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?

  • $\begingroup$ $K^B$ (maps from $B$ to $K$). $\endgroup$ Apr 25, 2017 at 12:08
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    $\begingroup$ You cannot do this "explicitly": it requires some form of the axiom of choice. $\endgroup$
    – Gro-Tsen
    Apr 25, 2017 at 12:23

1 Answer 1


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.

  • $\begingroup$ Out of interest, can you give a reference and/or sketch proof for this consistency result? $\endgroup$ Apr 25, 2017 at 19:25
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    $\begingroup$ 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!!! $\endgroup$
    – Rahman. M
    Apr 25, 2017 at 20:05
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    $\begingroup$ @Rahman: Unfortunately, that's all to typical for me to do. Nonetheless, it wasn't what I was doing at this time! :) $\endgroup$
    – Asaf Karagila
    Apr 25, 2017 at 20:16
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    $\begingroup$ I would be happy to learn more about the downvote! $\endgroup$
    – Asaf Karagila
    Apr 25, 2017 at 22:14
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    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Apr 26, 2017 at 9:52

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