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I need help with the following problem:

Let $\kappa$ be a weakly inaccessible cardinal.

Show there exists a closed and unbounded set of $\alpha < \kappa$ such that $L_\alpha\vDash ZFC$.

I know that if $\kappa$ is weakly inaccessible then $L_\kappa$ is a model of ZFC. But other than that I am not really sure now to proceed.

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Hint: We have $L_\kappa \models \mathrm{ZFC}$ and $\kappa$ is regular.

We will show something stronger, namely

$$ C := \{ \alpha < \kappa \mid L_{\alpha} \prec L_{\kappa} \} $$ is a club.

First show that $C$ is unbounded: Let $\beta < \kappa$. As in the proof of the Reflection Theorem recursively construct a strictly increasing sequence $(\alpha_n \mid n < \omega)$ with $\alpha_0 = \beta$ such that $L_{\sup_{n <\omega} \alpha_n} \prec L_\kappa$. Note that $\beta < \sup_{n <\omega} \alpha_n < \kappa$.

Now show that $C$ is closed. This follows pretty much immediately from the Tarski-Vaught test (/ criterion):

Lemma. (Tarski-Vaught)

Let $M \subseteq N$ be $\mathcal{L}$-models for some first order language $\mathcal{L}$. Then the following are equivalent:

  1. $M \prec N$ and
  2. For every $x_0, \ldots, x_n \in M$ and every $\mathcal{L}$-formula $\phi$. If $$ N \models \exists x \colon \phi(x, x_0, \ldots, x_n), $$ then there is some $x \in M$ (!) such that $$ N \models \phi(x, x_0, \ldots, x_n) $$

(You can find a proof of this here.)

Corollary. Let $N$ be a $\mathcal{L}$ structure and $(M_n \mid n < \omega)$ be a strictly increasing sequence such that $M_n \prec M_{n+1} \prec N$ for all $n < \omega$. Then $M := \bigcup_{n < \omega} M_n \prec N$.

Proof. Let $x_0, \ldots, x_n \in M$ and let $\phi$ be a $\mathcal{L}$-formula such that $$ N \models \exists x \colon \phi(x, x_0, \ldots, x_n). $$

Fix $k$ large enough such that $x_0, \ldots, x_n$ in $M_n$. Since $M_n \prec N$ we have that there is some $x \in M_n$ such that $$ N \models \phi(x, x_0, \ldots, x_n) $$ But $x \in M_n \subseteq M$, hence the Tarski-Vaught criterion is satisfied and we have $$ M \prec N. $$ Q.E.D.

Let us show that $C$ is closed: Fix a strictly increasing sequence $(\alpha_\beta \mid \beta < \gamma)$ of $\alpha_\beta \in C$, for some $\gamma < \kappa$. By definition of $C$ we have $$ L_{\alpha_\beta} \prec L_\kappa $$ for all $\beta < \gamma$.

Claim. We also have $$L_{\alpha_\beta} \prec L_{\alpha_{\beta^*}}$$ for $\beta < \beta^* < \gamma$.

Proof. For all $x_0, \ldots, x_n \in L_{\alpha_\beta}$, we have $$L_{\alpha_\beta} \models \phi(x_0, \ldots, x_n) \iff L_{\kappa} \models \phi(x_0, \ldots, x_n) \iff L_{\alpha_{\beta^*}} \models \phi(x_0, \ldots, x_n)$$ Q.E.D.

It now follows from the corollary that $$ L_{\sup_{\beta < \gamma} \alpha_{\beta}} \prec L_{\kappa}, $$ so that $\sup_{\beta < \gamma} \alpha_{\beta} \in C$, so that $C$ is indeed closed.


Since you don't know the reflection theorem (or rather its proof), you can also use the Tarski-Vaught criterion to show that $C$ is unbounded:

Let $\beta_0 < \kappa$. If $L_{\beta_0} \prec L_{\kappa}$, we are done. Otherwise fix a strict wellorder $<^*$ of $L_{\kappa}$ and select, for all formulae $\phi$ and all $x_0, \ldots, x_n \in L_{\beta_0}$ the $<^*$-least $x_{\phi, x_0, \ldots, x_n} \in L_{\kappa}$ such that $$ L_{\kappa} \models \exists x \phi(x,x_0, \ldots, x_n) \implies L_{\kappa} \models \phi(x_{\phi, x_0, \ldots, x_n},x_0, \ldots, x_n). $$

Let $\beta_0 < \beta_1$ be minimal such that $x_{\phi, x_0, \ldots, x_n} \in L_{\beta_1}$ for all $\phi$ and all $x_0, \ldots, x_n \in L_{\beta_0}$.

Now continue doing this (replacing $\beta_0$ with $\beta_1$ and creating $\beta_2$ and so on) to produce an increasing sequence $(\beta_k \mid k < \omega)$. Let $\alpha := \sup_{k < \omega} \beta_k$ and verify that $L_{\alpha} \prec L_{\kappa}$ with Tarski-Vaught (the proof is basically the same as in the Corollary I've proved above).

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  • $\begingroup$ Please excuse my incompetence, I am new to set theory/model theory. I don't follow the proof of unboundedness (as I do not know what the reflection theorem is). Could you also please also explain more on how the Tarski-Vaught test shows closure? $\endgroup$ – A. Collins Mar 8 at 0:25
  • $\begingroup$ @A.Collins Sure, I'll expand my answer. $\endgroup$ – Stefan Mesken Mar 8 at 0:31
  • $\begingroup$ @A.Collins Does the recent edit answer your questions? $\endgroup$ – Stefan Mesken Mar 8 at 0:50
  • $\begingroup$ Thank you for your expanded answer. I am slowing parsing my way thought it. I am not quite done yet, but I think it should. $\endgroup$ – A. Collins Mar 8 at 1:10
  • $\begingroup$ @A.Collins If you have further questions, just leave a comment. I may not answer quickly (because I'm currently traveling) but I (or maybe someone else) will address it eventually. $\endgroup$ – Stefan Mesken Mar 8 at 1:17

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