Notes:Hereditary sigma-ring

I'm writing down some "facts" so I don't keep redoing them on paper.

What I want to show

• [ilmath]\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))[/ilmath] for a system of sets, [ilmath]S[/ilmath].

Both are hereditary, and both are [ilmath]\sigma[/ilmath]-rings.

The result is true!

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TODO: Make a theorem out of this!

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Showing [ilmath]\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))[/ilmath]

Which way first?

[ilmath]\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))[/ilmath]

• Let [ilmath]A\in\mathcal{H}(\sigma_R(S))[/ilmath] be given. Then:
  • [ilmath]\exists B\in\sigma_R(S)[A\in\mathcal{P}(B)][/ilmath]
    • Notice that [ilmath]S\subseteq\mathcal{H}(S)[/ilmath] thus:
      • [ilmath]\sigma_R(S)\subseteq\sigma_R(\mathcal{H}(S))[/ilmath]
    • So [ilmath]B\in\sigma_R(\mathcal{H}(S))[/ilmath]
  • As [ilmath]\sigma_R(\mathcal{H}(S))[/ilmath] is hereditary [ilmath]\forall C\in\mathcal{P}(B)[C\in\sigma_R(\mathcal{H}(S))[/ilmath]
    • So [ilmath]A\in\sigma_R(\mathcal{H}(S))[/ilmath]
• This shows [ilmath]\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))[/ilmath]

[ilmath]\sigma_R(\mathcal{H}(S))\subseteq\mathcal{H}(\sigma_R(S))[/ilmath]

• Let [ilmath]A\in\sigma_R(\mathcal{H}(S))[/ilmath], then, either (by fact 3):
  1. [ilmath]A\in\mathcal{H}(S)[/ilmath]
     • This means: [ilmath]\exists B\in S[A\in\mathcal{P}(B)][/ilmath] (also said as: [ilmath]\exists B\in S[A\subseteq B][/ilmath])
     • But [ilmath]S\subseteq\sigma_R(S)[/ilmath], thus:
       • [ilmath]B\in\sigma_R(S)[/ilmath]
     • But [ilmath]\sigma_R(S)\subseteq\mathcal{H}(\sigma_R(S))[/ilmath] thus:
       • [ilmath]B\in\mathcal{H}(\sigma_R(S))[/ilmath]
     • As [ilmath]\mathcal{H}(\sigma_R(S))[/ilmath] is hereditary, all subsets of [ilmath]B[/ilmath] are in it.
     • As [ilmath]A\in\mathcal{P}(B)[/ilmath] we see [ilmath]A\in\mathcal{H}(\sigma_R(S))[/ilmath] - this completes this half of the proof.
  2. [ilmath] \exists ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}(S) [\bigcup_{n=1}^\infty A_n=A] [/ilmath]
     • Using part 1 we see that: [ilmath]\forall i\in\mathbb{N}[A_i\in\mathcal{H}(\sigma_R(S))][/ilmath]
       • But we know [ilmath]\mathcal{H}(\sigma_R(S))[/ilmath] is a [ilmath]\sigma[/ilmath]-ring, thus closed under countable union
     • Thus [ilmath]\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\sigma_R(S))[/ilmath]
       • But [ilmath]A:=\bigcup_{n=1}^\infty A_n[/ilmath] so
     • [ilmath]A\in\mathcal{H}(\sigma_R(S))[/ilmath]
• We have shown that in either case, [ilmath]A\in\mathcal{H}(\sigma_R(S))[/ilmath]

Facts

1. An hereditary system is a sigma-ring [ilmath]\iff[/ilmath] it is closed under countable unions.
   • Thus [ilmath]\sigma_R(\mathcal{H}(S))[/ilmath] is just [ilmath]\mathcal{H}(S)[/ilmath] with the additional property:
     • [ilmath]\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}(S)\left[\bigcup_{n=1}^\infty A_n\in\sigma_R(\mathcal{H}(S))\right][/ilmath]
2. [ilmath]\mathcal{H}(\mathcal{R})[/ilmath] is a [ilmath]\sigma[/ilmath]-ring (for any [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath])
   • This means [ilmath]\sigma_R(\mathcal{H}(\mathcal{R}))=\mathcal{H}(\mathcal{R})[/ilmath]
   • It also means [ilmath]\mathcal{H}(\sigma_R(S))[/ilmath] is a [ilmath]\sigma[/ilmath]-ring
3. [ilmath]\sigma_R(\mathcal{H}(S))[/ilmath] is just [ilmath]\mathcal{H}(S)[/ilmath] closed under countable union.
4. [ilmath]\sigma_R(\mathcal{H}(S))[/ilmath] is hereditary

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

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TODO: It seems, "hereditary sigma-ring" is the same as "sigma-ideal".

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