This question actually consists 3 related pieces of text, which I've gathered under this title about which I would like your opinion (they rather contain the implicit question "is this the right way to think about this issue"). I'm willing to offer bounty for this.

$1.$ Consider the following theorem: There is no function $f:\mathcal{P}\left(\mathbb{R}\right)\rightarrow\left[0,\infty\right]$ that is translation-invariant and $\sigma$-additive such that $0<f\left(\left(0,1\right)\right) < \infty$.

Its proof goes like this: We first define a equivalence relation $a\sim b:\Leftrightarrow a-b\in\mathbb{Q}$ on $\mathbb{R}$, then select a system $S$ of representatives from it that lies in $\left(0,1\right)$. This $S$ is then the key to prove the theorem.

What I'm concerned with is how we obtain $S$. The proof I know says "we have to use the axiom of choice to do that'': The choice function picks member from every an equivalence class. As I understand it the use of the axiom of choice comes from the fact that we can't determine explicitly the form of the set all the equivalence classes (if we have one representative we can easily get one that lies in the open unit interval).

Each equivalence class has to have the form $a+\mathbb{Q}$ for some real $a$ and it is clear that the equivalence class of any rational number is $\mathbb{Q}$. But the irrational numbers cause problem: We have no means to explicitly describe all equivalence classes of irrational numbers, since for a given $b,c\in\mathbb{R}\setminus\mathbb{Q}$, we don't know if they are in the same equivalence class or not: $1+\sqrt{2}$ and $2+\sqrt{2}$ obviously are; but for $\pi$ and $e$ we don't know, since currently it is unknown whether $\pi-e$ is rational or not. Now the axiom of choice circumvents this issue. But the need to use the axiom of choice seems to me to depend in this case only on present state of research: Maybe in 100 years we will settle the case for $\pi$ and $e$ and moreover devise some methods such that for given $b,c\in\mathbb{R}\setminus\mathbb{Q}$ we know if they are in the same equivalence class or not (for example if it turns out that we can somehow classify them we can distinguish by cases). Therefore saying we have to use the axiom of choice seems bad to me; saying "presently we cannot do without the axiom of choice" seems better.

$2.$ The above seems to me to be in contrast to saying "we have to use the axiom of choice to exhibit a functions that picks an element from every subset of the reals" since I read, that there there are strong model-theoretic arguments that imply that indeed no one will ever be able to explicitly construct such a function. But even in this case it seems to me, that the word "have" is too strong, since it implies that one can somehow prove that without the axiom of choice it is impossible to prove that statement.

$3.$ Is proving, that a statement $T$ cannot be proven without the axiom of choice the same as proving that $T$ is equivalent to the axiom of choice ?

Do you agree with me ?

  • $\begingroup$ Obviously, we do not have to assume anything near as powerful as the full axiom of choice to pick an element from each subset of the reals (we can for example, assume choice for just sets of a suitable size, or indeed choice specifically for the powerset of the reals, without the empty set of course). $\endgroup$ – Tobias Kildetoft Feb 25 '13 at 14:48
  • 6
    $\begingroup$ Your question, especially part 1, seems to presuppose that the axiom of choice is somehow about our ability to perform certain tasks --- check whether two reals are equivalent, or choose a representative from each equivalence class (note that these are not the same task). Despite the "choice" in its name, the axiom is not about our abilities but simply about the existence of some sets --- in this case, a set containing exactly one member of each equivalence class. $\endgroup$ – Andreas Blass Feb 25 '13 at 15:10
  • $\begingroup$ @AndreasBlass Yes, that's true. But the existence of such a set is just a formal way of capturing the idea of being able to make such a choice, I believe (correct me, if I'm wrong). And since thinking about sets is less natural (at least to me) than thinking about making choices, this probably transpires in my writing. Aren't mathematicians constantly thinking in a "lively" manner about the things they do, although all they is just manipulating sets ? $\endgroup$ – temo Feb 25 '13 at 16:52
  • 3
    $\begingroup$ @temo It is indeed possible to think and speak of using AC informally as "making choices" but first you have to understand what the axiom literally says, otherwise your intuition regarding real world notions may lead you astray (and probably will.) $\endgroup$ – Trevor Wilson Feb 25 '13 at 17:24
  • $\begingroup$ Regarding your question 2, this story about George Dantzig might justify your feelings. $\endgroup$ – Daniel Dinnyes Mar 6 '13 at 13:04

Let me answer in three parts.

