1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ord well-ordered by ∈

  1. Dec 25, 2011 #1

    jgens

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Let Ord denote the class of ordinals. Prove that Ord is is transitive and well-ordered by ∈.

    2. Relevant equations

    • A set X is called transitive if x ∈ X and y ∈ x imply y ∈ X.
    • A set X is called ordinal if it is transitive and well-ordered by ∈.
    • A total order (X,<) is well-ordered if and only if there does not exist an infinite descending sequence.

    3. The attempt at a solution

    Suppose X ∈ Ord and fix x ∈ X. If y ∈ x, then y ∈ X since X is an ordinal. If z ∈ y, then z ∈ X by the previous sentence. Since z ∈ y and y ∈ x, it follows that z ∈ x since X is well-ordered by ∈. This proves x is transitive.

    If y ∈ x, we clearly have ~(y ∈ y). If y1,y2,y3 ∈ x are such that y3 ∈ y2 and y2 ∈ y1, then y3 ∈ y1 since ∈ is a well-order on X. If z1,z2 ∈ x, then since ∈ is well-order on X, one and only one of the following holds: 1) z1 ∈ z2 2) z1 = z2 3) z2 ∈ z1. Lastly, suppose there is some infinite descending sequence a0 ∋ a1 ∋ ... in x. But since x ∋ a0 ∋ a1 ∋ ... is an infinite descending sequence in X and X is well-order by ∈, this is a contradiction. This proves x is well-ordered by ∈. In particular, x ∈ Ord, proving that Ord is transitive.

    If X ∈ Ord, then we clearly have ~(X ∈ X). If X,Y,Z ∈ Ord are such that X ∈ Y and Y ∈ Z, then X ∈ Z since Z is transitive. We show that ∈ is a total order with the following argument:

    If X,Y ∈ Ord, then Z = X∩Y is transitive. Let x denote the least element of X-Z and let y denote the least element of Y-Z. Suppose w ∈ x but ~(w ∈ Z). Then w ∈ X since X is transitive. But then w is the least element of X-Z, which is a contradiction. This means w ∈ Z, so x is a subset of Z. Conversely, if w ∈ Z but ~(w ∈ x), then x ∈ w. But then x ∈ Z which is a contradiction. This means x = Z. Arguing similarly shows that x = Z = y. Since x = y, we should have x ∈ z, which is absurd. This means Z = X or Z = Y. In particular, we have that one and only one of the following holds: 1) X ⊂ Y 2) X = Y 3) Y ⊂ X. Now suppose X ⊂ Y = X and let y denote the least element of Y-X. Since the elements of y are precisely the elements of x, this means X ∈ Y. This proves that Ord is well-ordered by ∈.

    Now suppose X0 ∋ X1 ∋ ... is an infinite descending sequence in Ord. But then X1 ∋ X2 ∋ ... is an infinite descending sequence in X0 which contradicts the fact that X is well-ordered. This completes the proof.


    My primary question is whether or not my argument for why Ord is well-ordered can be simplified. If it can, I would love suggestions.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Ord well-ordered by ∈
  1. Order Statistics (Replies: 0)

Loading...