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Usually the notion of cofinality is used for cardinals, though there seems to be no problem defining it for non-cardinal limit ordinals also: the cofinality of a limit $\alpha$ is the least $\gamma$ such that there is a less than $\gamma$ sequence of ordinals less than $\gamma$ such that the $\operatorname{sup}$ of the sequence is $\alpha$. But running through some examples leaves me wondering whether the notion is interesting: $\operatorname{cf} (\omega \cdot n) = \omega$, and $\operatorname{cf}(\epsilon_0) = \omega$, since $\omega, \omega^\omega, \omega^{\omega^\omega}, \dots$ is a confinal sequence.

My question is, can we prove that some non-cardinal ordinals are regular? More particularly, are there any countable ordinals that can be proven regular?

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The answer is no. Every regular ordinal is a cardinal.

The reason is simple, if $\alpha$ is not a cardinal, then it has a bijection with some smaller $\gamma$. This bijection is a sequence of order type $\gamma$ which is cofinal in $\alpha$. Therefore the cofinality of $\alpha$ is at most $\gamma<\alpha$.

If you want this sequence to be strictly increasing you can inductively thin out the sequence resulting from the bijection, at each step either you have finished and have an order type smaller than $\gamma$ or you will have a strictly increasing sequence of order type $\gamma$. Either way, we have $<\alpha$.

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  • $\begingroup$ thanks for the answer Asaf! $\endgroup$ – lesnikow May 18 '12 at 5:20
  • $\begingroup$ Wouldn't it be more correct to say "This bijection is a sequence $\gamma \to \alpha$ whose image is a cofinal subset of $\alpha$"? $\endgroup$ – John McClane May 6 '19 at 22:09
  • $\begingroup$ Well, what is a sequence? It's a function from some set. And since the image of the bijection is $\alpha$ itself, of course it's cofinal. $\endgroup$ – Asaf Karagila May 6 '19 at 22:12
  • $\begingroup$ I think that there is a substitution of concepts here. We're looking for a sequence (or a function) whose image has order type $\gamma$, but the bijection you're using is a function whose domain has order type $\gamma$, which is not the same. Thus in terms of the definition of the cofinality of ordinals, the conclusion "Therefore the cofinality of $\alpha$ is at most $\gamma$" in the second paragraph is premature. $\endgroup$ – John McClane May 7 '19 at 6:47
  • $\begingroup$ @John: Have you read the entire answer? $\endgroup$ – Asaf Karagila May 7 '19 at 6:49

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