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Let $M$ be a saturated model of a theory $T$ in a first-order language $\mathcal{L}$, and let $N$ be a submodel of $M$.

Is it possible to have a substructure $A\neq N$ of $M$ such that $N \subset A \subset M$ and every element of $A$ is definable by a formula in $L$ with parameters from $N$?

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Yes. Let $T$ be the theory of an endless discrete order, which is a complete theory. Let $M=\mathbb{Z}\cdot\mathbb{Q}$ consist of $\mathbb{Q}$ copies of the $\mathbb{Z}$ order, which is a countable saturated model of $T$, and let $N$ consist of only the even elements in each copy of $\mathbb{Z}$, which still forms a model of $T$ (though not an elementary substructure of $M$), with $A$ adding in some or all of the odd elements. Each of these new elements in $A$ is definable in $M$ from their immediate predecessors, which are in $N$.

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  • $\begingroup$ Wouldn't it work with taking any $N$ which is not an elementary submodel, and considering $A$ to be the elementary submodel generated by $N$? $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 14, 2014 at 12:14
  • $\begingroup$ Yes, that's right. The non-elementary submodels must miss some elements in one of their $\mathbb{Z}$ chains, and then $A$ would fill in those missing elements, which would be definable from the other elements of that $\mathbb{Z}$ chain. $\endgroup$
    – Joel David Hamkins
    Commented Oct 14, 2014 at 12:27
  • $\begingroup$ (I was actually talking about arbitrary languages and models, not just this one. For example $\Bbb N$ with $\leq$ and $\Bbb N\setminus\{6\}$ with $\leq$.) $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 14, 2014 at 12:33
  • $\begingroup$ Well, in general there might not be "the" elementary submodel generated by a given submodel, unless you have definable Skolem functions, which are not always available. But when they are, as here, then yes, that is what is going on with this example. $\endgroup$
    – Joel David Hamkins
    Commented Oct 14, 2014 at 12:36
  • $\begingroup$ For example, consider DLO and $\mathbb{Z}$ sitting inside $\mathbb{Q}$. This is not an elementary substructure, but we cannot add any new points definably from integers. $\endgroup$
    – Joel David Hamkins
    Commented Oct 14, 2014 at 12:37

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