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

  • $\begingroup$ Can you at least see why $(\mathbb{N},<)$ itself doesn't realize all $1$-types? $\endgroup$ – Chris Eagle Oct 16 '12 at 21:10
  • $\begingroup$ I know that every $n\in\mathbb{N}$ has his own type (it is self a principal type and because there are infinity many prinical type there exists a non-principal type. $\endgroup$ – Chris Oct 16 '12 at 21:14
  • $\begingroup$ I think it would be easier to understand if you just said that $\oplus$ is just gluing two partial orders one after the other (I doubt the $\oplus$ as you defined makes much sense if $R_0,R_1$ arbitrary relations). $\endgroup$ – tomasz Oct 17 '12 at 15:49
  • $\begingroup$ @ChrisEagle: I noticed that to show that the model ${\bf N}\oplus {\bf Z}^{\oplus \omega}$ is universal you probably need to show some form of q.e. anyway. I think you should post your comment to my (now deleted) answer as an answer. $\endgroup$ – tomasz Oct 17 '12 at 18:10
  • $\begingroup$ @tomasz: If I was willing to put in the work to do the details, I'd've answered already. $\endgroup$ – Chris Eagle Oct 17 '12 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.