In mathematics, particularly in functional analysis, a bornological space is a type of space which, in some sense, possesses the minimum amount of structure needed to address questions of boundedness of sets and linear maps, in the same way that a topological space possesses the minimum amount of structure needed to address questions of continuity.
Bornological spaces are distinguished by the property that a linear map from a bornological space into any locally convex spaces is continuous if and only if it is a bounded linear operator.
Bornological spaces were first studied by George Mackey.^{[citation needed]} The name was coined by Bourbaki^{[citation needed]} after borné, the French word for "bounded".
Bornologies and bounded maps
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A bornology on a set $X$ is a collection ${\mathcal {B}}$ of subsets of $X$ that satisfy all the following conditions:
${\mathcal {B}}$ covers $X;$ that is, $X=\cup {\mathcal {B}}$;
${\mathcal {B}}$ is stable under inclusions; that is, if $B\in {\mathcal {B}}$ and $A\subseteq B,$ then $A\in {\mathcal {B}}$;
${\mathcal {B}}$ is stable under finite unions; that is, if $B_{1},\ldots ,B_{n}\in {\mathcal {B}}$ then $B_{1}\cup \cdots \cup B_{n}\in {\mathcal {B}}$;
Elements of the collection ${\mathcal {B}}$ are called ${\mathcal {B}}$-bounded or simply bounded sets if ${\mathcal {B}}$ is understood.^{[1]}
The pair $(X,{\mathcal {B}})$ is called a bounded structure or a bornological set.^{[1]}
A base or fundamental system of a bornology ${\mathcal {B}}$ is a subset ${\mathcal {B}}_{0}$ of ${\mathcal {B}}$ such that each element of ${\mathcal {B}}$ is a subset of some element of ${\mathcal {B}}_{0}.$ Given a collection ${\mathcal {S}}$ of subsets of $X,$ the smallest bornology containing ${\mathcal {S}}$ is called the bornology generated by ${\mathcal {S}}.$^{[2]}
If $(X,{\mathcal {B}})$ and $(Y,{\mathcal {C}})$ are bornological sets then their product bornology on $X\times Y$ is the bornology having as a base the collection of all sets of the form $B\times C,$ where $B\in {\mathcal {B}}$ and $C\in {\mathcal {C}}.$^{[2]}
A subset of $X\times Y$ is bounded in the product bornology if and only if its image under the canonical projections onto $X$ and $Y$ are both bounded.
Bounded maps
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If $(X,{\mathcal {B}})$ and $(Y,{\mathcal {C}})$ are bornological sets then a function $f:X\to Y$ is said to be a locally bounded map or a bounded map (with respect to these bornologies) if it maps ${\mathcal {B}}$-bounded subsets of $X$ to ${\mathcal {C}}$-bounded subsets of $Y;$ that is, if $f({\mathcal {B}})\subseteq {\mathcal {C}}.$^{[2]}
If in addition $f$ is a bijection and $f^{-1}$ is also bounded then $f$ is called a bornological isomorphism.
Vector bornologies
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Let $X$ be a vector space over a field$\mathbb {K}$ where $\mathbb {K}$ has a bornology ${\mathcal {B}}_{\mathbb {K} }.$
A bornology ${\mathcal {B}}$ on $X$ is called a vector bornology on $X$ if it is stable under vector addition, scalar multiplication, and the formation of balanced hulls (i.e. if the sum of two bounded sets is bounded, etc.).
If $X$ is a topological vector space (TVS) and ${\mathcal {B}}$ is a bornology on $X,$ then the following are equivalent:
${\mathcal {B}}$ is a vector bornology;
Finite sums and balanced hulls of ${\mathcal {B}}$-bounded sets are ${\mathcal {B}}$-bounded;^{[2]}
The scalar multiplication map $\mathbb {K} \times X\to X$ defined by $(s,x)\mapsto sx$ and the addition map $X\times X\to X$ defined by $(x,y)\mapsto x+y,$ are both bounded when their domains carry their product bornologies (i.e. they map bounded subsets to bounded subsets).^{[2]}
A vector bornology ${\mathcal {B}}$ is called a convex vector bornology if it is stable under the formation of convex hulls (i.e. the convex hull of a bounded set is bounded) then ${\mathcal {B}}.$
And a vector bornology ${\mathcal {B}}$ is called separated if the only bounded vector subspace of $X$ is the 0-dimensional trivial space $\{0\}.$
Usually, $\mathbb {K}$ is either the real or complex numbers, in which case a vector bornology ${\mathcal {B}}$ on $X$ will be called a convex vector bornology if ${\mathcal {B}}$ has a base consisting of convex sets.
