Is the powerset of the real numbers well orderable?

Question

I know that $\mathbb N$ is well orderable and $\mathscr{P}(\mathbb N)$ (the real numbers) are well orderable, so I was wondering if $\mathscr{P}(\mathscr{P}(\mathbb N))$ is also well orderable. I tried thinking about the elements as vectors so that I could order them by magnitude or direction, but of course, some vectors have the same magnitude and/or direction. Any help on this would be greatly appreciated.

Answer 1

I think there are a few confusions here.

First of all, both well-orderedness and linearly-orderedness are properties of sets equipped with specific orders. It doesn't really make sense to say "$\mathbb{N}$ is well-ordered," for example; what we should say is "$\mathbb{N}$ with the usual ordering is well-ordered."

Second, and more substantively, there is a difference between well- and linearly-orderedness. In a linear order every (nonempty) finite set has a minimum since all elements are comparable; the defining property of a well-order, however, is that every subset has a least element.

This, then, is the complete situation:

• In $\mathsf{ZFC}$, every set is well-orderable, although of course not all linear orders are well-orders.

• In $\mathsf{ZF}$ (= set theory without choice), we can prove that $\mathbb{N}$ is well-orderable (using the usual ordering) and $\mathscr{P}(\mathbb{N})$ is linearly orderable (by bijecting with $\mathbb{R}$ and using the usual ordering on the latter).

• However, it is consistent with $\mathsf{ZF}$ that $\mathscr{P}(\mathbb{N})$ is not well-orderable. It is also consistent with $\mathsf{ZF}$ that $\mathscr{P}(\mathscr{P}(\mathbb{N}))$ is not even linearly orderable. Interestingly, these situations are related: in $\mathsf{ZF}$ we can prove that if $A$ is well-orderable then $\mathscr{P}(A)$ is linearly orderable (think about the phrase "least point of difference").

These issues have been discussed at various points on this site and MO; see e.g. here.