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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.

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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.)

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    $\begingroup$ Thanks for this nice answer, could you please just tell me why pointed implies full dimensionality of dual and vice versa? Is it a theorem of sorts? $\endgroup$
    – Croc2Alpha
    Oct 15, 2016 at 19:19
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    $\begingroup$ and could you please tell me why the inclusion you wrote implies inclusion in relative interior? $\endgroup$
    – Croc2Alpha
    Oct 15, 2016 at 19:28
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    $\begingroup$ Take a line $L$ contained in a cone $C$. Dualizing reverses inclusion, so $ L\subset C \iff C^*\subset L^*$ and $L^*$ is a hyperplane given by $\{x\in(\mathbb{R}^*)^d\mid \langle x,l\rangle=0\}$ where $l\in L$ is any nonzero point. $\endgroup$
    – Tachyon
    Oct 15, 2016 at 19:48
  • $\begingroup$ thanks could you please tell me how do we obtain from your answer specifically that $ K \subseteq ri((K^*)') $? Because I need a pointed cone whose relative interior contains ours $\endgroup$
    – Croc2Alpha
    Oct 15, 2016 at 19:50
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    $\begingroup$ We know $K\subset (K')^*$. Thus the relative interior of $K$ is contained in the relative interior of $(K')^*$. So if there is a point $p$ in $K\backslash ri((K')^*)$, then it must be a boundary point of both, so in particular there is some $x\in\mathbb{R}^d$ such that $\langle x,p\rangle =0$. But such a point is not in the interior of $K^*$, contrary to our choice of generators for $K'$. $\endgroup$
    – Tachyon
    Oct 15, 2016 at 19:57

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