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I'm learning set theory at the moment and there are a lost of gaps in my knowledge right now. I'm struggling to understand how the real numbers are constructed from equivalence classes of the set of almost homomorphisms on $\mathbb Z$.

First of all, I feel like I should have a problem with equivalence classes even existing in ZFC.

My main problem is that I don't really understand what an almost homomorphism is, and what it represents. As I currently understand, an almost homomorphism is a function such that $f(x+y)-f(x)-f(y)$ is finite, so when we construct the reals we set each real number to be an equivalence class of the set of all those functions. I guess I'm having a hard time because I haven't seen any examples of almost homomorphisms and I just can't believe that there would be so many functions that are almost homomorphisms. I mean, what equivalence class would represent e.g. $1$ or $\frac{3}{2}$? And how are the reals ordered if they are representations of equivalence classes of functions? (I am aware there are other ways to construct the reals).

Also, this is probably a stupid question, but seeing as there is a rational number between any two irrational numbers, why can't we find a bijection between $\mathbb Q$ and the irrationals, and thus between $\mathbb N$ and $\mathbb R$?

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    $\begingroup$ See this related question. Besides, note that your final question is not related to the rest. $\endgroup$ – José Carlos Santos Jun 19 '18 at 12:38
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That's a lot of questions, but if you want an example of an almost homomorphism, take a real number $x$ and consider the map$$\begin{array}{ccc}\mathbb{Z}&\longrightarrow&\mathbb Z\\n&\mapsto&\lfloor nx\rfloor.\end{array}$$Actually the equivalence class of this almost homomorphism corresponds to the real number $x$. In particular, the identity function (which is a homomorphism) corresponds to the real number $1$ and the map $n\mapsto\left\lfloor\frac{3n}2\right\rfloor$ corresponds to the real number $\frac32$.

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