Precise reason why a non standard model of arithmetic is not isomorphic to the natural numbers So I'm looking for a (more) precise reason why a non standard model of arithmetic is not isomorphic to the natural numbers. Under the usual language of arithmetics, +, <, S, $0$, $\times$
I.E) Let $M\vDash\ $Th$(\mathbb{N})$, where $M$ is non standard model obtained by the canonical way using compactness. Show that $M$ is not isomorphic to $\mathbb{N}$
Intuitively it makes (too much) sense that they are not isomorphic because $M$ has a nonstandard number $c$ that is bigger than all of $\mathbb{N}$. But I'm having some trouble penning it down into words.
Thus any easy but convincing proof is deeply appreciated. 
Cheers 
 A: There are a couple ways to do this; they all boil down to the same mathematics, but phrase it a bit differently. The key fact is that the standard model is well-ordered: any nonempty set of natural numbers has a least element. By contrast, if $M$ is a model of $Th(\mathbb{N})$, all we know is that every nonempty definable-in-$M$ subset of $M$ has a least element.

The simplest way is probably the following: in a hypothetical isomorphism from a nonstandard model to the standard model, where can nonstandard elements go? Specifically, suppose $f:M\rightarrow \mathbb{N}$ were an isomorphism. Let $X=\{f(c): c\in M$ is nonstandard$\}$ be the set of images of nonstandard elements. Since $\mathbb{N}$ is well-ordered, $X$ has a least element $x$. But:

*

*$x$ must be $>0$, since $f$ must send $0^M$ and only $0^M$ to $0$, and $0^M$ is not nonstandard.


*If $x$ is $>n$, then $x>n+1$: if $f(c)=n+1$, then $f(c-1)=n$, and $c-1$ is nonstandard if $c$ is.
And this shows that $x$ cannot exist.

Another approach, a bit more abstract, is to think about embeddings of the standard model in the nonstandard model. If $M\models Th(\mathbb{N})$, there's an obvious (necessarily injective) homomorphism $h_M$ from the standard model $\mathcal{N}=(\mathbb{N}; +,\times, <, S, 0)$ to $M$, and if $c\in M$ is nonstandard then $c\not\in ran(h_M)$. It's thus enough to show that if $M\models Th(\mathbb{N})$ (or indeed even much weaker theories of arithmetic), $h_M$ is the only homomorphism from $\mathcal{N}$ to $M$, since any isomorphism $i$ from $\mathcal{N}$ to $M$ would have $c$ in its range, and hence be different from $h_M$.
