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can anybody help me please? Is there a good way to prove that given a set of points, say $S = \{x_1, x_2, ..., x_n\}$, then show that the convex hull of $S$, that is, $conv(S)$ contains all the extreme points in $S$?

Is this equivalent to saying that taking the convex hull of the set does not add any extra extreme points?

Thanks a lot.

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  • $\begingroup$ $conv(S)$ contains all element of $S$, including the extreme points in $S$. Perhaps you mean that the extreme points of $conv(S)$ are in $S$? $\endgroup$ – lhf Jul 25 '14 at 10:08
  • $\begingroup$ Yes, sorry for not stating it clear enough. So I believe I do mean that. So for example, given $S$ same as defined above, maybe say $\{x_1, x_2\}$ are extreme points. Then $conv(S)$ will also only have extreme points $\{x_1, x_2\}$. $\endgroup$ – TLR Jul 28 '14 at 1:46
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The easiest way to prove your first question is to show the characterization $$\mathrm{conv}(S) = \bigg\{\sum_{i=1}^n \alpha_i \, x_i : \sum_{i=1}^n \alpha_i = 1, \alpha_i \ge 0\bigg\}.$$ That is, the convex hull just consists of all convex combinations of elements from $S$. Using this characterization, one can easily show that the extreme points of $\mathrm{conv}(S)$ belong to $S$.

To address your second question: as far as I known, extreme points are only defined for convex sets.

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