If $V=L$ and $\alpha<\aleph_1$, find a countable $\beta >\alpha$ s.t. when $x\in L_{\beta+1}$, $x\in L_{\beta}$

Question

Assume $V=L$. I want to show that for any countable ordinal $\alpha$, there is a countable $\beta >\alpha$ such that whenever a set of integers $x\in L_{\beta+1}$, it follows $x\in L_{\beta}$.

This is the question I should have asked last time (at least I hope).

Answer 1

Let me expand on Jonathan's excellent answer.

First note (working under $V=L$) that if $x\subseteq\omega$ then $x\in L_{\omega_1}$. This is how we prove that $\sf CH$ holds in $L$. And this is true by a condensation argument.

Now. to utilize this, we want practically any limit ordinal above $\omega_1$. If $\delta>\omega_1$ is a limit ordinal, then:

1. $L_\delta$ is a model of $V=L$.
2. $L_\delta$ satisfies "There is an ordinal $\beta$ such that no new subsets of $\omega$ are added after $L_\beta$".
3. $L_\delta$ satisfies $\alpha$ is countable (and in particular $\alpha<\beta$ where $\beta$ is the least ordinal mentioned in item (2)).

So by taking a countable elementary submodel $M$ such that $L_\alpha\in M$, and then collapsing $M$ to a transitive set we get some $L_\gamma$ satisfying (2) and (3) at the same time.

Finally, $L_\gamma$ and $L_\beta$ have the same power set of $\omega$, and $\gamma$ is limit. So we get that $L_{\beta+1}$ did not add subsets to $\omega$.

Answer 2

Consider the collapse of a countable elementary submodel of $H(\omega_2) = L_{\omega_2}$ containing $L_\alpha$. This has the form $M = L_\gamma$ for some $\alpha < \gamma < \omega_1$. Now let $\beta = \omega_1^{M}$ and follow that $\beta$ has the required property.