Is submodel of atomic model atomic

Question

I think the submodel (elementary or arbitrary) of an atomic model may not be atomic. But I do not know an example.

Answer 1

Recall for readers that an atomic model is a structure such that each type realized in the structure is generated by a single formula.

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Suppose $M\preccurlyeq N$, $N$ is atomic, and $m\in M$. By elementarity we have $tp^M(m)=tp^N(m)$, and by atomicity of $N$ we have that $tp^N(m)$ is generated by a single formula $\varphi$. Putting these together we get that $\varphi$ generates the type of $m$ in $M$ as well. So elementary submodels of atomic models are atomic.

If we look merely at substructures, however, everything breaks down immediately. It's a good exercise to show that every countable graph is an induced subgraph of an atomic countable graph. (HINT: think about gluing the starting graph to a copy of $\mathbb{N}$.)

Answer 2

Noah had shown that (1) every elementary substructure of an atomic model is atomic, (2) not every substructure of an atomic model is atomic.

One might ask: What about a substructure of an atomic model which is elementarily equivalent to it? That is, for a complete theory $T$, can a non-atomic model of $T$ embed (non-elementarily) in an atomic model of $T$?

The answer is yes. Consider the theory $T$ of forests (simple graphs with no cycles) augmented by axioms asserting that every vertex has infinitely many neighbors. This theory is complete.

A countable model of $T$ consists of a disjoint union of copies of the complete countably branching tree (at least one copy and at most countably many copies). The atomic model $M$ is the connected one (just one copy of the tree). Now pick an arbitrary vertex $v\in M$ and remove it (and all edges out of it). The resulting graph $N$ is disconnected - each of the neighbors of $v$ is now in its own connected component, and each connected component is a copy of the complete countably branching tree.

So $N$ is a non-atomic model of $T$ which is a substructure of the atomic model $M$.