Finite galleries and separating hyperplanes

If $T$ is empty there are no chambers, so the assertions about chambers and galleries are vacuous. If $\dim E=0$ and $T$ is nonempty, then $T=E=\{0\}$, there are no hyperplanes, and there is just one chamber. Its one-chamber gallery has length zero, and no adjacency of distinct chambers is possible. All conclusions follow. Henceforth assume $T\mathrm{\ne }\mathrm{\varnothing }$ and $n=\dim E\ge 1$, and use the coordinate Euclidean norm $\mathrm{\parallel }\cdot \mathrm{\parallel }$ on $E$.

We will use the following interpolation fact, proved here. If $V\subseteq E$ is nonempty, open and convex, $z\in \bar{V}$, $b\in V$ and $0<t\le 1$, then $(1-t)z+tb\in V$. The case $t=1$ is immediate. For $0<t<1$, take $r>0$ with $B(b,r)\subseteq V$ and, by the definition of closure, take $v\in V$ with $\mathrm{\parallel }v-z\mathrm{\parallel }<tr\mathrm{/}(1-t)$. Then $b^{\mathrm{\prime }}=b+(1-t)(z-v)\mathrm{/}t$ lies in $B(b,r)$, and $(1-t)z+tb=(1-t)v+t{b}^{\mathrm{\prime }}\in V$ by convexity. As $t$ tends to zero these points approach $z$, since their distance from $z$ is $t\mathrm{\parallel }b-z\mathrm{\parallel }$.

Let $e_1,\dots ,e_n$ be the coordinate basis. For each hyperplane $H$, let $j$ be the least index for which $e_j\mathrm{\notin }H$. Since $E\mathrm{/}H$ is one-dimensional, every $z\in E$ has a unique expression $z=h+t{e}_j$ with $h\in H$ and $t\in \mathbb{R}$. Set ${\alpha }_H(z)=t$. Uniqueness gives linearity, ${\alpha }_H(e_j)=1$, and $\ker {\alpha }_H=H$. With $a_i={\alpha }_H(e_i)$ and $z={\sum }_i{z}_i{e}_i$, we have ${\alpha }_H(z)={\sum }_i{a}_i{z}_i$ and $\mathrm{\mid }{\alpha }_H(z)-{\alpha }_H(w)\mathrm{\mid }\le ({\sum }_i\mathrm{\mid }{a}_i\mathrm{\mid })\mathrm{\parallel }z-w\mathrm{\parallel }$. The positive constant ${\sum }_i\mathrm{\mid }{a}_i\mathrm{\mid }$ therefore gives continuity. The preimage of $\{0\}$ is closed, and the two strict sign sets are open half-spaces: for a nonzero value, a ball of radius less than $\mathrm{\mid }{\alpha }_H(z)\mathrm{\mid }\mathrm{/}({\sum }_i\mathrm{\mid }{a}_i\mathrm{\mid })$ preserves its sign. Thus $H$ is closed. The least-index rule specifies these forms without an arbitrary family of choices.

Every convex subset $S\subseteq E$ is connected. Indeed, for $u,v\in S$ the segment map $\gamma :[0,1]\to E$, $\gamma (t)=(1-t)u+tv$, has image in $S$, and $\mathrm{\parallel }\gamma (t)-\gamma (s)\mathrm{\parallel }=\mathrm{\mid }t-s\mathrm{\mid }\mathrm{\parallel }v-u\mathrm{\parallel }$ proves continuity. Its image is connected by [F6] and [F7]. If $S$ had a separation into two nonempty relatively open parts, choosing $u$ and $v$ in different parts would give a separation of this segment by their traces, a contradiction. Empty sets and singletons have no such separation. This applies in particular to balls, half-balls, and the convex sign sets below.

Local finiteness and closedness of the hyperplanes show that their union is relatively closed in $T$: near a point outside the union, take an open neighbourhood meeting only finitely many hyperplanes and remove those closed sets. The complement is therefore open in $E$, since $T$ is open. Any connected subset of the complement has a constant sign for every ${\alpha }_H$, since the two open half-spaces would otherwise separate it. Conversely, every realizable sign set in $T$ is convex, hence connected by step 2.2, and lies in the complement. Thus chambers are exactly these sign sets and are convex. Each point of the complement has a ball inside it; that ball is connected and hence lies in the point's chamber. Consequently chambers are open.

