Euclidean Ramsey set

A finite set $X\subseteq {\mathbb{R}}^d$ is Euclidean Ramsey when, for every number of colors, some finite-dimensional Euclidean space has the property that every coloring of it contains a monochromatic isometric copy of $X$.

A finite set $X=\{x_0,\dots ,x_m\}\subseteq {\mathbb{R}}^d$ is spherical if all its points lie on one sphere: there are a center $c$ and radius $r$ with $\Vert x_i-c\Vert =r$ for every $i$. Subtracting the squared-distance equation for $x_0$ gives the equivalent linear equationsThus consistency of a finite linear system tests whether a common center exists. A nondegenerate triangle is spherical, with center its circumcenter. Three distinct collinear points are not spherical: the equal-distance equations along their line are inconsistent. This is the geometric obstruction behind three-term unit arithmetic progression is not Euclidean Ramsey.

Every Euclidean Ramsey set is spherical. The converse was historically conjectured. A September2026 preprint by Dömötör Pálvölgyi reports the seven-point configurationas a counterexample. These points do lie on the circle $(x-r{)}^2+y^2=r^2$; the non-Ramsey assertion is Theorem1 of that dated preprint.^[ref]

If finite point sets $X$ and $Y$ are Euclidean Ramsey, then their orthogonal Cartesian product $X\times Y$ is Euclidean Ramsey. The proof first chooses a finite Ramsey witness for $X$, colors a witness for $Y$ by the complete color pattern it induces on the first witness, and then applies the two Ramsey properties in succession.

The collinear set $\{0,1,2\}$ is not a Euclidean Ramsey set. In every dimension, coloring $x$ by $\lfloor 2\Vert x{\Vert }^2\rfloor$ modulo ten prevents a monochromatic congruent copy; the parallelogram law would force the corresponding three integer parts to have a second difference strictly between two and six, which cannot be divisible by ten.

Every nondegenerate triangle is a Euclidean Ramsey set.

The two endpoints of a nonzero line segment form a Euclidean Ramsey set. The finite configuration consists of those two points, rather than every point of the continuous line segment.

For a finite colouring with $q$ colours and a desired distance $d>0$, take the $q+1$ vertices of a regular simplex of side length $d$ in ${\mathbb{R}}^q$. The pigeonhole principle gives two vertices of the same colour. Every pair of distinct vertices has distance $d$, so this pair is an isometric copy of the prescribed endpoint pair. This also gives an explicit sufficient ambient dimension for every number of colours.

Every regular polygon is a Euclidean Ramsey set. Its cyclic rotation acts transitively on the vertices, so this is a special case of the Kriz theorem for cyclic transitive point sets.

A finite Euclidean set $X$ is approximately Ramsey if, for every number of colors and every $\epsilon >0$, some finite set has a monochromatic subset whose corresponding pairwise distances differ from those of $X$ by less than $\epsilon$.

A finite Euclidean configuration is edge Ramsey when every finite edge coloring of a suitable finite Euclidean set contains a monochromatic isometric copy. The edge Ramsey configurations are exactly the equidistant sets, equivalently the vertex sets of regular simplices.

A finite point set is cyclic transitive when a cyclic group of isometries acts transitively on it.

Every finite cyclic transitive point set is a Euclidean Ramsey set. In particular every regular polygon is Euclidean Ramsey.

For $A\subseteq X$, a coloring of $X^n$ is $A$-invariant when changing any coordinates from one member of $A$ to another does not change the color. A product argument upgrades the existence of a copy of $X$ on which $A$ is monochromatic to an $A$-invariant copy of every finite Cartesian power $X^n$.