Modell-Theory, Isomorphism

Question

Let $\mathcal{L}$ be a language of the first-order logic with a binary relationsymbol $\stackrel{\cdot}{<}$ (and no other symbols).

Observe the following set $T$ of theorems:

$T:=\{\forall v_0\forall v_1 (v_0<v_1\to\exists v_2(v_0<v_2\wedge v_2<v_1)), \forall v_0\exists v_1\quad v_1<v_0,\forall v_0\exists v_1\quad v_0< v_1\}$

Also $\stackrel{\cdot}{<0}$ is supposed to be a linear order. It is missing, that $T$ has to include these axioms:

$\forall v_0\quad \neg v_0<v_0, \forall v_0\forall v_1(v_0<v_1\vee v_0=v_1\vee v_1<v_0), \forall v_0\forall v_1\forall v_2 (v_0<v_1\wedge v_1<v_2\to v_0<v_2)$

Task:

Let $\mathcal{M}=(M,<_M)$ and $\mathcal{M}'=(M',<_M)$ be countable models of $T$. Show, that $\mathcal{M}\cong\mathcal{M}'$

Hello,

I have a question to this task, and I would like to solve it with your help. To show that $\mathcal{M}\cong\mathcal{M}'$

I have to show several things:

First of all there needs to exist a bijection

$\pi: M\to M'$ which states the following points:

a) $m_1<_M m_2\Leftrightarrow\pi(m_1)<_{M'}\pi(m_2)$

b) Let $c$ be a constant. Than holds $\pi(c^{\mathcal{M}})=c^{\mathcal{M}'}$

Normally this also includes function-symbols. But $T$ only contains the binary relation-symbol $\stackrel{\cdot}{<}$. Therefore I can skip this.

Is this everything what has to be done? Or do I miss something. How do I include the set $T$? The first thing I have to do now, is to give a suitable function $\pi$ and than show it holds a) and b).

Thanks in advance for your comments and tips.

The standard proof uses the Cantor Back and Forth argument. — André Nicolas, Jun 05 '16 at 19:07
Thank you. I did not know this method yet, but I looked it up. Do you have a source with a proof? — Mr.Topology, Jun 05 '16 at 19:10
The link in the above comment provides a good outline of a proof for your case (densely ordered sets with no first or last element). Also, I have mentioned a bunch of key words that will help locate other proofs. I am pretty sure for example that the question has come up on MSE before. — André Nicolas, Jun 05 '16 at 19:14
Therefore the key point of the proof is to give a construction for the bijection $\pi$ using the properties of $T$, as hinted in the link. And $\pi$ will satisfy the mentioned points a) and b) then by construction of $\pi$. — Mr.Topology, Jun 05 '16 at 19:22
Yes, or you can do it in two steps, an order preserving bijection from $M$ to the rationals, and one from the rationals to $M'$. And the constant symbol interpretations you mentioned as b) are irrelevant, there are no constant symbols. — André Nicolas, Jun 05 '16 at 19:38

Modell-Theory, Isomorphism

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