Someone claims $\mathbb{R}$ can be constructed as the following intriguing quotient, which is related to Gromov's bounded cohomology. I want to find out if it is true.

$$\frac{\bigl\{f:\mathbb{Z} \to \mathbb{Z} \mathrel| \mbox{ the set } \{f(m+n)-f(m)-f(n) \mathrel| m, n \in \mathbb{Z}\} \mbox{ is bounded}\bigr\}}{\{f:\mathbb{Z} \to \mathbb{Z} \mathrel| f \mbox{ is bounded}\} }.$$

EDIT See KConrad's comment below. A similar construction, described in Hermans - An elementary construction of the real numbers, the $p$-adic numbers and the rational adele ring, yields $\mathbb{Q}_p$ and the rational adele ring $\mathbb{A}$.

Main text

In A'Campo - A natural construction for the real numbers, a natural construction of the real numbers is given as follows. (EDIT: In this post I only address the bijection and the ring structure. The correspondence is complete; for that please refer to the paper.)

Definition (Bounded cochains)

Define $C^{n} = C^{n}(\mathbb{Z})$ to be $\operatorname{Map}(\mathbb{Z}^{\times n}, \mathbb{Z})$ and $C^n_b = C^{n}_b(\mathbb{Z})$ to be the subset consisting of functions $f$ having bounded image, i.e. $\operatorname{Card}(\operatorname{Im}(f)) < \infty$.

Definition (Differentials)

Define $d: C^n \to C^{n+1}$ to be such that $$df(x_1,\dotsc,x_{n+1}) = f(x_2,\dotsc,x_{n+1}) + \sum_{k=1}^{n}(-1)^{k} f(x_1, \dotsc, x_{k-1}, x_k+x_{k+1}, \dotsc, x_{n+1}) + (-1)^{n+1}f(x_1,\dotsc,x_n).$$

Obviously, $d(C^n_b) \subseteq C^{n+1}_b$, so $C^n_b \subseteq d^{-1}(C^{n+1}_b)$.

Algebraic Operations

Clearly, $C^1$ has a ring structure, where addition is given by point-wise addition, and multiplication is given by function composition.

Claim. $\mathbb{R} \simeq d^{-1}(C^2_b)/(C^1_b)$

This claim is made in page 1 (definition of $\mathbb{R}$) and page 6 (that $\mathbb{R}$ is the usual $\mathbb{R}$) of the paper. An explicit map $\Phi: d^{-1}(C^2_b) \to \mathbb{R}$ is given in page 4 as $$\lambda \mapsto \left[\left(\frac{\lambda (n+1)}{n+1}\right)_{n \in \mathbb{N}}\right]$$ using Cauchy sequences.

Question Why is $\ker(\Phi) = C^1_b$? By the definition of the equivalence on the set of Cauchy sequences, $\Phi(\lambda)$ represents $0 \in \mathbb{R}$ if and only if

For each $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\frac{|\lambda(n+1)|}{(n+1)} < \epsilon$ whenever $n > N$.

However, $\lambda: \mathbb{Z} \to \mathbb{Z}$ that sends $n$ to $\lfloor\sqrt{|n|}\rfloor$ is one such element that is not in $C^1_b$ (namely, not bounded).

EDIT As Anthony Quas points out below, such $\lambda$ isn't in the preimage of $d$. You can see this by taking $m = n \to \infty$. Still, I'm curious about a direct proof for the kernel being $C^1_b$. This is given in Anthony Quas's answer.


  • 6
    $\begingroup$ Who is 'someone'? Is it A'Campo? (Also, TeX note: $\Sigma_{k = 1}^n$ \Sigma_{k = 1}^n doesn't behave as an operator; prefer $\sum_{k = 1}^n$ \sum_{k = 1}^n. If you want to suppress the limits placement of \sum, then you can write \sum\nolimits_{k = 1}^n.) $\endgroup$
    – LSpice
    Aug 10, 2021 at 20:05
  • 1
    $\begingroup$ Is your $\lambda$ in $d^{-1}(C^2_b)$? I think not. $\endgroup$ Aug 10, 2021 at 20:10
  • 4
    $\begingroup$ Naive question: do you want to construct R as abelian group as a field or as a topological field? $\endgroup$
    – GSM
    Aug 10, 2021 at 20:11
  • 3
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – aorq
    Aug 10, 2021 at 20:17
  • 15
    $\begingroup$ The most useful first reference for this construction of $\mathbb{R}$ is, I think, Ross Street's "Update on the efficient reals"(maths.mq.edu.au/~street/reals.pdf). He gives some history of the construction, which has been independently discovered a few times, including by A'Campo. As I understand it, the first time it was done in print was an earlier (1985) paper of Street, cited in the link I just gave but not in the Wikipedia article. $\endgroup$ Aug 10, 2021 at 20:27

2 Answers 2


So here is my attempt to reconstruct the construction...

