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I saw this post by Carl Mummert, which describes how one can 'see' the reasonableness of ZFC, via iterating the power-set operation starting from the empty set. However, this iterations proceed in stages where each stage is from some well-founded collection. We can intuitively grasp finitely many iterations, and given our intuitive assumption of natural numbers we can also grasp their union, and we can repeat this through all the computable ordinals, but I can't see how one can justify that anything beyond that is reasonable besides simply appealing to the ordinals already existing, which would be circular if we base our understanding of ordinals on ZFC.

Specifically, to reach the first uncomputable ordinal we have to have all the computable ordinals, but these are in a sense not all there in the above 'construction', because if our goal is to justify the cumulative hierarchy we can't talk about any uncomputable ordinal since we have not constructed them yet, nor can we make use of the collection of all computable transitive well-orderings because it is merely a concept. Sorry if this is quite vague, but I don't at the moment know of a clearer way to express this meta-theoretic concept.

Also, if ZFC has a (countable) transitive model, then there already is a countable ordinal that is a model of ZFC, whose existence therefore ZFC alone cannot prove. But I think this is far beyond the first uncomputable ordinal, though it seems to me that it prevents non-circular justification of the cumulative hierarchy beyond it, since we don't even 'know' whether it exists without assuming more than ZFC.

So my main question is:

Is there some meta-system that isn't stronger than ZFC (but might be incomparable) but can construct the cumulative hierarchy and hence provide some kind of 'justification' for ZFC?

If not, what is the furthest we can go (or know how to) up the hierarchy?

I apologize that I don't even know precisely what meta-system can allow my description of the computable ordinals to go through.

Finally, the post by François G. Dorais that says that we cannot obtain the first uncountable ordinal without the use of the power-set axiom, which would present a severe obstacle if we do not have the power-set operation (for infinite sets). To me the full power-set is also conceptually dubious, but in the above discussion we are already taking for granted the power-set operation. Nevertheless this raises the question:

Does the answer to my question change if at each successor stage we merely add all definable subsets from the previous stage?

Does my question make sense? One motivation was that the consistency of PA can be proven in PRA plus quantifier-free transfinite induction up to $ε_0$, which is incomparable in strength to PA.

