Proposition 6

Every point of E is in the closure of at least one chamber.

If E reduces to a single point, this is clear. Otherwise, let $a \in E$ and let $H_1, \ldots, H_m$ be the hyperplanes of $\mathfrak{H}$ containing $a$. Since $\mathfrak{H}$ is locally finite, there exists a neighbourhood V of $a$ that does not meet any hyperplane of $\mathfrak{H}$ other than $H_1, \ldots, H_m$. Let D be a straight line passing through $a$ and not contained in any of the $H_i$; if $x \in D, x \neq a$, and $x$ is sufficiently close to $a$, the open segment $]ax[$ is contained in V and does not meet any of the $H_i$. Then $]ax[ \subset U$; since $]ax[$ is connected, it is contained in a chamber C, hence $a \in \overline{C}$.