Convex hull decomposition

Question

I recently encountered the following problem:
Given (d+2) points $x_1, ..., x_{d+2}$ in $\mathbb{R}^d$.
Is it true that

conv($x_1, ..., x_{d+2}$) = $\bigcup_{i=1}^{d+2}$ conv($x_1, ..., \hat{x_i}, ..., x_{d+2}$),

where the symbol $\hat{}$ means omission ?

I cannot find a counterexample and also cannot prove it. I tried to prove the RHS is convex but cannot proceed after writing down the formula of convex hull.

Any counterexample, proof, or any helpful theorem is welcome.

Answer 1

Yes, that's the content of Carathéodory's theorem.

If $S \subseteq \mathbb R^d$ and $x \in \operatorname{conv}(S)$, then there are $x_1, \dots, x_{d+1} \in S$ such that $x \in \operatorname{conv}(x_1, \dots, x_{d+1})$.