Algebra of sets

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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)

Definition

An algebra of sets is a collection of sets, [ilmath]\mathcal{A} [/ilmath] such that:

• [ilmath]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/ilmath]
  • 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

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TODO: Do this as a list of inline theorem boxes

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• [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

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This is trivial. In order to show [ilmath]\mathcal{A} [/ilmath] is a ring of sets we require two properties:

1. [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
2. [ilmath]\forall A,B\in\mathcal{A}[A-B\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

• Types of set algebras
• ring of sets
  • [ilmath]\sigma[/ilmath]-ring
• [ilmath]\sigma[/ilmath]-algebra

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

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TODO: what could this mean?

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Definition

An class [ilmath]R[/ilmath] of sets is an Algebra of sets if:

• [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]A-B=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

• [ilmath]\sigma[/ilmath]-algebra
• Ring of sets
• Types of set algebras