Show that $\delta_\eta=\aleph_{\delta_\eta}$ for all limit ordinals Define $\delta_\eta$ by transfinite recursion as follows.
$$
\delta_0=0,\delta_{t+1}=\aleph_{\delta_t}\\
\delta_\eta=\sup\{\delta_\xi:\xi<\eta\}\text{ for limit cardinal }\eta.
$$
How to show that $\delta_\eta=\aleph_{\delta_\eta}$ for all limit $\eta$?

In general, we have $\alpha \leq \aleph_\alpha$ by induction on $\alpha$.


*

*Base Case: for $\alpha=0$ this is obvious.

*Successor Case:  if $\alpha \leq \aleph_\alpha$ then either $\alpha$ is infinite in which case $|\alpha+1|=|\alpha| \leq \aleph_\alpha < \aleph_{\alpha+1}$, or else $\alpha$ is finite in which case $\alpha+1<\aleph_0 < \aleph_{\alpha+1}$. 

*Limit Case: Suppose $\lambda$ is a limit ordinal and $\alpha \leq \aleph_\alpha$ for all $\alpha<\lambda$. Then by definition $$\aleph_\lambda = \sup\{\aleph_\alpha \mid \alpha <\lambda\} \geq \sup\{\alpha \mid \alpha < \lambda\} = \lambda$$
So we get $\delta_\eta \leq \aleph_{\delta_\eta}$. For the other direction, by definition 
$$\aleph_{\delta_\eta} = \sup \{\aleph_\beta \mid \beta < \delta_\eta\} \leq \sup \{ \aleph_{\delta_\alpha} \mid \alpha<\eta\} = \sup\{\delta_{\alpha+1} \mid \alpha<\eta\} = \delta_\eta$$ 
