# Regions of hyperplane arrangements and their faces

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...

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 $$B(I)=\emptyset$$.