Let $\alpha$ be a limit ordinal and let $\lambda = cf(\alpha)$ be its cofinality. This means that there exists a strictly increasing $\lambda$-sequence $\langle \alpha_\xi \mid \xi < \lambda \rangle$ such that $\sup_{\xi < \lambda} \alpha_\xi = \alpha$.

I wonder if we can assume without loss of generality that the sequence is continuous, i.e. that the following holds: $$\forall \xi < \lambda \ \ [ \xi \text{ limit } \Rightarrow \alpha_\xi = \sup_{\gamma < \xi} \alpha_\gamma ].$$

My guess is that this is not true, but I can't come up with a proof.

  • 1
    $\begingroup$ I believe that, if $S$ is a set of ordinals, then the closure of $S$ has the same cardinality as $S.$ Does that help? $\endgroup$
    – bof
    Apr 18 '16 at 10:49

Yes -- if you have a sequence where some of the intermediate limits are missing, you can just replace $\alpha_\xi$ by the limit you want it to be.

This cannot possibly make $\alpha_\xi$ larger, so the sequence is still strictly increasing -- and the supremum of the sequence is the same as the supremum of the sequence elements with successor indices, so that doesn't change either.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.