most recent 30 from http://mathoverflow.net 2013-05-20T03:44:49Z http://mathoverflow.net/feeds/question/50165 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50165/compactness-theorem-with-preserved-substructure Bobby Kleinberg 2010-12-22T15:30:24Z 2010-12-22T16:06:43Z

<p>Suppose $T$ is a first-order theory whose signature contains $(+,\cdot,0,1,&lt;)$ 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?</p>

http://mathoverflow.net/questions/50165/compactness-theorem-with-preserved-substructure/50169#50169 Chris Eagle 2010-12-22T16:06:25Z 2010-12-22T16:06:25Z

<p>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.</p>

http://mathoverflow.net/questions/50165/compactness-theorem-with-preserved-substructure/50170#50170 Henry Towsner 2010-12-22T16:06:43Z 2010-12-22T16:06:43Z

<p>No. Suppose the signature of T contains a distinguished symbol $\omega$, and $T$ contains the statements $R(\omega)$ and the infinitely many statements $1+\cdots+1&lt;\omega$. 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.</p>