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Maybe it's well known to experts, but is there always an embedding $f$ of the standard model of Peano arithmetic into a nonstandard model? By Peano arithmetic I mean its first-order version, with the axiom schema of induction for each formula. I'm stuck at the step how to show $f$ is well defined. We can define $f(0)$ and if $f(n)$ is defined we can define $f(n+1)$. But then how can I show $\forall n, f(n)$ is well defined. More specifically, is the formula "$f(n)$ has a unique value" a valid one to use induction on?

Mikhail Katz
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Fan Zheng
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2 Answers2

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You need to remember that the embedding is not "definable" in the models you work with, it lives in the set theoretic universe outside the models.

So by verifying that $f(0)=0$ and $f(n+1)=f(n)+1$ only requires you to check that indeed this is a closed term (it has no variables), in which case its interpretation in the nonstandard model is unique.

It is also worth noting that (1) this embedding is not necessarily an elementary embedding, since $\sf PA$ is not a complete theory; and (2) the image of this embedding is not internally definable in the nonstandard model, since it is a subset which is inductive (includes $0$ and closed under successor) while $\sf PA$ proves the only definable inductive subset is the entire model.

Asaf Karagila
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    In note (1), you mean that the embedding need not be elementary. Some nonstandard models are elementary extensions of the standard model. – Andreas Blass Mar 28 '15 at 14:18
  • Now I sort of know where I'm stuck: if the domain of $f$ is the standard model, we can apply induction to sets to show $f$ is well defined. If we attempt the same thing with the domain replaced by a nonstandard model, we cannot show $f$ is well defined, for now we can only apply induction to internally defined predicates, but $f$ as a function is not internally defined in a first order theory, right? – Fan Zheng Mar 28 '15 at 14:35
  • @Andreas: Of course. I wrote this just as I woke up, from my phone. :-) – Asaf Karagila Mar 28 '15 at 14:43
  • @Fan: The embedding is not even a subset of the universe of the standard model. Of course it is not definable. – Asaf Karagila Mar 28 '15 at 14:44
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Suppose $\mathfrak A$ is your nonstandard model. You can define $f:\mathbb N\rightarrow\mathfrak A$ by recursion: $f(n+1)=$ the least element of $S_n:=\{x\in |\mathfrak A|: x>f(n)\}$.

It is a general fact that you define a function by (transfinite) recursion over a well-order such as $\mathbb N$, see e.g. Enderton, Elements of Set Theory.

Bjørn Kjos-Hanssen
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