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 April 20th, 2016, 01:37 PM #1 Newbie   Joined: Apr 2016 From: Pol Posts: 2 Thanks: 0 Faces of a convex set I can't prove the following property of extreme subsets: Let A be a convex subset of $\displaystyle R^n$ 1) Prove that: G(x)={ $\displaystyle z \in$ A :$\displaystyle [x-a(z-x),x+a(z-x)] \subset A$, for some $\displaystyle a>0$} Where G(x) is the intersection of the faces of A I have the following hints: This set is convex, If D is a convex subset of A and if F is a face of A such that ri(D) $\displaystyle \cap$ F is nonempty then D$\displaystyle \subset$F and for every x$\displaystyle \in$A if G is the intersection of the faces of A containing x, then x$\displaystyle \in$ri(G(x)). Where ri means relative interior. April 23rd, 2016, 03:44 AM   #2
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Quote:
 Originally Posted by Int G(x) is the intersection of the faces of A.
what is x then?

besides, if A is bounded and its interior not empty, then the intersection of the faces of A is empty. May 5th, 2016, 12:46 PM #3 Newbie   Joined: Apr 2016 From: Pol Posts: 2 Thanks: 0 G(x) is the intersection of the faces of A which containing x and x is random point which belongs to A. Tags convex, faces, set Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post honzik Geometry 1 April 10th, 2015 12:58 AM sososasa Algebra 6 March 22nd, 2014 09:52 AM cr34m3 Computer Science 1 April 20th, 2010 07:59 AM frederico Real Analysis 0 April 6th, 2009 11:31 AM

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