Bornivorous set

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Short description: A set that can absorb any bounded subset

In functional analysis, a subset of a real or complex vector space [math]\displaystyle{ X }[/math] that has an associated vector bornology [math]\displaystyle{ \mathcal{B} }[/math] is called bornivorous and a bornivore if it absorbs every element of [math]\displaystyle{ \mathcal{B}. }[/math] If [math]\displaystyle{ X }[/math] is a topological vector space (TVS) then a subset [math]\displaystyle{ S }[/math] of [math]\displaystyle{ X }[/math] is bornivorous if it is bornivorous with respect to the von-Neumann bornology of [math]\displaystyle{ X }[/math].

Bornivorous sets play an important role in the definitions of many classes of topological vector spaces, particularly bornological spaces.

Definitions

If [math]\displaystyle{ X }[/math] is a TVS then a subset [math]\displaystyle{ S }[/math] of [math]\displaystyle{ X }[/math] is called bornivorous[1] and a bornivore if [math]\displaystyle{ S }[/math] absorbs every bounded subset of [math]\displaystyle{ X. }[/math]

An absorbing disk in a locally convex space is bornivorous if and only if its Minkowski functional is locally bounded (i.e. maps bounded sets to bounded sets).[1]

Infrabornivorous sets and infrabounded maps

A linear map between two TVSs is called infrabounded if it maps Banach disks to bounded disks.[2]

A disk in [math]\displaystyle{ X }[/math] is called infrabornivorous if it absorbs every Banach disk.[3]

An absorbing disk in a locally convex space is infrabornivorous if and only if its Minkowski functional is infrabounded.[1] A disk in a Hausdorff locally convex space is infrabornivorous if and only if it absorbs all compact disks (that is, if it is "compactivorous").[1]

Properties

Every bornivorous and infrabornivorous subset of a TVS is absorbing. In a pseudometrizable TVS, every bornivore is a neighborhood of the origin.[4]

Two TVS topologies on the same vector space have that same bounded subsets if and only if they have the same bornivores.[5]

Suppose [math]\displaystyle{ M }[/math] is a vector subspace of finite codimension in a locally convex space [math]\displaystyle{ X }[/math] and [math]\displaystyle{ B \subseteq M. }[/math] If [math]\displaystyle{ B }[/math] is a barrel (resp. bornivorous barrel, bornivorous disk) in [math]\displaystyle{ M }[/math] then there exists a barrel (resp. bornivorous barrel, bornivorous disk) [math]\displaystyle{ C }[/math] in [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ B = C \cap M. }[/math][6]

Examples and sufficient conditions

Every neighborhood of the origin in a TVS is bornivorous. The convex hull, closed convex hull, and balanced hull of a bornivorous set is again bornivorous. The preimage of a bornivore under a bounded linear map is a bornivore.[7]

If [math]\displaystyle{ X }[/math] is a TVS in which every bounded subset is contained in a finite dimensional vector subspace, then every absorbing set is a bornivore.[5]

Counter-examples

Let [math]\displaystyle{ X }[/math] be [math]\displaystyle{ \mathbb{R}^2 }[/math] as a vector space over the reals. If [math]\displaystyle{ S }[/math] is the balanced hull of the closed line segment between [math]\displaystyle{ (-1, 1) }[/math] and [math]\displaystyle{ (1, 1) }[/math] then [math]\displaystyle{ S }[/math] is not bornivorous but the convex hull of [math]\displaystyle{ S }[/math] is bornivorous. If [math]\displaystyle{ T }[/math] is the closed and "filled" triangle with vertices [math]\displaystyle{ (-1, -1), (-1, 1), }[/math] and [math]\displaystyle{ (1, 1) }[/math] then [math]\displaystyle{ T }[/math] is a convex set that is not bornivorous but its balanced hull is bornivorous.

See also

References

Bibliography