1

Apologies if this is an elementary question that should have been obvious to me. I am learning about these topics very much from the perspective of an outside hobbyist, and am not a wizard of logic.

Do all non-standard models of PA add extra axioms? And if they do, doesn't that mean that the PA axioms do define a unique model in some sense? It's just that other models also satisfy those axioms (?).

Noam Hudson
  • 13
  • 2
    "Do all non-standard models of PA add extra axioms? " No; non-standard models of arithmetic are structures that satisfy all PA axioms but have additional elements outside the initial segment. This is due to the fact that first-order axioms for arithmetic do not "identify" the corresponding model up to isomorphism i.e. the first-order theory of arithmetic is not categorical. – Mauro ALLEGRANZA Mar 28 '23 at 10:32
  • But from that same Wiki page, the way it proves the existence of at least one non-standard model via the Compactness Theorem seems to involve adding infinite axioms. That's why I had this impression. – Noam Hudson Mar 28 '23 at 10:35
  • I think I understand though that this isn't necessary. Thanks for your clarification – Noam Hudson Mar 28 '23 at 10:36
  • Does this answer your question? What is an example of a non standard model of Peano Arithmetic? It contains, for example, the ultra power construction. – Lee Mosher Mar 28 '23 at 11:45
  • You may be able to add additional axioms to PA that exclude the standard model of arithmetic. Not sure what those might be though. – eyeballfrog Mar 28 '23 at 12:56
  • @eyeballfrog The usual example: not(Con(PA)). By Gödel's second incompleteness theorem, adding this to PA gives a consistent theory. But the standard model certainly doesn't satisfy it. (Because if you believe there is a standard model--or any model at all of PA--then you believe PA is consistent.) – Michael Weiss Mar 28 '23 at 14:38
  • " doesn't that mean that the PA axioms do define a unique model in some sense?" We know that there are other models (see linked post), from which we conclude that PA axioms do not define a unique model. Having said that, in order to "describe" how those "rogue" models looks like we have to work in a suitable "environment" with further assumptions, simply because all that is provable according to PA cannot do it. – Mauro ALLEGRANZA Mar 28 '23 at 14:46
  • In general, the existence of non-standard models of arithmetic has virtually nothing to do with the incompleteness of PA. Using compactness or ultrapowers, you can construct non-standard models of complete arithmetic which satisfy exactly the same first order sentences as the standard model despite not being (anywhere near) isomorphic to the standard model. – Ned Mar 28 '23 at 14:59

1 Answers1

0

As the comments indicate, there are several ways to show the existence of a nonstandard model of PA. The proof you've seen adds a new constant $c$ plus axioms $c>1$, $c>1+1$, etc. The ultraproduct construction takes a different route.

I don't know that I'd really describe the "$c>1+\ldots+1$" argument as adding new axioms, though. Technically that's true. But the new axioms are all about a brand new constant, and don't say anything about the standard integers.

Another way to prove the existence of non-standard models is to add an axiom like $\neg\text{Con(PA)}$ to PA. The resulting system is consistent by Gödel's second incompleteness theorem, and so has a model $N$. Since the standard model $\mathbb{N}$ satisfies Con(PA), $N$ must be nonstandard. That's more clearly a case of "adding a new axiom".

One says that $\mathbb{N}$ and $N$ are not elementarily equivalent, i.e., don't satisfy the sentences of L(PA) (the language of PA). (Note that L(PA) does not include the extra constant $c$.) So $\mathbb{N}$ and $N$ do not "satisfy the same axioms" in any sense.

The comments also mentioned that PA is not $\aleph_0$-categorical. This is a somewhat different (but related) issue. $\text{Th}(\mathbb{N})$ is so-called "true arithmetic" (or "full arithmetic"): the set of all sentences of L(PA) that are true for $\mathbb{N}$. (While we have no way of deciding, in general, exactly which sentences belong to $\text{Th}(\mathbb{N})$, it's still interesting theoretically.)

Both the "$c>1+\ldots+1$" argument and the ultraproduct construction show that there exists a model, call it $M$, of $\text{Th}(\mathbb{N})$ that is nonstandard. So $M$ and $\mathbb{N}$ are elementarily equivalent but not isomorphic.

We can also insist that $M$ is countable. That's what "not $\aleph_0$-categorical" means: two non-isomorphic countable models. So not just PA, but even $\text{Th}(\mathbb{N})$ is not $\aleph_0$-categorical.

So PA fails to define a unique model in two senses: there are other countable non-isomorphic models, and even stronger, there are other non-elementarily equivalent models.

Michael Weiss
  • 9,212