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edited Nov 18, 2018 at 21:20

Noah Schweber

ZFC doesn't even prove that universes exist in the first place. However:

ZFC proves that universes are exactly sets of the form $V_\kappa$ for $\kappa$ inaccessible. In particular, "any two universes are $\in$-comparable" is provable in ZFC alone; what's not provable is that there are many universes, or indeed any.
The successor universe of $\mathcal{U}$ is therefore a well-defined concept, and every universe has a successor universe assuming enough universes exist in the first place: ZFC proves that there is exactly one universe of height $\kappa$ for each inaccessible cardinal $\kappa$, so to get the successor universe we just "go up to the next inaccessible." ZFC can't prove that there always is a next inaccessible, but it does prove that "every set is contained in a universe" is equivalent to "there is a proper class of inaccessibles," and that each of these implies "every universe has a successor universe."

ZFC doesn't even prove that universes exist in the first place. However:

ZFC proves that universes are exactly sets of the form $V_\kappa$ for $\kappa$ inaccessible. In particular, "any two universes are $\in$-comparable" is provable in ZFC alone; what's not provable is that there are many universes, or indeed any.
The successor universe of $\mathcal{U}$ is therefore a well-defined concept, assuming enough universes exist in the first place: ZFC proves that there is exactly one universe of height $\kappa$ for each inaccessible cardinal $\kappa$, so to get the successor universe we just "go up to the next inaccessible." ZFC can't prove that there always is a next inaccessible, but it does prove that "every set is contained in a universe" is equivalent to "there is a proper class of inaccessibles," and that each of these implies "every universe has a successor universe."

ZFC doesn't even prove that universes exist in the first place. However:

ZFC proves that universes are exactly sets of the form $V_\kappa$ for $\kappa$ inaccessible. In particular, "any two universes are $\in$-comparable" is provable in ZFC alone; what's not provable is that there are many universes, or indeed any.
The successor universe of $\mathcal{U}$ is therefore a well-defined concept, and every universe has a successor universe assuming enough universes exist in the first place: ZFC proves that there is exactly one universe of height $\kappa$ for each inaccessible cardinal $\kappa$, so to get the successor universe we just "go up to the next inaccessible." ZFC can't prove that there always is a next inaccessible, but it does prove that "every set is contained in a universe" is equivalent to "there is a proper class of inaccessibles," and that each of these implies "every universe has a successor universe."

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answered Nov 18, 2018 at 20:37

Noah Schweber

ZFC doesn't even prove that universes exist in the first place. However:

ZFC proves that universes are exactly sets of the form $V_\kappa$ for $\kappa$ inaccessible. In particular, "any two universes are $\in$-comparable" is provable in ZFC alone; what's not provable is that there are many universes, or indeed any.
The successor universe of $\mathcal{U}$ is therefore a well-defined concept, assuming enough universes exist in the first place: ZFC proves that there is exactly one universe of height $\kappa$ for each inaccessible cardinal $\kappa$, so to get the successor universe we just "go up to the next inaccessible." ZFC can't prove that there always is a next inaccessible, but it does prove that "every set is contained in a universe" is equivalent to "there is a proper class of inaccessibles," and that each of these implies "every universe has a successor universe."