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}

 

Advertisements
This entry was posted in IMT Exercises, Measure Theory. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s