Two questions about the boolean algebra $P(\kappa)/Cub^*$

Question

I have two questions based on exercises in Kunen's set theory. Let $\kappa = cf(\lambda) > \omega$. Why is there a c.u.b. $C \subseteq \lambda$ of order type $\kappa$? I thought we just choose $C$ as the image of an increasing unbounded function $\kappa \to \lambda$, but I doubt that this has to be closed.

Also, why can we use this $C$ to get an isomorphism of boolean algebras $P(\kappa)/Cub^*(\kappa) \cong P(\lambda)/Cub^*(\lambda)$? If we just pull back with $\kappa \to \lambda$, I don't see why this will be well-defined. Note that $Cub^*$ is the ideal of non-stationary subsets.

The second problem is the following: Let $\kappa$ be an uncountable regular cardinal. I want to prove that there is a decreasing sequence of stationary sets $S_\alpha, \alpha < \kappa$, whose diagonal intersection is $\{0\}$. This is an exercise in Kunen's set theory, and there is a hint that one should use the preceding exercise, which says that the boolean algebra $B=P(\kappa)/Cub^*(\kappa)$ has infima indexed over $\kappa$, which correspond to the diagonal intersection in $P(\kappa)$.

Here's what I've done so far: Construct a decreasing sequence in $B$: Let $x_0=1$. If $x_\alpha$ is already defined, define $x_{\alpha+1} = x_\alpha$ if $x_\alpha$ is minimal and otherwise choose some $x_{\alpha+1} < x_\alpha$. If $\alpha$ is a limit and $x_\gamma$ is defined for all $\gamma < \alpha$, let $x_\alpha$ be the infimum of these $x_\gamma$.

Now if $x_\alpha = [S_\alpha]$, then $S_\alpha$ is stationary iff $x_\alpha \neq 0$; is this the case? For $\alpha < \beta < \kappa$, we have $x_\beta \leq x_\alpha$, i.e. there is a c.u.b. $C_{\alpha,\beta}$ such that $S_\beta \cap C_{\alpha,\beta} \subseteq S_\alpha$. Now perhaps there is some double-index diagonal intersection $C$ of these $C_{\alpha,\beta}$ (?) which is c.u.b. again and which we may intersect with every $S_\alpha$, so that we may assume $S_\beta \subseteq S_\alpha$, as desired.

Finally we have to ensure the infimum of the $x_\alpha$ is $0$. I wonder if this is true at all with this naive construction.

Answer 1

First question: Take the image of your increasing function from $\kappa$ to $\lambda$ and close it off. This will be of order type $\kappa$.

For the second part of the question, indeed, pullback along the embedding $e$ from $\kappa$ to $\lambda$ (with club range). Why does this give an isomorphism of the quotient algebras? Because the complement of the range of the embedding is in the nonstationary ideal on $\lambda$.

Answer 2

Second question: You pick a stationary $S \subset \kappa$. Define a sequence recursively like this:

$S_0 := S$, $S_{\alpha +1}:= S_{\alpha} - A_{\alpha}$ where $A_{\alpha}$ is defined as the set of the first $\alpha +1$ many elements of $S_{\alpha}$ (Note here that $S_{\alpha +1}$ remains stationary). And finally $S_{\gamma} := \bigcap_{\beta < \gamma} S_{\beta}$ for each limit $\gamma < \kappa$.

This gives you the desired sequence.

Answer 3

Reluctantly (not wanting to be more explicit than the book is) I decided to sketch some proof. 1st, get a maximal almost disjoint (modulo non-stationary) family of $\kappa$ stationary subsets of $\kappa$. (Easy to get: take 1st a disjoint partition of $\kappa$ into $\kappa$ stationary sets. – May need to add the transversal set of the minimal ordinals from this family, if this set happens to be stationary). Let $\mathcal{S} = { S_\alpha : \alpha < \kappa }$ be such a family.

Then define $\mathcal{T} = \{ T_\delta : \delta < \kappa \}$ by $T_\delta = \kappa \setminus \bigcup \{S_\alpha : \alpha < \delta \}$, and let $T = \triangle \{ T_\delta : \delta < \kappa \}$.

Since for every $ \delta$ $T \setminus T_\delta \subseteq \delta + 1$, $T$ is almost disjoint from all $S_\alpha$’s. Therefore, by maximality of $\mathcal{S}$, $T$ is non-stationary.

Finally, let $R_\delta = T_\delta \setminus T$. Then $\mathcal{R} = \{R_\delta : \delta < \kappa \}$ is as required, i.e. $\triangle \mathcal{R} =$ {0}.

Answer 4

Just a brief remark that the next to last solution is false.