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Hi,

I've been wondering about the following :

Is it possible, without the axiom of choice, to have two inequivalent complete norms on a vector space?

All the examples of inequivalent complete norms I've seen rely on the existence of Hamel bases...

This is most likely well-known, but I'd be glad if someone could provide a good reference.

Thank you, Malik

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No, it is not possible. There is a model due to Shelah of ZF+ dependent choice+every set of reals has the Baire property. There is a result of Garnir and Wright that implies that in such model, any two complete norms are equivalent. The Handbook of Analysis and Its Foundations by Eric Schechter has a chapter on this result and its consequences. The chapter is "The Dream Universe of Garnir and Wright". The result in that chapter is supposed to be slightly stronger than the one in the original papers, with which I'm not familiar.

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    $\begingroup$ The original approach by Garnir, as well Wright, and Brunner which followed a bit, all used Solovay's model in which all sets are Lebesgue measurable, which is a stronger demand than merely Baire property. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 1, 2012 at 15:56
  • $\begingroup$ A side remark: if we drop excluded middle then it is consistent with dependent choice that, up to topological equivalence, there is at most one complete separable metric on a given set. So intuitionistically one can even have the "uniqueness of (complete separable) metric", not just "uniqueness of norms". $\endgroup$
    – Andrej Bauer
    Commented Oct 1, 2012 at 16:10
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    $\begingroup$ @Michael: $\:$ Am I missing something, or does that model only show that, without the axiom of choice, it is possible to not have two inequivalent complete norms on a vectors, rather than that it is not possible to have two inequivalent norms on a vector space? $\;\;$ $\endgroup$
    – user5810
    Commented Oct 1, 2012 at 16:13
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    $\begingroup$ In that model, every two complete norms are equivalent. This shows that every proof in ZFC that inequivalent complete norms exist must make nontrivial use of AC. And dependent choice is essentially the part of the axiom of choice that can still be intuitively grasped and that amount to somewhat explicit constructions. If you read the question as whether the existence of inequivalent complete norms imply AC, the answer is surely no, the existence of a Hamel base for some given vector space involved in a construction is weaker than AC. $\endgroup$
    – Michael Greinecker
    Commented Oct 1, 2012 at 16:26
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    $\begingroup$ This is similar to the same question with "complete separable metric groups" ... for example, with Axiom of Choice we prove that the groups $\mathbb R$ and $\mathbb R^2$ are isomorphic. But since they are not homeomorphic, in fact AC cannot be omitted in that proof. $\endgroup$
    – Gerald Edgar
    Commented Oct 1, 2012 at 17:35

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