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I am surprised at the language used in connection with the axiom of choice. From the answer to a question a made (which turned out to be duplicate) about involvement of AC in Wiles’ proof of Fermat’s last theorem I note that the axiom (AC) is treated like a conjecture that (according to some) might be - or obviously is (according to some) – true.

My question is: In what sense could an axiom be true if it is independent of the Peano axioms (supplemented with addition and multiplication), or are there several contexts available in which the axiom may be true in one and not (ed: or unprovable) in another?

Mikael Jensen
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  • Your question is very unclear to me. Also, what does the axiom of choice have to do with Peano axioms? – Asaf Karagila Dec 26 '13 at 15:31
  • I thought they were closely related to the Zermelo–Fraenkel axioms. Perhaps I should have replaced Peano with the those names. Would it help any? – Mikael Jensen Dec 26 '13 at 15:39
  • AC is independent of ZF, not only of Peano arithmetic. So you can accept it or reject it. Some mathematicians reject it because it gives only "existence proofs", without any construction. Regarding Wiles'proof, see http://math.stackexchange.com/questions/151895/number-theory-in-a-choice-less-world – Jean-Claude Arbaut Dec 26 '13 at 15:41
  • Read is very interesting question which was asked in MathOverFlow: http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice – Yiorgos S. Smyrlis Dec 26 '13 at 15:47
  • @smyrlis I saw, on the same site, a question about whether or not people were worrying about the AC. That confused me just in the same way. Perhaps I should have asked about axioms in general and how they might be true in any sense. If they could be true it seems to me that they would be unnecessary. – Mikael Jensen Dec 26 '13 at 16:19
  • @Mikael: Truth is a semantic property. An axiom is true in an interpretation of the language. Unprovability is syntactical, it means that there is no proof; and so is independence (there is no proof, and no proof of the negation). – Asaf Karagila Dec 26 '13 at 16:36
  • I also wonder about “rejection” or acceptance (of the AC). If the AC can generate some separate theorems (from ZF) it seems to me similar to a natural number a, which can generate a subset of the natural numbers n, such as the products a*n. If I study this subset I don’t necessarily “reject” all the natural numbers (or vice versa). Both activities seem equally acceptable. In physics a model most often is intended to describe reality, which leads to the standard of verisimilitude. It is as if we had a mathematical “reality” and a corresponding standard. – Mikael Jensen Dec 29 '13 at 10:17

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The axiom of choice have nothing to do with number theory an Peano axioms. The language in which the axiom of choice is formalized has nothing in common with the language in which the Peano axioms and number theory are written.

However, when proving a statement one sometimes uses set theory as a formal theory in the background (even without referring to it explicitly. Simply by referring to sets, models, and so on), such theory is called "meta-theory", as say that we work "in that theory". The axiom of choice is a part of our meta-theory.

One can ask, if so, do we need the axiom of choice as part of our meta-theory for a certain proof to hold? Sometimes the answer is yes. For example if one wants to prove that every field has an algebraic closure. But it turns out that for statements about the natural numbers which are "simple enough", e.g. first-order statements in the language of Peano axioms, the axiom of choice does not enhance our power to prove things. So if we proved Fermat's last theorem with $\sf ZFC$ as a meta-theory, we can in fact prove it within $\sf ZF$.

(One caveat is when proving statement in set theory, one can talk about the axiom choice in the theory, in which case one proves a statement in the language of the theory, with the axioms of the theory which may or may not include the statement known as "the axiom of choice".)

Asaf Karagila
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  • It is but not absolutely obvious that "the axiom of choice does not enhance our power to prove things". Cf. for example Goodstein's "hydra" theorem – Hagen von Eitzen Dec 26 '13 at 17:35
  • Hagen, I never said it is obvious. But it is true. Any first-order statement (and simple enough second-order statements) about the integers which is provable with the axiom of choice, is also provable without it. – Asaf Karagila Dec 26 '13 at 17:36
  • I know, and indeed you specifically said "it turns out". I just wanted to make a warning to readers who might uncautiously nod and think that e.g. just because the language of PA does not allow us to reference "complicated" sets (or in Goodsteins's case: general ordinals) those "advenced" tools can automatically be gotten rid of. – Hagen von Eitzen Dec 26 '13 at 17:52