Choose $x\in C$ and $y_0\in D$ for any two chambers $C,D$. Their closed bounded segment $K=[x,y_0]$ lies in $T$. In coordinates it is compact by [F4], with $n\ge 1$ by step 1.1; transporting a cover along the coordinate isometry gives compactness in $E$. Local finiteness and openness of $T$ cover $K$ by open subsets of $E$, contained in $T$, each meeting only finitely many arrangement hyperplanes. The ambient-cover assertion of [F4] gives a finite subcover. Its union is an open neighbourhood $U\subseteq T$ of $K$ meeting only finitely many hyperplanes. Cover $K$ also by all balls $B(p,r\mathrm{/}2)$ with $p\in K$, $r>0$, and $B(p,r)\subseteq U$. The same ambient-cover assertion yields finitely many such balls. The minimum $\epsilon >0$ of their half-radii ensures that every point at distance less than $\epsilon$ from some point of $K$ lies in $U$, by the triangle inequality. Since $\mathrm{\parallel }(1-t)x+ty-((1-t)x+t{y}_0)\mathrm{\parallel }\le \mathrm{\parallel }y-y_0\mathrm{\parallel }$ for $0\le t\le 1$, every $y\in D$ sufficiently close to $y_0$ has its whole segment from $x$ inside $U$.

Suppose distinct chambers $C,D$ share a wall $W=\bar{C}\cap H=\bar{D}\cap K$. Since $W$ has affine dimension $n-1$ and is contained in both hyperplanes, its affine hull is both $H$ and $K$, so $H=K$. The set $\bar{C}$ is convex: given $u,v\in \bar{C}$, $0\le s\le 1$ and $\delta >0$, choose $c,d\in C$ within $\delta$ of $u,v$, respectively. Then $(1-s)c+sd\in C$ by step 3.1 and is within $\delta$ of $(1-s)u+sv$, proving that this last point is in $\bar{C}$. Hence $W$ is convex. Put $d=n-1$. Starting with $w_0\in W$, repeatedly choose a point of $W$ outside the affine span of the points already chosen until it is $H$; this gives $w_0,\dots ,w_d\in W$ with $w_i-w_0$ a basis of $H$. For $d>0$, the set $O=\{w_0+{\sum }_{i=1}^d{\lambda }_i(w_i-w_0):{\lambda }_i>0,{\sum }_{i=1}^d{\lambda }_i<1\}$ is contained in $W$ by repeated convex combinations. The coefficient functions ${\lambda }_i$ on $H$ are continuous: extend the basis of $H$ to one of $E$; each coefficient of $z-w_0$ is a linear combination of its Euclidean coordinates, hence an affine function of $z$ whose continuity follows from the finite-sum bound of step 2.1. Thus the strict inequalities define a relatively open subset of $H$, nonempty because it contains the barycentre of $w_0,\dots ,w_d$. For $d=0$, take $O=W=H=\{0\}$. This proves locally that $W$ contains a nonempty relatively open subset $O$ of $H$.

Choose $q\in H\cap T$, which exists because $H$ is an arrangement hyperplane. For every $z\in W$, we have $z\in \bar{T}$, so step 1.2 with $V=T$ and $b=q$ gives $(1-t)z+tq\in H\cap T$ for $0<t\le 1$. These points approach $z$. Therefore $W$ is contained in the closure in $H$ of $H\cap T$. In particular the nonempty relatively open set $O\subseteq W$ of step 4.2 meets $H\cap T$ by the definition of closure. Thus $O\cap T$ is nonempty and relatively open in $H$. Choose $a\in O\cap T$ and an open neighbourhood $N\subseteq T$ of $a$ meeting only finitely many arrangement hyperplanes. Each such hyperplane other than $H$ intersects $H$ in a proper linear subspace. If there are any, [F5] gives a direction in $H$ outside their union; a sufficiently short line through $a$ in this direction stays in $O\cap N$ and meets each excluded subspace in at most one parameter, so choose a parameter avoiding them all. If there are none, use $a$ itself. This yields $p\in O\cap T$ lying on no arrangement hyperplane other than $H$. When $\dim H=0$, there is no other distinct linear hyperplane in $E$, so this is precisely the latter case. Shrinking inside $N$ and avoiding its finitely many other closed hyperplanes gives a ball about $p$ contained in $T$ and meeting no arrangement hyperplane except $H$.

