Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking for an answer to the question:

"Show there exists an infinite cardinal $\kappa$ with $2^{cf(\kappa)}$ < $\kappa$ "

Where $cf(\kappa)$ is defined as the least $\alpha$ such that there is a map from $\alpha$ cofinally into $\beta$ and if f: $\alpha$ $\rightarrow$ $\beta$, f maps $\alpha$ cofinally iff ran(f) is unbounded in $\beta$.

I have looked in both Kunen's book and Jech's, but although there's lots of helpful things about the topic I can't quite get to the answer.

I know that $\kappa$ cannot be regular for the inequality to hold. The question is from a past exam, and the examiner's report indicates there's quite an easy proof but I can't seem to find it.

On a related note, is there a simplified expression for $cf(\kappa)^{cf(\kappa)}$?

share|improve this question

2 Answers 2

up vote 4 down vote accepted

All you need is a cardinal $\kappa>2^\omega$ with $\operatorname{cf}\kappa=\omega$. Start with $\kappa_0=2^\omega$, let $\kappa_{n+1}=\kappa_n^+$, and let $\kappa=\sup_n\kappa_n$. The same idea can be used to get $\kappa$ with any desired cofinality: you just need to make it the limit of a longer sequence.


share|improve this answer

Define $\beth_0=\aleph_0$, $\beth_{n+1}=2^{\beth_n}$, and $\beth_\omega=\sup\{\beth_n | n \in \omega \}$. Clearly $\beth_\omega$ has cofinality $\omega$ but is larger than $\beth_1=2^\omega$.

share|improve this answer

Your Answer


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.