39
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Are you asking a clarification of the previous answers in your discussion link? I didn't take a great look at it, but if it describes a construction between $\mathbb R$ and $\mathbb R \backslash S$ where $S$ is countable, how about you take $S = \mathbb Q$? $\endgroup$
    – Patrick Da Silva
    Nov 24, 2012 at 5:07
  • $\begingroup$ Sorry, I should have made it clear. I mean the cosets of $\mathbb{Q}$ in $\mathbb{R}$, not simple "set minus". $\endgroup$
    – KSackel
    Nov 24, 2012 at 5:09
  • $\begingroup$ @user : Oh! Quotient. Seriously, my bad. I just read it wrong. $\endgroup$
    – Patrick Da Silva
    Nov 24, 2012 at 5:09

1 Answer 1

Reset to default
49
$\begingroup$

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.]

$\endgroup$
14
  • 5
    $\begingroup$ It doesn't look counter-intuitive, it looks false... but of course, everything looks wrong without choice. But if we have choice, then of course the cardinality of the quotient is the same as that of the reals. Right? $\endgroup$
    – Patrick Da Silva
    Nov 24, 2012 at 5:15
  • 2
    $\begingroup$ Right, by the argument you indicated. $\endgroup$
    – Andrés E. Caicedo
    Nov 24, 2012 at 5:17
  • 3
    $\begingroup$ @PatrickDaSilva it's not that counter-intuitive - we have in most categories that we study that "onto" maps do not have one-sided inverses. Essentially, set theory without choice can be set theory with "structure" to your sets. Often that structure is complexity, and I think this is an obvious case where a quotient space is "more complex" than the parent spaces. $\endgroup$
    – Thomas Andrews
    Nov 24, 2012 at 6:40
  • 1
    $\begingroup$ @PatrickDaSilva That has nothing to do with category theory: it's like saying "haven't done enough English" when someone uses a word you don't know. What Thomas is saying is actually something very simple and you surely know examples of this phenomenon: there are surjective group homomorphisms with no left inverse, there are surjective continuous maps with no left inverse, etc. etc. $\endgroup$
    – Zhen Lin
    Nov 24, 2012 at 8:42
  • 1
    $\begingroup$ @Zhen : I guess "I didn't understand English well enough so that his comment would make sense to me" then! :P $\endgroup$
    – Patrick Da Silva
    Nov 24, 2012 at 8:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .