Why is a certain subset of a regular uncountable cardinal stationary?

Question

this is an excerpt from Jech's Set Theory (page 94).

For a regular uncountable cardinal $\kappa$ and a regular $\lambda<\kappa$ let $$E^\kappa_\lambda= \{\alpha<\kappa:\mbox{cf}\ \alpha=\lambda \}$$ It is easy to see that each $E^\kappa_\lambda$ is a stationary subset of $\kappa$

Well, I don't see so easily why this should hold.

Essentially, given a set which is disjoint to a set of this form, there is no reason for it to be bound, so I tried proving it can not be closed. However, I have no idea why this should hold either, as it might not have any ordinal with cofinality $\lambda$ as a limit point as well.

Thanks in advance

Answer 1

Note that if $C$ is a club, then $C$ is unbounded, and therefore has order type $\kappa$. Since $\lambda<\kappa$ we have some initial segment of $C$ of order type $\lambda$.

Show that this initial segment cannot have a last element. Its limit is in $C$ and by the regularity of $\lambda$ must have cofinality $\lambda$.

Therefore $C\cap E^\kappa_\lambda\neq\varnothing$.

Answer 2

Here is a sketch of the proof (I'll add more details later if necessary): Define $f: \kappa \rightarrow \kappa$ to be normal if $f$ is increasing and continuous. Now prove that $C\subseteq \kappa$ is a club iff $C$ is the image of a normal function: In the first direction, if $(x_{\alpha} : \alpha<\kappa)$ enumerates the elements of $C$ in increasing order, then $f(\alpha)=x_{\alpha}$ is the desired normal function. The other direction is immediate. Now show that for every normal function $f:\kappa \rightarrow \kappa$, $cf(f(\lambda))=\lambda$. Now combining these results, every club $C\subseteq \kappa$ is of the form $Range(f)$ for some normal $f:\kappa \rightarrow \kappa$, and $f(\lambda) \in E_{\lambda}^{\kappa} \cap C$