Archimedean ordered vector space
# Archimedean ordered vector space

In mathematics, specifically in order theory, a binary relation $\,\leq \,$ on a vector space $X$ over the real or complex numbers is called Archimedean if for all $x\in X,$ whenever there exists some $y\in X$ such that $nx\leq y$ for all positive integers $n,$ then necessarily $x\leq 0.$ An Archimedean (pre)ordered vector space is a (pre)ordered vector space whose order is Archimedean. A preordered vector space $X$ is called almost Archimedean if for all $x\in X,$ whenever there exists a $y\in X$ such that $-n^{-1}y\leq x\leq n^{-1}y$ for all positive integers $n,$ then$x=0.$

## Characterizations

A preordered vector space $(X,\leq )$ with an order unit $u$ is Archimedean preordered if and only if $nx\leq u$ for all non-negative integers $n$ implies $x\leq 0.$

## Properties

Let $X$ be an ordered vector space over the reals that is finite-dimensional. Then the order of $X$ is Archimedean if and only if the positive cone of $X$ is closed for the unique topology under which $X$ is a Hausdorff TVS.

## Order unit norm

Suppose $(X,\leq )$ is an ordered vector space over the reals with an order unit $u$ whose order is Archimedean and let $U=[-u,u].$ Then the Minkowski functional $p_{U}$ of $U$ (defined by $p_{U}(x):=\inf \left\{r>0:x\in r[-u,u]\right\}$) is a norm called the order unit norm. It satisfies $p_{U}(u)=1$ and the closed unit ball determined by $p_{U}$ is equal to $[-u,u]$ (that is, $[-u,u]=\{x\in X:p_{U}(x)\leq 1\}.$

### Examples

The space $l_{\infty }(S,\mathbb {R} )$ of bounded real-valued maps on a set $S$ with the pointwise order is Archimedean ordered with an order unit $u:=1$ (that is, the function that is identically $1$ on $S$). The order unit norm on $l_{\infty }(S,\mathbb {R} )$ is identical to the usual sup norm: $\|f\|:=\sup _|f(S)|.$

## Examples

Every order complete vector lattice is Archimedean ordered. A finite-dimensional vector lattice of dimension $n$ is Archimedean ordered if and only if it is isomorphic to $\mathbb {R} ^{n}$ with its canonical order. However, a totally ordered vector order of dimension $\,>1$ can not be Archimedean ordered. There exist ordered vector spaces that are almost Archimedean but not Archimedean.

The Euclidean space $\mathbb {R} ^{2}$ over the reals with the lexicographic order is not Archimedean ordered since $r(0,1)\leq (1,1)$ for every $r>0$ but $(0,1)\neq (0,0).$

