Timeline for answer to Reflection principle vs universes by Rodrigo Freire
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| Feb 3, 2021 at 1:58 | comment | added | Asaf Karagila♦ | @Andreas: Yes, there's a lot of subtle points here. In general, though, I think that most people will agree that the existence of a worldly cardinal is a large cardinal axiom, which is $V_\alpha$ satisfying ZFC. Of course, in the case of a worldly cardinal we are talking about the internal version of ZFC. But if the internal and external versions agree... | |
| Feb 3, 2021 at 1:32 | comment | added | Andreas Blass | @AsafKaragila Some of your comments seem to equate "large cardinal assumption" with "anything that provides standard models of ZFC" or "anything that provides $V_\alpha$'s satisfying ZFC". One of my favorite metatheories is ZFC plus a satisfaction predicate $S$ for $\in$-formulas, subject to the usual clauses characterizing satisfaction plus replacement for formulas involving $S$. This gives a club of $\alpha$'s such that $V_\alpha\models$ ZFC, but I wouldn't call it a large cardinal axiom. | |
| Jan 30, 2021 at 22:05 | comment | added | Mike Shulman | @TimCampion Yes, that's absolutely true: generality is an important consideration too. | |
| Jan 30, 2021 at 16:03 | comment | added | Tim Campion | @MikeShulman I mostly agree with your last comment. To nitpick: there's a difference between caring about strength of hypotheses versus caring about consistency strength of hypotheses. For a silly example, I believe that $ZFC$ and $ZFC+\neg Con(ZFC)$ are equiconsistent, but you don't need to think that ZFC "is the real world" to prefer theorems of the former to the latter. In the present case, you might prefer ZFC theorems to ZFC+Universes theorems for portability: to interpret the former, you just need a model of ZFC, which can be constructed with "less overhead" than a ZFC+Universes model. | |
| Jan 30, 2021 at 5:44 | history | edited | Rodrigo Freire | CC BY-SA 4.0 |
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| Jan 30, 2021 at 1:46 | comment | added | Mike Shulman | The main reason that I see for someone to be asking a question like this is not that they believe ZFC is the "real world" but don't believe that large cardinals are "real", but rather that they don't want to assume stronger hypotheses than necessary, i.e. that what they care about is not truth but consistency strength. For that purpose, the conservativity of ZFC/S over ZFC means we can feel free to use it without increasing the consistency strength, but also without worrying in practice about relating it back to ZFC at all or even talking about any models. | |
| Jan 30, 2021 at 1:44 | comment | added | Mike Shulman | Anyway, the problem of ill-foundedness only arises if you (1) believe that the "real world" is ZFC and (2) want to use ZFC/S and (3) want to think about a model of ZFC/S rather than just its syntax. My guess would be that if there really were a shift to ZFC/S, then the average mathematician would just start to think of ZFC/S as the "real world" instead, in which case they wouldn't see any problem. | |
| Jan 30, 2021 at 1:41 | comment | added | Mike Shulman | @AsafKaragila Well, someone who's just interested in "proving theorems" wouldn't be asking this question to begin with, would they? They'd just assume some Grothendieck universes and get on with it. | |
| Jan 29, 2021 at 22:32 | comment | added | Peter Scholze | I don't know about the HoTT community, but the number theory community (who is now using HTT) has a (bad?) habit of caring about ZFC. But they often appreciate cheats, so I'm sure they'd appreciate such a cheaty way of knowing that the main theorems of HTT are theorems of ZFC. | |
| Jan 29, 2021 at 22:24 | comment | added | Asaf Karagila♦ | @Mike: I always thought that people outside of set theory (especially those espousing different foundations, e.g HTT) are not interested in proving "theorems in ZFC", but rather just proving theorems. To that end, I am a bit worried if there is a shift to ZFC/S that even more of those silly "set theory is a bad foundations because it proves junk theorems" arguments will be raised, and this time they will be "more well-founded" about the ill-foundedness of the universe (pun not intended, but a happy accident). | |
