1
$\begingroup$

I have been trying to solve the following problem from Marker's Model Theory.

Find a pair of models $\mathcal{M}$ and $\mathcal{N}$, and a subset $A\subseteq \mathcal{M}$ so that

1) $\mathcal{M}\subseteq \mathcal{N}$

2) $\mathcal{M} \equiv \mathcal{N}$

3) $\text{dcl}_\mathcal{M}(A)\neq \text{dcl}_\mathcal{N}(A)$

Where dcl$_\mathcal{M}(A) = \{a\in |\mathcal{M}|\,:\, $there is a formula $\phi(x,\overline{y})$ and parameters $\overline{b}\in A$ so that $\mathcal{M}\models \exists ! x\phi(x,\overline{b}) \land \phi(a,\overline{b})\}$ i.e. it is the set of things definable in $\mathcal{M}$ with parameters from $A$.

I know that if $\mathcal{M}\prec\mathcal{N}$ then 3) is false, so I need $\mathcal{M}\nprec\mathcal{N}$ but I've been struggling to think of such an example. Can anyone help me think about sub-models which are elementary equivalent, but are not elementary models?

$\endgroup$
3
$\begingroup$

Consider $\mathcal{N} = ( \mathbb{N} , < )$, $\mathcal{M} = ( \mathbb{N}_{\text{even}} = \{ 0,2,4,\ldots \} , < )$, and let $A = \varnothing$. Clearly $\mathcal{M} \cong \mathcal{N}$, and so they are elementarily equivalent. Note that $\mathrm{dcl}_{\mathcal{N}} (A) = \mathbb{N}$ (each $n \in \mathbb{N}$ is the unique $n$th successor of the minimum element of the order, and we can write a formula saying this). Similarly, $\mathrm{dcl}_{\mathcal{M}} ( A ) = \mathbb{N}_{\text{even}}$.

$\endgroup$
  • $\begingroup$ I was going to suggest $\Bbb N$ and $\Bbb N\setminus\{0\}$. $\endgroup$ – Asaf Karagila Feb 7 '14 at 19:53
  • $\begingroup$ @Asaf: Ahhh.... yes, that would have been a bit simpler. Dammit. $\endgroup$ – user642796 Feb 7 '14 at 19:53
  • $\begingroup$ Ok so this correct but it feels like cheating as the two models are isomorphic. $\endgroup$ – James Feb 10 '14 at 17:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.