Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a set $A\subset \mathbb{R}^n$ such that $A\cap (x+\mathbb{Z}^n)\ne \emptyset$ for any $x\in \mathbb{R}^n$ (that is, $p(A)=\mathbb{T}^n$ for the projection $p:\mathbb{R}^n\rightarrow \mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$). Is it true that supremum of Eucledian distances between points of $A$ is not less then $\sqrt{n}$? (equality holds for the unit cube)

Or maybe even two points with $x=(x_1,\dots,x_n)$, $y=(y_1,\dots,y_n)$ with $|x_i-y_i|\geq 1-\epsilon$ for each coordinate?

It is not hard to check both claims for $n=2$, but already for $n=3$ I do not know.

share|improve this question
1  
A small correction, you should put "for any $x\in \mathbb{R}^n$" in the first line. Also, are you assuming that $A$ is closed? Otherwise, the second question is false even with $n=1$. –  rpotrie Oct 19 '10 at 12:29
1  
Remark: $\sqrt{n}/2$ is easy, by simply taking both $x = (0,0,\ldots,0)$ and $x = (\frac12,\frac12,\ldots,\frac12)$. –  Greg Kuperberg Oct 19 '10 at 12:34
add comment

1 Answer

The second claim is false for $n=3$. Choose $\varepsilon$ small and $\delta\ll\varepsilon$. Let $A$ be the set of all points $(x,y,z)\in\mathbb R^3$ satisfying the following inequalities: $$ \begin{cases} -1.5+\varepsilon+\delta &\le x+y+z &\le 1.5+\varepsilon \\ -1.5+\delta &\le x+y-z &\le 1.5 \\ -1.5+\delta &\le x-y+z &\le 1.5 \\ -1.5+\delta &\le -x+y+z &\le 1.5 \\ \end{cases} $$ The integer translates of this set cover the space, but its $\ell_1$-diameter is no greater than $3-\delta$.

Added. The first claim is false too. In the above example, fix $\delta=\varepsilon/10$ and add the inequality $$\max\{|x|,|y|,|z|\}\le 0.5+10\varepsilon$$ to the system.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.