Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. For every there is a set of points in the plane that contains no empty convex heptagon: no seven of its points are the vertices of a convex heptagon whose interior is free of the set. Hence does not exist, and neither does for any , since an empty convex -gon contains an empty convex heptagon among its vertices. The problem asks whether exists for every ; Horton's construction answers no. The problem's wording and the site's page omit the general-position convention; Horton defines over sets with no three collinear, as the literature on does, and is such a set.
The construction. Horton's set consists of the points for , where is read off the binary digits of with a base . Splitting on the lowest binary digit cuts into a bottom half and a top half that are scaled translates of each other, and every point of lies above every line through two points of . An empty convex polygon lying in one half maps affinely onto an empty convex polygon of the smaller set, so one may assume it meets both halves. Such a polygon has at most three vertices in and three in , hence at most six. The source card records the paper's observations and the counting step.
What remains of the question. The function exists for small : (Erdős), (Harborth), and exists by the independent proofs of Nicolás and Gerken, with established by Heule and Scheucher. Horton's note already records and leaves the hexagon case open.
Acceptance. The paper is refereed: J. D. Horton, Sets with no empty convex 7-gons, Canad. Math. Bull. 26 (1983), no. 4, 482–484. The curator of erdosproblems.com, Thomas Bloom, labels the problem disproved and credits Horton's paper for the nonexistence of for .