Contractibility and intersection components of chamber open sets

Prove openness from the formulas. For each arrangement hyperplane $H$, a nonzero linear functional defining $H$ is continuous: in finite-dimensional coordinates it is a finite linear combination of the coordinate functions. Thus $H$ is closed and $D_H(C)$ is open. The condition $x\in H\Rightarrow y\in D_H(C)$ describes the open subset $((E\setminus H)\times E)\cup (E\times D_H(C))$ of $E\times E$. The set $X_C$ is the intersection of the complexified complement with these open subsets, one for each member of the finite arrangement. It is therefore open in the complement. For an empty arrangement this intersection is the entire complement, also open.

Contract a chamber set. Fix $c\in C$. For $t\in [0,1]$, put $s(t)=\min (2t,1)$ and $r(t)=\max (2t-1,0)$, and define $K_c(x+iy,t)=((1-r(t))x+r(t)c)+i((1-s(t))y+s(t)c)$. These continuous scalar functions make the formula jointly continuous in the coordinate product topology. It first moves the imaginary coordinate linearly from $y$ to $c$, keeping $x$ fixed, and then moves the real coordinate linearly from $x$ to $c$, keeping the imaginary coordinate equal to $c$. If $x\in H$ during the first motion, both $y$ and $c$ lie in the convex half-space $D_H(C)$, so the moving imaginary coordinate does too. During the second motion the imaginary coordinate $c$ lies in $D_H(C)$ for every $H$, regardless of the real coordinate. Thus both motions stay in $X_C$; in particular they avoid every $H\times H$. With the subspace topology, $K_c$ is a homotopy from the identity to the constant map at $c+ic$. This point belongs to $X_C$, so the space is nonempty; [F4] now gives contractibility.

Split an adjacent intersection. Let $C,C^{\mathrm{\prime }}$ be adjacent and let $H_0$ be their unique separating hyperplane. Write $I=X_C\cap X_{C^{\mathrm{\prime }}}$. In $I$ the real coordinate cannot lie in $H_0$: otherwise the imaginary coordinate would have to lie in both opposite half-spaces $D_{H_0}(C)$ and $D_{H_0}(C^{\mathrm{\prime }})$. For every other hyperplane the two required half-spaces agree. Consequently the points whose real coordinate lies in $D_{H_0}(C)$ are exactly $X_{C,C^{\mathrm{\prime }}}$, and those whose real coordinate lies in $D_{H_0}(C^{\mathrm{\prime }})$ are exactly $X_{C^{\mathrm{\prime }},C}$. Equivalently these pieces are $I\cap (D_{H_0}(C)\times E)$ and $I\cap (D_{H_0}(C^{\mathrm{\prime }})\times E)$. They are disjoint and cover $I$. By step 1.1 and openness of the two half-spaces they are open in the complement, and in particular in $I$.

Connect both pieces. Fix $c\in C$ and restrict $K_c$ from step 1.2 to $X_{C,C^{\mathrm{\prime }}}$. Throughout the deformation the real coordinate remains in the convex half-space $D_{H_0}(C)$, since both its initial value and $c$ are in that half-space. During the first motion, for any other hyperplane containing the fixed real coordinate, the original imaginary coordinate and $c$ lie in the common required half-space, so their line segment does too. During the second motion the imaginary coordinate is $c$, which lies in every common required half-space for $H\mathrm{\ne }{H}_0$. The real coordinate never lies in $H_0$, so that hyperplane imposes no imaginary-coordinate condition. Thus this is a continuous contraction of $X_{C,C^{\mathrm{\prime }}}$ to $c+ic$. Interchanging $C$ and $C^{\mathrm{\prime }}$ gives a contraction of $X_{C^{\mathrm{\prime }},C}$. Both pieces are nonempty, since they contain $c+ic$ and $c^{\mathrm{\prime }}+i{c}^{\mathrm{\prime }}$ respectively for choices $c\in C$ and $c^{\mathrm{\prime }}\in C^{\mathrm{\prime }}$. By [F4] and [F5], each piece is contractible, path-connected and connected.

Identify the components. A connected subset of $I$ cannot meet both pieces: its intersections with the two disjoint open pieces from step 2.1 would give a partition into two nonempty relatively open subsets, contrary to [F6]. For a point in either piece, that entire piece is a connected subset containing the point by step 3.1, while every connected subset containing the point lies in that same piece. Its union, the component in [F6], is therefore exactly that piece. Hence $I$ has exactly the two stated connected components.