Difference between revisions of "Notes:Hereditary sigma-ring"

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==What I want to show==
 
==What I want to show==
 
* {{M|1=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}} for a system of sets, {{M|S}}.
 
* {{M|1=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}} for a system of sets, {{M|S}}.
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==Ideas==
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Both are hereditary, and both are {{sigma|rings}}.
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* Maybe just try showing {{M|\subseteq}} for each side on paper
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==Facts==
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# An hereditary system is a sigma-ring {{M|\iff}} it is closed under countable unions.
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#* Thus {{M|\sigma_R(\mathcal{H}(S))}} is just {{M|\mathcal{H}(S)}} with the additional property:
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#** {{M|1=\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}(S)\left[\bigcup_{n=1}^\infty A_n\in\sigma_R(\mathcal{H}(S))\right]}}<!--
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FACT 2
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{{Note|The result is true!}}
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-->
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# {{M|\mathcal{H}(\mathcal{R})}} is a {{sigma|ring}} (for any {{sigma|ring}}, {{M|\mathcal{R} }})
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#* This means {{M|1=\sigma_R(\mathcal{H}(\mathcal{R}))=\mathcal{H}(\mathcal{R})}}
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#* It also means {{M|\mathcal{H}(\sigma_R(S))}} is a {{sigma|ring}}<!--
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FACT 3
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{{Todo|Make a theorem out of this!}}
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-->
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==Showing {{M|1=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}}==
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# {{M|1=\sigma_R(\mathcal{H}(S))}} is just {{M|\mathcal{H}(S)}} closed under countable union.
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Which way first?
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==Proof of facts==
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==={{M|1=\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))}}===
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# An hereditary system is a sigma-ring {{M|\iff}} it is closed under countable unions.
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* Let {{M|A\in\mathcal{H}(\sigma_R(S))}} be given. Then:
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## Hereditary system is a sigma-ring {{M|\implies}} closed under countable unions
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** {{M|1=\exists B\in\sigma_R(S)[A\in\mathcal{P}(B)]}}
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##* It ''is a {{sigma|ring}}'' which means it is closed under countable unions. Done
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*** Notice that {{M|S\subseteq\mathcal{H}(S)}} thus:
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## A hereditary system closed under countable union {{M|\implies}} it is a {{sigma|ring}}
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**** {{M|\sigma_R(S)\subseteq\sigma_R(\mathcal{H}(S))}}
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### closed under set-subtraction
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*** So {{M|B\in\sigma_R(\mathcal{H}(S))}}
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###* Let {{M|A,B\in\mathcal{H} }} for some hereditary system {{M|\mathcal{H} }}. Then:
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** As {{M|\sigma_R(\mathcal{H}(S))}} is hereditary {{M|\forall C\in\mathcal{P}(B)[C\in\sigma_R(\mathcal{H}(S))}}
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###** {{M|A-B\subseteq A}}, but {{M|\mathcal{H} }} contains {{M|A}} and therefore all subsets of {{M|A}}
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*** So {{M|A\in\sigma_R(\mathcal{H}(S))}}
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###* Thus {{M|\mathcal{H} }} is closed under set subtraction.
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* This shows {{M|1=\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))}}
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### Closed under countable union is given.<!--
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==={{M|1=\sigma_R(\mathcal{H}(S))\subseteq\mathcal{H}(\sigma_R(S))}}===
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* Let {{M|A\in\sigma_R(\mathcal{H}(S))}}, then, either (by fact 3):
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*# {{M|A\in\mathcal{H}(S)}}
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*#* This means: {{M|1=\exists B\in S[A\in\mathcal{P}(B)]}} (also said as: {{M|\exists B\in S[A\subseteq B]}})
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*#* But {{M|S\subseteq\sigma_R(S)}}, thus:
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*#** {{M|B\in\sigma_R(S)}}
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*#* But {{M|\sigma_R(S)\subseteq\mathcal{H}(\sigma_R(S))}} thus:
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*#** {{M|B\in\mathcal{H}(\sigma_R(S))}}
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*#* As {{M|\mathcal{H}(\sigma_R(S))}} is hereditary, all subsets of {{M|B}} are in it.
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*#* As {{M|A\in\mathcal{P}(B)}} we see {{M|A\in\mathcal{H}(\sigma_R(S))}} - this completes this half of the proof.
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*# {{MSeq|A_n|in=\mathcal{H}(S)|pre=\exists|post=[\bigcup_{n=1}^\infty A_n=A]}}
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*#* Using part 1 we see that: {{M|1=\forall i\in\mathbb{N}[A_i\in\mathcal{H}(\sigma_R(S))]}}
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*#** But we know {{M|\mathcal{H}(\sigma_R(S))}} is a {{sigma|ring}}, thus closed under countable union
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*#* Thus {{M|1=\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\sigma_R(S))}}
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*#** But {{M|1=A:=\bigcup_{n=1}^\infty A_n}} so
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*#* {{M|A\in\mathcal{H}(\sigma_R(S))}}
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* We have shown that in either case, {{M|A\in\mathcal{H}(\sigma_R(S))}}
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==[[Notes:Hereditary sigma-ring/Facts|Facts]]==
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{{:Notes:Hereditary sigma-ring/Facts}}
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==[[Notes:Hereditary sigma-ring/Proof of facts|Proof of facts]]==
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{{:Notes:Hereditary sigma-ring/Proof of facts}}
  
