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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.

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up vote 6 down vote accepted

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 $\beth$.

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