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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 $\cal P(\Bbb R)$.

Akiva Weinberger
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    Asaf has already answered the question, but let me comment on your guess, at the end of the question, about reducing the problem of well-ordering $\mathbb R$ to the problem of ordering $\mathcal P(\mathbb R)$. There is no such reduction. It is consistent with ZF that every set can be linearly ordered but $\mathbb R$ cannot be well-ordered. The basic Cohen model has these properties. – Andreas Blass Nov 24 '15 at 16:33
  • cc @AndreasBlass In Jech's Axiom of Choice he shows that what he calls the Ordering Principle ("every set can be linearly ordered") is weaker than AC by showing that it holds in the basic Cohen model Andreas refers to (Jech calls it just that), in which a countable set of reals has enumeration. – BrianO Nov 24 '15 at 18:10
  • @Brian: Well a countable set of reals always have enumerations. But you mean that you add a countable set of reals, but then forget about its enumeration. – Asaf Karagila Nov 25 '15 at 08:28
  • @AsafKaragila Seeing what I wrote, I lost a crucial word "no", as in "no enumeration". Right, by definition of 'countable' there is an enumeration of the set of reals somewhere, but there no such enumeration in the symmetric extension. – BrianO Nov 25 '15 at 08:38
  • @Brian: Yes, but the key point is that the set of reals is only countable in the full generic extension, not in the intermediate symmetric model. So calling it countable without an enumeration is a bit of a misnomer (unless you're talking with someone who is already comfortable enough with these constructions and familiar enough with them to begin with). – Asaf Karagila Nov 25 '15 at 08:40
  • @AsafKaragila You're right to insist. In the symmetric model, it's confusing/misleading to call the set of reals "countable", as in that model it just isn't. In the outer model, yes, there's an enumeration, but in the intermediate model, the automorphisms have made sure that the members are as indistinguishable/indefinable as are the members of a Vitali set. – BrianO Nov 25 '15 at 08:46

1 Answers1

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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.

Community
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Asaf Karagila
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  • As a footnote for Akiva, this "Russell set" so-called because it's a countably infinite set of pairs of socks, rather than of shoes — the latter has definable choice functions. – BrianO Nov 24 '15 at 17:52