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Let $n\in\omega$. Call a real $x$ $n$-special if $x\in L_{\omega_{n}^{CK,x}}$. Are the $n$-special reals bounded in $L_{\omega_{1}}$? That is, is there $\beta<\omega_{1}$ such that $L_{\beta}$ contains all $n$-special reals?

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Let me assume first that by $\omega_1$ you meant $\omega_1^L$, that is, the $\omega_1$ of $L$. In this case, no, even the $1$-special reals are not bounded. To see this, note first that there are unboundedly many $\gamma\lt\omega_1^L$ with $\gamma$ countable in $L_{\gamma+1}$. (Any stage where a new real is added collapses everything to $\omega$.) For such a $\gamma$, let $x$ be the $L$-least code of $\gamma$, so that $x\in L_{\gamma+1}\subset L_{\omega_1^x}$, and so $x$ is $1$-special. Thus, in $L$ there are uncountably many distinct $1$-special reals, and so they cannot be bounded.

But if you meant $\omega_1$ to mean $\omega_1^V$, then of course it is consistent that $\omega_1^L\lt\omega_1^V$ and in this case all the constructible reals, including all the $n$-special reals, appear in $L_\beta$ where $\beta=\omega_1^L\lt\omega_1$, which would make them bounded in the sense you asked.

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  • $\begingroup$ Yes, $\omega_{1}^{L}$ was what I meant. Thanks for your very helpful response. $\endgroup$
    – user27352
    Oct 18, 2012 at 15:31

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