I previously asked a question about this problem. I think I found the answer myself, and I want to know how I did. I'm pretty new at this proof thing and have been working through the textbook Real Mathematical Analysis by Charles Chapman Pugh. Any criticism would be appreciated!(adsbygoogle = window.adsbygoogle || []).push({});

Problem (Chapter 1, Problem 22a):

Given ##ϵ>0##, show that the unit disc contains finitely many dyadic squares whose total area exceeds ##π−ϵ##, and which intersect each other only along their boundaries.

Notes

The first issue is the definition of ##\pi##. For the purpose of this problem, I will assume ##\pi## is the area of the unit disc and thus the least upper bound for the area of all subsets thereof. Secondly, the ##\text{area}## function and its properties are used freely without definition or proof.

Proof

Let

##P_m## be the set of dyadic squares with edge-length ##\frac{1}{2^m}## in ##\mathbb{R}^2##. These squares are either identical or intersect only at their boundaries (previously proved) and have an area of ##\frac{1}{4^m}##;

##D## be the unit disk;

##S_m = \{s \in P_m : s \subseteq D\}## be the squares in ##P_m## that are contained in the unit disk;

and ##A = \{a: m \in \mathbb{N} \text{ and } a = \text{area}(\bigcup S_m)\}##.

##A## is non-empty [since ##\text{area}(\bigcup S_1) = 1##] and ##\pi## is an upper bound for ##A##, and so ##c=\text{l.u.b. }A## exists. Choose any ##\epsilon > 0##. There exists (by virtue of c being the l.u.b.) an ##m## such that ##c - \epsilon < \text{area}(\bigcup S_m)##.

Lemma##c = pi##.

Clearly ##c \leq \pi## since otherwise there would exist an ##m## and ##\lambda > 0## such that ## \pi = c - \lambda < \text{area}(\bigcup S_m)\}##, and ##D## contains squares whose area total more than ##D## itself.

Suppose then that ##c < \pi##. Then every ##S^c_m = D\backslash \bigcup S_m## has a positive area less than ##\pi-c##. There must exist then a common dyadic square ##d \subseteq \bigcap_{m \in \mathbb{N}} S^c_m## with edge length ##\frac{1}{2^n}## and area ##\delta## such that ## 0<\delta = \frac{1}{4^n} < \pi - c## (otherwise, how does every ##S^c_m## have a positive area?). Now choose some ##m## such that ##c < \text{area}(\bigcup S_n) + \delta##. Now, ##n>m## since ##S_n## contains at least one square from ##S^c_m##. Since each square in ##S_m## is partitioned into smaller squares in ##S_n##, it is clear that ##\bigcup S_m \subseteq \bigcup S_n##, but ##S_n## also contains (at least) the extra square with the area ##\frac{1}{4^n}## and so ##\text{area}(\bigcup S_m) + \frac{1}{4^n} \leq\text{area}(\bigcup S_n)##. Thus,

##c < \text{area}(\bigcup S_m) + \frac{1}{4^n} \leq\text{area}(\bigcup S_n) \leq c##,

which is a contradiction. Therefore, ##c## cannot be less than ##\pi## and must instead be equal to ##\pi##.

and so ##c=\pi##.

Proof Cont.

By the lemma, ##\pi - \epsilon < \text{area}(\bigcup S_m)##. If ##S_m## were infinite, then ##\text{area}(\bigcup S_m)## would also be infinite (since all squares are of equal size); this is false because the area must be less than or equal to ##\pi##. Therefore for any ##\epsilon## a finite number of dyadic squares can be choosen whose area exceeds ##\pi-\epsilon##. Each dyadic square is the same size and unique by construction; therefore (by a previous theorem), the squares intersect only at their boundaries or not at all.

End of Proof

Thanks in advance!

EDITED

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# Dyadic Cubes in a Unit Disc

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