In my studies of convex cones I have recently met the following problem:
A non-empty set $ K\subseteq \mathbb{R}^d $ is called a convex cone if for all members $ x,y\in K $ and all non-negative real scalars $ 0 \leq\alpha,\beta \in \mathbb{R} $ we have $ \alpha x + \beta y \in K $. A convex cone is called pointed if we have $ K \cap -K = \{0\} $ and we denote by "ri" the relative interior. I am asked to prove that if $ K $ is a closed pointed cone, then there exists a pointed cone $ K' $ such that $ K \backslash \{0\} \subseteq ri(K') $ and of course $ K-K $ is the Minkowski difference meaning $ K-K = \{ k_1 -k_2 | k_1,k_2 \in K \} $. Now I have no idea how to proceed on this and have no idea how to formulate a solution for this, I can see intuitively why this mght hold maybe using hyperplane separation but not formally, could someone please assist on this? I thank all helpers.
I assume it suffices to work with full dimensional $K\subseteq\mathbb{R}^d$, so that the dual $K^*$ is pointed. Since $K$ is pointed, $K^*$ is full dimensional, so pick any set of $d$ linearly independent vectors $v_1,v_2,\dots,v_d$ in $K^*$ that lie in the relative interior of $K^*$ to get a subcone $K'=\text{cone}\{v_1,v_2,\dots,v_d\}$ that is still full dimensional.
Then $K'\subsetneq K^*\implies (K')^*\supsetneq K^{**}=K$ since dualizing is inclusion reversing and $K'$ is pointed since $K^*$ was full dimensional. (I guess the full dimensional assumption is not strictly necessary, but then we need to be careful about the vectors we choose.)