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If $K$ is a field, the dual of $K^{({\mathbb N})}$ is $K^{\mathbb N}$, and axiom of choice implies that the natural map from $K^{({\mathbb N})}$ to the dual of $K^{\mathbb N}$ is far from being surjective. Is there any field for which one can prove this without using AC?

(A related result that I found surprising is that ${\rm Hom}({\mathbb Z}^{({\mathbb N})},{\mathbb Z})={\mathbb Z}^{\mathbb N}$ and ${\rm Hom}({\mathbb Z}^{\mathbb N},{\mathbb Z})={\mathbb Z}^{({\mathbb N})}$.)

Ann o'Nymous
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    See related questions: http://mathoverflow.net/questions/49351/does-the-fact-that-this-vector-space-is-not-isomorphic-to-its-double-dual-require and http://mathoverflow.net/questions/49388/is-the-non-triviality-of-the-algebraic-dual-of-an-infinite-dimensional-vector-spa – Amit Kumar Gupta Apr 24 '11 at 21:53
  • Given the two questions Amit linked to, I'm going to vote to close this question as duplicate. Note that I do think it is a good question, just that it's already been answered. – Theo Johnson-Freyd Apr 25 '11 at 20:42

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