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ring of sets
Ring of Sets
Let $S$ be a set and $2^{S}$ be the power set of $S$. A subset $\mathcal{R}$ of $2^{S}$ is said to be a ring of sets of $S$ if it is a lattice under the intersection and union operations. In other words, $\mathcal{R}$ is a ring of sets if

for any $A,B\in\mathcal{R}$, then $A\cap B\in\mathcal{R}$,

for any $A,B\in\mathcal{R}$, then $A\cup B\in\mathcal{R}$.
A ring of sets is a distributive lattice. The word “ring” in the name has nothing to do with the ordinary ring found in algebra. Rather, it is an abelian semigroup with respect to each of the binary set operations. If $S\in\mathcal{R}$, then $(\mathcal{R},\cap,S)$ becomes an abelian monoid. Similarly, if $\varnothing\in\mathcal{R}$, then $(\mathcal{R},\cup,\varnothing)$ is an abelian monoid. If both $S,\varnothing\in\mathcal{R}$, then $(\mathcal{R},\cup,\cap)$ is a commutative semiring, since $\varnothing\cap A=A\cap\varnothing=\varnothing$, and $\cap$ distributes over $\cup$. Dualizing, we see that $(\mathcal{R},\cap,\cup)$ is also a commutative semiring. It is perhaps with this connection that the name “ring of sets” is so chosen.
Since $S$ is not required to be in $\mathcal{R}$, a ring of sets can in theory be the empty set. Even if $\mathcal{R}$ may be nonempty, it may be a singleton. Both cases are not very interesting to study. To avoid such examples, some authors, particularly measure theorists, define a ring of sets to be a nonempty set with the first condition above replaced by

for any $A,B\in\mathcal{R}$, then $AB\in\mathcal{R}$.
This is indeed a stronger condition, as $A\cap B=A(AB)\in\mathcal{R}$. However, we shall stick with the more general definition here.
Field of Sets
An even stronger condition is to insist that not only is $\mathcal{R}$ nonempty, but that $S\in\mathcal{R}$. Such a ring of sets is called a field, or algebra of sets. Formally, given a set $S$, a field of sets $\mathcal{F}$ of $S$ satisfies the following criteria

$\mathcal{F}$ is a ring of sets of $S$,

$S\in\mathcal{F}$, and

if $A\in\mathcal{F}$, then the complement $\overline{A}\in\mathcal{F}$.
The three conditions above are equivalent to the following three conditions:

$\varnothing\in\mathcal{F}$,

if $A,B\in\mathcal{F}$, then $A\cup B\in\mathcal{F}$, and

if $A\in\mathcal{F}$, then $\overline{A}\in\mathcal{F}$.
A field of sets is also known as an algebra of sets.
It is easy to see that $\mathcal{F}$ is a distributive complemented lattice, and hence a Boolean lattice. From the discussion earlier, we also see that $\mathcal{F}$ (of $S$) is a commutative semiring, with $S$ acting as the multiplicative identity and $\varnothing$ both the additive identity and the multiplicative absorbing element.
Remark. Two remarkable theorems relating to representations of certain lattices as rings or fields of sets are the following:
1. a lattice is distributive iff it is lattice isomorphic to a ring of sets (G. Birkhoff and M. Stone);
2. a lattice is Boolean iff it is lattice isomorphic to a field of sets (M. Stone).
References
 1 P. R. Halmos: Lectures on Boolean Algebras, SpringerVerlag (1970).
 2 P. R. Halmos: Measure Theory, SpringerVerlag (1974).
 3 G. Grätzer: General Lattice Theory, Birkhäuser, (1998).
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Comments
Definition
Reading various books and lectures on Measure Theory, I have never been able to find your definition of ring of sets.
Namely, I find always two definitions,
a) A ring is a (nonempty) collection of sets which is closed under finite difference (A\B) and finite union.
b) A ring is a (nonempty) collection of sets which is closed under finite symmetric difference (A /\ B) and finite intersection.
In both cases it's easy to prove that a ring is closed also for finite intersection (def a) and for finite union (def b).
On the other side, in order to prove the equivalence of the definition, I tried to use your definition to prove that a ring is closed under any type of difference (ordinary and symmetric), but I wasn't able to, and sincerely I think it's not possible.
Am I missing something? Could you clarify please?
Thank you and best regards,
Andrea
Re: Definition
No you're not. In this definition, you can easily find examples (see the examples attached to the entry) such that the ring is not closed under any of the difference operators. The definition you have corresponds to the definition of a field of sets in the same entry.
I will add some verbiage to clarify the different interpretations and also include some references where I found the definition of a ring of sets in the entry.
Chi
Re: Definition
Dear Chi,
thank you very much for your reply, which confirms my thought that our definitions are not equivalent.
Actually, the definitions I have do not correspond exactly to yours of a field of sets; the latter in fact, including S and the complement of each set, refers to what is usually called an algebra of sets.
My definitions of a ring of sets say in a nutshell that a ring contains the finite
a)union
b)difference
(as a consequence)
c)intersection
d)symmetric difference
of any of its sets, and that it is non empty (i.e., if it contains one only set, this must be the empty one, because of A\A).
It does contain however neither the whole set S, nor the complement of each set (if it actually does, it is called an algebra, or a field, which should be synonimous).
The reason of my questions is that, as far as I know, the whole powerful Measure Extension theory, which is one of the basis of Real Analysis, deeply relies on the concept of the ring generated by a semiring, which is in turn the most rudimental collection of sets on which a Measure has all its usual properties.
Any further comment of yours will be much appreciated.
All my best,
Andrea
Re: Definition
The point of my entry is to spell out the two fundamental theorems in lattice theory. But to state them, I need a definition for both a ring of sets and a field of sets, and none of them were defined on Planet Math at the time.
But I do see your point of view on how a term can be defined and interpreted differently from different perspectives.
I have made additional clarifications in the entry to try to clear out any confusions.
Thanks for pointing this out.
Chi
Re: Definition
Dear Chi,
thank you for your clarification; now the entry is very complete!
Best regards,
Andrea