## IMT Exercise 1.1.1

(Boolean closure). Show that if ${E,F\subset \mathbb{R}^d}$ are elementary sets, then the union ${E\cup F}$, the intersection ${E\cap F}$, and the set theoretic difference ${E\setminus F:=\{x\in E: x\notin F\}}$ and the symmetric difference ${E\Delta F:=(E\setminus F)\cup(F\setminus E)}$ are also elementary. If ${x\in\mathbb{R}^d}$, show that the translate ${E+x:=\{y+x: y\in E\}}$ is also an elementary set.

An elementary set is any subset of ${\mathbb{R}^d}$ which is the union of a finite number of boxes. Given that there is no requirement for the boxes to be disjoint, the first part, that ${E\cup F}$ is elementary, follows automatically.

Let ${E}$ be the finite union of boxes ${E_1\cup\ldots\cup E_n}$ and ${F}$ the finite union of boxes ${F_1\cup\ldots\cup F_m}$, then

$\displaystyle E\cap F=\bigcup_{\substack{i=1\ldots n\\ j=1\ldots m}}E_i\cap F_j$

So we just need to show that ${E_i\cap F_j}$ is a box (or the null set). Possibilities for ${E_i\cap F_j}$ are that it is: the null set, ${E_i}$, ${F_j}$ or a proper subset of both ${E_i}$ and ${F_j}$. In this last case let the intervals of intersection on the axes be ${I_1\ldots I_d}$, the intersection ${E_i\cap F_j}$ is the box given by the Cartesian product ${I_1\times\ldots\times I_d}$.

Next we prove that ${E\setminus F}$ is elementary. Let ${D_i=E_i\setminus F}$ then

$\displaystyle E\setminus F=\bigcup_{i=1\ldots n}D_i$

with

$\displaystyle D_i=\bigcap_{j=1\ldots m}E_i\setminus F_j$

Using the previous results for union and intersection it is enough to show that ${E_i\setminus F_j}$ is elementary. If ${E_i\subset F_j}$, ${E_i\setminus F_j=\emptyset}$; if ${E_i\cap F_j=\emptyset}$, ${E_i\setminus F_j=E_i}$; otherwise we have two boxes which partially intersect. For ease of notation consider ${A\setminus B}$ and consider the intervals ${I_{A,1}\ldots I_{A,d}}$ corresponding to box ${A}$ and how they are partitioned by the intersection with ${B}$. Consider ${I_{A,k}}$ the intersection with ${I_{B,k}}$ partitions this interval into

• ${I_{A,k}^M}$ the part of the interval ${I_{A,k}}$ which intersects with the corresponding interval of ${B}$, ${I_{B,k}}$.
• ${I_{A,k}^L}$ the lower interval, this is the part of the interval ${I_{A,k}}$ which is less than the interval ${I_{B,k}}$, or just the lower endpoint of ${I_{A,k}}$ (if the ${I_{B,k}}$ interval overlaps the lower end of the ${I_{A,k}}$ interval).
• ${I_{A,k}^U}$ the upper interval, this is the part of the interval ${I_{A,k}}$ which is greater than the interval ${I_{B,k}}$, or just the upper endpoint of ${I_{A,k}}$ (if the ${I_{B,k}}$ interval overlaps the upper end of the ${I_{A,k}}$ interval).

These intervals can be used to build a set of boxes, ${S}$, ${E\setminus F}$ is the union of the boxes in ${S\setminus (I_{A,1}^M\times \ldots\times I_{A,d}^M)}$. So ${E\setminus F}$ is elementary.

Given this result, the symmetric difference ${E\Delta F}$ is the intersection of two elementary sets and is therefore also elementary.

The translation of a box is another box and so the translation of the set ${E}$ is also elementary.

${\Box}$