1
$\begingroup$

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 $\mathsf{AC}$.

$\endgroup$
9
  • 1
    $\begingroup$ @LSpice, thanks! $\endgroup$ 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. $\endgroup$ Jul 22, 2022 at 20:12
  • $\begingroup$ @NoahSchweber, you mean $\zeta_{i-1} +1 = \zeta_i$. $\endgroup$ 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 $\zeta_i$. $\endgroup$ 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. $\endgroup$ Jul 22, 2022 at 21:19

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.