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It is consistent with ZFC that the universe is well-ordered, e.g. in $V=L$ where global choice holds. I also know that it is consistent that global choice fails (although I have no immediate example from the top of my head).

However one can try and ask a slightly weaker question, much like the axiom of choice implies every set can be linearly ordered, but the reversed implication fails; is it the same with global choice? The immediate answer is yes, it is consistent that the universe can be linearly ordered but the axiom of choice fails and therefore global choice has to fail too. But just how far can this be pushed?

Can we construct a model of ZFC in which the axiom of choice holds, but there is no linear ordering of the universe? Or does ZFC prove that the universe is linearly ordered?

One would expect that it would, because if we assume AC then the sets of ordinals already decide the universe in its entire form, and sets of ordinals can always be linearly ordered by: $$A\prec B\iff\min (A\triangle B)\in A.$$ (where $\triangle$ denotes the symmetric difference.)

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In Solovay's model, there is no ordinal definable non measurable set of reals. Hence there cannot be an ordinal definable linear ordering of power set of reals because one can pass, in a definable way, from such an ordering to a non measurable set as follows: Work in Cantor space. Write x~y iff they agree on all but finitely many bits. Restrict your ordering to the ~ equivalence classes. Let X be the set of those reals x such that the eq class containing x precedes the one containing x' where x' is obtained by changing all bits of x. Then X is not Lebesgue measurable. – Ashutosh Nov 2 '12 at 19:46
Ashutosh, while you are correct, you should also note that I asked about ZFC, not about ZF. – Asaf Karagila Nov 2 '12 at 20:46
I should also point out that there are simpler ways to prove there is no definable linear ordering of the universe without the axiom of choice. Simply add a non-linearly ordered set (e.g. amorphous sets). – Asaf Karagila Nov 2 '12 at 20:59
By Solovay's model I meant the model obtained by Levy collapsing an inaccessible - AC holds here; the choice free model is obtained by passing to HOD(R). – Ashutosh Nov 2 '12 at 21:17
See related question on whether there can be a global linear order without a global well-order:… – Joel David Hamkins Nov 3 '12 at 12:46
up vote 11 down vote accepted

Update. I've repaired the argument. The idea was to use the analogue of the usual non-AC arguments, but using a class forcing instead of just Cohen reals.

Theorem. Every model of ZFC has a class forcing extension that is a model of ZFC, in which there is no class global linear ordering of the universe that is definable from parameters.

Proof. Let $\mathbb{P}$ be the Easton support class forcing product $\mathbb{P}=\Pi_{\gamma\text{ reg}}\text{Add}(\gamma,\gamma\cdot 2)$, which forces to add $\gamma\cdot 2$ many Cohen subsets to every uncountable regular cardinal $\gamma$ (this is of course isomorphic to adding one). Suppose that $G\subset\mathbb{P}$ is $V$-generic, and consider the extension $V[G]$. The standard forcing arguments show that $V[G]$ is a model of ZFC, and in particular, of the axiom of choice.

