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Let $\mathfrak{A}=(\mathbb{N},S,0)$ be a structure where $S$ is the sucessor function.

Let $\mathfrak{B} =(\mathbb{N}\times \{0\} \cup \mathbb{Z} \times\{1\} ,S, 0)$ with $0 = (0,0)$ and

$$ S(k,i) = S(k+1,i) $$

the structure where informally speaking "first the natural numbers and then the integers".

Obviously, $|\mathbb{N}| = |\mathbb{N} \times \{0\} \cup \mathbb{Z}\times \{1\}|$.

But why are $\mathfrak{A}$ and $\mathfrak{B}$ not isomorphic? I am not looking for a proof but for an intuition.

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up vote 3 down vote accepted

If $f : \mathfrak{A} \to \mathfrak{B}$ preserves the successor function, then an easy induction argument shows that $f(n) = (n,0)$ for every natural number $n$. Therefore $f$ is not surjective.

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Hint: For all $n\in \mathbb N$, $S^n(0)=n$, but there is $a\in \mathfrak B$ such that $s^n(0)\neq a$ for all $n\in \mathbb N$.

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