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user21820
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  • I also saw http://mathoverflow.net/q/235896 but don't know how relevant it is. – user21820 May 13 '16 at 10:56
  • Related: http://math.stackexchange.com/questions/370240/does-mathsfzfc-neg-mathrmcon-mathsfzfc-suffice-as-a-foundations-of – Asaf Karagila May 13 '16 at 11:02
  • @AsafKaragila: Haha! At least in that case we could argue that it is silly because ZFC proves that PA is ω-consistent but ( ZFC + not Con(ZFC) ) does not have an ω-model... – user21820 May 13 '16 at 11:04
  • I have to admit that I'm not entirely sure what sort of answer you expect. Your question is $\Sigma_1$, and assuming that ZFC is consistent your question is not $\Delta_1$. – Asaf Karagila May 13 '16 at 11:08
  • @AsafKaragila: Well, I was wondering if there is something like Gentzen's consistency result, but for ZFC instead of PA. Gentzen himself thought that his proof was useful in not relying on the full induction schema of PA, thus providing 'independent' justification of PA. Since I'm asking for similar justification of ZFC, I don't see how my question can be expressed as a purely mathematical statement either. After all, some people disagree with Gentzen, so an answer to my question would be equally subjective. – user21820 May 13 '16 at 11:40
  • I had actually asked someone that same question recently over email. Perhaps because of their answer I felt a bit at odds with this question here... – Asaf Karagila May 13 '16 at 11:41
  • @AsafKaragila: Kleene said "To what extent the Gentzen proof can be accepted as securing classical number theory in the sense of that problem formulation is in the present state of affairs a matter for individual judgement, depending on how ready one is to accept induction up to ε0 as a finitary method." (according to wikipedia). For my question using the (standard) ordinals from ZFC is of course not acceptable in justifying the cumulative hierarchy unless we first have some 'independent' justification for the ordinals. – user21820 May 13 '16 at 11:43
  • @AsafKaragila: Ohh what was their answer, do they and you mind sharing? =) – user21820 May 13 '16 at 11:43
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    My question, in a nutshell, was if there is an ordinal $\alpha$ that a Gentzen-style argument with induction up to $\alpha$ can prove Con(ZFC). The answer was that we don't know, and the Gentzen-style arguments are a byproduct of the proof, and not the other way around. It is not clear to me, even if such Gentzen-style proof exists, that it is enough to "construct the von Neumann hierarchy" in a meaningful way. Namely, we construct the $V_\alpha$'s internally to ZF, regardless of the meta-theory (as long as it's "reasonable"). Your modified question yield V=L, so it shouldn't matter. – Asaf Karagila May 13 '16 at 14:23
  • @AsafKaragila: Ah I see. Do you know of any known result in any other direction, namely not restricted to Gentzen-style arguments? I just saw in the wikipedia article on NF that NFU plus Rosser's Axiom of Counting reaches $\beth_{\beth_n}$ for every finite $n$. I don't know how much that counts, since it seemed to me the purpose of stratification in NFU was to have implicit type hierarchy and Counting slightly violates that. – user21820 May 13 '16 at 16:53
  • I'm far out of my usual slice of ocean here. So no, I can't be of any further assistance, sorry. :-) – Asaf Karagila May 14 '16 at 08:59
  • Why is $\textrm{PRA}+\textrm{TI}(\varepsilon_0)$ considered incomparable with $\textrm{PA}$? $\textrm{PRA}+\textrm{TI}(\varepsilon_0)$ has a greater proof-theoretic ordinal using the $\vert, .,\vert_{Con}$ formulation: https://mathoverflow.net/a/420467 – C7X Apr 29 '22 at 14:58
  • @C7X: Incomparable in the philosophical sense, because PRA+TI(ε0) has very little inductive strength. – user21820 Apr 29 '22 at 15:03

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I think the informal argument that the cumulative hierarchy is a model of ZFC is meant to appeal to some kind of direct perception of the cumulative hierarchy, much like the informal argument that $\mathbb{N}$ is a model of PA relies on some kind of direct perception of $\mathbb{N}$. I don't expect the informal argument to be formalizable in weak theories that can only work with computable ordinals. Perhaps it could be formalized in some second-order set theory - the sort of "direct perception" seems to be related to second-order systems in my experience. But that would likely seem to be stronger than ZFC from a formal perspective.

Gentzen-style arguments are often called "ordinal analysis" or "computing the proof theoretic ordinal" these days. A nice survey is "The art of ordinal analysis" by Rathjen. The farthest progress has been that the proof theoretic ordinals of Kripke-Platek set theory and of $\Pi^1_2$ comprehension (a subsystem of second-order arithmetic) have been computed. There is still an enormous "gap", so to speak, between these theories and ZFC, so I don't think anyone is close to an ordinal analysis of ZFC (presuming that one is possible).

Carl Mummert
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  • The problem I have with the concept of "direct perception of the cumulative hierarchy" is that there really is no such thing except after one knows about and works in ZFC for a while. This is in contrast to the natural numbers, where repetitions of a certain fixed operation (of any sort) in the real-world form a 'directly perceptible' structure that seems to satisfy PA. We can therefore say, "look this real-world structure seems to show that PA is consistent, at least up to the limits we have tried so far". I see no such thing for the cumulative hierarchy beyond the computable ordinals. – user21820 May 19 '16 at 13:08
  • Yes, the informal argument may not be convincing to everyone, and perhaps not completely convincing to anyone. It is just a particular argument that has been of interest in the foundations of set theory. – Carl Mummert May 19 '16 at 13:33