1
$\begingroup$

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...

$\endgroup$
1
$\begingroup$

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 $B(I)=\emptyset$.

$\endgroup$
  • $\begingroup$ Thanks. This is very helpful! $\endgroup$ – Lemma1 Nov 29 '19 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.