Let $M$ be a structure in some language $L$, and $p$ be a nonprincipal complete type with respect to the full theory $T$ of $M$. If you apply the omitting type theorem, we obtain another $L$-structure $N$ omitting $p$.

Usually, the omitting type theorem is proved by expanding $L$ with constants and extending $T$ in stages, ensuring that the constants both name elements of the Skolem hull and avoid realizing $p$. We use compactness at the end. So at least prima facie the only similarity between $N$ and $M$ are that they are models of the same complete theory.

Can we omit $p$ so we obtain $N$ that is "related" to $M$, e.g., in terms of other types that they realize? For instance, can we take $N$ to be an elementary substructure of $M$ under some circumstance?


1 Answer 1


The question is a bit vague (what does "related" mean?), so I'm not sure it has a definitive answer. But I'll make an easy observation on the positive side and then give you an example on the negative side.

If the isolated types are dense relative to $T$, then $T$ has a prime model, which omits all the non-isolated types and embeds elementarily into every model of $T$. So in this case yes, you can start with a model $M$ and a non-isolated type $p$ and omit $p$ in an elementary substructure.

But you can't do this in general. Consider the language $\{P_n\mid n\in\omega\}\cup \{f_n\mid n\in \omega\}$, where each $P_i$ is a unary relation symbol and each $f_i$ is a unary function symbol. Let $T$ consist of the following axioms:

  • $\exists x\, \top$ (just to exclude the empty structure)
  • $\forall x\, \left(P_n(x)\leftrightarrow \lnot P_n(f_n(x)) \right)$, for each $n$
  • $\forall x\, \left(P_n(x)\leftrightarrow P_n(f_m(x))\right)$ for each $n$ and each $m\neq n$.
  • $\forall x\, (f_n(f_n(x)) = x \land f_n(f_m(x)) = f_m(f_n(x)))$

So each element $a$ of a model $M$ of $T$ encodes a binary sequence $\eta_a$ (by which of the $P_n$ holds of $a$), and the function $f_n$ flips the $n^{th}$ bit. In fact, the substructure $\langle a\rangle$ of $M$ generated by $a$ contains exactly one element $b$ encoding each binary string $\eta_b$ which agrees with $\eta_a$ eventually.

It shouldn't be too hard to check that $T$ is complete and has quantifier elimination. As a consequence, the type of an element $a$ is determined by $\eta_a$. And the structure $M_a = \langle a\rangle$ is prime over $a$ (in fact prime over any element of $M_a$) and realizes exactly those types $\eta_b$ which agree with $\eta_a$ eventually. So no model omitting $\eta_a$ shares any realized type in common with $M_a$. There are continuum-many models of $T$ up to isomorphism which are minimal for elementary embeddings (one for each equivalence class in $2^\omega$ under the eventual agreement equivalence relation), and there are no embeddings between them.

  • $\begingroup$ Thank you; it was exactly a negative example like yours that I wanted! $\endgroup$
    – Pteromys
    Apr 20, 2017 at 4:25

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