Uncountably Many Equivalence Classes

Question

I wrote this down in my notes: Can we have $\mu: P(\mathbb{R})\rightarrow [0,\infty] $ such that:

1. If $E_1...$ is a finite or infinite sequence of disjoint sets then $\mu(E_1 \cup \cup E_2...)=\mu(E_1)+\mu(E_2)+...$
2. If $E$ is congruent to $F$, then $\mu(E)=\mu(F)$.
3. $\mu(Q)=1$.

Lemma: There exists a set $N \subset[0,1)$ such that $\mu(N)$ is ill-defined.

Proof: Consider the following equivalence relation on $\mathbb{R}$: $x\sim y \leftrightarrow x-y \in \mathbb{Q}$ Then $\mathbb{R}$ decomposes into uncountably many equivalence classes.

This is the part that I don't really understand. How does $\mathbb{R}$ decompose into uncountably many equivalence classes? How does each equivalence class look like? Wouldn't all the integers and rational numbers be in the same equivalence class: $\{0,1,2,3,1/2,1/3,1/4,-1,1/5,1/6,\dots\}$. But what about the irrational numbers? Do each of them form their own equivalence class and since there are uncountably many of them, there are an uncountable number of equivalence classes? Also, can a single number be in two different equivalence classes? Thanks for the help.

Answer 1

It's not exactly that each irrational form its own equivalence class, but that's almost right, up to addition of a rational.

Every rational is trivially in the class of $0$. The diference of an irrational $x$ and another real number $y$ is irrational unless $y$ is of the form $x+r$, with $r$ rational. Therefore, the classes are of the form $[x]=\{x+r\mid r \in\Bbb{Q}\}$ for each $x\in \Bbb{R}\setminus \Bbb{Q}$ and the class $[0]$ for the rationals.

These are uncountable many classes, since each equivalence class has only countably many elements, but they cover the whole $\Bbb{R}$.

No number can be in two different equivalence classes, and this is a general property of equivalence relations. If $x\sim y$ and $x\sim z$, then $y\sim z$ (using symmetric and transitive properties).