A group $G$ is defined to have the Bergman property if for any subset $X=X^{-1}$ generating $G$ there exists $n$ such that $X^n=G$.
By a result of Bergman, the permutation group of any set has the Bergman property. Later this result was extended to some other automorphism groups (of sufficiently homogeneous structures).
Very often the Bergman property appears in combination with uncountable cofinality.
A group $G$ is defined to be of uncountable cofinality if it can not be written as the countable union of a strictly increasing sequence of proper subgroups.
An example of a group of uncountable cofinality without the Bergman property is any infinite finitely-generated group (for example, $\mathbb Z$).
Problem. Is there a group of countable cofinality possessing the Bergman property?
I came to this question studying macro-uniform functions on finitary balleans of groups, but quickly have found that it was asked by Bergman himself in 2004 or earlier.
In 2006 Anatole Khelif announced a counterexample to this problem and in his paper in C.R. Acad. Paris promised: Les démonstrations des résultats ci-dessus seront publiées ultérieurement.
But I cannot google any papers of Khelif containing the promised proof. What does it mean? Is there a counterexample or not? Does anybody know the real situation?