I have a question about a proof I have seen that there is a $\mathcal{L}_<$ model $\mathcal{M}$ of Th$(\mathbb{N})$ with an infinite descending chain $\dots a_3<a_2<a_1<a_0$ with $a_i\in M$ for all $i\in\mathbb{N}$.
My question is about a detail in the proof of this proposition. I am hoping that is is just a minor detail, and also hopefully a minor point of misunderstanding. I abbreviate some of the proof, but will mostly be faithful.
Construct a formula $\theta(v_0)$ that says that $v_0$ is minimal in $\mathcal{M}$. Construct another formula $\psi(v_0,v_1)$ that says that $v_1$ is the immediate successor to $v_0$. Let $\mathcal{L}'=\mathcal{L}\cup\{c\}$, where $c$ is a new constant symbol. For $n\in\mathbb{N}$ let $\phi_n$ be the $\mathcal{L}'$-formula $$ \exists v_0\exists v_1\exists\dots\exists v_{n-1}\theta(v_0)\land\psi(v_0,v_1)\land\psi(v_1,v_2)\land\cdots\land\psi(v_{n-2},v_{n-1})\land v_{n-1}<c $$ Let $T=\text{Th}(\mathbb{N}_<)\cup\{\phi_n|n\in\mathbb{N}\}$. For any finite subset $S$ of $T$, we can interpret $c$ to satisfy $S$. So by the compactness theorem $T$ is satisfiable.
So far I understand everything well in the proof, but when it comes time to showing that this means we can construct an infinite descending chain, I started to doubt myself. I will continue with the proof up to the point that I started to get confused:
Let $\mathcal{N}$ be a model of $T$ and let $a\in N$ be such that $a=c^\mathcal{N}$.
Clearly, for all $n\in\mathbb{N}$ we have $n<a$.
Further, the following sentence, $\phi$, is true in $\mathbb{N}$ and so is an element of Th$(\mathbb{N})$: $$ \forall v_0\theta(v_0)\lor\exists v_1\psi(v_1,v_0) $$ Let $a_0=a$. Then, since $\phi$ is true there is $a_1\in\mathcal{M}$ such that $\mathcal{M}\vDash\psi(a_1,a_0)$.
Now this is where I start to get confused. I don't see why the fact that $a_1\notin\mathbb{N}$ allows us to iterate.
Moreover, $a_1\notin\mathbb{N}$ since for any $n\in\mathbb{N}$ if $\mathbb{N}\vDash\psi(n,m)$ then $m\in\mathbb{N}$.
Consequently we can iterate, obtaining $a_2\in M\backslash\mathbb{N}$ such that $\mathcal{M}\vDash a_2<a_1$, and so on. This gives us $\{a_i|i\in\mathbb{N}\}$ such that for all $i\in\mathbb{N}$ we have $a_i\in M\backslash\mathbb{N}$ and $\mathcal{M}\vDash a_{i+1}<a_i$. $\quad\square$
My question comes right at the end of the proof. Why do we have to take the time to assert that these $a_i\notin\mathbb{N}$? Why would this not work if $a_i\in\mathbb{N}$? Can anybody help me to see how this last part of the argument is working? Any help would be appreciated!