TITLE: Is the boundary of a compact convex set given by the union of its proper faces?
QUESTION [6 upvotes]: 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 $V$.

REPLY [3 votes]: 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 $C$.