I came across an argument in a book, and I'm wondering why we need this proof.
Let $T \subset \mathbb{N}$ where:
- $0 \in T$
- If $n-1 \in T$ then $n \in T$
Let $A = \mathbb{N}\backslash T$, we claim that $A$ is the empty set.
We argue by contradiction. Let $A = \mathbb{N}\backslash T$ and assume $A$ is non-empty. By the well ordering principle, there exists a least element, $s \in A$ for which $s < a$ for all $a \in A$, $a \neq s$. If $s \in A$ then $s-1 \notin A$ since $s-1 < s$, and $s$ was the least element of $A$. If $s-1 \notin A$ then $s-1 \in T$. If $s-1 \in T$ then, by axiom 2, $s \in T$. This is a contradiction because $s$ can't be in both $A$ and $T$ since $A=\mathbb{N}\backslash T$. It follows that $A$ is empty.
I would have thought that the first two axioms could be used to define $\mathbb{N}$, and that $A=\mathbb{N}\backslash T$ would obviously be empty. Could someone please explain the necessity of this proof? I believe it is needed to show that mathematical induction is valid, but seems like a tautology to me.
One can go either way. If you start by assuming the well-ordering principle, then you prove the principle of induction in this way. Conversely, if you start, as you suggest, by assuming the principle of induction, then you can prove the well-orderedness of $\mathbb{N}$. Thus it's inefficient to assume both, so that people pick one to take as evident and prove the other from it.