Notes:Hereditary sigma-ring/Proof of facts

1. An hereditary system is a sigma-ring [ilmath]\iff[/ilmath] it is closed under countable unions.
   1. 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
   2. A hereditary system closed under countable union [ilmath]\implies[/ilmath] it is a [ilmath]\sigma[/ilmath]-ring
      1. 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.
      2. Closed under countable union is given.
2. [ilmath]\mathcal{H}(\mathcal{R})[/ilmath] is a [ilmath]\sigma[/ilmath]-ring (for any [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath])
   1. It is already shown that a hereditary system is closed under set subtraction, only remains to be shown closed under countable union
   2. 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].
        • Thus [ilmath]\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R})[/ilmath]
      • Thus [ilmath]\mathcal{H}(\mathcal{R})[/ilmath] is closed under countable union.
3. [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)
4. [ilmath]\sigma_R(\mathcal{H}(S))[/ilmath] is hereditary
   • Let [ilmath]A\in\sigma_R(\mathcal{H}(S))[/ilmath] be given. We want to show that [ilmath]\forall B\in\mathcal{P}(A)[/ilmath] that [ilmath]B\in\sigma_R(\mathcal{H}(S))[/ilmath].
     1. If [ilmath]A\in\mathcal{H}(S)[/ilmath], then [ilmath]\forall B\in\mathcal{P}(A)[B\in\mathcal{H}(S)[/ilmath] but [ilmath]B\in\mathcal{H}(S)\implies B\in\sigma_R(\mathcal{H}(S))[/ilmath]
        • We're done in this case.
     2. OTHERWISE: [ilmath]\exists(A_n)_{n=1}^\infty\subseteq\mathcal{H}(S)\left[\bigcup_{n=1}^\infty A_n=A\right][/ilmath] (by fact 3)
        • Let [ilmath]B\in\mathcal{P}(A)[/ilmath] be given.
          • Define a new sequence, [ilmath] ({ B_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}(S) [/ilmath], where [ilmath]B_i:=A_i\cap B[/ilmath]
            • [ilmath]A_i\cap B[/ilmath] is a subset of [ilmath]A_i[/ilmath] and [ilmath]A_i\in\mathcal{H}(S)[/ilmath], as "hereditary" means "contains all subsets of" [ilmath]A_i\cap B\subseteq A_i[/ilmath] thus [ilmath]A_i\cap B:=B_i\in\mathcal{H}(S)[/ilmath]
          • Clearly [ilmath]B=\bigcup_{n=1}^\infty B_n[/ilmath] (as [ilmath]B\subseteq A[/ilmath] and [ilmath]A=\bigcup_{n=1}^\infty A_n[/ilmath])
          • As [ilmath]\sigma_R(\mathcal{H}(S)[/ilmath] contains all countable unions of things in [ilmath]\mathcal{H}(S)[/ilmath] we know:
            • [ilmath]\bigcup_{n=1}^\infty B_n=B\in\sigma_R(\mathcal{H}(S))[/ilmath]
        • We have shown [ilmath]B\in\sigma_R(\mathcal{H}(S))[/ilmath]
   • We have completed the proof