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# 3.3: Intervals in Eⁿ

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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.)