Peano and induction according to Schaum's Outline As some of you may have noticed, I've been coming at Peano from a few different angles. This time I'm stuck on what the Schaum's Outline Abstract Algebra version might mean. Here's Schaum's Peano:


$1 \in \mathbb{N}$.
For each $n \in \mathbb{N},\:$ $\exists!\; n^* \in \mathbb{N}$ called the successor of $n$.
For each $n \in \mathbb{N}$ we have $n^* \neq 1$.
If $m, n \in \mathbb{N}$ and $m^* = n^* \implies m = n.$
Any subset $K$ of $\mathbb{N}$ having the properties (i) $1 \in K$, (ii) $k^* \in K$ whenever $k \in K$ is equal to $\mathbb{N}.$


It then says 


  
  states that the first natural number is $1$
  


which I can only assume is because 2. says there is a successor to a member of $\mathbb{N}$ called $n^*$ and nothing in $\mathbb{N}$ may have $1$ as a successor. That implies that $1$ cannot have as a successor $1$. To me this only establishes that $1^*$ of $1$ is something else in $\mathbb{N}$, but yes, there is an order to $\mathbb{N}$ where $1$ must be first. Then, 


  
  states that distinct natural numbers $m$ and $n$ have distinct
  successors $m + 1$ and $n + 1$.
  


Not sure where it is getting $m + 1$ and $n + 1$. This seems like an unwarranted supposition based on a semantic understanding of the word successor, as yet not stated or proven.

Then it goes on with a proposition aimed at establishing induction.

$P(m): m^* \neq m, \quad \forall\, m \in \mathbb{N}$

Then with the five postulates above, it wants to establish the $P$ proposition above. So, define

$K = [k : k \in \mathbb{N}, P(k) \;is\: true]$


  Then  $1 \in \mathbb{N}$ by 1., and $1^* \neq 1$ by 3. Thus, $P(1)$ is true and $1 \in K$. Next let $k$ be any element of $K$, then (a) $P(k): k^* \in k$ is true.


I guess this comes from simply repeating the proposition using items from $K$?


  Now if $(k^*)^* = k^*$ it follows from 4. that $k^* = k$, a contradiction of (a). Hence $P(k^*): (k^*)^* \neq k^*$ is true and so $k^* \in K$. Now $K$ has the two properties stated in postulate 5, thus, $K = \mathbb{N}$ and the proposition is true for all $m \in \mathbb{N}$.


What's eating me is whether the $k^* \neq k$ then $(k^*)^* \neq k^*$ really means "next", i.e., is this enough to really establish a successor as "the next number?" I don't see the original postulates really establishing this -- or are they?
$1 \in \mathbb{N}$.
  For each $n \in \mathbb{N},\:$ $\exists!\; n^* \in \mathbb{N}$ called the successor of $n$.
  For each $n \in \mathbb{N}$ we have $n^* \neq 1$.
  If $m, n \in \mathbb{N}$ and $m^* = n^* \implies m = n.$
  Any subset $K$ of $\mathbb{N}$ having the properties (i) $1 \in K$, (ii) $k^* \in K$ whenever $k \in K$ is equal to $\mathbb{N}.$
  


Maybe it would help to visualize the set of natural numbers with arrows indicating succession.

$1\to 2 \to 3 \to 4 \to \cdots$

In this chain of numbers:


There is exactly one outgoing arrow from each number, i.e. the successor relation is a function (axiom 2). 
$1$ has no incoming arrow, i.e. no number has a successor of $1$ (axiom 3). 
Every number has at most one incoming arrow, i.e. the successor function is one-to-one or injective (axiom 4).
There are no natural numbers that will not be included in the chain of numbers pictured above. In other words, you would be able to reach any number if you start at $1$ and repeatedly went from one number to the next (axiom 5).
$1 \in \mathbb{N}$.
  For each $n \in \mathbb{N},\:$ $\exists!\; n^* \in \mathbb{N}$ called the successor of $n$.
  For each $n \in \mathbb{N}$ we have $n^* \neq 1$.
  If $m, n \in \mathbb{N}$ and $m^* = n^* \implies m = n.$
  Any subset $K$ of $\mathbb{N}$ having the properties (i) $1 \in K$, (ii) $k^* \in K$ whenever $k \in K$ is equal to $\mathbb{N}.$
  


Maybe it would help to visualize the set of natural numbers with arrows indicating succession.

$1\to 2 \to 3 \to 4 \to \cdots$

In this chain of numbers:


There is exactly one outgoing arrow from each number, i.e. the successor relation is a function (axiom 2). 
$1$ has no incoming arrow, i.e. no number has a successor of $1$ (axiom 3). 
Every number has at most one incoming arrow, i.e. the successor function is one-to-one or injective (axiom 4).
There are no natural numbers that will not be included in the chain of numbers pictured above. In other words, you would be able to reach any number if you start at $1$ and repeatedly went from one number to the next (axiom 5).