Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\alpha$?

share|cite|improve this question
You really need to learn some TeX, you know. :) – Asaf Karagila Jun 30 '11 at 17:45
up vote 8 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.

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.