I have been reading the proof for the implication that if two $L$-structures are isomorphic then they are elementary equivalent. I've been wondering under what conditions we might prove some version of the converse ( I know it does not hold in general ).

Suppose T is the set of satsifable formulas by $\mathcal{A} $ . Then $ \mathcal{A} \models \phi_i(a_1, ... , a_n) \quad \forall \phi \in T$

Then I assume there exists $\mathcal{B}$ , which also has elements in its universe such that $ \mathcal{B} \models \phi_i(b_1,...,b_n) \quad \forall \phi\in T$.

Could I go about proving that that $ \exists \mathcal{ C} \subset \mathcal{B}$ such that $ \mathcal{C} \cong \mathcal{A} $ ?

I think I would do it based on the induction of the complexity of the formulas, where in order to satisfy all these formulas, the interpretations of everything would have to end up being the same?


1 Answer 1


Not even if $T$ is a complete theory.

Consider the theory of real-closed fields. It's a complete theory which has a smallest model $\Bbb Q_{\rm alg}\cap\Bbb R$, so you'd argue we can take another countable model which is not isomorphic to it, and then argue that by completeness of the theory they are elementary equivalent, but there is no substructure of the minimal model which is isomorphic to the larger one.

What about the parameters, you'd argue? Well, if $\mathcal A\models\varphi(a)$, then in particular $\mathcal A\models\exists x\varphi(x)$.

In any case, one can talk about $\Bbb R$ vs. some sufficiently saturated countable model of RCF (e.g. one that has many transcendental numbers of all sort of types), which then becomes an elementary equivalent to $\Bbb R$ even if you try to come up with some convoluted properties outside the language (but still algebraic in their core), but since the smaller field is countable, $\Bbb R$ is not isomorphic to any of its substructures at all.


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