Let $X\subseteq\mathbb{R}^n$ and let $F$ be a face of $\mathop{\rm conv} X$. Then $F=\mathop{\rm conv}(X\cap F)$

Question

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 $F$.

Answer 1

$\def\conv{\mathop{\mathrm{conv}}}$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 $f\in\conv(X\cap F)$.

Answer 2

This follows from the

Lemma. If $F$ is a face of $conv(X)$, $x\in F$ and a finite subset $A\subset X$ is inclusion-minimal subset for which $x\in conv(A)$, then $A\subset F$.

Proof. Induction in $|A|$. Base $n=2$ is given in the definition of a face. Induction step: write $x$ as a convex combination of $a\in A$ and some convex combination of the set $A\setminus a$. Use induction proposition and the base.