Let $\mathcal{A}$ be a central hyperplane arrangement in a (finite dimensional) real vector space $V$. Assume for each hyperplane $H\in\mathcal{A}$ that we're given a labelling $H^+$, $H^-$ of the connected components of $V\setminus H$.

  1. Given a subset $\mathcal{B}\subseteq \mathcal{A}$, is it possible for the set $$ \bigcap_{H\in \mathcal{B}} H^+ \cap \bigcap_{H\in \mathcal{A}\setminus\mathcal{B}} H^- $$ to be empty while the set $$ \bigcap_{H\in \mathcal{B}} H^+ \cap \bigcap_{H\in \mathcal{A}\setminus\mathcal{B}} (H\cup H^-) $$ is not?

  2. If 1. is false in general, what if we restrict to the following case?: Let $C$ be a strongly convex full-dimensional polyhedral cone in $V$. Set $\mathcal{A}=\{\mathbb{R}F\mid F\text{ a facet of }C\}$, and for each $H\in \mathcal{A}$, let $H^+$ be the connected component of $V\setminus H$ containing the interior of $C$.


2 Answers 2

  1. Yes: let $V=\mathbb{R}$, let $\mathcal{B}$ be empty, and let $\mathcal{A}$ be two copies of the origin, with $H^+$ being the positive numbers once, and the negative numbers once (this is a pretty degenerate example, but you can also get 3 hyperplanes in $\mathbb{R}^2$ giving you a line, etc).

  2. Yes: the problem with the example above is that taking the closure of these half-spaces doesn't necessarily commute with intersection. The set $D=\bigcap_{H\in \mathcal B} H^+$ is open, so its complement is closed. So, if the first intersection is empty, then $E=\bigcap_{H\in \mathcal A\setminus \mathcal B} H^-$ is contained in the complement of $D$, so its closure is as well. If $E$ contains a point, then $E'=\bigcap_{H\in \mathcal A\setminus \mathcal B} (H\cup H^-)$ will be its closure (by convexity: fix any point in $E$, and consider the straight path to any point of $E'$. All but one point of this segment lies in $E$, so this shows that the point in $E'$ is in the closure of $E$). But there are situations where $E$ is empty, but $E'$ is not.

  • $\begingroup$ Which open set are you referring to in the phrase "is contained in this open set"? $\endgroup$ Aug 6, 2016 at 0:07
  • $\begingroup$ Sorry, got mixed up there. I meant the complementary closed set. $\endgroup$
    – Ben Webster
    Aug 6, 2016 at 0:08
  • $\begingroup$ So, if I understand your argument correctly, then 1. is true under the additional assumption that your set $C$ is non-empty. Is this right? $\endgroup$ Aug 6, 2016 at 0:14
  • $\begingroup$ Also, do you have an example of a cone $C$ and (using my notation from 2.) a subset $\mathcal B\subseteq \mathcal A$ such that $\bigcap_{H\in\mathcal A\setminus\mathcal B} H^-=\emptyset$? $\endgroup$ Aug 6, 2016 at 0:18
  • $\begingroup$ I just realized that for 2. (as I originally wrote it and using my notation), $-\operatorname{Int}(C)$ is always contained in $\bigcap_{H\in\mathcal{A}\setminus\mathcal{B}} H^-$, and therefore the answer to 2. is in fact No. $\endgroup$ Aug 6, 2016 at 22:46

This is a follow-up/correction to Ben's answer.

Based on Ben's answer, it seems that the answer to 2. (as I originally wrote it) is no:

By Ben's answer, it suffices to show that $\bigcap_{H\in\mathcal{B}}H^-\neq\emptyset$ for every $\mathcal{B}\subseteq\mathcal{A}$. In fact, it's enough to show that this is true when $\mathcal{B}=\mathcal{A}$. But since we're dealing with a (full-dimensional) cone $C$, the intersection of all the $H^-$'s is exactly $-\operatorname{Int}(C)$. As this is non-empty, we win.


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