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...



Regions of hyperplane arrangements and their faces
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 $B(I)=\emptyset$.


