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For all S, for all a in S, |a|<>|S|. Proof?

  1. Dec 17, 2013 #1
    Obviously an element of a set S cannot have the same cardinality as S. I would be grateful for an elegant proof of this.
    Thanks
     
  2. jcsd
  3. Dec 17, 2013 #2
    What you're trying to prove is not true. For example: ##S=\{\{0\}\}##.
     
  4. Dec 17, 2013 #3
    Thanks, R136a1. Oops. Silly of me. I was trying to figure out why any class which has the same cardinality as a proper class is also a proper class. So this attempt was a bad one; perhaps you (or someone) can send me along a better path? Thanks again.
     
  5. Dec 18, 2013 #4
    The replacement axiom should be helpful here. It says (roughly) that any image of a set under a "function" is also a set.
     
  6. Dec 18, 2013 #5
    Thanks again, R136a1. I presume you are indicating the contrapositive of Replacement: from "if S is a set then its image will be a set" to "if the image of S under f is not a set, then S is not a set." On the surface that would seem to work, except that one could not talk about proper classes in ZFC: one would either need some theory where one could talk of classes, or rephrase my question something like this: "Suppose we have two models N and M such that N is a (non-conservative) extension of M, such that the class S is a set under N but not under M, with respect to which S is a proper class. Then we can talk about the cardinality of S (having it understood that we are working under N). So, if we have such a class S, and another class T which is a set under N such that |S|=|T|, then T is also a proper class with respect to M." I'm not sure which would be better; in any case I am still not sure how to prove it in these contexts. Any suggestions?
     
  7. Dec 18, 2013 #6
    I think it would be much easier to talk about an axiomatic theory that allows classes, such as NBG.

    But besides that, you should be able to state your theorem in ZFC too. Proper classes makes sense in ZFC, but they are not actual objects, rather they are shorthands for formulas. For example, see Jech's set theory: in the first chapter he shows how to work with classes in ZFC. In that sense, you can phrase and prove your question.
     
  8. Dec 20, 2013 #7
    Thanks for the suggestions, R136a1. I worked on it and came up with the conclusion that the statement is false, since, for example, ZFC (which dictates that the class of natural numbers is a set) has countable models (thanks for the Löwenheim-Skolem downward Theorem) and of course the universe of such a model is also countable. So my question is answered in the negative, as far as I can see.
     
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