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Is there a canonical way to define on any vector space over $\mathbb{K}=\mathbb{R},\mathbb{C}$ a norm ? (Or, if there isn't, can someone give me an example of a vector space over $\mathbb{K}$ that is not normable ?)

I have now looked through several books on the subject but nowhere is something like this mentioned and I also wasn't able to find a way to construct such norm (or to find a counterexample).

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    @PSeudoNeo The questions, answers, and links starting at http://math.stackexchange.com/questions/207990/vector-spaces-and-ac seem to indicate that under AC, every vector space has a Hamel basis, and without AC, there exists a vector space without a Hamel basis2013-03-10

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Pick a basis $B$ (in the algebraic sense, also known as a Hamel basis), so any vector can be uniquely written as $\sum_{b\in B}\lambda_b b$, with only finitely many of the $\lambda_b$ being nonzero. Define for instance $\left \|\sum_{b\in B}\lambda_b b\right \| := \max _{b\in B} |\lambda_b|$ (another possibility would be $\sum_{b\in B} |\lambda_b|$ instead of taking the maximum).

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    Here we are aussuming each vector space has a basis. Can we define norm on each vector space without assuming AC?2015-06-10
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Try books on the topic of "topological vector spaces": It is a theorem that every finite dimensional real or complex vector space has a norm, and that all norms are equivalent.

Correspondingly, there are infinite dimensional topological vector spaces that don't have a norm that induces the topology.

Canonical literature:

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    It is also a theorem that every finite-dimensional real or complex vector space has exactly one Hausdorff topology that makes it a nondiscrete topological vector space (meaning vector addition and scalar multiplication are continuous and the topology is not the discrete topology). This is harder than the theorem that all vector space norms on a finite-dimensional real or complex vector space define the same topology on the vector space, since a priori maybe a finite-dimensional space has a nondiscrete Hausdorff topology not coming from a norm. In fact it doesn't.2011-09-09
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The first axiom for a norm space is N1: $\left\Vert \mathbf{x}\right\Vert =0\implies \mathbf{x}=\mathbf{0}$. Now, if we take this away and have the other two viz. homogenity and the triangle inequality, then we're left with what is called a seminorm. Any vector space over $\mathbb{R}$ or $\mathbb{C}$ can be converted into a seminorm space as follows: take any functional $f$. This gives rise to a seminorm such that $\left\Vert \mathbf{x% }\right\Vert =\left\vert f\left( \mathbf{x}\right) \right\vert $ where $% \left\vert .\right\vert $ is the absolute value of an element $f\left( \mathbf{x}\right) $ of the underlying field. You may want to look up for how to define absolute values over an integral domain/field in aglebra. Now, the answer to your question, a seminorm space can $N$ be converted to a norm space $N/W$ by taking a collection $W$ of all vectors $% \mathbf{v}$ such that $\left\Vert \mathbf{v}\right\Vert =0$.\ This is a subspace since it trivially satisfies the axioms of a vector space and, furthermore, this function is a seminorm, given the properties of the absolute value. The new norm so defined is $\left\Vert \mathbf{x}+W\right\Vert _{N/W}=\left\Vert \mathbf{x}% \right\Vert _{N}$ for all $x+W\in N/W$. Thus every vector space can be converted into a norm space. For finite fields, however, the only absolute value definable is the trivial absolute value, making the freshly made norm space uninteresting.