Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the following true:

Conjecture A uniform space $(U;F)$ is totally bounded iff for every entourage $E$ of this space there exists a finite set $B\subseteq U$ and a natural $n$ such that $E^n[B] = U$.

If not, could you provide a counter-example?

share|cite|improve this question
I've corrected the formula (it was erroneous). Now it is $E^n[B]=U$. – porton May 25 '13 at 15:24

We are dealing with topological groups. I heard the following. Banaszczyk called a topological group $G$ to be weakly precompact (or weakly bounded), if for each neighborhood $U$ of the unit of the group $G$ there exist a finite subset $F$ of $G$ and a number $n$ such that $U^nF=G$. And the group of all monotonically increasing homeomorphisms of the unit segment (probably, endowed with the pointwise topology) is weakly bounded, but not totally bounded.

PS. We are writing a paper on the relations of different types of boundness in topological groups. If you are interested in it, then I can post here a link to the paper after we shall write it.

share|cite|improve this answer
Why are we dealing with topological groups? I think $E^n$ means the $n$-fold composition $E \circ \cdots \circ E$ of the entourage. – Martin May 26 '13 at 17:23
@Martin I think so. But personally I am not mainly dealing with pure uniform spaces, but with uniform spaces of topological groups (in the sentence “we” means I and some persons from our topological school :-) ). So it is naturally to me to propose a counterexample in the form of a topological group. – Alex Ravsky May 26 '13 at 18:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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