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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.

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    $\begingroup$ 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
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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.

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