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Transitivity of Element of

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Let A, B, and C be sets. Assume the standard ZFC axioms.

    Please see below for my updated question.


    Thanks.
     
    Last edited: Sep 26, 2012
  2. jcsd
  3. Sep 26, 2012 #2

    Dick

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    A ε B means the set A is a element of the set B, right? That's not the same thing as A being a subset of B. You have sets that are elements of other sets. Try to construct a counterexample.
     
  4. Sep 26, 2012 #3
    That's what I was afraid of. How then can I show that if A ε B and B ε C and C ε A, that C ε C (yields the contradiction I require).
     
  5. Sep 26, 2012 #4

    Dick

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    Pick A={a}. Pick B={A}={{a}}. So A ε B. Correct so far? Pick C={B}={{{a}}}. So B ε C. Is A ε C?? This is a little tricky. a is not equal to the set consisting of a.
     
  6. Sep 26, 2012 #5
    Yes I understand this. But how does the chain of set inclusions I mention lead to the fact that C ε C?

    (Thanks for all your help!)
     
  7. Sep 26, 2012 #6

    Dick

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    You seem to have deleted the question. I thought I suggested you try to prove that A ε B and B ε C doesn't necessarily imply A ε C? Since inclusion is not transitive?
     
  8. Sep 26, 2012 #7
    I agree. I am now asking modified question: How then can I show that if A ε B and B ε C and C ε A, that C ε C (yields the contradiction I require).

    Apologies for the confusion.
     
  9. Sep 26, 2012 #8

    Dick

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    I think you are reaching way too far for a contradiction. C ε C puts you squarely in Bertrand-Russell paradox territory. A simple example of A ε B and B ε C with simple sets and A not an element of C will serve nicely. I basically gave you one. Follow it up. It shows inclusion isn't transitive. That's all you need.
     
  10. Sep 26, 2012 #9
    This is a separate question. I absolutely want to be in such territory. I need to show that the aforementioned set inclusions yield C ε C and then I can invoke the axiom of regularity/foundation to show it is a contradiction.
     
  11. Sep 26, 2012 #10

    Dick

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    I don't think so. Inclusion ISN'T transitive. You can't say A ε B and B ε C implies A ε C. At all.
     
  12. Sep 26, 2012 #11
    Hmm....

    Then how do I go about using the axiom of regularity to prove that no set membership loops like that I described exist?
     
  13. Sep 26, 2012 #12

    Dick

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    C ε C violates regularity. But the statement C ε C doesn't follow from anything you've said before because inclusion isn't transitive.
     
  14. Sep 26, 2012 #13
    Yes but I want to show that A ε B and B ε C and C ε A violates regularity.
     
  15. Sep 26, 2012 #14

    Dick

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    Mmm. I'm not all that hot with set axiomatics. But you can simplify that. Suppose A ε B and B ε A, can you show that violates regularity?? Like I say, I don't have ZFC axioms at my fingertips.
     
  16. Sep 26, 2012 #15
    Then I can, but that doesn't help me.

    See http://en.wikipedia.org/wiki/Axiom_of_regularity

    It states:

    Every non-empty set A contains an element B which is disjoint from A.

    I appreciate all of your help.
     
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