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Suppose $\mathbb{V} = \mathbb{L}$ and there is a countable transitive model $\mathbb{M}$ of $ZFC$.

Let $\rho$ be the $\mathbb{L}$-rank, i.e. for all $a \in \mathbb{V}$, $\rho(a) = $the least $\alpha$ with $a \in L_{\alpha+1}$.

Define a pre-order $<'$ on $M$ by $a <' b$ iff $\rho(a) < \rho(b)$.

Then my first question is: under what circumstances is $<'$ first-order definable over $\mathbb{M}$?

My second question is: supposing $(\mathbb{V} \not= \mathbb{L})^\mathbb{M}$, and given $a, b \in M \backslash \mathbb{L}^\mathbb{M}$, when does $\mathbb{M}$ "know" that $a <' b$?

Formally, when is there a formula $\phi(x, y)$ (without parameters) such that $\mathbb{M} \models \phi(a, b)$ and for all $a', b'$ with $\mathbb{M} \models \phi(a', b')$, we have $a' <' b'$?

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Thinking about this has led me to the question: Suppose in $\mathbb{M}$ there is a class well-order of order type $\beta$ larger than $ORD^\mathbb{M}$. Can we (in $\mathbb{M}$) carry out the $L$-construction up to level $\beta$? I realize we will not be able to define classes-of-classes-of-classes within $\mathbb{M}$, but I wonder if we could still define 'small' objects (e.g. reals) that arise in $L_\beta$. –  jonasreitz Feb 2 '13 at 3:52
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If so, then I propose that we try to force over $\mathbb{M}$ to make $<'$ have very large order type $\beta$, so large that there is, in $L_\beta$, a function witnessing the countability of $ORD^\mathbb{M}$. Forcing to make the order type of $<'$ large presents its own challenge, but I envision adding $\omega$-many Cohen reals, carefully selected from appropriate levels of the $L$-hierarchy to give the correct order type. –  jonasreitz Feb 2 '13 at 3:52
    
I don't know about the answer to the first question, although in $M$ we can define "names" for elements of $L_\beta$: namely, supposing $\phi$ is a well-order of $ORD$, let each $\alpha$ be a name of $L_\gamma$ where $\gamma$ is the order type of $\alpha$ with respect to $\gamma$; then, using these names and Godel's 8 $L$-generating functions, we can represent the formation of any $x \in L_\beta$ as a finite tree of ordinals. With these particular names, however, it is not obvious that determining whether one name contains another (say) is first-order definable. –  Douglas Ulrich Feb 3 '13 at 1:59
    
Sorry, I meant, $\alpha$ should be a name of $L_\gamma$, where $\gamma$ is the order type of $\alpha$ with respect to $\phi$. –  Douglas Ulrich Feb 3 '13 at 2:00

2 Answers 2

This is a great question, and very subtle.

Here is a way to see that the relation cannot be uniformly definable. Suppose $L_\alpha$ is a countable model of ZFC, and $\phi$ works as you say in every model of the form $L_\alpha[c]$, a forcing extension to add a Cohen real $c$, chosen from $L$. (Note that in $L$ we may easily find such $L_\alpha$-generic Cohen reals, since $\alpha$ is countable.)

Fix any such extension $M=L_\alpha[c]$, which is a model of ZFC. Let $c_0$ and $c_1$ be the even and odd digits of $c$, respectively. Suppose without loss that $c_0\leq'c_1$, meaning that $c_0$ appears in $L$ before or at the same rank as $c_1$. If $M$ can define this relation, then there must be some condition forcing this instance of it, and so there must be some finite initial segment $p\subset c$, such that any $L_\alpha$-generic Cohen real $d$ extending $p$ will have $\phi(d_0,d_1)$. But fix some $L_\alpha$-generic $d_1$ extending its part of $p$, and then find $d_0$ that is $L_\alpha[d_1]$-generic and extending its part of $p$, and very high in $\omega_1^L$. This is possible because there are continuum many different $L_\alpha[d_1]$-generic Cohen reals in $L$, and so some of them must have high rank; and changing a finite part of such a real does not affect rank. So now we have $L_\alpha[d]$ with $d_1\lt' d_0$ and $d$ extends $p$, a contradiction.

So it isn't uniformly definable. But as Andres points out, this approach doesn't even answer whether it might be definable nevertheless in a non-uniform way.

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Joel, I thought of this, but one needs to argue that in $L_\alpha[c]$ will be Cohen reals that come from different $L$-levels, as $<'$ does not distinguish between sets of the same rank. –  Andres Caicedo Jan 29 '13 at 16:45
    
(Also, $<'$ needs not be the relation defined in $L_\alpha[d]$ by the formula that happens to define it in $L_\alpha[c]$.) –  Andres Caicedo Jan 29 '13 at 16:47
    
But there are continuum many Cohen reals, and so some of them must come from higher levels. The point is that you can make either $d_0$ or $d_1$ have very high rank, while respecting any finite condition $p$. –  Joel David Hamkins Jan 29 '13 at 16:48
    
About your second remark, you are right. What my argument shows is that there is no one formula that will work in all $L_\alpha[c]$'s. –  Joel David Hamkins Jan 29 '13 at 16:49
    
Joel, about the first issue, yes, of course; I only meant that the argument does not rule out all $L_\alpha[c]$. Nice question... –  Andres Caicedo Jan 29 '13 at 16:59

[Edit: The question is more subtle than I originally understood. I am leaving this here so as to avoid it being repeated by others.]

You can define $<'$ internally only if $M$ is a model of $V=L$, that is, only if $M$ is an $L_\alpha$. For example, $M$ could be (in $L$) a forcing extension of some $L_\alpha$. Being in $L$, every point in $M$ has a rank, but we only see in $M$ the rank of points in $L^M=L_\alpha$. However, $<'$ restricted to $L^M$ is definable in $M$. The usual definition ($a,b\in L$ and $\rho(a)<\rho(b)$) relativizes, so its definition from the point of view of $M$ gives the same relation as the definition of $<'$ in $L$ restricted to elements of $L_\alpha$.

A decent reference to see how $<'$ relativizes and the amount of absoluteness involved is Devlin's book on "Constructibility".

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Oops, of course. So that's the first question. I'm going to expand upon the second question; I didn't expect the first question to be so trivial. Sorry :( –  Douglas Ulrich Jan 25 '13 at 20:27
    
Wait, so the proof I was thinking of for your claim doesn't work. There are definable (proper class) well-orderings in $M$ of height greater than $ORD^M$. So I actually don't see your argument –  Douglas Ulrich Jan 29 '13 at 13:54
    
Hi Douglas, yes, you are right, I misunderstood the question when I first read it (your rephrasing clarifies the subtlety). –  Andres Caicedo Jan 29 '13 at 15:54
    
Somebody posted a similar question mathoverflow.net/questions/120173/can-a-model-of-v-neq-l-contain-a-class-giving-‌​the-l-ordering-on-all-its-sets/120212#120212 here, and they also were misunderstood. So there must be something confusing in the question –  Douglas Ulrich Jan 29 '13 at 15:57
    
@Andres, your answer still gives a useful starting point, since it shows that usual definition of the $L$-order will not suffice to define $<_L$ on all of $M$ -- but it's not clear whether another definition may be more successful. –  jonasreitz Jan 29 '13 at 16:13

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