Finding, for each $\alpha \alpha$ such that $L_{\beta + 1} \models \beta \text{ is countable}$

set-theory

I would like to prove the following statement:

Assume $V = L$. Then for each ordinal $\alpha < \omega_1$, there exists an ordinal $\beta > \alpha$ such that $L_{\beta + 1} \models \beta \text{ is countable}$.

For my first pass at a proof, I considered a countable elementary substructure $M$ of $L_{\omega_1}$ such that $\alpha \subseteq M$. Since $L_{\omega_1} \models \text{all ordinals are countable}$, $M \models \text{all ordinals are countable}$; further, since it can be shown via a condensation argument that $M$ is transitive, there exists a limit ordinal $\beta < \omega_1$ such that $\alpha < \beta$ and $M = L_{\beta}$. Thus, in $L_{\beta}$, all ordinals $\beta'$ are countable, and so for each $\beta'$, there exists a surjection $f_{\beta'} : \omega \rightarrow \beta'$. My hope was then to combine these $f_{\beta'}$ into a definable subset $f$ of $L_{\beta}$ such that $f : \omega \rightarrow \beta$ is surjective. Then this $f$ would belong to $L_{\beta + 1}$, and so $L_{\beta + 1}$ would think that $\beta$ is countable. However, because the union over the $f_{\beta'}$ is not necessarily a function and because I can't come up with any other way to combine the functions, I am stumped. Is there a better way to use the $f_{\beta'}$? Is there a better way to approach this question entirely?

Best Answer

Having $L_{\alpha+1}\models$ "$\alpha$ is countable" is a strong countability property - this means that we don't want $\alpha$ to be $\omega_1$-like, so we emphatically don't want to look at things like elementary submodels of $L_{\omega_1}$. In particular, a club of countable ordinals will not have the desired property (take e.g. set of Mostowski collapse images of $\omega_1^M$ for $M$ a countable transitive model of $L_{\omega_2}$).

Instead, note that the desired condition is the same as "$L_\alpha$ has a definable bijection between $L_\alpha$ and $\omega$" (since the stuff in $L_{\alpha+1}$ is exactly the stuff definable in $L_\alpha$, and $L_\alpha$ has a bijection between itself and $\alpha$). One nice way for a level of $L$ to see its own countability is for it to$^1$ be the first level of $L$ satisfying some sentence with hereditarily countable parameters:

Suppose $L_\eta$ satisfies enough of $\mathsf{ZFC}$, $a\in \mathsf{HC}^{L_\eta}$, and there is a formula $\varphi$ such that $L_\eta$ is the least level of $L$ containing $a$ as an element and satisfying $\varphi(a)$. Let $f_1,...,f_n$ be definable-over-$L_\eta$ Skolem functions for $\varphi(a)$, and - in $L_\eta$ - let $C$ be the Mostowski collapse of the closure of $tc(\{a\})$ under the Godel operations and the $f_i$s We have that $C$ is definable (scince we've only used bounded-quantifier-rank "tools" to build $C$ - regardless of how complicated $\varphi$ is, there are only finitely many Skolem functions we need), $a\in C$ (since we folded in the transitive closure), and $L_\eta\models$ "$C$ is countable" (since $a\in\mathsf{HC}^{L_\eta}$), so by assumption we get $C=L_\eta$.

And we can now apply this in a very silly way:

For each $a\in L_{\omega_1}$ consider the least level of $L$ which sees that $a$ is hereditarily countable.

$^1$Actually, we also need $L_\eta$ to satisfy a small fragment of $\mathsf{ZFC}$ as well. But I do mean a small fragment - we just need $L_\eta$ to be able to perform basic recursive constructions, a la Lowenheim-Skolem and Mostowski. So in partcular, $\mathsf{KP}$ is enough.

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