Concrete exanples of nonstandard models of $ZF$ $-$ _Infinity_ $+$ $\not$ _Infinity_

Question

Since it is known that $ZF$ $-$ Infinity $+$ $\lnot$ Infinity is bi-interpretable with $PA$, it seems reasonable (possibly) to infer that since $PA$ has non-standard models, $ZF$ $-$ Infinity $+$ $\lnot$ Infinity will also have non-standard models, i.e. will have models with infinite elements (sets?). Is it possible for someone to construct an example of a non-standard model of $ZF$ $-$ Infinity $+$ $\lnot$ Infinity? I am currently reading Vitezslav Svejdar's paper "Infinite natural numbers: an unwanted phenomenon, or a useful concept?", and it got me wondering what, if anything, corresponded to the notion of "infinite natural number" in a nonstandard model of $ZF$ $-$ Infinity $+$ $\lnot$ Infinity. Such an analogue would, of course, appear as finite to the model constructed, so it would be nice for anyone providing such a construction to show why it should appear so (my guess is that since the 'nonstandard elements' satisfy the same axioms as the 'standard elements' there is, in the model, no criterion by which to distinguish them). There is also the possibility that the interpretation of $PA$ as $ZF$ $-$ Infinity $+$ $\lnot$ Infinity will not admit any nonstandard models, so if this is the case, a proof of this fact would be an acceptable answer, also. Thanks in advance for any help given,

Answer 1

I claim that if you're comfortable with nonstandard models of PA, then you should be comfortable with nonstandard models of ZF-Inf+$\neg$Inf (call this theory T).

Suppose $M$ is a nonstandard model of PA. Let $c\in M$ be nonstandard - that is, $M$ thinks $c$ is finite (since $M$ thinks everything is finite), but externally the set $\{d\in M: M\models d<c\}$ is infinite. Then think about the element $a=2^{c+1}-1$ - that is, the element whose binary expansion is a string of $c$-many "$1$"s - and let's think about what $a$ represents in the Ackermann structure $Ack(M)$ assigned to $M$ (that is, the model of T built from $M$ via the usual Ackermann coding).

Remember that, basically, "elements"= "$1$s in the binary expansion". Since $a$'s binary expansion has (externally) infinitely many $1$s, we have $\{b: Ack(M)\models b\in a\}$ is (externally) infinite. Again, there's no difference between this phenomenon, and the usual way a theory like PA (which "thinks every element is finite") has nonstandard models.

And note that - just like $\mathbb{N}$ and $PA$ - the "standard model" $V_\omega$ of T is an initial segment of any model $N$ of $T$. And this fact "commutes with the Ackermann interpretation": the "standard" elements of an $M\models PA$ are exactly the ones which code "standard" elements of the associated $Ack(M)$. The situation really is the same; it's just that we're conditioned to think of models of set theories as more complicated than models of arithmetics.

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And, to address your title question, just like for PA we're not going to have any "nice" nonstandard models, by Tennenbaum's Theorem. We use the other direction of Ackermann coding: since ZF-Inf proves that $\mathbb{N}$ satisfies PA, associated to any model $V$ of ZF-Inf we have a model $Kca(V)$ of PA. Restrict attention to models of the stronger theory T=ZF-Inf+$\neg$Inf; then $Kca(V)$ is nonstandard iff $V$ was nonstandard. Nonstandard models of $T$ are going to be exactly as hard to describe as nonstandard models of PA.