3
$\begingroup$

I am new to abstract elementary classes (AEC) and am reading the early chapters of Baldwin's Categoricity text. I have shown that the disjoint amalgamation property holds for an AEC obtained by looking at the models of a first order theory (this is a straight forward compactness argument). However I have not been able to come up with an AEC for which the disjoint amalgamation property fails. I was hoping to have an example in $L_{\omega_1 \omega}$. Does anyone know of such an example?

Also, it is straightforward to come up with classes for which the joint embedding property (JEP) fails but the amalgamation property (AP) holds (For example: models of ACF will have this property). However what about AEC in which you have the JEP but not AP? The only example for a class of structures where this is true, that I know of, involves finite structures, so I would like to know of an example where all the structures are infinite.

$\endgroup$
  • $\begingroup$ Exercise 12, p. 332 of Hodges's Model Theory asks you to show that a certain class (the description is there) has the JEP property, but not AP. Perhaps it's worth a look? $\endgroup$ – Nagase Nov 13 '15 at 0:46
  • $\begingroup$ For what it's worth, I can't make sense of Hodges Exercise 12. It seems to me that the theory described there has only the empty model; there must be a typo. Also, as noted in the question, it's easy to come up with classes of finite structures which have HP and JEP but not AP, e.g. the class of acyclic graphs together with a unary predicate. kav11 wanted an example in the setting of infinite models of a sentence in $L_{\omega_1,\omega}$. $\endgroup$ – Alex Kruckman Nov 13 '15 at 19:51
  • $\begingroup$ @AlexKruckman - I suppose the third axiom should state that the relation is not transitive? $\endgroup$ – Nagase Nov 14 '15 at 12:40
3
$\begingroup$

Here's an example in $L_{\omega_1,\omega}$ for your first question. Let $L$ be the vocabulary with countably many unary predicates, $\{P_i\mid i\in \omega\}$. Let $T$ be the first-order theory asserting that the predicates interact generically (so for all $n$ and all $f\colon n\to 2$, there is an element $x$ such that for all $0\leq i < n$, $P_i(x)$ iff $f(i) = 1$.) Let $\varphi$ be the conjunction of the sentences in $T$, together with the following infinitary sentence: $$\forall x\, \forall y\, \left(\bigwedge_{i\in \omega}P_i(x) \leftrightarrow P_i(y)\right)\rightarrow (x = y).$$ Now models of $T$ are in bijection with dense subsets of $2^\omega$, and all embeddings are elementary for the fragment of $L_{\omega_1,\omega}$ generated by $\varphi$. The corresponding AEC clearly has JEP and AP (just take unions of dense subsets), but it fails to have the disjoint AP, since if $a\in A$ and $b\in B$ satisfy the same predicates, they must be identified in any amalgamation.


For your second question, it's easy to come up with combinatorial examples, and not just dealing with finite structures (e.g. take infinite graphs with no cycles, plus a unary predicate picking out a subset of the vertices, together with all embeddings). So you probably want another example in $L_{\omega_1,\omega}$.

Let's take the above example, but break amalgamation entirely. So add an additional predicate $Q$, and adjust $T$ to say that $Q$ is dense/co-dense. Now when you extend a structure by a new element $a$, you can choose whether to set $Q(a)$ or $\lnot Q(a)$, and structures in which you make different choices can't be amalgamated.

But we want JEP. To do this, add an equivalence relation $E$ and say that each equivalence class is a model of the theory described so far. Explicitly, this changes the infinitary sentence to: $$\forall x\, \forall y\, \left((xEy) \land\bigwedge_{i\in \omega}P_i(x) \leftrightarrow P_i(y)\right)\rightarrow (x = y).$$

Now given two models $A$ and $B$, we can put them together (JEP) by just making everything in $B$ inequivalent to everything in $A$, and there's no need to identify any elements. But AP fails, since we can take two extensions $A$ and $B$ of a model $C$ which both add an element to an existing equivalence class but color it differently by $Q$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy