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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,

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    There certainly are non-standard models of finite ZF: Each nonstandard model of PA can interpret finite ZF (with the binary coding), which leads to a similarly nonstandard model of finite ZF (whose ordinals are isomorphic to the original nonstandard model of PA). – 2017-01-15
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    @HenningMakholm: Could you give me a concrete example of a non-standard element of the non-standard model of finite $ZF$? – 2017-01-15
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    x @Thomas: Each non-standard element of the original non-standard PA model will produce such a non-standard set. (Tennenbaum's theorem seems to severely restrict how concretely one can describe a non-standard model, since the binary-coding interpretation is definably reversible). – 2017-01-15
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    @HenningMakholm: Have you a reference to Tennenbaum's theorem? In what way does it severely restrict how concretely one can describe a non-standard model? I understand that, relative to finite $ZF$, $V_{\omega}$ is a proper class (and consequently, all infinite subclasses of $V_{\omega}$ will be proper classes as well), but from outside the model, would the non-standard element appear to be 'infinite', just as in the non-standard model of $PA$? – 2017-01-15
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    [Tennenbaum's theorem](https://en.wikipedia.org/wiki/Tennenbaum's_theorem) says that a non-standard model of PA cannot have a computable addition or multiplication operation. (Since you're reading about non-standard models, I thought you would have come across it). – 2017-01-15
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    And yes, a non-standard element of a ZFfin model will have to be "infinite" in the sense that it sits on top of an infinite descending $\in$-chain (though that can, of course, only be seen from the outside). – 2017-01-15
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    @HenningMakholm: What does the non-standard model of finite $ZF$ 'see' when it 'sees' the infinite descending $\in$-chain? Also, regarding Tennenbaum's theorem--I am familiar with the fact but did not know that it 's proof was attributed to Tennenbaum. – 2017-01-15
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    @ThomasBenjamin The same thing a nonstandard model of PA sees: nothing. From its perspective, every descending chain of sets is nice and finite. – 2017-01-16
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    Also, just like with PA, there is no distinguished nonstandard model: so writing e.g. "the non-standard model" is missing the point. Just like with PA, there are *lots* of nonstandard models. – 2017-01-16

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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

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.

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.