Arbitrarily large sized gaps in the $L$ hierarchy

Question

Question: Assume $V=L$. Let $\xi < \omega_1$ be arbitrary. Then there is $\alpha < \omega_1$ such that $L_\alpha \cap P(\omega) = L_{\alpha+\xi} \cap P(\omega)$.

Here's my idea:

When $V = L$, $L_{\omega_2} \vDash ZFC-PowerSet$ (since when $V=L$ the $L$ and $H$ hierarchy coincide for uncountable cardinals (basically the proof of $GCH$)).

Since $L_{\omega_2} \vDash ZFC-Powerset$ it can develop the notions of formulas, definability, constructibility, ordinal addition, etc, and will moreover be correct about it (i.e, it's absolute). And since $P(\omega) \subseteq L_{\omega_1} \subseteq L_{\omega_2}$, we have that

$L_{\omega_2} \vDash L_{\xi + \omega_1} \cap P(\omega) = L_{\omega_1} \cap P(\omega)$

Then take a countable $X \prec L_{\omega_2}$ such that $trcl(\{\xi\}) \subseteq X$ and apply the Mostowski collapse on $X$. By the condensation theorem this is $L_\beta$ for some $\beta \in \omega_2$. $\beta \in \omega_1$ since $L_\beta$ is countable. Moreover $\pi(\xi) = \xi \in L_\beta$ because $X$ contains the transitive closure. Finally $\omega_1$ is definable in $L_{\omega_2}$ (as the least uncountable ordinal) and thus $\omega_1 \in X$. Let $\alpha = \pi(\omega_1) \in L_\beta$. $\alpha < \omega_1$ So we then have:

$L_\beta \vDash L_{\xi + \alpha} \cap P(\omega) = L_{\alpha} \cap P(\omega)$

And again $L_\beta \vDash ZFC-Powerset$ and is absolute about the $L$-hierarchies, etc. So then we have that $L_{\xi + \alpha} \cap P(\omega) = L_\alpha \cap P(\omega)$ as desired.

I think this is mostly fine except for the absoluteness/correctness parts which I handwaved. I am usually uncomfortable and unsure when I appeal to absoluteness (feels too handwavey) like above so I just wanted to see if I made any mistakes. I think I'm mostly fine about the absoluteness of the definability and constructibility stuff. However, I'm not sure about the complexity and absoluteness of ordinal addition. Is it absolute?

Answer 1

Yes, your proof is right. Ordinal addition is absolute: the ordinal successor operation is absolute, unions are absolute, and a transfinite recursion defined from absolute operations is absolute.

A couple extra things:

1. Your question about absoluteness of ordinal addition suggests a generalization, since this is pretty much the only aspect of ordinal addition you used. Let $F(\alpha,\xi)$ be any absolute operation on the ordinals with $F(\alpha, \xi)\ge \alpha$. Then for any $\xi<\omega_1$ there is an $\alpha<\omega_1$ such that $L_{F(\alpha,\xi)}\cap P(\omega) = L_{\alpha}\cap P(\omega).$
2. Even though this is under the assumption of V=L, it can be rephrased without it and the proof is similar: For any $\xi < \omega_1^L$ there is an $\alpha < \omega_1^L$ such that $L_{\alpha+\xi}\cap P(\omega) = L_\alpha\cap P(\omega).$
3. Also note that by just increasing how many ordinals we put in the elementary submodel, we can replace "there is an $\alpha<\omega_1$" with "there are unboundedly many $\alpha < \omega_1$".