Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer

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 $\lambda$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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