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Is there a plain vector space (not a topological vector space) such that it is impossible to define a function that serves as a norm for this space? Similarly, is there such space for inner product?

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  • $\begingroup$ See this. $\endgroup$ – David Mitra Jan 30 '17 at 10:47
  • $\begingroup$ And this. $\endgroup$ – Glitch Jan 30 '17 at 13:58
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Any vector space over any field is isomorphic to
the direct sum of copies of its field over any of its Hamel bases: $$V\cong\{f:B\to\mathbb{K}:f(x)=0\text{ a.e.}\}=\bigoplus_B\mathbb{K}$$

For a normed field one obtains
a tremendous variety of pairwise inequivalent norms: $$\|f\|_p:=\left(\sum_{b\in B}|f(b)|^p\right)^{1/p}\quad(1\leq p\leq\infty)$$

Among these an inner product,
as well as a C*-norm.

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  • $\begingroup$ How about without AC? :D $\endgroup$ – s.harp Jan 30 '17 at 17:40
  • $\begingroup$ @s.harp: I guess that could be just another of the many equivalences: AC iff Hamel Basis iff Norm iff ... ^^ $\endgroup$ – C-Star-W-Star Jan 30 '17 at 17:47
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I assume you are talking about real or complex vector spaces (although these observations can be generalised to vector spaces over non-archimedean fields).

At least assuming axiom of choice, a pure vector space over a given field is fully characterised by its dimension. Since there are normed and inner product of every dimension, it follows that every vector space is isomorphic (as a vector space) to a normed or inner product space, and we can import the norm via this isomorphism.

Furthermore, since there are Banach and Hilbert spaces of every (Hamel) dimension except infinite and smaller than ${\mathfrak c}$, every vector space of finite dimension or of dimension at least ${\mathfrak c}$ admits a Banach space, and in fact a Hilbert space structure (in fact, the existence of such a structure is equivalent to having dimension finite or at least ${\mathfrak c}$).

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  • $\begingroup$ Nice addition about the cardinality!! (+1) $\endgroup$ – C-Star-W-Star Jan 31 '17 at 11:21

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