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
  • 7
    You cannot do this "explicitly": it requires some form of the axiom of choice. – Gro-Tsen Apr 25 '17 at 12:23
up vote 28 down vote accepted

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.)

It follows that there is no explicit way to specify how you get a basis of the dual space. You have to appeal to Zorn's lemma.

  • Out of interest, can you give a reference and/or sketch proof for this consistency result? – Peter LeFanu Lumsdaine Apr 25 '17 at 19:25
  • 1
    I was reading your comment just before you edited it, I thought you were waiting Peter asks you for for giving the reference/sketch, of course you can!!! – Rahman. M Apr 25 '17 at 20:05
  • 2
    @Rahman: Unfortunately, that's all to typical for me to do. Nonetheless, it wasn't what I was doing at this time! :) – Asaf Karagila Apr 25 '17 at 20:16
  • 3
    I would be happy to learn more about the downvote! – Asaf Karagila Apr 25 '17 at 22:14
  • 2
    @spin: Well, not exactly. Here we use the fact that we have a topological vector space and a topological field. Over an arbitrary field, it's going to play differently. But there's no general way to construct a basis for a dual space, that's the important part. – Asaf Karagila Apr 26 '17 at 9:52

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.