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Statement

Lemma 2 (p. 3). Stated inside the proof of Theorem 1, for the sets $X,Y\subset\mathbb R^d$ and the finite group $G$ of that theorem, which stay fixed throughout: "Suppose that there exists $\lambda'>0$ for which Theorem 1 is true. Then, for any $\lambda''\geq\lambda'$ Theorem 1 is also true. In other words, the set $Z'=(X,0)\cup(Y,\lambda)$ [sic] is subtransitive, and if $G$ is soluble, then $Z'$ is subsoluble."

The $\lambda$ in the definition of $Z'$ is a misprint for $\lambda''$; the proof uses $Z'=(X,0)\cup(Y,\lambda'')$. In the corpus's words: if the prism $(X\times\{0\})\cup(Y\times\{\lambda'\})$ is subtransitive (subsoluble when $G$ is soluble) for some $\lambda'>0$, then so is $(X\times\{0\})\cup(Y\times\{\lambda''\})$ for every $\lambda''\ge\lambda'$.

Source. M.-R. Ivan, I. Leader and M. Walters, Generalised Prisms and Euclidean Ramsey Theory, arXiv:2606.13472v1 (11 June 2026), Lemma 2, p. 3; proof pp. 3–4.

Read depth. Claims checked: statement read clause by clause on the PDF; the proof read in full.

Proof pointer

Take a finite transitive set $U$ containing a copy of the prism at height $\lambda'$, and form the two-level product $U\times\{0,a\}$ with $a^2=(\lambda'')^2-(\lambda')^2$. It is transitive under the product of $U$'s group with $C_2$, and it contains a copy of the prism at height $\lambda''$, because the cross distances gain exactly $a^2$.

Two details of the printed proof (pp. 3–4) are read here as follows. The proof takes $a\ne0$, so the case $\lambda''=\lambda'$ is the hypothesis itself. In the soluble case it says "$H$ is soluble" of the full symmetry group $H$ of $U$; what the argument needs, and what subsolubility supplies, is some soluble group acting transitively on $U$, and the product of that group with $C_2$ is soluble. Neither point changes the statement. The proof never uses $G$ beyond the hypothesis on $\lambda'$.

Dependencies

None outside the paper; the group facts used 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.