Is this coordinate-free definition of convex hull correct? This is just for the convex hull of finitely many points in Euclidean space, without a choice of coordinates. The axiomatization of Euclidean geometry I have in mind is Tarski's axioms, for which betweenness is a primitive notion.


*

*Define the convex $0$-skeleton to be the finite set of points we are starting with.

*Define the convex 1-skeleton to be the set of all points which are between any two points of the convex 0-skeleton.

*Define the convex 2-skeleton to be the set of all points which are between any two points of the convex 1-skeleton.

*...

*Define the convex $n$-skeleton to be the set of all points which are between any two points of the convex $(n-1)$-skeleton.

*...

Then it follows (I believe) from the upper dimension axiom that these sets have to stabilize for some $n$ (i.e. the convex $m$-skeleton equals the convex $n$-skeleton for all $m \ge n$ for some $n$).

Question: Then is the convex hull is just the convex $n$-skeleton for the smallest $n$ for which the convex $m$-skeletons equal the convex $n$-skeleton for all $m \ge n$?
Is there any better coordinate-free definition which doesn't rely upon induction?
A pointer to a reference would be appreciated.
Attempt: Using coordinates, we just have that, given finitely (an integer $p$) many points , their convex hull is the set of all points which are a convex combination of the $p$ points.
Using the coordinate realization/model of Tarski's axioms, betweenness is equivalent to being a convex combination of 2 points (i.e. the point $b$ is between the points $a$ and $c$ if and only if $b$ is a convex combination of $a$ and $c$).
So trying to verify whether this inductive definition is correct comes down to verifying that any convex combination of $p$ points can be constructed inductively from convex combinations of two points at a time. I am pretty certain this follows from distributivity and associativity.
(Not that certain though or else I wouldn't be asking this question.)
 A: This is true, even for infinite sets, as a consequence of Carathéodory's theorem.
Namely, if the dimension of the space is $n$, and $x$ is a point in the convex hull of your set $S$ (finite or not), then you can select points $p_1, p_2, \dots, p_{s} \in S$, $s \leq n + 1$, and constants $a_1, a_2, \dots, a_{s} > 0$, such that
$$x = a_1 p_1 + \dots + a_s p_s, \quad a_1 + \dots + a_s = 1.$$ 
Then you can prove that $\frac{a_1}{a_1 + a_2} p_1 + \frac{a_2}{a_1 + a_2} p_2$ is in your $1$-skeleton (since it is between $p_1$ and $p_2$), that $\frac{a_1}{a_1 + a_2 + a_3} p_1 + \frac{a_2}{a_1 + a_2 + a_3} p_2 + \frac{a_3}{a_1 + a_2 + a_3} p_3$ is in your $2$-skeleton (since it is between $p_3$ and the last point), and so on by induction.
However, you can do better than this. By grouping the $p_i$'s first in twos, then in fours, in eights, etc., you can show that the whole convex hull is in your $k$-skeleton as soon as $2^k \geq n + 1$. So calling it a "$k$-skeleton" is misleading, as its dimension will typically be $2^k - 1$. (For example, the set of points between two opposite edges of a tetrahedron is the tetrahedron.) If you want to have a $k$-dimensional object at each stage, it might be preferable to only add points that are between a point in the $(k-1)$-skeleton and a point that is actually in $S$. The argument using Carathéodory's theorem shows that this works.
If you don't use Caratheodory's theorem, it's only clear that the second construction will terminate after $|S|$ steps, rather than $n + 1$.
