I want to understand in what sense topology is dual to bornology at a most basic level. Therefore, I rephrased the definition of a bornology in the following way:

Let $X$ be a set and let $\mathcal{P}(X)$ be the bounded lattice induced by the power set $P(X)$ together with unions, intersections, $X$ and $\emptyset$ as joints, meets, $1$ and $0$. An ideal $\beta \subseteq \mathcal{P}(X)$ is called a bornology if $\bigvee_{B \in \beta} \, B=1.$

This should be equivalent to the usual definition. An obvious dualization of this is a filter $\nu \subseteq \mathcal{P}(X)$, s.t. $\bigwedge_{N \in \nu}=0$.

This should be seen as the set of all neighbourhoods with respect to some topology on $X$ and the question is now how to extract a topology from it. I know that there is a unique topology for every neighbourhood system $\{N(x)\}_{x \in X}$ and of course I can define a trivial neighbourhood system from a given filter $\nu \subseteq \mathcal{P}(X)$ by setting $N(x)$ to be the subfilter of sets in $\nu$ containing $x$. This leads to the finest possible topology, but what I actually want is a unique coarsest neighbourhood system (s.t. $\nu = \bigcup_x \, N(x) $)! Unfortunately, I don't really have an idea how to start showing the existence of such a thing… do you? Of course, any other idea how to define open sets is welcome!

EDIT 1: it seems that coarsest is not the right property here, since for a given topology, the neighbourhoodfilter is in general not the coarsest neighbourhood system.

EDIT 2: I guess I made a really stupid mistake here: the filter $\nu$ cannot distinguish between topologies $\tau$, $\tilde{\tau}$ with neighbourhood filters $N(x)$, $\tilde{N}(x)$, s.t. $\bigcup_x \, N(x) =\bigcup_x \, \tilde{N}(x)= \nu$….

EDIT 3: For topological vector spaces $(X,\tau)$, $(X,\tilde{\tau})$ it should be true that from $\bigcup_x \, N(x) =\bigcup_x \, \tilde{N}(x)$ follows $N(0)=\tilde{N}(0)$, thus $\tau = \tilde{\tau}$. I used both continuity of multiplication and addition. Moreover, if $\tau$ is nontrivial, we have $\bigcap_{N \in \bigcup_x N(x)} \, N = \emptyset$. I didn't use that $\tau$ is Hausdorff.


3 Answers 3


It was not clear to me at first what your question has to do with bornologies, but now (EDITED) I see it. Any collection $\nu$ of subsets of $X$ (I assume that $X$ is nonvoid through the remainder of the answer since $X=\varnothing$ has no filters) is a subbasis of a unique topology $\tau_\nu$ on $X$ - to wit, the intersection of all topologies on $X$ containing $\nu$. The fact that $\nu$ is a filter will ensure that $\tau_\nu=\nu\cup\{\varnothing\}$. More precisely:

  • Since $\nu$ is closed under finite intersections, we conclude that $\nu$ is even a basis of $\tau_\nu$. This implies that $\tau_\nu$ is the unique topology on $X$ for which $\nu$ is a basis and thus the coarsest topology on $X$ containing $\nu$;

  • Since $\nu$ is closed under taking supersets (which, by the way, entails the fact that $1=X\in\nu$, implicitly used above), we conclude that any nonvoid open subset $U\in\tau_\nu$, being an union of members of $\nu$, belongs itself to $\nu$. Therefore, $\nu\cup\{\varnothing\}$ satisfies all the axioms for a topology on $X$. Since $\tau_\nu$ is the coarsest topology on $X$ containing $\nu$, we must have $\tau_\nu=\nu\cup\{\varnothing\}$, as claimed.

The property $\bigcap_{U\in\nu}U=\varnothing$ dual to a bornology being a cover of $X$ amounts to $\tau_\nu$ being $T_1$ and $X$ having at least two points (EDIT: in a previous version of the answer, it was suggested that $\tau_\nu$ was in this case only $T_0$ at best), since it means that for all $p\in X$ there is a $B\in\beta$ such that $x\not\in B$ (if $X=\{p\}$ then necessarily $\nu=\{X\}$ and therefore this property fails. Likewise, a singleton set admits no bornology which covers it). Therefore, given $X\ni p_j\not\in B_j\in\nu$, $j=1,2$ with $p_1\neq p_2$, we have that $B'_j=(B_1\cap B_2)\cup\{p_j\}\in\nu$ for $j=1,2$ with $p_1\not\in B'_2$, $p_2\not\in B'_1$. Conversely, if $\tau_\nu$ is $T_1$ and $X$ has at least two points, this implies that for all $p\in X$ we have at least one $B\in\nu$ which does not contain $p$, as claimed. On the other hand, $\nu=$ the collection of all nonvoid open subsets of $X$ being closed under taking supersets is considerably stronger than just being closed under unions, so this $\tau_\nu$ may be a very fine and disconnected topology, as the extreme case of Stone duality for complete Boolean algebras shows.

