# 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

• 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.) Aug 10 at 20:05
• Is your $\lambda$ in $d^{-1}(C^2_b)$? I think not. Aug 10 at 20:10
• Naive question: do you want to construct R as abelian group as a field or as a topological field?
– GSM
Aug 10 at 20:11
• en.wikipedia.org/wiki/…
– aorq
Aug 10 at 20:17
• 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. 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 $$f(n)/n\to 0$$.

• The last paragraph is exactly what I was trying to get! It makes my day, thanks :) 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.

• 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/… Aug 11 at 14:55
• @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. Aug 12 at 7:00
• 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 $\text{End}(\mathbb{Z}[1/p]/\mathbb{Z})$. Aug 12 at 7:32