Can one prove without the Axiom of Choice that for every normed vector space $X$ there exist a nonzero continuous linear functional on $X$?


2 Answers 2


No, this is equivalent to the Hahn–Banach theorem already in the case of Banach spaces.

See in my write up: https://arxiv.org/abs/2010.15632

  • $\begingroup$ Isn’t the answer, then, that you need BPIT but not necessarily AC? $\endgroup$ Oct 15, 2021 at 2:28
  • 3
    $\begingroup$ It is an answer, yes. But Hahn–Banach is weaker still, and it is exactly what you need, so that is a better answer. But when people ask "can you prove this and that without the axiom of choice", normally the question is read as "can you prove this and that in ZF", rather than "is this and that equivalent to AC". $\endgroup$
    – Asaf Karagila
    Oct 15, 2021 at 5:29

The answer is no. Let $\ell^\infty$ be the space of bounded sequences equipped with the $\sup$ norm, and $c_0$ its closed subspace of sequences convergent to zero. The quotient $\ell^\infty/c_0$ is then a Banach space when equipped with the quotient norm, and it is consistent that it does not have any nonzero continuous functionals - see a discussion in this answer.

  • $\begingroup$ I see that this specific example is also mentioned in Asaf's write up, in Theorem 49. $\endgroup$
    – Wojowu
    Oct 14, 2021 at 21:31
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    $\begingroup$ See also Martin Väth's paper "The dual space of $L_\infty$ is $L_1$", Indagationes Math. 9, No. 4, 619--625 (1998); Zbl 0922.46066. $\endgroup$ Oct 16, 2021 at 9:39

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