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I found this question here. The answer here is obviously not the answer I need. The question is:

Given $\epsilon > 0$, show that the unit disc contains finitely many dyadic squares whose total area exceeds $\pi - \epsilon$, and which intersect each other only along their boundaries.

I tried to solve it but a misunderstanding still remains.

My solution:

First I put the center of disc at the origin of Cartesian coordinates. Then I tile the number plane with dyadic squares each of side length $2^{-n}$. Now I need to come up with a formula that gives "an upper bound for the total area inside the disc that is not covered by a square and lies entirely within the disc". I came up with the formula by chatting with a friend online:

His key insight: The greatest possible distance between any two points in small square is its hypotenuse which is $\dfrac{\sqrt{2}}{2^n}$

Therefore draw a concentric circle of radius $1-\dfrac{\sqrt{2}}{2^n}$ inside the unit disc. Then any point on any square that "lies both inside and outside the disc" will always be outside the inner concentric circle.

Then it can be easily shown that the area between two concentric circles is:

$$\dfrac{2 \sqrt{2}}{2^n}- \dfrac{2 \pi}{2^{2n}}$$

This is "an upper bound for the total area inside the disc that is not covered by a square and lies entirely within the disc". This also is a decreasing function of $n$ whose limit goes to zero when $n \rightarrow \infty$.

Thus the total area "inside the disc that is not covered by a square and lies entirely within the disc" goes to zero when $n \rightarrow \infty$.

Thus the result of theorem follows.

My confusion:

Why is the bold sentence in "His key insight" true?

lorilori
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  • You can write out everything in a set theoretic fashion; I guess that should help; you can write $B(0,1):= {(x,y): x^2 + y^2 \leq 1}$ to denote the ball with the interior with radius $1$, maybe that will convince you – Rohan Didmishe Dec 21 '22 at 12:03

1 Answers1

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Let $S$ be a square with side length $2^{-n}$ that “lies both inside and outside the disc.” Then there is a point $a \in S$ with $\Vert a\Vert > 1$.

For all $b \in S$ is (using the triangle inequality) $$ \Vert b \Vert = \Vert a - (a-b)\Vert \ge \underbrace{\Vert a \Vert}_{> 1} - \underbrace{\Vert a-b\Vert}_{\le \sqrt 2/2^n} > 1 - \frac{\sqrt{2}}{2^n} $$ and that proves that $S$ is outside the inner circle.

Martin R
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