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It is well-known that the Axiom of Choice entails a number of counter-intuitive results, Banach-Tarski paradox is only one example. In this MSE question, Martin Sleziak asked about similar undesirable consequences under the Axiom of Determinacy. I found this question very interesting and, for example, an equivalence relation on the real line with strictly more than continuum many equivalence classes really surprised me!

I've been wondering whether one can also find some problematic results under the Axiom of Projective Determinacy - the axiom considered to be "true" by some mathematicians. Are there any "disasters" with Projective Determinacy?

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  • $\begingroup$ I mean, Banach–Tarski still holds. Just the partition is not projective. $\endgroup$
    – Asaf Karagila ♦
    Jul 3, 2019 at 21:07
  • $\begingroup$ @AsafKaragila Indeed, haven't thought about this while writing the question. $\endgroup$
    – Wojciech Aleksander Wołoszyn
    Jul 3, 2019 at 21:22

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I would say no, there are no "projective determinacy disasters" (other than the already-present ZFC disasters, of course). The point is that our ideas about how every set should behave don't really refine to ideas about how every projective set should behave - the class of projective sets is sufficiently technical that we approach it with a fairly open mind - and so we don't get the same "kick" as we do from full determinacy.

In a bit more detail:

Every determinacy disaster has a corresponding projective analogue - e.g. determinacy implies that the Vitali relation $$x\sim y\iff x-y\in\mathbb{Q}$$ has no transversal, and projective determinacy implies that it has no projective transversal. However, this analogy robs the disasters of a lot of their strength: our intuition that the Vitali relation should have a transversal doesn't in my opinion extend to an intuition that it should have a simple transversal, so this implication of projective determinacy isn't really counterintuitive.

Indeed, in my opinion this robs the disaster of all of its nonintuitiveness, and this pattern continues for all the other determinacy disasters (again, in my opinion).

The only disaster I can think of here is consistency implications: if you believe that ZFC + a measurable cardinal (for example) is likely to be inconsistent, then (once you see the relevant theorem) you also believe that ZFC + projective determinacy is inconsistent. And personally I don't think there is a truly compelling consistency argument at this level; I do think these theories are consistent, but I can easily imagine being wrong. This, of course, is not an issue with choice - Godel showed that ZFC is consistent if ZF is.

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  • $\begingroup$ Well, a measurable implies lightface PD, if my memory serves me right, and in fact much more than just that. That's kind of a plausibility argument from a measurable to "$n$ Woodin cardinals" for any $n$. $\endgroup$
    – Asaf Karagila ♦
    Jul 3, 2019 at 22:29
  • $\begingroup$ @AsafKaragila I don't really see how - lightface is much weaker than boldface. And even if I granted that, we'd just be reducing to a justification for a measurable. Honestly, I don't even think there's a truly compelling reason for ZF to be consistent - again, I'm just fairly confident that it is. $\endgroup$
    – Noah Schweber
    Jul 3, 2019 at 22:30
  • $\begingroup$ I don't know how Platonism work. I don't get that whole "belief that I'm correct". But if you believe Lightface PD, then I don't see why you wouldn't believe PD itself. Yes, that reduces to $0^\#$ or $x^\#$ for every real $x$, if I remember correctly. But if you really think about that, believing those already justifies a measurable. I mean, embeddings, man. Embeddings!!! $\endgroup$
    – Asaf Karagila ♦
    Jul 3, 2019 at 22:33
  • $\begingroup$ @NoahSchweber Thank you for your answer, very interesting; I wish to think about this. $\endgroup$
    – Wojciech Aleksander Wołoszyn
    Jul 3, 2019 at 22:45
  • $\begingroup$ @Asaf Your first comment is off. The consistency strength of lightface $\Delta^1_2$ determinacy is one Woodin cardinal. And one Woodin cardinal does not imply this amount of determinacy, you obtain it in an appropriate forcing extension. $\endgroup$
    – Andrés E. Caicedo
    Jul 5, 2019 at 13:31

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