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 :)


This is a homework, so I will just drop some hints and leave the details to you.

For the first part, notice that the sequence $1,2,\ldots$ can be cofinal in a model or not, and if it is, it may have a smallest element larger than all of $1,2,\ldots$, or it can have none. Showing that these are all the possibilities can be done using the fact that dense linear orderings without endpoints are $\omega$-categorical. Just take any model and construct an isomorphism with one of the three canonical models (which one depends on whether or not there are infinite elements in the model in question and if so, then whether or not there is a smallest infinite element), inductively. To show that these are all possible for models of the theory, it's probably best to find a reasonable axiomatization of the theory.

For the second part, notice that ${\Bbb Q}_0$ is just a dense-codense subset (that's all the theory says about it). Any model of the theory is also the model of the previous theory. The only thing that we can possibly use to differentiate between models is that the smallest element larger than all natural numbers, if it exists, may or may not belong to the dense-codense set. Showing that these are all possibilities can be done in pretty much the same way -- I believe that dense linear orderings without endpoints with a predicate for a dense-codense set are still $\omega$-categorical.

For arbitrary $n$, try to mix up the two constructions: adding a predicate for a dense-codense set (or multiple such sets, probably best chosen to be disjoint), and adding countably many constants for elements larger than all the previous constants.

  • $\begingroup$ Do you mean with a cofinite set $A$ in $X$ the following:$\forall x\in X\exists a\in A: x\leq a$ ? If yes, i don't understand the meaning of "smallest element larger than aal 1,2,... do you mean infinity? In addition it has to be cofinite because of the fact that the sequence 1,2,... is a countable sequence which is not bounded, or not? If i follow the definition the third model have to be $(\mathbb{Q}*,<,0,1,2,...)$ is this correct? Thank you for answer :) $\endgroup$ – Student of Mathematics Nov 27 '12 at 17:11
  • $\begingroup$ @StudentofMathematics: Not cofinite, cofinal. $\endgroup$ – tomasz Nov 27 '12 at 17:13
  • $\begingroup$ oh yes sorry i have make a mistake ... but i want to write cofinal :D $\endgroup$ – Student of Mathematics Nov 27 '12 at 17:18
  • $\begingroup$ @StudentofMathematics: I mean smallest among the elements larger than each of $1,2,3,\ldots$. This only needs to be considered if there are any elements larger than those, that is, if the sequence is not cofinal. I'm not sure what you mean by $\Bbb Q*$. $\endgroup$ – tomasz Nov 27 '12 at 17:22
  • $\begingroup$ By $\mathbb{Q}*$ i mean $\mathbb{Q}$ with infinity and minus infinity, because only the element $\infty$ is bigger then each of the elements in the sequence 1,2,3,... and thus the smallest element lager than every 1,2,3,... $\endgroup$ – Student of Mathematics Nov 27 '12 at 18:30

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