$\aleph$ function fixed points below a weakly inaccessible cardinal are a club set I am throwing yet another one of my solutions out here for the internets to debug and for future set-theory students. 

Let $\aleph_\delta$ a weakly inaccessible cardinal. Prove that $A =\{\alpha < \aleph_\delta \mid \alpha = \aleph_\alpha\}$ is a club set.

Any comments on my solution are welcome.
Thanks!
 A: First we will show that $A$ is closed.
Let $\langle\alpha_i \mid i < \lambda\rangle \subset A$, and let $\alpha_\lambda = \bigcup_{i < \lambda}\alpha_i$. We want to show $\alpha_\lambda = \aleph_{\alpha_\lambda}$. If $\alpha_\lambda < \aleph_{\alpha_\lambda}$, there is $\beta < \alpha_\lambda$ such that $|\alpha_\lambda| = \aleph_\beta$. Hence there is $\alpha_i$ such that $\alpha_\lambda \leq \aleph_{\alpha_i}$. But $\alpha_\lambda \leq \aleph_{\alpha_i} = \alpha_i < \alpha_\lambda$.
Now we will show that $A$ is not bounded. To show that, I will try to apply what I learned from this question.
Let $\gamma < \aleph_\delta$. $\aleph_\delta$ is weakly inaccessible, so it is a fixed point of the $\aleph$ function, hence: $\gamma \leq \aleph_\gamma < \aleph_\delta$. We define the following series:


*

*$\gamma_0 = \aleph_\gamma$ 

*$\gamma_{n+1} = \aleph_{\gamma_n}$

*$\gamma_\omega = \bigcup_{n < \omega}\gamma_n$


From the same reasons as before, for every $n$, $\gamma_n < \aleph_\delta$. Also, $\aleph_0 < \aleph_\delta$ and $\aleph_\delta$ is regular, therefor $\gamma_\omega < \aleph_\delta$, so if we show that $\gamma_\omega = \aleph_{\gamma_\omega}$, we win.
But this is true because if $\gamma_\omega < \aleph_{\gamma_\omega}$, we get that for some $n$, $\gamma_\omega \leq \aleph_{\gamma_n} < \aleph_{\gamma_{n+1}} = \gamma_{n+2} < \gamma_\omega$.
Victory.
