In $R^n,$ convex cone of any compact set is closed? cone of any compact set is closed?

Question

In $R^n,$ convex cone of a compact set is closed? cone of a compact set is closed?

Answer 1

Assume that $x_n\to x$ and $x_n$ are points of the convex cone.

Then there are $t_n$ such that $t_n\geq0$ and $t_nx_n$ belong to the compact set. Then there is a convergent subsequence $t_{n_k}x_{n_k}\to y$ and $y$ is in the compact. If $t_{n_k}\to\infty$ then $x_{n_k}=t_{n_k}x_{n_k}/t_{n_k}\to0$, since $t_{n_k}x_{n_k}$ is bounded for being convergent. Therefore, $x=0$, which is in the cone. If $t_{n_k}$ doesn't tend to infinity, then it has a convergent subsequence $t_{n_{k_s}}\to t\geq 0$. Therefore, $x=y/t$, which is in the cone.