Dyadic Squares in the Unit Disc: A Proof and Critique

[Someone2841]

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!

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

 

[Someone2841]

It might be important to proof that every rectangle in $\mathbb{R}^2$ contains a dyadic square and also more explicitly that $\bigcup S_m \subset \bigcup S_n \iff n>m$. Are these obvious/immediate enough to assume? (for instance, in a real analysis course)