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A model in laymen's terms is a mathematical structure (sets when within set theory) that satisfies a theory (or alternatively a set of axioms). For example the model Vω is a model of a theory of purely finite sets, sets like the set of all natural numbers N cannot be exist in this model (In what sense are inaccessible cardinals inaccessible? - second answer). What if we took the universe of sets VΩ (defined as the union of all of the cumulative hierarchy V1 ⋃ V2...) as a model for a theory (e.g. ZFC)? What are the implications of this? What would this theory be like? Is this even possible? I ask this because it appears we would have a theory that proves ALL sets (including inaccessible and large cardinals as from a Platonist point of view they are sets). I have only recently begun researching 'models' and 'consistency' so the entire idea may introduce several problems and paradoxes I am unaware of (please feel free to inform me if so).

As for clarification, since a theory is based off it's axioms, the theory ZFC (or a theory which has the cumulative hierarchy) will do for this example.

Asaf Karagila
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SI J
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  • You cannot define "$V_\Omega$" in a model of set theory, since the class of all ordinals is a proper class. So your question doesn't really make sense from the start. – Clement Yung Sep 10 '22 at 18:58
  • @ClementYung So does this mean a proper - class cannot be taken as an ordinal? Also VΩ is the proper class of sets not ordinals (it's instead indexed by the proper class of ordinal ORD) – SI J Sep 11 '22 at 01:12
  • What does it mean to "take something as a model for a theory (e.g. ZFC)"? – spaceisdarkgreen Sep 11 '22 at 01:29
  • @spaceisdarkgreen What if a theory (with ZFC as an example) had VΩ as a model? – SI J Sep 11 '22 at 01:32
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    $V_{\Omega}$ is a model of ZFC... (modulo the standard caveats about it being a proper class) – spaceisdarkgreen Sep 11 '22 at 01:35
  • If it is why? As the comment above disagrees that the class of all sets is a model of ZFC – SI J Sep 11 '22 at 01:36
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    The whole premise of the standard development of set theory is that the universe of sets satisfies the ZFC axioms. It's the foundational assumption of the subject. The complaint in the above comment is that the universe is a proper class (and there are real big caveats here, as I alluded to), but inasmuch as it makes sense to say a proper class is a model of a theory, the set theoretical universe is a model of ZFC. – spaceisdarkgreen Sep 11 '22 at 01:45
  • You can also profitably assume fewer or more axioms for your universe, depending on your goals. – spaceisdarkgreen Sep 11 '22 at 01:50
  • @spaceisdarkgreen My understanding is that the 'universe' of a theory was some segment of V. As for the case of ZFC if we took VΩ as a model doesn't that mean inaccessible or even large cardinals are provable (and even apart of the ZFC universe) in ZFC, which they are meant to be inconsistent too. – SI J Sep 11 '22 at 01:51
  • This is why I asked you "What does it mean to "take something as a model for a theory (e.g. ZFC)"?" in my first comment. You didn't really answer and we went on a tangent, and now it's back and it's clear now this is a source of misconception. – spaceisdarkgreen Sep 11 '22 at 01:55
  • I apologize for the misconception. I just thought It was obvious what it meant to take a model as theory. I included the example of finite set set theory (with it's model being Vω as it sets are purely finite) in order to state that perhaps VΩ is a similar case. – SI J Sep 11 '22 at 02:00
  • It sounds like you're asking to connect to a canonical axiomatizable theory to a given model. There's perhaps a loose sense in which ZFC-Inf (or probably more convincingly ZFC - Inf + $\lnot$ Inf) is the canonical theory associated to $V_\omega$, ( $V_\kappa$ for the least inaccessible $\kappa$ amd "ZFC + no innaccessibles" correspond similarly) but it's not really precise (unless we use a second-order axiomitization). In the case of the whole universe, it's much less clear what that theory should look like, but that doesn't stop people from trying... "V = Ultimate L" comes to mind. – spaceisdarkgreen Sep 11 '22 at 02:38
