Bornology

If ${\displaystyle {\mathcal {A}}}$ and ${\displaystyle {\mathcal {B}}}$ are bornologies on ${\displaystyle X}$ then ${\displaystyle {\mathcal {B}}}$ is said to be finer or stronger than ${\displaystyle {\mathcal {A}}}$ and also ${\displaystyle {\mathcal {A}}}$ is said to be coarser or weaker than ${\displaystyle {\mathcal {B}}}$ if ${\displaystyle {\mathcal {A}}\subseteq {\mathcal {B}}.}$[5]

A family of sets ${\displaystyle {\mathcal {A}}}$ is called a base or fundamental system of a bornology ${\displaystyle {\mathcal {B}}}$ if ${\displaystyle {\mathcal {A}}\subseteq {\mathcal {B}}}$ and for every ${\displaystyle B\in {\mathcal {B}},}$ there exists an ${\displaystyle A\in {\mathcal {A}}}$ such that ${\displaystyle B\subseteq A.}$

A family of sets ${\displaystyle {\mathcal {S}}}$ is called a subbase of a bornology ${\displaystyle {\mathcal {B}}}$ if ${\displaystyle {\mathcal {S}}\subseteq {\mathcal {B}}}$ and the collection of all finite unions of sets in ${\displaystyle {\mathcal {S}}}$ forms a base for ${\displaystyle {\mathcal {B}}.}$[5]

Every base for a bornology is also a subbase for it.

The intersection of any collection of (one or more) bornologies on ${\displaystyle X}$ is once again a bornology on ${\displaystyle X.}$ Such an intersection of bornologies will cover ${\displaystyle X}$ because every bornology on ${\displaystyle X}$ contains every finite subset of ${\displaystyle X}$ (that is, if ${\displaystyle {\mathcal {B}}}$ is a bornology on ${\displaystyle X}$ and ${\displaystyle F\subseteq X}$ is finite then ${\displaystyle F\in {\mathcal {B}}}$). It is readily verified that such an intersection will also be closed under (subset) inclusion and finite unions and thus will be a bornology on ${\displaystyle X.}$

Given a collection ${\displaystyle {\mathcal {S}}}$ of subsets of ${\displaystyle X,}$ the smallest bornology on ${\displaystyle X}$ containing ${\displaystyle {\mathcal {S}}}$ is called the bornology generated by ${\displaystyle {\mathcal {S}}}$.[5] It is equal to the intersection of all bornologies on ${\displaystyle X}$ that contain ${\displaystyle {\mathcal {S}}}$ as a subset. This intersection is well-defined because the power set ${\displaystyle \wp (X)}$ of ${\displaystyle X}$ is always a bornology on ${\displaystyle X,}$ so every family ${\displaystyle {\mathcal {S}}}$ of subsets of ${\displaystyle X}$ is always contained in at least one bornology on ${\displaystyle X.}$

Suppose that ${\displaystyle X}$ and ${\displaystyle Y}$ are topological vector spaces (TVSs) and ${\displaystyle f:X\to Y}$ is a linear map. Then the following statements are equivalent:

1. ${\displaystyle f}$ is a (locally) bounded map;
2. For every bornivorous (that is, bounded in the bornological sense) disk ${\displaystyle D}$ in ${\displaystyle Y,}$ ${\displaystyle f^{-1}(D)}$ is also bornivorous.[5]

If ${\displaystyle X}$ and ${\displaystyle Y}$ are locally convex then this list may be extended to include:

3. ${\displaystyle f}$ takes bounded disks to bounded disks;

If ${\displaystyle X}$ is a seminormed space and ${\displaystyle Y}$ is locally convex then this list may be extended to include:

4. ${\displaystyle f}$ maps null sequences (that is, sequences converging to the origin ${\displaystyle 0}$) into bounded subsets of ${\displaystyle Y.}$[5]

If ${\displaystyle f:X\to Y}$ is a continuous linear operator between two topological vector spaces (they need not even be Hausdorff), then it is a bounded linear operator (when ${\displaystyle X}$ and ${\displaystyle Y}$ have their von-Neumann bornologies). The converse is in general false.

