Ring of sets

In mathematics, there are two different notions of a ring of sets, both referring to certain families of sets.

In order theory, a nonempty family of sets $$\mathcal{R}$$ is called a ring (of sets) if it is closed under union and intersection. That is, the following two statements are true for all sets $$A$$ and $$B$$,
 * $$A,B\in\mathcal{R}$$ implies $$A \cup B \in \mathcal{R}$$ and
 * $$A,B\in\mathcal{R}$$ implies $$A \cap B \in \mathcal{R}.$$

In measure theory, a nonempty family of sets $$\mathcal{R}$$ is called a ring (of sets) if it is closed under union and relative complement (set-theoretic difference). That is, the following two statements are true for all sets $$A$$ and $$B$$, This implies that a ring in the measure-theoretic sense always contains the empty set. Furthermore, for all sets $A$ and $B$,
 * $$A, B \in \mathcal{R}$$ implies $$A \cup B \in \mathcal{R}$$ and
 * $$A, B \in \mathcal{R}$$ implies $$A \setminus B \in \mathcal{R}.$$
 * $$A\cap B=A\setminus(A\setminus B),$$

which shows that a family of sets closed under relative complement is also closed under intersection, so that a ring in the measure-theoretic sense is also a ring in the order-theoretic sense.

Examples
If $X$ is any set, then the power set of $X$ (the family of all subsets of $X$) forms a ring of sets in either sense.

If $(X, ≤)$ is a partially ordered set, then its upper sets (the subsets of $X$ with the additional property that if $x$ belongs to an upper set U and $x ≤ y$, then $y$ must also belong to $U$) are closed under both intersections and unions. However, in general it will not be closed under differences of sets.

The open sets and closed sets of any topological space are closed under both unions and intersections.

On the real line $R$, the family of sets consisting of the empty set and all finite unions of half-open intervals of the form $(a, b]$, with $a, b ∈ R$ is a ring in the measure-theoretic sense.

If $T$ is any transformation defined on a space, then the sets that are mapped into themselves by $T$ are closed under both unions and intersections.

If two rings of sets are both defined on the same elements, then the sets that belong to both rings themselves form a ring of sets.

Related structures
A ring of sets in the order-theoretic sense forms a distributive lattice in which the intersection and union operations correspond to the lattice's meet and join operations, respectively. Conversely, every distributive lattice is isomorphic to a ring of sets; in the case of finite distributive lattices, this is Birkhoff's representation theorem and the sets may be taken as the lower sets of a partially ordered set.

A family of sets closed under union and relative complement is also closed under symmetric difference and intersection. Conversely, every family of sets closed under both symmetric difference and intersection is also closed under union and relative complement. This is due to the identities Symmetric difference and intersection together give a ring in the measure-theoretic sense the structure of a boolean ring.
 * 1) $$A \cup B = (A\, \triangle\, B)\, \triangle\, (A \cap B)$$ and
 * 2) $$A \setminus B = A\, \triangle\, (A \cap B).$$

In the measure-theoretic sense, a is a ring closed under  unions, and a δ-ring is a ring closed under countable intersections. Explicitly, a σ-ring over $$X$$ is a set $$\mathcal{F}$$ such that for any sequence $$\{A_k\}_{k=1}^\infty \subseteq \mathcal{F},$$ we have $ \bigcup_{k=1}^\infty A_k \in \mathcal{F}.$

Given a set $$X,$$ a − also called an  − is a ring that contains $$X.$$ This definition entails that an algebra is closed under absolute complement $$A^c = X \setminus A.$$ A σ-algebra is an algebra that is also closed under countable unions, or equivalently a σ-ring that contains $$X.$$ In fact, by de Morgan's laws, a δ-ring that contains $$X$$ is necessarily a σ-algebra as well. Fields of sets, and especially σ-algebras, are central to the modern theory of probability and the definition of measures.

A  is a family of sets $$\mathcal{S}$$ with the properties  $$\varnothing \in \mathcal{S},$$ $$A, B \in \mathcal{S}$$ implies $$A \cap B \in \mathcal{S},$$ and $$A, B \in \mathcal{S}$$ implies $$A \setminus B = \bigcup_{i=1}^n C_i$$ for some disjoint $$ C_1, \ldots, C_n \in \mathcal{S}.$$  Every ring (in the measure theory sense) is a semi-ring. On the other hand, $$\mathcal{S} := \{\emptyset,\{x\},\{y\}\}$$ on $$X = \{x,y\}$$ is a semi-ring but not a ring, since it is not closed under unions.
 * If (3) holds, then $$\varnothing \in \mathcal{S}$$ if and only if $$\mathcal{S} \neq \varnothing.$$

A ' or ' is a collection $$\mathcal{S}$$ of subsets of $$X$$ satisfying the semiring properties except with (3) replaced with:
 * If $$E \in \mathcal{S}$$ then there exists a finite number of mutually disjoint sets $$C_1, \ldots, C_n \in \mathcal{S}$$ such that $$X \setminus E = \bigcup_{i=1}^n C_i.$$

This condition is stronger than (3), which can be seen as follows. If $$\mathcal{S}$$ is a semialgebra and $$E, F \in \mathcal{S}$$, then we can write $$F^c = F_1 \cup \ldots \cup F_n$$ for disjoint $$F_i \in S$$. Then: $$E \setminus F = E \cap F^c = E \cap (F_1 \cup \ldots \cup F_n) = (E \cap F_1) \cup \ldots \cup (E \cap F_n)$$

and every $$E \cap F_i \in S$$ since it is closed under intersection, and disjoint since they are contained in the disjoint $$F_i$$'s. Moreover the condition is strictly stronger: any $$S$$ that is both a ring and a semialgebra is an algebra, hence any ring that is not an algebra is also not a semialgebra (e.g. the collection of finite sets on an infinite set $$X$$).