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I am trying to prove $K^{**} = cl(K)$, where $K^{**}$ is the double dual of K, cl(K) is the closure of K, and K is a convex cone.

I was able to show that $cl(K)\subseteq K^{**}$ and am trying to show the other direction: $K^{**} \subseteq cl(K)$. I am thinking of proving by contradiction and argue that suppose there exists y $\in K^{**}$ such that $y \notin cl(K)$, then .... From posts online that give hints, I think I should use the separation theorem, and let the two non-empty disjoint convex sets be $\{y\}$ and cl(K). However, I am not sure how to continue from here and was hoping for some hints.

Thanks a lot.

Teemo
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  • Duplicate: https://math.stackexchange.com/questions/28727/dual-of-a-dual-cone – Ben W Jan 27 '20 at 23:20
  • This reminds me of the similar statement in functional analysis that if $X$ is a "nice enough" normed linear space then $X^{**}$ is isomorphic to its completion. (Here, "nice enough" means that the completion of $X$ is reflexive.) – Ben W Jan 27 '20 at 23:23
  • @BenW Thanks. In your linked url, under the answer of t.b. (the top endorsed answer), after he introduced the Hahn-Banach separation theorem, do you know why he was able to claim that m = inf ${ f(k), k \in \bar{k} }$? That is confusing to me since it's not part of the separation theorem. Thanks a lot. – Teemo Jan 27 '20 at 23:47

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