There is of course the standard construction of the reals by considering the set of sequences that are Cauchy with respect to the standard metric and taking the quotient by sequences that converge to 0, but in many other cases of the completion of a ring or group, we can complete by taking an inverse limit of factor rings/groups corresponding to some predetermined ideals/subgroups. For example, the $\mathfrak{p}$-adic integers are constructed using precisely this type of inverse limit construction, where the ideals are the powers of $\mathfrak{p}$, where $\mathfrak{p}$ is the ideal $(p)$ for $p$ a prime.

Is there any way to construct the reals using a similar inverse limit construction?

It seems like this might be related to the so-called "infinite prime", but I don't know if there is really a way to complete with respect to an "infinite prime", which doesn't seem like it is strictly a prime integer or ideal, so that doesn't really seem to help..

(If anyone can think of better tags, please go ahead and change the ones I've used. I wasn't really sure how to classify this).

  • $\begingroup$ It appears to me that it is not the arithmetic but the ordering that plays the game here. The question seems to still make sense if we ignore the arithmetic structure of rational number and only look at it as a countable dense linear ordering without end points. My first reaction is no, but I don't have a good argument for it. $\endgroup$
    – abcdxyz
    Mar 30, 2010 at 10:04
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    $\begingroup$ The infinite prime is certainly a mathematical object in the context of Arakelov geometry, although I don't know if that's relevant here. $\endgroup$ Mar 30, 2010 at 10:07
  • $\begingroup$ Rather, it's not strictly a prime. $\endgroup$ Mar 30, 2010 at 10:10
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    $\begingroup$ I think the things are the other way round. Since both usual primes and the Euclidean metrics define absolute values, and you can do many similar constructions with them (e. g. completion) it is sometimes more convenient to think of these objects together. So, given a global field K with a maximal order O, one often consider the set of places of K (equivalence classes of absolute values) as a better behaved replacement of Spec(O). In this sense the absolute value itself is a "generalized prime" tautologically, but this surely does not answer your question. $\endgroup$ Mar 30, 2010 at 10:15
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    $\begingroup$ In the category of archimedean ordered fields, $\mathbb{Q}$ is initial and $\mathbb{R}$ is terminal. I think this mirrors what happens at finite primes pretty well (a little too well, in fact). $\endgroup$ Mar 30, 2010 at 11:44

4 Answers 4


Although your question is not at all vague, there are a few completely different ways to interpret what a good answer would be. Knowing a bit about your personal preferences, I suspect the following is not at all what you wanted, but it is still of interest to the community.

The fact that $\mathbb{Q}_p$, for example, can be seen as an inverse limit boils down to the fact that the series representations $$\sum_{n \in \mathbb{Z}} a_n p^{-n}$$ give a homeomorphism between $\mathbb{Q}_p$ and the subspace of $\{0,1,\dots,p-1\}^{\mathbb{Z}}$ of eventually zero sequences (on the positive side). At first glance, it would seem that the same works for $\mathbb{R}$ since every element of $[0,\infty)$ has a very similar binary expansion $$\sum_{n \in \mathbb{Z}} a_n 2^n$$ where $(a_n)_{n \in \mathbb{Z}}$ is again in the space of all eventually zero sequences in $\{0,1\}^{\mathbb{Z}}$. Unfortunately this is not a bijection. In fact, there is no way to select such a sequence $(a_n)_{n\in\mathbb{Z}}$ continuously — binary rationals are always points of discontinuity no matter how the selection is made.

There is a funny way to remedy this which is commonly used in Computable Analysis and Reverse Mathematics. The idea is to represent real numbers again with series of the form $$\sum_{n \in \mathbb{Z}} a_n 2^n$$ but where $(a_n)_{n \in \mathbb{Z}}$ is now an eventually zero sequence from the extended set $\{-1,0,1\}^{\mathbb{Z}}$. The obvious cost is that no number has a unique representation in this form, but the side benefit is that the binary rationals are no longer 'special' in this way. In the end, this representation is extremely well behaved compared to ordinary binary expansions. This can be seen from the fact that any invariant continuous function between representations (see note) gives rise to a continuous function $\mathbb{R}\to\mathbb{R}$, and every continuous function $\mathbb{R}\to\mathbb{R}$ admits such a lifting.

