3
$\begingroup$

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 $F$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.