Are we in a nonstandard model of arithmetic?

Question

My question is: is $\mathbb{N}$ a nonstandard model of arithmetic for someone else?

Let $\mathbb{N}^*$ be a nonstandard model of Peano Arithmetic. Then that consists of a copy of $\mathbb{N}$ followed by a dense linear order of copies of $\mathbb{Z}$.

Is the reverse also true? Is our familiar $\mathbb{N}$ nonstandard with respect to someone else?

If this question is ill-posed, it is because I do not fully understand the nuances of model theory, not to mention non-standard model theory. So feel free to re-pose this precisely and answer it.

I would like to understand whether some models of arithmetic are inherently richer than others, or whether they are all equally rich.

Answer 1

Your question is not ill-posed and in fact admits a rather precise answer in the context of Joel David Hamkins' multiverse. The technical details of this may be beyond the level of this question but the basic assumption, on this view, is that there is not a single set-theoretic universe (governed by ZFC or whatever) but rather a multitude of such universes, in precise relationship to each other governed by the rules set out by Hamkins and collaborators.

On this view, whichever universe a particular observer happens to be in, his particular version of $\mathbb N$ is indeed nonstandard with respect to another universe in the Hamkins multiverse.

This recent MO thread provides additional information relevant to this question.

For a discussion see this paper.

Answer 2

The natural numbers are the prime model of PA: in a precise sense, they are the least rich model. There are lots of notions of richness of models; these are usually expressed in terms of the realized types, essentially what sort of behavior occurs in the model. E.g. the natural numbers are not very rich since "is divisible by $2$ and $3$ and $5$ and ..." is a very easily-describable behavior which is not realized in $\mathbb{N}$.

As to whether $\mathbb{N}$ is nonstandard in any sense: certainly not in the sense that we use the word when speaking of nonstandard models. Since $\mathbb{N}$ is the smallest model of PA, there's no way to shrink it further and still have a model of PA.

Incidentally, there's an interesting set-theoretic difficulty here: conceivably the thing we think is $\mathbb{N}$ could actually be a very clever nonstandard model. This has been investigated by a number of mathematicians and philosophers, and has had lots of discussion on this site (see e.g. this question). But that wouldn't mean "$\mathbb{N}$ is nonstandard," that would mean "we were wrong about what $\mathbb{N}$ is" - $\mathbb{N}$ is defined as the smallest thing it could be (this can be made completely precise via second-order logic (this question says a bit about that), but since second-order logic is intimately bound up in set theory this isn't entirely satisfying).