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My intuition is that co-finality is a non-decreasing function on the cardinals. If that's true, it seems to follow that all infinite cardinals are regular. In particular, $\aleph_0$ is clearly regular and its a fact that successor cardinals are regular. For the limit case, if $\kappa=\sup\{\lambda^+:\lambda<\kappa\}$, then $cf(\kappa)\ge \sup\{cf(\lambda^+):\lambda<\kappa\}=\sup\{\lambda^+:\lambda<\kappa\}=\kappa$ since co-finality is non-decreasing.

Because of this, I take it that my intuition that co-finality is non-decreasing is wrong. Can anyone confirm this?

In particular, I'm wondering if its possible that $cf(\kappa)=\aleph_0$ for some uncounbtable cardinal $\kappa$.

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Cofinality is a very much non-increasing function, and for every regular cardinal $\kappa$ we can find a limit cardinal $\lambda$ such that $\kappa=\operatorname{cf}(\lambda)\leq\lambda$.

In fact, we cannot prove the case of equality. That is, if a regular limit cardinal exists then we can prove the consistency of $\sf ZFC$, so adding to $\sf ZFC$ the assumption that such cardinal exists creates a strictly stronger theory than $\sf ZFC$ itself.

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  • $\begingroup$ That sorts that out. Thanks. $\endgroup$ – Elbu Dec 15 '13 at 23:26
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In fact cofinality is not non-decreasing. $\aleph_\omega = \sup\{\aleph_n | n\in \omega\}$ has cofinality $\aleph_0$, for instance.

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