3.3: Intervals in Eⁿ

Consider the rectangle in $E^{2}$ shown in Figure 2. Its interior (without the perimeter consists of all points $(x, y) \in E^{2}$ such that

$$a_{1}<x<b_{1}\text{ and } a_{2}<y<b_{2};$$

i.e.,

$$x \in\left(a_{1}, b_{1}\right)\text{ and } y \in\left(a_{2}, b_{2}\right).$$

Thus it is the Cartesian product of two line intervals, $\left(a_{1}, b_{1}\right)$ and $\left(a_{2}, b_{2}\right) .$ To include also all or some sides, we would have to replace open intervals by closed, half-closed, or half-open ones. Similarly, Cartesian products of three line intervals yield rectangular parallelepipeds in $E^{3} .$ We call such sets in $E^{n}$ intervals.

Definition

1. By an interval in $E^{n}$ we mean the Cartesian product of any $n$ intervals $\quad$ in $E^{1}$ (some may be open, some closed or half-open, etc.).

2. In particular, given

$$\overline{a}=\left(a_{1}, \ldots, a_{n}\right)\text{ and } \overline{b}=\left(b_{1}, \ldots, b_{n}\right)$$

with

$$a_{k} \leq b_{k}, \quad k=1,2, \ldots, n,$$

we define the open interval $(\overline{a}, \overline{b}),$ the closed interval $[\overline{a}, \overline{b}],$ the half-open interval $(\overline{a}, \overline{b}],$ and the half-closed interval $[\overline{a}, \overline{b})$ as follows:

$$\begin{aligned}(\overline{a}, \overline{b}) &=\left\{\overline{x} | a_{k}<x_{k}<b_{k}, k=1,2, \ldots, n\right\} \\ &=\left(a_{1}, b_{1}\right) \times\left(a_{2}, b_{2}\right) \times \cdots \times\left(a_{n}, b_{n}\right) \\ [\overline{a}, \overline{b}] &=\left\{\overline{x} | a_{k} \leq x_{k} \leq b_{k}, k=1,2, \ldots, n\right\} \\ &=\left[a_{1}, b_{1}\right] \times\left[a_{2}, b_{2}\right] \times \cdots \times\left[a_{n}, b_{n}\right] \\ (\overline{a}, \overline{b}] &=\left\{\overline{x} | a_{k}<x_{k} \leq b_{k}, k=1,2, \ldots, n\right\} \\ &=\left(a_{1}, b_{1}\right] \times\left(a_{2}, b_{2}\right] \times \cdots \times\left(a_{n}, b_{n}\right] \\ [a, b) &=\left\{\overline{x} | a_{k} \leq x_{k}<b_{k}, k=1,2, \ldots, n\right\} \\ &=\left[a_{1}, b_{1}\right) \times\left[a_{2}, b_{2}\right) \times \cdots \times\left[a_{n}, b_{n}\right) \end{aligned}$$

In all cases, $\overline{a}$ and $\overline{b}$ are called the endpoints of the interval. Their distance

$$\rho(\overline{a}, \overline{b})=|\overline{b}-\overline{a}|$$

is called its diagonal. The $n$ differences

$$b_{k}-a_{k}=\ell_{k} \quad(k=1, \ldots, n)$$

are called its $n$ edge-lengths. Their product

$$\prod_{k=1}^{n} \ell_{k}=\prod_{k=1}^{n}\left(b_{k}-a_{k}\right)$$

is called the volume of the interval (in $E^{2}$ it is its area, in $E^{1}$ its length) .\) The point

$$\overline{c}=\frac{1}{2}(\overline{a}+\overline{b})$$

is called its center or midpoint. The set difference

$$[\overline{a}, \overline{b}]-(\overline{a}, \overline{b})$$

is called the boundary of any interval with endpoints $\overline{a}$ and $\vec{b} ;$ it consists of 2$n$ "faces" defined in a natural manner. (How?)

