Let $\mathcal{M}$ be an $\mathcal{L}$-structure and define $S_\omega^\mathcal{M}(A)$ be the set of complete $\omega$-types with parameters in $A$ (ie the set of complete types in $\omega$-many variables). Is it true that if $\mathcal{M}$ is $\kappa$-saturated for some $\kappa>|A|$ then every type in $S_\omega^\mathcal{M}(A)$ will be realized in $\mathcal{M}$?

I think the answer is yes. Since the case where $A$ is nonempty is similar, for simplicity we will just consider $A=\emptyset$, so suppose $\mathcal{M}$ is weakly saturated and $\Sigma(v_n)_{n\in\omega}\in S_\omega^\mathcal{M}(\emptyset)$. We define a sequence of elements $m_l\in M$ for all $l\in\omega$ by induction, such that each $\Sigma(m_0,...,m_l,v_n)_{l<n\in\omega}$ is consistent with $\mathcal{M}$:

Let $\Delta_0(v_0)=\exists(v_n)_{0<n\in\omega}\Sigma(v_n)_{n\in\omega}:=\{\exists v_{n_1}...v_{n_k}\varphi(v_0,v_{n_1},...,v_{n_k}):\varphi(v_0,v_{n_1},...,v_{n_k})\in\Sigma(v_n)_{n\in\omega}\text{ and each }n_i>0\}$. ($v_0$ may or may not appear in each $\varphi$, of course.) Certainly $\Delta_0(v_0)$ is finitely satisfiable, since $\Sigma(v_n)_{n\in\omega}$ is, and so by weak-saturation there is $m_0\in M$ such that $\mathcal{M}\models \Delta_0(m_0)$. But this means that $\Sigma(m_0,v_n)_{0<n\in\omega}$ is consistent with $\mathcal{M}$, as desired.

Now suppose we have constructed $m_i$ for all $i<l$. We define $\Delta_l(v_l)=\exists(v_n)_{l<n\in\omega}\Sigma(m_0,...,m_{l-1},v_n)_{l\leqslant n\in\omega}$ in the obvious way, as $\{\exists v_{n_1}...v_{n_k}\varphi(m_0,...,m_{l-1},v_l,v_{n_1},...,v_{n_k}):\varphi(v_0,...,v_l,v_{n_1},...,v_{n_k})\in\Sigma(v_n)_{n\in\omega}\text{ and each }n_i>l\}$. (Apologies for the horrendous notation.) Now, $\Delta_l(v_l)$ is finitely satisfiable by construction of the $m_i$, else (taking conjunctions) there would be a formula $\varphi(v_0,...,v_l,v_{n_1},...,v_{n_k})\in\Sigma(v_n)_{n\in\omega}$ such that $\mathcal{M}\nvDash\exists v_l\exists v_{n_1}...v_{n_k}\varphi(m_0,...,m_{l-1},v_l,v_{n_1},...,v_{n_k})$, contradicting the induction hypothesis. Thus $\Delta_l(v_l)$ is finitely satisfiable and hence by weak saturation is realized, so there is an $m_l\in M$ such that $\mathcal{M}\models\Delta_l(m_l)$, and thus $\Sigma(m_0,...,m_l,v_n)_{l<n\in\omega}$ is consistent with $\mathcal{M}$, as desired.

Now, since any element of $\Sigma(v_n)_{n\in\omega}$ has only finitely many variables, it is clear that $\mathcal{M}\models \Sigma(m_n)_{n\in\omega}$, and so we are done.

I have two questions: first, is this proof correct? And second, if so, to what extent can we obtain similar results for general $\alpha$-types, where $\alpha$ is an infinite ordinal? Attempting to replicate the proof above for the general case using transfinite induction fails: although the successor ordinal case works fine, the limit ordinal case cannot be handled by this kind of argument.

(And this definitely makes sense, for it is clear that the result does not hold for sufficiently large $\alpha$... for instance, let $\mathcal{M}$ be any saturated structure of cardinality $\kappa$, and let $\alpha$ have cardinality strictly greater than $\kappa$. Letting $\Sigma(v_\delta)_{\delta<\alpha}=\{v_\delta\neq v_\gamma\}_{\{\delta\neq\gamma\in\alpha\}}$ we see that $\Sigma(v_\delta)_{\delta<\alpha}$ is clearly finitely satisfiable in $\mathcal{M}$ but not realizable in $\mathcal{M}$.)

So we certainly can't expect any kind of naive generalization of the above proof to work for general infinite ordinals $\alpha$, as the result does not hold in general. However, can we get any kind of analog provided that $|\alpha|\leqslant|M|$? Alternatively, can we get any kind of analog by restricting our attention to a more specific class of $\mathcal{L}$-formulas?

