# How to construct a basis for the dual space of an infinite dimensional vector space?

Let $V$ be an infinite-dimensional vector space over a field $K$. Then it is known that $\dim V < \dim V^*$. More precisely, by a result attributed to Kaplansky and Erdos, we have $\dim V^* = |K|^{\dim V}$.

I have not seen an actual construction of a basis of $V^*$. My question is: given a basis $B$ of $V$, is there an explicit description of a basis of $V^*$ in terms of $B$? Can you do this at least in the case where $\dim V$ is countable?

• $K^B$ (maps from $B$ to $K$). – Mikhail Katz Apr 25 '17 at 12:08
• You cannot do this "explicitly": it requires some form of the axiom of choice. – Gro-Tsen Apr 25 '17 at 12:23

It is consistent with the axioms of $\sf ZF$ that this is impossible. Specifically, if you consider $\Bbb R[x]$, then its dual space is just $\Bbb{R^N}$. And it is consistent with $\sf ZF$ that $\Bbb{R^N}$ does not have a Hamel basis.
(Under $\sf ZF+DC$, if all sets are Lebesgue measurable, or have the Baire property, then every group homomorphism between Polish groups is continuous. It follows that if $\Bbb{R^N}$ has a basis, then there is a discontinuous functional from $\Bbb{R^N}$ to $\Bbb R$ simply by cardinality arguments. And therefore such theories prove that $\Bbb{R^N}$ does not have a Hamel basis.)