Carathéodory's Theorem (Convex Analysis)

Theorem
If $E\subset \mathbb{R}^\ell$ and $\mathbb{x}\in \mbox{co}(E)$, then $\mathbf{x}$ is a convex combination of affinely independent points of $E$.

In particular, $\mathbf{x}$ is a convex combination of at most $\ell+1$ points of $E$.

Indirect Proof
Since $\mathbf{x}\in \mbox{co}(E)$, $\mathbf{x}$ is a convex combination of points in $E$.

From the definition of convex combination, $\mathbf{x}=\sum^k_{i=1}\gamma_i \mathbf{y}_i$ with $\gamma_i\geq0$, $\sum^k_{i=1}\gamma_i=1$, and $\mathbf{y}_i\in E$.

Let $K\subset \mathbb{N}$ be the set of all possible $k$ such that $\mathbf{x}$ is a convex combination of $k$ elements of $E$.

From the definition of natural numbers, $\mathbb{N}$ is well ordered.

$K$ has a smallest element $k_s$ by the definition of well ordered sets.

Suppose that the set $\lbrace \mathbf{y}_i: i\leq k_s\rbrace$ is affinely dependent.

From the condition for affinely dependent set, there exists a set $\lbrace \alpha_i: i\leq k_s\rbrace$ such that some $\alpha_i>0$, $\sum^{k_s}_{i=1}\alpha_i\mathbf{y}_i=0$, and $\sum^{k_s}_{i=1}\alpha_i=0$.

Pick the smallest $\gamma_j/\alpha_j>0$.

$\displaystyle \mathbf{x}=\left(\sum^{k_s}_{i=1}\gamma_i\mathbf{y}_i\right)+0=\left(\sum^{k_s}_{i=1}\gamma_i\mathbf{y}_i\right)-\left(\frac{\gamma_j}{\alpha_j}\sum^{k_s}_{i=1}\alpha_i\mathbf{y}_i\right)$

Rearrange to get:

$\displaystyle \mathbf{x}=\left(\sum^{k_s}_{i=1}\left(\gamma_i-\frac{\gamma_j}{\alpha_j}\alpha_i\right)\mathbf{y}_i\right)$

Since $\gamma_j>0$ is made the smallest and $\alpha_j>0$ is made the greatest, $\frac{\alpha_i}{\alpha_j}\gamma_j\leq \gamma_j\leq \gamma_i$ for all $i\leq k_s$, thus $\left(\gamma_i-\frac{\gamma_j}{\alpha_j}\alpha_i\right)\geq 0$.

$\left(\gamma_i-\frac{\gamma_j}{\alpha_j}\alpha_i\right)= 0$ for $i=j$.

Let $\gamma'_{i'}\equiv\gamma_i-\frac{\gamma_j}{\alpha_j}\alpha_i$ for $i\neq j$.

We can express $x=\sum^{k_s-1}_{i'=1}\gamma'_{i'} \mathbf{y}_i$, a contradiction to the assumption that $k_s=\min K$.