  • Thank you for your take on this question. I also did notice that the implications of this would be somewhat connected to V = Ultimate L. As previously stated before I am only a beginner at model theory so there may be gaps in my knowledge. – SI J Sep 11 '22 at 04:22
  • @spaceisdarkgreen Also is the universe for ZFC a segment of V (Vk for 'standard' ZFC) or V. What is the difference? – SI J Sep 22 '22 at 10:08
  • @SIJ what is “standard” zfc? Again, impossible to precisely answer a question without a precise meaning (“the universe for ZFC”, as opposed to “the universe for for some other axiomatization for set theory” is not a well-defined concept.) – spaceisdarkgreen Sep 22 '22 at 13:45
  • @spaceisdarkgreen Zermerlo-Frankel set theory with axiom of choice. Also it has no add-ons (e.g. + there is an inaccessible cardinal or - infinity). Pretty much the ZFC universe would be equivalent to Vk (a segment of V. It can only construct sets with powersets, so inaccessible cardinals would not be included). I am confused as this seems to be the universe for ZFC but people also say V (which includes all sets) is the universe for ZFC. I want to know the relation between these two 'universes' as they both appear to be a universe for ZFC. – SI J Sep 22 '22 at 23:48
  • @SIJ This whole thing is premised on a one-to-one relationship between models and axiomatizable theories that just doesn't exist. It's normal to talk about "the universe", but people don't say "the universe for ZFC"... it's not a thing. That said, I'd note that not including an axiom that says "there are inaccessibles" is much different than including an axiom $\lnot I$ that says "there are no inaccessibles". As I mentioned above, one could, plausibly, explaining clearly what they meant, argue that $V_\kappa$ for least inaccessible $\kappa$ is a canonical model for $ZFC+\lnot I$.. – spaceisdarkgreen Sep 23 '22 at 01:55
  • @SIJ One way of making this precise is the idea of quasi-categoricity, which is the observation that certain second-order set theories have a unique models up to isomorphism. For instance, ZFC$2$ (the natural second-order version of ZFC) + $\lnot I$ has $V\kappa$ for the least inaccessible $\kappa$ as a unique model. First-order theories never have anything like this due to Lowenheim-Skolem. But it's worse than that: even just considering initial segments, there are many $\alpha$ less than the least inaccessible such that $V_\alpha$ satisfies ZFC (cf "worldly cardinals"). – spaceisdarkgreen Sep 23 '22 at 02:13
  • @spaceisdarkgreen So from what I've gathered (specifically from your one to one relationship comment). V is the universe of ZFC but models of ZFC can only prove a portion of the universe (up to Va), depending on the model. In short both V and the model satisfy the ZFC but only a certain part of the universe can be proven in a theory (the model) even though V is satisfied. Is this correct? – SI J Sep 23 '22 at 06:47
  • @SIJ No, none of that makes sense. You are repeatedly conflating models and theories and continue to use the phrase “universe of ZFC” as if it means something. You need to learn the basic taxonomy of mathematical logic before you can communicate well on matters like this. – spaceisdarkgreen Sep 23 '22 at 15:29
  • @spaceisdarkgreen My apologies, that's just what I've gathered from your comment. I'll phrase my question in another way "why can ZFC prove sets up to Vk, when V is a model of ZFC?" V the universe, is a model of ZFC. Meaning, doesn't it prove the existence every set? yet ZFC is closed under Vk and can't prove further (e.g. Vk+1...). That's why I gathered something like my comment above. I guess the problem is around the difference between what it means to be a model, and what it means to be closed under something. – SI J Sep 23 '22 at 23:08
  • @SIJ "ZFC proves sets up to Vk", "V proves the existence of every set", "ZFC is closed under Vk and can't prove further". None of these three notions makes sense. 1) ZFC doesn't prove sets, it proves statements. Even if the statements are of the form "a unique set with property P exists", in which case you might say ZFC has proved a certain set exists, there are only countably many statements... certainly not enough to provide a proof for every set in Vk. 2) V doesn't prove things. 3) ZFC a a theory... I'd say it makes no sense for it to be "closed under Vk", if that phrase made any sense. – spaceisdarkgreen Sep 24 '22 at 00:27
