# Can a model of $V\neq L$ contain a class giving the $L$-ordering on all its sets?

This question is inspired by the excellent question by Douglas Ulrich When is $L$-Rank definable in inner models of $V=L$?

Suppose $M \in L$ is a countable model of $ZFC$, and furthermore suppose $M \vDash V \neq L$. These models have the funny property that, although every set in them is constructible, the model does not recognize this fact -- the "nonconstructible" sets simply arise at a level of the $L$-hierarchy that is greater than the ordinals of $M$. Provided any countable $ZFC$ model exists, then such models abound -- for example, we can easily build forcing extensions of countable models by building an $M$-generic filter directly, and the forcing extension will not recognize that the generic is constructible.

In the answer to the question linked above it is shown that such a model $M$ cannot define the $L$-order on all of its members, at least not with the usual definition -- it can only define $<_L$ for those elements that it recognizes as constructible (that is, for members of $L^M$). However, I am interested whether this limitation is inherent, or simply a limitation of definability over $M$. In particular, is it consistent for us to add the $L$-order on all of $M$, as a class, without destroying $ZFC$? I will phrase the question in terms of $GBC$ models, to make the use of classes explicit.

Is it consistent that $M \in L$ is a countable model of $GBC$ with $M \vDash V \neq L$, and there is class $U \in M$ such that $U$ gives the $L$-order on $M$? (That is, $\langle x,y \rangle \in U$ if and only if $x <_L y$, for all sets $x, y \in M$).

Note: In the context of $GBC$ the statement $V \neq L$ may be ambiguous -- I intend it to refer only to sets, and not to classes, e.g. $V \neq L$ means "there exists a nonconstructible set."

Such a $U$ would give us a proper class well order with order type larger than $ORD^M$, but this does not seem immediately problematic - many such well orders are definable over any model of $ZFC$. One way in which such a $U$ might be inconsistent is if, from $U$, we could show how to "construct" every set in $M$, but these constructions would be of more-than-$ORD^M$ length, and so this might not directly contradict $V \neq L$.

If such a $U$ is inconsistent, it would be nice to see why this limitation exists. On the other hand, if we can have such a $U$, is it universally possible?

If such a $U$ is consistent, are there any restrictions on the models $M$ that can have them?

For example, it might be consistent only if every member of $M$ has $L$-rank "not too much greater" than $ORD^M$.

EDIT: changed a stray occurence of "GBC" to "ZFC" in paragraph 1, for clarity.

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This is not an answer- this would be a comment if I had enough reputation points for comments. I just wanted to mention that I am no longer convinced by the answer to my question (to which you linked). – Douglas Ulrich Jan 29 '13 at 14:00
@Emil: Yes, Jonas wants the ordering of all of $M$ in order of constructibility. That ordering exists because he assumed $M\in L$. – Andreas Blass Jan 29 '13 at 14:05
Never mind, I misread it thinking it referred to $<_L$ inside the model rather than the external one. – Emil JeÅ™ábek Jan 29 '13 at 14:53
Douglas, just wanted to say I love your question. In contemplating these odd models in which every element is constructible, but $V \neq L$, I am always struck by the possibility that every set is, in fact, constructible, if we are willing to continue the $L$-construction far enough beyond $ORD$. This idea is addressed more rigorously by Joel Hamkins in his A Multiverse Perspective on the Axiom of Constructibility – jonasreitz Jan 29 '13 at 16:40
Nice question, Jonas! I assume throughout that you want $M$ to be transitive? – Joel David Hamkins Jan 29 '13 at 16:45