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Exercise 6.1.2
is not isomorphic to (in the well-ordering by ).
Answers
Proof. Since and clearly is a proper subset of (since but ). Now consider any and any . Then clearly also so . Thus is an initial segment of . Then, since it has already been shown that both and are well-ordered sets, it follows from Corollary 6.1.5a that they cannot be isomorphic. □