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== | + | ==[[/Definition|Definition]]== |
| − | A | + | {{/Definition}} |
| − | + | ==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. | |
| + | |||
| + | This ring is important because it means we may talk of a "[[Ring generated by|ring generated by]]" | ||
==First theorems== | ==First theorems== | ||
| Line 14: | Line 16: | ||
{{End Proof}} | {{End Proof}} | ||
{{End Theorem}} | {{End Theorem}} | ||
| − | |||
{{Begin Theorem}} | {{Begin Theorem}} | ||
Given any two rings, {{M|R_1}} and {{M|R_2}}, the intersection of the rings, {{M|R_1\cap R_2}} is a ring | Given any two rings, {{M|R_1}} and {{M|R_2}}, the intersection of the rings, {{M|R_1\cap R_2}} is a ring | ||
| Line 20: | Line 21: | ||
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. | 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> | + | 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 {{M|R_1}} is a ring | ||
| + | * <math>A-B\in R_1</math> as {{M|R_1}} is a ring | ||
| + | * <math>A\cup B\in R_2</math> as {{M|R_2}} is a ring | ||
| + | * <math>A-B\in R_2</math> as {{M|R_2}} 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. | ||
| + | {{End Proof}} | ||
| + | {{End Theorem}} | ||
==References== | ==References== | ||
| − | + | <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
- ↑ Page 19 -Measure Theory - Paul R. Halmos
