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Let $\langle L_\alpha \rangle$ denote the hierarchy of constructible sets namely $$L_0 = \emptyset$$ $$L_{\alpha+1} = \text{def}(L_\alpha)$$ $$L_{\gamma} = \bigcup_{\alpha<\gamma}{L_\alpha}$$ for $\gamma$ being limit ordinals and $$L = \bigcup_{\alpha \in \text{Ord}}{L_\alpha}$$ be the Godel constructible universe. It is well known that the ordinals are all in $L$.

In $L$, one can also construct this hierarchy and we call it the relativized constructible hierarchy, denoted by $\langle L_\alpha^L \rangle$.

It is easy to see that $\alpha^L = \alpha$ for any ordinal $\alpha$ and $L^L = L$ (i.e. $V=L$ holds in $L$). I want to ask whether it is true that $$L_\alpha^L = L_\alpha.$$

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How does one see that $L^L=L$ except by proving that $L_\alpha^L=L_\alpha$? –  Joel David Hamkins Apr 28 '11 at 14:22
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What do you mean when you write $\alpha^L=\alpha$? $L_\alpha^L=L_\alpha$ makes sense, since there is a definition of $L_\alpha$ in terms of $\alpha$ which you can apply inside $L$ or in $V$. But what is $\alpha^L$ supposed to mean? –  Stefan Geschke Apr 28 '11 at 15:40
    
@Joel: I take axiom of constructibility for granted (i.e. I never prove it as it is intuitively true). One can obtain L^L = L directly from the fact that the property "$x$ is constructible" is $\Delta_1$ and is absolute. –  An Hoa Apr 29 '11 at 1:29
    
@Stefan: The only trouble seems to be from the ambiguity of $\alpha^L$. My explanation is: take $\alpha$ in $V$ as an absolute concept. Let's take an example: consider the structure $(\omega\backslash\{0\},\in)$. In this structure, the set that has the role of the $0$-th ordinal is $1$. (Note that $\omega_1^L=\omega_1$ does not say that $\aleph_1^L = \aleph_1$! $\omega_1$ belongs to $L$, but it might not be the order type of $\aleph_1^L$) In case of $L$, $\alpha^L = \alpha$ is just obvious: $\Ord^L = \Ord$ and $\Ord$ is well-ordered so the only isomorphism between them is identity map. –  An Hoa Apr 29 '11 at 1:50
    
An Hoa, that's fine, but then the same argument showing that "$x$ is constructible" is $\Delta_1$ shows also that "$x=L_\alpha$" is $\Delta_1$. –  Joel David Hamkins May 12 '11 at 3:02
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1 Answer

up vote 5 down vote accepted

Yes. The construction relativizes level by level.

This can be verified by a straightforward induction. The point is that if $M$ is a transitive model of ZF and $D\in M$ is transitive, then ${\rm def}(D)\subset M$ and (therefore) ${\rm def}(D)={\rm def}(D)^M$.

To see this, either use that definability is $\Delta_1$ (and therefore absolute), or use that ${\rm def}(D)$ is the intersection of ${\mathcal P}(D)$ with the closure of $D\cup\{D\}$ under the Gödel operations. (If you are not familiar with this approach, Jech's "Set Theory" provides the details, in Chapter 13.)

The argument is "local", and it is easy to see that requiring $M$ to be a model of ZF is an overkill. In fact, for each limit $\alpha$, $L_\beta^{L_\alpha}=L_\beta$ for $\beta\lt\alpha$ (and even stronger resuts are possible).

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Thanks Andres. It is interesting to know that many stronger results are possible in this case. There is one point I want to clarify though: you meant $\text{def}(D)$ is the intersection of $\mathcal{P}(D)$ and the closure of $D \cup \{D\}$ under G\"odel operations, didn't you? –  An Hoa Apr 28 '11 at 8:51
    
Sure. I've edited the text accordingly. –  Andres Caicedo Apr 28 '11 at 14:18
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