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Is there a measurable set $U\subset \Bbb R$ of Lebesgue measure 0 satisfying the property: For any two points $p, q\in \Bbb R$, there exists $a\in \Bbb R$ such that $\{p+a, q+a\}\subset U$.

Is there a measurable set $U\subset \Bbb R^2$ of measure 0 satisfying the property: For any two points $P, Q\in \Bbb R^2$, there exists $a\in \Bbb R$ such that $\{P, Q\} +(a, a)\subset U$.

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The first question is equivalent to the question whether a set $U \subset \mathbb{R}$ of Lebesgue measure $0$ can have the property that its set of differences $U-U = \{ u-v: u,v \in U\}$ can be the whole real line $\mathbb{R}$. The answer is yes, and there are a few ways to construct sets like this, e.g.: Let $U_0$ be the set of real numbers in $[0,1]$ whose base $5$ expansion does not contain the digit $2$. Then $U_0$ is a regular Cantor set, obtained by subdividing $[0,1]$ into $5$ intervals and removing the middle one, then repeating this construction with the remaining $4$ intervals etc. Let $U=\mathbb{Z}+U_0$ be the $1$-periodic extension of $U$. Then both $U_0$ and $U$ are sets of measure zero, and it is easy to see that $U-U = \mathbb{R}$: If $a=A+ 0.a_1 a_2 a_3 \ldots$ with $A \in \mathbb{Z}$ and the expansion $0.a_1a_2a_3\ldots$ in base 5, then $a = b-c$, where $b=A+0.b_1b_2b_3\ldots$ and $c=0.c_1c_2c_3\ldots$ in base 5 with $b_k = a_k$ and $c_k = 0$ whenever $a_k \ne 2$, and $b_k = 3$ and $c_k=1$ whenever $b_k = 2$.

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Very nice. $\,$ –  Brian M. Scott Dec 12 '12 at 4:23
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