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 $C$.



Chambers of central hyperplane arrangements
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.


