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A $\beta$-model of a set theory (or higher-order-arithmetic theory) is a model $M$ of that theory which is correct about well-foundedness: if $x\in M$ is an illfounded relation, then there is some $a\in M$ which is a subset of the domain of $x$ with no minimal element.

I know a bit about $\beta$-models of ZFC- and $Z_2$-like theories - in particular, $\beta$-models of ZFC are just well-founded models of ZFC - but I've realized that embarrassingly I know nothing whatsoever about $\beta$-models of NF-like theories. The existence of a "reasonable" set theory without $\beta$-models would be amazing, so I'm sure that $(i)$ there are easy ways to construct $\beta$-models of NFU (or even strengthenings like NFU + Infinity + Choice) and $(ii)$ there are no major reasons to be more skeptical of the existence of a $\beta$-model of NF than of the mere consistency of NF.

That said, I still don't see how to whip them up. So:

How does one construct a $\beta$-model of NFU?

I'm especially interested in $\beta$-models of strong extensions of NFU (like NFU + Choice + Infinity + "Cantorian sets"). I'm also interested in heuristic arguments about why (I assume!) the $\beta$-consistency of NF should be equiplausible with the consistency of NF.

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  • $\begingroup$ Hopefully I can persuade Holmes or Forster to pop over and weigh in, because I would really love to know the answer to this, too. $\endgroup$ – Malice Vidrine Jan 14 at 8:08
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There are no $\beta$ models of NFU. The natural order relation on the ordinals of NFU in a given model M is ill-founded, externally, but M thinks it is well-founded. So there are collections of ordinals of M which have no minimal element, but none of these are sets of the model M. This is a classic result: there is a paper about it, to which I should supply a reference when I am not on vacation away from my office.

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(Thank you, Alice, for getting me to look at this) Before i answer i think i need a bit of clarification on what exactly a $\beta$-model is. On the face of it NFU cannot have a model which speaks about wellfoundedness without forked tongue, beco's the ordinals of any model of NF(U) are illfounded. That is to say, it is a theorem of NF(U) that there is (an explicitly definable) proper class of ordinals with no least member. (It may be that Rosser-Wang ``Nonstandard models for formal logics'' addresses your interests .. JSL some time early 1950's - they discuss NF in some detail) But it may be that you mean something subtly different.

  best wishes

     tf
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  • $\begingroup$ I think i can express my concern more specifically: i need to be told the difference between a $\beta$-model and an $\omega$-model. It is known since at least Rosser-and-Wang that the Quine systems have no $\omega$-models. $\endgroup$ – Thomas Forster Jan 14 at 16:01
  • $\begingroup$ @ThomasForster An $\omega$-model has its $\omega$ isomorphic to the usual $\omega$. To see the difference between this and $\beta$-model-ness, note that there are $\omega$-models of ZFC which are not $\beta$-models: e.g. via Gandy's basis theorem we can build a model of ZFC with well-founded part of height $\omega_1^{CK}$. $\endgroup$ – Noah Schweber Jan 14 at 18:44
  • $\begingroup$ The definition of well-founded I'm using is the one given in my question, which seems self-contained unless I'm missing something? $\endgroup$ – Noah Schweber Jan 14 at 18:45
  • $\begingroup$ Also, Randall Holmes' answer below states "The natural order relation on the ordinals of NFU in a given model M is ill-founded, externally, but M thinks it is well-founded," which seems to contradict your statement " it is a theorem of NF(U) that there is (an explicitly definable) proper class of ordinals with no least member" - am I missing something, or is one statement in error? $\endgroup$ – Noah Schweber Jan 14 at 18:46
  • $\begingroup$ And I'm confused by your claim that NFU has no $\omega$-models; it seems to me that if we run the Jensen construction starting with an $\omega$-model of ZFC with an automorphism $j$ (which necessarily moves an ordinal), we get an $\omega$-model of NFU? $\endgroup$ – Noah Schweber Jan 14 at 18:49

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