Constructible universe within constructible universe

2011-04-28T04:21:10Z

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 $$L_\alpha^L = L_\alpha.$$

Answer by Andres Caicedo for Constructible universe within constructible universe

2011-04-28T05:17:21Z

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).

Constructible universe within constructible universe - MathOverflow

Constructible universe within constructible universe

Answer by Andres Caicedo for Constructible universe within constructible universe