  • @SIJ I think a straightened-out version of what you're trying to ask is "ZFC doesn't prove there are inaccessibles, and yet there are inaccessibles in V, so how can V satisfy ZFC?" The answer is that ZFC doesn't prove there aren't any inaccessibles either. Whether there are any inaccessibles or not is undecided by ZFC. – spaceisdarkgreen Sep 24 '22 at 00:35
  • @spaceisdarkgreen Ok, thank you for confirming. That's kind of like what my above comment. That although V satisfies ZFC, Inaccessible cannot be proven nor disproven, almost like they are 'invisible' to ZFC. Sorry for the wrong terminology, I meant that ZFC can only 'construct' sets up to Vk. – SI J Sep 24 '22 at 00:57
  • @SIJ Well.. that last part is not really accurate either, as I indicated. But anyway, note also that the same phenomenon could be said about Vk... e.g. Vk satisfies "ZFC is consistent" and "there are worldly cardinials", but neither of these sentences is provable in ZFC. – spaceisdarkgreen Sep 24 '22 at 01:08
  • @spaceisdarkgreen The statement can't be proven or disproven. But can't Vk be proven via powerset construction. Or does it require us to assume that an inaccessible cardinal exists (k implies Vk). – SI J Sep 24 '22 at 02:03
  • @SIJ Yeah, you get $V_\kappa$ by iterating the power set operation $\kappa$ times (and using replacement and union at limit stages to take the union)... that is a definition in terms of $\kappa$. – spaceisdarkgreen Sep 24 '22 at 02:25
  • @spaceisdarkgreen It's important to note the difference between V and VΩ. V is the cumulative hierarchy (a hierarchy that cumulates and increases overtime) while VΩ is the union of every segment of V, making it the 'class' of all sets. In a comment above you stated that axiomatizable theory's cannot have a one - to - one relation to models. Does this mean you cannot take VΩ (not V) as a model. – SI J Sep 24 '22 at 02:42
  • @SIJ When people say "$V$ is the union of $V_\alpha$ for all ordinals $\alpha$", what they mean is that a set is in $V$ iff for some ordinal $\alpha$, it's in $V_\alpha$. ZFC proves that every set is a member of some $V_\alpha$, which is why if the author and the reader agree to use ZFC to work out their claims, $V$ is described as "the universe of all sets". If you believe $V_\Omega$ is a Platonic world of all sets which exists independently of human interaction, and you also believe it satisfies ZFC, then $V_\Omega$ is the same thing as $V$. – C7X Sep 25 '22 at 04:36
  • @C7X So then I assume from a constructible view (V isn't completed but rather constructed), V would be different from VΩ. From a constructible how would having V being a model look like (if this can happen) – SI J Sep 26 '22 at 09:59
  • @SIJ In both views "V is a model of ZFC" means "V satisfies each axiom of ZFC". I think this constructible view is called set-theoretic potentialism (not sure about the name), I'll call it this here. Assuming such a view and that regularity axiom holds, the construction of V is carried out along the ordinals, and manages to encompass all the sets there are. With some abuse of concept, this would look like "Ord is a worldly cardinal" to a potentialist: the ordinals go high enough for our construction of V to model replacement, even for formulae claiming e.g, powersets iterated many times exist. – C7X Sep 27 '22 at 02:41
  • @C7X You mentioned in another comment that it's hard to connect V to ZFC (as a model of ZFC). Is this in-part because an implication of this would prove that 'all' sets exists (including inaccessible and large cardinals)? Causing a case similar to V = ultimate L (which comes with it's own set of problems) which spaceisdarkgreen pointed out. – SI J Sep 30 '22 at 07:31
  • @SIJ By "connect V to ZFC (as a model of ZFC)" I think you mean associate ZFC as the canonical or "obvious" theory that axiomatizes how the members of V behave? (I.e. "ZFC is the canonical set theory.") You can't formalize "x set exists", so you can't nontrivially say "all sets exist", I think what you mean is "if V models ZFC, why doesn't ZFC prove all true statements about V"? There are two reasons why we can say the answer is no: (1) by assuming V models ZFC we assumed ZFC is sound. Then since ZFC is recursive the incompleteness theorem applies to all models of ZFC, such as V. – C7X Sep 30 '22 at 11:31
  • (contd, 2) This one is even more direct than incompleteness. Like spaceisdarkgreen said, there are uncountably many sets in V, and only countably many formulae. Even if we skip over the "true statements about V" fix I proposed and imagine there was some nontrivial way of formalizing the "x exists" that you want, this would have to be a formula without parameters, so there can only be countably many formulae of that form. But V is uncountably large – C7X Sep 30 '22 at 11:34