A sequentially continuous map ${\displaystyle f:X\to Y}$ between two TVSs is necessarily locally bounded.[5]

If ${\displaystyle f:S\to X}$ is a map and ${\displaystyle {\mathcal {B}}}$ is a bornology on ${\displaystyle X,}$ then ${\displaystyle \left[f^{-1}({\mathcal {B}})\right]}$ denotes the bornology generated by ${\displaystyle f^{-1}({\mathcal {B}}):=\left\{f^{-1}(B):B\in {\mathcal {B}}\right\},}$ which is called it the inverse image bornology or the initial bornology induced by ${\displaystyle f}$ on ${\displaystyle S.}$[5]

Let ${\displaystyle S}$ be a set, ${\displaystyle \left(T_{i},{\mathcal {B}}_{i}\right)_{i\in I}}$ be an ${\displaystyle I}$-indexed family of bounded structures, and let ${\displaystyle \left(f_{i}\right)_{i\in I}}$ be an ${\displaystyle I}$-indexed family of maps where ${\displaystyle f_{i}:S\to T_{i}}$ for every ${\displaystyle i\in I.}$ The inverse image bornology ${\displaystyle {\mathcal {A}}}$ on ${\displaystyle S}$ determined by these maps is the strongest bornology on ${\displaystyle S}$ making each ${\displaystyle f_{i}:(S,{\mathcal {A}})\to \left(T_{i},{\mathcal {B}}_{i}\right)}$ locally bounded. This bornology is equal to[5] ${\displaystyle {\textstyle \bigcap \limits _{i\in I}\left[f^{-1}\left({\mathcal {B}}_{i}\right)\right]}.}$

Let ${\displaystyle S}$ be a set, ${\displaystyle \left(T_{i},{\mathcal {B}}_{i}\right)_{i\in I}}$ be an ${\displaystyle I}$-indexed family of bounded structures, and let ${\displaystyle \left(f_{i}\right)_{i\in I}}$ be an ${\displaystyle I}$-indexed family of maps where ${\displaystyle f_{i}:T_{i}\to S}$ for every ${\displaystyle i\in I.}$ The direct image bornology ${\displaystyle {\mathcal {A}}}$ on ${\displaystyle S}$ determined by these maps is the weakest bornology on ${\displaystyle S}$ making each ${\displaystyle f_{i}:\left(T_{i},{\mathcal {B}}_{i}\right)\to (S,{\mathcal {A}})}$ locally bounded. If for each ${\displaystyle i\in I,}$ ${\displaystyle {\mathcal {A}}_{i}}$ denotes the bornology generated by ${\displaystyle f\left({\mathcal {B}}_{i}\right),}$ then this bornology is equal to the collection of all subsets ${\displaystyle A}$ of ${\displaystyle S}$ of the form ${\displaystyle \cup _{i\in I}A_{i}}$ where each ${\displaystyle A_{i}\in {\mathcal {A}}_{i}}$ and all but finitely many ${\displaystyle A_{i}}$ are empty.[5]

Suppose that ${\displaystyle (X,{\mathcal {B}})}$ is a bounded structure and ${\displaystyle S}$ be a subset of ${\displaystyle X.}$ The subspace bornology ${\displaystyle {\mathcal {A}}}$ on ${\displaystyle S}$ is the finest bornology on ${\displaystyle S}$ making the inclusion map ${\displaystyle (S,{\mathcal {A}})\to (X,{\mathcal {B}})}$ of ${\displaystyle S}$ into ${\displaystyle X}$ (defined by ${\displaystyle s\mapsto s}$) locally bounded.[5]

Let ${\displaystyle \left(X_{i},{\mathcal {B}}_{i}\right)_{i\in I}}$ be an ${\displaystyle I}$-indexed family of bounded structures, let ${\displaystyle X={\textstyle \prod \limits _{i\in I}X_{i}},}$ and for each ${\displaystyle i\in I,}$ let ${\displaystyle f_{i}:X\to X_{i}}$ denote the canonical projection. The product bornology on ${\displaystyle X}$ is the inverse image bornology determined by the canonical projections ${\displaystyle f_{i}:X\to X_{i}.}$ That is, it is the strongest bornology on ${\displaystyle X}$ making each of the canonical projections locally bounded. A base for the product bornology is given by ${\displaystyle {\textstyle \left\{\prod \limits _{i\in I}B_{i}~:~B_{i}\in {\mathcal {B}}_{i}{\text{ for all }}i\in I\right\}}.}$[5]

A subset of a topological space ${\displaystyle X}$ is called relatively compact if its closure is a compact subspace of ${\displaystyle X.}$ For any topological space ${\displaystyle X}$ in which singleton subsets are relatively compact (such as a T1 space), the set of all relatively compact subsets of ${\displaystyle X}$ form a bornology on ${\displaystyle X}$ called the compact bornology on ${\displaystyle X.}$[5] Every continuous map between T1 spaces is bounded with respect to their compact bornologies.

