Prove that the natural numbers are present on an inductive definition of another set

Question

If I give you the following definition of the set $A$, how could you prove it is equal the set of the natural numbers without an explicit definiton for the latter?

The set $A$ is inductively defined as follows:

i) $0 \in A$; and
ii) $\forall n$, a natural number, if $n \in A$, then $n+1 \in A$.

I can easily prove that $A$ is contained in the natural numbers, but I'm failing to see how to prove the converse without a similar definition for the natural numbers.

Thanks for taking the time to read me.

Answer 1

An inductive set $A$ is any set with the following properties:

• $\left\{\right\}\in A$
• $\forall x\left(x\in A\rightarrow S\left(x\right)\in A\right)$, where $S\left(x\right)$ is the successor of $x$

You can prove that $\mathbb{N}$ is an inductive set, and that if $A$ is any inductive set, then $\mathbb{N}\subseteq A$, but the other way around is false, that is, you can't prove that $A\subseteq\mathbb{N}$. Take for example $A=\mathbb{N}\cup\left\{a\right\}$, where $a$ is anything except a natural number. $A$ is inductive, but it is obvious that $A\subseteq\mathbb{N}$ is false.

Anyway, as far as I know, the best definition of the set of natural numbers is the following one: $\forall x\left(x\in\mathbb{N}\leftrightarrow\forall A\left(x\in A\right)\right)$, where $A$ is any inductive set.

Answer 2

It seems to me that your 'proof' that the set $A \subseteq \mathbb{N}$ must have a mistake because basically you're just defining what Tom Apostol calls an inductive set in his Calculus book. I mean, $A$ can just be a set of real numbers and could perfectly well satisfy your two propierties.

For instance you can take $A = [0, \infty[$ and then $A$ satisfies both properties, but nonetheless $A$ is not contained in $\mathbb{N}$.

In fact Apostol's book defines $\mathbb{N}$ as the set of all numbers that belong to every inductive set.

Answer 3

That's not an inductive definition of $A\:$. Rather, it's an inductive proof that $\ A \supseteq\mathbb N\:$. If you know additionally that $\: A \subseteq \mathbb N\:$ then you can conclude that $\:A = \mathbb N\:$. That's standard mathematical induction.