- #1

nateHI

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## 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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