Local finiteness of the interior relative arrangement

Let $u\in T^{\circ }$. By [F1], there is $r>0$ such that $B_E(u,r)\subseteq T$ and this ball meets only finitely many members of ${\mathcal{H}}^J$. Since it is open in $E$ and contained in $T$, it is an open ball in the subspace $T$. This proves the first assertion.

The intersection of the convex full Tits cone with the linear subspace $E$ is convex, so $T$ is convex. If $x,y\in T^{\circ }$, choose positive radii $a,b$ with $B_E(x,a),B_E(y,b)\subseteq T$. For $0<t<1$, every vector of norm less than $(1-t)a+tb$ can be written as $(1-t)h+tk$ with $\mathrm{\parallel }h\mathrm{\parallel }<a$ and $\mathrm{\parallel }k\mathrm{\parallel }<b$, by taking $h=k$ to be that vector divided by $(1-t)a+tb$ and multiplied by the respective radius. Convexity therefore gives a ball about $(1-t)x+ty$ contained in $T$; the endpoints are already interior. Thus $T^{\circ }$ is convex.

For each $u\in T^{\circ }$, the ball in step 1.1 is open in $E$ and contained in $T$, hence is contained in $T^{\circ }$. It is consequently an open neighbourhood of $u$ in $T^{\circ }$ and meets only finitely many hyperplanes in the restricted set $\{H\in {\mathcal{H}}^J:H\cap T^{\circ }\mathrm{\ne }\mathrm{\varnothing }\}$. By [F3], the restricted arrangement is locally finite.