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

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

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 (topological vector space).⁴

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 ${\displaystyle p_{U}(x):=\inf \left\{r>0:x\in r[-u,u]\right\}}$) 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

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

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