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Statement

Remark (§5, p. 18, unnumbered). For a finite $S\subset\mathbb R^n$, let the graph of $S$ join two points at unit distance. For $X=\{0,1\}$, the minimal $2$-Ramsey sets for $X$ are exactly the sets whose graph is an odd cycle; the minimal $k$-Ramsey sets are the sets whose graphs are $(k+1)$-critical (chromatic number $k+1$, and deleting any vertex lowers it); and the unique minimum-sized one is the regular simplex on $k+1$ vertices. For the odd-cycle sets ($k=2$) the paper notes that such a set may have no isometries, but can be transformed, keeping its unit distances, into a transitive set, and that a minimum-sized one is an equilateral triangle.

Derivation

The paper states these facts without proof. A coloring of $S$ without a monochromatic unit pair is a proper coloring of its graph, so $S$ is $k$-Ramsey for $X$ exactly when its graph has chromatic number above $k$; minimality is vertex-criticality; and a graph on $k+1$ vertices with chromatic number $k+1$ is complete, which forces a regular unit simplex. For $k=2$, an inclusion-minimal nonbipartite unit-distance graph is a chordless odd cycle, and such a cycle can be redrawn as a regular odd polygon of side one.

Source. Imre Leader, Paul A. Russell and Mark Walters, Transitive sets in Euclidean Ramsey theory, J. Combin. Theory Ser. A 119 (2012), no. 2, 382--396, doi:10.1016/j.jcta.2011.09.005; pages from the arXiv version 1012.1350v1 identified in the source digest.

Read depth. Claims checked: the remark (p. 18) was read against the print; the derivation above is the corpus's own.

Bears on

• Problem 174: the motivating example for the paper's Problem J (p. 19), on the "only if" direction of Conjecture A, for which the paper has no results (p. 18). It proves nothing about general Ramsey sets.