Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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.

share|improve this answer
add comment

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 $n\in \mathbb N$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.