Let $\Phi$ be a root system in a finite dimensional Euclidean space.
Let $S\subset \Phi$ a non-empty subset such that if $\alpha$ is in $S$ then $-\alpha$ is not in $S$.
Is it true that there is a base $\Delta$ of $\Phi$ such that each element of $S$ is positive root w.r.t. $\Delta$ (i.e. every $\alpha\in S$ is sum of some members of $\Delta$?)
I was trying to prove this affirmatively, using transitivity of Weyl group on Weyl chambers (and also on bases of $\Phi$), but I didn't get technically the main step in proceeding for the proof.
The answer is very easy, which I initially didn't think; however, I came to the question when understanding some intermediate step in the proof of conjugacy of Borels in the book of Humphreys.
Consider root system $A_2$; it has six vectors lying on corners of regular hexagon. Take $S$ be the set of three vectors which are inclined at angle 120 from each other. This set $S$ can not be positive w.r.t. any base, because there is no line through origin which keeps vectors of $S$ on one side.
After relooking in book, I came to know that with some assumption on question, the answer would become affirmative. It is an exercise in Humphreys (after proof of conjugacy of Borels).
If $\Psi$ is a subset of root system $\Phi$ such that
$\alpha,\beta\in\Psi$ and $\alpha+\beta$ is root implies $\alpha+\beta\in \Psi$, and
$\Psi$ is disjoint from $-\Psi$
then $\Psi$ is contained in $\Phi^+$ (positive root system) w.r.t. a base $\Delta$.