Every finite-dimensional vector space is isomorphic to its dual. 
However for an infinite-dimensional vector space $E$ over a field $K$ this is always false since its dual $E^\ast$ is a vector space of strictly larger dimension: $dim\_KE \lt dim\_K E^\ast $ (dimensions are cardinals of course). This is a non-trivial statement for which our friend Andrea Ferretti has given an astonishingly unexpected proof [here](https://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-it/13372#13372). This implies for example that a vector space of countably infinite dimension over a field $K$, like the polynomial ring $K[X]$, cannot be the dual of any $K$-vector space whatsoever. 


So an infinite-dimensional vector space is not isomorphic to its dual but it could be isomorphic to the dual of another vector space and my question is: which vector spaces are isomorphic to the dual of some other vector space and which are not?


In order to make the question a little more precise, let me remind you of an amazing theorem, ascribed by [Jacobson (page 246)](http://rads.stackoverflow.com/amzn/click/038790123X) to Kaplanski and Erdős: 


**The Kaplanski-Erdős theorem :** Let $K$ be a field and $E$ an infinite-dimensional $K$-vector space . Then for the dual $E^\ast$ of $E$ the formula $dim\_K (E^\ast) = card (E^\ast)$ obtains. 


So now I can ask


**A precise question :** Is there a converse to the Kaplanski-Erdős condition i.e. if a $K$- vector space $V$ (automatically infinite dimensional !) satisfies $dim\_K (V) = card (V)$ , is it the dual of some other vector space : $V \simeq E^\ast$? For example, is
 $\mathbb R ^{(\mathbb R)}$ - which satisfies the Kaplanski-Erdős condition ( cf. the "useful formula" below) - a dual?


**A vague request :** Could you please give "concrete" examples of duals and non-duals among infinite-dimensional vector spaces?


**A useful formula :** In this context we have the pleasant formula for the cardinality of an infinite-dimensional $K$-vector space $V$ ( for which you can find a proof by another of our friends, Todd Trimble, [here](https://mathoverflow.net/questions/49551/dimension-of-infinite-product-of-vector-spaces/49572#49572) )


$$ card \: V= (card \: K) \; . (dim\_K V) $$ 



Which vector spaces are duals ?
The $\mathbf{R}$-vector space $\mathbf{R}^{(\mathbf{R})}$ has dimension $\operatorname{card} \mathbf{R}$ by definition, so it is isomorphic to $\mathbf{R}^{\mathbf{N}}$ (by the Erdős-Kaplansky theorem and because $\operatorname{card}(\mathbf{R}^{\mathbf{N}}) = \operatorname{card} \mathbf{R}$). So $\mathbf{R}^{(\mathbf{R})}$ is isomorphic to the dual of $\mathbf{R}[X]$.


In general, your precise question is equivalent to the following purely set-theoretical question (which seems difficult). By the useful formula, the identity $\operatorname{card} V = \operatorname{dim}\_K V$ is equivalent to $\operatorname{card} K \leq \operatorname{dim}\_K V$. Let $\kappa = \operatorname{card} K$ and $\lambda = \dim\_K V$, and assume $\kappa \leq \lambda$. Does there always exist an infinite cardinal $\alpha$ such that $\kappa^\alpha = \lambda$? (here $\alpha$ is meant to be the dimension of the vector space whose dual is $V$). In general, Stephen's answer shows that there are counterexamples.


EDIT : in order to explain why the question is difficult, consider a field $K$ such that $\operatorname{card} K = \aleph\_1$ (for example, one could take $K=\mathbf{Q}((T\_i)\_{i \in I})$ with $\operatorname{card} I =\aleph\_1$). Take a $K$-vector space $V$ of dimension $\aleph\_1$. Then $V$ is a dual if and only if there exists $\alpha \geq \aleph\_0$ such that $\aleph\_1 = (\aleph\_1)^\alpha$, which amounts to say that $\aleph\_1 = 2^{\aleph\_0}$. In other words, $V$ is a dual if and only if the continuum hypothesis holds.


