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 $$n x \leq y$$ 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 $$n x \leq u$$ 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 $$l_{\infin}(S, \R)$$ 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 $$l_{\infin}(S, \R)$$ 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 $$\R^n$$ 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 $$\R^2$$ over the reals with the lexicographic order is Archimedean ordered since $$r(0, 1) \leq (1, 1)$$ for every $$r > 0$$ but $$(0, 1) \neq (0, 0).$$