About the definible sets $L_\alpha$

Question

Let $\alpha$ be an ordinal number. Is that true that $\alpha$ = $\beth_\alpha $ is equivalent to the statement $|L_\alpha|=|R_\alpha|$, where $L_\alpha $ is the $\alpha$-th stage of the constructible hierarchy?

Answer 1

For every infinite ordinal $\alpha$ it holds:

$|L_\alpha|=|\alpha|$ and $|R_\alpha|=\beth_{\omega+\alpha}$. If $\alpha$ is a $\beth$-fixed point then $\alpha=\omega+\alpha=\beth_\alpha$ (the addition there is ordinal addition). And it follows that equality holds.

On the other hand, if $\alpha$ is not a $\beth$-fixed point, then $\alpha\neq\beth_\alpha$. Clearly $|\alpha|\leq\beth_\alpha$, so in that case $|L_\alpha|<|R_\alpha|$.

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To see that $|L_\alpha|=|\alpha|$ note that $L_\omega$ is countable, and $|L_{\alpha+1}|=\aleph_0\cdot|L_\alpha|=|L_\alpha|=|\alpha|=|\alpha+1|$. For limit steps the proof is similar.

To see that $|R_\alpha|=\beth_{\omega+\alpha}$, use a similar induction and the definition of $\beth$ numbers.