Interior restriction preserves chambers and adjacency

By [F3], the complements in $T$ and $T^{\circ }$ are the same topological space. Their connected components, hence their chambers, are the same subsets of $E$. Their closures in $E$ and their affine dimensions therefore also agree.

The interior arrangement has open convex chambers. First put $n=\dim E$. If $n=0$, there are no hyperplanes, and $T^{\circ }$ is empty or a singleton; its chambers, if any, are open and convex directly. For $n\ge 1$, use the coordinate isomorphism $c:E\to {\mathbb{R}}^n$ and transported Euclidean norm $\mathrm{\parallel }\cdot \mathrm{\parallel }$ of [F2]. Each relative hyperplane is the kernel of a nonzero root-evaluation form $\mathrm{\ell }:E\to \mathbb{R}$ from [F1]. In coordinates write $\mathrm{\ell }(x)={\sum }_{i=1}^n{a}_{i}c(x{)}_i$ and $A={\sum }_{i=1}^n\mathrm{\mid }{a}_i\mathrm{\mid }>0$. Then $\mathrm{\mid }\mathrm{\ell }(x)-\mathrm{\ell }(y)\mathrm{\mid }\le A\mathrm{\parallel }x-y\mathrm{\parallel }$. At a point $x$ with $\mathrm{\ell }(x)\mathrm{\ne }0$, the ball of radius $\mathrm{\mid }\mathrm{\ell }(x)\mathrm{\mid }\mathrm{/}(2A)$ preserves its strict sign. Thus each hyperplane is closed in $E$, its strict half-spaces are open, and its trace on $T^{\circ }$ is closed in that subspace. These traces form a locally finite family by [F4]. Their union is therefore closed in $T^{\circ }$ by [F6]. Write $Q$ for its complement in $T^{\circ }$; since $T^{\circ }$ is open in $E$, so is $Q$.

The cone $T=U\cap E$ is convex: a convex combination of two points of $T$ belongs to $U$ by [F8] and to the linear subspace $E$ by [F1]. Its interior in $E$ is convex as well. Indeed, for $x,y\in T^{\circ }$, choose $r>0$ such that $B_E(x,r)$ and $B_E(y,r)$ are contained in $T$, using the coordinate Euclidean balls of [F2]. For $0\le t\le 1$ and $h\in E$ with $\mathrm{\parallel }h\mathrm{\parallel }<r$, convexity of $T$ gives $(1-t)(x+h)+t(y+h)=(1-t)x+ty+h\in T$. Thus $B_E((1-t)x+ty,r)\subseteq T$, so $(1-t)x+ty\in T^{\circ }$. If $T^{\circ }$ is empty the assertion is vacuous; this argument also applies when $\dim E=0$.

Continue with $n\ge 1$. The inequalities ${\max }_i\mathrm{\mid }{v}_i\mathrm{\mid }\le ({\sum }_i{v}_i^2{)}^{1\mathrm{/}2}\le \sqrt{n}{\max }_i\mathrm{\mid }{v}_i\mathrm{\mid }$ show that the Euclidean coordinate topology agrees with the product topology used in [F7]. Thus [F7] applies through $c$ to every convex subset of $E$. A connected subset of $Q$ has a constant strict sign for each root-evaluation form: otherwise the two relatively open sign parts would separate it. Conversely, the subset of $T^{\circ }$ with any fixed realizable set of strict signs is convex by linearity and step 1.3, is connected by [F7], and lies in $Q$. Every connected component of $Q$ is contained in one such region, and maximality makes it equal to that connected region. Hence the chambers are convex. At any point of $Q$ there is an $E$-ball contained in $Q$ by step 1.2. This ball is convex and connected by [F7], so all its points have the signs of its centre and lie in that centre's chamber. Hence every chamber is open. Together with the direct zero-dimensional case, this proves the assertion of step 1.2.

A shared wall in the interior arrangement is also a shared wall in the full arrangement, because restriction only discards hyperplanes and the chambers and their closures agree. Conversely, suppose distinct chambers $C,D$ share a full-arrangement wall $W$ supported by $H\in {\mathcal{H}}^J$. If $\dim E=0$ there are no hyperplanes and no adjacency, so assume positive dimension. The open convex sets $C,D$ are full-dimensional by steps 1.2 and 2.1, and $W$ has dimension $\dim E-1$. They must lie on opposite sides of $H$: if they lay on the same side, choose a point in the relative interior of the convex set $W$ and points of $C,D$. Convexity gives, at every sufficiently small common positive distance from $H$, relatively open slices of both chambers containing a common neighbourhood of that wall point; the chambers would intersect. A segment joining a point of $C$ to a point of $D$ therefore meets $H$, and it lies in $T^{\circ }$ by step 1.3. Thus $H$ is retained by the restriction, and the same set $W$ is a shared interior wall.

The interior is open by definition and convex by step 1.3, and its arrangement is locally finite by [F4]. Applying [F5] gives a finite gallery between any two interior chambers. Together with steps 1.1 and 3.1, this proves the stated chamber and adjacency identifications and connectivity.