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Problem #232: For $A\subset \mathbb{R}^2$ we define the upper density...

For $A\subset \mathbb{R}^2$ we define the upper density as\[\overline{\delta}(A)=\limsup_{R\to \infty}\frac{\lambda(A \cap...

Problem Statement

For $A\subset \mathbb{R}^2$ we define the upper density as\[\overline{\delta}(A)=\limsup_{R\to \infty}\frac{\lambda(A \cap B_R)}{\lambda(B_R)},\]where $\lambda$ is the Lebesgue measure and $B_R$ is the ball of radius $R$.

Estimate\[m_1=\sup \overline{\delta}(A),\]where $A$ ranges over all measurable subsets of $\mathbb{R}^2$ without two points distance $1$ apart. In particular, is $m_1\leq 1/4$?
Categories: Geometry Distances

Progress

A question of Moser [Mo66]. A lower bound of $m_1\geq \pi/8\sqrt{3}\approx 0.2267$ is given by taking the union of open circular discs of radius $1/2$ at a regular hexagonal lattice suitably spaced apart. Croft [Cr67] gives a small improvement of $m_1\geq 0.22936$.

The trivial upper bound is $m_1\leq 1/2$, since for any unit vector $u$ the sets $A$ and $A+u$ must be disjoint. Erdős' question was solved by Ambrus, Csiszárik, Matolcsi, Varga, and Zsámboki [ACMVZ23] who proved that $m_1\leq 0.247$.

Source: erdosproblems.com/232 | Last verified: January 14, 2026

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