I have recently been wondering if the following is consistent with ZFC:



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> For every infinite ordinal $\alpha$: $|V\_\alpha\cap L|=|\alpha|$.
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> 


Intuitively, this states that for $L$ is very "thin", in that it doesn't branch off too much from the set of ordinals as we climb cumulative hierarchy.


I have heard that it is possible to have it for $V\_{\omega+1}$, which is essentially having $\Bbb R^L$ countable. I suspect that this *is* consistent, and we can construct a model of this starting from $V=L$ and adding a proper class of generic sets giving bijections between parts of $L$ and ordinals, but I have no real background in forcing to see if that makes any sense.


Thanks in advance.



Can $L$ be thin?
**Claim:** $|V\_\alpha \cap L| = |\alpha|$ for every $\alpha \geq \omega$ implies that $0^\#$ exists. 


**Proof:** Let's assume, toward contradiction, that $0^\#$ doesn't exist that $|V\_\alpha \cap L| = |\alpha|$ for every $\alpha \geq \omega$.


Let $\mu$ be a singular cardinal in $V$. Let's apply the assumption $|V\_\alpha \cap L| = |\alpha|$ for $\alpha = \mu + 1$. In $V$, $\mu = |\mu + 1|=|V\_{\mu + 1} \cap L| \geq |\mathcal{P}^L(\mu)| \geq |(\mu^{+})^L|$.


In particular, $\mu^{+} > (\mu^{+})^L$. Therefore, the cofinality of $(\mu^{+})^L$ in $V$ is strictly below $\mu$. Applying Jensen's covering theorem, we conclude that the cofinality of $(\mu^{+})^L$ in $L$ is strictly below $\mu$ - a contradiction to the regularity of $(\mu^{+})^L$ in $L$. 


By the way, if we restrict the values of $\alpha$ for which we want $|V\_\alpha \cap L| = |\alpha|$ to be $\leq \aleph\_{\omega}$, the consistency strength drops considerably:


**Claim:** $|V\_\alpha \cap L| = |\alpha|$ for every $\omega \leq \alpha \leq \aleph\_\omega$ is equiconsistent with the existence of $\omega$ inaccessible cardinals. 


**Proof:** For the first direction, force over $L$ with an iteration of Levi collapses in order to make the $n$-th inaccessible in $L$ equal to the $\aleph\_n$ of the generic extension. For the second direction, note that the assumption implies that for all $1\leq n < \omega$, $\aleph\_n$ is a $\beth$-fixed point in $L$, since for every infinite $\beta < \aleph\_n$, $|V\_\beta \cap L| = |\beth\_\beta^L| < \aleph\_n$. In particular, $\aleph\_n$ is a limit cardinal in $L$. It is regular in $V$ and thus also in $L$ and therefore it is inaccessible. 


