3
$\begingroup$

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 $\kappa$ .

$\endgroup$
  • 1
    $\begingroup$ Please do not (essentially) delete your questions like you did. $\endgroup$ – user98602 Apr 5 '14 at 5:45
0
$\begingroup$

$S$ 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 $S_\eta'\cap S_{\eta'}'=\varnothing$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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