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Apr 18, 2018 at 21:32 comment added Monroe Eskew You couldn’t respond adequately.
Apr 18, 2018 at 20:03 comment added Mike Shulman This conversation doesn't seem to be going anywhere (we're each repeating ourselves), so let's give it up.
Apr 17, 2018 at 17:28 comment added Timothy Chow I agree with Monroe Eskew. Suppose I even grant Mike Shulman that the fact the two models have the same ordinals is not a "counterexample to the claim that a measurable cardinal is an induction principle." Fine. But on what grounds are you claiming that the existence of a measurable cardinal is an induction principle? It certainly doesn't look like an induction principle upon superficial examination. What makes something an induction principle? Can you give an example of an axiom that is definitely not an induction principle, and if so, why is it not an induction principle?
Apr 17, 2018 at 4:49 comment added Monroe Eskew I’m talking about a situation in which the ordinals between the two models are identitical (although isomorphic will do just fine). This means that they do inducitive constructions of exactly the same lengths. Furthermore, all the objects they have in common which are capable of some kind of inductive analysis in one model are this way in the other model. (This is the absoluteness theorem I mentioned.) Now, what do you mean by “inductive strength”?
Apr 16, 2018 at 23:03 comment added Mike Shulman Similarly, even though there is a bijection between the elements of M that M calls "ordinals" and the elements of N that N calls "ordinals", there's no reason to believe that a pair of elements that correspond under that bijection necessarily represent "the same level of inductive strength" inside of the models M and N.
Apr 16, 2018 at 23:03 comment added Mike Shulman M and N are two different models of set theory. What does it mean to say that they have "the same" levels? If I have two models $G,H$ of the theory of a group, even if there happens to be a "canonical" embedding of one into the other, an element of $G$ might have different properties qua element of $G$ than qua element of $H$.
Apr 16, 2018 at 21:40 comment added Monroe Eskew I have no idea what you mean by "internal ideas of well-foundedness." They have the same ordinals and so can do induction to exactly the same levels. What are you getting at?
Apr 16, 2018 at 19:17 comment added Mike Shulman In fact, it could just as easily go the other way: the fact that measurability is not preserved in inner models could be taken to imply that "having the same ordinals" in this external sense is not a characterization of "having the same notion of well-foundedness".
Apr 16, 2018 at 19:17 comment added Mike Shulman Yes, I know that; I'm objecting to the claim that that means that "M and N have the same notion of well-foundedness." Or, I suppose, if that's your definition of "the same notion of well-foundedness", then I don't see why it is any sort of counterexample to the claim that a measurable cardinal is an induction principle. Just because it so happens that M and N, as sets in some external metatheory, happen to agree about which elements of that metatheory are well-founded, doesn't mean that M and N have the same internal "ideas" about what well-foundedness means or how much induction is valid.
Apr 16, 2018 at 18:16 comment added Monroe Eskew Ok well this is quite well known. If M is a submodel of N with the same ordinals, both satisfying ZF, and R is a relation in M, then M and N agree on whether R is well-founded. (To be fair, this is an important and nontrivial absoluteness theorem.)
Apr 16, 2018 at 18:14 comment added Mike Shulman It's not clear to me that just because the ordinals in a model of ZF happen to be (externally to that model) isomorphic to the ordinals in some other model, it follows that the two models have the "same notion of well-foundedness" in whatever sense is relevant, since whatever that means it must be an internal notion. I don't even know what it would mean to say that two different models have "the same notion of well-foundedness".
Apr 16, 2018 at 16:59 comment added Monroe Eskew It's not at all clear to me that the existence of a measurable cardinal is an "induction principle." Measurable cardinals are not measurable in all inner models (classes satisfying ZF with the same ordinals and thus the same notion of well-foundedness), so this seems like a counterexample to the claim that they are some kind of "induction principle."
Apr 16, 2018 at 16:54 comment added Mike Shulman why does the fact that "if the universe is X high then it must also be Y wide" mean that we can't explain axioms about the height of the universe in terms of their inductive power? Obviously with a stronger induction principle you can prove more, and I don't see why statements about the "width" of the universe (whatever that means) should be considered any different from other consequences of such an induction principle that you can't prove without it.
Apr 16, 2018 at 13:36 comment added Monroe Eskew I don't think this idea explains large cardinals up to a certain level, like measurable cardinals. These have implications for the "width" of the universe, not merely "height," i.e. how far up induction takes you.
Apr 16, 2018 at 10:33 comment added Mike Shulman I think something can be said in the direction of why large cardinals measure consistency. Namely, essentially the only mathematical tool we have to prove things about infinite objects in finite language is induction, and so the strength of a theory consists in how many "inductive arguments" it asserts are well-founded. And as for the linear ordering, at least in classical logic any well-founded relation is bisimulable to a (linear) well-ordering. (Accordingly, in intuitionistic logic, I see no reason to expect "large cardinal axioms" to be analogously linearly ordered.)
Apr 16, 2018 at 9:54 history answered Monroe Eskew CC BY-SA 3.0