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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?

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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 $\rho(x)\leq\gamma$.

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  • $\begingroup$ Actually this is not totally correct. It's true that new reals are constructed arbitrarily high, but surprisingly, there are also arbitrarily long gaps where no new reals appear. Let $\alpha < \omega_1$ be arbitrary, and let $\omega_1 > \beta > \alpha$ be such that there is an elementary $j : L_\beta \to L_{\omega_2}$ with $cr(j) = \gamma \geq \alpha$. Then for all $\delta < \gamma$, $L_{\omega_2} \models$ "There are no new reals constructed between $\omega_1$ and $\omega_1 + \delta$." By elementarity and absoluteness, there are no new reals added between $\gamma$ and $\gamma + \delta$. $\endgroup$
    – Monroe Eskew
    Commented Nov 15, 2013 at 18:29
  • $\begingroup$ Thank you for your answer. Could you give me an explanation for your last statement? I can't see why equality can't hold if $\rho(x)=\gamma+1$. $\endgroup$
    – phil
    Commented Nov 15, 2013 at 18:33
  • $\begingroup$ @Monroe: which part of my answer is incorrect? $\endgroup$
    – Philip Welch
    Commented Nov 15, 2013 at 19:10
  • $\begingroup$ It seems like you said if $\alpha$ is a countable limit, then a new real appears in $L_{\alpha+1}$. $\endgroup$
    – Monroe Eskew
    Commented Nov 16, 2013 at 0:58
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    $\begingroup$ @Monroe: I was discussing in the parenthetical remark ``such $\alpha$'' with a new real appearing definably over $L_\alpha$. Then one will also appear over $L_{\alpha +1}$, ... , $L_{\alpha+\omega}$... :) $\endgroup$
    – Philip Welch
    Commented Nov 16, 2013 at 9:13

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