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Mathematics LibreTexts

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

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


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

    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.

    Screen Shot 2019-05-29 at 11.31.34 PM.png


    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

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


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

    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


    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

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

    Screen Shot 2019-05-29 at 11.47.41 PM.png

    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.)