### 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.

`\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$6more comments