Suppose you have a saturated model N of a complete theory T without finite models. How is it possibile to construct a proper saturated elementary substructure of N of the same cardinality of N ?
I tried to use a construction similar to the donward lowheneim-skolem theorem but I can not prove the saturation ...


closed as off-topic by Will Jagy, Emil Jeřábek, Stefan Kohl, Dima Pasechnik, Alex Degtyarev Apr 9 '15 at 19:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Emil Jeřábek, Stefan Kohl, Dima Pasechnik, Alex Degtyarev
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Try, instead, to give a proper elementary superstructure of the same cardinality which is also saturated. Then check why this is equivalent to what you want, using the fact that saturated models of the same cardinality must be isomorphic. $\endgroup$ – Richard Rast Apr 9 '15 at 16:11
  • $\begingroup$ I understand what you mean but I think that the existence of satured superstructure of the same cardinality is related to that the cardinality of N is an inaccessible cardinal using $\endgroup$ – eagle Apr 9 '15 at 16:35
  • 2
    $\begingroup$ You certainly don't need to use inaccessible cardinals. $\endgroup$ – Richard Rast Apr 9 '15 at 17:01
  • 4
    $\begingroup$ Take any elementary extension $M$ of $N$ of the same cardinaility. There is an elementary embedding of $M$ into $N$. Thus $N$ has an isomorphic proper elementary submodel, or, equivalently an isomorphic proper elementary extentsion. $\endgroup$ – Dave Marker Apr 9 '15 at 17:15
  • 1
    $\begingroup$ crossposted from math.stackexchange.com/questions/1227098/… $\endgroup$ – Will Jagy Apr 9 '15 at 18:48