If $\Phi$ is the root system of a complex semisimple Lie algebra $\mathfrak{g}$, then one way to choose a set $\Phi^+$ of positive roots is by choosing a hyperplane which does not contain any root, which will partition (by looking at the corresponding two half-spaces) the set of roots $\Phi$ into two subsets. Either one of these two subsets can be taken to be $\Phi^+$.
My question is about the converse. Can any set $\Phi^+$ of positive roots be obtained by this construction?
I am mostly interested in $A_n$ but the question makes sense in general.
Let $V$ be the real vector space spanned by the root system, and let $L: V \rightarrow \mathbb R$ be any linear functional with the property that $L(\Phi^+) \subset \mathbb R_{> 0}$; then $\ker(L)$ is a hyperplane that does the job.
One obvious choice for $L$ is as follows: From your $\Phi^+$, choose a set of simple roots $\alpha_i$, $1 \le i \le n$. It is well known that the $\alpha_i$ are a basis of $V$, and a root $\alpha = \sum_{i=1}^n c_i \alpha_i$ is in $\Phi^+$ if and only if all $c_i \in \mathbb Z_{\ge 0}$. Now set $L(\sum_{i=1}^n c_i \alpha_i) := \sum_{i=1}^n c_i$.
(With respect to the basis $(\alpha_i)_i$, this is the functional $(1,1, ...,1)^{tr}$. Notice that because of the fact that for all roots, the coefficients $c_i$ are either all $\ge 0$ or all $\le 0$, one actually has a lot of "wiggle room": any $(\ell_1, ..., \ell_n)^{tr}$ with all $\ell_i > 0$ gives another suitable hyperplane.)