
# 7.1: More on Intervals in Eⁿ. Semirings of Sets


I. As a prologue, we turn to intervals in $$E^{n}$$ (Chapter 3, §7).

## Theorem $$\PageIndex{1}$$

If $$A$$ and $$B$$ are intervals in $$E^{n},$$ then

(i) $$A \cap B$$ is an interval ($$\emptyset$$ counts as an interval);

(ii) $$A-B$$ is the union of finitely many disjoint intervals (but need not be an interval itself).

Proof

The easy proof for $$E^{1}$$ is left to the reader.

An interval in $$E^{2}$$ is the cross-product of two line intervals.

Let

$A=X \times Y \text { and } B=X^{\prime} \times Y^{\prime},$

where $$X, Y, X^{\prime},$$ and $$Y^{\prime}$$ are intervals in $$E^{1}.$$ Then (see Figure 29)

and

$A-B=\left[\left(X-X^{\prime}\right) \times Y\right] \cup\left[\left(X \cap X^{\prime}\right) \times\left(Y-Y^{\prime}\right)\right];$

see Problem 8 in Chapter 1, §§1-3.

As the theorem holds in $$E^{1}$$,

$X \cap X^{\prime} \text { and } Y \cap Y^{\prime}$

are intervals in $$E^{1},$$ while

$X-Y^{\prime} \text { and } Y-Y^{\prime}$

are finite unions of disjoint line intervals. (In Figure 29 they are just intervals, but in general they are not.)

It easily follows that $$A \cap B$$ is an interval in $$E^{2},$$ while $$A-B$$ splits into finitely many such intervals. (Verify!) Thus the theorem holds in $$E^{2}$$.

Finally, for $$E^{n},$$ use induction. An interval in $$E^{n}$$ is the cross-product of an interval in $$E^{n-1}$$ by a line interval. Thus if the theorem holds in $$E^{n-1},$$ the same argument shows that it holds in $$E^{n},$$ too. (Verify!)

This completes the inductive proof.$$\quad \square$$

Actually, Theorem 1 applies to many other families of sets (not necessarily intervals or sets in $$E^{n}).$$ We now give such families a name.

## Definition 1

A family $$\mathcal{C}$$ of arbitrary sets is called a semiring iff

(i) $$\emptyset \in \mathcal{C}$$ ($$\emptyset$$ is a member), and

(ii) for any sets $$A$$ and $$B$$ from $$\mathcal{C},$$ we have $$A \cap B \in \mathcal{C},$$ while $$A-B$$ is the union of finitely many disjoint sets from $$\mathcal{C}.$$

Briefly: $$\mathcal{C}$$ is a semiring iff it satisfies Theorem 1.

Note that here $$\mathcal{C}$$ is not just a set, but a whole family of sets. Recall (Chapter §§1-3) that a set family (family of sets) is a set $$\mathcal{M}$$ whose members are other sets. If $$A$$ is a member of $$\mathcal{M},$$ we call $$A$$ an $$\mathcal{M}$$-set and write $$A \in \mathcal{M}$$ (not $$A \subseteq \mathcal{M})$$.

Sometimes we use index notation:

$\mathcal{M}=\left\{X_{i} | i \in I\right\},$

briefly

$\mathcal{M}=\left\{X_{i}\right\},$

where the $$X_{i}$$ are $$\mathcal{M}$$-sets distinguished from each other by the subscripts $$i$$ varying over some index set $$I.$$

A set family $$\mathcal{M}=\left\{X_{i}\right\}$$ and its union

$\bigcup_{i} X_{i}$

are said to be disjoint iff

$X_{i} \cap X_{j}=\emptyset \text { whenever } i \neq j.$

Notation:

$\bigcup X_{i} \text{ (disjoint).}$

In our case, $$A \in \mathcal{C}$$ means that $$A$$ is a $$\mathcal{C}$$-set (a member of the semiring $$\mathcal{C})$$.

