When looking for non-measurable set here, the proof starts by taking the quotient group $\mathbb{R}/\mathbb{Q}$ which is the "modulo the rationals" operation. I get it until the sentence:
Each element of $\mathbb{R}/\mathbb{Q}$ intersects $[0, 1]$, and the axiom of choice guarantees the existence of a subset of $[0, 1]$ containing exactly one representative out of each element of $\mathbb{R}/\mathbb{Q}$.
where it loses me completely.
What I barely understand:
Each element of $\mathbb{R}/\mathbb{Q}$ intersects $[0, 1]$
First, is an element of $\mathbb{R}/\mathbb{Q}$ a representative of a set of the form $A_x = \{x+q, q \in \mathbb{Q} \}$ where $x \in \mathbb{R}$?
Is $A_x$ the same as all $v \in \mathbb{R}$ such that $x-v \in \mathbb{Q}$?
If so, is the folloxing true :
Since $\mathbb{Q}$ is dense in $\mathbb{R}$ then $\exists y \in [0,1]$ such that $A_y = A_x$
?
Next, what I completely do not understand:
the axiom of choice guarantees the existence of a subset of $[0, 1]$ containing exactly one representative out of each element of $\mathbb{R}/\mathbb{Q}$
I don't even know how to phrase my question for that part.
Why is the Axiom of Choice a necessity (I must warn that AC is also something I don't understand)?
If $\mathbb{R}/\mathbb{Q}$ intersects $[0,1]$, why can't we just "take" one element of each equivalence class that is in $[0,1]$ (like taking the $y$ used for $A_y$) ? Why do we need AC to do that?
The elements in the set $\mathbb{R}/\mathbb{Q}$ are equivalence classes, i.e sets of the form $A_x=\{y\in\mathbb{R}: x-y\in\mathbb{Q}\}$ when $x\in\mathbb{R}$. Now let $x\in\mathbb{R}$. There is some rational number $y\in (x-1,x)$. Let $z=x-y\in [0,1]$. Then $x-z=x-(x-y)=y\in\mathbb{Q}$ and hence $A_x=A_z$.
As for the second part, note that for a given $A_y$ there are infinitely many numbers $z\in [0,1]$ that satisfy $A_y=A_z$. So which one will you choose every time? It's really not clear how you can build a specific choice function. Yes, you must use the axiom of choice, a very strong version of this axiom. The axiom tells us that a choice function exists, even though we don't really see how can we find such a function.