For a set $X$ a bornology $\mathcal{B}$ is essentially an ideal in the power set $\mathcal{P}(X)$. Many sources including Wikipedia state additional property that $X = \bigcup \mathcal{B}$. Call it a sup-property. As $\mathcal{B}$ is an ideal this property can be restated as having $\{x\} \in \mathcal{B}$ for any $x \in X$.
However, I found an article by Daniel Heiss: Generalized Bornological Coarse Spaces And Coarse Motivic Spectra where it is shown that not requiring all singletons to be included into bornologies leads to some nice categorical properties. But I myself have pretty lax knowledge of coarse geometry.
So my question is: Do any results in abstract functional analysis or classical coarse geometry fail if one omits the sup-property? Would we lose something important without the sup-property?
It is evident that the sup-property comes from the fact the singletons are bounded in a topological vector space. And these bounded sets are a model object for the bornology. Moreover, the sup-property always holds for group or vector bornologies $\mathcal{B} \neq \{\emptyset\}$ as they must be translation invariant.
The omitted points can be conceptualized as representations of infinity. And one can always restrict the bornology $\mathcal{B}$ to the set of "finite points" $X_{\mathcal{B}} =\{x \in X : \{x\} \in \mathcal{B} \}$. In a point-free setting it seems the bornology becomes a study of ideals in a complete Boolean algebras. So it seems that from point-free point of view the behavior of $\mathcal{B}$ over $X$ and $X_{\mathcal{B}}$ will be equivalent.
At this point my main interest is exploring "non-standard" bornologies arising in analysis. By "non-standard", I mean not bounded sets per se. Many of them are obvious from the ideal characterization. For examples:
Meager sets of a topological space $X$ form a bornology if the space $X$ has no isolated points. If we omit the sup-property, then meager sets always form a bornology
Null sets form a bornology for a measurable space $(X,\Sigma,\mu)$ if $\hat\mu\{x\} = 0$ for any $x \in X$. If we omit the sup-property, then null sets always form a bornology.