  1. We when we say that we need to use the axiom is not because for some numbers do know and for some with don't know what is their equivalence class. There are models where the axiom of choice fails and it is impossible to choose representatives from $\Bbb{R/Q}$.

    This is why we need the axiom of choice, and will always need it, in order to construct a non-measurable set such as a Vitali set. It is not about our lack of knowledge about the explicit real numbers and so on, it is simply because we know that it is consistent that it is impossible to do if the axiom of choice fails.

  2. Yes. We can prove that in some models it is impossible to make this choice, for example there are models where all the sets are in fact Lebesgue measurable, and there are others where all the sets are Borel. In both these models we have that (1) the axiom of choice fails, and (2) there is no choice function from $\Bbb{R/Q}$.

  3. No. It is not the same thing. To see it more clearly observe that the axiom of choice is much stronger than the axiom of countable choice (choice from countable families of non-empty sets). But it is fine to say that you cannot prove the axiom of countable without some form of the axiom of choice. That is to say, we cannot prove it from the rest of the axioms of ZF. It does not mean that we can prove the axiom of choice if we have the axiom of countable choice at our disposal.

    Similarly the fact that we cannot prove the existence of a non-measurable set, or in particular a Vitali set, without some form of the axiom of choice does not imply that we can prove the axiom of choice from the existence of such set.

    The rule of thumb is that if a statement is talking about a concrete set (in our case, $\Bbb R$, or $\cal P(\Bbb R)$ if you prefer) then it will not imply the axiom of choice. Why? Because we can arrange a model where the axiom of choice holds for sets far larger than this one, and then breaks in the most acute way possible.

You might also be interested in: Axiom of choice, non-measurable sets, countable unions

  • $\begingroup$ Interesting. I did not know it was possible break choice only for "small" sets but make it hold for large enough sets. $\endgroup$ – Tobias Kildetoft Feb 25 '13 at 15:06
  • $\begingroup$ @Tobias Kildetoft: he was saying the opposite: it is possible for AC to hold for sufficiently small sets and fail for larger ones, where the line of "sufficiently small" can be placed arbitrarily high. $\endgroup$ – Carl Mummert Feb 25 '13 at 15:43
  • $\begingroup$ @CarlMummert Ahh, I read it wrong then (it did sound like a strange thing) $\endgroup$ – Tobias Kildetoft Feb 25 '13 at 15:45
  • 2
    $\begingroup$ @temo The failure of AC does not imply the impossibility of choosing representatives for $\mathbb{R}/\mathbb{Q}$. Choosing representatives for $\mathbb{R}/\mathbb{Q}$ is only one consequence of AC among many. $\endgroup$ – Trevor Wilson Feb 25 '13 at 17:20
  • 1
    $\begingroup$ @temo: It has been proven that it is possible that the axiom of choice fails and that we cannot choose representatives. It is also consistent that the axiom of choice fails and we can. I am telling you that $x\in\Bbb R$ and $x\nless 0$. It is possible that $x=0$, and it is possible that $x>43.1$, and it is possible that $x<500$. Just assuming AC fails is not enough information, we need to know how it fails. And we have proved that it is possible that such choice is impossible to make in some models of ZF. $\endgroup$ – Asaf Karagila Feb 25 '13 at 20:01

Asaf's answer is of course correct but I think there may be a misconception by the OP that it does not fully address. Not knowing whether $\pi + \mathbb{Q}$ and $e + \mathbb{Q}$ are the same is not the problem, because we could just split into cases: if they are the same, put $\pi-3$ into $S$ and if they are not the same, put both $\pi-3$ and $e-2$ into $S$. This is an ok thing to do in classical logic, and in either case that "step" of the Vitali set construction succeeds.

There are only finitely many mathematical constants that anyone has ever named, so we could just split into cases for each one as above, but that doesn't get us very far. The problem is that every equivalence class $C$ has the form $r + \mathbb{Q}$ for many different $r$ (infinitely many) and when we choose an element of $C$ we are only given $C$—we are not given the term "$r + \mathbb{Q}$" so we can't just make sure to pick $r$ itself based on the notation. The vast majority of equivalence classes $C$ don't contain any named constants like $e$ or $\pi$. So given a "typical" $C$, we know that it is nonempty by definition but have no rule to pick any element at all (which is the situation that the Axiom of Choice is designed to address.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.