# Why is this set stationary?

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.

-
Wait - is this valid? By Lemma 3.7 in Jech, $\operatorname{cf} \lambda = \operatorname{cf}\alpha$ since the sequence is increasing. But $\lambda = \operatorname{cf}\lambda$. – Paul Slevin Feb 15 '12 at 16:37

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.

-

The set $E^\kappa_\lambda$ is stationary, because every club subset $C\subset\kappa$ will have a $\lambda^{th}$ element, this will be an element in $E^\kappa_\lambda\cap C$. So the set meets every club, and hence is stationary. The reason every club subset of $\kappa$ has a $\lambda^{th}$ element is that $\lambda\lt\text{cof}(\kappa)$. The reason why the $\lambda^{th}$ element of a club is in $E^\kappa_\lambda$ is that the $\lambda^{th}$ element of the club has cofinality $\lambda$, being the limit of the $\lambda$ many prior elements of the club, and since $\lambda$ is regular, this means that the cofinality will be $\lambda$.

-