Bornivorous subsets
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A subset $A$ of $X$ is called bornivorous and a bornivore if it absorbs every bounded set.
In a vector bornology, $A$ is bornivorous if it absorbs every bounded balanced set and in a convex vector bornology $A$ is bornivorous if it absorbs every bounded disk.
Two TVS topologies on the same vector space have that same bounded subsets if and only if they have the same bornivores.^{[3]}
A sequence $x_{\bullet }=(x_{i})_{i=1}^{\infty }$ in a TVS $X$ is said to be Mackey convergent to $0$ if there exists a sequence of positive real numbers $r_{\bullet }=(r_{i})_{i=1}^{\infty }$ diverging to $\infty$ such that $(r_{i}x_{i})_{i=1}^{\infty }$ converges to $0$ in $X.$^{[5]}
Bornology of a topological vector space
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Every topological vector space$X,$ at least on a non discrete valued field gives a bornology on $X$ by defining a subset $B\subseteq X$ to be bounded (or von-Neumann bounded), if and only if for all open sets $U\subseteq X$ containing zero there exists a $r>0$ with $B\subseteq rU.$
If $X$ is a locally convextopological vector space then $B\subseteq X$ is bounded if and only if all continuous semi-norms on $X$ are bounded on $B.$
The set of all bounded subsets of a topological vector space $X$ is called the bornology or the von Neumann bornology of $X.$
If ${\mathcal {B}}$ is a convex vector bornology on a vector space $X,$ then the collection ${\mathcal {N}}_{\mathcal {B}}(0)$ of all convex balanced subsets of $X$ that are bornivorous forms a neighborhood basis at the origin for a locally convex topology on $X$ called the topology induced by ${\mathcal {B}}$.^{[4]}
If $(X,\tau )$ is a TVS then the bornological space associated with $X$ is the vector space $X$ endowed with the locally convex topology induced by the von Neumann bornology of $(X,\tau ).$^{[4]}
Theorem^{[4]} — Let $X$ and $Y$ be locally convex TVS and let $X_{b}$ denote $X$ endowed with the topology induced by von Neumann bornology of $X.$ Define $Y_{b}$ similarly. Then a linear map $L:X\to Y$ is a bounded linear operator if and only if $L:X_{b}\to Y$ is continuous.
Moreover, if $X$ is bornological, $Y$ is Hausdorff, and $L:X\to Y$ is continuous linear map then so is $L:X\to Y_{b}.$ If in addition $X$ is also ultrabornological, then the continuity of $L:X\to Y$ implies the continuity of $L:X\to Y_{ub},$ where $Y_{ub}$ is the ultrabornological space associated with $Y.$
Quasi-bornological spaces
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Quasi-bornological spaces where introduced by S. Iyahen in 1968.^{[6]}
A topological vector space (TVS) $(X,\tau )$ with a continuous dual$X^{\prime }$ is called a quasi-bornological space^{[6]} if any of the following equivalent conditions holds:
Every bounded linear operator from $X$ into a complete metrizable TVS is continuous.^{[6]}^{[7]}
Every knot in a bornivorous string is a neighborhood of the origin.^{[6]}
Every pseudometrizable TVS is quasi-bornological. ^{[6]}
A TVS $(X,\tau )$ in which every bornivorous set is a neighborhood of the origin is a quasi-bornological space.^{[8]}
If $X$ is a quasi-bornological TVS then the finest locally convex topology on $X$ that is coarser than $\tau$ makes $X$ into a locally convex bornological space.
Bornological space
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In functional analysis, a locally convex topological vector space is a bornological space if its topology can be recovered from its bornology in a natural way.
Every locally convex quasi-bornological space is bornological but there exist bornological spaces that are not quasi-bornological.^{[6]}
A topological vector space (TVS) $(X,\tau )$ with a continuous dual$X^{\prime }$ is called a bornological space if it is locally convex and any of the following equivalent conditions holds:
Every convex, balanced, and bornivorous set in $X$ is a neighborhood of zero.^{[4]}
Recall that a linear map is bounded if and only if it maps any sequence converging to $0$ in the domain to a bounded subset of the codomain.^{[4]} In particular, any linear map that is sequentially continuous at the origin is bounded.
