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Do we have: $${\sf ZF-AoI}\vdash\forall X,(X\text{ is a model of }{\sf ZF-AoI})\implies(X\text{ is at least countable})$$ AoI is the axiom of infinity.

I know that some models of $\sf ZF-AoI$ have no at-least-countable sets, but I'm fairly certain that those models contain no models of $\sf ZF-AoI$, making the statement vacuously true in those models.

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    By "countable" you mean "countably infinite," presumably? – 2017-01-20
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    @NoahSchweber Yes. – 2017-01-20
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    How do you define "countably infinite" in this context? Keeping in mind that there may be no $\omega$ to embed . . . – 2017-01-20
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    @NoahSchweber There exists an inductive set, and that inductive set injects into $X$. In models without inductive sets, nothing is at-least-countable. – 2017-01-20

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Yes, it does.

Working in a model $V$ of ZF-AoI, suppose $M$ is a set model of ZF-AoI. Then consider the set $X\subset M$ of $M$-ordinals, ordered by $\in^M$. This is a linear order.

Of course, it may not be well-founded, but it has a well-founded part. Let $$Y=\{x\in X: \mbox{the suborder of $M$-ordinals $

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    Thanks! But if we have $(\exists\text{ a model of }{\sf ZF-AoI})\implies{}$$(\exists\text{ an inductive set})$, and we also have ${\rm Con}({\sf ZF-AoI})\implies{}$$(\exists\text{ a model of }{\sf ZF-AoI})$ by Gödel's completeness theorem, do we have that $\sf ZF-AoI$ proves ${\rm Con}({\sf ZF-AoI})\implies{}$$(\exists\text{ an inductive set})$? That is, do we have that $\sf ZF-AoI+\rm Con(\sf ZF-AoI)$ proves the axiom of infinity? – 2017-01-20
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    @AkivaWeinberger No - because ZF-AoI doesn't prove the Completeness Theorem! There's an easy way to see this: take the usual model $V_\omega$ of ZF-AoI. Then since this model has the "right" natural numbers, it satisfies "ZF-AoI is consistent" (since ZF-AoI actually *is* consistent); but clearly $V_\omega$ contains no model of ZF-AoI. – 2017-01-20
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    There *is* a version of the Completeness Theorem for ZF-AoI (or rather, for PA) - the **arithmetized completeness theorem** - but it's quite a different thing. Roughly speaking, it asserts the existence of reasonable *class* models of certain consistent theories. – 2017-01-20
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    Would $\sf ZF-AoI$ be able to prove that all "_class_ models" of $\sf ZF-AoI$ are at-least-countable? – 2017-01-20
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    @AkivaWeinberger Actually, now that I think about it I suspect that my previous (deleted) comments about ACT were nonsense - I'm not sure there is a good version of ACT provable inside PA, rather than inside $RCA_0$ (or similar). I'll double-check that. However, re: your most recent comment, remember that there's no way to talk directly about class models - so how would you even *phrase* the statement "all class models are at-least-countable" in the language of set theory? (Also, the definition of "countable" would need to change to "embeds an inductive **class**" to have any hope of working.) – 2017-01-20
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In ZF$-$Infinity, you can prove if $X$ is a model of just Extensionality, Pairing, and Union, then $X$ contains a unique element that is isomorphic to each finite ordinal (where a "finite" ordinal is one that is not a superset of any limit ordinal), and these elements are different for different finite ordinals.

By the Axiom of Replacement, we can then (at the metalevel) make a set of all finite ordinals by mapping each element of $X$ to the unique ordinal it represents, if any, and to $\varnothing$ otherwise. This set will then be $\omega$ and inject into $X$.

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    @AndrésE.Caicedo: How is that not good enough? To me what the question asks is whether $X$ is countable, which is a question to be answered from the perspective of the external ZF$-$Inf universe, not a question about which formulas are true _in_ $X$. – 2017-01-20
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    @AndrésE.Caicedo: It's also slightly unclear to me what you mean by the "such an element" you say that $X$ does not contain. Could you clarify? – 2017-01-20
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    @AndrésE.Caicedo: OK. – 2017-01-20