Disjoint dyadic squares in unit disk

alanlu
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Homework Statement



Given ε > 0, show that there is a collection of disjoint dyadic squares in the unit disk that has a total area which exceeds π - ε.

Homework Equations



Define a dyadic interval as an interval of the form [a, b] such that a = p/2k and b = (p + 1)/2k, p and k are integers. A dyadic square is the product of two of these intervals.

This problem is ch1 22b of Pugh's Real Mathematical Analysis.

The Attempt at a Solution



I figured this had something to do with aliasing on an circular arc, so I wrote A(n) = { 1 \over 4^n } \sum_{i=1}^{2^n} \lfloor \sqrt{4^n - i^2} \rfloor as an expression where A is the area of a quadrant of the unit circle aliased down to a { 1 \over 2^n } granular dyadic grid. Then I noted that A → π/4 as n → ∞. What I am stuck on is the disjoint condition, as that makes requiring larger chunks of squares mandatory for the collection's total area to approach π I think, but it feels like to me that the supremum I want is less than π.
 
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I don't understand your difficulty. The squares in A(n) are disjoint. What do you mean by "makes requiring larger chunks of squares mandatory"?
 
I believe the squares include their boundaries, and by definition of the disjoint condition two adjacent squares would have a line/vertex (nonempty) intersection and so would not be disjoint. If I interpreted the question wrong and things are fine, then ... haha yeah, there isn't much difficulty after all. On the other hand, if the disjoint condition is that strict, then intuitively, I'm not even sure the question can be resolved.
 
I don't think anyone's going to worry about line and point overlaps, which are clearly sets of measure zero. But if it bothers you, use intervals like [a, b) or (a, b).
 
Alright, thank you haruspex. I'm starting to think the stricter interpretation has an interesting upper bound in its own right, but I don't think this would not be the forum for that discussion. :)
 
I asked the author for some clarification and he has chosen to grace me with a response! Here's the quote:

Me: ...I asked online for some help and the advice given seems to indicate that perhaps the problem is worded too strictly...

Charles: ...Nope, the problem is correct as stated. Think of a pile of sand. If you can take away 1/16 of the pile, and then take 1/16 of the remaining pile, and continue doing that indefinitely, how much will be left in the limit?

In short, I'm not quite done yet.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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