Weyl group reflects root to a base? I am reading Humphreys' Introduction to Lie Algebras and Representation Theory. In Theorem 10.3(c), it is stated that if $\alpha$ is a root then there exists a Weyl group reflection $\sigma$ such that $\sigma(\alpha) \in \Delta$, where here $\Delta$ is a base for the root system.  The proof begins by selecting $\gamma' \in E$ ($E$ is the vector space for our root system) such that $(\gamma',\alpha) = \epsilon > 0$ and $|(\gamma',\beta)| > \epsilon$ for all other roots $\beta \neq \pm \alpha$. I can follow the proof up to this point. From there, it is then "evident" that $\alpha$ lies in some base $\Delta$ (specifically, in the notation of Humphreys', in $\Delta(\gamma')$).  I don't understand why this is "evident".
 A: Ok, I think figured out the answer to this. 
For $\gamma' \in E$ a regular point, Humphreys' defines 
$$\Phi_+(\gamma') = \{\beta \in \Phi:(\beta,\gamma')>0\}.$$
Thus $(\alpha,\gamma')>0$ shows that $\alpha \in \Phi_+(\gamma')$.  By Theorem 10.1 of Humphreys' book, we know that there exists a base $\Delta(\gamma')$ consisting of roots in $\Phi_+(\gamma')$. Suppose $\alpha \notin \Delta(\gamma')$. Thus 
$$\Delta(\gamma') = \{\beta_1, \ldots, \beta_r\}$$
with $\beta_i \neq \pm \alpha$ (we know $\beta_i \neq -\alpha$ since $-\alpha \notin \Phi_+(\gamma')$). Since $\beta_i \in \Phi_+(\gamma')$, it follows from the statement that $|(\gamma',\beta)|>\epsilon$ for all $\beta \in \Phi$ that we have $(\gamma',\beta_i)>\epsilon$ for all $i$.  Since $\Delta(\gamma')$ is a base, it follows that there exists $k_i \in \mathbb{Z}_{\geq 0}$ such that $\alpha = \sum k_i \beta_i$ with at least one $k_j >0$.  But then 
$$\epsilon = (\gamma',\alpha) \geq (\gamma',\beta_j) > \epsilon,$$
a contradiction.  Thus $\alpha \in \Delta(\gamma')$.
