Showing existence of a non-standard model of arithmetic elementarily equivalent to standard model of arithmetic

Question

Let $\mathcal{M}_A=\langle\mathbb N, 0^{\mathcal{M}_A}, s^{\mathcal{M}_A}, +^{\mathcal{M}_A}, \times^{\mathcal{M}_A}, <^{\mathcal{M}_A}\rangle$ be the standard model for the language of arithmetic $\mathcal L_A$. Define theory $\text{Th}(\mathcal{M}_A)=\{\alpha : \mathcal{M}_A\models\alpha\}$.

I am trying to show the existence of model $\mathcal M$ that is elementarily equivalent to $\mathcal{M}_A$ and contains at least one non-standard element.

I can start the proof as follows: Extend $\mathcal{L}_A$ with a new constant symbol $c$, and call this such language $\mathcal{L}'$. Define $\Gamma=\text{Th}(\mathcal{M}_A)\cup\{\underbrace{s\dots s}_{\text{$n$ times}}(0)<c : n\in\mathbb N\}$. One can use compactness to demonstrate that $\Gamma$ is finitely satisfiable, and hence satisfiable. So there is some model $\mathcal M$ so that $\mathcal{M}\models\Gamma$ and contains non-standard element in its domain.

I am not sure whether $\mathcal M$ is elementarily equivalent to $\mathcal M_A$ (with respect to $\mathcal L_A$). I think the answer is yes but I am not sure how to prove it. To show that $\mathcal M$ is elementarily equivalent to $\mathcal M_A$, I need to demonstrate that $\mathcal M$ is the expansion of $\mathcal M_A$ to $\mathcal L'$. In other words, I need to show that for any constant, function and relation symbols in $\mathcal L_A$,

(a) $\text{Domain}(\mathcal M)=\text{Domain}(\mathcal M_A)$

(b) $c^{\mathcal M}=c^{\mathcal M_A}$

(c) $f^{\mathcal M}=f^{\mathcal M_A}$

(d) $R^{\mathcal M}=R^{\mathcal M_A}$

I am certain that (b) holds but I am not sure about (a), (c) and (d). Since $\mathcal M$ contains non-standard elements in its domain, I think (a) does not hold (?). The successor function $s$ in $\mathcal M$ is different from the one from $\mathcal M_A$ because when a non-standard element $c$ is involved, $s(c)$ will map to some bigger non-standard element (?). $<$ relation includes some non-standard elements as well.

Also, I wonder if there are sentences $\alpha$ in $\mathcal L_A$ that satisfies $\mathcal M\models\alpha$ but $\mathcal M_A\not\models\alpha$.

Answer 1

Every sentence $\alpha$ in $\mathcal{L}_A$ is either true in $\mathcal{A}$ or false in $\mathcal{A}$. Hence for every sentence $\alpha$ either $\alpha\in\text{Th}(\mathcal{M}_A)$ or $\neg\alpha\in\text{Th}(\mathcal{M}_A)$. Now note that $\mathcal{M}\models\text{Th}(\mathcal{M}_A)$. There is nothing more for $\mathcal{M}$ to add to $\text{Th}(\mathcal{M}_A)$. The addition of a single sentence will result in triviality.

Answer 2

Continuation of the proof:

... Since $\Gamma$ is satisfiable. There is some $\mathcal M'$ s.t. $\mathcal M'\models\Gamma$. Let $\mathcal M$ be reduct of $\mathcal M'$ by removing symbol $c$. $\mathcal M$ is a structure for $\mathcal L_A$ and its domain contains non-standard elements but they are not value of any term in $\mathcal L_A$. That is, $\mathcal M$ is not covered. To show that it is elementarily equivalent to the standard model $\mathcal M_A$, we need to demonstrate that for any $\mathcal L_A$ sentence $\alpha$, $\mathcal M_A\models\alpha$ iff $\mathcal M\models\alpha$. Let $\alpha$ be a $\mathcal L_A$ sentence.

($\Longrightarrow$) Trivial since $\mathcal M\models\text{Th}(\mathcal M_A)$, i.e. $\text{Th}(\mathcal M_A)\subseteq\text{Th}(\mathcal M)$.

($\Longleftarrow$) Suppose $\mathcal M\models\alpha$. Then, $\mathcal M\not\models\lnot\alpha$. Then, $\lnot\alpha\notin\text{Th}(\mathcal M)$. Since $\text{Th}(\mathcal M_A)\subseteq\text{Th}(\mathcal M)$, $\lnot\alpha\notin\text{Th}(\mathcal M_A)$. Since $\text{Th}(\mathcal M_A)$ is complete, $\alpha\in\text{Th}(\mathcal M_A)$.

Also, I wonder if there are sentences $\alpha$ in $\mathcal L_A$ that satisfies $\mathcal M\models\alpha$ but $\mathcal M_A\not\models\alpha$.

No. Let $\alpha$ be a sentence in $\mathcal L_A$. Suppose $\mathcal M\models\alpha$. This means $\alpha\in\text{Th}(\mathcal M)$. $\text{Th}(\mathcal M)$ consists of sentences made of symbols from $\mathcal L_A$ only (everything from $\text{Th}(\mathcal M_A)$ since $\mathcal M\models\text{Th}(\mathcal M_A)$) and sentences made of symbols from both $\mathcal L_A$ and $\mathcal L'$. Since $\alpha$ is a sentence in $\mathcal L_A$, we can conclude that $\alpha\in\text{Th}(\mathcal M_A)$, i.e. $\mathcal M_A\models\alpha$.