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

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("hereditary sigma-ring" is the same as "sigma-ideal")
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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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{{Note|The result is true!}}
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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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{{Todo|Make a theorem out of this!}}
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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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==Showing {{M|1=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}}==
 +
Which way first?
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==={{M|1=\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))}}===
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* Let {{M|A\in\mathcal{H}(\sigma_R(S))}} be given. Then:
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** {{M|1=\exists B\in\sigma_R(S)[A\in\mathcal{P}(B)]}}
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*** Notice that {{M|S\subseteq\mathcal{H}(S)}} thus:
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**** {{M|\sigma_R(S)\subseteq\sigma_R(\mathcal{H}(S))}}
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*** So {{M|B\in\sigma_R(\mathcal{H}(S))}}
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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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*** So {{M|A\in\sigma_R(\mathcal{H}(S))}}
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* This shows {{M|1=\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))}}
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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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FACT 2
 
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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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-->
 
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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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==Proof of facts==
 
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# An hereditary system is a sigma-ring {{M|\iff}} it is closed under countable unions.
 
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## Hereditary system is a sigma-ring {{M|\implies}} closed under countable unions
 
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##* It ''is a {{sigma|ring}}'' which means it is closed under countable unions. Done
 
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## A hereditary system closed under countable union {{M|\implies}} it is a {{sigma|ring}}
 
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### closed under set-subtraction
 
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###* Let {{M|A,B\in\mathcal{H} }} for some hereditary system {{M|\mathcal{H} }}. Then:
 
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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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###* Thus {{M|\mathcal{H} }} is closed under set subtraction.
 
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### Closed under countable union is given.<!--
 
  
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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)
 
  
 
{{Todo|It seems, "hereditary sigma-ring" is the same as "sigma-ideal".}}
 
{{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

  • H(σR(S))=σR(H(S)) for a system of sets, S.

Both are hereditary, and both are σ-rings.

The result is true!


TODO: Make a theorem out of this!

Showing H(σR(S))=σR(H(S))

Which way first?

H(σR(S))⊆σR(H(S))

  • Let A∈H(σR(S)) be given. Then:
    • ∃B∈σR(S)[A∈P(B)]
      • Notice that S⊆H(S) thus:
        • σR(S)⊆σR(H(S))
      • So B∈σR(H(S))
    • As σR(H(S)) is hereditary ∀C∈P(B)[C∈σR(H(S))
      • So A∈σR(H(S))
  • This shows H(σR(S))⊆σR(H(S))

σR(H(S))⊆H(σR(S))

  • Let A∈σR(H(S)), then, either (by fact 3):
    1. A∈H(S)
      • This means: ∃B∈S[A∈P(B)] (also said as: ∃B∈S[A⊆B])
      • But S⊆σR(S), thus:
        • B∈σR(S)
      • But σR(S)⊆H(σR(S)) thus:
        • B∈H(σR(S))
      • As H(σR(S)) is hereditary, all subsets of B are in it.
      • As A∈P(B) we see A∈H(σR(S)) - this completes this half of the proof.
    2. ∃(An)∞n=1⊆H(S)[⋃∞n=1An=A]
      • Using part 1 we see that: ∀i∈N[Ai∈H(σR(S))]
        • But we know H(σR(S)) is a σ-ring, thus closed under countable union
      • Thus ⋃∞n=1An∈H(σR(S))
        • But A:=⋃∞n=1An so
      • A∈H(σR(S))
  • We have shown that in either case, A∈H(σR(S))

Facts

  1. An hereditary system is a sigma-ring ⟺ it is closed under countable unions.
    • Thus σR(H(S)) is just H(S) with the additional property:
      • ∀(An)∞n=1⊆H(S)[⋃∞n=1An∈σR(H(S))]
  2. H(R) is a σ-ring (for any σ-ring, R)
    • This means σR(H(R))=H(R)
    • It also means H(σR(S)) is a σ-ring
  3. σR(H(S)) is just H(S) closed under countable union.
  4. σR(H(S)) is hereditary

Proof of facts

  1. An hereditary system is a sigma-ring ⟺ it is closed under countable unions.
    1. Hereditary system is a sigma-ring ⟹ closed under countable unions
      • It is a σ-ring which means it is closed under countable unions. Done
    2. A hereditary system closed under countable union ⟹ it is a σ-ring
      1. closed under set-subtraction
        • Let A,B∈H for some hereditary system H. Then:
          • A−B⊆A, but H contains A and therefore all subsets of A
        • Thus H is closed under set subtraction.
      2. Closed under countable union is given.
  2. H(R) is a σ-ring (for any σ-ring, R)
    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 (An)∞n=1⊆H(R) (we need to show ⟹⋃∞n=1An∈H(R))
        • This means, for each An∈H(R) there is a Bn∈R with An⊆Bn thus:
          • ∀(An)∞n=1⊆H(R)∃(Bn)∞n=1⊆R∀i∈N[Ai⊆Bi]
        • However R is a σ-ring, thus:
          • Define B:=⋃∞n=1Bn, notice B∈R
        • But a union of subsets is a subset of the union, thus:
          • ⋃∞n=1An⊆⋃∞n=1Bn:=B, thus
            • ⋃∞n=1An⊆B
          • BUT H(R) contains all subsets of all things in R, thus contains all subsets of B.
        • Thus ⋃∞n=1An∈H(R)
      • Thus H(R) is closed under countable union.
  3. σR(H(S)) is just H(S) closed under countable union.
    • Follows from fact 1. As H(S) is an hereditary system, the sigma-ring generated by it (the smallest sigma ring containing H(S) is just the set with whatever is needed to close it under the operators)
  4. σR(H(S)) is hereditary
    • Let A∈σR(H(S)) be given. We want to show that ∀B∈P(A) that B∈σR(H(S)).
      1. If A∈H(S), then ∀B∈P(A)[B∈H(S) but B∈H(S)⟹B∈σR(H(S))
        • We're done in this case.
      2. OTHERWISE: ∃(An)∞n=1⊆H(S)[⋃∞n=1An=A] (by fact 3)
        • Let B∈P(A) be given.
          • Define a new sequence, (Bn)∞n=1⊆H(S), where Bi:=Ai∩B
            • Ai∩B is a subset of Ai and Ai∈H(S), as "hereditary" means "contains all subsets of" Ai∩B⊆Ai thus Ai∩B:=Bi∈H(S)
          • Clearly B=⋃∞n=1Bn (as B⊆A and A=⋃∞n=1An)
          • As σR(H(S) contains all countable unions of things in H(S) we know:
            • ⋃∞n=1Bn=B∈σR(H(S))
        • We have shown B∈σR(H(S))
    • We have completed the proof




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

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