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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 $cf(\kappa)^{cf(\kappa)}$?

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2 Answers

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.

$(\operatorname{cf}\kappa)^{\operatorname{cf}\kappa}=2^{\operatorname{cf}\kappa}$.

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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 $\beth_1=2^\omega$.

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