A cardinal $\kappa$ is weakly Mahlo if it's weakly inaccessible and the set $\{\lambda\in\kappa:\,\lambda\,\text{weakly inaccessible}\}$ is stationary in $\kappa$.
Let's define $E_0=\{\lambda:\,\lambda\,\text{weakly inaccessible}\}$, $E_{\alpha+1}=\{\lambda\in E_\alpha:\,|\lambda\cap E_\alpha|=\lambda\}$ and $E_\lambda=\bigcap_{\alpha<\lambda}E_\alpha$ if $\lambda$ is a limit ordinal. Now, I'm interested in see that $\kappa\in E_\kappa$ and, in fact, that $\kappa$ is the $\kappa^{th}$-cardinal of $\{\alpha:\alpha\in E_\alpha\}$. Does anybody have any idea how to see this?
Thanks in advance.
It is enough to show that actually $E_\alpha\cap\kappa$ is stationary by induction on $\alpha<\kappa$.
The case $\alpha=0$ follows from the fact that $\kappa$ is weakly inaccesible, and thus the set of limit cardinals less than $\kappa$ is closed unbounded.
There are two cases:
Suppose $E_\alpha\cap\kappa$ is stationary. Then $(E_\alpha\cap\kappa)'$ is club and thus $(E_\alpha\cap\kappa)'\cap E_\alpha$ is stationary, and as this is a subset of $E_{\alpha+1}$, we get that $E_{\alpha+1}$ is stationary.
Now suppose $\alpha<\kappa$ is limit, and all $E_\beta$ are stationary for $\beta<\alpha$. Suppose $E_\alpha$ is not stationary. Then there is a club $C$ such that for each $\beta\in C$ there is some $\xi<\alpha$ with $\beta\notin E_\xi$. Let, for each $\beta\in C$, $f(\beta)$ be the least $\xi$ such that $\beta\notin E_\xi$; notice that each $f(\beta)$ is a successor ordinal and $f(\beta)<\alpha.$ There exists a stationary set $S\subseteq C\cap E_0$ such that $f$ is constant on $S$, with value $\xi+1$. Pick some $\alpha\in (E_\xi\cap\kappa)'\cap S$; $(E_\xi\cap\kappa)'$ is club since $E_\xi\cap\kappa$ is stationary , then $\alpha\in E_\xi$ as $\alpha\in S$ because $f(\alpha)=\xi+1$. Then as $\alpha$ is also a regular cardinal; $S\subseteq E_0$, we get that since $\alpha\in (E_\xi\cap\kappa)',$ $|\alpha\cap E_\xi|=\alpha$, therefore $\alpha\in E_{\xi+1}$. Contradiction.
Hence as $E_\alpha\cap\kappa$ is stationary for each $\alpha<\kappa$, we get $\kappa\in E_\kappa$.
Now let us see that in fact $\{\alpha<\kappa:\alpha\in E_\alpha\}$ is stationary. Suppose not, then there is a club $C$ such that for each $\alpha\in C$ there is some $\xi<\alpha$ with $\alpha\notin E_\xi$. Define $f:C\rightarrow \kappa$ as in (2), then arguing similarly there is a stationary set $S\subseteq C\cap E_0$ such that $f$ has constant value $\xi+1$, and we get a similar contradiction.
The argument I had initially given, that is, to use that there is a stationary set $S\subseteq\kappa$ of weakly inaccesibles such that $V_\alpha\preceq V_\kappa$ for all $\alpha\in S$, is shorter but only works if $\kappa$ is also inaccesible though.