Let $A,B,C$ be models and suppose that $A \cong B \preccurlyeq C$. What simple additional requirement is sufficient to entail that $A \preccurlyeq C$?


  • To see that this does not always follow, recall that $(\mathbb{N}\setminus\{0\}) \cong \mathbb{N} \preccurlyeq \mathbb{N}$ and yet $(\mathbb{N} \setminus\{0\}) \not\preccurlyeq \mathbb{N}$
  • "$\cong$" denotes model isomorphism, "$\preccurlyeq$" denotes the elementary submodel relation.
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    $\begingroup$ What makes you think there is a simple (or natural, or non-trivial) sufficient condition? $\endgroup$ – Rob Arthan Feb 16 '16 at 22:50
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    $\begingroup$ The question doesn't really make sense unless you assume that $A$ and $B$ are both substructures of $C$ to begin with. $\endgroup$ – Alex Kruckman Feb 16 '16 at 23:48
  • $\begingroup$ I don't get it. I mean, $A \cong B$ means there exists an isomorphism $f : A \rightarrow B$, right? And $B \preccurlyeq C$ means there exists an elementary embedding $g:B \rightarrow C$, right? Well, just form the composite $g \circ f$. $\endgroup$ – goblin Feb 17 '16 at 8:18
  • $\begingroup$ @goblin Usually the notation $B\preccurlyeq C$ means that $B$ is an elementary substructure of $C$, so in particular the domain of $B$ is literally a subset of the domain of $C$. Of course, it's often conceptually useful to replace this notion with the existence of a specified embedding from $B$ to $C$. As you note, there is an embedding from $A$ to $C$ obtained by composing the isomorphism with the embedding from $B$ to $C$. But the OP clearly has a situation in mind where $A$ comes with a different specified embedding (not necessarily elementary) into $C$, as evidenced by the example. $\endgroup$ – Alex Kruckman Feb 17 '16 at 17:49
  • $\begingroup$ (1) This was the point of my comment - the setup of the question should either state that $A$ is a substructure of $C$ or that it comes with a specified embedding into $C$, different from the one coming from the isomorphism, which is trivially elementary. (2) You'll run into trouble reading model theory if you misinterpret $A\preccurlyeq B$ as "there exists an elementary embedding from $A$ to $B$". Rather, you should read it as the assertion "the embedding from $A$ to $B$ (which should be specified or clear from context, e.g. an inclusion) is elementary." $\endgroup$ – Alex Kruckman Feb 17 '16 at 17:50

Here's one possible answer: Model completeness of the theory $T$. Recall that $T$ is model complete if whenever $A$ and $B$ are models of $T$ and $A$ is a substructure of $B$, then $A$ is an elementary substructure of $B$.

In your case, if $T$ is model complete, $A\cong B$, $B\preccurlyeq C$, and $A$ is a substructure of $C$, then $A\preccurlyeq C$. Of course, this is a little silly: We didn't actually use $B$, already the fact that $A$ is a substructure of $C$ which is a model of $T$ was enough.

But if you ask for this conclusion to hold for all models of $T$, model completeness is a necessary hypothesis. To see this, suppose that $T$ is not model complete. Then I can find some a model $A$ which is a substructure of a model $C$, such that $A\not\preccurlyeq C$. But by compactness, there is an elementary extension $C\preccurlyeq C'$ and $B\cong A$ such that $B\preccurlyeq C'$. But then $A$ is a substructure of $C'$ and $A\not\preccurlyeq C'$, but $A\cong B\preccurlyeq C'$, contradicting our condition on $T$.

Edit: As pointed out below in the comments, the argument above only works when $T$ is complete (the step where an isomorphic copy of $A$ is elementarily embedded in an elementary extension of $C$ uses $A\equiv C$). What the proof above actually shows is that the following are equivalent:

  1. Whenever $A$ and $B$ are substructures of $C\models T$ and $A\cong B\preccurlyeq C$, then $A\preccurlyeq C$.
  2. Every completion of $T$ is model complete. That is, whenever $A$ is a substructure of $C\models T$ and $A\equiv C$, then $A\preccurlyeq C$.
  • $\begingroup$ I can't quite follow the second part of your answer. A corollary of your argument is that $A\equiv C$. $\endgroup$ – Primo Petri Feb 19 '16 at 8:57
  • $\begingroup$ Oh, oops, I accidentally assumed that $T$ was complete! $\endgroup$ – Alex Kruckman Feb 19 '16 at 15:32

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