I have encountered this when I was thinking about differentiability in Banach spaces. There, for $x\in X$ we usually need functionals $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$. This is a simple consequence of Hahn-Banach theorem and enables one to convert the problem at hand into a problem in ordinary calculus. Now My question is:

Suppose we have a Banach space $X$ whose dual $X^*$ separates points, i.e. for every nonzero $x\in X$ there is $u\in X^*$ such that $u(x)\neq 0$. Can one prove in ZF that for all nonzero $x\in X$ there is $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$ ?

I know that Hahn-Banach theorem is strictly weaker than axiom of choice, but I'm looking for a proof without using any "choicy" argument.

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    $\begingroup$ This doesn't answer your question, but I'd just like to point out that you don't need any choice principle to prove the Hahn-Banach theorem for separable Banach spaces (and one can make a case that this is all we care about in practice). $\endgroup$
    – Nik Weaver
    Aug 16, 2013 at 17:31
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    $\begingroup$ A suggestion of a possible counter-example: take X to be $\ell^\infty$ (trivial that dual separates points, even in ZF, because we can just use the point evaluations) and take your $x$ to be any sequence that does not attain its bound. Then it seems that to get a functional which attains its norm on $x$ one is going to need some choice or ultrafilter argument $\endgroup$
    – Yemon Choi
    Aug 16, 2013 at 17:31
  • $\begingroup$ @Yemon My own guess was also that it's not provable in ZF, but my knowledge of set theory is nowhere near proving an independence result. $\endgroup$ Aug 16, 2013 at 18:02
  • $\begingroup$ @Nik That's interesting, can you give me a reference please? $\endgroup$ Aug 16, 2013 at 18:04
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    $\begingroup$ @Mohammad: it's easy, just run through the usual proof of H-B, extending the functional one dimension at a time. At each step choose the maximum allowed value of the functional on the new vector --- so, no choice. If the space was separable to start with, that means we have a basis $(x_n)$ indexed by $\mathbb{N}$, so if we're extending from the subspace $E_0$, pass through the sequence of subspaces ${\rm span}(E_0, x_1, \ldots, x_n)$. No choice there either. $\endgroup$
    – Nik Weaver
    Aug 16, 2013 at 20:20

1 Answer 1


Yemon Choi suggested a counterexample in the comments which can easily be made into an actual counterexample.

It is consistent with ZF (see Asaf Karagila's answer here for a few references) that the dual space of $\ell_\infty$ is $\ell_1$ with the usual norm. The duality pairing is $$ \langle x,y \rangle = \sum x_n y_n. $$ Since the coordinate functionals separate points in $\ell_\infty$, the hypothesis in the question is satisfied.

On the other hand, for the sequence $x_n = 1-1/n$ in $\ell_\infty$ there obviously is no sequence $y \in \ell_1$ of norm one such that $\sum x_n y_n = 1 = \lVert x_n \rVert_\infty$.

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    $\begingroup$ See also mathoverflow.net/q/5351/45 on this site for the relationship between choice and the dual space of $\ell_\infty$. $\endgroup$ Aug 16, 2013 at 19:27
  • $\begingroup$ Thanks. I'll accept this later to see if I can get any more information in the meantime. $\endgroup$ Aug 16, 2013 at 19:52
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    $\begingroup$ @MohammadSafdari: You're welcome. There's surely a lot more that can be said, so feel free to wait :-) For further reading I recommend Eric Schechter's Handbook of Analysis and its Foundations for a thorough discussion of numerous weak forms of the axiom of choice and their uses in functional analysis. It is readable with only minimal background in set theory. The case of $(\ell_\infty)^\ast$ is discussed in the section on Pincus's Pathology. $\endgroup$
    – Martin
    Aug 16, 2013 at 21:45

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