Return to Question

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 isomorphic to the Skolem model. What is the strongest sense in which one can take the word "isomorphic" here?

Eligible editors who think this is a legitimate question can vote to reopen it by clicking on the "reopen" button in the line below.

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?

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, apparently 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). Is this a correct application of the theorem?

Second, 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.

Eligible editors who think this is a legitimate question can vote to reopen it by clicking on the "reopen" button in the line below.

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 isomorphic to the Skolem model. What is the strongest sense in which one can take the word "isomorphic" here?

Eligible editors who think this is a legitimate question can vote to reopen it by clicking on the "reopen" button in the line below.