The formula

$(\forall A, B \in \mathcal{C}) \quad A \cap B \in \mathcal{C}$

means that the intersection of two $$\mathcal{C}$$-sets in a $$\mathcal{C}$$-set itself.

Henceforth, we will often speak of semirings $$\mathcal{C}$$ in general. In particular, this will apply to the case $$\mathcal{C}=$${intervals}. Always keep this case in mind!

Note 1. By Theorem 1, the intervals in $$E^{n}$$ form a semiring. So also do the half-open and the half-closed intervals separately (same proof!), but not the open (or closed) ones. (Why?)

Caution. The union and difference of two $$\mathcal{C}$$-sets need not be a $$\mathcal{C}$$-set. To remedy this, we now enlarge $$\mathcal{C}.$$

## Definition 2

We say that a set $$A$$ (from $$\mathcal{C}$$ or not) is $$\mathcal{C}$$-simple and write

$A \in \mathcal{C}_{s}^{\prime}$

iff $$A$$ is a finite union of disjoint $$\mathcal{C}$$-sets (such as $$A-B$$ in Theorem 1).

Thus $$\mathcal{C}_{s}^{\prime}$$ is the family of all $$\mathcal{C}$$-simple sets.

Every $$\mathcal{C}$$-set is also a $$\mathcal{C}_{s}^{\prime}$$-set, i.e., a $$\mathcal{C}$$-simple one. (Why?) Briefly:

$\mathcal{C} \subseteq \mathcal{C}_{s}^{\prime}.$

If $$\mathcal{C}$$ is the set of all intervals, a $$\mathcal{C}$$-simple set may look as in Figure 30.

## Theorem $$\PageIndex{2}$$

If $$\mathcal{C}$$ is a semiring, and if $$A$$ and $$B$$ are $$\mathcal{C}$$-simple, so also are

$A \cap B, A-B, \text { and } A \cup B.$

In symbols,

$\left(\forall A, B \in \mathcal{C}_{s}^{\prime}\right) \quad A \cap B \in \mathcal{C}_{s}^{\prime}, A-B \in \mathcal{C}_{s}^{\prime}, \text { and } A \cup B \in \mathcal{C}_{s}^{\prime}.$

Proof

We give a proof outline and suggest the proof as an exercise. Before attempting it, the reader should thoroughly review the laws and problems of Chapter 1, §§1-3.

(1) To prove $$A \cap B \in \mathcal{C}_{s}^{\prime},$$ let

$A=\bigcup_{i=1}^{m} A_{i}(\text {disjoint}) \text { and } B=\bigcup_{k=1}^{n} B_{k} \text { (disjoint),}$

with $$A_{i}, B_{k} \in \mathcal{C}.$$ Verify that

$A \cap B=\bigcup_{k=1}^{n} \bigcup_{i=1}^{m}\left(A_{i} \cap B_{k}\right) \text { (disjoint),}$

and so $$A \cap B \in \mathcal{C}_{s}^{\prime}$$.

(2) Next prove that $$A-B \in \mathcal{C}_{s}^{\prime}$$ if $$A \in \mathcal{C}_{s}^{\prime}$$ and $$B \in \mathcal{C}$$.

Indeed, if

$A=\bigcup_{i=1}^{m} A_{i} \text { (disjoint),}$

then

$A-B=\bigcup_{i=1}^{m} A_{i}-B=\bigcup_{i=1}^{m}\left(A_{i}-B\right) \text { (disjoint).}$

Verify and use Definition 2.

(3) Prove that

$\left(\forall A, B \in \mathcal{C}_{s}^{\prime}\right) \quad A-B \in \mathcal{C}_{s}^{\prime};$

we suggest the following argument.

