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It is a nice theorem of Zermelo that if we replace the Replacement schema with its second-order counterpart, "The image of a set under any function is again a set", then we necessarily get a model isomorphic to some $V_\kappa$ for an inaccessible $\kappa$.

What happens when we take $\sf Z$, i.e. $\sf ZFC$ without Replacement (but with Separation), and replace the Separation schema with its second-order counterpart: "If $A$ is any collection of sets, then for every $x$, $\{u\in x\mid u\in A\}$ is a set"?

If we add first-order Replacement back, one can see immediately that any model has to be well-founded. Once we have a rank function, an ill-founded model implies there is a set of ordinals of the model without a minimum. Since this set is internally bounded by some ordinal, the second-order axiom implies that this witness is in fact internal to the model.

So in particular, every model of this $\sf ZFC_{1.5}$-hybrid, is isomorphic to a unique transitive model. But this is not all. If $x$ is any set in such transitive model, and $y\subseteq x$, then applying second-order Separation we get that $y$ is also a member of the model. So power sets are computed correctly. Therefore it is necessarily the case that the model is some $V_\alpha$, and since it satisfies $\sf ZFC$ this is necessarily a worldly $\alpha$.

(And easily it can be seen that if $\alpha$ is a worldly cardinal, then $V_\alpha$ satisfies this $\sf ZFC_{1.5}$.)

What happened with the second-order Zermelo (which I will avoid denoting by $\sf Z_2$ for obvious reasons)?

Specifically, since there is no rank function what can we say about this theory?

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  • $\begingroup$ Is your first paragraph about ZFC? $\endgroup$
    – David Roberts
    Aug 12, 2017 at 21:47
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    $\begingroup$ Yes, it is about ZFC. The proof, by the way, is as follows: (1) Show the model is well-founded, essentially as I have shown here, but using Replacement to directly pick a sequence of ordinals; now we can assume the model is transitive (2) Show that power sets are computed correctly by showing that if $A$ is in the model, every subset of $A$ is the image of some class function applied to $A$; so our model is some $V_\alpha$, finally (3) show that the ordinals have to be regular, and thus $\alpha$ is inaccessible. $\endgroup$
    – Asaf Karagila
    Aug 12, 2017 at 21:47
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    $\begingroup$ Huh. Thanks. It works with Separation as well, at least steps (1) and (2), so we do get a worldly cardinal. $\endgroup$
    – Asaf Karagila
    Aug 12, 2017 at 21:52
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    $\begingroup$ "What's this here?", he says, pointing to a random point on the page. In the ensuing excitement... $\endgroup$
    – David Roberts
    Aug 12, 2017 at 22:01
  • $\begingroup$ Do you have the foundation axiom in your version of Zermelo? The way you described it, it seems as if you want to include it, but many versions of Z do not. If you don't have it, then it would seem to open the door to all kinds of strange models. $\endgroup$ Aug 12, 2017 at 23:40

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