Metric on the profinite completion of the integers?

$\newcommand{\Z}{{\mathbb Z}}\newcommand{\N}{{\mathbb N}}$Please excuse that I write many details that can be found elsewhere (many, for example, in the German wikipedia article on profinite integers) and that are not all directly related to the question of a metric. My intention is to present my current understanding of $\hat\Z$. I am obviously not a specialist of topology.

One way of representing $\hat\Z$ is to write its elements as sequences $(r_i+i\Z)_{i\in\N}$, where the sequence $r_i$, $i\in\N$ satisfies the compatibility condition $r_i\equiv r_j \mod j$ whenever $j|i$. The elements of $\Z$ are identified with the constant sequences. As $r_{k!}+k!\,\Z$ determines $r_1,\dots,r_k$, we can also represent $\hat\Z$ as the set of sequences $(s_k+k!\Z)_{k\in\N}$, where $s_{k+1}\equiv s_k\mod k!$ for all $k$. In both representations the topology on $\hat\Z$ is the (restriction of the) product topology of the discrete topologies of the $\Z/i\Z$ or $\Z/k!\Z$, respectively. In the second representation, this means that a base is given by the sets $O_{m,t}=\{(s_k+k!\Z)_{k\in\N}\in\hat\Z\mid s_k\equiv m\mod k!\mbox{ for }k\leq t\}$, $t\in\N$, $m\in\{0,\dots,t!-1\}$. (So the open sets are precisely the unions of any number of sets $O_{m,t}$.) It is well known that a countable product of metric spaces is metrizable. So we define on the discrete spaces $\Z/n\Z$ the trivial metric $d_n(a,b)=0$ or $1$ if $a=b$ or $a\neq b$, respectively, and a metric on $\hat\Z$ in the first representation can be defined as $$d(s,\tilde s)=\sum_{i\in\N}2^{-i}\frac{d_i(r_i+i\Z,\tilde r_i+i\Z)} {1+d_i(r_i+i\Z,\tilde r_i+i\Z)}\mbox{ if }s=(r_i+i\Z)_{i\in\N}\mbox{ etc.}$$ Observe also that the product topology is the topology of pointwise convergence; here the convergence in the factors is with repect to the discrete topology.

The second representation shows that every element of $\hat\Z$ can be written uniquely as a formal series $\sum_{k\in\N}a_k\,k!$, where $a_k\in\{0,\dots,k\}$ and that every such formal series represents some element of $\hat\Z$. This shows (as for the real numbers) that $\hat\Z$ is uncountable. It also suggests to introduce a second metric based on the second representation: $d(s,\tilde s)=1/N$ if $s=(s_k+k!\Z)_{k\in\N}$, $\tilde s=(\tilde s_k+k!\Z)_{k\in\N}$ and $N\in\N$ is maximal such that $\tilde s_k\equiv s_k$ for $k<N$. If there is no such $N$, i.e. $s=\tilde s$ then we put $d(s,\tilde s)=0$. It is straightforward to verify the axioms of a metric. We even have $d(s,\bar s)\leq \max(d(s,\tilde s),d(\tilde s,\bar s)$. Clearly $O_{m,t}=\{s\in\Z\mid d(m.s)\leq 1/t\}$. Therefore the topology of $\hat\Z$ is the one generated by $d$. It is also left to the reader to show that $\hat\Z$ is complete and that $\Z$ is dense in $\hat\Z$ with respect to this metric. By construction, addition in $\hat\Z$ (which is defined componentwise) is continuous. Also the multiplication (again defined componentwise) is continuous since $d(s\tilde s,0)\leq \min(d(s,0),d(\tilde s,0))$.

Observe that $\hat Z$ contains zero divisors, i.e. $s\tilde s$ can be zero even if both $s$, $\tilde s$ are nonzero. This is best seen in a third representation of $\hat\Z$. It is based on the Chinese remainder theorem which implies that $\Z/n\Z\simeq (\Z/p_1^{n_1}\Z)\times\cdots\times(\Z/p_k^{n_k}\Z)$ if $n=p_1^{n_1}\cdots p_k^{n_k}$ is the prime factor decomposition of $n$. We find that $\hat\Z\simeq\prod_{p\in{\mathcal P}}\Z_p$, where $\mathcal P$ is the set of all primes and, for a prime $p$, $\Z_p$ is that set of all sequences $(u_i+p^i\Z)_{i\in\N}$ where $u_{i+1}\equiv u_i\mod p^i$ for all $i$. These last sets $\Z_p$ are the well known $p$-adic integers. Now consider $s\in\hat\Z$ for which all components are 0 in this third representation except the one for a certain prime $p$ and consider an analogous $\tilde s$ corresponding to some different prime $\tilde p$. Both are not zero, but their product is.

The topology on $\hat\Z$ corresponds to the product topology of the $p$-adic topologies on $\Z_p$ which are coming from the $p$-adic metric $d_p$ and so a third metric on $\hat\Z$ can be defined by $$d(s,\tilde s)=\sum_{p\in\mathcal P}2^{-p}\frac{d_p(a_p,\tilde a_p)} {1+d_p(a_p,\tilde a_p})\mbox{ if }s=(a_p)_{p\in\mathcal P},\tilde s=(\tilde a_p)_{p\in\mathcal P}$$ in the third representation. This is, I think, what Mathmo indicated in his comment.

Summary: One metric on $\Z$ that leads to $\hat\Z$ is the following: $$d(m,n)=1/\sup\{N\in\N\mid m\equiv n\mod N!\}.$$ $\hat\Z$ can be seen as the completion of $\Z$ with respect to this metric. This is best seen if the elements of $\hat\Z$ are written as formal series $$\sum_{k\in\N}a_k\,k!\mbox{ with certain }a_k\in\{0,\dots,k\}.$$ The open sets of $\Z$ in the topology generated by $d$ are those $O\subseteq \Z$ such that for every $m\in O$ there exists a $N$ such that $O$ contains the set of all $n$ with $n\equiv m\mod N!$.