Algebra of sets
Algebra of sets  
[ilmath]\mathcal{A}\subseteq\mathcal{P}(X)[/ilmath] For an algebra of sets, [ilmath]\mathcal{A} [/ilmath] on [ilmath]X[/ilmath]  
Defining properties:  

1)  [ilmath]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/ilmath] 
2)  [ilmath]\forall A,B\in\mathcal{A}[A\cup B\in\mathcal{A} ][/ilmath] 
 Note: Every algebra of sets is a ring of sets (see below)
Contents
Definition
An algebra of sets is a collection of sets, [ilmath]\mathcal{A} [/ilmath] such that^{[1]}:
 [ilmath]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/ilmath]^{[Note 1]}
 In words: For all [ilmath]A[/ilmath] in [ilmath]\mathcal{A} [/ilmath] the complement of [ilmath]A[/ilmath] (with respect to [ilmath]X[/ilmath]) is also in [ilmath]\mathcal{A} [/ilmath]
 [ilmath]\forall A,B\in\mathcal{A}[A\cup B\in\mathcal{A}][/ilmath]
 In words: For all [ilmath]A[/ilmath] and [ilmath]B[/ilmath] in [ilmath]\mathcal{A} [/ilmath] their union is also in [ilmath]\mathcal{A} [/ilmath]
Claim 1: Every algebra of sets is also a ring of sets
Immediate properties
TODO: Do this as a list of inline theorem boxes
 [ilmath]\mathcal{A} [/ilmath] is [ilmath]\setminus[/ilmath]closed
 [ilmath]\emptyset\in\mathcal{A} [/ilmath]
 [ilmath]X\in\mathcal{A} [/ilmath]
 [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]closed
Proof of claims
Claim 1: Every algebra of sets is also a ring of sets
This is trivial. In order to show [ilmath]\mathcal{A} [/ilmath] is a ring of sets we require two properties:
 [ilmath]\forall A,B\in\mathcal{A}[A\cup B\in\mathcal{A}][/ilmath]  this is clearly satisfied by definition of an algebra of sets
 [ilmath]\forall A,B\in\mathcal{A}[AB\in\mathcal{A}][/ilmath]  that is [ilmath]\mathcal{A} [/ilmath] must be [ilmath]\setminus[/ilmath]closed
 But we've already shown this in the immediate properties section above!
This completes the proof
See also
Notes
 ↑ Recall [ilmath]A^C:=XA[/ilmath]  the complement of [ilmath]A[/ilmath] in [ilmath]X[/ilmath]
References

OLD PAGE
An Algebra of sets is sometimes called a Boolean algebra
We will show later that every Algebra of sets is an Algebra of sets
TODO: what could this mean?
Definition
An class [ilmath]R[/ilmath] of sets is an Algebra of sets if^{[1]}:
 [math][A\in R\wedge B\in R]\implies A\cup B\in R[/math]
 [math]A\in R\implies A^c\in R[/math]
So an Algebra of sets is just a Ring of sets containing the entire set it is a set of subsets of!
Every Algebra is also a Ring
Since for [math]A\in R[/math] and [math]B\in R[/math] we have:
[math]AB=A\cap B' = (A'\cup B)'[/math] we see that being closed under Complement and Union means it is closed under Set subtraction
Thus it is a Ring of sets
See also
References
 ↑ p21  Halmos  Measure Theory  Graduate Texts In Mathematics  Springer  #18