A cardinal $\kappa$ is weakly inaccessible iff $\kappa > \omega$, $\kappa$ is regular, and $\forall\lambda<\kappa(\lambda^+<\kappa)$


(here $\lambda^+$ is the successor cardinal)


A cardinal $\kappa$ is strongly inaccessible iff $\kappa > \omega$, $\kappa$ is regular, and $\forall\lambda<\kappa(2^\lambda<\kappa)$


My question is how to prove if $\kappa$ is weakly inaccessible, then it is the $\kappa$-th $\aleph$ fixed point, also if $\kappa$ is strongly inaccessible, then it is the $\kappa$-th $\beth$ fixed point?


This question is found in I.13.17 of The Foundations of Mathematics by Kenneth Kunen.



If $\kappa$ is weakly inaccessible, then is it the $\kappa$-th aleph fixed point
If $\kappa$ is weakly inaccessible, then it is a limit cardinal and hence $\kappa=\aleph\_\lambda$
for some limit ordinal $\lambda$. Since the cofinality of $\aleph\_\lambda$ is
the same as the cofinality of $\lambda$, it follows by the regularity of $\kappa$ that
$\lambda=\kappa$, and so $\kappa=\aleph\_\kappa$, an $\aleph$-fixed point. 


The next
$\aleph$-fixed point after any ordinal $\beta\_0$ must have
cofinality $\omega$, since it is $\sup\_n\beta\_n$, where
$\beta\_{n+1}=\aleph\_{\beta\_n}$. So if a weakly inaccessible
$\kappa$ is the $\delta$-th $\aleph$-fixed point, it cannot be
that $\delta$ is a successor ordinal, and so $\delta$ is a limit ordinal. Since the
$\aleph$-fixed points are closed, this implies $\kappa$ has the same cofinality as $\delta$, and so by regularity it follows that $\kappa=\delta$ and thus, $\kappa$ is the $\kappa$-th fixed point.


Essentially the same argument works with $\beth$ and strongly
inaccessible cardinals, simply by replacing $\aleph$ everywhere with $\beth$.


