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Let $C$ be a compact convex subset of a finite-dimensional real vector space $V$ with non-empty interior (where $V$ is equipped with the unique Hausdorff linear topology, i.e. with the standard topology on $\mathbb{R}^n \cong V$). Is the boundary of $C$ given by the union of all proper faces of $C$?

Recall that a convex subset $F$ of $C$ is a face of $C$ if $\lambda x + (1-\lambda) y \in F$ for some $x, y \in C$ and for some $0 < \lambda < 1$ implies $x, y \in F$. A face $F$ of $C$ is a proper face if $F \neq C$.

If the above is true, I'd like to know how to prove it. Moreover, I wonder whether the statement is still true if the set is only closed and what one can say about the more general case of any (not necessarily finite-dimensional) locally convex space $V$.

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How do you define proper face? –  leo Apr 15 '12 at 5:33
    
I have edited the question, now the definition of a proper face is included. –  Tom Jonathan Apr 15 '12 at 5:39
    
Let $C$ a convex subset of such a space $V$. For each $x,y\in V$ define $S(x,y)=\{\lambda x + (1-\lambda) y \in F:\lambda\in[0,1]\}$ the segment that join $x$ with $y$. So, given $x,y\in C$ any $S(x,y)$ satisfies your definition of Proper Face, however $\bigcup_{x,y\in C} S(x,y)=C$ and $C$ might be different of what one have in mind by union of Proper Faces. For example consider a closed cube $C$ in $\mathbb{R}^3$. One think that the union of proper faces of $C$ must be a "box" $B$, but with that definition, it is all $C$ –  leo Apr 15 '12 at 6:04
    
@leo: No, $S(x,y)$ is not a face of $C$ in general. –  Robert Israel Apr 15 '12 at 6:07
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@leo For all $x, y \in C$ and all $\lambda \in (0, 1)$: If $\lambda x + (1-\lambda) y \in F$ then $x, y \in F$. –  WimC Apr 15 '12 at 6:32
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1 Answer

up vote 3 down vote accepted

A convex set $C$ in ${\mathbb R}^n$ has a supporting hyperplane at each boundary point, and the intersection of a supporting hyperplane with $C$ is a proper face of $C$.

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Yeah, that's an easy and sufficient argument, thanks! –  Tom Jonathan Apr 15 '12 at 7:12
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