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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 $L_alphavDash ZFC$.

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 $L_kappa models mathrmZFC$ 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 $mathcalL$-models for some first order language $mathcalL$. Then the following are equivalent:

1. 
   $M prec N$ and


2. For every $x_0, ldots, x_n in M$ and every $mathcalL$-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 $mathcalL$ 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 $mathcalL$-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).