I am working with Peano Axioms as follows:
- $0 \in \mathbb{N}$.
- $n \in \mathbb{N} \implies S(n) \in \mathbb{N}$.
- $(\forall n \in \mathbb{N})(S(n) \neq 0)$.
- $n \neq m \implies S(n) \neq S(m)$.
- Let $A$ be a subset of $\mathbb{N}$. If $0 \in A$ and $n \in A \implies S(n) \in A$, then $A = \mathbb{N}$.
I would like to motivate the presence of axiom 5 by presenting an example in which there are unnecessary elements. For instance the set $A = \{0,1,2,\star,3,4,...\}$ such that $\star \neq 0$ and $S(\star) = \star$. As far as I can see set $A$ satisfies the first 4 axioms and fails the fifth. In order to prove this it should be sufficient to prove that no natural number is equal to its successor. Thus that $\star$ cannot belong to $\mathbb{N}$. A proof might look like this:
Let $A$ be the set of natural numbers that are not equal to their successor. By axiom 3 $S(0) \neq 0$. Thus $0 \in A$. Suppose $n \in A$. Thus $S(n) \neq n$. Suppose towards contradiction that $S(S(n)) = S(n)$. By axiom 4 this implies $S(n) = n$. Thus $n$ should not be in $A$. This is a contradiction. Thus $S(S(n)) \neq S(n)$. In other words $n \in A \rightarrow S(n) \in A$ and the induction is complete.
I would like to ask two things:
- Is my example and proof valid?
- What could be a more general example of unnecessary elements in $\mathbb{N}$ that are ruled out by axiom 5? There are of course infinitely many ways in which one can build a set which contains the naturals as a subset, I am looking for the most general example or reasoning one can make rather than my very particular one.
The set $A$ you've given isn't even a subset of $\mathbb{N}$. Already by the way you defined $S(*)=*$ it is ruled out by second axiom since $* \notin \mathbb{N}$.
The most basic example of a set that would fail to satisfy the fifth axiom would be a set of all even numbers from $\mathbb{N}$ including a zero, $A=\{2n|n\in\mathbb{N}\}\cup\{0\}$.