Consider the equivalence relation on $\mathbb{R}$ given by $x\sim y \iff x-y\in\mathbb{Q}$.
Several questions about this have been asked before, but I could not find anywhere what the disjoint equivalence classes of $\mathbb{R}/\sim$ are.
I know that the equivalence classes are of the form $$[a]=\{a+q:q\in\mathbb{Q}\}$$ where $a$ is irrational.
But how would I go about writing down $\mathbb{R}$ as a union of disjoint equivalence classes? (I believe this is possible for any equivalence relation on any set).
I want to write $$\mathbb{R}=\mathbb{Q}\sqcup \bigsqcup_{\text{some condition on $a$}}[a]\,.$$
I know that the condition cannot be $a\notin\mathbb{Q}$, because then, for example, $[\pi+1]=[\pi]$, so the union is not disjoint.
What condition on $a$ makes this work? Is it $\{a\in[0,1):a\notin\mathbb{Q}\}$? If so, can anyone give a hint for how to prove that these are all disjoint?
It's easily shown that the specified relation is indeed an equivalence relation. Your goal is to choose a unique representative for each equivalence class by some kind of formula or algorithm.
Unfortunately, that's not possible.
Suppose $A$ is a set of unique representatives for the set of equivalence classes.
If you apply the Axiom of Choice (inherently non-constructive), such a set $A$ exists.
But one can show that for any such set $A$, the set $A\cap \mathbb{R}$ is non-measurable.
Since the assertion that there exist non-measurable subsets of $\mathbb{R}$ is known to be independent of the standard axioms of Set Theory without the Axiom of Choice, that effectively blocks your goal of "constructing" such a set $A$.