Peano Axioms.
Let $\mathbb N \neq \emptyset$ be a set and $S:\mathbb N \to \mathbb N$ a function. The elements of $\mathbb N$ are the natural numbers. If $n \in \mathbb N$ then $S(n)$ is the successor of $n$. $\mathbb N$ and $S$ satisfy the following axioms:
* $A_1$: $S$ is onte-to-one.
* $A_2$: $\mathscr R(S) \neq \mathbb N$ i.e $S$ is not surjective.
* $A_3$: If $u \notin \mathscr R(S)$ and $M \subseteq \mathbb N$ such that:
$i)$ $u \in M$.
$ii)$ If $n \in M$ then $S(n) \in M$.
Then $M = \mathbb N.$
And
Theorem. $\exists! u \in \mathbb N: u \notin \mathscr R(S).$
We denote $u$ by $1$.
The problem.
Let $P = \mathbb Z$ and $S:P \to P$ defined by: $$ S(a) = \begin{cases} -a & \text{if $a > 0$} \\ 1 & \text{if $a=0$} \\ -(a-1) & \text{if $a<0$} \end{cases} $$
I already prove that $(P,S,0)$ satisfy axioms $A_1$ and $A_2$ but I have no idea how to prove $A_3$.
My try:
Let $A \subset P$ such that:
- 1. $0 \in A$.
- 2. If $a \in A$ then $S(a) \in A$.
I need to prove that $A = P$. In order to do that I define $B = P-A$ and trying to show that $B = \emptyset$. Lets prove this by contradiction.
If $B \neq \emptyset$ there is $a \in B$ then we have the following cases:
- $a=0$:
- If $a=0$ then $0 \in P-A \Rightarrow 0 \in P$ and $0 \notin A$. This is a contradiction by definition of A. Then $0 \notin B$.
- $a>0$
- $a<0$
I have no clue how to prove the other cases.
If $a\in B$ with $a>0$ we have $a\notin A$, let $b$ be such that $S(b)=a=-(b-1)$ and $b<0$. Here we have either $b\in A$ or $b\notin A$, clearly we cannot have $b\in A$ as that entails that $a\in A$ which we had already excluded. so $b\notin A$. We can continue this predeccesor process and will eventually reach that $S^n(0)=a$ for some $n$, but we already have that $0\in A$ and hence $S^n(0)=a\in A$ as well and we have a contradiction. This is similarly done for the negative case