Is it possible to extend well ordering principle/induction to all well ordered sets? Today I was thinking about well ordering of naturals,and how by induction we can prove some properties of natural numbers.Now I started wondering if this is property of natural numbers,which are well ordered by membership relation,then does this apply to all well ordered sets?
Namely what I say is following:
If $T$ is a nonempty infinite set,well ordered by some relation $\leq$,and let P be property such that :
1)$P(x_0) $ holds where $x_0$ is least element of $T$
2)When $P(x)$ holds for some $x\in T$ then it holds for least element of set $T -\{p\in T | p \leq x\}$ 
then I assume it should hold for all elements.
Only issue I see so far is maybe that we should consider only sets which do not have greatest element,but I do not see clearly how could presence of greatest element destroy this hypothesis.
I happen to see this as generalization of induction to all well ordered sets,but I might be just naive because I am not very experienced on this level.
Is it possible to continue down this path and formalize such general induction,just for the fun of it?If not why,and if yes also why?What kinds of problems would one have to consider to define this,or what kind of paradox could arise?
Thanks in advance
 A: There is an induction principle for well-ordered set, but it's not quite the one you have formulated. The right condition is


*

*When $P(y)$ holds for every $y<x$, then $P(x)$ holds too.


Like long induction for the integers, this variant doesn't need a distinction between base cases and induction steps. When $x_0$ is the least element, the assumption "$P(y)$ for all $y<x_0$" is just vacuous.
To prove this principle, let's assume that the condition holds, and there is some $x\in T$ that $P$ doesn't hold for. Because $T$ is well, ordered, there is a first $x$ that fails $P$. But that means that every $y<x$ must meet $P$, and by the induction condition this means that $P(x)$ holds. So the first $x$ to fail $P$ doesn't fail it at all, which is absurd. Thus there cannot be any $x$ that fails $P$.

As a counterexample to your formulation take $T=\{-\frac1n\mid n\in\mathbb N_+\}\cup\{0\}$ with the usual ordering and let $P(x)$ mean $x\ne 0$.
This $T$ is well-ordered, and $P(x)$ satisfies both your conditions, but plainly isn't true for all of $T$.
