Suppose $T$ is a first-order theory whose signature contains $(+,\cdot,0,1,<)$ as well as a unary predicate $R(x)$. Suppose every finite subset $S \subseteq T$ has a model in which the set of elements satisfying $R(x)$ forms a substructure isomorphic to the field of real numbers. Does it follow that $T$ itself has a model in which the set of elements satisfying $R(x)$ forms a substructure isomorphic to the field of real numbers?
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No. Suppose the signature of T contains a distinguished symbol $\omega$, and $T$ contains the statements $R(\omega)$ and the infinitely many statements $1+\cdots+1<\omega$. Then any finite subset of $T$ has a model where $R$ is isomorphic to the reals and $\omega$ is interpreted as some large enough real. But in any model of the entire theory $T$, $\omega$ has to be interpreted as something larger than any real number. |
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No. Just let the signature contain lots (more than cardinality continuum) of constants and let the axioms of T be that all the constants are different and R holds for all of them. |
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