Every bounded linear operator from $X$ into a seminormed space is continuous.^{[4]}
Every bounded linear operator from $X$ into a Banach space is continuous.^{[4]}
The locally convex topology induced by the von Neumann bornology on $X$ is the same as $\tau ,$$X$'s given topology.
Every bounded seminorm on $X$ is continuous.^{[4]}
Any other Hausdorff locally convex topological vector space topology on $X$ that has the same (von Neumann) bornology as $(X,\tau )$ is necessarily coarser than $\tau .$
$X$ is the inductive limit of normed spaces.^{[4]}
$X$ is the inductive limit of the normed spaces $X_{D}$ as $D$ varies over the closed and bounded disks of $X$ (or as $D$ varies over the bounded disks of $X$).^{[4]}
$X$ carries the Mackey topology $\tau (X,X^{\prime })$ and all bounded linear functionals on $X$ are continuous.^{[4]}
$X$ has both of the following properties:
$X$ is convex-sequential or C-sequential, which means that every convex sequentially open subset of $X$ is open,
$X$ is sequentially bornological or S-bornological, which means that every convex and bornivorous subset of $X$ is sequentially open.
where a subset $A$ of $X$ is called sequentially open if every sequence converging to $0$ eventually belongs to $A.$
Every sequentially continuous linear operator from a locally convex bornological space into a locally convex TVS is continuous,^{[4]} where recall that a linear operator is sequentially continuous if and only if it is sequentially continuous at the origin.
Thus for linear maps from a bornological space into a locally convex space, continuity is equivalent to sequential continuity at the origin. More generally, we even have the following:
Any linear map $F:X\to Y$ from a locally convex bornological space into a locally convex space $Y$ that maps null sequences in $X$ to bounded subsets of $Y$ is necessarily continuous.
Sufficient conditions
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Mackey–Ulam theorem^{[9]} — The product of a collection $X_{\bullet }=(X_{i})_{i\in I}$ locally convex bornological spaces is bornological if and only if $I$ does not admit an Ulam measure.
As a consequent of the Mackey–Ulam theorem, "for all practical purposes, the product of bornological spaces
is bornological."^{[9]}
The following topological vector spaces are all bornological:
There exists a bornological LB-space whose strong bidual is not bornological.^{[13]}
A closed vector subspace of a locally convex bornological space is not necessarily bornological.^{[4]}^{[14]}
There exists a closed vector subspace of a locally convex bornological space that is complete (and so sequentially complete) but neither barrelled nor bornological.^{[4]}
Bornological spaces need not be barrelled and barrelled spaces need not be bornological.^{[4]} Because every locally convex ultrabornological space is barrelled,^{[4]} it follows that a bornological space is not necessarily ultrabornological.
Every locally convex bornological space is infrabarrelled.^{[4]}
Every Hausdorff sequentially complete bornological TVS is ultrabornological.^{[4]}
Thus every complete Hausdorff bornological space is ultrabornological.
In particular, every Fréchet space is ultrabornological.^{[4]}
The finite product of locally convex ultrabornological spaces is ultrabornological.^{[4]}
Every Hausdorff bornological space is quasi-barrelled.^{[15]}
Given a bornological space $X$ with continuous dual$X^{\prime },$ the topology of $X$ coincides with the Mackey topology$\tau (X,X^{\prime }).$
In particular, bornological spaces are Mackey spaces.
Every quasi-complete (i.e. all closed and bounded subsets are complete) bornological space is barrelled. There exist, however, bornological spaces that are not barrelled.
Every bornological space is the inductive limit of normed spaces (and Banach spaces if the space is also quasi-complete).
Let $X$ be a metrizable locally convex space with continuous dual $X^{\prime }.$ Then the following are equivalent:
If $L:X\to Y$ is a linear map between locally convex spaces and if $X$ is bornological, then the following are equivalent:
$L:X\to Y$ is continuous.
$L:X\to Y$ is sequentially continuous.^{[4]}
For every set $B\subseteq X$ that's bounded in $X,$$L(B)$ is bounded.
If $x_{\bullet }=(x_{i})_{i=1}^{\infty }$ is a null sequence in $X$ then $L\circ x_{\bullet }=(L(x_{i}))_{i=1}^{\infty }$ is a null sequence in $Y.$
If $x_{\bullet }=(x_{i})_{i=1}^{\infty }$ is a Mackey convergent null sequence in $X$ then $L\circ x_{\bullet }=(L(x_{i}))_{i=1}^{\infty }$ is a bounded subset of $Y.$
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