Are the standard natural numbers an outstanding model of PA?

Is the model $\mathcal N$ of the standard natural numbers in any way outstanding from all the possible (non-standard) models of PA? For example, might it be that $\mathcal N$ is some kind of a minimal model, i.e. it is contained in every other model in some useful sense?

I am not very familiar with the terminology here. I heard about elementary submodels, prime models, atomic models, … . I heard that every interpretation of ZFC has its unique natural numbers, hence it seems there is a way to distinguish $\mathcal N$ in there. Answers here seem to imply that there is an embedding of $\mathcal N$ into each non-standard model.

Solutions Collecting From Web of "Are the standard natural numbers an outstanding model of PA?"

Expanding on Qiaochu’s comment, the standard model $\mathbb{N}$ is the unique (up to isomorphism) minimal model of PA in the sense of embeddings. Namely, the following is true:

  • $\mathbb{N}$ embeds uniquely into every model of PA. Moreover, if $\mathcal{M}$ is elementarily equivalent to $\mathbb{N}$, then the embedding of $\mathbb{N}$ into $\mathcal{M}$ is elementary.

  • If $\mathcal{M}$ is a model of PA not isomorphic to $\mathbb{N}$, then $\mathcal{M}$ does not embed into $\mathbb{N}$.

Each of these facts is a straightforward exercise in model theory – the key point being that $\mathbb{N}$ is characterized uniquely amongst models of PA (up to isomorphism) by the second-order principle of induction. Besides giving another characterization of $\mathbb{N}$ (phrased differently: $\mathbb{N}$ is the only well-ordered model of PA), second-order induction lets us reason, well, inductively when comparing arbitrary models of PA with $\mathbb{N}$. For instance, if $\mathcal{M}$ is a model of PA, we build an embedding $f_\mathcal{M}: \mathbb{N}\rightarrow\mathcal{M}$ by induction as:

  • $f_\mathcal{M}(0)=0^\mathcal{M}$, and

  • having defined $f_\mathcal{M}(n)$, we let $f_\mathcal{M}(n+1)=f_\mathcal{M}(n)+^\mathcal{M}1^\mathcal{M}$.

By the second-order induction principle for $\mathbb{N}$, this does in fact define a map from $\mathbb{N}$ to $\mathcal{M}$, and it’s not hard to show that this map is an embedding.

Note, by the way, that PA is overkill here: we may replace it by the theory of the natural numbers with successor. This theory, in contrast to PA, is complete and decidable!

Another very different characterization of $\mathbb{N}$ is via computability: the standard model of PA is the only countable model of PA with a computable presentation, by Tennenbaum’s theorem. This is a more involved result. In particular, while PA is still overkill, there are theories of arithmetic much stronger than arithmetic with successor which are too weak for the Tennenbaum phenomenon to hold for them.