Help proving that $\mathfrak{M}_0$ is not isomorphic to $\mathfrak{N}$, Help proving that $\mathfrak{M}_0$ is not isomorphic to $\mathfrak{N}$, where $\mathfrak{N}$ is the standard model of arithmetic and $\mathfrak{M}_0$ is constructed as follows:
To the language of arithmetic $\mathcal{L}_A$ we add a new constant symbol $c$. Let $\Sigma=\{0<c,S0<c,SS0<c,...\}$. And now consider the set $\Sigma\cup Th\mathfrak{N}$, where $Th\mathfrak{N}$ is the theory of arithmetic. Suppose $\mathfrak{M}$ is a model of $\Sigma\cup Th\mathfrak{N}$. And then finally, restrict $\mathfrak{M}$ to the symbols of $\mathcal{L}_A$ before the constant was added. Let $\mathfrak{M}_0$ be this restricted structure.
I think I have to show that while a bijection $h$ exists between the universes of $\mathfrak{M}_0$ and $\mathfrak{N}$ the bijection does not preserve structural properties. My thought was to show that $h(c^\mathfrak{M_0})\neq c^\mathfrak{N}$ for all constant symbols $c$ in $\mathcal{L}_A$. But then I am getting a bit confused:
There is only one constant symbol in $\mathcal{L}_A$, $0$, which in both $\mathfrak{N}$ and $\mathfrak{M}_0$ denote $0$. So, $h(0^\mathfrak{M_0})=0^\mathfrak{N}$.
My gut is that since there is one more element in the universe of $\mathfrak{M}_0$ than $\mathfrak{N}$ then there cannot be a bijection between the two. But since there is no symbol for $c^\mathfrak{M_0}$ how do you show this?
Any and all help is greatly appreciated!
 A: (1) "there is one more element in the universe of $\mathfrak{M}_0$ than $\mathfrak{N}$".
Not so! The non-standard number which interprets $c$ in $\mathfrak{M}$, has a successor (because every number has a successor), which has another successor, etc., with all these successors being distinct non-standard elements in the domain. Moreover, one of the sentences of the theory of arithmetic is that zero is the only number which isn't a successor: so the non-standard number which interprets $c$ has to have a predecessor too, which has another predecessor, etc with all these predecessors being distinct non-standard elements in the domain.
And that's only the beginning of what else has to be in the domain of a non-standard model of arithmetic!
(2) But the fact that the domain of $\mathfrak{M}_0$ contains (not just one but) an infinite number of extra things not in the domain of $\mathfrak{N}$ does not by itself show that the two domains cannot be put in bijection. After all the domain of natural numbers contains an infinite number of extra things not in the domain of even numbers but famously there is a bijection between them!
(3) Having cleared the ground, suppose there IS a bijection $f$ between the domain of $\mathfrak{M}_0$ and the domain of standard model $\mathfrak{N}$. Now ask yourself, where $f$ can send the non-standard elements. There must be a least natural number which is the $f$-image of   a non-standard number: what could it be?? .... and for the answer here jump to Noah Schweber's answer to Precise reason why a non standard model of arithmetic is not isomorphic to the natural numbers