The set of relatively compact subsets of ${\displaystyle \mathbb {R} }$ form a bornology on ${\displaystyle \mathbb {R} .}$ A base for this bornology is given by all closed intervals of the form ${\displaystyle [-n,n]}$ for ${\displaystyle n=1,2,3,\ldots .}$

Given a metric space ${\displaystyle (X,d),}$ the metric bornology consists of all subsets ${\displaystyle S\subseteq X}$ such that the supremum ${\displaystyle \sup _{s,t\in S}d(s,t)<\infty }$ is finite.

Similarly, given a measure space ${\displaystyle (X,\Omega ,\mu ),}$ the family of all measurable subsets ${\displaystyle S\in \Omega }$ of finite measure (meaning ${\displaystyle \mu (S)<\infty }$) form a bornology on ${\displaystyle X.}$

Suppose that ${\displaystyle X}$ is a topological space and ${\displaystyle {\mathcal {B}}}$ is a bornology on ${\displaystyle X.}$

The bornology generated by the set of all topological interiors of sets in ${\displaystyle {\mathcal {B}}}$ (that is, generated by ${\displaystyle \{\operatorname {int} B:B\in {\mathcal {B}}\}}$ is called the interior of ${\displaystyle {\mathcal {B}}}$ and is denoted by ${\displaystyle \operatorname {int} {\mathcal {B}}.}$[5] The bornology ${\displaystyle {\mathcal {B}}}$ is called open if ${\displaystyle {\mathcal {B}}=\operatorname {int} {\mathcal {B}}.}$

The bornology generated by the set of all topological closures of sets in ${\displaystyle {\mathcal {B}}}$ (that is, generated by ${\displaystyle \{\operatorname {cl} B:B\in {\mathcal {B}}\}}$) is called the closure of ${\displaystyle {\mathcal {B}}}$ and is denoted by ${\displaystyle \operatorname {cl} {\mathcal {B}}.}$[5] We necessarily have ${\displaystyle \operatorname {int} {\mathcal {B}}\subseteq {\mathcal {B}}\subseteq \operatorname {cl} {\mathcal {B}}.}$

The bornology ${\displaystyle {\mathcal {B}}}$ is called closed if it satisfies any of the following equivalent conditions:

1. ${\displaystyle {\mathcal {B}}=\operatorname {cl} {\mathcal {B}};}$
2. the closed subsets of ${\displaystyle X}$ generate ${\displaystyle {\mathcal {B}}}$;[5]
3. the closure of every ${\displaystyle B\in {\mathcal {B}}}$ belongs to ${\displaystyle {\mathcal {B}}.}$[5]

The bornology ${\displaystyle {\mathcal {B}}}$ is called proper if ${\displaystyle {\mathcal {B}}}$ is both open and closed.[5]

The topological space ${\displaystyle X}$ is called locally ${\displaystyle {\mathcal {B}}}$-bounded or just locally bounded if every ${\displaystyle x\in X}$ has a neighborhood that belongs to ${\displaystyle {\mathcal {B}}.}$ Every compact subset of a locally bounded topological space is bounded.[5]

If ${\displaystyle X}$ is a topological vector space (TVS) then the set of all bounded subsets of ${\displaystyle X}$ form a bornology (indeed, even a vector bornology) on ${\displaystyle X}$ called the von Neumann bornology of ${\displaystyle X}$, the usual bornology, or simply the bornology of ${\displaystyle X}$ and is referred to as natural boundedness.[5] In any locally convex TVS ${\displaystyle X,}$ the set of all closed bounded disks forms a base for the usual bornology of ${\displaystyle X.}$[5]

A linear map between two bornological spaces is continuous if and only if it is bounded (with respect to the usual bornologies).

Suppose that ${\displaystyle X}$ is a commutative topological ring. A subset ${\displaystyle S}$ of ${\displaystyle X}$ is called a bounded set if for each neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X,}$ there exists a neighborhood ${\displaystyle V}$ of the origin in ${\displaystyle X}$ such that ${\displaystyle SV\subseteq U.}$[5]