Difference between revisions of "Ring of sets"

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Note that every [[Algebra of sets]] is also a ring, and that an [[Algebra of sets]] is sometimes called a '''Boolean algebra'''
 
Note that every [[Algebra of sets]] is also a ring, and that an [[Algebra of sets]] is sometimes called a '''Boolean algebra'''
==Definition==
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==[[/Definition|Definition]]==
A Ring of sets is a non-empty class {{M|R}}<ref>Page 19 - Halmos - Measure Theory - Springer - Graduate Texts in Mathematics (18)</ref> of sets such that:
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{{/Definition}}
* <math>\forall A\in R\forall B\in R(A\cup B\in R)</math>
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* <math>\forall A\in R\forall B\in R(A-B\in R)</math>
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==A ring that exists==
 
==A ring that exists==
 
Take a set {{M|X}}, the [[Power set|power set]] of {{M|X}}, {{M|\mathcal{P}(X)}} is a ring (further still, an [[Algebra of sets|algebra]]) - the proof of this is trivial.  
 
Take a set {{M|X}}, the [[Power set|power set]] of {{M|X}}, {{M|\mathcal{P}(X)}} is a ring (further still, an [[Algebra of sets|algebra]]) - the proof of this is trivial.  
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{{End Proof}}
 
{{End Proof}}
 
{{End Theorem}}
 
{{End Theorem}}
 
 
 
 
==References==
 
==References==
 
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<references/>
 
{{Definition|Measure Theory}}
 
{{Definition|Measure Theory}}

Latest revision as of 17:21, 18 August 2016

A Ring of sets is also known as a Boolean ring

Note that every Algebra of sets is also a ring, and that an Algebra of sets is sometimes called a Boolean algebra

Definition

A Ring of sets is a non-empty class [ilmath]R[/ilmath][1] of sets such that:

  • [math]\forall A\in R\forall B\in R[A\cup B\in R][/math]
  • [math]\forall A\in R\forall B\in R[A-B\in R][/math]

A ring that exists

Take a set [ilmath]X[/ilmath], the power set of [ilmath]X[/ilmath], [ilmath]\mathcal{P}(X)[/ilmath] is a ring (further still, an algebra) - the proof of this is trivial.

This ring is important because it means we may talk of a "ring generated by"

First theorems

The empty set belongs to every ring


Take any [math]A\in R[/math] then [math]A-A\in R[/math] but [math]A-A=\emptyset[/math] so [math]\emptyset\in R[/math]

Given any two rings, [ilmath]R_1[/ilmath] and [ilmath]R_2[/ilmath], the intersection of the rings, [ilmath]R_1\cap R_2[/ilmath] is a ring


We know [math]\emptyset\in R[/math], this means we know at least [math]\{\emptyset\}\subseteq R_1\cap R_2[/math] - it is non empty.

Take any [math]A,B\in R_1\cap R_2[/math] (which may be the empty set, as shown above)

Then:

  • [math]A,B\in R_1[/math]
  • [math]A,B\in R_2[/math]


This means:

  • [math]A\cup B\in R_1[/math] as [ilmath]R_1[/ilmath] is a ring
  • [math]A-B\in R_1[/math] as [ilmath]R_1[/ilmath] is a ring
  • [math]A\cup B\in R_2[/math] as [ilmath]R_2[/ilmath] is a ring
  • [math]A-B\in R_2[/math] as [ilmath]R_2[/ilmath] is a ring

But then:

  • As [math]A\cup B\in R_1[/math] and [math]A\cup B\in R_2[/math] we have [math]A\cup B\in R_1\cap R_2[/math]
  • As [math]A- B\in R_1[/math] and [math]A- B\in R_2[/math] we have [math]A- B\in R_1\cap R_2[/math]

Thus [math]R_1\cap R_2[/math] is a ring.

References

  1. Page 19 -Measure Theory - Paul R. Halmos