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Statement

Lemma 3 (p. 4). Stated inside the proof of Theorem 1, for its fixed $X$, $Y$ and $G$: "Let $x\in X$, $y\in Y$, and $n\in\mathbb N$. Then Theorem 1 is true for $\lambda_n=\frac{1}{\sqrt{n}}\|x-y\|$."

In the corpus's words: for any $x\in X$, $y\in Y$ and integer $n\ge1$, the prism $(X\times\{0\})\cup(Y\times\{\lambda_n\})$ with $\lambda_n=\|x-y\|/\sqrt n$ is subtransitive, and subsoluble when $G$ is soluble. When $x=y$ the height is zero; then $X$ and $Y$ are the same orbit and the "prism" is $X$ itself, which is transitive. The use made of the lemma in Theorem 1 needs $x\ne y$, so that the heights are positive and tend to zero.

Source. M.-R. Ivan, I. Leader and M. Walters, Generalised Prisms and Euclidean Ramsey Theory, arXiv:2606.13472v1 (11 June 2026), Lemma 3, p. 4; proof pp. 4–6.

Read depth. Claims checked: statement read clause by clause on the PDF; the proof read in full, its cross-distance step recomputed here.

Proof pointer

After moving a fixed point of $G$ to the origin, the proof (pp. 4–6) builds $n+1$ intermediate $G$-orbits interpolating linearly from $X$ to $Y$, takes the union of the cyclic rotations of their product, and shows that the wreath product $G\wr C_{n+1}$ acts transitively on it, soluble when $G$ is. Two parallel copies of $X$ and $Y$ inside this set sit at distance $\lambda_n$ in the perpendicular directions.

On p. 6 the proof says that $X$ and $Y$ "have the same centre". That is not true for every pair of orbits of a common group (a trivial group has any two points as singleton orbits), and it is not needed: the two copies differ by a translation perpendicular to the first factor $\mathbb R^d$ of length $\|x-y\|/\sqrt n$, so every cross squared distance is $\|u-v\|^2+\|x-y\|^2/n$ for $u\in X$, $v\in Y$, which is exactly what the lemma requires.

Dependencies

None outside the paper; the group facts used (a fixed point of a finite isometry group, solubility of $G\wr C_{n+1}$ for soluble $G$) are recorded on the elementary-facts page.

Bears on

• Problem 174: a step in the proof of Theorem 1; it bears on the problem only through that theorem.