# Archimedean ordered vector space

In mathematics, specifically in order theory, a binary relation ${\displaystyle \,\leq \,}$ on a vector space ${\displaystyle X}$ over the real or complex numbers is called Archimedean if for all ${\displaystyle x\in X,}$ whenever there exists some ${\displaystyle y\in X}$ such that ${\displaystyle nx\leq y}$ for all positive integers ${\displaystyle n,}$ then necessarily ${\displaystyle x\leq 0.}$ An Archimedean (pre)ordered vector space is a (pre)ordered vector space whose order is Archimedean.[1] A preordered vector space ${\displaystyle X}$ is called almost Archimedean if for all ${\displaystyle x\in X,}$ whenever there exists a ${\displaystyle y\in X}$ such that ${\displaystyle -n^{-1}y\leq x\leq n^{-1}y}$ for all positive integers ${\displaystyle n,}$ then${\displaystyle x=0.}$[2]

## Characterizations

A preordered vector space ${\displaystyle (X,\leq )}$  with an order unit ${\displaystyle u}$  is Archimedean preordered if and only if ${\displaystyle nx\leq u}$  for all non-negative integers ${\displaystyle n}$  implies ${\displaystyle x\leq 0.}$ [3]

## Properties

Let ${\displaystyle X}$  be an ordered vector space over the reals that is finite-dimensional. Then the order of ${\displaystyle X}$  is Archimedean if and only if the positive cone of ${\displaystyle X}$  is closed for the unique topology under which ${\displaystyle X}$  is a Hausdorff TVS.[4]

## Order unit norm

Suppose ${\displaystyle (X,\leq )}$  is an ordered vector space over the reals with an order unit ${\displaystyle u}$  whose order is Archimedean and let ${\displaystyle U=[-u,u].}$  Then the Minkowski functional ${\displaystyle p_{U}}$  of ${\displaystyle U}$  (defined by ${\displaystyle p_{U}(x):=\left\{r>0:x\in r[-u,u]\right\}}$ ) is a norm called the order unit norm. It satisfies ${\displaystyle p_{U}(u)=1}$  and the closed unit ball determined by ${\displaystyle p_{U}}$  is equal to ${\displaystyle [-u,u]}$  (that is, ${\displaystyle [-u,u]=\{x\in X:p_{U}(x)\leq 1\}.}$ [3]

### Examples

The space ${\displaystyle l_{\infty }(S,\mathbb {R} )}$  of bounded real-valued maps on a set ${\displaystyle S}$  with the pointwise order is Archimedean ordered with an order unit ${\displaystyle u:=1}$  (that is, the function that is identically ${\displaystyle 1}$  on ${\displaystyle S}$ ). The order unit norm on ${\displaystyle l_{\infty }(S,\mathbb {R} )}$  is identical to the usual sup norm: ${\displaystyle \|f\|:=\sup _{}|f(S)|.}$ [3]

## Examples

Every order complete vector lattice is Archimedean ordered.[5] A finite-dimensional vector lattice of dimension ${\displaystyle n}$  is Archimedean ordered if and only if it is isomorphic to ${\displaystyle \mathbb {R} ^{n}}$  with its canonical order.[5] However, a totally ordered vector order of dimension ${\displaystyle \,>1}$  can not be Archimedean ordered.[5] There exist ordered vector spaces that are almost Archimedean but not Archimedean.

The Euclidean space ${\displaystyle \mathbb {R} ^{2}}$  over the reals with the lexicographic order is not Archimedean ordered since ${\displaystyle r(0,1)\leq (1,1)}$  for every ${\displaystyle r>0}$  but ${\displaystyle (0,1)\neq (0,0).}$ [3]