The context of this question is that one of Hassett's famous compactifications of $M_{0,n}$ by means of weighted stable marked curves. I imagine the answer to my question is well known, but I haven't been able to locate a reference or come up with a proof.
Consider the set of weights $\mathcal{D}_{0,n}=\{(a_1,\dots,a_n): a_i\in\mathbb{Q},\,0<a_i\leq 1\text{ and }\sum a_i>2\}$.
This set admits a chamber decomposition. For our purposes, we will parametrize each chamber as follows. Given a weight $A\in\mathcal{D}_{0,n}$, the (fine) chamber containing it is uniquely determined by the set $$ S_A = \{I\subsetneq[n]:\sum_{i\in I}a_i>1\}. $$ Definition. We say that $A$ and $A'$ live in the same (fine) chamber if $S_A=S_{A'}$.
In this context, my question is the following:
Question: Given two weights $A,B$ such that $S_A\subsetneq S_B$, is it always possible to find weights $A',B'$ in the same chambers as $A,B$, respectively, such that $A'\leq B'$ (this is $a_i'\leq b_i'$ for all $i$)?
Edit: Using induction it is possible to reduce the problem to the case where $|S_A|+1=|S_B|$. In this case there is a unique set $I_0\in S_B\setminus S_A$ such that $\sum_{j\in J}a_j\leq 1$ for every subset $J\subsetneq I_0$. Then, the question becomes whether the wall $W:=W_{I_0}$ defined by $I_0$ is a facet for the chambers associated to $A$ and $B$.
Now, to check $W$ is a facet we need to check that the hyperplane defined by $I_0$ is not redundant for either of the chambers. I think I might have an argument for this, while I convince myself that it is accurate any advice or references are welcome.
