Principle of Proof by Induction on a Well-ordering

Question

Let $(X,\leq)$ be a woset (well ordered set). Let $E$ be a subset of $X$ such that:

(i) the smallest element of $X$ is a member of $E$

(ii) for any $x\in X$, if $y<x\rightarrow y\in E$, then $x\in E$

Then $E=X$.

proof Suppose $E\neq X$. Then $X-E$ is a nonempty subset of $X$. Since $(X,\leq)$ is a woset, there exists the smallest element of $X-E$, call it $x_0$. Then $\forall y(y<x_0\rightarrow y\in E)$. By (ii), $x_0\in E$. A contraddiction.

I can't see the role of (i) in the proof.

Answer 1

(i) has no role in the proof. It's not actually necessary at all.

Answer 2

You can observe that (i) follows immediately from (ii), because the statement holds vacuously for the minimal element of $X$.

The statement you wrote is analogue to complete (or strong) induction on the natural numbers. You could also state this in the following way:

If $E\subseteq X$ such that:

1. $\min X\in E$,
2. For every $x\in X$, if $x\in E$ then its successor (if it exists) is in $E$,
3. For every limit (non-minimal and non-successor) element $x$ of $X$, if all its predecessors are in $E$, then $x\in E$.

Then $E=X$.