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This is 1.22a in Pugh's Real Mathematical Analysis (p. 44):

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.

The desired result goes against my intuition: It seems that I should be able to fit as many squares as I want inside the circle, assuming the squares are small enough.

To try to get an idea of what was going on, I started filling the first quadrant with squares — viz dyadic squares along the $x$-axis: $[0, 1/2]\times[0,1/2], [2/4, 3/4]\times[0, 1/4], [6/8, 7/8]\times[0, 1/8], \ldots$ I figured that, if only a finite number fit inside, then eventually the corner furthest from the origin would fall outside the circle — i.e., its distance would be greater than $1$. And so I calculated

$$\lim_{n\rightarrow \infty} \sqrt{\left( \sum_{k=1}^n \frac{1}{2^k}\right)^2+\left( \frac{1}{2^n}\right)^2}=1$$

which suggests that, indeed, I can fit as many dyadic squares as I want in this direction — and so why not in any direction?

I'm guessing I'm missing something here. Could someone please shed some light? — on what I'm misunderstanding, how else I might start, or both? Beyond what I've tried, I'm not sure I have any idea where to start with this problem. (Not to mention the one that follows, in which the squares are disjoint.)

Micah
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dmk
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1 Answers1

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I think you're misinterpreting the question. It's not asking you to show that only finitely many dyadic squares can fit into the circle. Rather, it's asking you to show that you can get arbitrarily close to filling the circle using only a finite number of squares. Of course you could then go on to add even more squares (and get even closer to filling it!)

Micah
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