Archimedean ordered vector space

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Short description: A binary relation on a vector space

In mathematics, specifically in order theory, a binary relation [math]\displaystyle{ \,\leq\, }[/math] on a vector space [math]\displaystyle{ X }[/math] over the real or complex numbers is called Archimedean if for all [math]\displaystyle{ x \in X, }[/math] whenever there exists some [math]\displaystyle{ y \in X }[/math] such that [math]\displaystyle{ n x \leq y }[/math] for all positive integers [math]\displaystyle{ n, }[/math] then necessarily [math]\displaystyle{ x \leq 0. }[/math] An Archimedean (pre)ordered vector space is a (pre)ordered vector space whose order is Archimedean.[1] A preordered vector space [math]\displaystyle{ X }[/math] is called almost Archimedean if for all [math]\displaystyle{ x \in X, }[/math] whenever there exists a [math]\displaystyle{ y \in X }[/math] such that [math]\displaystyle{ -n^{-1} y \leq x \leq n^{-1} y }[/math] for all positive integers [math]\displaystyle{ n, }[/math] then[math]\displaystyle{ x = 0. }[/math][2]

Characterizations

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

Properties

Let [math]\displaystyle{ X }[/math] be an ordered vector space over the reals that is finite-dimensional. Then the order of [math]\displaystyle{ X }[/math] is Archimedean if and only if the positive cone of [math]\displaystyle{ X }[/math] is closed for the unique topology under which [math]\displaystyle{ X }[/math] is a Hausdorff TVS.[4]

Order unit norm

Suppose [math]\displaystyle{ (X, \leq) }[/math] is an ordered vector space over the reals with an order unit [math]\displaystyle{ u }[/math] whose order is Archimedean and let [math]\displaystyle{ U = [-u, u]. }[/math] Then the Minkowski functional [math]\displaystyle{ p_U }[/math] of [math]\displaystyle{ U }[/math] (defined by [math]\displaystyle{ p_{U}(x) := \inf\left\{ r \gt 0 : x \in r [-u, u] \right\} }[/math]) is a norm called the order unit norm. It satisfies [math]\displaystyle{ p_U(u) = 1 }[/math] and the closed unit ball determined by [math]\displaystyle{ p_U }[/math] is equal to [math]\displaystyle{ [-u, u] }[/math] (that is, [math]\displaystyle{ [-u, u] = \{ x\in X : p_U(x) \leq 1 \}. }[/math][3]

Examples

The space [math]\displaystyle{ l_{\infin}(S, \R) }[/math] of bounded real-valued maps on a set [math]\displaystyle{ S }[/math] with the pointwise order is Archimedean ordered with an order unit [math]\displaystyle{ u := 1 }[/math] (that is, the function that is identically [math]\displaystyle{ 1 }[/math] on [math]\displaystyle{ S }[/math]). The order unit norm on [math]\displaystyle{ l_{\infin}(S, \R) }[/math] is identical to the usual sup norm: [math]\displaystyle{ \|f\| := \sup_{} |f(S)|. }[/math][3]

Examples

Every order complete vector lattice is Archimedean ordered.[5] A finite-dimensional vector lattice of dimension [math]\displaystyle{ n }[/math] is Archimedean ordered if and only if it is isomorphic to [math]\displaystyle{ \R^n }[/math] with its canonical order.[5] However, a totally ordered vector order of dimension [math]\displaystyle{ \,\gt 1 }[/math] can not be Archimedean ordered.[5] There exist ordered vector spaces that are almost Archimedean but not Archimedean.

The Euclidean space [math]\displaystyle{ \R^2 }[/math] over the reals with the lexicographic order is not Archimedean ordered since [math]\displaystyle{ r(0, 1) \leq (1, 1) }[/math] for every [math]\displaystyle{ r \gt 0 }[/math] but [math]\displaystyle{ (0, 1) \neq (0, 0). }[/math][3]

See also

References

  1. Schaefer & Wolff 1999, pp. 204–214.
  2. Schaefer & Wolff 1999, p. 254.
  3. 3.0 3.1 3.2 3.3 Narici & Beckenstein 2011, pp. 139-153.
  4. Schaefer & Wolff 1999, pp. 222–225.
  5. 5.0 5.1 5.2 Schaefer & Wolff 1999, pp. 250–257.

Bibliography