Archimedean ordered vector space

Template:Short description In mathematics, specifically in order theory, a binary relation $\le$ 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 $nx\le y$ for all positive integers $n,$ then necessarily $x\le 0.$ An Archimedean (pre)ordered vector space is a (pre)ordered vector space whose order is Archimedean.Template:Sfn A preordered vector space $X$ is called almost Archimedean if for all $x\in X,$ whenever there exists a $y\in X$ such that $-n^{-1}y\le x\le n^{-1}y$ for all positive integers $n,$ then $x=0.$Template:Sfn

Characterizations

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

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 (topological vector space).Template:Sfn

Order unit norm

Suppose $(X,\le )$ 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 $p_U(x):=inf\{r>0:x\in r[-u,u]\}$) 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, $[-u,u]=\{x\in X:p_U(x)\le 1\}.$Template:Sfn

Examples

The space $l_{\mathrm{\infty }}(S,ℝ)$ 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 $l_{\mathrm{\infty }}(S,ℝ)$ is identical to the usual sup norm: $\Vert f\Vert :=\underset{}{sup}|f(S)|.$Template:Sfn

Examples

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

The Euclidean space ${ℝ}^2$ over the reals with the lexicographic order is Template:Em Archimedean ordered since $r(0,1)\le (1,1)$ for every $r>0$ but $(0,1)\ne (0,0).$Template:Sfn

See also

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Bibliography

• Template:Narici Beckenstein Topological Vector Spaces
• Template:Schaefer Wolff Topological Vector Spaces