Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 5 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 $\beth$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.