  • @C7X Since there is not enough formulae for every set in V ('there are only countably many formulae, but uncountably many sets in V'), this causes there to be 'unprovable' sets (in the sense that there is no formulae to define the set or prove the set's existence). This is quite similar to being 'invisible' to a theory and it's models (as in the case of large cardinals) as their existence aren't confirmed. Although, ignoring this issue, would a hypothetical theory of sets which does prove the existence of all sets within it's universe look something like V = ultimate L? – SI J Oct 04 '22 at 11:05
  • @SIJ I think you're asking about undefinable sets, which do exist in an uncountable model, since there are uncountably many sets and only countably many formulae. These are the sets not defined by some formula, without reference to proof. (Again, what does "prove a set exists" mean?) If you would like "sets that have definitions where ZFC proves exactly one set satisfies that definition", these are called "ZFC-good definitions", I know some research has been done into those that are $\Sigma_1$. – C7X Oct 05 '22 at 00:29
  • @C7X 'Proving a set exists' goes hand in hand with 'defining' the set (with the use of formulae). I'm guessing that these 'undefinable' sets are 'invisible' to a countable model as they only exist (or can be defined) in uncountable models. Also how do we prove if an uncountable models satisfies a theory if there are only countably many formulae? – SI J Oct 05 '22 at 00:38
  • " 'Proving a set exists' goes hand in hand with 'defining' the set (with the use of formulae)" Why is that? There are definitions of a set which are not necessarily ZFC-good, for example if we assume Con(ZFC), there is a least $\alpha$ such that $L_\alpha\vDash ZFC$, this can be expressed as a first-order formula of $\alpha$, but ZFC can't prove there is a least $\alpha$ where this is true (i.e. it doesn't prove said formula is a "good definition"). Such $\alpha$ exists in a countable model too: just take a countable model of ZFC+"exists an inacc", this necessarily contains such an $\alpha$. – C7X Oct 05 '22 at 01:31
  • "Also how do we prove if an uncountable models satisfies a theory if there are only countably many formulae?" I don't see why this is a problem, where κ is the least inacc cardinal we can prove V_κ is a model of ZFC by showing it satisfies replacement, powerset, etc., and that's uncountable. – C7X Oct 05 '22 at 01:32
  • @SIJ Maybe you are also curious about uncountable pointwise-definable models of ZFC instead of just definability in V, there are indeed none due to there being countably many formulae. – C7X Oct 05 '22 at 01:39
  • @C7X So we don't necessarily have to define sets (using formulae) in uncountable models like V or V_k? Apologies again for using a wrong definition of 'proving a set exists'. What I actually mean is that set doesn't exist within that model, and from the perspective of that model you cannot prove it exists (as it is not a member of it and therefore 'invisible'). – SI J Oct 05 '22 at 01:55
  • @SIJ Just because formulae are unable to define all the sets doesn't mean they're a useless tool for defining some of the sets. In fact with the traditional foundations of logic, formulae are our only tool for unambiguously defining an object. (I mean this in contrast to, e.g., outlining in English what a set contains.) – C7X Oct 05 '22 at 03:57
  • @C7X Off topic but are Reinhardt (& Berkeley) cardinals apart of V (or conversely V_Ω) . Or do they transcend the cumulative hierarchy as a whole? – SI J Oct 07 '22 at 02:19
  • @SIJ Since "a Reinhardt + choice" is inconsistent, you'll need to formulate cardinality the AC-less way it's done most by non-set-theorists, which is as an equivalence class: if two sets have a bijection between them, they're in the same class. One drawback of having to do this is that all non0 cardinals (including 1!) are proper classes and therefore not in V. So yes, speaking strictly they are not in the cumulative hierarchy, but if you reject choice this isn't any sort of special property of Reinhards being "too big for V", since all cardinals have to be formulated in this different way. – C7X Oct 07 '22 at 05:20
