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If one has a nonstandard model $\mathcal{N}$ of PA and adjoins to the first-order theory the countable list of axioms $1<H,\, 2<H,\, 3<H, \ldots$ (satisfied in $\mathcal{N}$) for all the "standard" natural numbers, an application of downward Lowenheim-Skolem theorem should yield a countable nonstandard model $\mathcal{N}'$ of PA (such as the one described by Skolem in 1933/34).

If one wants to get more of the structure available in Skolem's model, such as transfer for all definable functions, how would one do this exactly using downward L-S? Joel David made some suggestions in this direction in his "comments" below, and I am interested in a possible reference that may fill in some of the details.

More specifically, in Robinson's framework $(\mathbb{R}, {}^{\ast}\mathbb{R}, \ast)$, each hyperinteger $H$ can be included in a countable substructure elementarily equivalent to the Skolem model. What is the strongest sense in which one can take the word "equivalent" here?

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Yes‌‌‌‌‌‌‌‌‌‌‌. –  Emil Jeřábek Aug 22 '13 at 13:24
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One doesn't need a theory to apply the downward LS theorem, but only a structure. So if you have a nonstandard model $\mathcal{N}$ and a nonstandard element $H$ in it, then by LS there is a countable elementary substructure $\mathcal{M}\prec\mathcal{N}$ with $H\in\mathcal{M}$. In contrast, if you take the theory of true arithmetic (true in the standard model $\mathbb{N}$), and add those assertions as you described, then the theory is consistent and hence has a (countable) model by the completeness theorem (rather than by LS). –  Joel David Hamkins Aug 22 '13 at 13:34
    
@Emil and Joel: thanks very much. –  katz Aug 22 '13 at 14:02
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Yes, you can do this also. Just take any situation where you already have full transfer, and take a countable elementary substructure. (You could even add the transfer relation in the formal language of the structure here.) Every definable function will still have its transfer analogue in this countable substructure, by elementarity. –  Joel David Hamkins Aug 23 '13 at 11:40
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My point with that parenthetical remark was that the transfer relation for a given structure is itself a kind of structure (between two structures), and one can make a larger structure in which that transfer structure is explicitly part of the structure, and then get a nonstandard version of that. But this isn't necessary for the rest of your question. Two different structures, say $R$ and $R^\ast$, can be thought of as a single two-sorted structure, which admits non-standard extensions. –  Joel David Hamkins Aug 25 '13 at 11:36

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