Let

$B=\bigcup_{k=1}^{n} B_{k}, \quad B_{k} \in \mathcal{C}.$

Then

$A-B=A-\bigcup_{k=1}^{n} B_{k}=\bigcap_{k=1}^{n}\left(A-B_{k}\right)$

by duality laws. But $$A-B_{k}$$ is $$\mathcal{C}$$-simple by step (2). Hence so is

$A-B=\bigcap_{k=1}^{m}\left(A-B_{k}\right)$

by step (1) plus induction.

(4) To prove $$A \cup B \in \mathcal{C}_{s}^{\prime},$$ verify that

$A \cup B=A \cup(B-A),$

where $$B-A \in \mathcal{C}_{s}^{\prime},$$ by (3).

Note 2. By induction, Theorem 2 extends to any finite number of $$\mathcal{C}_{s}^{\prime}$$-sets. It is a kind of "closure law."

We thus briefly say that $$\mathcal{C}_{s}^{\prime}$$ is closed under finite unions, intersections, and set differences. Any (nonempty) set family with these properties is called a set ring (see also §3).

Thus Theorem 2 states that if $$\mathcal{C}$$ is a semiring, then $$\mathcal{C}_{s}^{\prime}$$ is a ring.

Caution. An infinite union of $$\mathcal{C}$$-simple sets need not be $$\mathcal{C}$$-simple. Yet we may consider such unions, as we do next.

In Corollary 1 below, $$\mathcal{C}_{s}^{\prime}$$ may be replaced by any set ring $$\mathcal{M}$$.

## Corollary $$\PageIndex{1}$$

If $$\left\{A_{n}\right\}$$ is a finite or infinite sequence of sets from a semiring $$\mathcal{C}$$ (or from a ring $$\mathcal{M}$$ such as $$\mathcal{C}_{s}^{\prime}$$), then there is a disjoint sequence of $$\mathcal{C}$$-simple sets (or $$\mathcal{M}$$-sets) $$B_{n} \subseteq A_{n}$$ such that

$\bigcup_{n} A_{n}=\bigcup_{n} B_{n}.$

Proof

Let $$B_{1}=A_{1}$$ and for $$n=1,2, \ldots$$,

$B_{n+1}=A_{n+1}-\bigcup_{k=1}^{n} A_{k}, \quad A_{k} \in \mathcal{C}.$

By Theorem 2, the $$B_{n}$$ are $$\mathcal{C}$$-simple (as are $$A_{n+1}$$ and $$\bigcup_{k=1}^{n} A_{k}).$$ Show that they are disjoint (assume the opposite and find a contradiction) and verify that $$\bigcup A_{n}=\bigcup B_{n}:$$ If $$x \in \bigcup A_{n},$$ take the least $$n$$ for which $$x \in A_{n}.$$ Then $$n>1$$ and

$x \in A_{n}-\bigcup_{k=1}^{n-1} A_{k}=B_{n},$

or $$n=1$$ and $$x \in A_{1}=B_{1}. \quad \square$$

Note 3. In Corollary 1, $$B_{n} \in \mathcal{C}_{s}^{\prime},$$ i.e., $$B_{n}=\bigcup_{i=1}^{m_{n}} C_{n i}$$ for some disjoint sets $$C_{n i} \in \mathcal{C}.$$ Thus

$\bigcup_{n} A_{n}=\bigcup_{n} \bigcup_{i=1}^{m_{n}} C_{n i}$

is also a countable disjoint union of $$\mathcal{C}$$-sets.

II. Recall that the volume of intervals is additive (Problem 9 in Chapter 3, §7). That is, if $$A \in \mathcal{C}$$ is split into finitely many disjoint subintervals, then $$v A$$ (the volume of $$A)$$ equals the sum of the volumes of the parts.

We shall need the following lemma.

