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Let $\mathfrak{A}_0=(A_0,R_0)$ and $\mathfrak{A}_1=(A_1,R_1)$ be structures such that for each $i<2$, $A_i$ is a nonempty set and $R_i\subseteq A_i\times A_i$. We define the structure $\mathfrak{A}=\mathfrak{A}_0\oplus\mathfrak{A}_1=(A,R)$ as follows: $A=A_0\dot{\cup}A_1=(A_0\times\left\{0\right\})\cup(A_1\times\left\{1\right\}) \\ R = \left\{((a,i),(a',i'))\in A \times A:(i=0\ \text{and}\ i'=1)\ \text{or}\ (i=i'\ \text{and} \ (a,a')\in R_i)\right\}$


Consider the complete theory $T=Th((\mathbb{N},<))$.\

Prove that the structure $(\mathbb{N},<)\oplus(\mathbb{Z},<)$ realizes every 1-type of T.\

Prove that the structure $(\mathbb{N},<)\oplus(\mathbb{Z},<)\oplus(\mathbb{Z},<)$ realizes every 2-type of T but not every 3-type of T.


For the first part: I think i have an idea. I have to use an elementary monomorphism/embedding from the structure $(\mathbb{N},<)$ to $(\mathbb{N},<)\oplus(\mathbb{Z},<)$ because than all formulas are preserved. Such a embedding exists on a natural way: $f(n)=(n,0)$. But then?!

For the second part: i think i also have to make such an embedding to preserve the formulas, but i dont know how to do this. i think i have to make the map $f(n,i)=(n,i)$ for $i=0,1$. For the 3-type i have to find a counterexample.

Can someone help me?! Thank you Chris

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    @tomasz: If I was willing to put in the work to do the details, I'd've answered already.2012-10-17

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