Axiom equivalent to the induction one I have been studying by myself set theory, natural numbers
 and properties of the real numbers, because i will enroll in a course of introduction to real analysis next semester in my university. I live in Brazil, and i am using a book written by a respected local author. I was capable of doing some exercises, but there are some that i cannot seem to have success. This one is an example that i have been struggling for more than a week. Here is the problem and my thoughts on how to solve it:
Let $P1$ be the axiom that says " $s: \mathbb N \rightarrow\ \mathbb N$ is injective , $s$ being the succesor function" .
Let $P2$ be the axiom that says " $\mathbb N \setminus s(\mathbb N)$ consists of one element, the element ${1}$. "
Prove that in the presence of the first two axioms, the following statements are equivalent:
$(i)$ Let $X \subset \mathbb N$ be a subset that ${1} \in X$ and $\forall {n} \in X, s({n}) \in X$. Then $X= \mathbb N$.
$(ii)$For every $ A \subset \mathbb N$, $ A \neq \emptyset  ,$ it follows  that $  A \setminus s(A) \neq \emptyset $  
I tried to use contradiction, letting $Y= \mathbb N \setminus X$ , and assuming $Y$ is not empty. If ,$Y$ is not empty, then $Y-s(Y) \neq \emptyset$. I was trying to somehow get to the conclusion that $Y-s(Y)$ was empty, and get to a contradiction, but i couldn't grasp what would the elements of $Y$ be.
Yet another attempt, was to try to prove that $X=N$ by showing that $ X \subset \mathbb N $ and $ \mathbb N \subset X $. $X $ is defined to be a subset of $\mathbb N$ but i could not get to the other part using the $(ii)$ condition.
I can't see why the $(ii)$ is needed or revelant. It seems to me that is a much weaker "axiom" than the first one.
This is not part of any homework assigment, i just honestly really like math.
 A: $\left(i\right)\implies\left(ii\right)$
Let $A\subseteq\mathbb{N}$ and $A\neq\varnothing$.
Define $B_{1}=A$ and $B_{s\left(n\right)}=s\left(B_{n}\right)$ for
every $n\in\mathbb{N}$. 
Let $B$ be the union of these sets.
Then $\mathbb{N}-B\neq\mathbb{N}$ so $\left(i\right)$ tells us that
$1\in B\vee\exists n\in\mathbb{N}\left[s\left(n\right)\in B\wedge n\notin B\right]$.
If $1\in B$ then also $1\in A$. This because $1\in B-A$ leads to $1\in s(\mathbb N)$ which is not true.
Then $1\in A-s\left(A\right)$ and we are ready.
If $1\notin B$ then $\exists n\in\mathbb{N}\left[s\left(n\right)\in B\wedge n\notin B\right]$.
If $s\left(n\right)\in B_{s\left(k\right)}=s\left(B_{k}\right)$ for
some $k\in\mathbb{N}$ then $s\left(n\right)=s\left(m\right)$ for
some $m\in B_{k}$. 
Then the injectivity of $s$ tells us that $n=m\in B_{k}\subseteq B$
and we arrived at a contradiction. 
So we conclude that $s\left(n\right)\in B_{1}=A$.
The injectivity of $s$ assures us that $s\left(n\right)\in s\left(A\right)$
if and only if $n\in A$ which is not the case. 
So $s\left(n\right)\in A-s\left(A\right)$.

edit
$\left(ii\right)\implies\left(i\right)$
Taking $X=A^{c}$ we can write $\left(ii\right)$ as:
$s\left(X^{c}\right)^{c}\subseteq X\implies X=\mathbb{N}$
It is obvious that we can write $\left(i\right)$ as: 
$1\in X\wedge s\left(X\right)\subseteq X\implies X=\mathbb N$.
So we are ready if we can show that $1\in X\wedge s\left(X\right)\subseteq X\implies s\left(X^{c}\right)^{c}\subseteq X$.
For $X\subseteq\mathbb{N}$ have $\mathbb{N}-\left\{ 1\right\} =s\left(\mathbb{N}\right)=s\left(X\cup X^{c}\right)=s\left(X\right)\cup s\left(X^{c}\right)$
which is a union of disjoint sets since $s$ is injective.
So actually we have $s\left(X^{c}\right)^{c}=s\left(X\right)\cup\left\{ 1\right\} $
which is exactly what we needed.
