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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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    @leo For all $x, y \in C$ and all $\lambda \in (0, 1)$: If $\lambda x + (1-\lambda) y \in F$ then $x, y \in F$.2012-04-15

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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!2012-04-15