Regarding this answer. We can take $K=\{(0,0)\}\cup\{(a,b):~a>0,b>0\}$. We have $K^*=\{(a,b):~a\geq 0,b\geq 0\}$. Let us take $C=\{(0,0)\}\cup\{(a,b):~a\geq 0,b>0\}$ and notice $K \subset C$. But we do not have $K^{**}\subset C$ as stated in your answer, because $K^{**}=K^*$ that is larger than $C$.

  • $\begingroup$ "$K^{**} = K^*$ that is larger than $C$" --- no, $K^{**} = K^* = C$. $\endgroup$ – Zach Teitler Jun 23 '17 at 14:20
  • $\begingroup$ Thanks for your quick response. C is not $K^*$, because $K^*$ has both $a,b\geq 0$, while $C$ has $b>0$. $\endgroup$ – Daniel Porumbel Jun 23 '17 at 22:12
  • $\begingroup$ Oh, right! Sorry. The answer you linked to needs $C$ to be a closed convex cone. If $K \subseteq C$ and $C$ is closed and convex, then $K^{**} \subseteq C^{**} = C$. The inclusion is more or less automatic, the equality uses the hyperplane separation and that $C$ is closed. $\endgroup$ – Zach Teitler Jun 23 '17 at 22:44
  • $\begingroup$ Ok, thanks; everything is clear now. $\endgroup$ – Daniel Porumbel Jun 23 '17 at 22:53

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