Complete theory with exactly n countable models?

You can refine Ehrenfeucht’s example getting rid of the constants.

Here is what John Baldwin suggested:

Consider the theory in the language $L=\{\le\}$, saying

• $\le$ is a total preorder (transitive, total [hence reflexive], not necessarily anti-symmetric) without least or last element. (Notice that the binary relation defined by $x\le y \land y\le x$ is an equivalence relation. Call it $E$.)
• For each $n$, $E$ has exactly one class of size $n$. Call it $C_n$.
• $C_i\le C_j$ (for $i\le j$) setwise.
• $E$-classes are densely ordered: for any two points there is a point $\le$-between them and not $E$-equivalent to any of them.

Check that this theory is complete.

Note that each finite equivalence class in this new theory plays the role of one of the constants in the classical example, so you get three countable models the same way.

See Vaught's theorem on wikipedia. It says:

Ehrenfeucht gave the following example of a theory with 3 countable models: the language has a relation ≥ and a countable number of constants c0, c1, ...with axioms stating that ≥ is a dense unbounded total order, and c0< c1