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END OF PROOF OF FACT 1
 
  
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-->
 
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# {{M|\mathcal{H}(\mathcal{R})}} is a {{sigma|ring}} (for any {{sigma|ring}}, {{M|\mathcal{R} }})
 
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## It is already shown that a hereditary system is closed under set subtraction, only remains to be shown closed under countable union
 
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## Closed under countable union
 
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##* Let {{M|1=(A_n)_{n=1}^\infty\subseteq\mathcal{H}(\mathcal{R})}} (we need to show {{M|1=\implies\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R}) }})
 
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##** This means, for each {{M|A_n\in\mathcal{H}(\mathcal{R})}} there is a {{M|B_n\in\mathcal{R} }} with {{M|A_n\subseteq B_n}} thus:
 
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##*** {{M|1=\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]}}
 
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##** However {{M|\mathcal{R} }} is a {{sigma|ring}}, thus:
 
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##*** Define {{M|1=B:=\bigcup_{n=1}^\infty B_n}}, notice {{M|B\in\mathcal{R} }}
 
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##** But a [[union of subsets is a subset of the union]], thus:
 
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##*** {{M|1=\bigcup_{n=1}^\infty A_n\subseteq\bigcup_{n=1}^\infty B_n:=B}}, thus
 
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##**** {{M|1=\bigcup_{n=1}^\infty A_n\subseteq B}}
 
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##*** BUT {{M|\mathcal{H}(\mathcal{R})}} contains all subsets of all things in {{M|\mathcal{R} }}, thus contains all subsets of {{M|B}}.
 
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##** Thus {{M|1=\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R})}}
 
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##* Thus {{M|\mathcal{H}(\mathcal{R})}} is closed under countable union.<!--
 
  
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END OF PROOF OF FACT 2
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-->
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# {{M|1=\sigma_R(\mathcal{H}(S))}} is just {{M|\mathcal{H}(S)}} closed under countable union.
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#* Follows from fact 1. As {{M|\mathcal{H}(S)}} is an hereditary system, the sigma-ring generated by it (the smallest sigma ring containing {{M|\mathcal{H}(S)}} is just the set with whatever is needed to close it under the operators)
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{{Todo|It seems, "hereditary sigma-ring" is the same as "sigma-ideal".}}
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{{Notes|Measure Theory}}[[Category:Finished notes]]

Latest revision as of 19:26, 24 May 2016

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!


TODO: Make a theorem out of this!

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




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

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