I'm now on page 40 of a set theory book and I've hit the natural numbers. I think the book has oversimplified some things.
The successor of a set $x$ is defined to be $S(x)=x\cup\{x\}$
A set $I$ is inductive if:
- $0\in I$
- $n\in I\implies (n+1)\in I$ where $(n+1)=S(n)$ (it's just notational niceness)
The natural numbers are defined as follows:
$\mathbb{N}=\{x|x\in I \forall\text{ inductive sets }I\}$
APPARENTLY this leaves "the only remaining question: whether there are any inductive sets at all"
This leads to "the axiom of infinity: AN inductive set exists"
This is where I get confused, surely we have at least one inductive set already, we have axioms that state: there is an empty set, what makes sets equal, and the axiom of union, from the axiom of schema of comprehensions we can define inductiveness as a property surely
So I think we actually have one inductive set from the axioms already
Then Lemma 1.4 happens: "$\mathbb{N}$ is inductive, and if $I$ is any inductive set, then $\mathbb{N}\subset I$."
This I have a problem with.
My approach
Right now we have a fixed definition of $S$, let us generalise this. (I see S functions as iterators really (programming terminology))
Lets consider $S_1=S\circ S$ ("even numbers") (I wanted to define the odd but you can "pretend 0" is the first odd, then it all follows.
I want to see a proof that "Any two (increasing) recursively defined sequences that do not converge and start at zero have an inductive set of points in common"
As without this only 0 would be in all inductive sets.
Please help me solve this.