ring of setsRingOfSets2006-03-21 19:39:412007-12-18 10:56:25Definitionfield of sets\usepackage{amssymb,amscd}
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% define commands here\subsection*{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 \emph{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
\begin{itemize}
\item for any $A,B\in \mathcal{R}$, then $A\cap B\in \mathcal{R}$,
\item for any $A,B\in \mathcal{R}$, then $A\cup B\in \mathcal{R}$.
\end{itemize}
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 non-empty, 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 non-empty set with the first condition above replaced by
\begin{itemize}
\item for any $A,B\in \mathcal{R}$, then $A-B\in \mathcal{R}$.
\end{itemize}
This is indeed a stronger condition, as $A\cap B=A-(A-B)\in \mathcal{R}$. However, we shall stick with the more general definition here.
\subsection*{Field of Sets}
An even stronger condition is to insist that not only is $\mathcal{R}$ non-empty, 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 \emph{field of sets} $\mathcal{F}$ of $S$ satisfies the following criteria
\begin{itemize}
\item $\mathcal{F}$ is a ring of sets of $S$,
\item $S\in\mathcal{F}$, and
\item if $A\in\mathcal{F}$, then the complement $\overline{A}\in\mathcal{F}$.
\end{itemize}
The three conditions above are equivalent to the following three conditions:
\begin{itemize}
\item $\varnothing\in\mathcal{F}$,
\item if $A,B\in \mathcal{F}$, then $A\cup B\in \mathcal{F}$, and
\item if $A\in\mathcal{F}$, then $\overline{A}\in\mathcal{F}$.
\end{itemize}
A field of sets is also known as an \emph{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.
\textbf{Remark}. Two remarkable theorems relating to \PMlinkescapetext{representations} of certain lattices as rings or fields of sets are the following:
\begin{enumerate}
\item a lattice is distributive iff it is \PMlinkname{lattice isomorphic}{LatticeIsomorphism} to a ring of sets (G. Birkhoff and M. Stone);
\item a lattice is \PMlinkname{Boolean}{BooleanLattice} iff it is lattice \PMlinkescapetext{isomorphic} to a field of sets (M. Stone).
\end{enumerate}
\begin{thebibliography}{7}
\bibitem{prh} P. R. Halmos: {\em Lectures on Boolean Algebras}, Springer-Verlag (1970).
\bibitem{prh1} P. R. Halmos: {\em Measure Theory}, Springer-Verlag (1974).
\bibitem{gg} G. Gr\"atzer: {\em General Lattice Theory}, Birkh\"auser, (1998).
\end{thebibliography}