I posted this question on Math StackExchange but did not get a full answer. I hope it's not a problem if I ask again here.


Is there a way to compute explicitly the $L$−rank $\rho(\bigcup x)$ of $\bigcup x$ in terms of $\rho(x)$? I know it's necessarily $\rho(\bigcup x)\leq\rho(x)$ and that both $\rho(\bigcup x)<\rho(x)$ and $\rho(\bigcup x)=\rho(x)$ can hold, but what else can be said, maybe distinguishing between $\rho(x)$ limit ordinal and successor ordinal? For example, is it possible that equality holds if $\rho(x)$ is a successor ordinal? If so, could you give an example?



Computing $L$-rank (constructible universe)
There is no way to compute explicitly $\rho(\bigcup x)$ in terms of $\rho(x)$, in any meaningful fashion: e.g. (working in $L$) for arbitrarily large countable $\alpha$ there are reals $x\subseteq \omega$ with $\rho(x)$ a limit or successor greater than $\alpha$ but with $\rho(\bigcup x) =\omega$ of course. (Because for arbitrarily large such $\alpha$ there is a new real $x$ definable over $L\_\alpha$. If $\alpha$ is not a successor, then there will always be a new real definable over $L\_{\alpha +1 } $ too; for example the real that codes $L\_\alpha$; similarly if $\alpha$ is not a limit there will in any case be a new real definable over $L\_{\alpha + \omega}$.) Thus the gap between the two ranks under either assumption can be as wide as it could conceivably be.


If $\rho(x)=\gamma+1$ then $\rho(\bigcup x)= \rho(x)$ can happen. Work in $L$: Suppose $x\subseteq \omega$ has $\rho(x)=\gamma +1$. Let $y = \{ \{u\} \mid u \in x \}$. Then $\rho(y)\leq\gamma +1 $; $\rho(\bigcup y) =\rho(x)$, but $\rho(y)\leq\gamma$ would imply that $\rho(x)\leq\gamma$. 


