I have come across a proof for this statement(link to paper at end), which however I do not understand. It makes use of Lemma 5.5, which I've also included.
Lemma 1. The interior of the complement of a proper convex set $C \subset \mathbb{R}^n$ is non-empty.
Theorem: Every proper convex cone $K \neq \mathbb{R}^n$ is a subset of a half-space.
Proof:
By lemma 1, we know $K^c$ has a non-empty interior. If there is a vector $v \in K^c$, such that $-v \in K^c$ as well, then it is possible to take the conic hull of $\{v\}$ and $K$. After this operation, $-v$ cannot belong to conic-hull $(\{v\} \cup K)$. Therefore, a convex cone $K$ can always be expanded unless there are no vectors $v \in K^c$ such that $-v \in K^c$.
The fact that for every vector in $K^c$ we have its opposite inside $K$ makes $-K^c$ identical to int $(K)$ which means $K^c$ is now itself a convex cone. As a result, the boundary $\operatorname{cl}(K) \cap \operatorname{cl}\left(K^c\right)$ is the intersection of two convex cones and is therefore convex. The boundary or the set of points $\{v \in K \mid-v \in K\}$ is closed under scalar action on top of being closed under addition by virtue of being a convex cone. This means that the boundary is an $(n-1)$-dimensional vector subspace, i.e. a hyperplane. $\blacksquare$
I understand that we are trying, in this proof, to show that $K^c$ is also a convex cone, but I don't understand how we get there. Would very much appreciate your help!
link: original paper here: https://arxiv.org/pdf/2303.05423.pdf
