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Apr 13, 2017 at 12:19 history edited CommunityBot
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Nov 16, 2016 at 22:27 vote accept Damian
Nov 1, 2016 at 16:24 comment added Nik Weaver $\Sigma_1$ constructions generally are not seen as predicative, for obvious reasons. $\Delta_0$ replacement would certainly be accepted. But if you adopt intuitionistic semantics, for any $\phi$, having a proof that $(\forall x \in A)(\exists!y)\phi(x,y)$ would entail knowing how to construct such a $y$ for every $x \in A$, and this would allow the predicative construction of the set of all $y$'s, assuming $A$ was predicatively constructible.
Nov 1, 2016 at 16:02 comment added Asaf Karagila I would imagine that perhaps some limited form of Replacement is considered predicative. Maybe something like $\Sigma_1$?
Nov 1, 2016 at 14:54 comment added Nik Weaver Powerset is certainly not considered predicatively acceptable, but I don't recall ever seeing any doubts about replacement expressed in the predicativist literature.
Oct 31, 2016 at 19:57 history edited Noah Schweber CC BY-SA 3.0
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Oct 31, 2016 at 19:50 history edited Noah Schweber CC BY-SA 3.0
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Oct 31, 2016 at 19:44 history edited Noah Schweber CC BY-SA 3.0
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Oct 31, 2016 at 0:40 comment added Noah Schweber On the philosophical side of things, I believe that neither Replacement nor Powerset are usually viewed as predicatively acceptable. (I could be wrong.) But to address the question "Is there a predicative proof that $\omega_1$ exists?", I would need a precise definition of predicativity. (In particular, I don't think getting rid of Replacement entirely is a good idea - without Replacement, we can't even prove that the ordinal $\omega+\omega$ exists! So it's really a sensitive question, axiomatically speaking.)
Oct 31, 2016 at 0:35 history edited Noah Schweber CC BY-SA 3.0
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Oct 31, 2016 at 0:26 history answered Noah Schweber CC BY-SA 3.0