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

Let $X=\{v_1,\dots ,v_m\}$ be the vertices of a regular polygon, indexed cyclically so that $v_i\mapsto v_{i+1}$ is an isometry. We first prove the product lemma needed for the cyclic-symmetry argument.

Suppose $A\subseteq X$ has the property that, for every number of colors, a finite witness forces a copy of $X$ on which the vertices corresponding to $A$ are monochromatic. We claim that for every $N,k$ there is a finite witness forcing an $A$-invariant copy of $X^N$. Proceed by mathematical induction on $N$. The case $N=1$ is the assumption. For the step, choose a finite set $S$ that forces an $A$-invariant $X^{N-1}$ for a coloring with $k^{|X|-|A|+1}$ colors, and choose $T$ that forces an $A$-monochromatic $X$ for a coloring with $k^{|S|}$ colors.

Given a $k$-coloring $c$ of $S\times T$, color $t\in T$ by its complete vector $(c(s,t){)}_{s\in S}$. In the resulting copy of $X$ in $T$, that vector is the same for every member of $A$. Each $s\in S$ can therefore be colored by the vector consisting of $c(s,a)$ for one $a\in A$ and the colors $c(s,x)$ for $x\in X\setminus A$. Applying the choice of $S$ gives an $A$-invariant copy of $X^{N-1}$, and its product with the chosen copy of $X$ in $T$ is the desired $A$-invariant copy of $X^N$. This proves the lemma rather than assuming it.

We now prove by induction on $r$ that a finite witness forces a copy of $X$ whose first $r$ vertices have one color. The case $r=1$ is trivial. For $r=2$, use that a line segment is a Euclidean Ramsey set; each occurrence of a segment congruent to $v_1{v}_2$ can be extended to a copy of the regular polygon, and only finitely many extensions are needed for a finite witness.

Assume the assertion for $r$, and put $A=\{v_1,\dots ,v_r\}$. The product lemma lets us work inside an $A$-invariantly colored copy of $X^N$, with $N$ as large as required. For an increasing $(m-1)$-element subset $H=\{a_1<\cdots <a_{m-1}\}$ of $[N]$ and $1\le i\le r$, define a word $w_i(H)$ by putting $v_{i+j}$ in coordinate $a_j$, with subscripts read modulo $m$, and putting $v_1$ in every other coordinate. Color $H$ by the $r$-tupleBy the Finite Ramsey theorem, for sufficiently large $N$ there are coordinates $b_1<\cdots <b_m$ on which all $(m-1)$-subsets have the same tuple color.

For $1\le i\le r+1$, let $z_i$ be the word whose entries in coordinates $b_1,\dots ,b_m$ arecyclically, and whose other entries are $v_1$. For $i\le r$, the word $z_i$ differs from $w_i(\{b_2,\dots ,b_m\})$ only in coordinate $b_1$, where the two entries $v_i,v_1$ lie in $A$. Similarly $z_{i+1}$ differs from $w_i(\{b_1,\dots ,b_{m-1}\})$ only in coordinate $b_m$, again by two members of $A$. The $A$-invariance and the homogeneous choice of the $b_j$ therefore giveHence $z_1,\dots ,z_{r+1}$ have one color.

The cyclic words $z_1,\dots ,z_m$ form an isometric copy of $\sqrt{m}X$: in each of the $m$ varying coordinates the cyclic shift is an isometry, so every squared distance is multiplied by $m$. Rescaling the finite witness by $1/\sqrt{m}$ gives a copy of $X$. The induction reaches $r=m$, proving that every regular polygon is a Euclidean Ramsey set.

For three consecutive vertices $p_0,p_1,p_2$ of a regular $M$-gon scaled to have side length one,Since the polygon is Euclidean Ramsey, choosing $M$ large enough proves that $\{0,1,2\}$ is an approximately Euclidean Ramsey set.

In fact every finite $X\subseteq {\mathbb{R}}^d$ is approximately Ramsey. First perturb each coordinate of each point onto a sufficiently fine lattice $\delta \mathbb{Z}$, changing all pairwise distances by less than $\epsilon /2$. For each coordinate, map the finitely many required integers to consecutive vertices of a very large regular polygon, scaled so that one angular step has arc length $\delta$. If the step angle is $\theta$, the chord replacing a difference of $q$ lattice steps has lengthas $\theta \to 0$, uniformly over the finitely many differences involved. Taking the orthogonal Cartesian product of $d$ such polygons therefore embeds the perturbed points with a further distance error below $\epsilon /2$. Each polygon is Euclidean Ramsey, and their product is Euclidean Ramsey by the product theorem for Euclidean Ramsey sets. A monochromatic copy of that product contains the corresponding approximate copy of $X$, completing the proof.

The vertices satisfyAfter removing the root $z=0$, the product of the other roots has modulus equal to the constant term of $((z+1{)}^N-1)/z$, namely $N$. Hence the product of the $N-1$ chord lengths from one vertex of a unit regular polygon is $N$. Multiplying this identity over all $N$ vertices counts every chord twice, so the product of all chord lengths is $N^{N/2}$. Scaling the circle by $R$ scales each of its $N(N-1)/2$ chords by $R$, giving

Let $X$ be the vertex set of a regular polygon and let the cyclic rotation group $A$ act transitively on it. The finite invariant-colouring lemma says that for every number of colours there are positive weights $w_1,ldots,w_N$ with ${\sum }_i{w}_i^2=1$ such that every colouring of the weighted Cartesian powercontains a monochromatic setFor completeness, prove the lemma along a cyclic composition series for $A$. For a prime cyclic quotient, refine each colour to the finite vector of colours obtained by applying the quotient elements in every active coordinate. The Hales-Jewett theorem supplies a variable block on which this vector is constant. Assign that block squared weights summing to the squared weight of the coordinate it replaces. This makes the quotient orbit monochromatic without changing any orbit distance. Iterating through the cyclic factors proves the lemma for the finite cyclic group $A$.

Because rotations commute, for $a,b\in A$ each coordinate satisfiesfor a fixed vertex $v$. Hence the squared distance between the two corresponding product points isThe monochromatic orbit is therefore isometric to $X$. This proves that every regular polygon is a Euclidean Ramsey set.

Now consider the edge-colouring definition. If all pairwise distances in $X$ are equal, then $X$ is a regular simplex. Take a sufficiently large regular simplex $S$ of the same side length. The ordinary finite Ramsey theorem gives a monochromatic $|X|$-vertex complete subgraph, and every such vertex set is isometric to $X$. Thus $X$ is an edge Ramsey set.

Conversely, if $X$ has two distances, let $a$ and $b$ be its least and greatest distances. Colour every Euclidean edge by whether its length equals $a$. Every isometric copy of $X$ contains both an edge of length $a$ and one of length $b$, so no copy is monochromatic. Hence the edge Ramsey sets are exactly the equidistant finite sets.

Allowing similar copies does not change the answer. If $r=b/a>1$, colour an edge of length $d$ by the parity ofIn every similar copy, the images of a shortest and a longest edge have lengths $ta$ and $tb=rta$, whose displayed integers differ by one. They receive opposite colours. Thus no non-equidistant $X$ is edge Ramsey even up to similarity, while the regular-simplex argument already supplies an isometric copy.

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.