Models of ZFC & Club Sets.

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.

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).

• 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? Mar 8, 2019 at 0:25
• @A.Collins Sure, I'll expand my answer. Mar 8, 2019 at 0:31
• @A.Collins Does the recent edit answer your questions? Mar 8, 2019 at 0:50
• 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. Mar 8, 2019 at 1:10
• @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. Mar 8, 2019 at 1:17