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    $\begingroup$ Well, there is Quine's New Foundations in set theory, in which the diagonal argument is blocked from disproving the existence of a set of all sets, because of the inability to express the predicate $x \not \in x$. But I gather that NF does not block the diagonal argument from demonstrating the uncountability of the reals, so this isn't quite an answer to the problem at hand... $\endgroup$ Commented Nov 22, 2010 at 23:28
  • 17
    $\begingroup$ Reverse mathematics (en.wikipedia.org/wiki/Reverse_mathematics) is almost exactly about studying which axioms and arguments are necessary for certain theorems. If you want to know whether axiom X is necessary for a theorem Y, you can try to see if there's a model of Y in which X doesn't hold. It's not as easy to see whether a certain argument is necessary, but often you can axiomatize what it means to be able to do a certain argument, e.g. there are systems which capture what it means to be able to use a compactness argument, or induction, or transfinite recursion, etc. $\endgroup$ Commented Nov 23, 2010 at 18:27
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    $\begingroup$ @Amit: Yes, I'm familiar with reverse mathematics. But let me repeat what I said above: Think closely! How would you axiomatize what it means to be able to diagonalize? What candidate do you have in mind for a model in which the reals are countable? I stand by what I said; nobody has a clue. $\endgroup$ Commented Nov 24, 2010 at 2:28
  • 6
    $\begingroup$ 1. Are you saying reverse math never proves results of the form "Argument A is necessary for Theorem T" (in some reasonable sense of the word "necessary")? 2. You presume that diagonalization is necessary for Cantor's theorem - someone who believes not wouldn't need a model where the reals are countable. 3. Why is it that "being able to diagonalize" can't be axiomatized? A huge variety of arguments get referred to as compactness arguments, so naively one would think you couldn't axiomatize "being able to do a compactness argument", but doesn't Weak Konig's Lemma sort of do just that? $\endgroup$ Commented Nov 25, 2010 at 8:44
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    $\begingroup$ @TimothyChow From what I've seen, I agree with you that this sort of mathematics hasn't been studied at nearly the level required to prove such a theorem, but "nobody has any idea", I don't think is a very good way to surmise the situation. In particular, for set theory developed over a certain paraconsistent logic, Cantor's theorem is unprovable. See "What is wrong with Cantor's diagonal argument?" by Ross Brady and Penelope Rush. So, if one developed enough of reverse mathematics in such a context, one could I think meaningfully ask this question. $\endgroup$ Commented Oct 25, 2017 at 12:50