Difference between revisions of "Properties of classes of sets closed under set-subtraction"
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m (Alec moved page Class of sets closed under set-subtraction properties to Properties of classes of sets closed under set-subtraction: Obscure name changed) |
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+ | {{Refactor notice|grade=A|msg=The old page was crap, it is preserved below but it doesn't provide anything useful}} | ||
+ | ==Statement== | ||
+ | Suppose {{M|\mathcal{A} }} is an arbitrary class of [[set|sets]] with the property that: | ||
+ | * {{M|1=\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}]}} where {{M|A-B}} denotes "[[set subtraction]]" ({{AKA}}: [[relative complement]]) | ||
+ | Then we claim the following are true of {{M|\mathcal{A} }}: | ||
+ | # {{M|\mathcal{A} }} is {{M|\cap}}-closed. That is: {{M|1=\forall A,B\in\mathcal{A}[A\cap B\in\mathcal{A}]}} | ||
+ | ==Proof of claims== | ||
+ | {{Requires proof|easy=1|grade=C|msg=Outlines follow for each claim: | ||
+ | # Prove {{M|1=A\cap B=A-(A-B)}} or indeed {{M|B-(B-A)}}. Include picture of [[Venn diagram]] | ||
+ | }} | ||
+ | ==Notes for further claims== | ||
+ | I know from a reference (see the old page below) that: | ||
+ | # If {{M|\mathcal{A} }} is {{sigma|{{M|\cup}}-closed}} then it is {{sigma|{{M|\cap}}-closed}} | ||
+ | # Any countable union of sets in {{M|\mathcal{A} }} can be expressed as a countable disjoint union of sets in {{M|\mathcal{A} }}<ref group="Note">Note that this doesn't require {{M|\mathcal{A} }} to be closed under union, we can still talk about unions we just cannot know that the result of a union is in {{M|\mathcal{A} }}. This simply says IF there is a [[sequence]] of elements of {{M|\mathcal{A} }}, say {{MSeq|A_n|in=\mathcal{A} }}, where {{M|1=\bigcup_{n=1}^\infty A_n}} just happens to be in {{M|\mathcal{A} }} then there exists a sequence of [[pairwise disjoint]] sets whose union is the same.</ref> | ||
+ | These feel very "measure theory specific" though and might be true in the case of being closed under arbitrary union or for an arbitrary collection of sets whose union is {{M|\mathcal{A} }} | ||
+ | ==Notes== | ||
+ | <references group="Note"/> | ||
+ | ==References== | ||
+ | <references/> | ||
+ | {{Elementary set theory navbox|plain}} | ||
+ | {{Measure theory navbox|plain}} | ||
+ | {{Theorem Of|Measure Theory|Elementary Set Theory}} | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | =OLD PAGE= | ||
==Theorem statement== | ==Theorem statement== | ||
If {{M|\mathcal{A} }} is a class of subsets of {{M|\Omega}} such that<ref name="PTACC">Probability Theory - A comprehensive course - Second Edition - Achim Klenke</ref> | If {{M|\mathcal{A} }} is a class of subsets of {{M|\Omega}} such that<ref name="PTACC">Probability Theory - A comprehensive course - Second Edition - Achim Klenke</ref> |
Latest revision as of 11:38, 21 August 2016
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Contents
Statement
Suppose [ilmath]\mathcal{A} [/ilmath] is an arbitrary class of sets with the property that:
- [ilmath]\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}][/ilmath] where [ilmath]A-B[/ilmath] denotes "set subtraction" (AKA: relative complement)
Then we claim the following are true of [ilmath]\mathcal{A} [/ilmath]:
- [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]-closed. That is: [ilmath]\forall A,B\in\mathcal{A}[A\cap B\in\mathcal{A}][/ilmath]
Proof of claims
Grade: C
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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The message provided is:
Outlines follow for each claim:
- Prove [ilmath]A\cap B=A-(A-B)[/ilmath] or indeed [ilmath]B-(B-A)[/ilmath]. Include picture of Venn diagram
This proof has been marked as an page requiring an easy proof
Notes for further claims
I know from a reference (see the old page below) that:
- If [ilmath]\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-[ilmath]\cup[/ilmath]-closed then it is [ilmath]\sigma[/ilmath]-[ilmath]\cap[/ilmath]-closed
- Any countable union of sets in [ilmath]\mathcal{A} [/ilmath] can be expressed as a countable disjoint union of sets in [ilmath]\mathcal{A} [/ilmath][Note 1]
These feel very "measure theory specific" though and might be true in the case of being closed under arbitrary union or for an arbitrary collection of sets whose union is [ilmath]\mathcal{A} [/ilmath]
Notes
- ↑ Note that this doesn't require [ilmath]\mathcal{A} [/ilmath] to be closed under union, we can still talk about unions we just cannot know that the result of a union is in [ilmath]\mathcal{A} [/ilmath]. This simply says IF there is a sequence of elements of [ilmath]\mathcal{A} [/ilmath], say [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{A} [/ilmath], where [ilmath]\bigcup_{n=1}^\infty A_n[/ilmath] just happens to be in [ilmath]\mathcal{A} [/ilmath] then there exists a sequence of pairwise disjoint sets whose union is the same.
References
Template:Elementary set theory navbox
|
OLD PAGE
Theorem statement
If [ilmath]\mathcal{A} [/ilmath] is a class of subsets of [ilmath]\Omega[/ilmath] such that[1]
- [math]\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}][/math] - that is closed under set-subtraction (or [ilmath]\backslash[/ilmath]-closed)
Then we have:
- [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]-closed
- [ilmath]\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-[ilmath]\cup[/ilmath]-closed[ilmath]\implies[/ilmath] [ilmath]\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-[ilmath]\cap[/ilmath]-closed
- Any countable union of sets in [ilmath]\mathcal{A} [/ilmath] can be expressed as a countable disjoint union of sets in [ilmath]\mathcal{A} [/ilmath][Note 1]
Proof:
Notes
- ↑ Note that this doesn't require [ilmath]\mathcal{A} [/ilmath] to be closed under union, we can still talk about unions we just cannot know that the result of a union is in [ilmath]\mathcal{A} [/ilmath]
References
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