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Simply set theory question: all standard forms of ZFC imply power set of {naturals}?

  1. May 23, 2012 #1
    Is it true that for every standard formulation T of ZFC, T ⊢ the power set of {naturals}?

    After all, the empty set axiom and the pairing axiom are in T, and so we get N. Then by the power set axiom we get P(N).
     
  2. jcsd
  3. May 23, 2012 #2
    Re: Simply set theory question: all standard forms of ZFC imply power set of {natural

    The existence of the empty set and the pairing axiom does not give us the existence of the natural numbers. Indeed: all natural numbers may exist that way, but perhaps they will not be contained in a set!
    For the existence of a set of natural numbers, ZFC has included a special axioms that gives us that: the existence of an infinite set. Together with that, we can prove that the natural numbers exist. And by the power set axiom, also P(N) exists. So the answer to your question is yes.
     
  4. May 23, 2012 #3
    Re: Simply set theory question: all standard forms of ZFC imply power set of {natural

    So Infinity, Empty-Set and Pairing are jointly sufficient and individually necessary for P(N)?
     
  5. May 23, 2012 #4
    Re: Simply set theory question: all standard forms of ZFC imply power set of {natural

    And the power set axiom, of course.
     
  6. May 23, 2012 #5
    Re: Simply set theory question: all standard forms of ZFC imply power set of {natural

    Whoops. Right, thanks.
     
  7. May 29, 2012 #6
    Re: Simply set theory question: all standard forms of ZFC imply power set of {natural

    One thing to note here is that the power set [itex]P(\mathbb{N})[/itex] might not be the same in all models, some may contain only some of the subsets.

    (indeed, it's also possible to create models where [itex]\mathbb{N}[/itex] is different, but that's much less common)
     
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