I know that every element of $\mathbb{N}$ is definable the standard model of Peano Arithmetic. Does there exist a countable non-standard model of PA where the same is true?
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10$\begingroup$ Yes. PA has definable Skolem functions, hence the set of definable elements of any model is an elementary submodel, which is nonstandard as long as the original model is not elementarily equivalent to $\mathbb N$. $\endgroup$– Emil JeřábekCommented Apr 23, 2020 at 8:30
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$\begingroup$ Why does having definable Skolem functions imply that the set of definable elements of any model is an elementary submodel? $\endgroup$– Marcus DubiousCommented Apr 23, 2020 at 20:01
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4$\begingroup$ @MarcusDubious To see that the definable elements form an elementray submodel (in the presence of definable Skolem functions), apply the so-called Tarski-Vaught test of elementarity, the test is explained on: math.stackexchange.com/questions/3226832/… $\endgroup$– Ali EnayatCommented Apr 23, 2020 at 20:35
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To move this off the unanswered queue, the answer is yes. For any $\mathcal{M}\models\mathsf{PA}$, the set $\mathcal{M}_{\mathit{def}}$ of parameter-freely-definable elements of $\mathcal{M}$ is an elementary substructure of $\mathcal{M}$ by the Tarski-Vaught test. (More generally this occurs in any theory with definable Skolem functions, another important example being $\mathsf{ZFC+V=L}$.)
This gives us the following quite nice result:
$Th(-)$ provides a bijection between isomorphism types of pointwise-definable models of $\mathsf{PA}$ and completions of $\mathsf{PA}$.
That is, every completion $T$ of $\mathsf{PA}$ has exactly one pointwise-definable model up to isomorphism, and this model is nonstandard iff $T\not=\mathsf{TA}$. A key step to proving the above is the following:
If $\mathcal{A},\mathcal{B}$ are elementarily equivalent and each is pointwise-definable, then $\mathcal{A}\cong\mathcal{B}$.