The conclusion to draw from this is that, up to some very nice blurring, $\mathbb{R}$ is indeed an inverse limit just like $\mathbb{Q}_p$.

Note: The topology on eventually zero sequences is not exactly the product topology, it is given by the metric $$d(\vec{x},\vec{y}) = \inf\{2^n : \forall m \geq n\,({x_m = y_m})\}.$$ The invariance of a function $f$ between eventually zero sequences from $\{-1,0,1\}^{\mathbb{Z}}$ means that if $\vec{x}$ and $\vec{y}$ represent the same real number then so do $f(\vec{x})$ and $f(\vec{y})$.

  • $\begingroup$ You can express the condition that $a_n$ is eventually zero as follows: $$\sum_{-\infty<n\ll \infty}a_n 2^n$$ although to my eyes $\sum_{n\gg -\infty}a_n 2^{-n}$ would appear more common. $\endgroup$ Sep 13, 2010 at 6:31

The main formal difference between completing $Q$ with respect to the infinite or to a finite prime is that in the second case, as you know, the completion $Q_p$ is a topological field equipped with an open, compact, local, subring $Z_p$ whose natural topology (as local ring) coincide with the induced topology from the completion. From this fact one obtains that the open subring of $Q_p$ (and not $Q_p$ itself) given by $Z_p$ can be described as the inverse limit of the $Z_p/p^m$, as $m$ increases indefintiely.

It seems to me that the lack of the existence of this open subring in the infinite prime case makes it very unlikely for $R$ to have a natural, inverse limit, non-trivial description.

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    $\begingroup$ also note that $\mathbb{R}$ cannot be written as a inverse limit of discrete spaces. $\endgroup$ Mar 30, 2010 at 11:51

First proposed answer:

See link http://en.wikipedia.org/wiki/Continued_fraction

Basically what we do is to correspond each rational number canonically to a continued fraction.(see the wiki link). We can denote $\mathbb{Q}^n$ the rational number that can be represented canonically by a continued fraction of length $n$. The restriction map (corresponding a rational number in $\mathbb{Q}$ with continuous fraction $[a_0;a_1, ..]$ to the rational number in $\mathbb{Q}^n$ with continued fraction expression $[a_0; a_1, \ldots , a_n]$ ) will then give a projection from $\mathbb{Q}$ to $\mathbb{Q}^n$. Notice that this map weakly preserve the ordering. Real number is then the inverse limit of this system. That is because an irrational number can be expressed uniquely as a continued fraction.

We do not have these projection map ring homomorphism, but we do have them ring homomorphism with some perturbation. This is consistent with the way the set theorist view real number as $ \omega^ \omega$ as well.

Second proposed answer:

Let me propose another answer which share some spirit with the previous one. The advantage here is that the quotient structure is also ring. But the disadvantage is the thing obtained is not exactly the inverse limit. I let you decide which one of them is better.

Let $ R $ be any discrete subring of $ \mathbb{Q}$ (e.g. $ R =\mathbb{Z} $, $ R = \mathbb{Z}[1/2]$). We define the map $ p_R: \mathbb{Q} \rightarrow R$ by $ q \mapsto r$ where $r$ is the greatest element in $R$ such that $r \leq q$. It can be seen that this is a projection. If $ R_1 $ is a subring of $R_2$, it can be seen that the above system of maps induces a projection from $R_2$ to $R_1$. Hence we have an inverse system of subset of $\mathbb{Q}$. The inverse limit of this system is something, in fact it is a ring . To get the $\mathbb{R}$ we need to quotient away the sequences that converge to 0.

Seems like nothing new, I am repeating the Cauchy construction and further more the projection that I was considering was not even a ring homomorphism. Let me explain why I think this construction is more reasonable than it first appears.

Let us observe what did we actually do for the case of primes $p$. A way to view $ \mathbb{Z} / p^n$ is to think of it as we are identifying every integer divisible by $p^n$. If we think of prime in term of valuation then this means ignoring the small difference ( $p^n$ is small in this case). That harmonize very well with the construction above (and also the construction by continuous fraction) where we throw away some small thing.

For the complain that the projection is not a ring homomorphism. It is true that it is not a ring homomorphism but it is a ring homomorphism up to a small perturbation. In fact, we can choose $ R$ approximates $ \mathbb{Q}$ close enough such that the projection map behave like a ring homomorphism (identity map) "locally". And at the limit stage we recover back the field structure.

