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I'm currently thinking about how different models of set theory view each other. In particular I'm looking at how well-foundedness behaves between different models.

So we have the Axiom of Regularity:

$\forall x [x \not = \emptyset \rightarrow \exists y \in x (y \cap x = \emptyset)]$

We know also (by your choice of Compactness for first-order theories, Mostowski Collapse, or Ultrapower construction), that this axiom (and indeed the whole of $ZFC$) can be satisfied where the particular relation being interpreted as membership is not, strictly speaking, well-founded (at least from the perspective of what we normally take to be standard models).

I'm intruiged by a principle of Joel Hamkins' Multiverse View (given in [Hamkins 2012]):

Well-foundedness Mirage. Every universe $V$ is non-well-founded from the perspective of another universe.

I'm just trying to get my hands on a model of $ZFC$ that thinks that `a standard model' (if I may be excused the absolutism for a minute) of $ZFC$ is non-well-founded. Thus far I just can't quite crack it. A compactness construction would seem to be easiest, but I'm struggling to generate one that thinks the standard model is non-well-founded. Any hints/references/examples would be really gratefully received!

References: [Hamkins 2012] Hamkins, Joel-David, `The Set-Theoretic Multiverse', THE REVIEW OF SYMBOLIC LOGIC, Volume 5, Number 3, September 2012

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I'm very glad to hear that you're intrigued by the ideas in my paper.

As Asaf says, the point of the well-foundedness mirage is to assert strongly that there is no "standard" model. The axiom is asserting baldly that every set-theoretic universe is non-standard with respect to some other (better) universe. Indeed, the strong form of the mirage says that even the natural numbers of the given universe of set theory are non-standard with respect to the larger, better universe of set theory.

This axiom is probably the most controversial of the multiverse axioms in that paper, and I had included it in order to emphasize the point that we actually have little reason to think that we've got an absolute notion of well-foundedness, or even of the finite.

The main result of

shows that the multiverse axioms are consistent, if ZFC itself is consistent, and so one cannot argue that they express an incoherent vision of set theory.

But to answer your question, the multiverse model of this paper explains how one might imagine the true multiverse. Namely, the main result of the paper is that the collection of countable computably saturated models of ZFC satisfies all the multiverse axioms, including the well-foundedness mirage. And so one might imagine that the actual set-theoretic universe is like a countable computably saturated model of set theory inside some large set-theoretic universe. In such a case, that large set-theoretic universe would look upon it as ill-founded. The point of the well-foundedness mirage is to assert that every set-theoretic universe is like that. We only believe a model is "standard" because we haven't yet moved to the set-theoretic context that reveals it as ill-founded.

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  • $\begingroup$ Great! Thanks very much Joel. That's a nice example, and I'll go away and look at it some more (I've read that paper but not digested it fully; it's particularly interesting that the collection of all models of $ZFC$ does not satisfy the multiverse axioms). There's a whole bunch of philosophical/foundational questions I'd love to ask you, but I'll leave them for personal correspondence. $\endgroup$ – Neil Barton Apr 21 '14 at 13:52
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You're misinterpreting the mirage.

Standard models are always well-founded. By definition. If $V$ is a universe of $\sf ZFC$ and $M\in V$ is a standard model then it means that $\in$ is the membership relation of $M$, and $V$ always thinks that $\in$ is well-founded due to the axiom of regularity.

The mirage really says that if $V$ is a universe of set theory, and it thinks it is well-founded, then there is some $V'$ such that $V'$ knows about a decreasing sequence in $V$.

If $M$ is a standard model of $\sf ZFC$ and $N\in M$ is a standard model of $\sf ZFC$ such that $M\models N\text{ is a standard model of }\sf ZFC$, then by taking an ultrapower of $M$, $M'$ we have a model $N'$ that $M'$ thinks is standard, but $V$ itself knows the truth.

So $N'$ is a standard model in some universe (internally to $M'$) but in $V$ we have that $N'$ is actually a non-standard model.

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  • $\begingroup$ Thanks for the response! I think there's an issue though, neither $V$ nor $\in$ have any absolute significance for Hamkins. Every model thinks that it is the standard model (and transitive et cetera). The Ultrapower is interesting, but I'm really looking for a model that thinks a standard model (in this case $V$) is non-standard. Your construction provides me with a non-standard model that is viewed as standard from another model. (Also, should there be a prime on the third $V$ of the second paragraph? Please put me right if I'm missing something obvious!). $\endgroup$ – Neil Barton Feb 4 '14 at 21:01
  • $\begingroup$ Neil, start $M$ as a universe, then $N$ is a standard model; therefore $M'$ is a universe thinking that $N'$ is standard, but then $V$ knows that $M'$ is wrong and $N'$ is non-standard. $\endgroup$ – Asaf Karagila Feb 4 '14 at 21:07
  • $\begingroup$ I feel like I'm being really slow here. So here we have a universe ($M'$) thinking that a nonstandard universe ($N'$) is standard. What I want is a non-standard universe thinking that a standard universe is non-standard. So does $N'$ think that $N$ is non-standard in the above example? (I really appreciate your time, if I don't get it after this attempt I'll just go mull it over for a bit by myself). $\endgroup$ – Neil Barton Feb 4 '14 at 21:14
  • $\begingroup$ But that's not the mirage you quote. And it seems to me as something which is impossible to boot. I'll give it some thought. $\endgroup$ – Asaf Karagila Feb 4 '14 at 21:28
  • $\begingroup$ Asaf, you seem to have a typo in your third paragraph, about $V$ and $V'$. It should say that "...$V'$ knows about a decreasing sequence in $V$". $\endgroup$ – JDH Apr 20 '14 at 2:18

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