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The book is Predicate Calculus by Goldrei. Given the hint and other similar exercises, this is the only way I know how to go about this:

Take the set $\text{Th}(\mathcal{N}) \cup \{ \textbf{c} \not = \underbrace{\textbf{1}+...+\textbf{1}}_{n} | n \in \mathbb{N}\}$. Then $\mathbb{N}$ models each finite subset since we can let $\textbf{c}^\mathcal{N}$ be any number that isn't of the form $\underbrace{\textbf{1}+...+\textbf{1}}_{n}$ for the values of $n$ in the subset. Therefore the union has a model, $\mathcal{M}$, lets say with domain $M$. Since the cardinality of this new language is finite, then by the DLS theorem $M$ has cardinality at most $\aleph_0$.

But now I have just changed the language by adding a constant symbol $\textbf{c}$. Is it valid to say that $\mathcal{M}$ models the original langauge because it is a subset of this new langauge?

Now for showing that the two models are not isomorphic. I know the argument show go something like this: Some $n$ in the standard model must get mapped to an element in $M$ that interprets the new constant symbol $\textbf{c}$, but we have $n = \underbrace{1+...+1}_{n}$ yet $\textbf{c}^{\mathcal{M}} \not = \underbrace{1+...+1}_{n}$ for any $n$.

I am having trouble formalizing this. Let $\theta : \mathbb{N} \to M$.

$$(n,\underbrace{1,...,1}_{n}) \in =^{\mathcal{N}}$$ iff

$$(\theta(n),\underbrace{\theta(1),...,\theta(1)}_{n}) \in =^{\mathcal{M}}$$

(actually this an abuse of notation, as the former should be $(n, +^{\mathcal{N}}(\textbf{1}^{\mathcal{N}}, +^{\mathcal{N}} (\textbf{1}^{\mathcal{N}}, +^{\mathcal{N}}(...)) ) )$, but this is far less readable)

Now, by the premise $\theta(n) = \textbf{c}^{\mathcal{M}}$, but what the heck should $\theta(1)$ be? If I am mapping into some general domain then how do I know what the elements are, let alone how $+^{\mathcal{M}}$ and $\cdot^{\mathcal{M}}$ are defined?

If we try to make it more concrete by using what should intuitively be a model, $\mathbb{N} \cup \{ \infty \}$, well first we would have to prove that this is a model of $\text{Th}(\mathcal{N})$ and how do we do that??

ngc1300
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    $\mathbb{N}\cup{\infty}$ is definitely not a model of $Th(\mathcal{N})$ in any natural way - think about the sentence "$\forall x\exists y(x<y)$" (fine, "$<$" isn't in your language, but it can be defined in terms of "$+$" which is). – Noah Schweber May 21 '22 at 02:26
  • @NoahSchweber Ok, but now I'm confused. See the last few comments https://math.stackexchange.com/a/1565/455950 – ngc1300 May 21 '22 at 02:29
  • Note the last sentence of Jason DeVito's second-to-last comment: "If you're thinking that (the interpretation of) $c$ gives a counterexample, note that $M$ contains something it thinks of as $c+1$, which is larger than $c$." – Noah Schweber May 21 '22 at 02:39
  • @NoahSchweber Ok so $\mathbb{N} \cup { \infty_n }$ where $\infty_{n+1} = 1+\infty_{n}$ would be a model? – ngc1300 May 21 '22 at 02:45
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    No - what about $\infty_0\cdot\infty_0$? Or $\infty_0-1$? Or $\lfloor {\infty_0\over 2}\rfloor$? Even worse, we must have something which our model thinks is the "best approximation" to $\sqrt[\infty_5-2]{{(\infty_3)^3\over \infty_1}}$; what is it? Models of arithmetic are incredibly complicated beasts. In fact, no model of $\mathsf{PA}$ (or even much less) can be easy to describe in a precise sense! (See also here for another example of something which turns out to not be a model.) – Noah Schweber May 21 '22 at 02:47
  • @NoahSchweber Intersting, thanks. So $M$ can never be recursively defined. So how do we go about showing that $M$ isn't isomorphic? – ngc1300 May 21 '22 at 02:54
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    We don't need to concretely describe $M$ completely in order to show that there is no isomorphism $\mathbb{N}\rightarrow M$. Use induction in $\mathbb{N}$: show that no element of $\mathbb{N}$ can be mapped to any element of $M$ with infinitely many predecessors (such as $c$, for example). – Noah Schweber May 21 '22 at 02:59
  • @NoahSchweber Here's my attempt. Define $x<y$ as $\exists z \ x+z = y$. $0$ can't be mapped to $\textbf{c}^\mathcal{M}$ because $\lnot \exists x \ x<0$ is true for $\mathbb{N}$ but $\lnot \exists x \ x<c$ false for $M$, since we know that any element that interprets $\underbrace{\textbf{1}+...+\textbf{1}}{n}$ would be a counter example. Any other $m$ cannot get mapped to $\textbf{c}^\mathcal{M}$ because $\underbrace{\textbf{1}+...+\textbf{1}}{n} < c$ is true in $M$ for all $n$ but $\underbrace{\textbf{1}+...+\textbf{1}}_{n} < m$ is only true in $\mathbb{N}$ when $n < m$. – ngc1300 May 21 '22 at 20:50
  • Not sure where induction comes in? Also why does DeVito's model guarantee that there is an element $c+1$ but $\mathbb{N} \cup { \infty }$ doesn't? – ngc1300 May 21 '22 at 20:52

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