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  1. removing query box from page

    +– {: .query} Mike Shulman: Is there a formal statement in some formal system along the lines of “a non-extensional choice operator does not imply AC”?

    Toby: I don't know about a formal statement, but I can give you an example.

    Recall: In Per Martin-Löf's Intuitionistic Type Theory (and many other systems along similar lines), the basic notion axiomatised is not really that of a set (even though it might be called ’set’) but instead a preset (or ’type’). Often one hears that the axiom of choice does hold in these systems, which doesn't imply classical logic due to a lack of quotient (pre)sets. However, if we define a set to be a preset equipped with an equivalence predicate, then the axiom of choice fails (although we have COSHEP if presets come with an identity predicate).

    A lot of these systems (including Martin-Löf's) use ’propositions as types’, in which x:AP(x)\exists_{x:A} P(x) is represented as x:AP(x)\sum_{x:A} P(x), which comes equipped with an operation π: x:AP(x)A\pi: \sum_{x:A} P(x) \to A. That is not going to get us our choice operator, but since a choice operator is constructively questionable anyway, then let's throw in excluded middle. This is known to not imply choice, but we do have, for every preset AA, an element ε A\varepsilon_A of A¬AA \vee \neg{A}, that is of A AA \uplus \empty^A. It's not literally true that ε A\varepsilon_A is of type AA, of course, but that would be unreasonable in a structural theory; what we do have is a fixed ε A\varepsilon_A such that, if AA is inhabited, then ε A=ι A(e)\varepsilon_A = \iota_A(e) for some (necessarily unique) ee of type AA (where ι A\iota_A is the natural inclusion AA AA \to A \uplus \empty^A), which I think should be considered good enough. This is for presets (types), but every set has a type of elements, so that gets us our operator.

    How is this nonextensional? We do have ε A=ε B\varepsilon_A = \varepsilon_B if A=BA = B (which is a meaningful statement to Martin-Löf, albeit not a proposition exactly), but if AA and BB are given as subsets of some UU, then we may well have A=BA = B as subsets of UU without A=BA = B in the sense of identity of their underlying (pre)sets. In particular, if f:UVf: U \to V is a surjection and AA and BB are the preimages of elements xx and yy of VV, then x= Vyx =_V y will not imply that ε A=ε B\varepsilon_A = \varepsilon_B, and the proof of the axiom of choice does not go through. It will go through if xx and yy are identical, that is if x=yx = y in the underlying preset of VV, so again we do get choice for presets (again), but not for sets.

    I'm not certain that a nonextensional global choice operator won't imply excluded middle in some other way, but I don't see how it would. You'd want to do something with the idea that ε A\varepsilon_A always exists but belongs to AA if and only if AA is inhabited, but I don't see how to parse it (just by assuming that it exists) to decide the question.

    Mike Shulman: That’s a very nice explanation/example, and it did help me to understand better what’s going on; thanks! (Did you mean to say “excluded middle” and not “AC” in your final paragraph?) What I would really like, though, is a statement like “the addition of a nonextensional global choice operator to ____ set theory is conservative” (i.e. doesn’t enable the proving of any new theorems that doen’t refer explicitly to the choice operator). Of course I am coming from this comment, wondering whether what you suggested really is a way to get a choice operator without implying the axiom of choice.

    Toby: Yeah, I really did mean to say ’excluded middle’; remembering that comment, I assume that the real question is whether the thing is OK for a constructivist. I just argued ITT+EMCO\mathbf{ITT} + EM \vDash CO, and I know the result ITT+EM¬AC\mathbf{ITT} + EM \not\vDash AC, so I conclude ITT+CO¬AC\mathbf{ITT} + CO \not\vDash AC; but I don't know ITT+CO¬EM\mathbf{ITT} + CO \not\vDash EM for certain. I certainly don't have ITT+CO\mathbf{ITT} + CO conservative over ITT\mathbf{ITT}, nor with any other theory (other than those that already model COCO, obviously).

    Mike Shulman: Where should I look for a proof that ITT+EM\mathbf{ITT} + EM doesn’t imply AC?

    Toby: I'm not sure, it's part of my folk knowledge now. Probably Michael J. Beeson's Foundations of Constructive Mathematics is the best bet. I'll try to get a look in there myself next week; I can see that it's not exactly obvious, and perhaps my memory is wrong now that I think about it.

    Mike Shulman: I’m trying to prove the sort of statement I want over at SEAR+?.

    Toby: No, I can't get anything at all out of Beeson (or other references) about full AC (for types equipped with equivalence relations) in ITT\mathbf{ITT}.

    Harry Gindi: I have references for this discussion that should settle the issue at hand:

    Bell, J. L., 1993a. ’Hilbert’s epsilon-operator and classical logic’, Journal of Philosophical Logic, 22:1-18

    Bell, J. L., 1993b. ’Hilbert’s epsilon operator in intuitionistic type theories’, Mathematical Logic Quarterly 39:323-337

    Meyer Viol, W., 1995a. ’A proof-theoretic treatment of assignments’, Bulletin of the IGPL, 3:223-243

    Toby: Thanks, Harry! Now I just have to find these journals at the library. =–


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  2. added section about the global choice operator in dependent type theory


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