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I know that, assuming Axiom of Choice, every set is well-orderable. I know also that the assertion that $\mathbb{R}$ is NOT well-orderable is consistent with ZF. How can I find other sets such that, in ZF, I can't prove their well-orderability? For example, which elements of Von Neumann Hierarchy

$V_0 = \emptyset \\ V_{\alpha+1}=P(V_{\alpha}) \\ V_\lambda = \cup_{\alpha<\lambda} V_\alpha $

(except $V_n$ for $n$ finite ordinal, and $V_\omega$ ) can be well-ordered?

Asaf Karagila
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Ervin
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    I do not know of any interesting theorems on propagation of well-orderability. For example, without choice we may have well-orderable sets $A$ such that $\mathcal P(A)$ is not well-orderable, and $\mathcal P^2(A)$ is not even linearly orderable. We may have countably many sets of size $2$ whose union does not contain countably infinite subsets. On the other hand, it is a theorem of Tarski that $\mathcal W(A)$, the collection of well-orderable subsets of $A$, is always larger than $A$ itself. – Andrés E. Caicedo Feb 01 '14 at 07:44
  • The collection of finite subsets of a well-orderable set is well-orderable. – bof Feb 01 '14 at 08:28

2 Answers2

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It is consistent that $V_{\omega + 1}$ that is not well-orderable.

$\omega \subset V_\omega$. So $\mathscr{P}(\omega) \subseteq P(V_\omega) = V_{\omega + 1}$. So if $V_{\omega + 1}$ could be well ordered, then $\mathscr{P}(\omega) = 2^\omega$ could be well-ordered. As you mentioned $\mathbb{R}$ which is isomorphic can be not well orderable..

As $V_\alpha \subseteq V_\beta$ for $\alpha < \beta$. Thus it is consistent that all $V_\alpha$ for $\alpha \geq \omega + 1$ are not well-orderable.

William
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It's somewhat difficult to give a complete characterization, for several reasons.

  1. $L$ is well-ordered, in fact $\rm HOD$ is well-ordered, so every subset of these models is always well-ordered. Since both of these are models of $\sf ZFC$ it guarantees that it's not as simple as you'd like it to be.

  2. If $V$ is a model of $\sf ZF+\lnot AC$ and $M$ is an inner model satisfying $\sf AC$ then there is a set of ordinals in $V\setminus M$. So models disagreeing on the truth of $\sf AC$ will also disagree on the sets of ordinals. But given such set $A$, if we consider $M[A]$ it is still a model of $\sf ZFC$. So just by considering sets of ordinals we know that there is a difference, but it's hard to estimate it.

  3. As you and William both note, it's consistent that $\Bbb R$ cannot be well-ordered. By a quick calculation, $|\Bbb R|=|V_{\omega+1}|$. Therefore in such model no $V_\alpha$ for $\alpha>\omega$ can be well-ordered. And we can using forcing and symmetric models ensure that the least non-well orderable $\alpha$ is pretty much anything.

  4. If $V$ is a model of $\sf ZF$, then for every $x\in V$ there is a generic extension of $V$ such that $x$ can be well-ordered in that extension (e.g. making $x$ countable), so being non well-orderable is not absolute between models, and we cannot really talk about sets which "absolutely cannot be well-ordered".

If we combine the last two points, we get that we can mess the axiom of choice in so many ways, and we can control these ways relatively easily and nicely (especially if all we want is violation of choice, and not some stronger properties a-la determinacy). Pretty much anything goes.

Asaf Karagila
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  • If $M$ is an inner model and $x$ is some set, what does $M[x]$ mean? – William Feb 01 '14 at 11:14
  • @William: It is the smallest model including $M$ having $x$ as an element. – Asaf Karagila Feb 01 '14 at 11:16
  • @William: At least assuming $x\subseteq M$ (e.g. if $x$ is a set of ordinals). – Asaf Karagila Feb 01 '14 at 11:20
  • is $L$ well ordered because the constructible universe always obeys the axiom of choice or is it the other way around? – Tim Seguine Feb 01 '14 at 11:29
  • @Tim: Not every model of $\sf ZFC$ is well-orderable. For a class model to be well-orderable one has to work slightly harder than just the axiom of choice. In $L$ we have a canonical well-order, which implies the axiom of choice holds. But there are models of $\sf ZFC$ which cannot be linearly ordered (as a whole model! each set can be ordered, of course). – Asaf Karagila Feb 01 '14 at 11:31
  • @AsafKaragila Okay, I didn't need the extra information actually, but thanks for answering, you cleared it up. – Tim Seguine Feb 01 '14 at 11:36