In the book "Numbers" by Ebbinghaus et. al, the Natural numbers are defined as:
The natural numbers form a set $\mathbb N$, containing a distinguished element $0$, called zero, together with a successor function $S:\mathbb N \to \mathbb N$, of $\mathbb N$ into itself, which satisfies the following axioms:
(S1) $S$ is injective,
(S2) $0 \notin S(\mathbb N)$,
(S3) If a subset $M \subset \mathbb N$ contains zero and is mapped into itself by $ S $, then $M = \mathbb N$.
Does this definition puts any strict requirement that only $0$ can be starting element? In other words, does (S2) excludes any possibility of another starting element, say $w_0 \notin S(\mathbb N)$? If it was strict requirement, I think the author would have explicitly mentioned something like :(S2) only $0 \notin \mathbb N$
Also, as (S1) says $S$ is just injective (and not surjective), so there can be some other elements outside of range of $S(\mathbb N)$ similar to $0$.
The reason I ask, as if there exists such element $w_0 \notin S(\mathbb N)$, we can have a chain of elements in sequence $ w_0 \to w_1 \to w_2 $ and so on with starting element $w_0$ independent of the one with the $0$ (zero). Then, there can be two different sets, $M_1$ and $M_2$ for S(3), where $M_1$ include $w_0$ and $M_2$ doesn't, but both still include $0$.
then, $M_1 = \mathbb N$
and also, $M_2 = \mathbb N$
giving two different sets for $\mathbb N$ with different sizes, which seems wrong to me.
Can you please let me know what I understood wrong here?
Suppose that $\mathbb N$ contains an element $w_0$ like you describe, that is, $w_0 \neq 0$ yet $w_0$ is not the successor of any element. Then consider the set $\mathbb N \setminus \{w_0\}$. Clearly this set contains 0. And it is also mapped into itself by $S$, since by assumption $w_0$ is not the successor of any element. It follows by (S3) that $\mathbb N \setminus \{w_0\} = \mathbb N$. But this is a contradiction. So no such element $w_0$ exists.
More informally, consider the set $$ \{0, S(0), S(S(0)), S(S(S(0))), \ldots\}. $$ This set satisfies the requirements of (S3), and therefore it is precisely this set that is $\mathbb N$, and there cannot be any other elements.