Based on the examples given in https://en.wikipedia.org/wiki/Conical_combination, we know that the convex cone generated by a set containing the origin point does not necessarily a closed set, even if the convex cone is generated by convex compact set.
In addition, we also have the following known result.
S is a non-empty convex compact set which does not contain the origin, the convex conical hull of S is a closed set.
I am wondering if we relax the condition of convexity, is there a case such that the convex conical hull of compact set in $\mathbb{R}^n$ not including the origin is not closed.
Best Answer
This is not true and here is a counterexample in $\mathbb R^3$ (It is obviously true in $\mathbb R^1$ and I have the impression that it might be true in $\mathbb R^2$).
Let $$ C_\pm := \{ (x,y,z) \in \mathbb R^3 \mid x = \pm 1 \text{ and } y^2 + (z-1)^2 = 1 \} $$ and $$ K := C_+ \cup C_-. $$ Obviously, $K$ is compact. Let us check that its convex, conical hull $D$ is not closed. It is clear that the points $$ x_n := n \,(0, 1/n, 1 - \sqrt{1 - 1/n^2}) $$ belong to $D$ for $n \in \mathbb N$ and $x_n \to (0, 1, 0) =: x$. However, $x$ does not belong to $D$: Indeed, since the third component is zero, it can only be a convex conical combination of points of the form $(\pm 1, 0, 0)$ and this is not possible.