| Jan 29, 2021 at 20:17 | comment | added | Mike Shulman | @AsafKaragila Okay, I think I finally get it; thanks for your patience. I think my attitude toward this would be the same as (what I took to be) Rodrigo's though: the goal is to prove theorems of ZFC, for which purpose we can just work syntactically with ZFC/S and then apply conservativity, rather than worry about any actual models of ZFC/S. | |
| Jan 29, 2021 at 19:06 | comment | added | Asaf Karagila♦ | @Mike: That is correct. I didn't claim that all models of ZFC/S are non-standard, of course. But if there is a well-founded model, then it had to involve large cardinal axioms, which is what we are trying to avoid. (Note that Rodrigo's answer formulates the model as a $V_\alpha$, rather than any transitive model.) For example, assuming $0^\#$ exists, let $\kappa=\omega_1$ in $V$, then $L_\kappa=V_\kappa^L$ is an elementary submodel of $L$, witnessing ZFC/S as a standard model. | |
| Jan 29, 2021 at 19:03 | comment | added | Mike Shulman | But it seems to me that there could still be some models of ZFC/S that do satisfy Con(ZFC) (at least, if Con(ZFC) is true). Were you claiming that all models of ZFC/S must contain nonstandard integers? | |
| Jan 29, 2021 at 19:01 | comment | added | Mike Shulman | @AsafKaragila Okay, maybe I almost get it. Say from a model $M$ of ZFC we build a model $M'$ of ZFC/S. If in $M'$ it were true that $V_\alpha \vDash$ ZFC, then we would have $M' \vDash$ Con(ZFC), and we can't expect to construct from any model of ZFC a model of ZFC+Con(ZFC). But otherwise, in $M'$ there must be an axiom of ZFC that $V_\alpha$ doesn't satisfy, which must be a non-standard axiom since $V_\alpha$ satisfies all the standard axioms. Right? | |
| Jan 29, 2021 at 18:19 | comment | added | Asaf Karagila♦ | @Mike: In the ZFC/S universe, $V_\alpha$ is still a set, it still has a truth predicate, and it still provides us the consistency of whatever theory is true there. So to avoid having proved Con(ZFC) internally to that universe the only solution is that internal-ZFC now have new axioms, i.e. non-standard integers, relative to its meta-theory. | |
| Jan 29, 2021 at 18:07 | comment | added | Mike Shulman | @AsafKaragila Sorry, I'm lost. The proof of relative consistency doesn't go by constructing a model. I thought the whole point was that you can't prove internally in ZFC/S that $V_\alpha$ is a model of ZFC since it only satisfies all the axioms as a schema. What does that have to do with standardness of integers? (And can you please talk about "integers in model X" so that I don't get confused about which model you're calling "standard"?) | |
| Jan 29, 2021 at 18:01 | comment | added | Asaf Karagila♦ | @Mike: If the integers are standard, then the model you have is a model of ZFC, internally, so that means that ZFC is consistent. Since the extension is conservative, it can't prove something like that. So if you didn't happen to already have this model in your ZFC universe to begin with, a universe of ZFC/S is going to be non-standard. | |
| Jan 29, 2021 at 15:59 | comment | added | Mike Shulman | @AsafKaragila So in this particular situation, which universe are you calling "the" universe and which are you calling the "meta-universe"? Are you saying that when you prove in ZFC the relative consistency of ZFC/S, you can also prove in ZFC that any model of ZFC/S will contain integers that are nonstandard relative to the ambient ZFC? | |
| Jan 29, 2021 at 15:03 | comment | added | Asaf Karagila♦ | @Mike: Yes, but the universe has its own meta-theory and meta-universe. It's turtles all the way down. | |
