Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property?

Question

Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property?

The question can be stated in a fashion not requiring much background:

Let $M$ be a countable ultra-homogeneous relational structure - namely, a countable set equipped with a bunch of relations on its finite cartesian powers, such that any isomorphism between two finite substructures of $M$ extends to an automorphism of $M$. It is classical (and easy to see) that the age of $M$, namely the class $K$ of finite substructures of $M$ (and isomorphic copies thereof) is a Fraïssé class, and in particular has the Amalgamation Property:

Amalgamation Property (AP) - whenever $A,B_i \in K$ for $i=0,1$ and $f_i\colon A \to B_i$ are embeddings, there is $C \in K$ and further embeddings $g_i\colon B_i \to C$ such that $g_0f_0 = g_1f_1$ (i.e., $B_0$ and $B_1$ can be amalgamated in $K$ over $A$).

In all examples I am familiar with, the class of all countable substructures of $M$ also has the same property, but I see no reason why this should be true in general. Any counterexample (or proof that this does hold in general) will be welcome.

[Of course, there are usually going to be in $M$ countable substructures $A_0,A_1$ with an isomorphism $f\colon A_0 \to A_1$ which does not arise from an automorphism - but this does not exclude the possibility of proper embeddings $g_i\colon M \to M$ such that $g_1^{-1} g_0$ extends $f$.]

ADDENDUM: Notice that if $M$ is saturated then its countable substructures have AP, so a counter-example will have to be non saturated, and in particular non $\aleph_0$-categorical, with an infinite language.

ADDENDUM #2: see also A restatement, in terms of the semi-group product of the left-invariant completion of a Polish group, of http://mathoverflow.net/questions/71389

Answer 1

EDIT NOTE: Thanks to Emil Jeřábek's comment, (1) has been modified; $X$ in the theorem has been quantified, and the bold sentence in (4) has been added.

I will first present a counterexample using a structure that has (infinitely many) functions; then I will explain how this functional counterexample can be turned into a relational one.

We begin with some preliminaries:

(1) Recursively saturated models that have elimination of quantifiers are ultra-homogeneous. This is a basic result in model theory.

(2) If $M_0$ and $M_1$ are models of $PA$ (Peano arithmetic), and $M_0$ is a submodel of $M_1$, then $SSy(M_{0})\subseteq SSy(M)$. This follows from the definition of $SSy(M)$ (the standard system of $M$). Recall that for a model $M$ of $PA$, $SSy(M)$ is the collection of subsets of $\omega$ that are "coded" by some element of $M$, where "coded" can be defined in various ways, e.g., as: $X \subseteq \omega$ is coded by $c \in M$ if for all $n \in \omega$, $M \models$ “the $n$-th prime divides $c$” iff $n \in X$.

(3) The heart of this counterexample is the following theorem [it is Theorem 2.3.1 (p.40) of the Kossak-Schmerl text on models of Peano arithmetic].

Theorem. Let $M_0$ be a countable recursively saturated model of $PA$, and suppose $X$ is some fixed subset of $\omega$. Then $M_0$ has elementary end extensions $M_1$ and $M_2$, such that $M_0 \cong M_{1} \cong M_2$, and whenever $M_{3}\models PA$ is an amalgamation of $M_1$ and $M_2$, then $X\in SSy(M_3)$.

(4) Given $M \models PA$, let $M^{+}$ be the EXPANSION of $M$ by the first-order definable functions of $M$. We observe that if $N^{+}$ is a substructure of $M^{+}$, then the reduct $N$ is a model of $PA$ since the universe of $N$ is closed under the functions available in $M^{+}$, and therefore $N$ is an elementary submodel of $M$ because $PA$ has definable Skolem functions. Note, furthermore, that $M^{+}$ eliminates quantifiers, and is also recursively saturated, hence ultrahomogeous.

(1)-(4) show that for a countable recursively saturated model $M$ of $PA$, the collection of substructures of $M^{+}$ do not satisfy amalgamation.

More specifically, thanks to the aforementioned theorem in (3), by first choosing some subset $X$ of $\omega$ that is missing from the standard system of $M$, we can be assured of the existence of (end) embeddings $f_{i}:M^{+}\rightarrow M^{+}$ for $i=0,1$ with the property that if there is a structure $N^{+}$, and embeddings $g_{i}:M\rightarrow N^{+}$ for $i=0,1$, with $g_{0}f_{0}=g_{1}f_{1}$, then by (2) and (4) $N^{+}$ is not a substructure of $M^{+}$.

Now we explain how to obtain a relational counterexample.

Given a model $A$ in a language with functions, let $\cal{A}$ be the relational structure obtained by replacing each $n$-ary function $f$ in $A$ by the usual $(n+1)$-ary relation known as the graph of $f$.

