I'm working for Modeltheory and i have the follwoing information:
Define $T_0=Th((\mathbb{Q},<,0,1,2,\cdots))$ and $T_1=Th((\mathbb{Q},<,0,\frac{1}{2},\frac{3}{4},\cdots,1-\frac{1}{2^n},\cdots))$.
I proved that $T_0=T_1$ (because for every formula you can use only finitely many individual constants and then you can make a isomorfism, and this implies elementairy equivalence). I also proved that $(\mathbb{Q},<,0,1,2,\cdots)$ and $(\mathbb{Q},<,0,\frac{1}{2},\frac{3}{4},\cdots,1-\frac{1}{2^n},\cdots)$ are not isomorphic (this follows from the fact that such a fucntion form the first model to the second can not be surjective).
But then i must prove the following statement: Prove that $T_0$ has, up to isomorphism, exactly three countable models.
I can conclude that there must be at least 3 models, because the given to models of $T_0$ are not isomorphic (see above), and Vaught's Theorem says, that there is not a complete theory with exactly two models, thus there are at least 3 models of $T_0$, but than i can't imagine what the third model have to be and why there are not more.
The following question was (and here i also can not imagine the right way to prove it): Prove that $T:=Th((\mathbb{Q},\mathbb{Q}_0,<,0,1,2,\cdots))$ has, up to isomorphism, exactly four countable models. (Here we notate $\mathbb{Q}_0$ as a relation with one open variable, in fact $x$ in this relation iff $x\in\mathbb{Q}_0$ and $\mathbb{Q}_0:=\left\{\frac{s}{2^k}:s\in\mathbb{Z}, k>0\right\}$). Find a complete theory that has, up to isomorphism, exactly five countable models. Prove that for each $n\geq3$ there exists a complete theory that has up to isomorphism exactly $n$ countable models.
Note: $p_0=2,p_1=3,p_2,\cdots$ the sequence of prime numbers and for each n, let $\mathbb{Q}_n$ be the set of all reational numbers $\frac{s}{t}$ where $s\in\mathbb{Z}$ and there exists $k>0$ such that $t=(p_n)^k$ (thus here we get also the definition from $\mathbb{Q}_0$).
I hope that someone can help me. I know that this is a difficult topic, but i want to understand it completely. Thank you for help :)