Meanwhile, I claim that there is no class global linear ordering of $V[G]$ that is definable in the language of set theory in $V[G]$ using parameters. Suppose that there were, and that $\psi(x,y,z)$ is a formula forced by a condition $q\in G$ to define a linear order when used with the parameter named by $\dot z$. Let $\gamma$ be a regular cardinal far above the support of $q$ and any of the conditions appearing in $\dot z$. The forcing at stage $\gamma$ added the mutually generic sets $g_\alpha$ for $\alpha\lt\gamma+\gamma$. Let $A$ be the set of all $g_\alpha$ from the first block, that is, for $\alpha\lt\gamma$, and let $B$ be the set of $g_\beta$ from the second block, for $\gamma\leq\beta\lt\gamma+\gamma$. Fix the corresponding canonical names $\dot A$ and $\dot B$, as derived from $\dot G$. Suppose without loss of generality that $A$ precedes $B$ in the definable linear order, so that $\phi(A,B,z)$ holds in $V[G]$. Fix a condition $p\in G$ below $q$ such that $p\Vdash\phi(\dot A,\dot B,\dot z)$. The condition $p$ mentions less than $\gamma$ much information about the stage $\gamma$ forcing. Thus, it details fewer than $\gamma$ many bits of fewer than $\gamma$ many sets each in $A$ and $B$. Furthermore, a density argument shows that every possible initial segment of a subset of $\gamma$ occurs for $\gamma$ many of the Cohen sets in each block. So every set in $A$ whose digits are partially specified by $p$ can be matched by a set in $B$ that agrees with those digits, and vice versa. Since $p$ specifies fewer than $\gamma$ many bits, there is in $V$ an automorphism $\pi$ of the stage $\gamma$ forcing that carries out a permutation of the coordinates, carrying altogether all the $\alpha$s in the first block to $\beta$s in the second block and vice versa, in such a way that happens to have $\pi(p)$ and $p$ both in $G$. We may extend $\pi$ to an automorphism of $\mathbb{P}$, and consider the resulting transformation of names. Since $\pi$ swaps the two blocks altogether, we have $\dot A^\pi_G=B$ and $\dot B^\pi_G=A$. The choice of $\gamma$ ensures that $\dot z^\pi=\dot z$. But since $p\Vdash\phi(\dot A,\dot B,\dot z)$ we also have $\pi(p)\Vdash\phi(\dot A^\pi,\dot B^\pi,\dot z^\pi)$, and as $\pi(p)\in G$, this means that $\phi(B,A,z)$ in $V[G]$, which means that $B$ precedes $A$ in the linear order, a contradiction. QED

This argument can be used to show, as Ali mentions in his answer, that there can be a definable class of pairs, having no choice function. Specifically, consider all sets of pairs, which would include the pairs of the form $\{A,B\}$ as I denote them in the proof. If we could definably select one or the other, then we fix a condition $p$ forcing which one is selected, and then find an automorphism $\pi$ that swaps the two elements of the pair, while having $\pi(p)$ still in $G$. Thus, the other set must also be selected, a contradiction.

Corollary. Every model of set theory has a class forcing extension with a definable class of unordered pairs, such that no definable class (with parameters) selects exactly one set from each of those pairs.

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So if we allow $G$ as a predicate, we should be able to define a linear ordering of he universe in ZFC? Or would we still have to talk about it in NBG? – Asaf Karagila Oct 27 '12 at 12:02
Yes. If the ground model has a global well-order, then allowing $G$ as a predicate in $V[G]$ will allow us to define a well-order in $V[G]$: $x$ is before $y$ if $x$ has an earlier name than $y$. (For this, you have to define $V$ in $V[G]$, but this is possible by my theorem on the definability of the ground model after closure point class forcing.) – Joel David Hamkins Oct 27 '12 at 12:06
Yes, well-ordering I understand that is definable. I imagine this is about the same proof as the usual proof that AC holds in the ground model then it holds in the extension, but for GC. Your suggestion for a next question is good; I actually had in mind something else: can we prove this definable linear order is an extension of $(V,\in)$ or $(V,\subseteq)$? Essentially this is the same track as the usual independence of linear order theorems, but with the axiom of choice holding and global choice failing... – Asaf Karagila Oct 27 '12 at 12:18
Joel, I believe you have a couple of typos there. I will correct them now, but feel free to undo my doing if I'm just misunderstanding. – Asaf Karagila Oct 27 '12 at 12:30
Thanks for correcting... – Joel David Hamkins Oct 27 '12 at 12:35

Joel nicely answers Asaf's question. Here I just want to add some footnotes to his answer.

1. Joel' argument shows that $ZFC$ cannot even prove the definable class form of axiom of choice for pairs. Historically, this was first done by with a similar argument by Easton in his 1964-thesis (printed in the Annals of Mathematical Logic).

2. In the pre-forcing era, Mostowski had already shown that $ZFA$ + "the universe can be linearly ordered" does not imply "every set can be well-ordered" (see p.51 of Jech's book on the Axiom of Choice).

3. As shown in the proof of Theorem 5.21 (p.71) of Jech's text, Mostwoski's argument can be transplanted to the forcing context to show that indeed $V$ can be definably linearly ordered in Cohen's so-called basic model of the failure of the axiom of choice.

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Yes, to the second and third points I referred in my answer. One can observe that in the Halpern-Levy model (which is a direct and vastly more difficult Mostowski linearly ordered model) we have that the universe is linearly ordered, but the axiom of choice fails. – Asaf Karagila Oct 27 '12 at 18:41

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