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cody
cody
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Note that Reinhardt cardinals do not provably exist in ZFC. If they were, it would contradict Gödel's theorem. It is the largest cardinal currently defined which is believed to be consistent with ZFC.

As Mozibur notes, you can't have a largest such cardinal, since given a consistent extension of ZFC, you can always (in theory) find a stronger theory which proves the existence of larger cardinals. However, it may be found next week that Reinhard cardinals are actually not consistent with ZFC. That's the tragedy of the incompleteness theorem.

It turns out, however, that if ZFC is consistent, there is a smallest cardinal which is not provably such in ZFC. To show this, you can simply consider the set of all uniquely defined syntactic objects which ZFC proves to be cardinals, and take the smallest cardinal not in that set.

Edit: I missed the fact that Reinhardt cardinals are inconsistent with Choice. You can replace ZFC by ZF everywhere in my comment though, or Reinhardt cardinals by some smaller cardinal numbers (superhuge for instance).

Note that Reinhardt cardinals do not provably exist in ZFC. If they were, it would contradict Gödel's theorem. It is the largest cardinal currently defined which is believed to be consistent with ZFC.

As Mozibur notes, you can't have a largest such cardinal, since given a consistent extension of ZFC, you can always (in theory) find a stronger theory which proves the existence of larger cardinals. However, it may be found next week that Reinhard cardinals are actually not consistent with ZFC. That's the tragedy of the incompleteness theorem.

It turns out, however, that if ZFC is consistent, there is a smallest cardinal which is not provably such in ZFC. To show this, you can simply consider the set of all uniquely defined syntactic objects which ZFC proves to be cardinals, and take the smallest cardinal not in that set.

Note that Reinhardt cardinals do not provably exist in ZFC. If they were, it would contradict Gödel's theorem. It is the largest cardinal currently defined which is believed to be consistent with ZFC.

As Mozibur notes, you can't have a largest such cardinal, since given a consistent extension of ZFC, you can always (in theory) find a stronger theory which proves the existence of larger cardinals. However, it may be found next week that Reinhard cardinals are actually not consistent with ZFC. That's the tragedy of the incompleteness theorem.

It turns out, however, that if ZFC is consistent, there is a smallest cardinal which is not provably such in ZFC. To show this, you can simply consider the set of all uniquely defined syntactic objects which ZFC proves to be cardinals, and take the smallest cardinal not in that set.

Edit: I missed the fact that Reinhardt cardinals are inconsistent with Choice. You can replace ZFC by ZF everywhere in my comment though, or Reinhardt cardinals by some smaller cardinal numbers (superhuge for instance).

Source Link
cody
cody
  • 221
  • 2
  • 10

Note that Reinhardt cardinals do not provably exist in ZFC. If they were, it would contradict Gödel's theorem. It is the largest cardinal currently defined which is believed to be consistent with ZFC.

As Mozibur notes, you can't have a largest such cardinal, since given a consistent extension of ZFC, you can always (in theory) find a stronger theory which proves the existence of larger cardinals. However, it may be found next week that Reinhard cardinals are actually not consistent with ZFC. That's the tragedy of the incompleteness theorem.

It turns out, however, that if ZFC is consistent, there is a smallest cardinal which is not provably such in ZFC. To show this, you can simply consider the set of all uniquely defined syntactic objects which ZFC proves to be cardinals, and take the smallest cardinal not in that set.