# Equivalence relation on $\mathbb{R}$: $x\sim y \iff x-y\in\mathbb{Q}$ - disjoint equivalence classes

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?

• Note: If $a\in(0,1)$ with $a\notin \mathbb{Q}$, then $a\sim a-\large{\frac{1}n}$, for all positive integers $n$, and $a-\large{\frac{1}n}\in(0,1)$, for all sufficiently large positive integers $n$. – quasi Aug 4 '18 at 5:32
• It is quite tautological but if $I$ denotes the set of irrational numbers and $\sim$ your equivalence relation (on $I$), then I think that if you set each $a$ to belong to one equivalence class of $I/\sim$ you could get your goal. – Dog_69 Aug 4 '18 at 8:42

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 $A$.

• I think your argument is in the wrong direction. You used AC when you say: "Let $A$ be a set of unique representatives." – Gaston Burrull Aug 4 '18 at 6:11
• Well, I said "Suppose $A$ is ..." – quasi Aug 4 '18 at 6:14
• I mean, you have to invoke the axiom of choice in that point, not after proving that such a subset is a non-measurable set and then proving: "But it's not possible to construct a non-measurable subset of R, without invoking the Axiom of Choice." which is not necessary true. – Gaston Burrull Aug 4 '18 at 6:19
• I don't follow your objection. If we assume such a set $A$ exists, one can prove it's non-measurable without using the Axiom of Choice. – quasi Aug 4 '18 at 6:20
• Ok, I see your point. I'll edit my answer . . . – quasi Aug 4 '18 at 6:24