# A decreasing sequence of stationary sets.

I'm having trouble with exercise 45 page 91 of Kunen's book:

Let $\kappa > \omega$ be regular. Show that there are stationary sets $S_\alpha \subset \kappa$ for $\alpha < \kappa$ such that $\alpha < \beta \rightarrow S_\beta \subset S_\alpha$ , and the diagonal intersection of the $S_\alpha$ is $\{0\}$.

I don't know how to begin with this construction, help me please!

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Note the hint to use Exercise 44, which reads as follows:

Let $\kappa>\omega$ be regular, and let $\mathscr{B}$ be the Boolean algebra $\mathscr{P}(\kappa)/\mathrm{Cub}^*(\kappa)$. If $A\subset\kappa$, $[A]\in\mathscr{B}$ is its equivalence class. Show that if $A_\alpha\subset\kappa$ for $\alpha<\kappa$, then the inf, $\bigwedge_{\alpha<\kappa}[A_\alpha]$ exists and equals $[D]$, where $D$ is the diagonal intersection, $$D=\{\alpha<\kappa:\forall\beta<\alpha(\alpha\in A_\beta)\}\;.$$

(Here $\mathrm{Cub}^*(\kappa)$ is the ideal of non-stationary subsets of $\kappa$.)

If you could find the sets $S_\alpha$ required for your problem, their equivalence classes would be a descending $\kappa$-chain in $\mathscr{B}$ with infimum $[\{0\}]=\mathrm{Cub}^*(\kappa)$, the zero element of $\mathscr{B}$. The presence of the hint suggests that you should look for such a descending chain.

Note that if $S$ and $T$ are disjoint stationary subsets of $\kappa$, $[S]<[S\cup T]$ in $\mathscr{B}$: $S\subseteq S\cup T$ implies that $[S]\le[S\cup T]$, and the symmetric difference $S\triangle(S\cup T)=T\notin\mathrm{Cub}^*$, so $[S]\ne[S\cup T]$. Thus, if we had a family $\{T_\alpha:\alpha<\kappa\}$ of $\kappa$ pairwise disjoint stationary subsets of $\kappa$, it would be trivial to build a strictly ascending $k$-chain of stationary subsets of $\kappa$ by setting $S_\alpha'=\bigcup_{\xi<\alpha}T_\xi$. Getting a strictly descending chain may take a little more thought, but it’s just as easy: just let $S_\alpha'=\bigcup_{\alpha\le\xi<\kappa}T_\xi$ for $\alpha<\kappa$. Corollary 6.12 ensures the existence of the stationary sets $T_\alpha$, so we have our sets $S_\alpha'$.

Clearly for $\alpha<\beta<\kappa$ we have $S_\alpha'\triangle S_\beta'\supseteq S_\alpha'\notin\mathrm{Cub}^*(\kappa)$ and hence $[S_\beta']<[S_\alpha']$, and we know from Exercise 44 that if $D$ is the diagonal intersection of the sets $S_\alpha'$, then $[D]=\bigwedge_{\alpha<\kappa}[S_\alpha']$. We’d like this to be the zero element of $\mathscr{B}$.

This is where it’s nice to be working in a complete Boolean algebra. We don’t have to worry about whether $\bigwedge_{\alpha<\kappa}[S_\alpha']$ actually is $0_{\mathscr{B}}$: we simply replace each $[S_\alpha']$ by $\lnot[D]\land[S_\alpha']$, since $$\bigwedge_{\alpha<\kappa}\Big(\lnot[D]\land[S_\alpha']\Big)=\lnot[D]\land\bigwedge_{\alpha<\kappa}[S_\alpha']=\lnot[D]\land[D]=0_{\mathscr{B}}\;.$$

Now $\lnot[D]\land[S_\alpha']=[(\kappa\setminus D)\cap S_\alpha']=[S_\alpha'\setminus D]$, so $S_\alpha'\setminus D$ is almost what we want for $S_\alpha$, so let’s set $S_{\alpha}=S_\alpha'\setminus D$ and see how close the family $\{S_\alpha:\alpha<\kappa\}$ comes to fulfulling our requirements.

