Bijection between $\mathbb{R}$ and $\mathbb{R}/\mathbb{Q}$

I was wondering if it is possible to produce an explicit bijection $h\colon \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Q}$. If we can produce an explicit injection $i\colon \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Q}$, can the Cantor-Bernstein-Schroeder Theorem be used constructively?

It is clear that the two sets have the same cardinality, so the existence of such a bijection is trivial. What I am really looking for is a nice-looking bijection, or a proof that no such nice-looking bijection exists, for a definition of "nice-looking" which I cannot quite figure out.

One problem which I think makes finding such a bijection difficult is that any natural injection of $\mathbb{R}/\mathbb{Q}$ (ie, those injections in which one representative is chosen from each coset) produces a non-measurable set, specifically a Vitali set.

I hate to ask such a vague question, but I'm really not sure about whether the correct answer is constructive, or whether it is a proof that any such bijection is in some sense "very complicated."

As a final note, the motivation for this question came from this discussion, in which I was somewhat astonished to see such a clear, constructive bijection given between $\mathbb{R}$ and $\mathbb{R} \setminus S$, where $S$ is countable.

Solution 1:

It is not possible.

It is consistent with set theory without choice that $\mathbb R/\mathbb Q$ has strictly larger cardinality that $\mathbb R$. (This looks counter-intuitive, since $\mathbb R/\mathbb Q$ is a quotient.)

This is the case because, using a fact that goes back to Sierpiński (Sur une proposition qui entraîne l’existence des ensembles non mesurables, Fund. Math. 34, (1947), 157–162. MR0023318 (9,338i)), in any model of $\mathsf{ZF}$ where all sets of reals have the Baire property, it is not even possible to linearly order $\mathbb R/\mathbb Q$.

(Sierpiński argues in terms of Lebesgue measure. The argument in terms of the Baire property is analogous, and has the additional technical advantage of not requiring any discussion of consistency strength issues.)

A couple of years ago, Mike Oliver gave a nice talk on this topic (How to have more things by forgetting where you put them); he is not exactly using $\mathbb R/\mathbb Q$, but the arguments easily adapt. The point of the talk is precisely to give some intuition on why we expect the quotient to be "larger".

[Of course, in the presence of choice, the two sets have the same size. The argument above shows that the use of choice is unavoidable.]