Exercise 1.1.1 (Boolean closure)

Proof. By definition, we can represent both sets as unions $E={\bigcup <!--FUNCTION\; APPLICATION-->}_{i=1}^n{E}_k$ and $F={\bigcup <!--FUNCTION\; APPLICATION-->}_{j=1}^k{F}_j$ for some boxes $E_1,\dots ,E_k$ and $F_1,\dots ,F_k$ in $R^d$.

(i)

The first assertion is obvious. Denote $C_l:=E_l$ for $1\le l\le n$ and $C_{l+n}:=F_l$ for $1\le l\le k$. Then obviously $E\cup F={\bigcup <!--FUNCTION\; APPLICATION-->}_{l=1}^{n+k}{C}_l$ and since $C_l^{\prime }s$ are boxes, $E\cup F$ must be elementary by definition.

(ii)

The best way to prove the assertion "the intersection of two elementary sets is elementary" is to reduce it to the assertion "the intersection of two boxes is again a box". The validity of the reduction can be shown using several set-theoretic laws: $$\begin{array}{llll}\hfill E\cap F& =\left(\underset{i=1}{\overset{n}{\bigcup }}{E}_i\right)\cap \left(\underset{j=1}{\overset{k}{\bigcup }}{F}_j\right)=\left(\underset{i=1}{\overset{n}{\bigcup }}{E}_i\cap F_1\right)\cup \dots \cup \left(\underset{i=1}{\overset{n}{\bigcup }}{E}_i\cap F_k\right)& \hfill & \\ \hfill & =[(E_1\cap F_1)\cup \dots \cup (E_n\cap F_1)]\cup \dots \cup [(E_1\cap F_k)\cup \dots \cup (E_n\cap F_k)]& \hfill & \\ \hfill & =(E_1\cap F_1)\cup \dots \cup (E_n\cap F_1)\cup \dots \cup (E_1\cap F_k)\cup \dots \cup (E_n\cap F_k)& \hfill & \\ \hfill & =\underset{i=1}{\overset{n}{\bigcup }}\underset{j=1}{\overset{k}{\bigcup }}{E}_i\cap F_j& \hfill & \end{array}$$

Thus, $E\cap F$ is a finite union of box intersections. Now we demonstrate that the intersection of boxes remains a box.

Pick an arbitrary intersection $B\cap C$ of two boxes $B,C$. By definition

$$\begin{array}{llll}\hfill B& =\left\{(x_1,\dots ,x_d)\in {ℝ}^d:a_1\le x_1\le b_1,\dots ,a_d\le x_d\le b_d\right\}& \hfill & \\ \hfill C& =\left\{(x_1,\dots ,x_d)\in {ℝ}^d:c_1\le x_1\le g_1,\dots ,c_d\le x_d\le g_d\right\}& \hfill & \end{array}$$

for some $a_i,b_i,c_i,g_i\in ℝ$. By definition of intersection

$$\begin{array}{llll}\hfill B\cap C& =\left\{(x_1,\dots ,x_d)\in {ℝ}^d:(a_i\le x_i\le b_i)\text{ and }(c_i\le x_i\le g_i)\right\}& \hfill & \\ \hfill & =\left\{(x_1,\dots ,x_d)\in {ℝ}^d:\max <!--FUNCTION\; APPLICATION-->\{a_i,c_i\}\le x_i\le \min <!--FUNCTION\; APPLICATION-->\{b_i,g_i\}\right\}& \hfill & \\ \hfill & =\left[\max <!--FUNCTION\; APPLICATION-->\{a_1,c_1\},\min <!--FUNCTION\; APPLICATION-->\{b_1,g_1\}\right]\times \dots \times \left[\max <!--FUNCTION\; APPLICATION-->\{a_d,c_d\},\min <!--FUNCTION\; APPLICATION-->\{b_d,g_d\}\right]& \hfill & \end{array}$$

Since $\max <!--FUNCTION\; APPLICATION-->\{a_1,c_1\},\min <!--FUNCTION\; APPLICATION-->\{b_1,g_1\}\in ℝ$ the above set is a box by definition.

(iii)

We will use the strategy of reducing the given assertion about the elementary sets to an identical assertion made about boxes. In this case, we have $$\begin{array}{llll}\hfill E\setminus F& =\left(\underset{i=1}{\overset{n}{\bigcup }}{E}_i\right)\setminus \left(\underset{j=1}{\overset{k}{\bigcup }}{F}_j\right)=\left(\underset{i=1}{\overset{n}{\bigcup }}{E}_i\setminus F_1\right)\cap \dots \cap \left(\underset{i=1}{\overset{n}{\bigcup }}{E}_i\setminus F_k\right)& \hfill & \\ \hfill & =[(E_1\setminus F_1)\cup \dots \cup (E_n\setminus F_1)]\cap \dots \cap [(E_1\setminus F_k)\cup \dots \cup (E_n\setminus F_k)]& \hfill & \\ \hfill & =\underset{i=1}{\overset{n}{\bigcup }}\underset{j=1}{\overset{k}{\bigcap }}{E}_i\setminus F_j.& \hfill & \end{array}$$

By parts (i) and (ii) of this exercise, it suffices to prove the assertion for an arbitrary box difference $B\setminus C$.

$$\begin{array}{llll}\hfill B\setminus C& =\left\{(x_1,\dots ,x_d)\in {ℝ}^d:(a_i\le x_i\le b_i)\text{ and not }(c_i\le x_i\le g_i)\right\}& \hfill & \\ \hfill & =\left\{(x_1,\dots ,x_d)\in {ℝ}^d:x_i\in [a_i,b_i]\setminus [c_i,g_i]\right\}& \hfill & \\ \hfill & =\underset{i=1}{\overset{d}{\prod }}\left[a_i,b_i\right]\setminus \left[c_i,g_i\right].& \hfill & \end{array}$$ At last, if we manage to demonstrate that the set difference of two intervals is again a union of at most two intervals, then $B\setminus C$ must be elementary, considering how Cartesian products interact with set unions. This, however, follows immediately by tediously considering each of the six non-trivial cases $a\le c\le g\le b$, $\dots ,$ $c\le a\le b\le g$.

Thus, $B\setminus C$ can be represented as a union of Cartesian products taken over these at most $2d$ intervals, i.e., as a union of boxes.

(iv)

The fact that $E△F=(E\setminus F)\cup (F\setminus E)$ is elementary follows from the fact that $E\setminus F$ and $F\setminus E$ are elementary (by part 3), and that their union must therefore be elementary too (part 1).

(v)

As always, we reduce the assertion to a similar assertion made about boxes. Notice that $$\begin{array}{llll}\hfill x+E& =x+\underset{i=1}{\overset{n}{\bigcup }}{E}_i=\left\{x+y:y\in \underset{i=1}{\overset{n}{\bigcup }}{E}_i\right\}=\left\{x+y:y\in E_1\text{ or }\dots \text{ or }y\in E_n\right\}& \hfill & \\ \hfill & =\underset{i=1}{\overset{n}{\bigcup }}\left\{x+y:y\in E_i\right\}=\underset{i=1}{\overset{n}{\bigcup }}\left(x+E_i\right)& \hfill & \end{array}$$

Thus, $x+B$ is a box by definition, and we are done.