What is the cardinality of $\Bbb{R}^L$?

Question

By $\Bbb{R}^L$, I mean the set that is interpreted as $\Bbb{R}$ in $L$, Godel's constructible universe. For concreteness, and to avoid definitional questions about $\Bbb{R}$, I'm looking at the set ${\cal P}(\omega)$ as a proxy. I would think it needs to be countable, since only definable subsets of $\omega$ are being considered, but I don't see how Cantor's theorem would fail here, since $L$ is a model of $\sf{ZFC}$. (Actually, that's a lie: this probably means that there is no injection $f:\omega\to{\cal P}(\omega)^L$ in $L$, even if there is one in $V$. I'm still a little hazy on all the details, though.) But the ordinals are in $L$, so $L$ is not itself countable, and there must exist genuinely uncountable elements of $L$. What does the powerset operation (in $L$) do to the cardinality of sets (as viewed in $V$), then?

Answer 1

It is impossible to give a complete answer to this question just in $\sf ZFC$.

For once, it is possible that $V=L$ is true in the model you consider, so in fact $\Bbb R^L=\Bbb R$, so the cardinality is $\aleph_1$.

On the other hand it is possible that the universe is $L[A]$ where $A$ is a set of $\aleph_2$ Cohen reals, in which case $\Bbb R^L$ is still of size $\aleph_1$.

And it is also possible that the universe is $L[G]$ where $G$ is a function which collapses $\omega_1^L$, resulting in having $\Bbb R^L$ as a countable set.

There are other axioms, such as large cardinal axioms (e.g. "$0^\#$ exists") which imply that $\Bbb R^L$ is countable, and one can arrange all sort of crazy tricks, where the result is large or small compared to $\Bbb R$.

One thing is for certain, $|\Bbb R^L|=|\omega_1^L|\in\{\aleph_0,\aleph_1\}$.