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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 $X$ 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:

The property $\bigcap_{U\in\nu}U=\varnothing$ dual to a bornology being a cover of $X$ seems to amount to $\tau_\nu$ being $T_0$ (possibly weaker than this?). 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.

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.

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 topologies on (topological) dual spaces are defined in terms of 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 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. The book "Bornologies and Functional Analysis" by Henri Hogbe-Nlend (North-Holland, 1977) discusses such topics in depth.

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:

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.

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.

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 $X$ 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 $X$ seems to amount to $\tau_\nu$ being $T_0$ (possibly weaker than this?). 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 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.

The duality between (convex) bornology and (locally convex) topology acquires a deeper meaning in the theory of locally convex vector spaces, since compatible topologies on dual spaces are defined in terms of bornologies of the original spaces, and vice-versa. The book "Bornologies and Functional Analysis" by Henri Hogbe-Nlend (North-Holland, 1977) discusses such topics in depth.

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 $X$ 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 $X$ seems to amount to $\tau_\nu$ being $T_0$ (possibly weaker than this?). 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.

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 topologies on (topological) dual spaces are defined in terms of 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 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. The book "Bornologies and Functional Analysis" by Henri Hogbe-Nlend (North-Holland, 1977) discusses such topics in depth.

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 $X$ 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 $X$ seems to amount to $\tau_\nu$ being $T_0$ (possibly weaker than this?). 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.

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 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.

The duality between (convex) bornology and (locally convex) topology acquires a deeper meaning in the theory of locally convex vector spaces, since compatible topologies on dual spaces are defined in terms of bornologies of the original spaces, and vice-versa. The book "Bornologies and Functional Analysis" by Henri Hogbe-Nlend (North-Holland, 1977) discusses such topics in depth.

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 $X$ 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 $X$ seems to amount to $\tau_\nu$ being $T_0$ (possibly weaker than this?). 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 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.

The duality between (convex) bornology and (locally convex) topology acquires a deeper meaning in the theory of locally convex vector spaces, since compatible topologies on dual spaces are defined in terms of bornologies of the original spaces, and vice-versa. The book "Bornologies and Functional Analysis" by Henri Hogbe-Nlend (North-Holland, 1977) discusses such topics in depth.