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Given an arbitrary set $A\subset\mathbb R^n$, why do all extreme points of the convex hull $\operatorname{conv}(A)$ lie in $A$?

(An extreme point of a convex set is defined as one that cannot be written as a strictly convex combination of two distinct points of this set.)

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closed as off-topic by Claude Leibovici, Davide Giraudo, user91500, JonMark Perry, user223391 May 25 '17 at 0:51

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An element $x\in\operatorname{conv}A$ if and only if there are a positive natural number $n$, elements $x_1,\cdots,x_n\in A$ and non-negative real numbers $t_1,\cdots,t_n$ such that $$\begin{cases}x=\sum\limits_{k=1}^n t_kx_k\\ \sum\limits_{k=1}^n t_k=1\end{cases}$$

We can assign to each $x\in\operatorname{conv}A$ the least natural number $n(x)$ such that some $x_1,\cdots, x_{n(x)}$ and $t_1,\cdots,t_{n(x)}$ as above exist. Notice that $x\in A\iff n(x)=1$.

Suppose $n(x)\ge 2$. Then, $$x=t_1x_1+\sum_{k=2}^{n(x)} t_kx_k=t_1x_1+(1-t_1)\sum_{k=2}^{n(x)}\frac{t_k}{1-t_1}x_k$$ and by minimality all the $t_i$-s are $> 0$. Notice that, by hypothesis $1-t_1=\sum\limits_{k=2}^{n(x)}t_k>0$ and, thus $$x':=\sum_{k=2}^{n(x)}\frac{t_k}{1-t_1}x_k\in\operatorname{conv} A$$

and $x$ is a strictly convex combination of $x_1$ and $x'$. If $x'\ne x_1$, then $x_1$ is not extremal. On the other hand, if $x'=x_1$, then $x=x_1\in A$, against the assumption $n(x)\ge2$.

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Let $x\in \text{conv}(A)$ be extreme. By definition of the convex hull, $x=\sum_i \lambda_i a_i$ for $a_i\in A$, $\sum_i \lambda_i=1$, and $\lambda_i\ge 0$. Since $A\subset\text{conv}(A)$ and $x$ was extreme, only one $\lambda_i$ can be nonzero (and it must be equal to $1$), so $x=a_i$ for this $i$. Thus $x\in A$.

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