Proposition 2

Let $A$ be a non-empty subset of $E$, let $\mathfrak{H}_A$ be the set of hyperplanes $H \in \mathfrak{H}$ containing $A$, let $A'$ be the intersection of the $H \in \mathfrak{H}_A$, and let $F$ be a facet of $E$ open in $A'$ (§ 1, no. 3, Prop. 7). Then $W(A) = W(A') = W(F)$, and $W(A)$ is generated by the reflections with respect to the hyperplanes in $\mathfrak{H}_A$.

We first prove the following assertion:

(I) Let $C$ be a chamber, let $x$ and $y$ be two points of $\overline{C}$ and let $w \in W$ be such that $w(x) = y$. Then $x = y$ and $w$ belongs to the subgroup $W_{\mathfrak{N}}$, where $\mathfrak{N}$ is the set of walls of $C$ containing $x$.

We argue by induction on the length $q$ of $w$ (relative to the set $S$ of reflections with respect to the walls of $C$), the case $q = 0$ being obvious. If $q \geq 1$, there exist a wall $H$ of $C$ and an element $w' \in W$ such that $w = s_H w'$ and $l(w') = q - 1$. Since $l(s_H w) < l(w)$, the chambers $C$ and $w(C)$ are on opposite sides of $H$ by Th. 1 of no. 2. Thus, $\overline{C} \cap w(\overline{C}) \subset H$, so $y \in H$. Thus $y = w'(x)$ and the induction hypothesis implies that $x = y$ and $w' \in W_{\mathfrak{N}}$. Since $y \in H$, it follows that $H \in \mathfrak{N}$, and hence that $w = s_H w' \in W_{\mathfrak{N}}$, completing the proof of (I).

Proof of Theorem 2: this follows from (I) and Lemma 2.

Proof of Proposition 1: we know that two distinct facets are disjoint and have distinct closures (§ 1, no. 2, Cor. of Prop. 3). The equivalence of (i), (ii) and (iii) follows. On the other hand, it is clear that

and assertion (I) shows that (i) $\Longrightarrow$ (vii).

Proof of Proposition 2: let $A''$ be the affine subspace of $E$ generated by $A$. Clearly, $W(A) = W(A'')$. By Prop. 7 of § 1, no. 3, there exists a point $x \in A''$ that does not belong to any hyperplane $H \in \mathfrak{H} - \mathfrak{H}_A$. Let $F_x$ be the facet containing $x$: it is open in $A''$ and Prop. 1 shows that

hence $W(A) = W(A') = W(F_x)$. Replacing $A$ by $F$, we also have

hence the proposition.