for a set $X$ a bornology $\mathcal{B}$ is essentially an ideal in s 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'm 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, 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:

1. 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

2. 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