Why isn't there a total order of $\cal P(\Bbb R)$?

Question

I have heard that, in some models of ZF, $\cal P(\Bbb R)$ has no total order. How could one prove this?

I already know that $\Bbb R$ isn't necessarily well-ordered. I'm guessing that one could reduce the problem of well-ordering $\Bbb R$ to the problem of ordering $\cal P(\Bbb R)$.

Answer 1

This can be proved in several ways.

The easiest I know is to use Cohen's second model (you can find it in Jech The Axiom of Choice in Chapter 5) in which there is a sequence $\langle A_n\mid n<\omega\rangle$ such that:

1. $A_n\subseteq\mathcal P(\Bbb R)$.
2. $A_n$ has two elements.
3. If $n\neq m$, then $A_n\cap A_m=\varnothing$.
4. $\prod A_n=\varnothing$.

Namely, you can add a Russell set of sets of reals. And if $\mathcal P(\Bbb R)$ was provable linearly well-orderable, then this would have been impossible. You can actually do a bit worse too (e.g. add an amorphous set of sets of reals).

Other methods include showing that if $\mathcal P(\Bbb R)$ can be linearly ordered, then there is a Vitali set and a set without the Baire Property. For the Lebesgue measurability you need stronger consistency strength, but not for the Baire Property (as shown by Shelah). You can find the details in this wonderful answer by Andres Caicedo.