Prove Order Isomorphism: α=β If (α,∈) & (β,∈)

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SUMMARY

The discussion centers on proving the order isomorphism between ordinals α and β under the relation (α,∈) and (β,∈). It is established that these orderings are isomorphic if and only if α equals β. The user has successfully demonstrated that the class Ord is transitive and well-ordered by ∈. A lemma regarding order-preserving maps is introduced, and the user acknowledges the utility of transfinite induction in proving the lemma, specifically addressing the condition where z ≤ f(z) does not hold.

PREREQUISITES
  • Understanding of ordinal numbers and their properties
  • Familiarity with the concept of order-preserving maps
  • Knowledge of transfinite induction techniques
  • Basic comprehension of set theory and the relation ∈
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  • Study the properties of ordinal numbers in set theory
  • Learn about transfinite induction and its applications in proofs
  • Explore order-preserving maps and their significance in order theory
  • Investigate the implications of well-ordering in the context of ordinals
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Mathematicians, logicians, and students of set theory interested in understanding ordinal isomorphism and the foundational principles of order theory.

jgens
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I am trying to prove the following results: If α and β are ordinals, then the orderings (α,∈) and (β,∈) are isomorphic if and only if α = β.

So far, I have only proved that the class Ord is transitive and well-ordered by ∈. I can prove this result with the following lemma: If f:α→β is an order-preserving map, then z ≤ f(z). However, I am having difficulty proving this lemma without something like transfinite induction. Any help?
 
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Assume that z\leq f(z) does NOT hold. Then there is a least a such that f(a)<a. Take f of both sides.
 
That works perfectly! I am silly for not thinking of something like that. Thanks!
 

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