Regions of hyperplane arrangements and their faces

Question

Consider a finite hyperplane arrangement $\mathcal{A}$ over $\mathbb{R}^n$. Let the regions given by $\mathcal{A}$ be $\mathcal{R}(\mathcal{A})=\{A_1,\dots A_m\}$ for some $m$.

For any index set $I\subseteq \{1,\dots,m\}$, is it true that $$ B(I)=\left(\bigcap_{i\in I} \overline{A_i}\right) \backslash \left(\bigcup_{i\not\in I}\overline{A_i}\right), $$ is convex?

It is easy to see that $\bigcap_{i\in I}\overline{A_i}$ is convex. But I haven't found a way to show that $B(I)$ is convex. I have to say that I am rather unexperienced in geometry...

Answer 1

Choose two points $x,y\in B(I)$ and a point $z$ on the segment $xy$. We should prove that $z\in B(I)$. It reads as

(i) $z\in \overline{A_i}$ for all $i\in I$; and

(ii) $z\notin \overline{A_j}$ for $j\notin I$.

(i) follows from $x,y\in \overline{A_i}$ and the fact that $\overline{A_i}$ is convex.

For proving (ii), assume that $j\notin I$, but $z\in \overline{A_j}$. The set $\overline{A_j}$ is the intersection of closed subspaces $S_1,S_2,\ldots,S_n$, where $\partial S_i=H_i$ and $H_1,\ldots,H_n$ are our hyperplanes. Note that $x$ does not belong to $\overline{A_j}$, that is, for some index $\alpha$ we have $x\notin S_{\alpha}$. But $z\in S_\alpha$, thus $y$ lies in the interior of $S_{\alpha}$. Therefore $x,y$ lie in different open half-spaces with common boundary $H_\alpha$. It implies that $x,y$ can not belong to the closure of the same region. Therefore $I=\emptyset$ and $B(I)=\emptyset$.