So basically I want to find out what is the upper bound of the "number" of countable models of
Th($\mathbb{N})$ up to isomorphism.
In the book by Richard Kaye, on Peano arithmetics, he simply said
"clearly there are at most $2^{\aleph_0}$ of countable models of Th($\mathbb{N}$)".
But regrettably, it was not clear to me why is this the case.
Any insight on this is deeply appreciated.
Cheers
Up to isomorphism, a countable model can assumed to have underlying set $\mathbb{N}$. So the model is determined by a choice of two elements of $\mathbb{N}$ to be $0$ and $1$ and two maps $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ to be addition and multiplication. Thus there are at most $$|\mathbb{N}|\cdot|\mathbb{N}|\cdot|\mathbb{N}^{\mathbb{N}\times\mathbb{N}}|\cdot|\mathbb{N}^{\mathbb{N}\times\mathbb{N}}|=2^{\aleph_0}$$ countable models, up to isomorphism.