# Metric on the profinite completion of the integers?

The p-adic integers come with a metric and associated topology, both of which can be restricted down to the integers.

Does this also apply to the profinite completion of the integers? Do they have either an associated topology or metric, and if so, what does this induce on the integers? What do open sets of integers look like in this topology?

• There is a homeomorphism $\widehat{\mathbb Z}\cong \prod_p\mathbb Z_p$ – Mathmo123 Jun 7 '16 at 17:54
• You basically get a (space homeomorphic to the) Cantor set. – Henno Brandsma Jun 7 '16 at 17:55
• I'm having a hell of a time imagining what the product topology of all the p-adic numbers is. How does one visualize this? – Mike Battaglia Jun 8 '16 at 3:55
• @HennoBrandsma - are you saying the profinite completion of the integers is homeomorphic to the Cantor set, or the restriction to Z is? – Mike Battaglia Jun 8 '16 at 3:56
• The completetion is, as a compact metrisable zero-dimensional space without isolated points. And as $\mathbb{Z}$ is countable dense in it, it must then be homeomorphic to the rationals, by another standard theorem. – Henno Brandsma Jun 8 '16 at 4:22


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 $$k. If there is no such $$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 $$n\equiv m\mod N!$$.

• Thanks, very interesting! So strangely, we have $d(0,2)=2$, but $d(0,3)=1$, $d(0,5)=1$, $d(0,7)=1$, etc. That is, the largest factorial that makes any prime congruent to 0 is $1!$, except for 2 where it's $2!$. Very interesting! I am curious what convergence of some common series looks like with this metric and topology. – Mike Battaglia May 25 at 0:38
• @MikeBattaglia Precisely, $d(0,2)=1/2$. The formal series for elements of $\hat Z$ are quite analogous to those for $p$-adic integers. The principal difference is that they are much less suitable for expressing products than the latter. Also, the metric does not come from a valuation since $d(0,7!)=1/7$, but $d(0,2)=1/2,d(0,3)=\cdots=d(0,7)=1$. Therefore the definition with $1/...$ is as good as with the $2^{-...}$ used for the $p$-adic metric. – Helmut May 25 at 9:19

User Helmut makes good points in his answer. I just want to stress that

• The topology on $$\Bbb Z$$ with respect to which $$\hat{\Bbb Z}$$ is the completion can be described as follows: $$x \in \Bbb Z$$ has as basis of open neighbourhoods $$(x + n \Bbb Z)_{n \in \Bbb N}$$. In other words, a subset of $$\Bbb Z$$ is open iff together with each element $$x$$ in it, it contains some arithmetic progression containing $$x$$ (i.e. for some $$n$$, all $$x+n\Bbb Z$$); or in yet other words, a subset of $$\Bbb Z$$ is open iff it is a union of arithmetic progressions. Two integers $$x, y$$ are "close" to each other if (corrected, thanks Helmut:) $$x-y$$ has "many" prime divisors (with "high" exponents), or in other words, if there are "many" arithmetic progressions which contain both $$x$$ and $$y$$. E.g. $$360$$ is closer to $$0$$ than most numbers around it; and in the first $$10'000$$, no number comes closer to $$5053$$ than $$13$$. (NB 1: This is not strictly true for any metric which induces the topology, see below. NB 2: This describes the subspace topology on $$\Bbb Z$$ induced from the one on $$\hat{\Bbb Z}$$. The aforementioned sets are not open when viewed as subsets of $$\hat{\Bbb Z}$$; just like $$a+p^{17}\Bbb Z$$ is open in $$\Bbb Z$$ with the $$p$$-adic topology, but not in $$\Bbb Z_p$$.)

• There are many metrics which induce this topology, but none of them is given by a ring norm (different from the $$p$$-adic setting).

• Since on integers, the $$p$$-adic metrics $$d_p(x,y) := \lvert x-y\rvert_p$$ are already bounded by $$1$$, we don't need the extra construction in Helmut's "third metric", and e.g.

$$d(x,y) := \displaystyle\sum_{p \text{ prime}} \dfrac{d_p(x,y)}{2^{n(p)}}$$

where $$p$$ is the $$n(p)$$-th prime, is a kind of "visualisable" metric which induces the topology. $$x$$ and $$y$$ are close to each other if they are close $$p$$-adically for many $$p$$. A sequence $$x_n$$ converges to $$x$$ iff $$\lim_{n\to \infty} \lvert x_n-x\rvert_p =0$$ for all prime $$p$$.

• Note however that the metric has a lot of leeway. I think one could do something like

$$d(x,y) := \displaystyle\sum_{p \text{ prime}} w_p \cdot d_p(x,y)$$

where $$w_p$$ (the weight for the $$p$$-adic metric) just has to satisfy $$\sum_p w_p < \infty$$. (Compare Show that the countable product of metric spaces is metrizable.) So if we set $$w_p = 2^{-100}$$ for $$p=2,3,5,7$$ but $$w_{11} = 1$$, and $$w_p = 2^{-n(p)}$$ for all other primes as before, then contrary to what I said in my first point, $$5042$$ is closer to $$5053$$ than $$13$$ is, because I made this metric so that it puts a lot of weight on the difference being divisible by $$11$$. The topology alone does not really define "closeness". What is important in the end is that all $$p$$-adic differences "weigh in".

• These are interesting thoughts. One thing is, is there some kind of unique translation-invariant metric that induces the topology? If you think about it as a product of p-adic metrics, it seems like no, but if we're talking about arithmetic sequences, then those are translation invariant. Phrased differently, should 360 be closer to 0 than 361 is to 1? You get the same arithmetic sequences either way. – Mike Battaglia May 25 at 0:49
• @MikeBattaglia: I think all metrics in both answers are translation-invariant (i.e. $d(x,y) = d(x+z,y+z)$ for all $x,y,z$), in partic. $d(361,1)=d(360,0)$ for all of them. When I say that no ring norm induces any of the metrics, the problem with the obvious candidates $\lvert x \rvert_d := d(0,x)$ is that they cannnot be multiplicative, i.e. $d(0,xy) \neq d(0,x)\cdot d(0,y)$ for some $x,y$ (proof idea: $\hat{\Bbb Z}$ has zerodivisors which give counterexamples, then use density of $\Bbb Z$ to approximate those). It might be possible to achieve submultiplicativity though. – Torsten Schoeneberg May 25 at 5:15
• @TorstenSchoeneberg Very nice complements to my answer! I like in particular the characterisation of the open sets of $Z$. One small correction: $x,y$ are close if $x-y$ has "many prime divisors" of "high" exponent. Observe that for a sequence $x_k$ to converge to $y$, we also need that the set of prime divisors of $x_k-y$ contains every prime $p$ for large $k\geq k(p)$. – Helmut May 25 at 9:39