Remark: proper filters must satisfy $\varnothing\not\in\nu$, so including $0=\varnothing$ to $\tau_\nu$ by hand is necessary.

If $X$ is a vector space (over $\mathbb{R}$ or $\mathbb{C}$) and $\beta$ is a convex vector bornology (i.e. closed under addition, scalar multiplication and absolutely convex hulls), there is a different way to define a topology on $X$ - namely, you adopt as your neighborhood filter of zero the collection of all $\beta$-bornivorous disks (i.e. absolutely convex subsets that absorb all members of $\beta$). This defines a locally convex topology on $X$ - the finest such one w.r.t. which the bounded subsets are precisely the members of $\beta$. However, it is clear that the vector space structure of $X$ plays an essential role in this case. Interestingly, such a topology is Hausdorff if and only if $\beta$ covers $X$, and it is well known that a vector space topology in $X$ is Hausdorff if and only if it is $T_1$ (in fact, it is even completely regular in this case).

The duality between (convex vector) bornology and (locally convex vector) topology acquires a deeper meaning in the theory of locally convex vector spaces, since compatible (locally convex vector) topologies on (topological) dual spaces are defined in terms of (convex vector) bornologies of the original spaces, and vice-versa - more precisely, the seminorms in the dual space $E'$ defining such a topology may be taken as supremum seminorms over the members of a (convex vector) bornology $\beta$ on the original space $E$, provided $\beta$ is compatible with the topology of $E$ (i.e. consisting of bounded subsets of $E$). This can be seen as an extension of the concept of compact-open topology in function spaces. Again, this topology is $T_1$ (hence Hausdorff) if and only if $\beta$ covers $E$, since the former amounts to the above seminorms separating points in $E'$. Conversely, equicontinuous subsets of $E'$ form a (convex vector) bornology therein for each (locally convex vector) topology on $E$ for which $E'$ as the topological dual of $E$. However, since any singleton set in $E'$ is trivially equicontinuous, such a bornology always covers $E'$. The book "Bornologies and Functional Analysis" by Henri Hogbe-Nlend (North-Holland, 1977) discusses such topics in depth.

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    $\begingroup$ Thx for your answer! So, my intention was to see if the data of a topology are completely dual to that of a bornology.. the dual notion of an ideal is a filter, but not every topology is a filter so I interpreted $\nu$ as the neighbourhoods... as I mentioned in the edit, this is not enough to rebuild a unique topology $\endgroup$ May 12, 2016 at 6:54
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    $\begingroup$ It is enough if you require $\nu\cup\{X\}$ to be a topological basis of the sought-for topology (if you were dealing with Boolean algebras instead of just lattices, a filter should contain 1=$X$, so in this case it would amount to requiring that $\nu$ itself to be a topological basis). Specifying a topological basis for $X$ is just the same as specifying a neighbourhood basis for each point of $X$, so in this sense the choice of topology is unique. Of course, different topological bases may lead to the same topology - I'm assuming that's what you meant in your second edit. $\endgroup$ May 13, 2016 at 16:32
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    $\begingroup$ you're right, I can of course interpret $\nu$ directly as the topology (or basis of that), but that would be very unsatisfactory since for a topology it is very unnatural to be a filter (i know, who am I to decide..)... hence I decided to interpret it as the set of all neighbourhoods which is naturally a filter, even though in general I do not get back a unique topology... $\endgroup$ May 13, 2016 at 21:53
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    $\begingroup$ I still think you do get a unique topology, for even if you choose different neighborhood filters for each $x\in X$ from $\nu$ (subject only to the obvious criterion $U\in N(x)\Rightarrow x\in U$), the topological basis they form together is still $\nu\cup\{X\}$. Thinking better of it, since filters are closed under taking supersets, the union of any family of members of $\nu$ still belongs to $\nu$, so $\tau_\nu=\nu\cup\{\varnothing,X\}$ satisfies all axioms for a topology on $X$, and the property $\cap_{U\in\nu}U=\varnothing$ amounts to $\tau_\nu$ being $T_0$. I'll fix that in my answer. $\endgroup$ May 14, 2016 at 17:33
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    $\begingroup$ I really like the word bornivorous. $\endgroup$
    – mme
    Dec 5, 2018 at 16:04

Since this query has resurfaced, I would like to suggest that the pro and ind categories of Grothendieck are a suitable framework to discuss the topic. These are ways to extend a given categories by adding formally projective and inductive limits.

