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

For every infinite ordinal $\alpha$: $|V_\alpha\cap L|=|\alpha|$.

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.


2 Answers 2


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.

  • $\begingroup$ That a very straightforward argument, I also like the result about limiting the question to $\alpha\leq\aleph_\omega$. $\endgroup$
    – Wojowu
    Oct 7, 2015 at 11:29

This situation is a consequence of the existence of $0^\sharp$.

To see this, suppose that $0^\sharp$ exists. It follows that every uncountable cardinal is a Silver indiscernible and a limit of Silver indiscernibles. If $\alpha$ is any infinite ordinal, then $\beth_{\omega+\alpha}^L$, which is definable from $\alpha$, must appear before the next indiscernible (since they are closed under definability in $L$), which is before $(|\alpha|^+)^V$. Since $V_\alpha\cap L=(V_\alpha)^L$ has size $\beth_{\omega+\alpha}^L$, it follows that this will have size strictly less than $|\alpha|^+$ and hence size $|\alpha|$, as desired.

  • 10
    $\begingroup$ It is equivalent to $0^\#$: Take $\alpha = \mu + 1$ where $\mu = \aleph_{\omega}^V$ one can see that $|P(\mu )^L| = |{\mu^{+}}^L| = \mu$ so $V$ doesn't agree with $L$ on the successor to this singular and thus $0^\#$ exists. $\endgroup$
    – Yair Hayut
    Oct 6, 2015 at 20:11
  • $\begingroup$ @Yair: I was going to make that into an answer, but then I saw your comment. So you should make that into an answer. $\endgroup$
    – Asaf Karagila
    Oct 6, 2015 at 23:01
  • $\begingroup$ I also was writing up that as an extension to my answer when you commented. Please go ahead and post as an answer! $\endgroup$ Oct 6, 2015 at 23:58

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