Proving Properties of Dyadic Cubes in Real Analysis

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In summary, 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}##. It is proven that
  • #1
nateHI
146
4

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.
 
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  • #2
nateHI said:

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.
 
  • #3
gopher_p said:
##\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!
 

Related to Proving Properties of Dyadic Cubes in Real Analysis

1. What is a dyadic cube in real analysis?

A dyadic cube in real analysis is a geometric shape that is constructed by dividing a larger cube into smaller cubes that are all equal in size. The division is done by bisecting each side of the larger cube, resulting in 8 smaller cubes. This process can be repeated, resulting in a hierarchy of cubes with different sizes.

2. What is the importance of dyadic cubes in real analysis?

Dyadic cubes are useful in real analysis because they allow for the study and analysis of functions on a smaller, more manageable scale. By breaking down a larger region into smaller cubes, it becomes easier to analyze the behavior of a function within that region.

3. How are dyadic cubes related to the binary number system?

Dyadic cubes are closely related to the binary number system because the division of a cube into smaller cubes is based on the concept of binary counting. Each bisection of a cube results in 2 smaller cubes, similar to how binary numbers increase in value by a factor of 2 with each additional digit.

4. Can dyadic cubes be used to approximate functions?

Yes, dyadic cubes can be used to approximate functions. By dividing a region into smaller cubes and analyzing the function on each cube, we can get a more accurate representation of the function's behavior within that region. This can be useful in various applications, such as numerical integration or solving differential equations.

5. How are dyadic cubes used in the study of fractals?

Dyadic cubes are often used in the study of fractals because they can be used to model and analyze self-similar structures. By repeatedly dividing a larger cube into smaller cubes, we can create a fractal-like structure that exhibits similar patterns on different scales. This allows for a deeper understanding of fractals and their properties.

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