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It's true (in $V$) that for any infinite ordinal, $|L_\alpha|=|\alpha|$.

My question: Is it also true in $L$? i.e., does $L$ itself satisfies $|L_\alpha|=|\alpha|$ for any infinite ordinal $\alpha$?

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You really need to learn some TeX, you know. :) –  Asaf Karagila Jun 30 '11 at 17:45
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up vote 7 down vote accepted

Yes, this is true in $L$. This follows from what you said, since you could jump inside $L$ and then apply your general fact that it is true in $V$, since the $V$ of $L$ is $L$.

Alternatively, you could simply observe that the proof that $|L_\alpha|=|\alpha|$ in $V$ also shows that $|L_\alpha|^L=|\alpha|^L$ at the same time. The reason is that because every object in $L_{\alpha+1}$ is a definable-from-parameters subset of $L_\alpha$, it is determined by a formula (a natural number) and a finite subset of $L_\alpha$ (the parameters). Thus, at successor stages, $|L_{\alpha+1}|=\omega\cdot|L_\alpha|$, which is equal to $|L_\alpha|$, and so at each stage the cardinality does not increase. (And $L$ can observe that as well as $V$.) The general fact now follows by transfinite recursion.

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@Joel: Thanks, I thought this was it but for some reason wasn't 100% sure regarding this! –  Asaf Karagila Jun 30 '11 at 19:08
    
Thank you, Joel. –  mmm Jun 30 '11 at 21:30
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