4 ms·

I’m uncertain what your point is. Like I said, in ZFC, there is no bijection from A to its power set. I don’t think anything you’ve stated contradicts that. Are

I’m uncertain what your point is. Like I said, in ZFC, there is no bijection from A to its power set. I don’t think anything you’ve stated contradicts that. Are there alternative axiom systems in which you can construct such a bijection?

cubefox 3y ago
There is no bijection (surjection, to be more general) from A to P inside your ZFC model even if both A and P are assumed to be countable. The bijection we talk about here would be just another object inside the model. So you can't infer from the lack of such an object that P is uncountable.
- openasocket 3y ago
  Yeah, Cantors theorem is a theorem written in the language of ZFC set theory. In other axiom systems it may not be true. But you can also say that about literally every theorem beyond, like, modus ponens. Is that the point you are trying to make?
  - cubefox 3y ago
    The point is simply that it doesn't imply the existence of uncountable sets in ZFC. Modus ponens is different. The semantic entailment relation ⊨ between (P→Q, P) and Q is valid iff there is no model such that the former is true and the latter is false. Which is a sentence in plain English. It doesn't require the existence of some object that is mapping the premises to the conclusion inside the model. It doesn't even require the existence of a syntactic deduction rule (⊢) that tells you from (P→Q, P) to infer Q.
    - ginnungagap 3y ago
      It is a theorem of ZFC that uncountable sets exist and every model of ZFC will have a set that the model believes to be uncountable. It doesn't matter than the metatheory might believe that model to be countable (why should the metatheory have the correct notion of what it means to be countable anyway?).
      - cubefox 3y ago
        > It is a theorem of ZFC that uncountable sets exist This is simply false, as I already explained. > and every model of ZFC will have a set that the model believes to be uncountable. That is something else. (And I wouldn't use the nebulous term "believes" here, it's just that the model lacks an object which maps A to P.) > It doesn't matter than the metatheory might believe that model to be countable (why should the metatheory have the correct notion of what it means to be countable anyway?). "The meta theory" here is simply sentences expressed in natural language, or beliefs held by people expressing those sentences. It is the language in terms of which everything formal is ultimately defined. It's the only thing that ultimately matters.
        openasocket 3y ago
        It is absolutely not false! This is taught in every undergraduate set theory course. Please point to me the step in the above proof where there is an error.
        cubefox 3y ago
        Certainly not in every undergraduate class, though I don't doubt that the subtleties around these issues may often be taught wrong. I already did point you to the errors, and I included the reference to Shapiro's book.
        openasocket 3y ago
        You keep making vague references to concepts without showing how they even remotely contradict Cantors theorem. You explained to me the power set axiom, which, thanks, I guess? But I don’t understand what your point is. Are you claiming X is not in the power set of the natural numbers? X is, unambiguously, a set. And it is clearly a subset of A. Therefore, it is in the power set. If you don’t understand that, I think you need to review the power set axiom in ZFC. Then you said “Cantor's theorem stating that there is no mapping f from A onto P merely means that the mapping f itself can't exist inside a model of ZFC”. Which is literally identical to saying “under ZFC, there are uncountable sets”. You just don’t like it because that statement isn’t wrapped in eight layers of indirection with model theory.

cubefox 3y ago

There is no bijection (surjection, to be more general) from A to P inside your ZFC model even if both A and P are assumed to be countable. The bijection we talk about here would be just another object inside the model. So you can't infer from the lack of such an object that P is uncountable.

openasocket 3y ago
Yeah, Cantors theorem is a theorem written in the language of ZFC set theory. In other axiom systems it may not be true. But you can also say that about literally every theorem beyond, like, modus ponens. Is that the point you are trying to make?
- cubefox 3y ago
  The point is simply that it doesn't imply the existence of uncountable sets in ZFC. Modus ponens is different. The semantic entailment relation ⊨ between (P→Q, P) and Q is valid iff there is no model such that the former is true and the latter is false. Which is a sentence in plain English. It doesn't require the existence of some object that is mapping the premises to the conclusion inside the model. It doesn't even require the existence of a syntactic deduction rule (⊢) that tells you from (P→Q, P) to infer Q.
  - ginnungagap 3y ago
    It is a theorem of ZFC that uncountable sets exist and every model of ZFC will have a set that the model believes to be uncountable. It doesn't matter than the metatheory might believe that model to be countable (why should the metatheory have the correct notion of what it means to be countable anyway?).
    - cubefox 3y ago
      > It is a theorem of ZFC that uncountable sets exist This is simply false, as I already explained. > and every model of ZFC will have a set that the model believes to be uncountable. That is something else. (And I wouldn't use the nebulous term "believes" here, it's just that the model lacks an object which maps A to P.) > It doesn't matter than the metatheory might believe that model to be countable (why should the metatheory have the correct notion of what it means to be countable anyway?). "The meta theory" here is simply sentences expressed in natural language, or beliefs held by people expressing those sentences. It is the language in terms of which everything formal is ultimately defined. It's the only thing that ultimately matters.
      - openasocket 3y ago
        It is absolutely not false! This is taught in every undergraduate set theory course. Please point to me the step in the above proof where there is an error.
        cubefox 3y ago
        Certainly not in every undergraduate class, though I don't doubt that the subtleties around these issues may often be taught wrong. I already did point you to the errors, and I included the reference to Shapiro's book.
        openasocket 3y ago
        You keep making vague references to concepts without showing how they even remotely contradict Cantors theorem. You explained to me the power set axiom, which, thanks, I guess? But I don’t understand what your point is. Are you claiming X is not in the power set of the natural numbers? X is, unambiguously, a set. And it is clearly a subset of A. Therefore, it is in the power set. If you don’t understand that, I think you need to review the power set axiom in ZFC. Then you said “Cantor's theorem stating that there is no mapping f from A onto P merely means that the mapping f itself can't exist inside a model of ZFC”. Which is literally identical to saying “under ZFC, there are uncountable sets”. You just don’t like it because that statement isn’t wrapped in eight layers of indirection with model theory.