Constructible universe within constructible universe - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:58:21Z http://mathoverflow.net/feeds/question/63254 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63254/constructible-universe-within-constructible-universe Constructible universe within constructible universe An Hoa 2011-04-28T04:21:10Z 2011-04-28T14:18:51Z <p>Let $\langle L_\alpha \rangle$ denote the hierarchy of constructible sets namely $$L_0 = \emptyset$$ $$L_{\alpha+1} = \text{def}(L_\alpha)$$ $$L_{\gamma} = \bigcup_{\alpha&lt;\gamma}{L_\alpha}$$ 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 $L$.</p> <p>In $L$, one can also construct this hierarchy and we call it the relativized constructible hierarchy, denoted by $\langle L_\alpha^L \rangle$.</p> <p>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 $$L_\alpha^L = L_\alpha.$$</p> http://mathoverflow.net/questions/63254/constructible-universe-within-constructible-universe/63256#63256 Answer by Andres Caicedo for Constructible universe within constructible universe Andres Caicedo 2011-04-28T05:17:21Z 2011-04-28T14:18:51Z <p>Yes. The construction relativizes level by level. </p> <p>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) ${\rm def}(D)={\rm def}(D)^M$. </p> <p>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 <code>$D\cup\{D\}$</code> under the Gödel operations. (If you are not familiar with this approach, Jech's "Set Theory" provides the details, in Chapter 13.)</p> <p>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). </p>