Let me start with the case of vector spaces since the situation here is more transparent and this is the one which has been studied in most detail, at least for bornologies.

If we start with the category of Banach spaces, we obtain respectively that of locally convex spaces and of convex bornological spaces (Buchwalter, Hogbe Nlend). This suggests that the natural structure of a vector space (of functions, distributions) depends on whether they are natural projective or inductive limits of Banach spaces. Thus the smooth functions on $I$, the unit interval—a lcs, the distributions thereon—a cbs. Similarly, the continuous functions on the line—a lcs, the compactly supported measures thereon—a cbs.

These examples suggest continuing by incorporating duality in its functional analytic form. The basic example is the functor which assigns to each Banach space its dual. As a contravariant functor, this extends to one from the pro category into the ind category and vice versa, in our case form the category of lcs’s into cbs’s. The fact that one can regard the dual of a lcs as a cbs (and vice versa) in a natural way has already been mentioned in the above answers and there are many indications external to category theory that is the more natural approach to duality rather than the spandrel of regarding the dual of a lcs as a lcs with various candidate topologies. (Health warning: this is very much a minority opinion).

Turning to the non linear case, the basic category is that of metric spaces with Lipschitz functions as morphisms. (In order to keep things simple, I will the blanket assumption that everything is sight is complete). The corresponding pro category is that of uniform spaces. The ind category has not, to my knowledge, been discussed but it can be defined directly as a class of bornologies, but enriched by the fact that each bounded set is provided with a suitable metric. Everybody has a collection of interesting uniform spaces in his toolbox, most of the relevant bornological spaces have, in fact, suitable metrics.

  • $\begingroup$ When $C$ has finite limits and finite colimits there always exists an adjunction $\mathcal{O}: \mathsf{Ind}(C) \leftrightarrows \mathsf{Pro}(C) :\mathcal{S}$, which is induced by Isbell duality. Yet, I think your argument is not very well organized, Top is not an accessible category by any means. $\endgroup$ Mar 6, 2020 at 8:51
  • $\begingroup$ Top doesn’t occur im my post and this is not an argument but an informal discourse which I had hoped was in the spirit of the query. $\endgroup$
    – user131781
    Mar 6, 2020 at 9:31

This might not be the kind of answer you're looking for, but I think this comparison shows that topology and bornology are not in duality at the level of sets.

A topology (using the neighborhood system definition) specifies for each point a neighborhood filter, as well as compatibility conditions between these filters. This can be used to specify the convergence of filters to points.

A bornology specifies a single ideal (that covers the entire space), with no analogue of the compatibility conditions between points. This can be used to specify a notion of "convergence to infinity" for filters.

In the vector space setting, the ability to scale this asymptotic behavior using scalar sequences enables the duality between locally convex topologies and convex bornologies, which is what people usually refer to when they speak of this duality. But the duality falls apart even for quasi-convex topological abelian groups, e.g. the failure of a Grothendieck completeness theorem that you would expect if such a duality held in more generality.

As an aside, in a way one of the aims of condensed mathematics is to improve this state of affairs creating a topos where aspects of this duality are built in to every set, but by going the opposite direction as bornological vector spaces, i.e. tweaking sets using extremally disconnected spaces (and very, loosely adding more colimits) rather than tweaking bornologies using very connected scalars (and very loosely, adding more limits).

  • $\begingroup$ On the other hand, there must be some sort of duality between subalgebras and quotient algebras. Note that a topology can be specified by a ($\bigcup$, $\cap$)-subalgebra of the powerset, whereas specifying an ideal is equivalent to specifying a quotient algebra (by this ideal). $\endgroup$ Jan 8 at 5:00
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    $\begingroup$ I hadn't really thought of it that way, but the condition that the ideal be a cover would need to be reflected into a condition on the topology. As-is, there are 4 topologies on the 2-element set, but only 1 bornology. $\endgroup$ Jan 8 at 6:21

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