Lemma 3

Let $C$ be a chamber of $\mathbf{R}$ and $P$ a closed subset of $\mathbf{R}$ containing $R_+(C)$ (in the notation of no. 6). Let $\Sigma = B(C) \cap (-P)$, and let $Q$ be the set of roots that are linear combinations of elements of $\Sigma$ with non-positive integer coefficient. Then, $P = R_+(C) \cup Q$.

It is enough to show that $P \cap (-R_+(C)) = Q$. Let $-\alpha \in Q$. Then $\alpha$ is the sum of $n$ elements of $\Sigma$. We show, by induction on $n$, that $-\alpha \in P$. This is clear if $n = 1$. If $n > 1$, then by Prop. 19 of no. 6 we can write $\alpha = \beta + \gamma$ with $\gamma \in \Sigma$ and $\beta$ the sum of $n - 1$ elements of $\Sigma$. By the induction hypothesis, $-\beta \in P$; since $-\gamma \in P$ and since $P$ is closed, $-\alpha \in P$. Thus, $Q \subseteq P \cap (-R_+(C))$. Conversely, let $-\alpha \in P \cap (-R_+(C))$. Then $\alpha$ is the sum of $p$ elements of $B(C)$. We show, by induction on $p$, that $-\alpha \in Q$. This is clear if $p = 1$. If $p > 1$, then by Prop. 19, we can write $\alpha = \beta + \gamma$ with $\gamma \in B(C)$ and $\beta$ a root that is the sum of $p - 1$ elements of $B(C)$. Since $-\gamma = \beta + (-\alpha)$ and since $P$ is closed, $-\gamma \in P$, hence $\gamma \in \Sigma$. Moreover, $-\beta = \gamma + (-\alpha)$ so $-\beta \in P$ since $P$ is closed. By the induction hypothesis, $-\beta \in Q$, so $-\alpha = -\beta - \gamma \in Q$. Thus, $P \cap (-R_+(C)) \subseteq Q$.