GPT 6 Astra Ultra attempt on Math-5521
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Erdős Problem 232: density of measurable unit-distance-free sets
1) Statement and scope
Let $m_1$ be the supremum of the upper planar densities of measurable sets containing no two points at distance one. Is $m_1\leq1/4$? The question is known to have an affirmative answer. This attempt proves the elementary bounds \[ \frac{\pi}{8\sqrt3}\leq m_1\leq\frac27. \] The sharper step needed to reach $1/4$ is unresolved in this attempt.2) A periodic lower bound
Let $\Lambda$ be generated by $(2,0)$ and $(1,\sqrt3)$, and put $A=\bigcup_{\lambda\in\Lambda}B(\lambda,1/2)$, with open discs. Two points in one disc have distance less than one. Distinct lattice centres have distance at least two, so points in different open discs have distance strictly greater than one. Thus $A$ is admissible. The fundamental parallelogram has area $2\sqrt3$ and contains one disc of area $\pi/4$. Boundary cells meeting a radius-$R$ circle contribute $O(R)$ area, so the density exists and equals $\pi/(8\sqrt3)$.3) Translating a finite unit-distance graph
Suppose that a unit-distance graph $G$ is embedded at $v_1,\ldots,v_s$ and has independence number $\alpha$. For each $x\in\mathbb R^2$, the vertices for which $x+v_i\in A$ form an independent set. Consequently \[ \sum_{i=1}^s1_A(x+v_i)\leq\alpha. \] Integrating over $B_R$, and using $\lambda((B_R+v_i)\mathbin{\triangle}B_R)=O_{v_i}(R)$, gives \[ s\lambda(A\cap B_R)\leq\alpha\pi R^2+O_G(R). \] Hence $\overline\delta(A)\leq\alpha/s$. Extra unit-distance edges in an embedding can only strengthen the pointwise inequality.4) The seven-vertex bound
Use points \[ O=(0,0),\quad X=(\sqrt3,0),\quad U=(\sqrt3/2,1/2),\quad V=(\sqrt3/2,-1/2). \] The five edges $OU,OV,UV,UX,VX$ have length one. Let $R$ be the rotation with cosine $5/6$ and sine $\sqrt{11}/6$. Adjoin $Y=RX$, $U'=RU$, and $V'=RV$, together with the analogous five edges for the rotated copy and the edge $XY$. Indeed $|X-Y|^2=2\cdot3(1-5/6)=1$. These are seven distinct points.An independent set containing $O$ contains none of $U,V,U',V'$, and contains at most one of $X,Y$. It has size at most two. An independent set avoiding $O$ meets each of the triangles $\{X,U,V\}$ and $\{Y,U',V'\}$ at most once, and again has size at most two. The graph-averaging argument therefore gives $m_1\leq2/7$.
5) Remaining gap and sources
Since $2/7>1/4$, this does not prove the requested upper bound. Ambrus, Csiszárik, Matolcsi, Varga and Zsámboki prove $m_1\leq0.247$ in their paper; its Fourier and optimization certificate is not reconstructed here. The disc construction and Moser-spindle argument above are standard, not claimed as new.Sources: https://www.erdosproblems.com/232; G. Ambrus et al., The density of planar sets avoiding unit distances, https://arxiv.org/abs/2207.14179.