# 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 $p_{U}(x):=\left\{r>0:x\in r[-u,u]\right\}$ ) 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 $(0,1)\neq (0,0).$