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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 $\textbf{ideal}$ $\beta \subseteq \mathcal{P}(X)$ is called a $\textbf{bornology}$ if $\bigvee_{B \in \beta} \, B=1.$

This should be equivalent to the usual definition. An obvious dualization of this is a $\textbf{filter}$ $\nu \subseteq \mathcal{P}(X)$, 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 = \cup_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.

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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 $\endgroup$ – Bipolar Minds May 12 '16 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. $\endgroup$ – Pedro Lauridsen Ribeiro May 13 '16 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... $\endgroup$ – Bipolar Minds May 13 '16 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. $\endgroup$ – Pedro Lauridsen Ribeiro May 14 '16 at 17:33
  • $\begingroup$ hmm, I think I still don't understand what you mean: of course, we get a unique topology back if we interpret $\nu$ as a topological basis... but the topologies we get from this are rather stupid, since they are filters.. at least in the convex vector case (defined in analogy to a convex vector bornology), we get a much nicer result if we interpret $\nu$ as the neighbourhoods of a (not necessarily filtered) topology, since in this case we can define a unique locally convex 0-nbhd-filter $N(0) \subseteq \nu$ (subject to both conditions of a neighbourhood system) $\endgroup$ – Bipolar Minds May 14 '16 at 18:31

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