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Source. M. Grinsztajn, A 63-dimensional counterexample to Borsuk's conjecture, unpublished note, May 2026, as described on the source card. Lemma 6 is on p. 5 and its proof ends on p. 6.

Statement

For the set $X\subset\mathbb R^{63}$ of Lemma 4, Lemma 6 (p. 5) states: "Every subset $Y\subset X$ with $\operatorname{diam}(Y)<\operatorname{diam}(X)$ has at most 5 points."

Proof pointer

pp. 5--6. By Lemma 5, two points $x_c,x_{c'}$ are at full diameter exactly when $c\not\sim c'$, and $p$ is at full diameter from $x_c$ exactly when $b\not\sim c$. So if $p\notin Y$ the vertices behind $Y$ form a clique in $\Gamma$, and if $p\in Y$ then $b$ together with the vertices behind $Y\setminus\{p\}$ forms a clique; either way the clique number 5 of Lemma 1 gives $\lvert Y\rvert\le5$.

Dependencies and read depth

Depends on Lemma 1, item 2, and Lemma 5. Read depth: claims checked; the statement and proof were read on pp. 5--6. The clique bound it uses is the script's computational claim, unaudited here.

Bears on. E0505: the subset bound from which the note's Theorem 1 counts at least 65 parts.