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Do the infinite cardinals correspond to sets?

  1. Sep 26, 2011 #1
    For the infinite cardinal numbers (of which there are infinitely many), do they each necessarily correspond to some set? I mean we know that aleph-naught corresponds to N, c (aleph-one by continuum hypothesis) corresponds to R, but what about all the other infinite cardinals?

    Is it possible to construct a set of each of those cardinalities?

    Thanks for any help you can provide.
     
  2. jcsd
  3. Sep 26, 2011 #2

    Hurkyl

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    Yes, more or less by the definition of "cardinal number".
     
  4. Sep 26, 2011 #3

    disregardthat

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    Not necessarily, there are axioms of inaccessible cardinals which does not correspond to sets in ZFC which can be constructed.
     
  5. Sep 26, 2011 #4
    Perhaps we have chosen the wrong definition? Anyway I'm basically just wondering if this is still an open question in mathematics or if it has already been proven or disproven. Are the "sets" of these cardinalities something specific that we know to exist, or just something that exists merely in the realm of theory?
     
  6. Sep 26, 2011 #5
    Could you elaborate a little more please?
     
  7. Sep 26, 2011 #6

    Hurkyl

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    No -- the intent of cardinal numbers is to quantify sets up to bijection, so it is desirable that every cardinal number corresponds to a set, and every set has a cardinality.


    Without diving into whatever philosophical notions you're having, I'll just comment that our knowledge that a cardinal number exists is equivalent to our knowledge that a set exists of that cardinality.
     
  8. Sep 26, 2011 #7

    Fredrik

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    There's no difference. When mathematicians say that a set with property P "exists", it only means that by the axioms of some set theory, and the axioms of an associated proof theory, the sentence [itex]\exists x~~ P(x)[/itex] is a theorem.
     
  9. Sep 27, 2011 #8
    I thought existence implied that there is a model for the sentence, or so that

    (equivalently, I think), the existence does not contradict the rest of the theory.
     
  10. Sep 27, 2011 #9

    Fredrik

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    I don't have a lot of experience with this stuff, so I might have it wrong. But regarding your two statements: When you say "...there is a model...", you're using the word "is" in the sense "exists", so I don't see how what you said contains any information. It's like saying that "all we have to do to prove that there exists an odd integer, is to prove that there exists an odd integer". Can you rephrase your statement without using the word "is" like that?

    The second statement looks wrong to me. Consider "there exists a set such that the continuum hypothesis is true". This doesn't contradict ZFC, but that doesn't mean that there exists such a set in ZFC.
     
  11. Sep 27, 2011 #10

    Fredrik

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    According to Wikipedia an "inaccessible cardinal" is just a cardinal that can't be reached by repeated power set operations. Are you saying that this implies that they aren't sets?

    Actually, now that I think about it, the idea of a cardinality that's not associated with any set doesn't make sense. "A and B have the same cardinality" means that there's a bijection from A onto B. The concept is only defined for sets.
     
  12. Sep 28, 2011 #11

    lavinia

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    ordinals can be generated sequentially. This was Cantor's original method. This generating process can be done with sets starting with the null set.
     
  13. Sep 28, 2011 #12
    Note the OP's question:
    So disregardthat seems to be saying that we cannot construct a set (in ZFC) having an inaccessible cardinal as its cardinality.
     
  14. Sep 28, 2011 #13

    Hurkyl

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    There's a bit of a technicality that may or may not cause problems.

    Depending upon what specifically you mean by "constructible", it may be that inaccessibility means that ZFC can't prove a particular construction results in a set that is an ordinal, instead of a proper class that is the class of all ordinals.
     
  15. Oct 4, 2011 #14
    Not sure if this sheds light on the question of the OP but there are theorems of finite combinatorial mathematics which cannot be proved without an additional axiom of inaccessible large cardinals.

    "Finite functions and the neccessary use of large cardinals" Harvey M. Friedman, Annals of Mathematics, 148 (1998) 803-893. http://arxiv.org/abs/math/9811187

    Skippy

    PS I've tried reading it twice in the past ten years. Can't get past the first 20 pages or so without hurting my brain so I can't explain the argument.
     
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