Consider the class $K$ of well-ordered sets $(W,\leq)$. Although that class is not first-order axiomatizable, it has an associated first-order theory $Th(K)$. Now consider the class of ordinals $On$, where each ordinal $\alpha \in On$ is equipped with the $\leq$ relation. Is the first-order theory of $On$ the same as the first-order theory of $K$? I apologize if this question is too elementary.
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1$\begingroup$ I'm confused about what $Th(K)$ is. Does the language have a symbol $\le$ that compares well-ordered sets with each other? If so, how is it defined/interpreted? And does the language include equality? $\endgroup$– KarlCommented May 19 at 18:52
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Yes, since every ordinal is a well-ordered set, and every well-ordered set is isomorphic to an ordinal. So the classes are the same (up to isomorphism).
answered May 19 at 17:32
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1$\begingroup$ Naively, couldn't the theories differ because there's only one ordinal per isomorphism class? Are we considering a language with equality? $\endgroup$– KarlCommented May 19 at 18:46
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3$\begingroup$ @Karl we're not somehow thinking of the whole class as a single first-order structure. When $K$ is a class of $L$-structures, $\mathrm{Th}(K)$ is the set of all $L$-sentences true of all structures in $K$. In this case, $L=\{\leq\}$ is the language of order in first-order logic with equality. $\endgroup$ Commented May 19 at 19:05