Is the set of collections of directions satisfying the following convexity conditions "geometric"? Let $X_n$ denote the set of all collections $(v_{ij})$ of points on the sphere $S^2$, for $1 \leq i,j \leq n$, $i \neq j$, such that:


*

*$v_{ji} = -v_{ij}$,   

*the origin $O$ in $\mathbb{R}^3$ is in the convex hull of $v_{ab}$, $v_{bc}$ and $v_{ca}$ for all $1 \leq a < b < c \leq n$.  

*the origin $O$ in $\mathbb{R}^3$ is in the convex hull of $v_{ab}$, $v_{bc}$, $v_{cd}$ and $v_{da}$, for all $1 \leq a < b < c < d \leq n$.
and so on.
Let us call conditions 1, 2, etc. the convexity conditions.
Let $C_n$ be the subset of $X_n$ consisting of all collections $(v_{ij})$ arising from configurations $(\mathbf{x}_i)$, $1 \leq i \leq n$, of $n$ distinct points in $\mathbb{R}^3$, using the formulas:  
$$v_{ij} = \frac{\mathbf{x}_j-\mathbf{x}_i}{|\mathbf{x}_j-\mathbf{x}_i|}$$
for $1 \leq i,j \leq n$, $i \neq j$. One can indeed check that elements of $C_n$ satisfy the convexity conditions above.
Let us call the closure $\bar{C}_n$ of $C_n$ the set of "geometric" collections of pairwise directions. My question can now be stated. Is every element of $X_n$ "geometric"? It is easy to show that $\bar{C}_n \subseteq X_n$. My question is whether or not $X_n$ is equal to $\bar{C}_n$.
Note: I have simplified my post, compared to previous versions, as this is the crucial point remaining. I apologize for that. I hope this shorter version would be easier to read.
 A: First, let's show that conditions 3, 4, etc follow from 1 and 2.
For simplicity, I'll only consider "generic solutions"/"general positions": ie no two points are the same, and no four points lie in a plane. Resolving these cases should be a mere technical matter.
For a triple of points in condition 2, there must be a relation $rv_{ij}+sv_{jk}+tv_{ki}=0$ for some $r,s,t>0$. This leads to
$$
v_{ik} = -v_{ki} = r'v_{ij}+s'v_{jk}\quad\text{where}\quad
r'=r/t>0 \text{ and } s'=s/t>0.
$$
Also, note that condition 2 does not actually depend on the ordering $i<j<k$: it applies to distinct $i,j,k$ of any order.
For a quadruplet of points in condition 3, we may use the above result on the triplets $a,b,c$ and $c,d,a$ to obtain
$$
v_{ac}=pv_{ab}+qv_{bc}\quad\text{and}\quad v_{ca}=rv_{cd}+sv_{da}
\quad\text{for some}\quad p,q,r,s>0.
$$
Since $v_{ac}+v_{ca}=0$, this makes
$$
pv_{ab}+qv_{bc}+rv_{cd}+sv_{da}=0
$$
placing the origin in the convex hull. Again, the order of $a,b,c,d$ does not matter.
The same approach could be used to reduce the condition on $n+1$ points to that on $n$ points.

Next, we try to express the $v_{ij}$ in terms of vectors $x_i$. As observed in the problem statement, if we start with vectors $x_i$ and define the derived points
$$
x_{ij}=\frac{x_j-x_i}{|x_j-x_i|}
$$
on the plane, the set of points $v_{ij}=x_{ij}$ satisfy the convexity conditions.
Given a set of points, $v_{ij}$, that satisfy the convexity conditions, I'll construct vectors $x_i$ which result in $v_{ij}=x_{ij}$. 
Note that the vectors $x'_i=sx_i+u$, where $s>0$ and $u$ is a fixed vector, result in $x'_{ij}=x_{ij}$. So given $v_{12}$, we are free to select $x_1=0$ and $x_2=v_{12}$ without loss of generality, ensuring $x_{12}=v_{12}$.
For $k>2$, $x_{1k}=v_{1k}$ and $x_{2k}=v_{2k}$ uniquely determines $x_k$. This is basically just that given $x_1$ and $x_2$ and the directions $x_{1k}$ and $x_{2k}$ towards $x_k$, the point $x_k$ is determined.
So, we now have $x_1, \ldots, x_n$ (uniquely determined up to translation and scaling) resulting in $x_{12}=v_{12}$, $x_{1k}=v_{1k}$, and $x_{2k}=v_{2k}$, and need to show the remaining $x_{ij}=v_{ij}$. 
The point $v_{ij}$ lies on the plane spanned by the vectors $v_{1i}$, $v_{1j}$ since we can write $rv_{ij}+sv_{j1}+tv_{1i}=rv_{ij}-sv_{1j}+tv_{1i}=0$. Similarly, $v_{ij}$ lies on the plane spanned by $v_{2i}$, $v_{2j}$. These two planes intersect in a line, which intersects the sphere in two points, determining $v_{ij}$ uniquely up to change of sign. The requirement that $rv_{ij}-sv_{1j}+tv_{1i}=0$ for $r,s,t>0$, is enough to determine which of the two points.
However, since the $x_{ij}$ also satisfy the convexity conditions, the same result applies to $x_{ij}$: it lies on the same two planes as $x_{1i}=v_{1i}$, etc. So, as the set of points is uniquely determined from $v_{12}$, $v_{1k}$, and $v_{2k}$, and are those are identical for $v$ and $x$, the remaining must also be the same: ie, $v_{ij}=x_{ij}$ for all $i,j$.