Let $M$ be a countable recursively saturated model of $PA$. To see that the family of substructures of $\cal{M^{+}}$ do not satisfy amalgamation, we simply observe that if $(X,\cdot \cdot \cdot)$ is a substructure of $\cal{M^{+}}$, and $\overline {X}$ is the closure of $X$ under the functions available in $M^{+}$ , then the inclusion map $i_{X}:X\rightarrow \overline{X}$ is an embedding of the substructure of $\cal{M^{+}}$ determined by $X$ into the substructure of $\cal{M^{+}}$ determined by $\overline{X}$. Therefore, if $AP$ holds in this relational context for some amalgamating substructure with universe $X$, by composing each $g_i$ with $i_{X}$ then $AP$ would also have to hold in the functional context.

Answer 2

An example is the "complete rainbow graph without monochromatic triangles". Let $L = \{R_i : i \in \omega\}$ be a language consisting of $\omega$-many binary relation symbols $R_i$, and take the class $\mathcal{K}$ of finite $L$-structures $D$ such that

• each $R_i^D$ is a graph relation (that is, irreflexive and symmetric);
• the $R_i^D$s are disjoint;
• every ordered pair of distinct elements of $D$ lies in some $R_i^D$ (completeness of the graph);
• no $R_i^D$ contains a triangle.

We think of each $R_i^D$ as consisting of the edges of $D$ of colour $i$.

Then $\mathcal{K}$ is a strong amalgamation class (join the ears of the finite amalgam with edges of a colour that's not used in either ear). Let $M$ denote its Fraïssé limit, and let $\mathcal{U}$ denote the class of $L$-structures which embed into $M$ (so including countably infinite ones).

We will see that $\mathcal{U}$ does not have amalgamation.

Let $A \in \mathcal{U}$ be countably infinite. Then $A$ has two adjacent edges $a a_0, a a_1$ of different colours, as there are no monochromatic triangles in $A$ and $A$ is a complete graph. Enumerate $A$ as $a, a_0, a_1, a_2, \cdots$. Let $i_1$ be the colour of $a a_0$ and let $i_0$ be the colour of $a a_1$ (this is not a typo!). Enumerate the set $\omega$ of colours as $i_0, i_1, i_2, \cdots$.

Define structures $B, C$ as follows. Let $B$ consist of $A$ together with a new point $b$ such that $ba$ has colour $i_0$ and $ba_j$ has colour $i_j$ for $j \geq 0$. Let $C$ consist of $A$ together with a new point $c$ such that $ca$ has colour $i_1$ and $ca_j$ has colour $i_j$ for $j \geq 0$. Then $B$ and $C$ are complete and contain no monochromatic triangles, so lie in $\mathcal{U}$.

Then $B, C$ cannot amalgamate over $A$, as $b$ and $c$ cannot be identified in the amalgam due to their different coloured edges to $a$, and any coloured edge between $b$ and $c$ would form a monochromatic triangle.

(Note that $M$ is not $\omega$-categorical by the Ryll-Nardzewski theorem, as we have infinitely many isomorphism classes of substructures of size $2$. Also note that $M$ is not $\omega$-saturated: take the type which consists of formulae "$xy$ is not of colour $i$" for each $i \in \omega$.

In general, if a relational Fraïssé limit $M$ is $\omega$-categorical or has free amalgamation, then $\mathcal{U}$ has amalgamation. This follows for the $\omega$-categorical case by a König's lemma argument doing the amalgam finite piece by finite piece - for finite amalgams you only have finitely many possibilities by Ryll-Nardzewski - and for the free amalgamation case you take the canonical amalgam at each finite step.)

Answer 3

EDIT: the argument below assumes finite language, which I took for granted for no good reason.

$\DeclareMathOperator\Th{Th}\DeclareMathOperator\Diag{Diag}$ The class of countable substructures of $M$ does have the amalgamation property if the language of $M$ is finite.

First, ultrahomogeneity implies that for any $n$ there are only finitely many $n$-types realized in $M$, each of them principal (the type of any $n$-tuple is generated by the conjunction of its diagram). This implies that there are only finitely many nonequivalent formulas in $n$-variables (namely, disjunctions of the generators), hence $\Th(M)$ is $\omega$-categorical by the Ryll-Nardzewski theorem. Alternatively, $\mathrm{Aut}(M)$ is oligomorphic as two $n$-tuples with the same diagram are in the same orbit, which is also equivalent to $\omega$-categoricity by (another variant of) the Ryll-Nardzewski theorem.

Then, take $A,B_0,B_1\subseteq M$ and embeddings $f_i\colon A\to B_i$. Let $B_i^+$ be $B_i$ expanded with constants for every element $a\in A$ realized in $B_i^+$ by $f_i(a)$. Every finite subset of $T=\Th(M)\cup\Diag(B_0^+)\cup\Diag(B_1^+)$ is consistent because the class of finitely generated (= finite) substructures of $M$ has AP (or it is easily checked directly), hence there exists a countable model $C\models T$. Since $C\equiv M$, we may assume $C=M$ by categoricity, and then we can define $g_i\colon B_i\to C$ satisfying $g_0f_0=g_1f_1$ in the obvious way.