### Summary


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](https://mathoverflow.net/questions/401441/a-natural-construction-of-real-numbers#comment1026184_401507) below. A similar construction, described in [Hermans - An elementary construction of the real
numbers, the $p$-adic numbers and the
rational adele ring](https://www.universiteitleiden.nl/binaries/content/assets/science/mi/scripties/bachelor/2017-2018/hermans-bscthesis.pdf), yields $\mathbb{Q}\_p$ and the rational adele ring $\mathbb{A}$.




---


### Main text


In [A'Campo - A natural construction for the real numbers](https://arxiv.org/abs/math/0301015v1), 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](https://mathoverflow.net/questions/401441/a-natural-construction-of-real-numbers#comment1026025_401441), 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](https://mathoverflow.net/a/401445).


### Related


* [Category-theoretic description of the real numbers](https://math.stackexchange.com/q/839848) (Mathematics Stack Exchange)
* Gromov's bounded cohomology, see
[Ivanov - Notes on the bounded cohomology theory](https://arxiv.org/abs/1708.05150) and the 9th page
of A'Campo's paper.



A natural construction of real numbers?
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 $f(n)/n\to 0$.


