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?


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, $\mathrm{dcl}_{\mathcal{M}} ( A ) = \mathbb{N}_{\text{even}}$.

  • $\begingroup$ 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

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