Let $\kappa$ be a singular cardinal with $\operatorname{cf}\kappa = \lambda > \omega$.
Let $C = \{ \alpha_{\zeta} \mid \zeta < \lambda \}$ be a strictly increasing continuous sequence of cardinals with limit $\kappa$. I want to show this is a closed unbounded set in $\kappa$.
Obviously it is unbounded, because $\sup C = \kappa$, but how do I show it is closed? Is it to do with continuity? Also I am used to dealing with clubs in regular cardinals, are there any differences for clubs in singular cardinals?
Thanks very much.
Yes, you are correct. The fact that $C$ is closed has everything to do with the sequence $\{ \alpha_\zeta \}_{\zeta < \lambda}$ being continuous.
The main idea is that since the sequence is strictly increasing, then for any limit ordinal $\gamma < \lambda$ we have that $\lim_{\zeta < \gamma} \alpha_\zeta = \sup ( \{ \alpha_\zeta : \zeta < \gamma \} )$. Therefore, if $\alpha < \kappa$ is a limit ordinal such that $\sup ( C \cap \alpha ) = \alpha$, let $\gamma = \min \{ \zeta < \lambda : \alpha_\zeta \geq \alpha \}$. We can show that $\gamma$ must be a limit ordinal, and that $C \cap \alpha = \{ \alpha_\zeta : \zeta < \gamma \}$. Using continuity we have that $$\alpha_\gamma = \lim_{\zeta < \gamma} \alpha_\zeta = \sup ( \{ \alpha_\zeta : \zeta < \gamma \} ) = \sup ( C \cap \alpha ) = \alpha,$$ and so $\alpha \in C$.