I'm working through some discrete geometry and am trying to get a more rigorous grasp of convex geometry. In particular, I would like to try proving this theorem without the use of Radon's Lemma, which is the classic route. Is this proof valid?
Carathéodory's Theorem: Given a point set $X \subset \mathbb{R}^d$, any point in conv$(X)$ may be expressed as a convex combination of at most $d+1$ points in $X$.
Proof: Let $X$ be as in the statement; if $|X| \leq d+1$, we are done, so assume that $|X| > d+1$, and assume for contradiction that $c \in$ conv$(X)$ may not be expressed as a convex combination of $d+1$ points in $X$. Since the convex hull are those points which may be expressed as convex combination, we know that $c$ is a convex combination of $q>d+1$ points, say $\{x_i\}_{i=1}^{q} = Z \subseteq X$.
As the largest affine independent point set in $\mathbb{R}^d$ is of size $d+1$, we know that $c$ may be expressed as an affine combination of $d+1$ points $D$ in $Z$, since the convex hull is a subset of the affine hull. That is:
$$c = \sum_{n=1}^{d+1} b_n x_{i_n}$$
where $\sum_{n=1}^{d+1} b_n = 1$ and the $b_i \in \mathbb{R}$. By the supposition, at least one of the $b_i$ must be negative; suppose that it is $b_1$. I claim we can find a convex set which includes $D$ and excludes $c$. Let $\lambda_1 = b_1/2$ and let:
$$\lambda_n = b_n + \frac{b_1}{2d}$$
It follows that $\sum_{n=1}^{d+1} \lambda_n = 1$. Let $l = \sum_{n=1}^{d+1} \lambda_n x_{i_n}$; I claim that the set described above is conv$(D \cup\{l\}) = K$. It is clear that $K \supseteq$ conv$(D)$, we must show that $c \notin K$. Indeed, assume for contradiction that $c \in K$, which implies $(c+l)/2$ is a convex combination of $K$. This means that $c+l = 2\sum_{n=1}^{d+1} \Lambda_n x_{i_n} + 2\Lambda_{d+2}l$, where $\sum \Lambda_i =1$ and $\Lambda_i \geqslant 0$. Then:
$$c + (1-2\Lambda_{d+2})l = 2\sum_{n=1}^{d+1} \Lambda_n x_{i_n}$$
which implies:
$$\frac{b_1 (3-2\Lambda_{d+2})}{2} = 2\Lambda_1$$
but by the convexity condition, $0 \leqslant \Lambda_{d+2} \leqslant 1$, and the supposition gives $b_1 < 0$, hence we have:
$$0 > \frac{b_1}{2} \geqslant \frac{b_1 (3-2\Lambda_{d+2})}{2} = 2\Lambda_1 \geqslant 0$$
which is a contradiction. Then $c \notin K$, where $K$ contains $D$. In other words, there is a small-enough convex set that includes any $D$ but not $c$; by definition $c$ is not in the convex hull of $D$. Any decomposition of $\text{conv}(X)$ into simplices shows that this is impossible ($c$ must be in some simplex), so we are done. $\square$
EDIT: For clarity on my main concern, I do not know if the last sentence is justified, or is in fact identical to Caratheodory's Theorem in a different wording. Am I begging the question?
Best Answer
Carathéodory's theorem can be rephrased as "given a set $X \subseteq \mathbb{R}^d$, the set of simplices $\{\text{conv}(Y) : Y \subseteq X, |Y| \leq d+1\}$ covers $\text{conv}(X)$", i.e. the union if these simplices is exactly $\text{conv}(X)$. So it seems, indeed, that you use the result that you are trying to prove.