showing the natural numbers exist from axioms (help with making sense of book)

Question

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.

Answer 1

Actually, it is very unclear to me what you’re asking, but I have a feeling, this might be your problem which you want to be resolved:

One might wonder how this lemma can possibly be true since there are so many sets on which you can perform induction – not only $ℕ_0$.

Intuitively one recognizes not only $ℕ_0 = \{0,1, 2, 3, …\}$, but also $$\{0, 2, 4, …\} \quad\text{and}\quad \{0, 3, 6, 9, …\}$$ as inductive sets – probably because one is used to carry out inductions on such sets as well, not only on $ℕ_0$. One is inclined to think that these are inductive sets as well, just corresponding to $S^2$ and $S^3$ respectively.

But they are not. There is an issue to this:

Inductive sets are explicitly just defined only for the successor function $S(n) = n ∪ \{n\}$ and not any other successor functions. The lemma just doesn’t make any statements about more generalized notions of inductive sets with e.g. $S^2$ or some other successor function $S'$. It doesn’t care.

If that’s your problem, think of it that way: An inductive set has to contain $0 = ∅$, and therefore $1 := 0 ∪ \{0\} = \{∅\}$, and therefore $2 := 1 ∪ \{1\} = \{∅,\{∅\}\}$, and so on. So it has to contain at least $ℕ_0 = \{0, 1, 2, 3, …\}$ as a subset. But this isn’t a proof, of course. The lemma just states that this is really true and that $ℕ_0$ itself is such an inductive set.

And if this isn’t your problem, I think you should clarify your question.

(By the way, I write $ℕ_0$ because $0$ is not a natural number.)