  • @C7X Thank for clarifying. People made it sound like it is outside of the "scope" of V (implying that V has a limit which directly contradicts that 'there is no largest cardinal') and that it was 'too large' for V to contain. I'm also assuming, that V contains cardinals which are on Reinhardt cardinal level or larger, that is consistent with ZFC (which you could hypothetically define, although it would take a very experienced mathematician to do so). Reinhardt also created a theory of 'hyper-ordinals' (the universe had layers like V_Ω+1,etc) which Reinhardt cardinals were related too. – SI J Oct 07 '22 at 10:55
  • Interesting, I'd never heard of them having to do with Reinhardts, only huges. A better way to put what I meant is "Reinhardt cardinals are no more 'outside V' than any other cardinal". I'm not sure what you mean by "V contains cardinals which are on Reinhardt cardinal level or larger, that is consistent with ZFC", since consistency strength differs from actual size there's no guarantee that a stronger large cardinal property implies a literally larger cardinal. – C7X Oct 07 '22 at 20:24
  • @C7X So since V has no limit and can be extended forever forward, V would contain a large cardinal similar in size (or consistency) to a Reinhardt cardinal. This cardinal just isn't defined yet. That's how it appears to me from the platonic view, so feel free to inform me. Also when you say it's inconsistent with the axiom of choice, do you mean that axiom of choice inhibits it's construction (it's implication does not allow elementary embeddings of the universe onto itself)? Or do you mean that Reinhardt cardinals consistency is to 'strong' for the axiom of choice (implying a limit on it)? – SI J Oct 09 '22 at 09:00
  • @SIJ "V has no limit and can be extended forever forward, V would contain a large cardinal similar in size" You can argue this. "(or consistency)" As per your deleted comment, there is not a consensus that "consistency strength approximates size", in fact much research into large cardinals is into the many cases where this fails (called identity crises). Almost all resources about large cardinals online order them by consistency strength, and the tree drawings do not double as size comparisons. About ZFC+"a Reinhardt exists", Kunen proved that it is inconsistent, but I don't know the details – C7X Oct 09 '22 at 19:14
  • The problem with formulating "a Reinhardt exists" is that it can't be done by a sentence in the language of ZFC, we need an axiom schema. Not only that, we need a new function symbol to denote the embedding. This is why people usually talk about "ZFj+a Reinhardt exists", where we append a function symbol j to ZFC's language. Choice doesn't inhibit this construction (you can still throw in all the axioms saying that j is elementary), but when including both choice and the schema we can prove a contradiction. – C7X Oct 09 '22 at 19:16
  • @C7X So the problem is that it is incompatible with choice and not that Reinhardt cardinals are too "strong' for choice. I'm also assuming by your comment that not only V doesn't have a limit on how large it is, but also on how 'consistent' it can get (there is no largest and also a no 'strongest' cardinal). You could hypothetically create stronger large cardinals that satisfy ZFC. Consistency ≠ size, but it has been proven that SOME large cardinals are larger than others (e.g. a Mahlo cardinal has several inaccessible cardinals below it), This may or may not be with Reinhardt cardinals. – SI J Oct 11 '22 at 00:33
  • @SIJ Choice doesn't inhibit the construction, the problem is Reinhardts are "too strong". The Reinhardt schema implies consistency of some large cardinal axioms also proved (Kunen) to be inconsistent with choice, like "there is a nontrivial elementary embedding from some $V_{\lambda+2}\to V_{\lambda+2}$". The problem isn't using a schema to formulate it; ORD is Mahlo is a large cardinal schema and that's consistent with choice. ZFC+"there exists κ such that V_κ models ZFj+Reinhardt" proves consistency of ZFj+Reinhardt, but I'm not sure if this theory is consistent. – C7X Oct 11 '22 at 02:42
  • I'm a bit hesitant to say "Reinhardts are so strong they break choice", "measurables are so strong they break V=L", etc, because it gives the mental image that these cardinals are some sort of objects whose presence destroys deeply held assumptions about the universe, just because of how large they are. (This chain has reached >50 comments. If you don't want to get the requirement for SE chat, I recommend using some other service like Discord) – C7X Oct 11 '22 at 02:49

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