I am thinking about this without solutions. I would like you to give hints. Let $Q$ be a polyhedron with $Q=convex.hull (X)$ for some $X \subset R^n$. Let $E$ be a face of $Q$. Prove that $E \cap X \ne \empty$ and that $E=convex.hull (E \cap X)$. Thank you.
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It's general fact that a convex compact closed set in a Hausdorff locally convex topological vector space is generated by its extreme points, this is Krein-Milman theorem.
In particular, your X has to contain all the extreme points of Q, this is by the very definition of extreme points. Now the extreme points for a face( considered as a convex set) is the same as an extreme point for X that lies in the face. Apply Krein-Milman theorem again to the face gives you the desired result.
The wikipedia link for K-M theorem is too short, you may consult chapter section 1.? in
Fundamentals of the theory of operator algebra, Volume 1,by Kadison and Ringrose.
Hua
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Can you show me the page? Thank you. – user130121 Feb 21 '14 at 14:07
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Sure, it's on page 32, theorem 1.4.3. – Hua Feb 21 '14 at 14:39
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However, Krein-Milman theorem needs compact property. Is any face compact? – user130121 Feb 24 '14 at 02:57
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You said polyhedron, so I suppose you mean it's compact. – Hua Feb 24 '14 at 12:51
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Polyhedron means that ${x:Ax \leq b}$ not compact if unbounded. – user130121 Feb 24 '14 at 15:45