7
$\begingroup$

Can there be two closed cones $K_1$ and $K_2$ in $\mathbb{R}^3$ such that $K_1+K_2$ need not be closed?

$\endgroup$
2
  • $\begingroup$ see these notes. $\endgroup$
    – mookid
    Commented Mar 14, 2014 at 14:52
  • $\begingroup$ or rather this note from here $\endgroup$
    – reded
    Commented May 20 at 12:18

1 Answer 1

7
$\begingroup$

Yes this can happen. Let $K_1 = \{(x,y,z) : x^2+y^2\leq z^2 \}$ be the second order cone in $\mathbb{R}^3$, and let $K_2 = \{ t(1,0,-1) : t \geq 0\}$.

Then $(0,1,0)$ is not an element of $K_1 + K_2$. But $(0,1,0) = \lim_{t \to \infty} [t(1,0,-1) + (-t,1+1/t,\sqrt{t^2+(1+1/t)^2})]$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .