# Do you need the Axiom of Choice to assert that every real vector space has a norm?

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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– Andrés Caicedo Mar 10 '13 at 18:14