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I take the following to be a commonly-accepted definition of an inner model (in the context of ZFC):

Let (M,∈) be a model of ZFC. The pair (N,E) is an inner model of ZFC in M if:
(i) (N,E) is a model of ZFC
(ii) E is the restriction of ∈ to N
(iii) N is a transitive class of M
(iv) The class Ord is the same in N as it is in M.

If (N,E) satisfies just (i) and (ii), then we say that (N,E) is a standard model of ZFC in M. Sometimes we also say that (N,E) is a standard submodel of ZFC in (M,∈).

Now, as is explained here, in any given universe, the existence of a standard model of ZFC is stronger than simply the existence of a model of ZFC. This is also pointed out in the Wikipedia article on standard models (here). However, it is both stated later in that very same Wikipedia article and is mentioned in an article on Cantor's Attic (here) that Godel showed that any model of ZFC has a least inner model, called the constructible universe. This seems to give us that Con(ZFC) implies the existence of a model implies the existence of a standard submodel, since inner models are certainly standard submodels. This chain of implication is problematic insofar as it appears to contradict the fact that Con(ZFC) is strictly weaker than the existence of a standard model of ZFC.

I suspect that the issue at hand is caused by (a) sloppiness regarding just what universe our statements apply to in what situation, and (b) sloppiness regarding the conditions under which we assert Con(ZFC) in the first place. Here is what I see as the resolution: As is discussed by Asaf Karagila in the first link above, when we say that ZFC is consistent, we are doing so in some grand universe V of sets, which (together with some relation ∈) we typically take to be a model of ZFC. In the context of V, to say that ZFC is consistent is to say that there is some set M and some relation E on M, both in V, such that (M,E) is a model for ZFC. That is, Con(ZFC) tells us only about the existence of a certain object living in V with certain properties. L, on the other hand, is a proper class of our universe V, i.e is not a member of V. From the inside of V, then, L is not seen as existing; much less is it seen as being a standard model. From the outside of V, however, L looks like a standard submodel of V. But as soon as we're external to V, then our domain of discussion is some bigger universe V', which (together with some relation ∈') we take to be a model of ZFC. However, since ∈ need not be the restriction of ∈' to V, L is not necessarily a standard submodel of ZFC in V'.

That is, in order to know that something is an inner model of V, we have to know that it exists in the first place, and sometimes this requires that our domain of discussion be bigger than V. In the case of Godel's constructible universe L, we know that it is an inner (and hence standard) model of V only when our domain of discussion V' is bigger than V; and in this case, L need not be a standard model relative to our bigger universe V'.

Is this resolution plausible? Thank you in advance.

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Curtis Mason
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    https://math.stackexchange.com/q/1368436/622 https://math.stackexchange.com/q/1618204/622 https://math.stackexchange.com/q/557924/622 and probably a few others. In short, class models are not models in the internal sense. If they were, then the universe itself would witness that ZFC is consistent, and therefore ZFC proves its own consistency. – Asaf Karagila Jul 29 '19 at 09:28

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I’m not sure I understand everything you say in your resolution, but it captures the main ideas that we need to be careful about internal vs external, set vs proper class, and where things are proved.

The statement that is stronger than Con(ZFC) is that there is a transitive set model of ZFC. This statement, along with the weaker statement that a not-necessarily-transitive set model exists (which is equivalent to Con(ZFC)) are statements we can fully formalize as sentences in the language of set theory.

For proper class “models” like $V$ or $L$, on the other hand, we can’t necessarily internalize the idea that they are models, since there is generally no satisfaction relation for proper classes (certainly never for $V$ by Tarski’s theorem, though we can sometimes get one for $L$ under large cardinal assumptions). Instead, the idea that they are models must be understood as a scheme of relativized axioms proven one at s time. $V$ is quite trivially a transitive proper class model of ZFC by this standard, so we can see that the “existence” of a transitive proper class model doesn’t mean anything on its own.

What we get from the existence of proper class models are relative consistency results in the metatheory rather than consistency results in the theory. For instance the fact that every axiom of ZFC + V=L relativized to $L$ can be proved in ZF implies that the consistency of ZF implies the consistency of ZFC + V=L. This can be viewed as a purely proof theoretical result in a finitary metatheory, or as being about nested models, viewed externally as set models in a set theory.

