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+– {: .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 $\exists_{x:A} P(x)$ is represented as $\sum_{x:A} P(x)$, which comes equipped with an operation $\pi: \sum_{x:A} P(x) \to A$. 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 $A$, an element $\varepsilon_A$ of $A \vee \neg{A}$, that is of $A \uplus \empty^A$. It's not literally true that $\varepsilon_A$ is of type $A$, of course, but that would be unreasonable in a structural theory; what we do have is a fixed $\varepsilon_A$ such that, if $A$ is inhabited, then $\varepsilon_A = \iota_A(e)$ for some (necessarily unique) $e$ of type $A$ (where $\iota_A$ is the natural inclusion $A \to A \uplus \empty^A$), 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 $\varepsilon_A = \varepsilon_B$ if $A = B$ (which is a meaningful statement to Martin-Löf, albeit not a proposition exactly), but if $A$ and $B$ are given as subsets of some $U$, then we may well have $A = B$ as subsets of $U$ without $A = B$ in the sense of identity of their underlying (pre)sets. In particular, if $f: U \to V$ is a surjection and $A$ and $B$ are the preimages of elements $x$ and $y$ of $V$, then $x =_V y$ will not imply that $\varepsilon_A = \varepsilon_B$, and the proof of the axiom of choice does not go through. It will go through if $x$ and $y$ are identical, that is if $x = y$ in the underlying preset of $V$, 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 $\varepsilon_A$ always exists but belongs to $A$ if and only if $A$ 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 $\mathbf{ITT} + EM \vDash CO$, and I know the result $\mathbf{ITT} + EM \not\vDash AC$, so I conclude $\mathbf{ITT} + CO \not\vDash AC$; but I don't know $\mathbf{ITT} + CO \not\vDash EM$ for certain. I certainly don't have $\mathbf{ITT} + CO$ conservative over $\mathbf{ITT}$, nor with any other theory (other than those that already model $CO$, obviously).
Mike Shulman: Where should I look for a proof that $\mathbf{ITT} + EM$ 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 $\mathbf{ITT}$.
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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