# What is the consistency strength of this kind of iterating Berkeley cardinals?

[EDIT] After suggestion from Monroe Eskew, and after having an e-mail correspondence with Prof. Joan Bagaria, I'll re-present the older question as the second in a series of 4 questions. I've had an e-mail correspondence with Prof. Joan Bagaria, and asked him the first of the following 4 questions, and I'll present his answer to it [after taking his own permission on posting it here].

Question 1. I define Berkeley cardinal in a rank $$V_{\kappa}$$ as the same definition of Berkeley cardinal but the cardinal itself, the transitive sets and the embeddings involved in its definition all are restricted to be elements of $$V_{\kappa}$$.

Now if $$b$$ is a Berkeley cardinal in $$V_{\kappa}$$, does it necessarily follow that $$b$$ must be a Berkeley cardinal in the next rank $$V_{\kappa +1}$$, and more generally does it need to be still a Berkeley cardinal in every higher rank $$V_{\kappa^+}$$ ?

I personally think the answer is to the negative, but I don't know how to prove it?

Question 2: Is there a clear inconsistency to having an inaccessible set $$S$$ [i.e., $$\bigcup(S)=V_{\xi}$$ for an inaccessible $$\xi$$] of transitive universes $$V_{\kappa}$$ where each $$\kappa$$ is inaccessible [where generally $$\gamma$$ is inaccessible iff $$\gamma$$ is a limit ordinal such that $$V_{\gamma}$$ is not the set union of a subset of it that is not equinumerous to it], such that for each $$\kappa$$ we have: $$\langle V_{\kappa}, \in ^{V_{\kappa}}\rangle \models ZF + \exists a (a \text{ is a Berkeley cardinal})$$, and such that for each $$V_{\alpha}, V_{\beta} \in S: \alpha < \beta$$, the first Berkeley cardinal in $$V_{\beta}$$ is not a subset of $$V_{\alpha}$$?

If that is possible, then we can define a new large cardinal property $$\xi$$ that is the smallest ordinal that is a set of all ordinals in the union set of a set $$S$$ fulfilling the above qualifications. Call such a cardinal as a "Fluctuating Cardinal".

Question 3: Supposing that there is no clear inconsistency, then would it be interpretable in $$ZF$$ plus one of the large cardinal properties mentioned by Bagaria and Koellner?

Question 4: Can $$\bigcup(S)$$ be compatible with Choice?

I mean we can arrange for all of the $$V_{\alpha}$$ stages in $$S$$ to be non-supertransitive, so we can contemplate having external choice sets on all elements of these stages that do not have internal choice sets (i.e. the choice sets are elements of the next stage, but not of the stage itself), so although choice is not satisfied of course inside each stage $$V_{\alpha}$$ (because it contains a BC inside it), yet choice can be satisfied outside it, i.e. in the next stage $$V_{\beta}$$ that is an element of $$S$$.

By this our $$\xi$$ fluctuating cardinal would be bigger than all Berkeley cardinals and yet it is not a Berkeley cardinal and yet $$\xi$$ would be consistent with full choice?

• Berkeley cardinals are beyond anything that has been studied. So asking for a large cardinal property that does the work really makes no sense currently. One could answer the question by saying that the consistency strength is the existence of a proper class of inaccessibles $\kappa$ such that ..., just repeating precisely what you said, and that would be a perfectly reasonable acceptable large cardinal assumption. You need to clarify what you mean. Also, precisely what version of inaccessibility do you have in mind? (We are in a ZF setting, so there are several inequivalent such versions). – Andrés E. Caicedo Jan 4 at 19:32
• Yes, I am aware of all of that. There is one paper on the subject. My point is that your questions are rather premature given that state of affairs. It would be like asking whether supercompactness is consistent when people were barely coming to terms with measurability. – Andrés E. Caicedo Jan 4 at 19:49
• By strongly inaccessible do you mean that for each $\lambda<\kappa$, $2^\lambda<\kappa$? (This gives you choice and therefore a contradiction.) Maybe you should mean instead something about how the power set of $\lambda$ relates to $V_\kappa$? – Andrés E. Caicedo Jan 4 at 19:56
• @ZuhairAl-Johar— Your question would be more interesting and motivated if you could give examples of things being Berkeley cardinals in one rank, and then failing to be so in a higher rank. – Monroe Eskew Jan 5 at 15:31
• @MonroeEskew, hmm..., I'll try! Thanks – Zuhair Al-Johar Jan 5 at 15:34

Suppose $$b$$ is a BC in $$V_{\kappa}$$. Take the model $$L(V_{\kappa})$$, i.e., the least inner model of $$\text{ZF}$$ that contains $$V_{\kappa}$$. Then in this model there is no non-trivial elementary embedding from $$V_{\kappa}$$ into $$V_{\kappa}$$, and so $$b$$ is not a BC in $$V_{\kappa+1}$$. The reason is that any (non-trivial) elementary embedding $$j:V_{\kappa} \to V_{\kappa}$$ in $$L(V_{\kappa})$$ would extend naturally to an elementary embedding $$j:L(V_{\kappa}) \to L(V_{\kappa})$$, definable in $$L(V_{\kappa})$$ from $$j$$. But $$\text{ZF}$$ proves that there is no non-trivial elementary embedding from the universe into itself that is definable with parameters (by Suzuki).