## lemma 1

Let $$X_{1}, X_{2}, \ldots, X_{m} \in \mathcal{C}$$ (intervals in $$E^{n}).$$ If the $$X_{i}$$ are mutually disjoint, then

(i) $$\bigcup_{i=1}^{m} X_{i} \subseteq Y \in \mathcal{C}$$ implies $$\sum_{i=1}^{m} v X_{i} \leq v Y;$$ and

(ii) $$\bigcup_{i=1}^{m} X_{i} \subseteq \bigcup_{k=1}^{p} Y_{k}\left(\text {with } Y_{k} \in \mathcal{C}\right)$$ implies $$\sum_{i=1}^{m} v X_{i} \leq \sum_{k=1}^{p} v Y_{k}$$.

Proof

(i) By Theorem 2, the set

$Y-\bigcup_{i=1}^{m} X_{i}$

is $$\mathcal{C}$$-simple; so

$Y-\bigcup_{i=1}^{m} X_{i}=\bigcup_{j=1}^{q} C_{j}$

for some disjoint intervals $$C_{j}.$$ Hence

$Y=\bigcup X_{i} \cup \bigcup C_{j} \text { (all disjoint).}$

$v Y=\sum_{i=1}^{m} v X_{i}+\sum_{j=1}^{q} v C_{j} \geq \sum_{i=1}^{m} v X_{i},$

as claimed.

(ii) By set theory (Problem 9 in Chapter 1, §§1-3),

$X_{i} \subseteq \bigcup_{k=1}^{p} Y_{k}$

implies

$X_{i}=X_{i} \cap \bigcup_{k=1}^{p} Y_{k}=\bigcup_{k=1}^{p}\left(X_{i} \cap Y_{k}\right).$

If it happens that the $$Y_{k}$$ are mutually disjoint also, so certainly are the smaller intervals $$X_{i} \cap Y_{k};$$ so by additivity,

$v X_{i}=\sum_{k=1}^{p} v\left(X_{i} \cap Y_{k}\right).$

Hence

$\sum_{i=1}^{m} v X_{i}=\sum_{i=1}^{m} \sum_{k=1}^{p} v\left(X_{i} \cap Y_{k}\right)=\sum_{k=1}^{p}\left[\sum_{i=1}^{m} v\left(X_{i} \cap Y_{k}\right)\right].$

But by (i),

$\sum_{i=1}^{m} v\left(X_{i} \cap Y_{k}\right) \leq v Y_{k} \text { (why?);}$

so

$\sum_{i=1}^{m} v X_{i} \leq \sum_{k=1}^{p} v Y_{k},$

as required.

If, however, the $$Y_{k}$$ are not disjoint, Corollary 1 yields

$\bigcup Y_{k}=\bigcup B_{k} \text { (disjoint)},$

with

$Y_{k} \supseteq B_{k}=\bigcup_{j=1}^{m_{k}} C_{k j}(\text {disjoint}), \quad C_{k j} \in \mathcal{C}.$

By (i),

$\sum_{j=1}^{m_{k}} v C_{k j} \leq v Y_{k}.$

As

$\bigcup_{i=1}^{m} X_{i} \subseteq \bigcup_{k=1}^{p} Y_{k}=\bigcup_{k=1}^{p} B_{k}=\bigcup_{k=1}^{p} \bigcup_{j=1}^{m_{k}} C_{k j} \text { (disjoint),}$

all reduces to the previous disjoint case.$$\quad \square$$

## Corollary $$\PageIndex{2}$$

Let $$A \in \mathcal{C}_{s}^{\prime}$$ ($$\mathcal{C}=$$ intervals in $$E^{n}$$). If

$A=\bigcup_{i=1}^{m} X_{i}(\text {disjoint})=\bigcup_{k=1}^{p} Y_{k} \text { (disjoint)}$

with $$X_{i}, Y_{k} \in \mathcal{C},$$ then

$\sum_{i=1}^{m} v X_{i}=\sum_{k=1}^{p} v Y_{k}.$

(Use part (ii) of the lemma twice.)

Thus we can (and do) unambiguously define $$v A$$ to be either of these sums.