If we define a relation $\sim$ between real numbers so that $x \sim y$ holds precisely if $y - x$ is rational, then we need AC to prove that there exists a set of distinct representatives for the equivalence classes of $\sim$. Do we also need AC to prove the same for $\sim$ defined on just the algebraic numbers? After all, we then have only countably many classes. But it still seems hard?!

  • $\begingroup$ We don't need the full axiom of choice for proving the existence of distinct representatives. The full axiom of choice is much more powerful. We do need some choice-like axiom, though. I don't know about the algebraic case, but if we do need something, what we need is probably weaker still. $\endgroup$
    – Arthur
    Jul 4, 2018 at 21:30
  • $\begingroup$ @XanderHenderson That's a red herring - no amount of choice is needed here, at all. $\endgroup$ Jul 4, 2018 at 21:33
  • $\begingroup$ @Noah: No amount is some amount... what kind of a logician are you? :P $\endgroup$
    – Asaf Karagila
    Jul 4, 2018 at 21:47

2 Answers 2


No, AC is no longer needed.

The key point is that we can well-order the algebraic real numbers without using choice, as follows:

  • First, fix an enumeration $\{P_i:i\in\mathbb{N}\}$ of the nonconstant polynomials with rational coefficients.

  • Now given an algebraic number $r$, let $\#(r)$ be the least $i$ such that $r$ is a solution to $P_i$.

  • Finally, for algebraic numbers $r\not=s$ write $r\triangleleft s$ iff one of the following holds:

    • $\#(r)<\#(s)$, or

    • $\#(r)=\#(s)$ and $r<s$.

  • It's now not hard to show that $\triangleleft$ well-orders the set of algebraic numbers (in fact, with order-type $\mathbb{N}$). (The key point here is that any nonconstant polynomial has finitely many roots.)

We can use this well-ordering to build a Vitali set: given an algebraic number $r$, let $rep(r)$ be the $\triangleleft$-least algebraic real which differs from $r$ by a rational number. The set $$\{rep(r): r\mbox{ is algebraic}\}$$ is then a "$\mathbb{Q}$-Vitali set" as desired.

More generally, if we can prove without choice that $A$ is a well-orderable set of reals, then we can also prove without choice that there is an "$A$-Vitali set." Put another way:

"Every well-orderable set of reals has a Vitali subset" is provable without choice.


No. The Vitali set is given by choosing from the fibers of a surjection from $\Bbb R$ to $\Bbb R/\Bbb Q$.

If you restrict your surjection to a countable set, any countable set, then you can first enumerate it as $\{r_n\mid n\in\Bbb N\}$ and then choose the least indexed $r_n$ in each fiber.

  • $\begingroup$ A bit less overkill than talking about well orderings. $\endgroup$
    – Jack M
    Jul 5, 2018 at 11:31
  • 1
    $\begingroup$ Well, it's not really of an overkill to talk about well-orders. This is really an obvious generalization. $\endgroup$
    – Asaf Karagila
    Jul 5, 2018 at 11:31

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