Elementary equivalence of ordinals - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T09:38:36Zhttp://mathoverflow.net/feeds/question/35971http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/35971/elementary-equivalence-of-ordinalsElementary equivalence of ordinalsRicky Demer2010-08-18T14:20:11Z2010-08-18T17:25:14Z
<p>What is the smallest ordinal alpha which is elementarily equivalent to some smaller ordinal beta with the signature (<)?</p>
<p>What is the corresponding ordinal beta?</p>
<p>What if we instead require that beta be an elementary substructure of alpha?</p>
http://mathoverflow.net/questions/35971/elementary-equivalence-of-ordinals/35982#35982Answer by Andreas Blass for Elementary equivalence of ordinalsAndreas Blass2010-08-18T16:23:13Z2010-08-18T16:23:13Z<p>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 <code>$\omega^\omega$</code> and <code>$\omega^\omega\cdot2$</code>. 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). </p>