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Let $M$ be a non-standard model of $PA$, $a\in |M|$ be an arbitrary non-standard number and $T$ be a theory of arithmetic. We want to choose a subset $M'\subsetneq M$ such that:

  1. $M'\models PA^-$ (or $Q$)
  2. $a\not \in |M'|$
  3. there exists $b\in M'$ such that $M\models a<b$
  4. $M'\models T$

Q1. For which $M\models PA$, $a\in M$, and $T$ we can find a subset $M'$ satisfies above conditions?

Q2. For which $T$ is it true that: for every model $M\models PA$, for every nonstandard $a \in M$, there exists $M'$ satisfies above conditions?

Thanks.

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    $\begingroup$ If you take for $a$ a (parameter-free) $\Delta_0$-definable element of $M$, the conditions imply that $M'$ is not a $\Delta_0$-elementary submodel of $M$, hence it cannot satisfy the formalized MRDP theorem. That is, this property is false for any theory $T\supseteq I\Delta_0+\mathit{EXP}$. $\endgroup$
    – Emil Jeřábek
    Commented Jun 20, 2016 at 22:50
  • $\begingroup$ @EmilJeřábek: Thank you for your comment. What about weaker theories like $I\Delta_0$ or even weaker like $IE_1$? (I edited my post as you suggested.) $\endgroup$
    – Erfan Khaniki
    Commented Jun 21, 2016 at 0:05
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    $\begingroup$ @EmilJeřábek Could you explain your comment? In a nonstandard model of true arithmetic, for example, which extends that theory, there are no such definable nonstandard elements. $\endgroup$
    – Joel David Hamkins
    Commented Jun 21, 2016 at 0:32
  • $\begingroup$ But meanwhile, the theory TA of true arithmetic does not have the desired property, because of Gaifman's theorem on minimal extensions. That is, there is a minimal extension of the standard model, which would violate the desired properties. $\endgroup$
    – Joel David Hamkins
    Commented Jun 21, 2016 at 1:12
  • $\begingroup$ @Joel I read the question as "for which theories T is it true that: for every model M of PA, for every nonstandard a in M, there exists M' etc." So, my argument works as long as there exists at least one model of PA with a nonstandard $\Delta_0$-definable element, which it does. The existence of other models without such elements is irrelevant. $\endgroup$
    – Emil Jeřábek
    Commented Jun 21, 2016 at 8:28

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