TITLE: A decreasing sequence of stationary sets.
QUESTION [4 upvotes]: 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!

REPLY [3 votes]: 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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