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[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?

[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].

EDIT: a problem was found with the answer for now, so it'll be presented once its solved, the answer was deleted upon the request of Joan Bagaria himself. So as of now, the first question doesn't have an answer yet.

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?

[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 $S$ of transitive universes $V_{\kappa}$ where each $\kappa$ is inaccessible [i.e.; $\kappa$ is a limit ordinal such that $V_{\kappa}$ 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 $\lambda$ 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 $\lambda$ fluctuating cardinal would be bigger than all Berkeley cardinals and yet it is not a Berkeley cardinal and yet $\lambda$ would be consistent with full choice?

[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?

Is there a clear inconsistency to having a proper class $S$ of transitive universes $V_{\kappa}$ where each $\kappa$ is inaccessible [i.e.; $\kappa$ is a limit ordinal such that $V_{\kappa}$ 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 $V_{\alpha}$?

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?

[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 $S$ of transitive universes $V_{\kappa}$ where each $\kappa$ is inaccessible [i.e.; $\kappa$ is a limit ordinal such that $V_{\kappa}$ 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 $\lambda$ 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 $\lambda$ fluctuating cardinal would be bigger than all Berkeley cardinals and yet it is not a Berkeley cardinal and yet $\lambda$ would be consistent with full choice?