# Inequivalent complete norms and the axiom of choice

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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@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? $\;\;$ –  Ricky Demer Oct 1 '12 at 16:13
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. –  Gerald Edgar Oct 1 '12 at 17:35