We already know that its diagonal intersection is in $\mathrm{Cub}^*(\kappa)$, i.e., is non-stationary. In fact, its diagonal intersection is

$$\left\{\xi<\kappa:\forall\alpha<\xi\Big(\xi\in S_\alpha\Big)\right\}=\left\{\xi<\kappa:\forall\alpha<\xi\Big(\xi\in S_\alpha'\setminus D\Big)\right\}=\{0\}\;.$$

(The predicate $\forall\alpha<\xi\Big(\xi\in S_\alpha'\setminus D\Big)$ is vacuously true when $\xi=0$.)

It’s also clear that $S_\beta\subseteq S_\alpha$ whenever $\alpha<\beta<\kappa$. The real question is whether the $S_\alpha$ are still stationary.

Suppose that $[S_\alpha]$ is non-stationary. Then $0_{\mathscr{B}}=[S_\alpha]=\lnot[D]\land[S_\alpha']$, so $[S_\alpha']\le [D]\le[S_\alpha']$, and hence $[S_\alpha']=[D]$. If $\alpha<\beta<\kappa$, $S_\beta\subseteq S_\alpha$, so $S_\beta$ is also non-stationary, and therefore $[S_\beta']=[D]$ as well. In particular, $[S_\beta']=[S_\alpha']$, contradicting our earlier observation that $[S_\beta']<[S_\alpha']$, and it follows that the sets $S_\alpha$ must be stationary.

In other words, $\{S_\alpha:\alpha<\kappa\}$ is exactly what’s wanted.

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How do I know that $\neg[D] \wedge [S_{\alpha}\prime]=[(\kappa∖D)\bigcap S_{\alpha}\prime]$ ?, and it happens also with $[A]\wedge [B]=[A \bigcup B]$? – sanluc May 23 '12 at 23:48
@sanluc: (That should be $[A]\lor[B]=[A\cup B]$.) Because the map $\mathscr{P}(\kappa)\to\mathscr{B}:A\mapsto[A]$ is a quotient map. In the Boolean algebra $\mathscr{P}(\kappa)$, $\lnot D=\kappa\setminus D$, so $\lnot[D]=[\kappa\setminus D]$ in $\mathscr{B}$. – Brian M. Scott May 24 '12 at 3:07
yes, sorry I meant $[A] \vee [B]=[A \cup B]$. Thanks! – sanluc May 24 '12 at 22:36

Take the stationary set $X=\{\alpha\in\kappa : cf(\alpha)=\omega\}$ and for each element $\alpha$ of this set take a sequence $(\beta_{\alpha,n})_{n\in\omega}$ such that its limit is $\alpha$. Now, there is some $n_0\in\omega$ such that for all $\delta\in\kappa$ the sets $S_\delta=\{\alpha\in X : \beta_{\alpha,n_0}\geq\delta\}$ are stationary. To see this observe that if the contrary was true then for every $n\in\omega$ there is some $\delta_n$ and some club $C_n$ such that if $\alpha\in X\cap C_n$ then $\beta_{\alpha,n}<\delta_n$. Take then $\delta$ to be the limit of the $\delta_n$ and $C=\bigcap_{n\in\omega}C_n$. If $\alpha\in C\cap X$ then $\beta_{\alpha,n}<\delta$ for all $n$ and hence $\alpha<\delta$, which implies that $(C\setminus\delta)\cap X=\varnothing$ a contradiction.

Now I claim that these $S_\delta$ are the set we are looking for. Indeed, if $\alpha\in\triangle S_\delta$ this means that $\alpha\in S_\gamma$ for all $\gamma<\alpha$, i.e $\beta_{\alpha,n}\geq\gamma$ for all $\gamma<\alpha$. Hence $\beta_{\alpha,n}\geq\alpha$, a contraction, by the definition of the $\beta_{\alpha,n}$.

The argument in the first paragraph is used (at least in Jech's Set Theory) to show that the $X$ I defined can be written as a union of $\kappa$ disjoint stationary sets.

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I guess by $W$ you mean $X$. – azarel May 19 '12 at 1:53
@azarel Yes, thanks. – Apostolos May 19 '12 at 8:51