I am trying to see if the function $f: \mathbb{Z} \to \mathbb{C}$ defined by $$ f(n) = \frac{1}{n-z} $$ for some $z \in \mathbb{C}\backslash \mathbb{R}$ can be continuosly extended to the profinite completion of $\mathbb{Z}$, denoted by $\hat {\mathbb{Z}}$. Since $\hat{\mathbb{Z}}$ is compact this is possible if and only if this function is uniformly continuous in the metric that induces the topology of $\mathbb{\hat{Z}}$, i.e. $$ d(n,m)= \inf\ \{ k^{-1} \mid n - m = 0 \mod k!\}. $$
I have been able to prove that if $d(n,m)$ is small and $d(n,0) \not= d(m,0)$ then $\lvert f(n) - f(m) \rvert$ is small, but I don't have a clue for the general case.
In the profinite topology, $3 + n! \to 3$, as $n\to \infty$, but $$ f ( 3 + n!) \to 0 \not= f(3).$$ So $f$ is not continuous.