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math.CO — Combinatorics

Measurable Independence Density Equals the Finite Independence-Ratio Infimum in the Euclidean Plane

Contributed by Ákos Dúcz

Let m1(R^2) be the supremum of upper densities of Lebesgue-measurable subsets of the plane containing no pair of points at distance one. We show, using the spectral rigidity theorem in OpenAI's recent proof that the plane is not five-colorable, that m1(R^2) = inf_G alpha(G)/|V(G)|, where G ranges over nonempty finite unit-distance graphs in the plane. The argument constructs an isometry-invariant law on independent subsets of the countable algebraic plane, projects the occupancy indicator onto the continuous spectral factor without changing its expectation, and extracts a measurable independent set with arbitrarily small density loss. As a consequence, if f(n) is the least independence number among unit-distance graphs on n vertices, then f(n)/n converges to m1(R^2). The proof is nonquantitative and uses the cited spectral rigidity result as a black box.

Primarily AI-generated textHuman understanding: some partsSpectral RigidityUnit-distance graphsindependence ratioinvariant random independent setsmeasurable sets

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