Let $X$ be a closed convex subset of $\mathbb{R}^n$. Let $\cal F$ be the set of all faces of $X$ (that is, $\cal F$ is a set of subsets of $X$). Is the set $\bigcup{\cal E}=\bigcup_{F\in{\cal E}}F$ closed for every ${\cal E}\subseteq{\cal F}$?
Recall that a face of a convex set $X$ is a convex set $F\subseteq X$ such that every line segment from $X$ whose interior meets $F$ lies in $F$.
The answer is clearly yes for polyhedra. But non-polyhedral closed convex sets can have infinite (uncountable) number of faces, and the union of an infinite set of closed sets need not be closed. Despite this, I still believe that the answer is yes in general. Does anyone have a proof or a counterexample? Thanks.
Best Answer
Let's say that a closed convex subset $X\subset R^n$ has property C if the union of any collection of its faces is closed.
An easy example of $X$ which does not have property C is the closed unit disk $D$ in $R^2$: The set of faces consists of $D$ itself and of all singletons in the boundary circle of $D$. Hence, taking any non-closed subset of the boundary circle we see that $D$ does not satisfy property C. One can prove more:
Proposition. The following are equivalent for closed convex subsets $X\subset R^n$:
$X$ satisfies property C.
$X$ has only finitely many faces.
$X$ is a polyhedron.