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Let α and β be two von Neumann ordinals

  1. Jan 7, 2010 #1
    (a) Let α and β be two von Neumann ordinals. Show that α ⊂ β if and only if α ∈ β.

    (b) Show that the Axiom of Foundation implies that a transitive set which is linearly ordered by ∈ is an ordinal


    I can't seem to follow through this properly, any help?
     
  2. jcsd
  3. Jan 7, 2010 #2
    Re: ordinals

    What's your definition of an ordinal?
     
  4. Jan 7, 2010 #3

    EnumaElish

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    Re: ordinals

    (a) Assume the definition is "any ordinal is defined by the set of ordinals that precede it," as in Wikipedia (paraphrasing John von Neumann). Using the natural numbers as a special case, and using notation "[x]" to mean "do you think x is in there somewhere?" or (alternatively) "assume it is," I can write A = {0, 1, ..., A - 1} and B = {0, 1, ..., [A,] ..., B-1}.

    Suppose:
    (i) A is an element of B
    (ii) A is not a proper subset of B

    Show contradiction.

    Now suppose:
    (i) A is a proper subset of B
    (ii) A is not an element of B

    Show contradiction.

    (b) What is a transitive set? What does it mean that it is linearly ordered by ∈ ? Can you come up with an example? Now assume the set in your example is not an ordinal. How do you write "A is not an ordinal" using logical (set theoretic) symbols? Why does this contradict the Axiom of Foundation?
     
    Last edited: Jan 7, 2010
  5. Jan 8, 2010 #4
    Re: ordinals

    Don't forget that there are two distinct types of ordinal involved in the proof: successor ordinals and limit ordinals, and you must give distinct arguments for each.
     
  6. Jan 12, 2010 #5
    Re: ordinals

    For part i, is it a contradiction since von neumann ordinals are totally ordered by ∈?
     
  7. Jan 12, 2010 #6
    Re: ordinals

    For part ii i was thinking

    let x be a transitive set which is linearly ordered by ∈. We need to prove that the order is a well-ordering. If not then there is some subset y⊆x which has no ∈ minimal element. Then we have an infinite ∈ chain ∈ an ∈ an-1 ∈... ∈ a1 ∈ y which contradicts the axiom of foundation as it implies that there are no sets ai such that x ∈ a1 ∈ a2 ∈...∈ an and there are no infinitetly descending ∈ chains in x.

    Does that make sense?
     
  8. Jan 13, 2010 #7

    EnumaElish

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    Re: ordinals

    Suppose B = {0, 1, ..., A, ..., B-1}; why does this imply A is a subset of B?
     
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