spaceisdarkgreen
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  • What distinguishes a set model from a class model? From what I've seen, a set model is a set within a given model V of ZFC that is also a model of ZFC. What makes V or L a proper class model? Are the notions of "set" and "class" absolute notions, or do they depend on our ground model/universe? – Curtis Mason Jul 26 '19 at 22:31
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    @CurtisMason Probably best at first to avoid model-hopping and just work in ZFC (or inside an arbitrary fixed model of ZFC if you prefer). A set model is a set $(M,E)$ that satisfies all of the given axioms (this is unproblematic since the recursive definition of satisfaction in $(M,E)$ can be easily formalized in ZFC). A proper class model is a proper class $(M,E)$ such that the relativization $\varphi^{(M,E)}$ of each axiom holds, which is something that can only be understood as a scheme. For instance $(L,\in)$ is a proper class model of ZFC+V=L. – spaceisdarkgreen Jul 26 '19 at 23:18
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    @CurtisMason The key here being that we cannot in general even state "$(M,E)$ is a model of ZFC" in the language of set theory for $(M,E)$ a proper class. As for set models, $(V_\kappa,\in)$ for $\kappa$ inaccessible is a set model of ZFC. And yes, it is a relative thing. Satisfaction in $(V_\kappa,\in)$ cannot be defined in $V_\kappa$ itself, and from its perspective, it is a proper class (its own version of $V$). Same for $V_\kappa$'s constructible universe $L^{V_\kappa}$ (which happens to be $L_\kappa$) which is a set model of ZFC+V=L, but a proper class model from $V_\kappa$'s perspective. – spaceisdarkgreen Jul 27 '19 at 00:01
  • How do we know what is a set and what isn’t? This notion seems to be dependent on our ambient universe V. V usually denotes the Von Neumann universe, if I am correct. How is this V to be described? Is it the “true” universe of sets? I ask this because each model of ZFC internally looks like a cumulative hierarchy, just as the Von Neumann V does. Because of this, I am having difficulty making out what to think about the Von Neumann universe. – Curtis Mason Jul 27 '19 at 01:28
  • @CurtisMason For a given model, the elements of the model are its sets, it's as simple as that. However it's an open philosophical question whether the cumulative hierarchy is a sensible unique mathematical object that forms the standard model of set theory. According to a set-theoretical Platonist, yes, there is a real collection $V$ of all the well-founded sets, full stop. Things like the CH have absolute truth values, we just need to improve our intuition to find better axioms that give them. But inasmuch as one believes in many valid backgrounds, "what is a set" does indeed vary. – spaceisdarkgreen Jul 27 '19 at 02:46
  • @CurtisMason It is a bit misleading to identify $V$ with the Von-Neumann hierarchy. $V$ is the universe of sets under consideration. Most Platonists would say non-well-founded sets exist, but we don't have as good of an intuition for them and we will exclude them from study by assuming foundation. Then we prove this implies every set has a Von-Neumann rank, and the $V_\alpha$ are just some sets we define to express this organizing principle. So it is a bad choice of notation that makes $V=\bigcup_\alpha V_\alpha$ look like a definition of $V$ rather than a theorem that all sets have rank. – spaceisdarkgreen Jul 27 '19 at 02:53
  • @CurtisMason One can also take a formalist approach where we conceive of the operation as just studying the deductive consequences of ZFC and whatever other axioms we might feel like. Here there are no wrong answers other than axioms that are inconsistent, although there are aesthetic considerations as always. While a platonist may think measurable cardinals exist since it is plausible and the theory really hangs together nicely with them there, a formalist will simply think there are cool things you can prove by assuming their existence and they aren't obviously inconsistent, so let's assume. – spaceisdarkgreen Jul 27 '19 at 03:14
  • @CurtisMason Note in formalism the objects don't have to be 'real' in any way. Proper classes are just predicates and "$V_\kappa$ is a model of ZFC" is just a sentence in LST. (I'm actually not sure why I'm going on here, other than just liking to talk. You should read the intro of Kunen's set theory book and maybe others with a philosophical bent. There's also the "believing the axioms" papers by Maddy exploring Platonism and a lot of other philosophical papers if you dig... Feferman wrote a lot of skeptical stuff here and there. Plus there's a lot more on MSE/MO, as I'm sure you know.) – spaceisdarkgreen Jul 27 '19 at 03:26
  • So, just to be terribly clear, V denotes our given universe of sets. We take V to be a “model” for ZFC, not in the sense that there is a definable satisfaction relation on V, but rather, in the sense of a scheme of statements. Now, since we take the axiom of foundation to hold in V, V has the structure of the Von Neumann hierarchy. However, any set model (M,E) of ZFC in V will also internally appear to have the structure of the Von Neumann hierarchy. – Curtis Mason Jul 27 '19 at 04:16
  • @CurtisMason 'Yes' to everything except the second sentence, and 'kinda' to the second. I think I've overcomplicated things a bit here. We can develop model theory in ZFC... models are just sets with certain properties. We can talk about semantics of ZFC in ZFC in terms of this model theory, but none of these models will be the universe of all sets $V$... this is a proper class, a thing that we can't even refer to directly in ZFC, since one of the first things we learn exploring ZFC is that there is no set of all sets. – spaceisdarkgreen Jul 28 '19 at 01:34
  • @CurtisMason The sense in which $V$ is a model is that each of the axioms relativized to $V$ hold. This is trivial, since the axioms relativized to $V$ are straightforwardly equivalent to the axioms themselves. (Inner models like $L$ are less trivial... the relativizations of the axioms to $L$ as well as other relevant statements like GCH follow from the axioms themselves, but not trivially.) But they aren't models in the sense of the model theory we develop in ZFC. – spaceisdarkgreen Jul 28 '19 at 01:34