Is convex-hull operation an open mapping?

Question

Let $\mathcal{F}$ denote the set of all finite subsets of $\mathcal{R}^n$, endowed with Hausdorff metric, denoted by $d_H$.
Let $\mathcal{C}=\{co(F):F\in\mathcal{F}\}$, also endowed with Hausdorff metric.
($co(F)$ denotes the convex hull of $F$)

I have figured out $co(\cdot):\mathcal{F}\rightarrow\mathcal{C}$ is a continuous mapping. It is because $$d_H(co(A),co(B))\leq d_H(A,B).$$

(From this post: Hausdorff metric and convex hull )

My question: Is $co(\cdot)$ an open mapping? That is, maps open set to open set.

I try to prove in the following way.
Let $U\subset\mathcal{F}$ be an open set. I want to check $co(U)^c$ is closed. Let $\{G_n\}\subset co(U)^c$ and $G_n\rightarrow G$. Need to show $G\in co(U)^c$.

Suppose instead $G\in co(U)$. Then there exists $F\in U$ such that $co(F)=G$. I want to construct a sequence $\{F_n\}\subset\mathcal{F}$ such that $co(F_n)=G_n$ for all $n$, and $F_n\rightarrow F$. If such sequence exists, then when $n$ large enough we have $F_n\in U$. So $G_n\in co(U)$. A contradiction.

But I am not sure how to prove that such sequence $\{F_n\}$ exists. Each $G_n$ is a convex set in $\mathcal{R}^n$ and $G_n\rightarrow G$. Given a finite subset $F\subset\mathcal{R}^n$ such that its convex hull is $G$. I want to write each $G_n$ as a convex hull of $F_n$, and $F_n$ converges to $F$.

So far I try to define $F_n=ext(G_n)\cup[G_n\cap F]$, where $ext$ denote extreme points. $F_n$ is a finite set because $ext(G_n)$ and $F$ are both finite. Clearly, $co(F_n)=G_n$ because $F_n\subset G_n$ and $F_n\supset ext(G_n)$. I expect that $F_n$ converges to $ext(G)\cup[G\cap F]$, which equals to $F$. Is it true that $F_n\rightarrow F$ under Hausdorff metric?

Answer 1

No, the convex hull operator is not an open map. Let $A=\{a,b\}$ be a two-point set. Its convex hull is a line segment $L$. Let $c\notin L$ be a point near the midpoint $(a+b)/2$, and let $T$ be the convex hull of $\{a,b,c\}$ (it's a triangle). Observe that

• $d_H(T, L)\to 0$ as $c\to (a+b)/2$
• any set $B$ such that $co(B)=T$ must contain $c$, and therefore $d_H(A,B)\ge |a-b|/2$

Hence, the image of a neighborhood of $A$ of radius less than $|a-b|/2$ is not an open set: it does not contain any neighborhood of $L$.