Let $\mathcal{B}$ be the finite set of hyperplanes meeting $U$. For every distinct pair $H,K\in \mathcal{B}$, their intersection has codimension two. Since $x$ is in no arrangement hyperplane, the affine span of $x$ and $H\cap K$ is a proper affine hyperplane. Avoid the finitely many such affine hyperplanes when choosing $y$ close to $y_0$ in $D$; this is possible by applying [F5] to their direction subspaces and moving on a short line in a direction outside their union, avoiding finitely many intersection parameters. In dimension less than two there are no distinct pairs to avoid. The segment from $x$ to $y$ then meets no two distinct arrangement hyperplanes simultaneously. It meets each member of $\mathcal{B}$ at most once, because its endpoints avoid that hyperplane and a nonzero affine linear function on a segment has at most one zero.

The ball from step 5.1 meets both $C$ and $D$, since $p\in W$. Each of its two half-balls is connected by step 2.2 and lies in the complement, hence is contained in a single chamber. Each chamber lies on one side of $H$ by step 3.1, so the distinct chambers $C,D$ occupy opposite half-balls. Every other arrangement hyperplane misses the entire ball and has constant sign on it. Their signs therefore differ at $H$ and agree at every other hyperplane. The argument applies to any shared wall, so its defining hyperplane is the only sign change and is unique for this adjacency.

List the finitely many crossing points in order. Between successive crossing points the segment lies in a chamber by step 3.1. At a crossing point $p\in H$, local finiteness and the absence of a simultaneous crossing give a ball in $T$ meeting no arrangement hyperplane except $H$. Its two half-balls are connected by step 2.2, and contain the two neighbouring segment pieces, so each is contained in the corresponding chamber, say $A$ and $B$. These chambers are distinct because they lie on opposite sides of $H$. Their signs agree at every other arrangement hyperplane, which misses the entire ball. The entire slice of the ball by $H$ lies in both chamber closures.

To prove equality of the full wall intersections, take $z\in \bar{A}\cap H$ and $b\in B$. Then $z\in \bar{T}$, and step 1.2 puts $z_t=(1-t)z+tb$ in $T$ for $0<t\le 1$. For each arrangement hyperplane other than $H$, continuity of its defining form gives at $z$ the weak sign of $A$, which is the weak sign of $B$ by step 6.2; its value at $b$ has the strict sign of $B$. Linearity therefore gives the strict sign of $B$ at $z_t$. For $H$ the value at $z$ is zero, so its value at $z_t$ also has the strict sign of $B$. Thus $z_t\in B$ by step 3.1, and $z_t\to z$ proves $z\in \bar{B}\cap H$. Interchanging $A,B$ proves $\bar{A}\cap H=\bar{B}\cap H$. This intersection contains the relatively open ball slice of step 6.2, so its affine dimension is $n-1$: the slice contains $n$ affinely independent points in $H$ (one point when $n=1$), and the intersection is contained in $H$. Consequently it is a wall of both chambers under [F1]. The successive chambers therefore form a finite gallery. A hyperplane is crossed by this segment exactly when its endpoint signs differ, and then it is crossed once by the affine linearity from step 5.2. Thus this gallery crosses precisely the hyperplanes separating $C$ and $D$, once each; in particular their number is finite.

For any finite gallery from $C$ to $D$, step 6.1 shows that each adjacency changes exactly the sign of its defining hyperplane. Hence every separating hyperplane must be crossed an odd, and therefore positive, number of times, while every other hyperplane is crossed an even number of times. Every gallery has length at least the number of separating hyperplanes. The gallery of step 7.1 attains that bound. Equality holds exactly when each separating hyperplane is crossed once and no other hyperplane is crossed. Since $\mathcal{A}$ is a set of hyperplanes, coincident hyperplanes contribute only once.