Can a closed convex cone not containing a line passing through the origin contain a line?

Question

Suppose that $K$ is a closed convex cone in $\mathbb{R}^n$. We know that $K$ does not contain any line passing through the origin; that is, $K \cap -K = \{0\} $. Does it imply that $K$ does not contain any line of the form $x_{0} + \mathrm{span}\{y\}$, where $x_0 \ne 0$ and $y \ne 0$? I feel like the answer is yes, but couldn't come up with a proof for it. Any help is much appreciated!

Answer 1

It cannot contain such a line. Indeed, if a closed cone $K$ contains $x_0 + \operatorname{span}\{y\}$, then it contains $\operatorname{span}\{y\}$.

Note first that it suffices to show that $y \in K$; any non-zero multiple of $y$ will also produce the same span, so the same argument would show $\lambda y \in K$ for $\lambda \neq 0$. Of course, since $0 \in K$, we can remove the $\lambda \neq 0$ restriction.

Proving $y \in K$ is simple: $$x_0 + ny \in x_0 + \operatorname{span}\{y\} \subseteq K \implies \frac{1}{n}(x_0 + ny) = y + \frac{x_0}{n} \in K.$$ Since $K$ is closed, $\lim_{n \to \infty} y + \frac{x_0}{n} = y \in K$.

Note that closedness is very important here. The cone $$K = \{(x, y) : x > 0\} \cup \{0\}$$ is an example of a non-closed cone for which $K \cup -K = \{0\}$, but which contains a line, e.g. $y = 1$.

Answer 2

There is a hyperplane $H$ separating $K$ from $-K$. Any line not parallel to $H$ will meet both half spaces determined by $H$. That rules out all lines parallel to $H$.

To rule out the lines parallel to $H$ takes a little more work. Since $K$ is not a full half space, $H$ is not unique and the lines parallel to $H$ will not be parallel to all the other separating hyperplanes.