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What is the smallest ordinal alpha which is elementarily equivalent to some smaller ordinal beta with the signature (<)?

What is the corresponding ordinal beta?

What if we instead require that beta be an elementary substructure of alpha?

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2 Answers 2

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The first-order theory of well-orderings was studied in great detail in a paper of Doner, Mostowski, and Tarski, "The elementary theory of well-ordering -- a metamathematical study" [Logic Colloquium '77, edited by A. Macintyre, L. Pacholski, and J. Paris, North-Holland (1978) pp. 1-54]. In particular, their Corollary 44 characterizes (unless their notation is very non-standard --- I haven't checked carefully) when two ordinals are elementarily equivalent. Modulo an apparent typo in the definition just before the corollary (one of the strict inequalities should be non-strict), it seems that the first pair of distinct but elementarily equivalent ordinals is $\omega^\omega$ and $\omega^\omega\cdot2$. A thorough reading of the paper (which I don't have time for right now) should also reveal the answer to your second question, about elementary submodels (probably the same pair of ordinals).

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    $\begingroup$ I think that any ordinal is elementary equivalent to some ordinal less than $\omega^\omega 2,$ but I think none of these should be elementarily equivalent. $\endgroup$ Aug 18, 2010 at 17:45
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As far as I can see the problems are fully discussed in "The elementary theory of well orderings" by Doner, Mostowski and Tarski mentioned already by Andreas Blass. Lemma 44 covers as Andreas Blass explained the elementary equivalence and Lemma 47 there answers the part on elementary substructures.

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