Suppose $f\colon\mathbb Z\to\mathbb Z$ satisfies $|f(m+n)-f(m)-f(n)|\le M$ as $m,n$ run over $\mathbb Z$. Then setting $m=n=2^k$, we see $|f(2^{k+1})-2f(2^k)|\le M$, from which it follows that $f(2^k)/2^k$ is a Cauchy sequence, and so converges to some $\alpha\in\mathbb R$.

Now given $n\in (2^{k-1},2^k]$, let its binary expansion be $n=2^{k-1}+2^{j_1}+\ldots+2^{j_r}$ (with $r< k\le \log_2 n$). Inductively, we can show $\Big|f(n)-\big[f(2^{k-1})+\ldots +f(2^{j_r})\big]\Big|\le rM$, from which, together with the above, we can deduce that the full sequence $f(n)/n$ converges to $\alpha$.

On the other hand, given $\alpha\in\mathbb R$, if one defines $f(n)=\lfloor \alpha n\rfloor$, then it is easy to see that $f(n+m)-f(n)-f(m)$ takes values in $\{0,1\}$, so that the map is surjective.

Finally, suppose $f(n)/n\to 0$ and $|f(n+m)-f(n)-f(m)|\le M$ is bounded. We have to show that $f$ is bounded. If $|f(n_0)|\ge 2M$ for some $n_0$, then $|f(2n_0)|\ge 2|f(n_0)|-M$, from which we see inductively that $f(2^kn_0)\ge (2^k+1)M$, contradicting the assumption that $f(n)/n\to 0$.

  • 2
    $\begingroup$ The last paragraph is exactly what I was trying to get! It makes my day, thanks :) $\endgroup$
    – Student
    Aug 10, 2021 at 20:48

This is an extended comment.

A natural (in my mind) way to capture this construction is by the following categorical interpretation.

Consider the category $\text{ACG}$ of Abelian Coarse Groups, which are abelian group objects in the category of coarse spaces, as defined here*. This is an additive category.

Note that every abelian locally compact topological group $G$ could be seen as a coarse group, by setting $E\subset G\times G$ to be controlled iff $\{x-y\mid (x,y)\in E\}$ is precompact in $G$.

One sees easily that the inclusion $\mathbb{Z}\hookrightarrow \mathbb{R}$ is an isomorphism in $\text{ACG}$, thus we get a natural isomorphism of rings, $\text{End}_\text{ACG}(\mathbb{Z})\simeq \text{End}_\text{ACG}(\mathbb{R})$.

One also sees easily that the natural map of rings $\mathbb{R} \simeq \text{End}_\text{AG}(\mathbb{R}) \to \text{End}_\text{ACG}(\mathbb{R})$ is an isomorphism (here $\text{AG}$ stands for Abelian Groups).

One gets a natural ring isomorphism $\mathbb{R} \simeq \text{End}_\text{ACG}(\mathbb{Z})$ by composing the above maps. Further, Hom sets in the category of abelain coarse groups could be naturally topologized, and this is an isomorphism of topological rings.

Similarly, one gets that $\mathbb{Q}\hookrightarrow \mathbb{A}$ is an isomorphism in $\text{ACG}$ (here $\mathbb{A}$ stands for the rational adels) and concludes an isomorphism of topological rings $\mathbb{A}\simeq \text{End}_\text{ACG}(\mathbb{Q})$.

$*$ For our discussion here, we take for morphisms in the category of coarse spaces classes of controlled maps, that is maps $f:X\to Y$ such that for every controlled $E\subset X\times X$, $f\times f(E)\subset Y\times Y$ is controlled, and $f\sim g:X\to Y$ if $f\times g(X\times X)$ is bounded in $Y$. We note that the category thus obtained has products, so talking about group objects makes sense.

  • 7
    $\begingroup$ The 2018 bachelor's thesis in Leiden by Tessa Hermans gives a construction in this way for $\mathbf R$, $\mathbf Q_p$, and the adele ring of $\mathbf Q$: see universiteitleiden.nl/binaries/content/assets/science/mi/… $\endgroup$
    – KConrad
    Aug 11, 2021 at 14:55
  • 2
    $\begingroup$ @KConrad, thanks for the reference. This seems an excellent bachelor's thesis! Indeed, it puts things in a categorical framework. However, this framework is slightly different then the one I suggest. In particular, it does not allow isomorphisms such as $\mathbb{Z}\simeq\mathbb{R}$, which in my mind demystify the picture. $\endgroup$
    – Uri Bader
    Aug 12, 2021 at 7:00
  • $\begingroup$ Let me illustrate further by discussing the $p$-adics. It is easy to see that $\text{End}(\mathbb{Q}_p)\simeq \mathbb{Q}_p$ and $\mathbb{Q}_p\simeq \mathbb{Q}_p/\mathbb{Z}_p \simeq \mathbb{Z}[1/p]/\mathbb{Z}$. We thus construct the $p$-adics as $\text{End}(\mathbb{Z}[1/p]/\mathbb{Z})$. $\endgroup$
    – Uri Bader
    Aug 12, 2021 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.