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.


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    $\begingroup$ 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 at 20:05
  • 1
    $\begingroup$ Is your $\lambda$ in $d^{-1}(C^2_b)$? I think not. $\endgroup$ Aug 10 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 at 20:11
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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – aorq
    Aug 10 at 20:17
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    $\begingroup$ 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 at 20:27

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 $f(n)/n\to 0$.

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

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    $\begingroup$ 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 at 14:55
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    $\begingroup$ @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 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 at 7:32

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