# Ring generated by

Here ring refers to a ring of sets

## Definition

Given any class of sets [ilmath]A[/ilmath], there exists a unique ring [ilmath]R_0[/ilmath] such that [ilmath]E\subseteq R_0[/ilmath] and such that if [ilmath]R[/ilmath] is any ring with [ilmath]E\subseteq R[/ilmath] and [ilmath]R\ne R_0[/ilmath] then [ilmath]R_0\subset R[/ilmath]

We call [ilmath]R_0[/ilmath] the ring generated by [ilmath]A[/ilmath], often denoted [ilmath]R(A)[/ilmath]

### Proof

Since [ilmath]\mathcal{P}(X)[/ilmath] (where [ilmath]A[/ilmath] is a collection of subsets of [ilmath]X[/ilmath]) is a ring (infact an algebra) we know that a ring containing [ilmath]A[/ilmath] exists.

Since the intersection of any collection of rings is a ring (see the theorem here), it is clear that the intersection of all rings containing [ilmath]A[/ilmath] is the required ring [ilmath]R_0[/ilmath].

## Important theorems

### Every set in R(A) can be finitely covered by sets in A

If [ilmath]A[/ilmath] is any class of sets, then every set in [ilmath]R(A)[/ilmath] can be covered by a finite union of sets in [ilmath]A[/ilmath]

The class of all sets which may be covered by a finite union of sets in [ilmath]A[/ilmath] is a ring! (call it [ilmath]R_f[/ilmath]) Since [ilmath]A\subset R_f[/ilmath] we see that [ilmath]R(A)\subset R_f[/ilmath] (by The intersection of sets is a subset of each set)

Using the implies-subset relation we see that $S\in R(A)\implies S\in R_f$, that is, given a set [ilmath]S[/ilmath] in the generated ring, [ilmath]S[/ilmath] is in the ring of things that can be covered by a finite union of things in [ilmath]A[/ilmath]