  • $\begingroup$ I'm not really sure what inverse system you're considering, but even if this description gives you the real numbers as a set I don't think it's enough to tell you what the arithmetic on R looks like. $\endgroup$ Mar 30, 2010 at 11:44
  • $\begingroup$ True, but as I mentioned above. Is this about arithmetics? $\endgroup$
    – abcdxyz
    Mar 30, 2010 at 11:51
  • $\begingroup$ Thanks, I notice that the definition of the map just now was not clear. I also revisit whether we can put the arithmetic rule of not, but its still seems no. $\endgroup$
    – abcdxyz
    Mar 30, 2010 at 12:13
  • $\begingroup$ Interesting, this projection does not give a arithmetic structure, but it does give something like arithmetic structure up to some perturbation. However, at the limit stage the arithmetic structure is again recovered. $\endgroup$
    – abcdxyz
    Mar 30, 2010 at 12:23
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    $\begingroup$ Also note that developing a rational number, expressed in least terms as a quotient of integers, as continued fraction representation is the same as doing the Euclidean algorithm for the two integers in the quotient representation. $\endgroup$
    – Regenbogen
    Mar 30, 2010 at 15:50

In general, a completion of a normed field is naturally a colimit, since you can make a poset of normed fields generated by equivalence classes of sets of Cauchy sequences. I don't think there is an inverse limit construction of the reals, but it may be interesting to analyze the failure modes of some attempts.

First, we should take a closer look at the p-adic construction. We start by restricting our view to a valuation ring, namely the subring of elements of absolute value at most one. Then, we take an inverse limit over quotients by powers of the maximal ideal. Finally, we take the fraction field of the resulting complete ring.

The first step in the construction is already problematic with the archimedean valuation, because it is not an ultrametric, i.e., the set of rationals of norm at most 1 is not closed under addition. We can boldly press on, noting that we still have a continuous multiplicative monoid structure, and it has a maximal ideal $M = \{ x \in \mathbb{Q} : |x| < 1 \}$. Unfortunately, positive powers of M are equal to M itself, so the inverse limit of the quotients is just the monoid $\{-1,0,1\}$, which is not the closed unit disk. We could also try the inverse limit of quotients by the directed system of ideals $I_\alpha = \{x \in \mathbb{Q} : |x| < \alpha \}$, but all of the quotients are totally disconnected. This is bad, because (if I'm not mistaken) the inverse limit of a directed system of totally disconnected spaces is totally disconnected, while the interval $[-1,1]$ (often called the "ring of integers of $\mathbb{R}$") is what we really want, and its connected component(s) are far from singletons.

  • $\begingroup$ I probably tried to do something silly and completely different from what you have in mind, but I don't see how to compute the quotients in the third paragraph. $\endgroup$ Mar 30, 2010 at 19:12
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    $\begingroup$ Who calls $[-1,1]$ the "ring of integers of $\mathbb{R}$"? I want to have a talk with them. It's not a ring, and it doesn't contain the integers! $\endgroup$ Mar 30, 2010 at 21:07
  • $\begingroup$ @Pete: It is quite possible that no one actually calls it by that name, but lots of people say that other people call it that. I heard the term from Kiran Kedlaya (who seemed to be passing on what he had heard elsewhere), but there are close variants in print. For example, Durov calls it the completed local ring at infinity in his massive tome on Arakelov geometry, where it has a bit more structure than a monoid. $\endgroup$
    – S. Carnahan
    Apr 1, 2010 at 4:30
  • $\begingroup$ @François: A quotient of a (topological) commutative monoid M is constructed using a (closed) congruence relation, which is a (closed) equivalence relation invariant as a subset of $M \times M$ under translation by the diagonal action of M. Given a closed ideal I, the subset $I \times I$ forms a closed congruence relation. In particular, the ideals I described above should not have had strict inequalities. $\endgroup$
    – S. Carnahan
    Apr 1, 2010 at 4:38
  • $\begingroup$ Now that I'm thinking a little more about it, the maximal ideal is not closed, and this produces a strange topology on the "residue field" (namely, 0 is dense in {-1,0,1}). $\endgroup$
    – S. Carnahan
    Apr 1, 2010 at 4:42

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