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 $-n^ y \leq x \leq n^ y$ 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_(x) := \inf\left\$) 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/math> (that is, u, u= \.

Examples

The space $l_(S, \R)$ 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 $l_(S, \R)$ 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 $(0, 1) \neq (0, 0).$

See also

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References

Bibliography

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{{Ordered topological vector spaces

Functional analysis
Order theory