Prove $\operatorname{cf}(\alpha)\leq|\alpha|$ if $\alpha$ is a limit ordinal.

I know every surjection onto α has range unbounded in $α$, but I'm stuck on how to formalize this argument.Thanks!

  • 5
    $\begingroup$ Please don't delete your question if it has received an answer. $\endgroup$ – Asaf Karagila Dec 13 '17 at 12:46

This is easier if you prove that the cofinality is always a cardinal.

And it seems you have the tools for that. Let me outline the steps:

  1. $\operatorname{cf}(\operatorname{cf}(\alpha)) = \operatorname{cf}(\alpha)$. This one is easy.

  2. If $\alpha=\operatorname{cf}(\alpha)$, then $\alpha$ is a cardinal. Otherwise, take $f\colon|\alpha|\to\alpha$ to be some bijection, then it has an unbounded range. Depending on how you define cofinality in the first place this might give you a direct contradiction or require you to distill a strictly increasing function. But in either case, you're done.

  3. $|\alpha|$ is the largest cardinal which is $\leq\alpha$.

Combine these and you should be fine.


For a limit ordinal, $\alpha$, the cofinalty is the smallest ordinal $\delta$ such that there is a $\delta$-indexed increasing sequence whose limit is $\alpha$. There is an $\alpha$-indexed sequence whose limit is $\alpha$ because you can just take the sequence that lists every ordinal less than $\alpha$ in increasing order. Since $\alpha$ satisfies the rule and $cf(\alpha)$ is the smallest ordinal that satisfies the rule, we get that $cf(\alpha)\leq|\alpha|$.

  • $\begingroup$ But you haven't argued the key issue: that $\operatorname{cf}(\alpha)$ is a cardinal. As written, all the argument tells us is that $\operatorname{cf}(\alpha)\le\alpha$. $\endgroup$ – Andrés E. Caicedo Dec 13 '17 at 12:38

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