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Inspired by answers to Can squares of infinite area always cover a unit square? that involve randomly spaced squares covering every point in a larger square with probability $1$, I was trying to see whether I can come up with a deterministic proof. Along the way, the following problem occurred to me:

Let $\{a_i\}$ be an ordered set of real numbers such that $\forall i \in \Bbb Z^+: a_{i+1}\leq a_i$ and $\lim_{m\to\infty} \sum_{i=1}^ja_i^2 = \infty$.

Form a sequence of two dimensional points within the unit square $p_k$ such that $$\forall k : p_k = ( k \sqrt{2}-\left\lfloor k \sqrt{2} \right \rfloor , k \sqrt{3}-\left\lfloor k \sqrt{3} \right \rfloor )$$

Prove that for every every pair of reals $(x,y)$ with $0\leq x\leq 1$ and $0\leq y\leq 1$ there exists some $n$ such that $p_n = (s,t)$ and $$ \left\{ \begin{array}{c} s - \frac12 a_n < x < s +\frac12 a_n \\ t - \frac12 a_n < y < t +\frac12 a_n \end{array} \right. $$ Geometrically, this says that if $\lim_{i\to\infty} a_i^2 = \infty$ and you place squares of side $a_i$ centered at the deterministic locations $p_i$, then the entire unit square is covered.

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    $\begingroup$ Shouldn't the constraint be $\sum_{i\geq 1}a_i^2=+\infty$? $\endgroup$ – Jack D'Aurizio Dec 21 '17 at 0:24
  • $\begingroup$ The only way for $a_{i+1} \leq a_i$ for all $i$, but $\lim_{i\rightarrow\infty} a_i^2 = \infty$, is if $a_i\rightarrow -\infty$. Is this what you want? $\endgroup$ – Michael Dec 21 '17 at 3:22
  • $\begingroup$ My apologies, I wrote limit when meaning sum. I will correct the problem and title now. $\endgroup$ – Mark Fischler Dec 21 '17 at 14:58
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    $\begingroup$ Good job forcing the sequence to be monotone! Otherwise it would surely be possible to avoid a chosen point with rational coordinates. $\endgroup$ – Jyrki Lahtonen Dec 21 '17 at 15:07
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Edit: Oops, I missed the condition $\alpha_{i+1}\le\alpha_i$.

No. To save you some time, no other sequence works either:

If $(p_n)$ is any sequence of points in the unit square there exists a sequence $(\alpha_n)$ with $\sum\alpha_n^2=\infty$ such that the squares with center $p_n$ and side length $\alpha_n$ do not cover the unit square.

Let's write $$d((x,y),(s,t))=\max(|x-s|,|y-t|).$$Since the unit square is uncountable and a sequence has at most one limit we can find $p$ in the unit square such that $p\ne p_n$ for all $n$ and also $p_n\not\to p$. Let $$\alpha_n=d(p,p_n)/10.$$Then $p$ is not in the square with center $p_n$ and side length $\alpha_n$ for any $n$, and we certainly have $\sum\alpha_n^2=\infty$, because in fact $\alpha_n$ does not even tend to $0$ as $n\to\infty$.

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  • $\begingroup$ Actually, for your argument to work you would have to say that no subsequence of $p_n$ approaches $p$. For the sequence of $p_n$ I gave in the problem, every point in the square that is a limit point of some sub-sequence of $p_n$. $\endgroup$ – Mark Fischler Dec 22 '17 at 23:00
  • $\begingroup$ @MarkFischler Not that it matters, since I can't see how to get monotonicity, but no, I don't think so. Knowing just $p\ne p_n$ shows that $\alpha_n>0$ and $p$ is not in the $n$-th square, and knowing just $p_n\not\to p$ shows that $\alpha_n\not\to0$. Exactly what part of the argument fails if a subsequence approaches $p$? $\endgroup$ – David C. Ullrich Dec 22 '17 at 23:09

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