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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.

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    $\begingroup$ 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 $n=1$. $\endgroup$
    – rpotrie
    Commented Oct 19, 2010 at 12:29
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    $\begingroup$ Remark: $\sqrt{n}/2$ is easy, by simply taking both $x = (0,0,\ldots,0)$ and $x = (\frac12,\frac12,\ldots,\frac12)$. $\endgroup$
    – Greg Kuperberg
    Commented Oct 19, 2010 at 12:34

1 Answer 1

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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.

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