I was looking at the proof that the set of all $\alpha$ such that $L_{\alpha+1}\cap P(\omega) = L_\alpha\cap P(\omega)$ is unbounded in $\omega_1^L$ At the beginning of the paper Gaps in the Constructible Universe by Marek and Srebrny (see page 364). They appeal to a theorem by Devlin
If $M\prec L_{\omega_1^L}$ then $M$ is transtive. (And hence by condensation is $L_\alpha$ for some $\alpha$.)
The proof of the result from this theorem given in the article seems easy enough. Here it is, paraphrased:
If $\gamma<\omega_1^L,$ then take a minimum cardinality elementary submodel $L_\alpha\prec L_{\omega_1^L}$ containing $\gamma,$ which obviously has $\alpha\ge \gamma.$ Since we can do this construction relative to $L,$ $(|\alpha| = |\gamma|< \omega_1)^L,$ so $\alpha < \omega_1^L.$ If $x\in P(\omega)\cap L_{\alpha+1}$ then $x$ is a subset of $\omega,$ definable over $L_\alpha,$ but since $L_\alpha \prec L_{\omega_1^L},$ $L_\alpha$ satisfies separation, so subset of $\omega$ definable over $L_\alpha$ is already in $L_\alpha.$
My question is why was Devlin's theorem necessary? Can't we just take the Mostowski collapse of the above elementary submodel? This will be an $L_\alpha$ with $\gamma\le \alpha <\omega_1^L$ that satisfies separation. If not for Devlin’s theorem it would no longer necessarily be an elementary submodel of $L_{\omega_1^L},$ but I don't see where this fact is used... something isomorphic to a model of separation is still a model of separation, right?