Is it true that $L_\alpha \prec L_\beta$ implies that $V_\alpha\prec V_\beta$?

Question

Given two ordinals $\alpha<\beta$ such that $L_\alpha\prec L_\beta$, is it necessarily true that also $V_\alpha\prec V_\beta$? is the converse of this implication true? I'm working on a set theory homework and, if this is true, I got it. However, I don't know if this is actually true. I wouldn't know at first how to prove it or how to find any counter example, so any hint would be welcomed. Thanks in advance for any help.

Answer 1

No, this is not true - in fact, it's outright disprovable. For example, we have the following:

1. $V_\alpha\not\preccurlyeq V_{\omega_1}$ for any countable ordinal $\alpha$.

2. There is a countable $\alpha$ such that $L_\alpha\preccurlyeq L_{\omega_1}$.

Both of the facts above require proof, but they're good exercises; I've put spoilered hints below.

Hint for 1:

If $\alpha<\omega_1$ then (either $\alpha$ is finite and things are boring or) $V_\alpha$ contains a well-ordering of $\omega$ with ordertype $\alpha$ which implies $V_\alpha\models$ "There is a countable well-ordering not isomorphic to any ordinal."

Hint for 2:

This is actually a consequence of a much more general result: working in an at-most-countable language, if $\mathfrak{A}$ is any structure and $(\mathfrak{A}_\beta)_{\beta<\omega_1}$ is a sequence of countable structures such that $$\mathfrak{A}=\bigcup_{\beta<\omega_1}\mathfrak{A}_\beta,$$ $$\beta<\gamma\implies\mathfrak{A}_\beta\subseteq\mathfrak{A}_\gamma,$$ and $$\lambda\mbox{ limit}\implies \mathfrak{A}_\lambda=\bigcup_{\beta<\lambda}\mathfrak{A}_\beta,$$ then there is some $\eta<\omega_1$ such that $\mathfrak{A}_\eta\preccurlyeq \mathfrak{A}$. To apply this result here, note that - unlike for the $V$-hierarchy - we have $\vert L_\alpha\vert=\aleph_0$ for each countably infinite $\alpha$. (This hint can be pushed beyond countable "pieces;" the key new notion is cofinality.)