Is every convex fundamental region regularly open?

Question

Recall that a regular open set is an open set $U$ for which $U = \operatorname{int} \operatorname{cl} U$.

A fundamental region for a finite group $G\leq \mathcal{O}(V)$ is an open set $F\subseteq V$ such that nonidentity elements transport $F$ to a disjoint set and the closures $\{\operatorname{cl}(TF)\mid T\in G\}$ cover $V$.

In general, fundamental regions are not unique (up to conjugation): If we remove any point from $F$, the closure is not affected by this, so the covering property is still met. Unfortunately, $F\setminus \{f\}$ is not regularly open, which I used in an attempted proof that connected fundamental regions of Coxeter groups are unique up to conjugation.

Since many of the „nice“ fundamental regions are

1. connected
2. contractible
3. convex

(ordered by strength), and we can find fundamental regions for cases 1 and 2 which are not regularly open, my question would be:

Is every convex fundamental region regularly open?

Counterexamples

$\mathbb R^2 \setminus \left( \{0\} \times [0,\infty) \right)$ is a fundamental region for the trivial group $\{\mathrm{id}_{\mathbb R^2}\}$ which is contractible, but not regularly open, since it is a dense proper subset.

Answer 1

Every convex open subset of $\mathbb R^n$ is regularly open. The non-obvious part of the proof is the inclusion $\text{int}(\text{cl}(U)) \subset U$. I'll give the proof in $\mathbb R^2$; it's not too hard to generalize to $\mathbb R^n$.

Consider a point $p = (x,y) \in \text{int}(\text{cl}(U))$. Since open boxes are a basis for the topology on $\mathbb R^n$, there exists $r > 0$ such that $(x-r,x+r) \times (y-r,y+r) \subset \text{cl}(U)$. Slice this box in half horizontally and vertically to obtain four quadrants, each of which is a subset of $\text{cl}(U)$:

• the first quadrant $Q_1 = (x,x+r) \times (y,y+r)$
• the second quadrant $Q_2 = (x-r,x) \times (y,y+r)$
• the third quadrant $Q_3 = (x-r,x) \times (y-r,y)$
• the fourth quadrant $Q_4 = (x,x+r) \times (y-r,y)$

Since each of these quadrants is an open set that contains a point of $\text{cl}(U)$, each also contains a point of $U$, say $q_i \in Q_i \cap U$. The line segment $\overline{q_1 q_2}$ is therefore in the convex set $U$, and it crosses the "positive $y$-axis", i.e. it crosses the segment $\{x\} \times (y,y+r)$ at some point contained in $U$, call that point $p_+$. Also, the line segment $\overline{q_3 q_4}$ is in the convex set $U$, and it crosses the "negative $y$-axis", i.e. the segment $\{x\} \times (y-r,y)$, at some point contained in $U$, call that point $p_-$. The whole segment $\overline{p_- p_+}$ is therefore contained in $U$, and that segment contains $p$, proving that $p \in U$.