We often denote intervals by single letters, e.g.. $A=(\overline{a}, \overline{b}),$ and write $d A$ for "diagonal of $A^{\prime \prime}$ and $v A$ or vol $A$ for "volume of $A . "$ If all edge-lengths $b_{k}-a_{k}$ are equal, $A$ is called a cube (in $E^{2},$ a square). The interval $A$ is said to be degenerate iff $b_{k}=a_{k}$ for some $k,$ in which case, clearly,

$$\operatorname{vol} A=\prod_{k=1}^{n}\left(b_{k}-a_{k}\right)=0.$$

Note 1. We have $\overline{x} \in(\overline{a}, \overline{b})$ iff the inequalities $a_{k}<x_{k}<b_{k}$ hold simultaneously for all $k .$ This is impossible if $a_{k}=b_{k}$ for some $k ;$ similarly for the inequalities $a_{k}<x_{k} \leq b_{k}$ or $a_{k} \leq x_{k}<b_{k}$. Thus a degenerate interval is empty, unless it is closed (in which case it contains $\overline{a}$ and $\overline{b}$ at least).

Note 2. In any interval $A$,

$$d A=\rho(\overline{a}, \overline{b})=\sqrt{\sum_{k=1}^{n}\left(b_{k}-a_{k}\right)^{2}}=\sqrt{\sum_{k=1}^{n} \ell_{k}^{2}}.$$

In $E^{2},$ we can split an interval $A$ into two subintervals $P$ and $Q$ by drawing a line (see Figure 2$) .$ In $E^{3},$ this is done by a plane orthogonal to one of the axes of the form $x_{k}=c\left($ see §§4-6, Note 2$),$ with $a_{k}<c<b_{k} .$ In particular, if \right. $c=\frac{1}{2}\left(a_{k}+b_{k}\right),$ the plane bisects the $k$ th edge of $A ;$ and so the $k$ th edge-length of $P($ and $Q)$ equals $\frac{1}{2} \ell_{k}=\frac{1}{2}\left(b_{k}-a_{k}\right) .$ If $A$ is closed, so is $P$ or $Q,$ depending on our choice. (We may include the "partition" $x_{k}=c$ in $P$ or $Q . )^{1}$

Now, successively draw $n$ planes $x_{k}=c_{k}, \quad c_{k}=\frac{1}{2}\left(a_{k}+b_{k}\right), \quad k=1,2, \ldots, n .$ The first plane bisects $\ell_{j}$ leaving the other edges of $A \mathrm{un}-$ changed. The resulting two subintervals $P$ and $Q$ then are cut by the plane $x_{2}=c_{2},$ bisecting the second edge in each of them. Thus we get four subintervals (see Figure 3 for $E^{2}$. Each successive plane doubles the number of subintervals. After $n$ steps, we thus obtain $2^{n}$ disjoint intervals, with all edges $\ell_{k}$ bisected. Thus by Note $2,$ the diagonal of each of them is

$$\sqrt{\sum_{k=1}^{n}\left(\frac{1}{2} \ell_{k}\right)^{2}}=\frac{1}{2} \sqrt{\sum_{k=1}^{n} \ell_{k}^{2}}=\frac{1}{2} d A.$$

Note 3. If $A$ is closed then, as noted above, we can make any one (but only one $)$ of the $2^{n}$ subintervals closed by properly manipulating each step.

The proof of the following simple corollaries is left to the reader.

Corollary $\PageIndex{1}$

No distance between two points of an interval $A$ exceeds $d A,$ its diagonal. That is, $(\forall \overline{x}, \overline{y} \in A) \rho(\overline{x}, \overline{y}) \leq d A$

Corollary $\PageIndex{2}$

If an interval $A$ contains $\overline{p}$ and $\overline{q},$ then also $L[\overline{p}, \overline{q}] \subseteq A$.

corollary $\PageIndex{3}$

Every nondegenerate interval in $E^{n}$ contains rational points, i.e., points whose coordinates are all rational.

(Hint: Use the density of rationals in $E^{1}$ for each coordinate separately.)