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Math people:

This question is 95% answered (the first answer) at Does every $\mathbb{R},\mathbb{C}$ vector space have a norm? and Vector Spaces and AC . The questions, answers, and links found there seem to assert that if you assume the Axiom of Choice, then every vector space has a Hamel basis and hence a norm, and conversely, if you assume every vector space has Hamel basis, then AC follows.

But a norm doesn't have to be given by a Hamel basis. For example, on $L^2([0,1])$ you can use the standard norm, which I don't think can be defined using a Hamel basis, though it can be defined using a Schauder basis. So I think the question is still open.

EDIT: I want the underlying field to be the real numbers.

Stefan (STack Exchange FAN)

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Stefan Smith
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It is consistent that the axiom of choice fails and there is a vector space over the real numbers which cannot be a topological vector space in a nontrivial way, and in particular it means that it cannot be normed, since the norm would induce a nontrivial topology.

The construction is due to Läuchli, from 1963 (and the extension to real vector spaces follows from my masters thesis).

The more interesting question is whether the assertion "every vector space has a norm" implies the axiom of choice. The answer to that, I do not know at this time.

Asaf Karagila
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  • Thank you for your prompt reply. It seems that your answer is "yes" (?). I'm not sure what you mean by "nontrivial", or for a norm to be discrete. I want the field to be the reals (I edited the question to clarify). I would rather stick to the textbook definition of norm and avoid the term "topological vector space", both for my benefit and anyone else following this question. – Stefan Smith Mar 10 '13 at 19:10
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    @Stefan: Norms define metrics, which in turn define a topology. One can show that the topology induced by a norm has to obey certain rules which prevent it from being the discrete topology (i.e. every set is open), if the only topology fitting to a metric space on the vector space is the discrete one, then it cannot possibly be induced by a norm. Therefore there cannot be a norm compatible with the vector space structure. – Asaf Karagila Mar 10 '13 at 19:52
  • so essentially, Lauchli and you showed that under ZF (without AC), there exists a vector space without a norm? Did you assume the negation of AC? – Stefan Smith Mar 10 '13 at 20:13
  • @Stefan: No, of course not. He showed that it is consistent that the axiom of choice fails and such vector space exists. His proof was limited to countable fields, though. In my thesis I took it up and polished it to allow arbitrary (but fixed, of course) fields. – Asaf Karagila Mar 10 '13 at 20:39
  • Thanks. I am not a logician and I find such matters confusing. I only really care about real vector spaces. I'm not sure what you mean by "consistent". Consistent with what? Did you two show that there is a model of set theory in which AC fails and a real vector space without a norm exists? – Stefan Smith Mar 10 '13 at 22:47
  • @Stefan: It means that given a model of ZF, I can produce another model of ZF in which there is a real vector space which cannot be a normed space, of course the axiom of choice fails in such model. I don't know, however, whether or not in every model of ZF in which the axiom of choice fails there is a real vector which cannot be normed. That is an interesting, and probably open, question. – Asaf Karagila Mar 10 '13 at 23:57
  • what is the relation between the two models of ZF? – Stefan Smith Mar 11 '13 at 01:01
  • @Stefan: In the case the first one was a model of ZFC, then the second one is "slightly" larger (i.e. has strictly more sets); in the case the first was not a model of ZFC, but just ZF, it might be the case that we need to reduce to some smaller model (which is definable from the first) in which ZFC holds, and then extend that one. In any of the cases, they both have the same ordinals. – Asaf Karagila Mar 11 '13 at 01:05