A natural construction of real numbers? 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 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.
Related

*

*Category-theoretic description of the real numbers (Mathematics Stack Exchange)

*Gromov's bounded cohomology, see
Ivanov - Notes on the bounded cohomology theory and the 9th page
of A'Campo's paper.

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