# Do we adopt the term "normed space" which is over any ordered-field?

The definition of normed vector spaces, as far as I know, is only defined for vector space over $\Bbb R$ or $\Bbb C$, like this one:

However, it seems no harm to talk about whether a vector space over a general ordered field forms a normed space. So do we adopt such situation? If not, why?

I think that if you want to work with norms on vector spaces over fields in general, then you have to use the concept of valuation.

Valued field: Let $K$ be a field with valuation $|\cdot|:K\to\mathbb{R}$. This is, for all $x,y\in K$, $|\cdot|$ satisfies:

1. $|x|\geq0$,
2. $|x|=0$ iff $x=0$,
3. $|x+y|\leq|x|+|y|$,
4. $|xy|=|x||y|$.

The set $|K|:=\{|x|:x\in K-\{0\}\}$ is a multiplicative subgroup of $(0,+\infty)$ called the value group of $|\cdot|$. The valuation is called trivial, discrete or dense accordingly as its value group is $\{1\}$, a discrete subset of $(0,+\infty)$ or a dense subset of $(0,+\infty)$. For example, the usual valuations in $\mathbb{R}$ and $\mathbb{C}$ are dense valuations.

Norm: Let $(K,|\cdot|)$ be a valued field and $X$ be a vector space over $(K,|\cdot|)$. A function $p:X\to \mathbb{R}$ is a norm iff for each $a,b\in X$ and each $k\in K$, it satisfies:

1. $p(a)\geq0$ and $p(a)=0$ iff $a=0_X$,
2. $p(ka)=|k|p(a)$,
3. $p(a+b)\leq p(a)+p(b)$

There is a whole research area in which arbitrary valued fields are considered and these fields are not necessarily ordered fields. It is called non-Archimedean Functional Analysis. A comprehensive starting point to read about normed spaces in this context is the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - Dekker New York (1978).

For the study of more advanced stuff, like locally convex spaces over valued fields I recommend the book: Locally Convex Spaces over non-Arquimedean Valued Fields - [C.Perez-Garcia,W.H.Schikhof] - Cambridge Studies in Advanced Mathematics (2010).

Now if you wonder whether the concept of valuation can be generalized, the answer is yes. On a field $K$ you can take a map $|\cdot|:K\mapsto G\cup\{0\}$ satisfying

1. $|x|\geq0$,
2. $|x|=0$ iff $x=0$,
3. $|x+y|\leq max\{|x|,|y|\}$,
4. $|xy|=|x||y|$.

where $G$ is an arbitrary multiplicative ordered group and $0$ is an element such that $0<g$ for all $g\in G$. In this new setting, a norm can take values in an ordered set $Y$ in which $G$ acts making of $Y$ a $G$-module. For an introdution in this area I recommend the paper:

Banach spaces over fields with a infinite rank valuation, In J. Kakol, N. De Grande-De Kimpe, and C. Perez-Garcia, editors, p-adic Functional Analysis, volume 207 of Lecture Notes in Pure and Appl. Math., pages 233-293. Marcel Dekker - [H.Ochsenius A., W.H.Schikhof] - 1999

• Wow. So detailed answer!! Thanks.
– Eric
Dec 15, 2017 at 16:17
• amazing answer. This is exactly what I asked a year ago. Dec 15, 2017 at 20:09