Definition. Let $\kappa$ be an infinite cardinal. We say that an $L$-structure $\mathfrak{A}$ is $\kappa$-saturated iff all $1$-types over sets of cardinality less than $\kappa$ are realised in $\mathfrak{A}$.

Proposition. $\mathfrak{A}$ is $\kappa$-saturated iff all $n$-types over sets of cardinality less than $\kappa$ are realised in $\mathfrak{A}$.

In Marker's book "Model Theory: An Introduction" one can find the following proof for the proposition stated above. (To avoid confusion: For me, every type has to be complete by definition. This is different to Marker's definition of types.)

Proof. We prove this by induction on $n$. Assume that $\mathfrak{A}$ is $\kappa$-saturated and $X\subseteq A$ with $|X|<\kappa$. Let $p\in S_{n}^{\mathfrak{A}}(X)$. Let $q\in S_{n-1}^{\mathfrak{A}}$ be the type $\{\phi(v_1,\ldots,v_{n-1})\;:\;\phi\in p\}$. By induction, $q$ is realised by some $\bar a$ in $\mathfrak{A}$. Let $r\in S_{1}^{\mathfrak{A}}(X\bar a)$ be the type $\{\psi(\bar a,w)\;:\;\psi(v_1,\ldots,v_n)\in p\}$. Since $\mathfrak{A}$ is $\kappa$-saturated and $|X\bar a|<\kappa$, we can realise r by some $b$ in $\mathfrak{A}$. Then, $(\bar a,b)$ realises $p$.

Question. I do not immediately see why $r=\{\psi(\bar a,w)\;:\;\psi(v_1,\ldots,v_n)\in p\}$ is finitely satisfiable. I have to prove this. And I do this as follows:

Let $\psi_1(\bar a,w),\ldots,\psi_k(\bar a,w)\in r$. Then $\psi_1(\bar a,w)\land\ldots\land\psi_k(\bar a,w)\in r$. Thus, we only have to show that every $\psi(\bar a,w)\in r$ is satisfiable in $\mathfrak{A}$. So, let $\psi(v_1,\ldots,v_n)\in p$ and we assume that the variable $v_n$ does really occur in $\psi$. Then $\exists v_n\psi(v_1,\ldots,v_n)\in q$. Hence $\mathfrak{A}\models\exists v_n\psi(\bar a,v_n)$. Hence $\psi(\bar a,w)$ is satisfiable in $\mathfrak{A}$. Thus $r$ is finitely satisfiable. Since it is complete, it really is a type.

I wonder if one can see more easily that $r$ is really a type. The reason why I wonder is that the proof in Marker's book naturally claims that $r$ is a type.

  • $\begingroup$ Two marginal comments. As you state it (well, as Marker writes it), there seems to be something special about types with finitely many variables. But the theorem in fact holds also for $\kappa$-types. It is also unnecessarily restrictive to speak about cardinality of the type. You should instead refer to the cardinality of the parameters of the type. (This is relevant when the language has large cardinality.) $\endgroup$ Mar 28, 2015 at 14:27

1 Answer 1


I believe the proof you give is the most straightforward way to see that $r$ is a type.

As you move further into model theory, you'll find that often routine compactness arguments like this can be replaced by arguments using saturation of a "monster model". Such an argument would goes like this:

Fix a $\lambda$-saturated and $\lambda$-strongly homogeneous model $\mathbb{M}$, for some large $\lambda$ (larger than the cardinals we care about and the sizes of the models we care about: $\kappa$ and $|\mathfrak{A}|$ in this case). Then all the models we care about (just $\mathfrak{A}$ in this case) embed in $\mathbb{M}$.

Now since $p$ is a type over $X\subseteq \mathfrak{A}\preceq \mathbb{M}$ of size $<\kappa$, by $\lambda$-saturation it is realized in $\mathbb{M}$ by $\overline{a}b$. Here $\overline{a}$ is a tuple of length $n-1$, and $b$ is a singleton.

Now by induction, we can pick $\overline{a}'$ realizing $q$ in $\mathfrak{A}$. But $\overline{a}'$ and $\overline{a}$ have the same type over $X$, so by the strong homogeneity of $\mathbb{M}$, there is an automorphism $\sigma$ of $\mathbb{M}$ fixing $X$ with $\sigma(\overline{a}) = \overline{a}'$. Let $b' = \sigma(\overline{b})$. Then $b$ realizes $\{\varphi(\overline{a},x_n)\mid \varphi\in p\}$, so $b'$ realizes $r = \{\varphi(\overline{a}',x_n)\mid \varphi\in p\}$, hence $r$ is consistent.

The idea can be summarized in this way: $p$ is consistent, so $q$, its restriction to the first $n-1$ variables, is consistent. The claim that $r$ is consistent is the claim that for the particular realization $\overline{a}'$ of $q$, it's consistent to extend it to a realization of $p$. But any two realizations of $q$ "look the same" (captured by the notion of an automorphism of $\mathbb{M}$), so if one can be extended to a realization of $p$ (and it can, since $p$ is consistent), so can $\overline{a}'$.

Note that I'm not suggesting you use this argument here. For one thing, it rather puts the cart before the horse, assuming you have a very saturated and strongly homogeneous monster model around before you've even proven basic facts about saturated models. Also, in this case, the monster model proof is arguably no simpler than the compactness argument you gave. I just wanted to point out that there's another way of thinking about arguments like this, which can be substantially more efficient when you get to proofs which would require applying compactness several times.


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