A more rigorous version of Scott's answer: If a topology on a group $G$ is translation-invariant, then it also defines a uniformity on $G$, by definition a distinguished set of neighborhoods of the diagonal $G \times G$ that is analogous to a metric. Actually, in the present example with $G = \mathbb{Z}$, the uniformity comes from a metric. Like the metric spaces that they generalize, uniform spaces have completions. The completion of $\mathbb{Z}$ with respect to the uniformity cited by Furstenberg is exactly the adelic profinite completion of $\mathbb{Z}$. Or if $G$ is any group, there is a similar topology generated by finite-index subgroups, and a uniformity, and the completion is the profinite completion.