0
$\begingroup$

As I'm not allowed to ask a new question due to limit reached matter, I still want to EDIT this one as communicated with @Alex Kruckman in the comments below. I would like to understand the relationship of the two definitions relating AEC's notion of a type by assuming existence of amalgamation in Jarden's approach with realizing formulas in elementary extensions in Hodge's approach. What should I take for an AEC $(K,\preceq)$ to obtain exactly Hodge's notion and prove its equivalence with the amalgamation approach ?

OLD Question:

I have come across two different definitions of saturation and I would like to relate them. One is in Wilfrid Hodges book (first snippet below) and it says that an elementary extension $B$ realizes all types with less than $<\lambda$ elements. The other approach is the definition 1.0.25 in the second snippet below. It says that $M$ of cardinality $\lambda^+$ is saturated if $M$ realizes type of every sub-model of cardinality $\lambda$. I do not even know what is to be proved for an equivalence of these two approaches. It appears to me that in the second approach we are looking at smaller models and in the first approach at larger elementary extensions.

enter image description here

enter image description here

$\endgroup$
10
  • 4
    $\begingroup$ Voting to close as not research level - but while I'm here: your summaries of what these definitions say are all wrong. The Hodges definition doesn't say all types with less than $\lambda$ elements are realized, it says that all types over sets $X$ with $|X|<\lambda$ are realized. The Jarden-Shelah definition doesn't say all types of submodels of cardinality $\lambda$ are realized, it says all types over submodels $N$ with $|N|=\lambda$ are realized. $\endgroup$ Dec 29, 2020 at 14:27
  • $\begingroup$ @AlexKruckman I understand what you write, but still Hodges uses formulas while Jarden amalgamation; how can I relate these two concepts ? What should I take for $(K,\preceq)$ in Jarden's approach ? $\endgroup$ Dec 29, 2020 at 18:27
  • $\begingroup$ Well, that's a totally different question. It's because the Hodges definition is in the context of model theory of first-order logic, while the Jarden-Shelah definition is in the context of AECs, where there is no such thing as a formula. As for why the notion of Galois type in AECs is a suitable replacement for the notion of type in first-order logic, this deserves its own question. $\endgroup$ Dec 29, 2020 at 18:30
  • $\begingroup$ @AlexKruckman I will definitely write one with the hope that you will not downvote this time. $\endgroup$ Dec 29, 2020 at 18:34
  • $\begingroup$ Indeed, I will not downvote if your question is of good quality (e.g. not including images of text from books or links to the middle of long PDFs). $\endgroup$ Dec 29, 2020 at 18:35

1 Answer 1

5
$\begingroup$

First of all, the context of abstract elementary classes (AECs) is a generalization of the context of model theory of first-order logic. For any first-order theory $T$, we can let $K$ be the class of models of $T$, and let $\preceq$ be the elementary substructure relation between models of $T$. Then $(K,\preceq)$ is an AEC. [Thus the name: $(K,\preceq)$ is an elementary class, and the axioms of AECs are intended to abstract the properties of elementary classes].

For the rest of the answer, let's fix the first-order theory $T$, and the corresponding elementary class $(K,\preceq)$ as above.

Next, I'll explain how the notion of Galois type from AECs is a generalization of the notion of type (over a model) from model theory. Let $M\models T$. Let's write $S(M)$ for the set of first-order complete types over $M$ and $gS(M)$ for the set of Galois types over $M$ in $(K,\preceq)$. We'd like to establish a bijection between $S(M)$ and $gS(M)$. Let $p\in gS(M)$. Then $p$ is the $E_{K,\preceq}$ equivalence class of some triple $(M,N,a)$ where $M\preceq N$ and $a$ is a tuple from $N$. We map $p$ to $\text{tp}_N(a/M)\in S(M)$, the complete first-order type of $a$ over $M$ in the model $N$.

  • To check the map is well-defined, it suffices to show that if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M')$, since the equivalence relation $E_{K,\preceq}$ is the transitive closure of $E^*_{K,\preceq}$. By definition, if $(M,N,a)E^*_{K,\preceq}(M',N',a')$, then $M = M'$ and there exists $N''$ with $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ which fixes $M$ pointwise and has $f(a) = a'$. But then $$\text{tp}_{N}(a/M) = \text{tp}_{N''}(f(a)/M) = \text{tp}_{N'}(a'/M').$$
  • The fact that the map is injective is a consequence of a standard amalgamation result in model theory (see Theorem 6.4.1 in Hodges). If $\text{tp}_N(a/M) = \text{tp}_{N'}(a'/M)$, then there is an elementary extension $N'\preceq N''$ and an elementary embedding $f\colon N\to N''$ which fixes $M$ pointwise and has $f(a) = a'$. This exactly says that $(M,N,a)E^*_{K,\leq} (M,N',a')$, so the $E_{K,\leq}$-equivalence classes of $(M,N,a)$ and $(M,N',a')$ are equal.
  • The map is surjective, because for any complete first-order type $q(x)\in S(M)$, $q(x)$ is realized by some tuple $a$ in some elementary extension $M\preceq N$. Then $q(x)$ is the image of the equivalence class of the triple $(M,N,a)$.

Ok, so when we're working with an elementary class, a Galois type over a model $M$ is essentially the same thing as a complete first-order type over $M$.

Finally, we want to see that the notion of saturation as defined by Jarden-Shelah is a generalization of the notion of saturation in model theory. In your quoted passage, Jarden-Shelah define what it means for $M$ to be "saturated in $\lambda^+$ over $\lambda$". I claim that when we're working with an elementary class, $M$ is saturated in $\lambda^+$ over $\lambda$ if and only if $M$ has cardinality $\lambda^+$ and is saturated (i.e. $\lambda^+$-saturated).

One direction is trivial: If $M$ has cardinality $\lambda^+$ and is $\lambda^+$-saturated, then for any model $N\preceq M$ with $|N| = \lambda$, every complete $n$-type over $N$ is realized in $M$ by saturation (i.e. $M$ is full over $N$), so $M$ is saturated in $\lambda^+$ over $\lambda$.

In the other direction, suppose $M$ is saturated in $\lambda^+$ over $\lambda$. Let $X\subseteq M$ with $|X| <\lambda^+$, and let $p(x)\in S(X)$ be a complete type over $X$. By Löwenheim-Skolem, we can pick an elementary substructure $N\preceq M$ with $X\subseteq N$ and $|N| = \lambda$. Let $p'(x)\in S(N)$ be any extension of $p(x)$ to a complete type over $N$. Since $M$ realizes every type in $S(N)$, $p'(x)$ is realized in $M$, and hence so is $p(x)$. So $M$ is $\lambda^+$-saturated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.