# Does a norm have to map to $\mathbb R$?

Wikipedia defines a norm on a vector space $V$ as a function $p : V \mapsto \mathbb R$. I've seen this defined similarly elsewhere. However, it seems to me that a real codomain isn't always necessary.

Take as an example the vector space over $\mathbb Q$ with vectors $v \in \mathbb Q^n$. If we combine this with the $L^1$ norm, then the codomain is again $\mathbb Q$.

Am I missing something here? Is my example flawed?

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Isn't always necessary for what? – Qiaochu Yuan Feb 22 '13 at 0:56
I was thinking of representing the output of a norm function on a computer, where you wouldn't always need to represent real numbers to fully capture possible values. – geomnerd Feb 22 '13 at 1:14
Mathematical concepts are tools, and tools have uses. You can change definitions however you want, but then you get new tools which may not be as useful as the old ones. The uses I'm familiar with for normed vector spaces require that the codomain be $\mathbb{R}$ so that you get metric spaces and can talk about convergence, completeness, etc. – Qiaochu Yuan Feb 22 '13 at 1:19

## 3 Answers

The norm is some measure of size. It is very convenient to measure sizes by real numbers (actually, by non-negative real numbers). In case the vectors you consider all have rational entries then indeed the norm will be a non-negative rational number. Since $\mathbb Q\subseteq \mathbb R$, this case is already included in the more general case.

For analytic reasons it is very convenient that the codomain for the norm be complete. This is why don't see $\mathbb Q$ is a codomain for a norm. It is possible to consider norms taking values in complete lattices other than $\mathbb R$, depending on the problem at hand. However, usually, when vector spaces are over the field $\mathbb R$ or $\mathbb C$, it is most convenient to consider $\mathbb R$ as the codomain for a norm function.

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Just minor terminology/wording: "The norm is some measure of size" - I think it's more appropriate to think of norms as a measure of length, in particular, relative to a zero object (as opposed to metric spaces). Measure spaces more closely relate to quantifying size. – alancalvitti Feb 22 '13 at 1:08
I see what you mean @alancalvitti . I meant 'size' to be taken very broadly, so I will leave it as it is in the post and let your comment provide the nuance. – Ittay Weiss Feb 22 '13 at 1:19
Ok, in any case, +1. – alancalvitti Feb 22 '13 at 1:54

The fact that a norm maps to $\Bbb R$ does not mean that the norm is surjective. It can map to a proper subset of $\Bbb R$.

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Having the norm map to $\mathbb R$ is just a definition, and the map definitely need not be surjective. There's the example you gave, there's also the example of the $p$-adic metric on $\mathbb Q$ which has image given by all integral powers of $p$. One nice thing about having norms map to $\mathbb R$ is that $\mathbb R$ is complete, and a normed vector space in its own right, although the image of the norm map is just $\mathbb R^{\geq 0}$. Someone can probably add to this.

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