TITLE: Why is this set stationary?
QUESTION [7 upvotes]: Let $\kappa$ be a regular uncountable cardinal, and let $\lambda < \kappa$ be regular. Define the set $$E_{\lambda}^{\kappa} = \{\alpha < \kappa \mid \operatorname{cf}\alpha=\lambda \}.$$ I am trying to show that this is a stationary subset of $\kappa$, i.e. that it has non-zero intersection with every closed unbounded set $\subseteq \kappa$. 
My attempt at solving this was to take a club $C$ and then construct an increasing sequence $\langle \beta_{\xi} \mid \xi < \lambda\rangle$ in $C$. Since $\operatorname{cf}\lambda = \lambda$ then $\alpha:= \displaystyle\lim_{\xi \rightarrow \lambda}\beta_{\xi} < \kappa$, and hence $\alpha \in C$. If I could show that $\operatorname{cf}\alpha = \lambda$, then $E_{\lambda}^{\kappa} \cap C \not= \emptyset$, which is good.
Is this the right way to go about this? Thanks.

REPLY [5 votes]: Your approach works. You have an increasing sequence $\langle \beta_\xi:\xi<\lambda\rangle$ in $C$, with $\alpha=\sup\limits_{\xi<\lambda}\beta_\xi$. Since $\lambda<\kappa$, and $\kappa$ is regular, $\alpha<\kappa$, and since $C$ is closed, this implies that $\alpha\in C$. The sequence $\langle \beta_\xi:\xi<\lambda\rangle$ is evidently cofinal in $\alpha$, so $\operatorname{cf}\alpha=\operatorname{cf}\lambda=\lambda$, since $\lambda$ is regular. Hence $\alpha\in E_\lambda^\kappa\cap C$, as desired.