Finite pair separation yields a finite relative ball

If $E=0$, it has no proper hyperplanes, so the assertion is immediate. Otherwise put $d=\dim E>0$, choose vertices and $\epsilon$ by Coordinate vertices inside a relative open ball and set $r=\epsilon \mathrm{/}d$. All vertices lie in $T\subseteq \mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}$. For each of the finitely many pairs $(p_i^{+},p_i^{-})$ the ambient root hyperplanes strictly separating that pair form a finite set by Only finitely many hyperplanes separate two Tits-cone points. Let $\mathcal{F}$ be their finite union.

Every proper trace $K=H\cap E\in {\mathcal{H}}^J$ meeting $B_E(u,r)$ strictly separates one such pair by A hyperplane near the centre separates opposite coordinate vertices. Restricted evaluation and ambient evaluation agree on $E$, so its representative ambient hyperplane $H$ belongs to $\mathcal{F}$. Hence all these $K$ belong to the finite set $\{H\cap E:H\in \mathcal{F}\}$. Also $0<r=\epsilon \mathrm{/}d\le \epsilon$, so $B_E(u,r)\subseteq B_E(u,\epsilon )\subseteq T$ by [F1] and the choice of $\epsilon$. This proves the assertion, counting coincident traces once.