  • $\begingroup$ I've updated my answer with an alternative proof. $\endgroup$ Oct 31, 2020 at 0:00
  • $\begingroup$ @AlexKruckman oh yes, that is certainly cleaner – thank you!! $\endgroup$ Oct 31, 2020 at 5:38

1 Answer 1


First question. Your proof looks correct to me, except for one point: You really do need $\aleph_0$-saturation here, not just weak saturation. Weak saturation only tells you that every $n$-type over the empty set is realized, but you need to realize types over finitely many parameters.

Second question. A very natural generalization is true: If $M$ is $\kappa$-saturated and $|A|<\kappa$, then every type in $S^M_\kappa(A)$ is realized in $M$.

In fact, your proof generalizes immediately to show this. You wrote "although the successor ordinal case works fine, the limit ordinal case cannot be handled by this kind of argument." But the limit ordinal case is easy, by compactness. If $\lambda$ is a limit ordinal and $(m_\alpha)_{\alpha<\lambda}$ is a sequence from $M$ such that for all $\alpha<\lambda$, $\Sigma((m_\beta)_{\beta<\alpha},(v_\beta)_{\beta\geq \alpha})$ is consistent with the elementary diagram of $M$, then $\Sigma((m_\beta)_{\beta<\lambda},(v_\beta)_{\beta\geq \lambda})$ is consistent with the elementary diagram of $M$. The reason is that any finite subset of $\Sigma((m_\beta)_{\beta<\lambda},(v_\beta)_{\beta\geq \lambda})$ only mentions finitely many of the $(m_\beta)_{\beta<\lambda}$, and hence is contained in $\Sigma((m_\beta)_{\beta<\alpha},(v_\beta)_{\beta\geq \alpha})$ for some $\alpha<\lambda$.

The reason you can't continue the induction to arbitrary ordinals is that once you hit a sequence of length $\kappa$, you can't continue with the successor step - if $M$ is only $\kappa$-saturated, you can't realize a type over the entire sequence you've built so far in $M$.

Follow-up: Here's an alternative proof. It really has the same ideas as your proof, but it's a bit cleaner. First a lemma. Recall that for $A\subseteq M$, a map $f\colon A\to M$ is partial elementary if any tuple from $A$ has the same type in $M$ as its image under $f$. Given a type $p(x)$ over $A$, write $f_*p(x)$ (the pushforward of $p$ by $f$) for the set of formulas obtained by replacing elements in $a$ appearing in $p$ by their images under $f$. So $$\varphi(x,a_1,\dots,a_n)\in p(x) \text{ iff } \varphi(x,f(a_1),\dots,f(a_n))\in f_*p(x).$$

Lemma: Suppose $A\subseteq M$, $p(x)\in S_1^M(A)$, and $f\colon A\to M$ is a partial elementary map. Then $f_*p(x)\in S_1^M(f(A))$ (in particular, it is consistent).

The proof of this lemma is where you have to do some syntactic mucking around with existential quantifiers like in your proof. But it's a bit nicer, since we only have to deal with one free variable, not infinitely many!

Now suppose $M$ is $\kappa$-saturated and $A\subseteq M$ with $|A|<\kappa$. Let $\Sigma((x_\alpha)_{\alpha<\kappa})\in S^M_\kappa(A)$.

Since $\Sigma$ is consistent, there is some realization $(n_\alpha)_{\alpha<\kappa}$ in an elementary extension $M\preceq N$. We construct a sequence $(m_\alpha)_{\alpha<\kappa}$ in $M$ by transfinite induction. To define $m_\alpha$, we let $p_\alpha(x) = \text{tp}(n_\alpha/A(n_\beta)_{\beta<\alpha})$, define a partial elementary map $f_\alpha$ by $f_\alpha(n_\beta) = m_\beta$ for all $\beta<\alpha$, and realize $(f_\alpha)_*p_\alpha(x)$ in $M$ by saturation (since $|A\cup \{m_\beta\mid \beta<\alpha\}|<\kappa$).

To make this construction work, we just need to prove by transfinite induction on $\alpha\leq \kappa$ that each $f_\alpha$ is partial elementary, so we can apply the lemma. The case $\alpha = \kappa$ will establish that $M\models \Sigma((m_\alpha)_{\alpha<\kappa})$.

Base case: $f_0$ is the empty map, so it is trivially partial elementary.

Successor case: If $f_\alpha$ is partial elementary, use the fact that $m_{\alpha}$ realizes $(f_{\alpha})_*\text{tp}(n_{\alpha}/A(n_\beta)_{\beta<\alpha})$ to check that $f_{\alpha+1}$ is partial elementary.

Limit case: If $\lambda$ is a limit and $f_\alpha$ is partial elementary for all $\alpha<\lambda$, then $f_\lambda$ is partial elementary, since any finite tuple from the domain of $f_\lambda$ is actually in the domain of some $f_\alpha$.


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