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Let $X$ be a metrizable topological space, and let $B$ be a nontrivial bornology on $X$. Sze-Tsen Hu showed in 1949 that $B$ is the collection of bounded sets with respect to some metric for the topology on $X$ if and only if $B$ has a countable base and for any $S\in B$, there exists a $T\in B$ such that the closure of $S$ is a subset of the interior of $T$. (See this journal paper.)

I’m interested in the analogous result for uniformities. That is, $X$ be a uniformizable topological space, AKA a completely regular space, and let $B$ be a nontrivial bornology on $X$. My question is, under what circumstances is $B$ the collection of bounded sets with respect to some uniformity for the topology on $X$.

Note that a subset $A$ of a uniform space is said to be bounded if for each entourage $V$, $A$ is a subset of $V^n[F]$ for some natural number $n$ and some finite set $F$.

Keshav Srinivasan
  • 10,204

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It looks like this problem was solved by Tom Vroegrijk in this 2009 journal paper, 60 years after the analogous problem for metric spaces was solved.

Let $X$ be a uniformizable topological space, AKA a completely regular space, and let $B$ be a nontrivial bornology on $X$. Let's call a sequence $(U_n)$ of open sets a bounding sequence if the closure of $U_n$ is a subset of $U_{n+1}$ for all $n$ and every element of $B$ is a subset of some $U_n$. And let's call a set $S$ saturated if $S$ if for every bounding sequence $(U_n)$, $S$ is a subset of some $U_n$. Then $B$ is the collection of bounded sets with respect to some uniformity for the topology on $X$ if and only if $B$ contains every saturated set.

Keshav Srinivasan
  • 10,204