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The theory $\mathsf{TNT}$, introduced by Hao Wang in 1952, adds negative types to simple Type Set Theory $\mathsf{TST}$, so it's written exactly as $\mathsf{TST}$ but with the type indices ranging over $\mathbb Z$ instead of just $\mathbb N$.

Let $\mathsf{ZF}\text{-Reg.}$ be the milieu for models, let $M$ be a transitive non-well-founded model of $\mathsf{ZF}$, by that I mean $(M, \in_M)$ where $\in_M$ is not well founded. So as seen from the outside of $M$, there must exist a non-standard infinite ordinal $\zeta$ such that there exists an infinite descending sequence $V_\zeta, V_{\zeta-1}, V_{\zeta-2},\dotsc$ of stages of $\mathsf{ZF}$. Now, take $\mathcal M = \displaystyle\bigcup_{n \in \mathbb N} V_{\zeta \ \pm \ n}$ :

Can this this provide a model of $\mathsf{TNT}$, where each sort $i$ range over $V_{\zeta + i}$, and the membership relation from sort $i$ to sort $i+1$ is the membership relation restricted to $V_{\zeta+i} \times V_{\zeta + i + 1}$?

Can we have an omega model this way? I mean the set of naturals in $\mathcal M$ is standard, i.e. externally well-founded finite von Neumann ordinals. Which (if it models $\mathsf{TNT}$) is known to violate $\mathsf{AC}$.

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    $\begingroup$ @LSpice, thanks! $\endgroup$
    – Zuhair Al-Johar
    Jul 22, 2022 at 19:52
  • $\begingroup$ Your claim about $\zeta$ is not true: all we can say is that there are $M$-ordinals $\zeta_i$ ($i$ a (true) natural number) such that $\zeta_0>\zeta_1>...$ - we need not have $\zeta_{i+1}+1=\zeta_i$. In particular, if $M$ is an ill-founded $\omega$-model, no such chain will exist. $\endgroup$
    – Noah Schweber
    Jul 22, 2022 at 20:12
  • $\begingroup$ @NoahSchweber, you mean $\zeta_{i-1} +1 = \zeta_i$. $\endgroup$
    – Zuhair Al-Johar
    Jul 22, 2022 at 20:35
  • $\begingroup$ No, I don't: note that as indices increase, my $\zeta$s decrease. So $\zeta_{i+1}$ should be smaller than $\zeta_i$. $\endgroup$
    – Noah Schweber
    Jul 22, 2022 at 20:36
  • $\begingroup$ @NoahSchweber, is there a result that this cannot be done, I mean its always the case that the $\zeta_i$'s are not immediate predecessor chains. $\endgroup$
    – Zuhair Al-Johar
    Jul 22, 2022 at 21:19

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Per comments, the above conditions might not be enough to ensure the result of interpreting $\sf TNT$, however, the following line would work to answer the first quetion:

Suppose we work in $\sf ZF−Reg.$ on a transitive non-$\omega$-non-well-founded model $M$ of $\sf Finite \ \sf ZF$ (i.e. $\sf ZF -\text{ inf.+ every set is finite}$), so the rank of every nonempty set must be a successor rank, now since its non-well-founded then there must exist a non-standard ordinal, i.e. internally looks like a finite von Neumann ordinal but externally it has an infinite predecessor chain subset of it, now working externally [in $\sf ZF-Reg.$] since $M$ is just a set, then pick any such ordinal $\zeta$ and send $V_\zeta$ to $0$, then send its predecessor stage to $\mathbb N \setminus 1 $ (which captures the integer $−1$), and the predecessor stage of that to $\mathbb N \setminus 2 $, etc...; send each $V_{\zeta +n}$ to $n$ for each $n \in \mathbb N$. Notice that $\mathbb N$ is the set of all standard naturals in $M$. It's easy to define the restrictions on membership relations, the types are the stages of $M$ that are the preimages of the intergers under the above assignment, and the rest goes through easily.

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