The well ordering principle

Question

Here is the statement of The Well Ordering Principle: If $A$ is a nonempty set, then there exists a linear ordering of A such that the set is well ordered. In the book, it says that the chief advantage of the well ordering principle is that it enables us to extend the principle of mathematical induction for positive integers to any well ordered set. How to see this? An uncountable set like $\mathbb{R}$ can be well ordered by the well ordering principle, so the induction can be applied to a uncountable set like $\mathbb{R}$?

Answer 1

The first thing you should be aware of: The Well-Ordering-Theorem is equivalent to the Axiom of Choice, and is highly non-constructive. Deriving the principle of transfinite induction on well-ordered sets is rather easy, the problem with $\mathbb{R}$ is that such a well-ordering exists only non-constructively, that is, you cannot explicitly give it, you can just assume that it exists.

A possible usage of it is when proving that $\mathbb{R}$ as $\mathbb{Q}$ vector space has a base (even though this follows from the more general theorem that every vector space has a base, which is mostly proved by Zorn's lemma but can as well be proven by transfinite induction).

Assume $\sqsubseteq$ was such a well-ordering on $\mathbb{R}\backslash \{0\}$ (which we can trivially get from a well-ordering on $\mathbb{R}$). Define $$\begin{align*} A_0 &:= \{\inf_\sqsubseteq\; \mathbb{R}\}\\ A_{n+1} &:=\begin{cases} A_n \cup\{\inf_\sqsubseteq\{ x\in\mathbb{R}\mid x\mbox{ lin. indep. from } A_n\}\} & \mbox{ if well-defined}\\ A_n &\mbox{ otherwise} \end{cases}\\ A_\kappa &:= \bigcup_{n<\kappa} A_n \qquad\qquad\text{for limit ordinals }\kappa. \end{align*} $$

If for some $A_i$, no more element can be added to $A_{i+1}$, then we have a basis. If not, we can proceed. But trivially, we know that at least $A_\mu$ is such a basis, where $\mu$ is the ordinal isomorphic to $\sqsubseteq$.

(Sorry for the bad formatting, but this latex-thingy refuses to recognize my backslashes and curly brackets).