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I would like to see an explicit example of a non-trivial elementary extension of the structure $(\mathbb{N}, +, \cdot, 0, 1)$ where $\mathbb{N}$ includes zero. Most of all I am interested in countable ones.

Lisa Lokteva
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2 Answers2

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Unfortunately, Tennenbaum's theorem (https://en.wikipedia.org/wiki/Tennenbaum%27s_theorem) shows that any such extension is non-computable. So there isn't really an explicit example.

That said, if you are happy with ultrafilters, then fixing a nonprincipal ultrafilter $\mathcal{U}$ we can take the ultrapower of $\mathbb{N}$ along $\mathcal{U}$. As long as $\mathcal{U}$ is not countably closed (it's enough to assume for instance that $\mathcal{U}$ is an ultrafilter on $\mathbb{N}$), the result will be a nonstandard elementary extension of $\mathbb{N}$. Unfortunately, it will be uncountable - at least size continuum.

Note that the Henkinization proof of compactness is constructive: it actually builds you a model (countable, even!) of arithmetic which is not isomorphic to the standard model. This is mathematically much tamer in a variety of senses than the ultrapower construction.

Noah Schweber
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  • The proof of the compactness theorem a la Henkin is not constructive (see the first paragraph of your answer and/or look at the proof!). – Rob Arthan Dec 29 '15 at 02:10
  • @RobArthan I was using constructive in the informal sense. Note that the nonconstructivity is in passing to a complete theory extending $Th(\mathbb{N})\cup{c>n: n\in\omega}$; Henkinizing a complete theory is, I believe, constructive. And completing a theory can be done definably (e.g. greedy algorithm). – Noah Schweber Dec 29 '15 at 02:31
  • What informal sense of constructive do you have in mind? – Rob Arthan Dec 29 '15 at 02:38
  • @RobArthan The proof constructs a model, in a definable (if not computable) way. That's all I meant. – Noah Schweber Dec 29 '15 at 03:37
  • Out of interest, what do you mean when you say "completing a theory can be done definably (e.g., greedy algorithms)"? – Rob Arthan Jan 02 '16 at 01:01
  • @RobArthan If T is a recursive theory, then the set of (Godel numbers of) sentences consistent with T is definable in first-order arithmetic. Moreover, this is uniform, so we can iterate it: let $T_0=T$, $T_{n+1}$ be $T_n$ together with $\varphi$ for the (Godel number-)least $\varphi$ not in $T_n$ but consistent with $T_n$. Let $T'=\bigcup T_n$; then $T'$ is not recursive, but is arithmetically definable. – Noah Schweber Jan 09 '16 at 04:12
  • @RobArthan, I tried to give a slightly different answer to your question below. – Mikhail Katz Jan 13 '16 at 16:39
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It is worth pointing out is that nonstandard models of arithmetic are constructible in the logical theory WKL$_0$. Weak König's lemma is sometimes considered to be a form of the axiom of choice; however, it can be proven in classical Zermelo–Fraenkel set theory without the axiom of choice. In the sense of being constructible in ZF, such models can be called explicit. Such an explicit model was constructed by Skolem as early as 1933 using sequences of ordinary integers (i.e., sequences drawn from the so-called intended model).

A closely related discussion is taking place here.

Community
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Mikhail Katz
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