## Embedding of standard model of arithmetic to PA-model [closed]

Hi,

I am working on the following problem:

Let $S_{Arithmetic} = \{+, *, 0, 1\}, \mathfrak{M}$ a model for PA (first-order peano axioms) }, and $\mathbb{N} = (\mathbb{N},+ ^{\mathbb{N}}, *^{\mathbb{N}},0^{\mathbb{N}},1^{\mathbb{N}} )$.

Construct an embedding $f : \mathbb{N} \rightarrow |\mathfrak{M}|$ and show that f is unique.

So, for $f$ to be an embedding, the following has to be true (correct me if I'm wrong):

• $0^\mathfrak{M} = f(0^\mathbb{N})$
• $1^\mathfrak{M} = f(1^\mathbb{N})$
• $+^\mathfrak{M} (f(a_0),f(a_1)) = f(+^\mathbb{N}(a_0,a_1))$
• $*^\mathfrak{M} (f(a_0),f(a_1)) = f(*^\mathbb{N}(a_0,a_1))$
• $f$ injective

Now, if I set $f$ to $f(x) := x$, it seems to me that all these properties are satisfied, but I am not sure what to do to prove this assignment and what to do to show that $f$ is unique.

I would be glad if someone could point me in the right direction.

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Hint: Suppose that both $f,g$ are such embeddings and prove by induction that $f(n)=g(n)$ for all $n\in\mathbb N$. – Asaf Karagila Jun 20 at 21:13
Thank you for your answer! So $f(x):=x$ is the correct embedding? Do I need to show something else aside from the 5 points listed above? (and the uniqueness you just pointed out) – Siegler Jun 20 at 21:16
If PA is consistent then it has models as large as I would like. I can take, therefore, a model of size continuum and by transport of structure think of its elements as co-countable subsets of the real line. Why would $f(x)=x$ be an embedding from $\mathbb N$ to this model? It won't be because the identity is not a function from $\mathbb N$ into the power set of the real numbers. To construct an embedding do the obvious, use induction (again) to define $f(n)$. – Asaf Karagila Jun 20 at 21:29
I don't really understand what to show / why I have to use induction. So I have no clue how to define $f(n)$ inductively. – Siegler Jun 20 at 21:43
For future reference, a question like this which albeit being well-posed make a bad question for this site but a good fit for math.stackexchange.com – Asaf Karagila Jun 20 at 21:51
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