Article · Wikipedia archive · Last revised Jun 7, 2026

Bornivorous set

In functional analysis, a subset of a real or complex vector space that has an associated vector bornology is called bornivorous and a bornivore if it absorbs every element of If is a topological vector space (TVS) then a subset of is bornivorous if it is bornivorous with respect to the von-Neumann bornology of .

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

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

Definitions

If X {\displaystyle X} is a TVS then a subset S {\displaystyle S} of X {\displaystyle X} is called bornivorous1 and a bornivore if S {\displaystyle S} absorbs every bounded subset of X . {\displaystyle X.}

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 X {\displaystyle X} 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 M {\displaystyle M} is a vector subspace of finite codimension in a locally convex space X {\displaystyle X} and B M . {\displaystyle B\subseteq M.} If B {\displaystyle B} is a barrel (resp. bornivorous barrel, bornivorous disk) in M {\displaystyle M} then there exists a barrel (resp. bornivorous barrel, bornivorous disk) C {\displaystyle C} in X {\displaystyle X} such that B = C M . {\displaystyle B=C\cap M.} 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 X {\displaystyle X} 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 X {\displaystyle X} be R 2 {\displaystyle \mathbb {R} ^{2}} as a vector space over the reals. If S {\displaystyle S} is the balanced hull of the closed line segment between ( 1 , 1 ) {\displaystyle (-1,1)} and ( 1 , 1 ) {\displaystyle (1,1)} then S {\displaystyle S} is not bornivorous but the convex hull of S {\displaystyle S} is bornivorous. If T {\displaystyle T} is the closed and "filled" triangle with vertices ( 1 , 1 ) , ( 1 , 1 ) , {\displaystyle (-1,-1),(-1,1),} and ( 1 , 1 ) {\displaystyle (1,1)} then T {\displaystyle T} is a convex set that is not bornivorous but its balanced hull is bornivorous.

See also

See also

References

References

Bibliography

Bibliography