# Ring of sets

Jump to: navigation, search

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] of sets such that:

• $\forall A\in R\forall B\in R[A\cup B\in R]$
• $\forall A\in R\forall B\in R[A-B\in R]$

## 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 $A\in R$ then $A-A\in R$ but $A-A=\emptyset$ so $\emptyset\in R$

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 $\emptyset\in R$, this means we know at least $\{\emptyset\}\subseteq R_1\cap R_2$ - it is non empty.

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

Then:

• $A,B\in R_1$
• $A,B\in R_2$

This means:

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

But then:

• As $A\cup B\in R_1$ and $A\cup B\in R_2$ we have $A\cup B\in R_1\cap R_2$
• As $A- B\in R_1$ and $A- B\in R_2$ we have $A- B\in R_1\cap R_2$

Thus $R_1\cap R_2$ is a ring.