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Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional

Let $E$ be a vector space of infinite dimension over a (commutative) field $k$ and let $E^*=Hom_k(E,k)$ be its dual. It seems to be well-known that $E^*$ is never isomorphic to $E$ (in fact, much bigger than $E$). Can anyone give me a proof of this fact (with reference if possible) ? Put another way, after choosing a basis for $E$ we may ask : if $E=k^{(I)}$ with $I$ infinite, can you give a proof of the fact that the dimension of $E=k^I$ is bigger than $card(I)$ ?

NB : If $k$ and $I$ are countable then $k^{(I)}$ also is, whereas $k^I$ is not by Cantor's diagonal argument (recall that $I$ is infinite). This is a proof in that case.

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See mathoverflow.net/questions/13322/… – Jonas Meyer May 8 2011 at 19:35
I'm not sure I understand. Every separable Hilbert space is isomorphic to its dual. In the case that k is the real numbers see, for example, the Riesz Representation Theorem. – Aaron Hoffman May 8 2011 at 19:38
3 
 Aaron: That is a different type of dual. Here we are just talking about linear maps, while the Hilbert space dual you mention consists of continuous linear maps. – Jonas Meyer May 8 2011 at 19:39
Consider $E=\oplus_{n\geq 0}\mathbb{Q}$; it has a countable basis over $\mathbb{Q}$. Then $E^\ast=\prod_{n\geq 0}\mathbb{Q}$ which doesn't have a countable basis over $\mathbb{Q}$. – Somnath Basu May 8 2011 at 19:42
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 Surely the answer is no, in general. If $\vert k\vert=\kappa^{\vert I\vert}$ for some cardinal $\kappa$ then $\vert k^I\vert=\kappa^{\vert I\vert^2}=\kappa^{\vert I\vert}=\vert k\vert$. – George Lowther May 8 2011 at 20:02

closed as exact duplicate by Jonas Meyer, Qiaochu Yuan, Bruce Westbury, Pete L. Clark, Todd Trimble May 8 2011 at 20:25

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