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Wikipedia's article on ultrametric spaces seems to suggest that an ultrametic space can also be a normed vector space.

It seems to be impossible for an ultrametric to be induced by a vector space norm, because if $v$ is a nonzero vector, then $$d(0,2v)=\|2v\|=2\|v\|>\|v\|=\max(d(0,v),d(v,2v)) $$ contradicting $d$ being an ultrametric.

So what's going on there? Is there a concept of "normed vector space" that dispenses with the requirement that scalar multiplication scales the norm accordingly? (That would be more of a "normed abelian group", then).

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up vote 7 down vote accepted

The correct equality is $\|a v\|=|a|\|v\|$ for some absolute value $|\cdot|$ on the scalar field. Absolute values can be nonarchimedean and satisfy the ultrametric inequality. In such a case, $|k|\le 1$ for all $k\in\Bbb N$.

For instance, $\|(v_1,\cdots,v_n)\|:=\sqrt{|v_1|_p^2+\cdots+|v_n|_p^2}$ makes $\Bbb Q^n$ an ultrametric normed vector space over $\Bbb Q$, where $|\cdot|_p$ stands for the $p$-adic absolute value.

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Ah, that makes sense. Thanks. – Henning Makholm Aug 10 '14 at 11:59

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