{R} [/ilmath], thus contains all subsets of [ilmath]B[/ilmath].
- [ilmath]\bigcup_{n=1}^\infty A_n\subseteq\bigcup_{n=1}^\infty B_n:=B[/ilmath], thus
- Thus [ilmath]\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R})[/ilmath]
- This means, for each [ilmath]A_n\in\mathcal{H}(\mathcal{R})[/ilmath] there is a [ilmath]B_n\in\mathcal{R} [/ilmath] with [ilmath]A_n\subseteq B_n[/ilmath] thus:
- Thus [ilmath]\mathcal{H}(\mathcal{R})[/ilmath] is closed under countable union.
- Let [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{H}(\mathcal{R})[/ilmath] (we need to show [ilmath]\implies\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R})[/ilmath])
- [ilmath]\sigma_R(\mathcal{H}(S))[/ilmath] is just [ilmath]\mathcal{H}(S)[/ilmath] closed under countable union.
- Follows from fact 1. As [ilmath]\mathcal{H}(S)[/ilmath] is an hereditary system, the sigma-ring generated by it (the smallest sigma ring containing [ilmath]\mathcal{H}(S)[/ilmath] is just the set with whatever is needed to close it under the operators)