Suppose that $\mathcal{M} \preccurlyeq \mathcal{N}$. Then by definition we have that $\mathcal{M}$ is a substructure of $\mathcal{N}$ s.t. for any (possibly empty) tuple $\overline{a}$ from $M^n$ and for any $\phi \in L$, we have that $\mathcal{N} \models \phi(\overline{a})$ implies that $\mathcal{M} \models \phi(\overline{a})$.

Now we aim to show that $\mathcal{M} \equiv \mathcal{N}$. This is true iff $Th(\mathcal{M}) = Th(\mathcal{N})$. I can only think of how to show that $Th(\mathcal{M}) \supseteq Th(\mathcal{N})$. This is done as follows: let $\phi \in Th(\mathcal{N})$. Then for any tuple $\overline{a} \in M^n$, we have that $\mathcal{N} \models \phi(\overline{a})$. Among these tuples are the empty tuple, so that we can say

$$ \mathcal{N} \models \phi() $$

so that of course

$$ \mathcal{N} \models \phi $$

and hence applying the definition of $\preccurlyeq$ we get:

$$ \mathcal{M} \models \phi() \implies \mathcal{M} \models \phi $$

Then we've established that $Th(\mathcal{N}) \subseteq Th(\mathcal{M})$. But how do we show that $Th(\mathcal{M}) \subseteq Th(\mathcal{N})$?


Suppose $\phi\in\text{Th}(\mathcal{M})$. Then $\lnot\phi\not\in\text{Th}(\mathcal{M})$. So by your argument, $\lnot\phi\not\in\text{Th}(\mathcal{N})$, and hence $\phi\in\text{Th}(\mathcal{N})$.

More generally, if $T$ and $T'$ are complete theories, then $T\subseteq T'$ implies $T = T'$.

Comment: The definition of $\preccurlyeq$ is often taken to be $\mathcal{N}\models \phi(\overline{a})$ if and only if $\mathcal{M}\models\phi(\overline{a})$, for $\overline{a}$ from $\mathcal{M}$. This is equivalent, of course.

  • $\begingroup$ So we're here making use of the fact that $Th(\mathcal{N})$ is complete, correct? And this is because the theory of any one particular model is complete (but not necessarily the theory of a set of more than one models), correct? $\endgroup$ – user1770201 Nov 7 '13 at 20:18
  • $\begingroup$ @user1770201: this is a general fact: any extension of a complete theory (in the same language) is either trivial or inconsistent. $\endgroup$ – tomasz Nov 7 '13 at 23:44
  • $\begingroup$ @user1770201 That's right. $\endgroup$ – Alex Kruckman Nov 8 '13 at 2:49

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