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In our background universe $V$ - satisfying $\operatorname{ZFC}$ - we say that an ordinal $\delta$ is a Woodin cardinal iff it satisfies one of the following equivalent properties:

  1. For all $A \subseteq V_{\delta}$ there is a cardinal $\kappa < \delta$ such that for all $\nu < \delta$ there is a definable elementary embedding $j \colon V \prec M$, $M$ transitive, such that $\operatorname{crit}(j) = \kappa$, $j(\kappa) > \nu$ and $j(A) \cap V_{\nu} = A \cap V_{\nu}$. (Note that by reflecting $A^* := \{0\} \times A \cup \{1 \} \times V_{\delta}$ instead and choosing $\nu$ to be a cardinal, we can furthermore require that $V_{\nu} \subseteq M$.)
  2. For all $A \subseteq V_{\delta}$ there is some $\kappa < \delta$ such that for all $\nu < \delta$ there is an elementary embedding as in 1. which is the canonical embedding $j = j_{E}$ of some extender $E$.
  3. For every function $f \colon \delta \to \delta$ there is some $\kappa < \delta$ such that $f '' \kappa \subseteq \kappa$ and such that there is a definable elementary embedding $j \colon V \prec M$, $M$ transitive, with $\operatorname{crit}(j) = \kappa$ and $V_{j(f)(\kappa)} \subseteq M$.
  4. For every function $f \colon \delta \to \delta$ there is some $\kappa < \delta$ with $f '' \kappa \subseteq \kappa$ and some extender $E$ with associated embedding $j_E \colon V \prec M$, $M$ transitive, such that $\operatorname{crit}(j_{E}) = \kappa$ and $V_{j_{E}(f)(\kappa)} \subseteq M$.

In WHAT IS THE THEORY ZFC WITHOUT POWER SET? Victoria Gitman, Joel David Hamkins, and Thomas A. Johnstone showed - amonst other things - that in a model $N$ of $\operatorname{ZFC}_{-}$ (i.e. $\operatorname{ZFC}$ with the power set axiom removed) it is possible to have an embedding $j_{E} \colon N \to M$ associated to an $N$-extender $E \in N$ such that $M$ is transitive but $j_{E}$ is not elementary. I suspect that their construction can be generalized to produce a $\operatorname{ZFC}_{-}$-model $N$ in which 1. and 2. (resp. 3. and 4.) are not equivalent.

[This motivates a vague question, which is not the intended focus of this post, namely:

Q: Is there a model $N$ of $\operatorname{ZFC}_{-}$ in which 1.(/3.) holds but 2.(/4.) fail?]

I, however, am more interested in models of $\operatorname{ZFC}^{-}$ (i.e. $\operatorname{ZFC}$ with the power set axiom removed to which we add the axiom scheme of collection). Here (again demonstrated by Gitman, Hamkins and Thomas for $N$-measures - the same argument seems to work for $N$-extenders) Łoś's theorem does hold, so that a lack of elementarity doesn't stop us. Thinking through the proof it also seems that now 1. and 2. (resp. 3. and 4.) are equivalent. The remaining question is:

Q: Is it possible to have a transitive model $N$ of $\operatorname{ZFC}^{-}$ such that 2. and 4. are not equivalent? (All relevant extenders $E$ are supposed to be elements of $N$ and the elementarity of $j_{E}$ is interpreted as a meta-theorem in $N$.)


Edit: I'd like to thank Joel Hamkins for his helpful comments on the matter which made me aware of some issues in my original question. In order to avoid most of these subtleties, I've now narrowed the scope of my question. However, if the answer to it is negative, I will also welcome suggestions of more general interpretations of the Woodinness of $\delta$ in/over $N$ that allow for a positive answer.

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  • $\begingroup$ Can you clarify what you mean in 1 by saying "there is a definable j..."? The concept of "definable" is not expressible in first-order set theory, but I think you intend for this definition to be formalized in ZFC. I guess you mean that $j$ is a class, and you can make it first-order by going to the induced extender. But in a ZFC- context, it isn't clear what you would mean. Do you still insist the embedding is definable even when, say, the power set axiom fails badly? $\endgroup$ Commented Dec 25, 2016 at 20:50
  • $\begingroup$ I'm sorry, but I don't really understand your reply. How does it make sense to assume MK if the question is about the effects of assuming only ZFC-, a much weaker theory? My question was about the meaning of statement 1 in ZFC-, since it doesn't seem formalizable as you have stated it. $\endgroup$ Commented Dec 25, 2016 at 21:39

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