Archimedean ordered vector space

In mathematics, specifically in order theory, a binary relation ≤ 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 ≤ y for all positive integers n, then necessarily x ≤ 0. An Archimedean (pre)ordered vector space is a (pre)ordered vector space whose order is Archimedean.[1] 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⁻¹y ≤ x ≤ n⁻¹y for all positive integers n, then x = 0.[2]

Characterizations

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

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.[4]

Order unit norm

Suppose (X, ≤) 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 ${\displaystyle p_{U}(x):=\left\{r>0:x\in r[-u,u]\right\}}$) 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] (i.e. [-u, u] = \{ x \in X : p_U(x) ≤ 1 \}.[3]

Examples

The space 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 (i.e. the function that is identically 1 on S). The order unit norm on l_∞(S, ℝ) is identical to the usual sup norm: ${\displaystyle \left\|f\right\|:=\sup \left|f(S)\right|}$.[3]

Examples

Every order complete vector lattice is Archimedean ordered.[5] A finite-dimensional vector lattice of dimension 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 > 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 r(0, 1) ≤ (1, 1) for every r > 0 but (0, 1) ≠ (0, 0).[3]

See also

• Archimedean property
• Ordered vector space

References

1. Schaefer & Wolff 1999, pp. 204–214.
2. Schaefer & Wolff 1999, p. 254.
3. Narici 2011, pp. 139-153.
4. Schaefer & Wolff 1999, pp. 222–225.
5. Schaefer & Wolff 1999, pp. 250–257.

Sources

• Narici, Lawrence (2011). Topological vector spaces. Boca Raton, FL: CRC Press. ISBN 1-58488-866-0. OCLC 144216834.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. 3. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.CS1 maint: ref=harv (link)