Proving Properties of Dyadic Cubes in Real Analysis

[nateHI]

Homework Statement

A closed cube in ##\mathbb{R}^n## of the form
##
\left[\frac{i_1}{2^m}, \frac{i_1+1}{2^m}\right]\times
\left[\frac{i_2}{2^m}, \frac{i_2+1}{2^m}\right]\times
\cdots
\left[\frac{i_n}{2^m}, \frac{i_n+1}{2^m}\right],
##
where ##m, i_1,\ldots,i_n\in\mathbb{Z}##, is called a (closed) dyadic
cube of sidelength ##2^{-m}##. Prove the following:

1) Closed dyadic cubes of a fixed sidelength ##2^{-m}##
are almost disjoint and cover all of ##\mathbb{R}^n##.
2) Each dyadic cube of sidelength ##2^{-m}## is contained
in exactly one "parent'' dyadic cube of sidelength
##2^{-m+1}##, which, conversely, has exactly ##2^n##
"children'' of sidelength ##2^{-m}##.
3) A consequence of the above facts is the following
: given any two closed dyadic cubes (possibly of different sidelength), either
they are almost disjoint or one of them is contained in the other.

Homework Equations

It wasn't given in the problem statement from the instructor but ##\mathbb{R}^n## must be [0,1]x[0,1]...[0,1] an ##n## dimensional cube possibly translated.

The Attempt at a Solution

1) Let ##x\in\mathbb{R}^n## be an element, not on the boundary, of two dyadic cubes of the same generation. In other words, the two dyadic cubes in question are not almost disjoint. In this situation the sidelength of one of the cubes will not be ##2^{-m}##, contradicting the definition of a dyadic cube given in the problem statement.

Dyadic cubes cover ##R^n## as can be seen by simply taking the very first generation (##m=0##) and examining the interval of one of the dimensions. This interval will be ##[0,1]## and completely contain all of elements of that dimension. The same argument can be applied to each interval and associated dimension.

(EDIT: I just realized this part of my solution attempt is very wrong. No need to read it but I'll leave it up in case someone already started to reply.)
Another approach to showing ##\mathbb{R}^n## is covered by the dyadic cube given in the problem statement:
Let ##m=0## and let ##Q_n## be an ##n## dimensional dyadic cube that doesn't cover ##\mathbb{R^n}##. Then ##\mathbb{R^n}=Q_n\cup {Q_n}^c##. This union will be both closed and open ##\mathbb{R^n}=[...)x[...)...x[...)##implying the complement of ##\mathbb{R^n}## is closed. If the complement of ##\mathbb{R^n}## is closed then ##\mathbb{R^n}## is not finite, a contradiction.

2) Just help me with the first part then maybe I can do the 2nd completely on my own.

3) Just help me with the first part then maybe I can do the 3rd completely on my own.

 

[gopher_p]

Homework Statement

A closed cube in ##\mathbb{R}^n## of the form
##
\left[\frac{i_1}{2^m}, \frac{i_1+1}{2^m}\right]\times
\left[\frac{i_2}{2^m}, \frac{i_2+1}{2^m}\right]\times
\cdots
\left[\frac{i_n}{2^m}, \frac{i_n+1}{2^m}\right],
##
where ##m, i_1,\ldots,i_n\in\mathbb{Z}##, is called a (closed) dyadic
cube of sidelength ##2^{-m}##. Prove the following:

1) Closed dyadic cubes of a fixed sidelength ##2^{-m}##
are almost disjoint and cover all of ##\mathbb{R}^n##.
2) Each dyadic cube of sidelength ##2^{-m}## is contained
in exactly one "parent'' dyadic cube of sidelength
##2^{-m+1}##, which, conversely, has exactly ##2^n##
"children'' of sidelength ##2^{-m}##.
3) A consequence of the above facts is the following
: given any two closed dyadic cubes (possibly of different sidelength), either
they are almost disjoint or one of them is contained in the other.

Homework Equations

It wasn't given in the problem statement from the instructor but ##\mathbb{R}^n## must be [0,1]x[0,1]...[0,1] an ##n## dimensional cube possibly translated.

##\mathbb{R}^n## usually denotes the Cartesian product of ##n## copies of ##\mathbb{R}##; i.e. it's ##\mathbb{R}\times\mathbb{R}\times\dots\times\mathbb{R}##.

The Attempt at a Solution

1) Let ##x\in\mathbb{R}^n## be an element, not on the boundary, of two dyadic cubes of the same generation. In other words, the two dyadic cubes in question are not almost disjoint. In this situation the sidelength of one of the cubes will not be ##2^{-m}##, contradicting the definition of a dyadic cube given in the problem statement.

You haven't really proved anything here. Why must the side length of one of the cubes not be ##2^{-m}##? I know it seems obvious, but the statement you're trying to prove also seems obvious, right?

I recommend thinking about what must be true of the coordinates of a point not on the boundary of any of the cubes. Think about what the cubes and their boundaries look like in ##\mathbb{R}## and ##\mathbb{R}^2## if you need to, and then generalize.

Dyadic cubes cover ##R^n## as can be seen by simply taking the very first generation (##m=0##) and examining the interval of one of the dimensions. This interval will be ##[0,1]## and completely contain all of elements of that dimension. The same argument can be applied to each interval and associated dimension.

I'm not sure what you're trying to say here, but it doesn't look right.

A family ##\mathcal{F}## of subsets of a set ##X## is said to cover ##X## if and only if for every element ##x\in X## there is a set ##F\in\mathcal{F}## such that ##x\in F##. For your problem, you're being asked to show that every family of "##m##-cubes" in ##\mathbb{R}^n## (cubes of set side length ##2^{-m}##) is a cover of ##\mathbb{R}^n##.

In other words you need to show that for all ##x\in\mathbb{R}^n## and all ##m\in\mathbb{Z}## there is an ##m##-cube ##C## with ##x\in C##. Again, figure out how the proof works in ##\mathbb{R}##, and then try to generalize.

(EDIT: I just realized this part of my solution attempt is very wrong. No need to read it but I'll leave it up in case someone already started to reply.)
Another approach to showing ##\mathbb{R}^n## is covered by the dyadic cube given in the problem statement:
Let ##m=0## and let ##Q_n## be an ##n## dimensional dyadic cube that doesn't cover ##\mathbb{R^n}##. Then ##\mathbb{R^n}=Q_n\cup {Q_n}^c##. This union will be both closed and open ##\mathbb{R^n}=[...)x[...)...x[...)##implying the complement of ##\mathbb{R^n}## is closed. If the complement of ##\mathbb{R^n}## is closed then ##\mathbb{R^n}## is not finite, a contradiction.

2) Just help me with the first part then maybe I can do the 2nd completely on my own.

3) Just help me with the first part then maybe I can do the 3rd completely on my own.

 

[nateHI]

##\mathbb{R}^n## usually denotes the Cartesian product of ##n## copies of ##\mathbb{R}##; i.e. it's ##\mathbb{R}\times\mathbb{R}\times\dots\times\mathbb{R}##.

You're correct. This was the source of the issues I was having with this problem. I didn't understand the problem statement correctly. Now that I understand the problem correctly and with your hints I'll take another crack at it.

Thanks!