*All structures are countable with countable signature.*

Given a structure $\mathcal{A}$, the *age* of $\mathcal{A}$, $Age(\mathcal{A})$, is the set of structures isomorphic to finitely-generated substructures of $\mathcal{A}$ (see <https://en.wikipedia.org/wiki/Age_(model_theory)>). (This isn't really a set, but we can restrict attention to those structures contained in $H\_{\omega\_{17}}$, say.) $Age(\mathcal{A})$ automatically satisfies the Joint Embedding and Hereditariness properties, but may lack the Amalgamation property: an example of this is the undirected graph $\mathcal{G}$ with integers as vertices, and an edge between ever pair $(z, z+1)$.

Given a collection $\mathbb{K}$ of finite structures closed under isomorphism with the Joint Embedding, Hereditariness, and Amalgamation properties, we can form the *Fraïssé limit* of $\mathbb{K}$, the unique homogeneous structure $F(\mathbb{K})$ with $Age(F(\mathbb{K}))=\mathbb{K}$. Specifically, let $\mathbb{P}(\mathbb{K})=\mathbb{P}$ be the poset whose elements are finite sequences $\emptyset\prec \mathcal{A}\_0\prec . . . \prec\mathcal{A}\_n$ of structures in $\mathbb{K}$, ordered by reverse extension; then forcing with $\mathbb{P}$ produces the Fraïssé limit. Formally, there is a countable collection $\mathcal{D}$ of dense sets in $\mathbb{P}$ such that the direct limit of the terms of any $\mathcal{D}$-generic filter through $\mathbb{P}$ is isomorphic to the Fraïssé limit.

Now if $\mathbb{K}$ lacks the Amalgamation property, there will be no homogeneous structure with age $\mathbb{K}$. The poset $\mathbb{P}(\mathbb{K})$, however, still makes sense, and we can consider the collection $Fil(\mathbb{K})$ of all structures formed by (taking direct limits of terms of) maximal filters through $\mathbb{P}(\mathbb{K})$.

Moreover, we can compare structures in $Fil(\mathbb{K})$ in terms of relative genericity. For $D\subseteq \mathbb{P}(\mathbb{K})$ dense, let $\mathcal{O}\_D$ be the set of structures which can be built by filters hitting $D$. These sets $\mathcal{O}\_D$ then generate a topology $\tau\_\mathbb{K}$ on $Fil(\mathbb{K})$. This space is terrible, but seems reasonably natural if what we're interested in is how generic elements of $Fil(\mathbb{K})$ are.

As it turns out, elements of $Fil(\mathbb{K})$ can be extremely generic, even though the Fraïssé limit as such may not exist. Specifically, let $Gen(\mathbb{K})$ be the class of isomorphism classes of structures $\mathcal{A}$ such that in every open $\mathcal{O}\_D$ there is a copy $\mathcal{B}\cong\mathcal{A}$. Then, for example, consider the graph $\mathcal{G}$ from the first paragraph and let $\mathbb{K}=Age(\mathcal{G})$. $Fil(\mathbb{K})$ is precisely the class of acyclic infinite graphs in which each vertex has degree at most 2. However, the graph $\mathcal{G}$ is special among these graphs, since there is a countable collection $\mathcal{D}$ of dense subsets of $\mathbb{P}(\mathbb{K})$ such that any $\mathcal{D}$-generic filter corresponds to $\mathcal{G}$. This means $Gen(\mathbb{K})=\{\mathcal{G}\}$, and so in some sense $\mathcal{G}$ is the closest thing to a Fraisse limit of $Age(\mathcal{G})$ as we might reasonably expect to exist.

In general, if $Gen(Age(\mathcal{A}))$ has precisely one element, denote that element by $G(\mathcal{A})$, and call it the *generalized Fraïssé limit* of $Age(\mathbb{A})$.

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My questions are as follows. First, the obvious:

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> What are some sources for learning about this construction? In particular, what's it *actually* called?
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Second,

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> When is $Gen(Age(\mathcal{A}))$ nonempty? When does it have exactly one element?
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Finally, the fact that $G(\mathcal{G})=\mathcal{G}$ motivates me to ask:

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> Suppose $Gen(Age(\mathcal{A}))$ has exactly one element. What can we say about $G(\mathcal{A})$ in relation to $\mathcal{A}$? For example, what structures $\mathcal{A}$ do we have $G(\mathcal{A})=\mathcal{A}$?
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A couple quick observations: any homogeneous structure $\mathcal{A}$ will certainly have $G(\mathcal{A})=\mathcal{A}$. On the other hand, consider the graph $\mathcal{G}\_\omega$ consisting of the disjoint union of infinitely many copies of $\mathcal{G}$; then $G(\mathcal{G}\_\omega)=\{\mathcal{G}\}$, so not every structure has this property even if it has singleton $Gen$. Finally, note that we will always have $G(G(\mathcal{A}))=G(\mathcal{A})$ if $G(\mathcal{A})$ exists.