This theorem is obviously true if the set $X$ is finite (so that $\mathop{\rm conv} X$ is a convex polytope). I believe it is true for any set $X\subseteq\mathbb{R}^n$ but I cannot prove it. Can anybody please prove this or give a counter-example? Many thanks!


Notes: $\mathop{\rm conv}X$ denotes the convex hull of the set $X$.
A face of a convex set $C$ is defined to be a convex set $F\subseteq C$ such that every line segment from $C$ whose relative interior has a non-empty intersection with $F$ is contained in $F$.



Let $X\subseteq\mathbb{R}^n$ and let $F$ be a face of $\mathop{\rm conv} X$. Then $F=\mathop{\rm conv}(X\cap F)$
$\def\conv{\mathop{\mathrm{conv}}}$Recall that $\conv X$ is the set of all convex combinations of points from $X$. In a convex combination 
$$
 f=\sum\_{i=1}^k\alpha\_ix\_i, \qquad x\_i\in X, \quad \alpha\_i>0, \quad \sum\_{i=1}^k\alpha\_i=1,
$$
for a point $f\in F$, all the points $x\_i$ should lie in $F$ (and hence in $X\cap F$); indeed, if, say, $x\_k\notin F$, then $f$ is a relatice interior point of the segment between $x\_k$ and
$$
 \frac1{1-\alpha\_k}\sum\_{i=1}^{k-1}\alpha\_ix\_i
$$
which is impossible.
Thus $f\in\conv(X\cap F)$.


