# A criterion of total boundness of a uniform space

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

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.
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