Semirings of Sets and Standard Examples

What a Semiring of Sets Requires

A semiring is a useful set system when intersections stay within the system and differences can be broken into finitely many simple pieces. The difference condition does not require the difference itself to be a member; it requires a finite disjoint union of members. This distinction is important for intervals and rectangles: cutting one such set by another can produce multiple pieces.

All definitions below concern an ambient set \(X\). They are statements about families of subsets of \(X\); no measure or measure space is needed to verify them. When we use intervals, \(X=\mathbb{R}\); for rectangles, \(X=\mathbb{R}^d\) with \(d\) a positive integer.

The axioms do not say that \(X\in\mathcal{S}\), that arbitrary unions belong to \(\mathcal{S}\), or that \(A\setminus B\) must itself be in \(\mathcal{S}\). In particular, a semiring is not being defined as a sigma-algebra. We now verify the axioms for standard families, paying close attention to endpoint conventions.

Half-Open Intervals on the Real Line

Let \(\mathcal{I}\) consist of \(\varnothing\) and all bounded half-open intervals \([a,b)\subseteq\mathbb{R}\) with \(a<b\). The left endpoint belongs to \([a,b)\), while the right endpoint does not. The interval is empty when the lower endpoint is at least the upper endpoint, so it is convenient to use the convention \([u,v)=\varnothing\) whenever \(u\ge v\).

Proof. By definition, \(\varnothing\in\mathcal{I}\). For two nonempty members \([a,b)\) and \([c,d)\), their intersection is \([\max(a,c),\min(b,d))\), interpreted as empty if \(\max(a,c)\ge\min(b,d)\). Thus the intersection belongs to \(\mathcal{I}\); intersections involving the empty set are also empty.

It remains to decompose the difference. If either interval is empty, then \([a,b)\setminus\varnothing=[a,b)\), and \(\varnothing\setminus[c,d)=\varnothing\), so the required decompositions hold. Suppose both intervals are nonempty. The part of \([a,b)\) lying strictly to the left of \([c,d)\) is \([a,\min(b,c))\), and it is nonempty exactly when \(a<\min(b,c)\). The part lying to the right is \([\max(a,d),b)\), and it is nonempty exactly when \(\max(a,d)<b\). Omitting either piece when its lower endpoint is not less than its upper endpoint gives \[ [a,b)\setminus[c,d) = [a,\min(b,c))\ \cup\ [\max(a,d),b). \] The two displayed pieces are disjoint: every point in the first is less than \(c\), while every point in the second is at least \(d\), and \(c<d\). To see that they exhaust the difference, take \(x\in[a,b)\setminus[c,d)\). Since \(x\notin[c,d)\), either \(x<c\) or \(x\ge d\). In the first case \(x\) lies in the left piece, and in the second it lies in the right piece. Conversely, both pieces consist of points in \([a,b)\) outside \([c,d)\). Thus the difference is a union of at most two disjoint members of \(\mathcal{I}\), proving the final axiom. \(\square\)

Half-Open Rectangles in Finite Dimensions

Fix a positive integer \(d\). For each coordinate \(i\), take an interval \(I_i\in\mathcal{I}\), and form the rectangle \(I_1\times\cdots\times I_d\subseteq\mathbb{R}^d\). Let \(\mathcal{R}_d\) consist of all such products. This convention includes the empty set, since a product with an empty coordinate interval is empty. A nonempty member is a product \(\prod_{i=1}^d[a_i,b_i)\) with \(a_i<b_i\) for every \(i\).

Proof. The empty set belongs to \(\mathcal{R}_d\), for example as \(\varnothing\times\mathbb{R}\times\cdots\times\mathbb{R}\) would not be appropriate because \(\mathbb{R}\) is not a bounded interval; instead, the product of \(d\) members of \(\mathcal{I}\) with at least one coordinate \(\varnothing\) is the empty set and belongs to \(\mathcal{R}_d\). If \(A=\prod_{i=1}^d A_i\) and \(B=\prod_{i=1}^d B_i\), then \[ A\cap B=\prod_{i=1}^d(A_i\cap B_i). \] Each coordinate intersection belongs to \(\mathcal{I}\) by the interval result, so \(A\cap B\in\mathcal{R}_d\).

We prove the difference condition for nonempty \(A\) and \(B\); if \(A=\varnothing\), the difference is empty, and if \(B=\varnothing\), it is \(A\) itself. Write \(A_i=[a_i,b_i)\) and \(B_i=[c_i,d_i)\), and set \(J_i=A_i\cap B_i\). By the interval result, \(A_i\setminus B_i\) is a union of at most two disjoint half-open intervals; call its nonempty pieces \(K_{i,1},\ldots,K_{i,r_i}\), where \(0\le r_i\le2\).

For each coordinate \(i\) and each piece \(K_{i,q}\), form the rectangle \[ J_1\times\cdots\times J_{i-1}\times K_{i,q}\times A_{i+1}\times\cdots\times A_d, \] omitting any empty rectangle. Each is in \(\mathcal{R}_d\). A point of \(A\setminus B\) has at least one coordinate outside the corresponding interval \(B_i\). Let \(i\) be its first such coordinate. Its earlier coordinates belong to \(J_1,\ldots,J_{i-1}\), its \(i\)-th coordinate belongs to one of the \(K_{i,q}\), and its later coordinates belong to the corresponding \(A_j\). Thus the point lies in one of the displayed rectangles. Conversely, every point in one of those rectangles belongs to \(A\), and its \(i\)-th coordinate is outside \(B_i\), so it is outside \(B\). The rectangles are pairwise disjoint: rectangles assigned different indices \(i\) impose, at the earlier index, membership in \(J_i\) for the rectangle with the larger index and membership outside \(B_i\) for the rectangle with the smaller index. Those conditions cannot both hold. For a fixed \(i\), different \(K_{i,q}\) are disjoint by the interval result. There are at most \(2d\) rectangles. This proves the difference axiom and completes the proof. \(\square\)

Finite Subsets and a Boundary Pitfall

Let \(X\) be any set, possibly infinite, and let \(\mathcal{F}\) be the family of all finite subsets of \(X\). The empty set is finite, so it belongs to \(\mathcal{F}\). If \(A,B\in\mathcal{F}\), then \(A\cap B\) and \(A\setminus B\) are finite. In particular, \(A\cap B\in\mathcal{F}\), and the difference condition holds with the single piece \(A\setminus B\) when that set is nonempty, or with the empty union when it is empty. Thus \(\mathcal{F}\) is a semiring. This example differs from intervals and rectangles: here the difference is itself a member, so no splitting is needed.

Common mistake: treating a half-open endpoint as if it belonged to both adjacent pieces. For \(A=[0,4)\) and \(B=[2,3)\), a wrong calculation is \(A\setminus B=[0,2]\cup[3,4)\). It incorrectly includes \(2\), even though \(2\in B\), and \(4\), even though \(4\notin A\). The correct calculation is \([0,2)\cup[3,4)\): the first piece ends just before \(2\), while the second begins at \(3\), which is excluded from \(B\). Endpoint conventions are part of the set calculation, not cosmetic notation.

These examples show why the semiring conditions are stated in terms of finite disjoint decompositions rather than closure under differences. Intersections are computed coordinatewise for rectangles, while differences are organized by the first coordinate that leaves the second rectangle. For finite subsets, the same axioms hold for a simpler reason: intersections and differences remain finite.