Disjoint Union of Stationary Sets

Question

Let $\kappa$ be a regular uncountable cardinal, and let $\lambda < \kappa$ be a regular cardinal. Then the set $$E_{\lambda}^{\kappa} := \{ \alpha < \kappa \mid \operatorname{cf}\alpha = \lambda \}$$ is stationary in $\kappa$. It is a result that every stationary subset of $E_{\lambda}^{\kappa}$ is the disjoint union of $\kappa$ stationary sets. Now define $$S = \{\alpha < \kappa \mid \operatorname{cf}\alpha<\alpha\}.$$

In Chapter 8, Jech makes the following claim: "Every stationary subset $W \subseteq S$ is the disjoint union of $\kappa$ stationary sets: By Fodor’s Theorem, there exists some $λ < κ$ such that $W ∩ E_{\lambda}^{\kappa}$ is stationary."

I'm a bit confused by this. Here is what I managed to extract from this problem: Is it true that S is always stationary? In other words, is it always possible for it to contain stationary subsets $W$?

If so, then consider the function $\operatorname{cf}:W \to \kappa$ which gives the cofinality of $\alpha \in S$. By definition this is regressive (i.e. $\operatorname{cf}\alpha < \alpha$) so by Fodor's Theorem, there exists a $\lambda < \kappa$ and a stationary set $T \subseteq W$ such that $\operatorname{cf}\alpha = \lambda$ on $T$. If $\lambda$ is a regular cardinal then I can see that $T \subseteq E_{\lambda}^{\kappa} \cap W$ and so $W\cap E_{\lambda}^{\kappa}$ is stationary as a superset, and we're done. But what if $\lambda$ is not a regular cardinal, and then $E_{\lambda}^{\kappa}$ doesn't make sense?

Any help with this would be appreciated.

Answer 1

Suppose that $W\subseteq S$ is stationary. Then as you say, $\operatorname{cf}$ is a pressing-down function on $W$, so it is constant on a stationary subset of $W$, i.e., there are a stationary $T\subseteq W$ and a $\lambda\in\kappa$ such that $\operatorname{cf}\alpha=\lambda$ for each $\alpha\in T$. Cofinalities are always regular, so $\lambda$ is regular, $E_\kappa^\lambda$ is defined, and clearly $T\subseteq W\cap E_\kappa^\lambda$. Since $T$ is stationary, the possibly larger set $W\cap E_\kappa^\lambda$ is also stationary, and all’s well.

Added: Note that $\lambda$ cannot be $1$: that would mean that every $\alpha\in T$ was a successor ordinal, and $T$ would be disjoint from the cub of limit ordinals and hence not stationary after all.

Answer 2

Let $\Lambda$ denote the set of limit ordinals less than $\kappa$. Since $\Lambda$ is a club $W\cap \Lambda$ is stlationary. Now your argument holds applied to this stationary.

Answer 3

@Paul Slevin: I don't follow your argument. First of all, if $\operatorname{cf}(\alpha)=\lambda$, then $\lambda$ must be regular. The set of all $\alpha\in W$ with $\operatorname{cf}(\alpha)=\lambda$ is just $W\cap E_\lambda^\kappa$. By Fodor's lemma, this is stationary for some $\lambda<\kappa$. Also, as was pointed out before, the set of successor ordinals is not stationary and hence the $\lambda$ that you get from Fodor's lemma is not $1$. So, we know that for some regular $\lambda>1$, $W\cap E_\lambda^\kappa$ is stationary. But we don't know whether $E_\lambda^\kappa\subset W$, which you seem to claim.

The following hasn't been addressed yet, it seems: whether the set $S$ contains a stationary set.

Since $\kappa$ is uncountable, there is a regular (infinite) cardinal $\lambda<\kappa$. Now the set of all $\alpha<\kappa$ with $\operatorname{cf}(\alpha)=\lambda$ and $\alpha>\lambda$ is a stationary subset of $S$. Why? Because it is the intersection of a stationary set, namely $E_\lambda^\kappa$, and a club, namely $\kappa\setminus(\lambda+1)$.