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?
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I've corrected the formula (it was erroneous). Now it is $E^n[B]=U$. – porton May 25 '13 at 15:24
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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.
Alex Ravsky
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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
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@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