well-ordering principle

Question

I'm confused by the well-ordering principle . The proof is clear but I don't have any idea why is it true . It says that every non-empty set can be well-ordered but $C^{1}$ is a non-empty set but there's no linear order defined on it . Furthermore , I don't know how to define well-ordering relation on an open subset of $R^{1}$ e.g $(0,1)$

Answer 1

The well-ordering principle is non-constructive, in the sense that it doesn't tell you how to well-order a certain set, but rather that some well-order exists.

This well-order is often completely incoherent with the natural structure of the set you will consider, for example in the case of $(0,1)$. The well-order will disagree with the $<$ order in most cases.

But that's fine. Note that we can define a dense order on $\Bbb N$ by taking a bijection with $\Bbb Q$ and "pulling back" the order of the rationals. Does this order agree with the usual order on the natural numbers? Hardly. But that's fine.

Note that you can find a bijection between $\Bbb C$ and $\Bbb R$, and this bijection does define a linear ordering of $\Bbb C$. Again, this linear ordering is completely incoherent with the structure of $\Bbb C$, but such order does in fact exist -- even if it doesn't respect other structure of the set.

For the case of $(0,1)$, or any other set of cardinality $2^{\aleph_0}$ we can even prove that without using the well-ordering principle (or the axiom of choice, or some other equivalent of course) we might not be able to well-order $(0,1)$. These proofs are rather technical and difficult, but the key point is that we can produce models of set theory where the axiom of choice fails and $(0,1)$ cannot be well-ordered.