In proving the following theorem, I do not see why $S$ is the union of the pairwise disjoint stationary sets $S'_\eta$. It seems that for this to hold, you need every $\alpha_\xi$ to be equal to some $\gamma_\eta$?

Theorem. Suppose $\kappa$ is regular uncountable and $\lambda<\kappa$ is regular. Then the stationary set $S=\{\alpha<\kappa:\mbox{cf}(\alpha)=\lambda\}$ may be partitioned into $\kappa$ pairwise disjoint stationary sets.

proof. For each $\alpha\in S$ , let $(\alpha_{\xi})_{\xi<\lambda}$ be an increasing sequence in $\kappa$ with $\sup_{\xi<\lambda}\alpha_{\xi}=\alpha$ . For each $\eta<\kappa$ and $\xi<\lambda$ consider

$S_{\eta,\xi}=\{\alpha\in S:\eta\leq\alpha_{\xi}\}$.

Claim: There exists $\xi<\lambda$ such that $S_{\eta,\xi}$ is stationary in $\kappa$ for all $\eta<\kappa$ . Well, otherwise for all $\xi<\lambda$ there exists $\eta_{\xi}<\kappa$ and a club $C_{\xi}$ such that $C_{\xi}\cap S_{\eta_{\xi},\xi}=\varnothing$ , so that each element in $C_{\xi}$ has $\alpha_{\xi}<\eta_{\xi}$ . Then $C=\bigcap_{\xi<\lambda}C_{\xi}$ is club and $\alpha=\sup_{\xi<\lambda}\alpha_{\xi}\leq\sup_{\xi<\lambda}\eta_{\xi}<\kappa$ for each $\alpha\in C\cap S$ . But $C\cap S$ is stationary in $\kappa$ ; in particular it is unbounded in $\kappa$ . Contradiction.

Let $\xi<\lambda$ be given by the claim and $S_{\eta}=\{\alpha\in S:\eta\leq\alpha_{\xi}\}$ . Define $f(\alpha)=\alpha_{\xi}$ . Then $f$ looks down on each stationary set $S_{\eta}$ . For each $\eta<\kappa$ , using the Pressing Down Lemma, let $S'_{\eta}$ be a stationary subset of $S_{\eta}$ and $\eta\leq\gamma_{\eta}$ with $f(\alpha)=\gamma_{\eta}$ for all $\alpha\in S'_{\eta}$ . Then $\gamma_{\eta}\neq\gamma_{\eta'}$ implies $S'_{\eta}\cap S'_{\eta'}\neq\varnothing$ . In particular, $\left|\{S'_{\eta}:\eta<\kappa\}\right|=\left|\{\gamma_{n}:\eta<\kappa\}\right|$ . Since the $\gamma_{\eta}$ are cofinal in $\kappa$ and $\kappa$ is regular, this set has cardinality $\kappa$ .

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    $\begingroup$ Please do not (essentially) delete your questions like you did. $\endgroup$ – user98602 Apr 5 '14 at 5:45

$S$ need not be the union of the pairwise disjoint stationary sets $S_\eta'$, but it doesn’t matter: it’s sufficient that they are pairwise disjoint and stationary, and that there are $\kappa$ of them. To see this, enumerate them as $\mathscr{S}=\{T_\xi:\xi<\kappa\}$. Let $T=\bigcup\mathscr{S}$, and let $D=S\setminus T$. If $D$ is non-stationary, $$\{T_0\cup D\}\cup\big(\mathscr{S}\setminus\{T_0\}\big)\tag{1}$$ is a partition of $S$ into $\kappa$ stationary subsets, and if $D$ is stationary, $\{D\}\cup\mathscr{S}$ is a partition of $S$ into $\kappa$ stationary subsets. (Come to think of it, you could simply use $(1)$ in all cases, since $T_0\cup D$ will always be stationary and disjoint from each $T_\xi$ with $\xi>0$.)

Now a few comments on the argument as written:

$C_\xi\cap S_{\eta_\xi,\xi}=\varnothing$ doesn’t quite say that each element of $C_\xi$ has $\alpha_\xi<\eta_\xi$; it says that if $\alpha\in C_\xi\cap S$, then $\alpha_\xi<\eta_\xi$. The rest of that paragraph is stated rather unclearly. What you mean is that if $\alpha\in C\cap S$, then for each $\xi<\lambda$ we have $\alpha_\xi<\eta_\xi$, and therefore $$\alpha=\sup_{\xi<\lambda}\alpha_\xi\le\sup_{\xi<\lambda}\eta_\xi<\lambda\;.$$ Thus, the stationary set $C\cap S$ is bounded, which is absurd.

I assume that when you say that $f$ ‘looks down on’ each of the stationary sets $S_\eta$, you mean that it’s a pressing down (or regressive) function on each of them. Near the end I think that you meant to say that if $\gamma_\eta\ne\gamma_{\eta'}$, then $S_\eta'\cap S_{\eta'}'=\varnothing$.

  • $\begingroup$ Of course! Thanks $\endgroup$ – Forever Mozart Oct 12 '13 at 8:51
  • $\begingroup$ @Lusinghiero: You’re welcome. $\endgroup$ – Brian M. Scott Oct 12 '13 at 8:52

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