Continuum hypothesis in L I have a lemma that gives me that the power set (and intersection with L) of a countable ordinal $\alpha$ is subset $L_{\omega_1}$ with $\omega_1$ being minimal in that respect. I know that $L_\alpha$ is also countable.
Now I get that $P(\omega)\cap L$ is subset of $L_{\omega_1}$. But how do I know that $P(\omega)\cap L$  is not smaller than $\omega_1$? Do I already have some form of CH?
 A: We can prove that for all infinite ordinals $\alpha$, $|L_\alpha| = |\alpha|$. So we know that $|P(\omega) \cap L| \leq |L_{\omega_1}| = |\omega_1| = \omega_1$. All of this is done “externally”, if you will. The proof of this fact is done in ZF (actually, it can be done in a much weaker set theory, but let’s stick with ZF for concreteness).
However, to show the continuum hypothesis holds in $L$, we need to show that $L \models |P(\omega)| \leq \omega_1$. Strictly speaking, we would need to exhibit an injection $P(\omega) \cap L \to (\omega_1)^L$, where $(\omega_1)^L$ is what $L$ believes to be $\omega_1$. Formally, $(\omega_1)^L$ is the smallest ordinal such that there is no bijection $(\omega_1)^L \to \omega_0$ which is in $L$. We know that $(\omega_1)^L \leq \omega_1$, but we don’t necessarily know there is an equality. And we also don’t immediately know that the injection $P(\omega) \cap L \to \omega_1$ produced above is itself an element of $L$ (though in fact it is, but that would require inspecting the proof).
To actually use the result quoted in the first paragraph, we note that $L \models ZF$. Thus, we see that $L \models |P(\omega) \cap L| \leq \omega_1$, since the proof in paragraph 1 was carried out in ZF. And we also know that $L \models V = L$. So in particular, $L \models P(\omega) = P(\omega) \cap L$. Then $L \models |P(\omega)| \leq \omega_1$. So the Continuum Hypothesis does indeed hold in $L$ (and in fact the same approach can be applied to the generalised Continuum Hypothesis).
