In Calculus I, Apostol appeals to the definition of an inductive set given in previous pages to prove the PMI (it's somewhat funny because the proof is just two lines).
I have been trying to do it in another way. I would like to know whether there is something wrong with it.
Theorem (I.36): Principle of Mathematical Induction: Let $S$ be a set of positive integers which has the following properties:
- (a) The number $1$ is in the set $S$.
- (b) If an integer $k$ is in $S$, then so is $k+1$.
Proof:
Let suppose that there is a set $S \subset \mathbb{N}$ which contains $1$ and an integer $k > 1$. Let assume that $k+1 \notin S$ (i.e. property (b) does not hold). Then it follows that if $k$ is in $S$, then so is $k-1 = m$, which is the same as saying that, for every integer $m \geq 1$ in $S$, there is an integer $m+1 \in S$. Hence properties (a) and (b) are fulfilled, $\implies S = \mathbb{N}$.
We want to show that if (a) and (b) is satisfied by $S$ then $S=\mathbb{N}$.
Suppose to the contrary that $S \neq \mathbb{N}$. This means that there is another set of positive integers $T=S'$ (relative to $\mathbb{N}$) such that $S\cup T= \mathbb{N}$.
Since $T$ is a set of positive integers, then, by the Well Ordering Principle, it must have a least element say $a \neq 1$ since $1 \in S$. Now, we are sure that $a-1$ is in $S$, $a$ being the least element in $T$. But by (b), since $a-1\in S$, it must be that $a\in S$. A contradiction. Thus, $S= \mathbb{N}$.