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 $\mathbb{R}$?

1http://en.wikipedia.org/wiki/Transfinite_induction – Mar 27 '11 at 04:07

@Antonio: I think it is a bit misleading to say that "in practice" something that is rather standard in set theory is almost impossible. – Andrés E. Caicedo Mar 27 '11 at 04:19

@Andres: I *think* you're talking to me. I meant any of the "usual" properties of the reals that are related to the analytical or algebraic properties of the reals; but yes, you are right, that was somewhat misleading. I'll "fix" it. – Arturo Magidin Mar 27 '11 at 04:21

2@Yuan (corrected): See http://en.wikipedia.org/wiki/Transfinite_induction . Yes, you can do "induction on $\mathbb{R}$", and this is often done in set theory. But it is very hard to use this kind of argument to establish algebraic or analytical properties of $\mathbb{R}$, because the wellorderings that exist on $\mathbb{R}$ have little/nothing to do with the algebraic or analytical structure of $\mathbb{R}$, so there is no reason to expect us to be able to go from "$P(s)$ for all $s\prec r$" to "$P(r)$", when the property $P$ has little to do with the ordering. – Arturo Magidin Mar 27 '11 at 04:23

1@Yuan: If you want to deal with the usual properties of the reals related to algebra or analysis, it is usually better to use something like: http://math.stackexchange.com/questions/4202/inductiononrealnumbers – Arturo Magidin Mar 27 '11 at 04:26

@Arturo Magidin: I am trying to figure out the statement trying to prove that the submodule of a free module over a PID is also free. Hungerford used the transfinite induction to show this. – Yuan Mar 27 '11 at 04:37

@Yuan: But that's not "induction over $\mathbb{R}$". Here, the only property of the set you are interested in is the cardinality, which is settheoretic; when dealing with settheoretic properties, transfinite induction *does* tend to work well. – Arturo Magidin Mar 27 '11 at 04:39

@Arturo Magidin:Thanks for your comments. It does make sense to me. – Yuan Mar 27 '11 at 08:27
1 Answers
The first thing you should be aware of: The WellOrderingTheorem is equivalent to the Axiom of Choice, and is highly nonconstructive. Deriving the principle of transfinite induction on wellordered sets is rather easy, the problem with $\mathbb{R}$ is that such a wellordering exists only nonconstructively, 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 wellordering on $\mathbb{R}\backslash \{0\}$ (which we can trivially get from a wellordering on $\mathbb{R}$). Define $$\begin{align*} A_0 &:= \{\inf_\sqsubseteq\; \mathbb{R}\}\\ A_{n+1} &:=\begin{cases} A_n \cup\{\inf_\sqsubseteq\{ x\in\mathbb{R}\mid x\mbox{ lin. indep. from } A_n\}\} & \mbox{ if welldefined}\\ A_n &\mbox{ otherwise} \end{cases}\\ A_\kappa &:= \bigcup_{n<\kappa} A_n \qquad\qquad\text{for limit ordinals }\kappa. \end{align*} $$
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 latexthingy refuses to recognize my backslashes and curly brackets).
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According to one definition of a proof, it does not follow from the fact that every vector space over $\mathbb{R}$ has a basis that every vector space over $\mathbb{Q}$ has a basis. I realize that assuming the axiom of choice, it can probably still be proven another way that every vector space over $\mathbb{Q}$ has a basis and once you know how to prove that about vector spaces over $\mathbb{R}$, it's easy to figure out how to prove that about vector spaces over $\mathbb{Q}$. – Timothy Feb 06 '19 at 00:44