| Jan 29, 2021 at 14:50 | comment | added | Mike Shulman | @AsafKaragila Can you say exactly what you mean by "the universe" and "non-standard"? Under a naive reading I would think that "the universe has non-standard integers" would be a contradiction in terms, since the standard integers should by definition be those that exist in the universe. | |
| Jan 29, 2021 at 11:41 | comment | added | Rodrigo Freire | Thanks @AsafKaragila, I am big fan of Levy, it is good to hear this. I think of those conservative extensions as technical devices which are useful to prove ZFC theorems. These $\alpha$ and $\beta$ can be thought of as Hilbertian ideal objects (differently from inaccessibles) because the extension is conservative. It means that we may not commit ourselves with a strong notion of truth of these extensions. | |
| Jan 29, 2021 at 11:22 | comment | added | Asaf Karagila♦ | Rodrigo, @Mike: While I don't have the unfettered access to Azriel Levy that I had as his TA, I know that (1) he was a good friend of Feferman, and the two invariably influenced each other and shared ideas; and (2) he was extremely modest and hated the whole idea of getting credit for old ideas, when I'd ask him for who did what back in the early days of forcing he would usually say that it doesn't matter and who can remember anyway (although that's often a sign that it was him). My point is that Azriel would be okay with this being credited to Feferman, regardless to who came up with it. | |
| Jan 29, 2021 at 11:08 | comment | added | Asaf Karagila♦ | These conservative extensions do pose an ontological problem, though, since without large cardinal axioms they invariably mean that the universe has non-standard integers. Now this is not a problem per se, but it does mean that our "naive understanding of the integers" is incorrect as far as "the mathematical universe" is concerned. This is not necessarily a bad thing, of course, but it is something to contend when thinking about that. | |
| Jan 28, 2021 at 17:19 | history | edited | Gro-Tsen | CC BY-SA 4.0 |
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| Jan 28, 2021 at 17:04 | comment | added | Rodrigo Freire | I think this strategy goes back to Levy, Definability in axiomatic set theory I, 1965, but Levy adds a constant M and axioms saying that M is transitive, countable and reflects every sentence. I have not seen your comment, sorry. | |
| Jan 28, 2021 at 16:57 | comment | added | Mike Shulman | Isn't this what Feferman does in the paper “Set-theoretical foundations of category theory" that I mentioned in a comment? | |
| Jan 28, 2021 at 15:39 | comment | added | Rodrigo Freire | Yes, for each $n$ fixed in the metatheory, $V_{\alpha}$ is a $\Sigma_n$-universe. | |
| Jan 28, 2021 at 15:27 | comment | added | Peter Scholze | OK, great! So basically this is a version of the $\Sigma_{15}$-universes mentioned above, but tweaked so that the precise value of $15$ has become irrelevant? | |
| Jan 28, 2021 at 15:23 | comment | added | Rodrigo Freire | Yes, $\alpha$ may have countable cofinality and your explanation is right. $V_{\alpha}$ satisfies each instance of (first-order) replacement only: For each definable function $f$ from $V_{\alpha}$ to $V_{\alpha}$, the direct image of an element of $V_{\alpha}$ is also an element of $V_{\alpha}$. If it were closed under direct images in general, then $\alpha$ would be strongly inaccessible. | |
| Jan 28, 2021 at 15:09 | comment | added | Peter Scholze | Thanks for this proposal! I have to understand better what this means concretely. How is it avoided that $\alpha$ is strongly inaccessible? Is it again that when one takes unions of sets in $V_\alpha$, one has to make sure that everything is definable in $V_\alpha$? So $\alpha$ may have countable cofinality in $V$, because there are certain countable sequences of ordinals $<\alpha$ converging to $\alpha$, they are just not definable in $V_\alpha$? | |
| Jan 28, 2021 at 14:28 | history | edited | Rodrigo Freire | CC BY-SA 4.0 |
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| Jan 28, 2021 at 13:59 | history | answered | Rodrigo Freire | CC BY-SA 4.0 |