Inverse image (under convex-hull operator) of an open set of convex subsets is open?

Question

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

Suppose that $V\subset\mathcal{C}$ is an open set in $\mathcal{C}$.
Let $co^{-1}(V)=\{F\in\mathcal{F}:co(F)\in V\}$.
Is it true that $co^{-1}(V)$ is an open set in $\mathcal{F}$?

My intuition is yes, but not sure how to argue. I guess there should be an easy way to argue.

Answer 1

Use the fact that $$d_H (A, B) \geq d_H (\mbox{co }(A) , \mbox{co} (B) )$$

Answer 2

Hausdorff metric and convex hull

I just learned from this post the fact that $$h(co(A),co(B))\leq h(A,B),$$ where $h$ is Hausdorff metric.

Therefore if $h(F_n,F)\rightarrow 0$, then $h(co(F_n),co(F))\rightarrow 0$.
Hence $co(\cdot)$ is a continuous mapping.

Hence the inverse image of an open set under $co(\cdot)$ is open.

I think this argument is valid, is it?