Recently I try to figure out some facts about one specific way of "constructing" a non-standard model for (peano) arithmetic. I guess there are answer to my question already out there, but somehow I cannot find them...
The construction I've in mind goes like this: first let $S_N = (\bar 0, \overline{succ}, \bar +, \bar \cdot)$ be the signature of a first-order language $L(S_N)$ and let $\mathcal{N} = (\mathbb{N}, 0, succ, +, \cdot) $ be a structure of the natural numbers.
Then, set $S' = S \cup\{c\}$ and $\Gamma = Th_{L(S')}(\mathcal{N}) \cup \{ \neg c \equiv 0, \neg c \equiv \overline{succ}(0), \neg c \equiv \overline{succ}(\overline{succ}(0)), \ldots\}$ where $Th_L(\mathcal{M}) := \{ \phi \in Fm_L \mid \mathcal{M} \vDash \phi\}$.
Given the completeness theorem, $\Gamma$ is satisfiable and therefore let $\mathcal{N}^*$ such that $\mathcal{N}^*\vDash \Gamma$.
Does this construction work?
$S'$ does not contain any relation symbol. So, if one wants to talk about ordering within the model $\mathcal{N}^*$ one has to add $\leq$ to the structure $\mathcal{N}$ and to the signature $S_N$, right?
Given such a symbol $\leq$, how can one proof that $c^{\mathcal{N}^*} > n$ and not $n_0 < c^{\mathcal{N}^*} < n_1$ where $n, n_0, n_1 \in \mathbb{N}\subseteq |\mathcal{N}^*|$.
Is the set $Th_{L(S')}(\mathcal{N}) $ countable? Why?
Thanks for any help. : )