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. 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) 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 )

$$ card \: V= (card \: K) \; . (dim_K V) $$