3.3.E: Problems on Intervals in Eⁿ (Exercises)
(Here \(A\) and \(B\) denote intervals.)
Prove Corollaries 1-3.
Prove that if \(A \subseteq B,\) then \(d A \leq d B\) and \(v A \leq v B\).
Give an appropriate definition of a "face" and a "vertex" of \(A\).
Find the edge-lengths of \(A=(\overline{a}, \overline{b})\) in \(E^{4}\) if
\[
\overline{a}=(1,-2,4,0) \text { and } \overline{b}=(2,0,5,3).
\]
Is \(A\) a cube? Find some rational points in it. Find \(d A\) and \(v A\).
Show that the sets \(P\) and \(Q\) as defined in footnote 1 are intervals, indeed. In particular, they can be made half-open (half-closed) if \(A\) is half-open (half-closed).
\([\text { Hint: Let } A=(\overline{a}, \overline{b}]\),
\[
P=\left\{\overline{x} \in A | x_{k} \leq c\right\}, \text { and } Q=\left\{\overline{x} \in A | x_{k}>c\right\}.
\]
To fix ideas, let \(k=1,\) i.e., cut the first edge. Then let
\[
\overline{p}=\left(c, a_{2}, \ldots, a_{n}\right) \text { and } \overline{q}=\left(c, b_{2}, \ldots, b_{n}\right) \text { (see Figure } 2 ),
\]
and verify that \(P=(\overline{a}, \overline{q}]\) and \(Q=(\overline{p}, \overline{b}] .\) Give a proof. \(]\)
In Problem \(5,\) assume that \(A\) is closed, and make \(Q\) closed. (Prove it!)
In Problem 5 show that \((\text { with } k \text { fixed })\) the \(k\) th edge-lengths of \(P\) and \(Q\) equal \(c-a_{k}\) and \(b_{k}-c,\) respectively, while for \(i \neq k\) the edge-length \(\ell_{i}\) is the same in \(A, P,\) and \(Q,\) namely, \(\ell_{i}=b_{i}-a_{i}\).
[Hint: If \(k=1,\) define \(\overline{p}\) and \(\overline{q}\) as in Problem \(5 . ]\)
Prove that if an interval \(A\) is split into subintervals \(P\) and \(Q(P \cap Q=\emptyset)\), then \(v A=v P+v Q .\)
[Hint: Use Problem 7 to compute \(v A, v P,\) and \(v Q .\) Add up. \(]\)
Give an example. (Take \(A\) as in Problem 4 and split it by the plane \(x_{4}=1 . )\)
*9. Prove the additivity of the volume of intervals, namely, if \(A\) is subdivided, in any manner, into \(m\) mutually disjoint subintervals \(A_{1}, A_{2}, \ldots, A_{m}\) \(i n E^{n},\) then
\[
v A=\sum_{i=1}^{m} v A_{i}.
\]
(This is true also if some \(A_{i}\) contain common faces).
[Proof outline: For \(m=2,\) use Problem 8.
Then by induction, suppose additivity holds for any number of intervals smaller than a certain \(m\) \((m>1) .\) Now let
\[
A=\bigcup_{i=1}^{m} A_{i} \quad\left(A_{i} \text { disjoint }\right).
\]
One of the \(A_{i}\) (say, \(A_{1}=[\overline{a}, \overline{p}] )\) must have some edge-length smaller than the corresponding edge-length of \(A\left(\operatorname{say}, \ell_{1}\right) .\) Now cut all of \(A\) into \(P=[\overline{a}, \overline{d}]\) and \(Q=A-P(\text { Figure } 4)\) by the plane \(x_{1}=c\left(c=p_{1}\right)\) so that \(A_{1} \subseteq P\) while \(A_{2} \subseteq Q .\) For simplicity, assume that the plane cuts each \(A_{i}\) into two subintervals \(A_{i}^{\prime}\) and \(A_{i}^{\prime \prime} .\) (One of them may be empty.)
Then
\[
P=\bigcup_{i=1}^{m} A_{i}^{\prime} \text { and } Q=\bigcup_{i=1}^{m} A_{i}^{\prime \prime}.
\]
Actually, however, \(P\) and \(Q\) are split into fewer than \(m\) (nonempty) intervals since \(A_{1}^{\prime \prime}=\emptyset=A_{2}^{\prime}\) by construction. Thus, by our inductive assumption,
\[
v P=\sum_{i=1}^{m} v A_{i}^{\prime} \text { and } v Q=\sum_{i=1}^{m} v A_{i}^{\prime \prime},
\]
where \(v A_{1}^{\prime \prime}=0=v A_{2}^{\prime},\) and \(v A_{i}=v A_{i}^{\prime}+v A_{i}^{\prime \prime}\) by Problem \(8 .\) Complete the inductive proof by showing that
\[
v A=v P+v Q=\sum_{i=1}^{m} v A_{i} .]
\]