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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 preceeding 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.

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4 Answers

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 $\lambda$.

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But the image of $\kappa \to \lambda$ already has type; why does that not change after taking the closure? –  Martin Brandenburg Sep 3 '10 at 14:35
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By "close off" I mean add all sups of bounded subsets of the image. It is straight forward that this set is closed and unbounded in $\lambda$. Let $S$ be a proper initial segment of the image. This is of some size $\mu<\kappa$. How many proper initial segments $T$ of your initial segment $S$ are there? Well, $\mu$-many. You add a sup to every such $T$. Hence, below the sup of $S$ you only add $\mu<\kappa$ many new points. Hence, every proper initial subset of the club is of size $<\kappa$. Hence the club is of order type $\kappa$. –  Stefan Geschke Sep 3 '10 at 14:45
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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.

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You might want to add $0$ to all those sets, note that the intersection is not supposed to be empty but rather a singleton. But then of course you can add to each $S_\alpha$ anything you'd like to keep in the intersection :) –  Asaf Karagila Sep 3 '10 at 17:43
    
I don't understand; $0$ is always in the diagonal intersection. –  Martin Brandenburg Sep 3 '10 at 21:23
    
Yes you're right :). 0 is always in the diagonal intersection. But still 1 must not be an element of $S_0$. –  Stefan Hoffelner Sep 3 '10 at 22:12
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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}.

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Apparently the OP already knew (from an earlier exercise) that the Boolean algebra $\mathcal P(\kappa)/$club is $\kappa$-complete. So once you have an antichain $\{s_\alpha:\alpha<\kappa\}$ of size $\kappa$, your proof can be formulated succinctly by working entirely in this Boolean algebra. The (Boolean) joins of tails $t_\xi=\bigvee_{\alpha>\xi}s_\alpha$ form a strictly decreasing $\kappa$-sequence. If its infimum (which is known to exist) isn't 0, subtract it from all the $t_\xi$'s. –  Andreas Blass Feb 16 '12 at 17:25
    
Dear Andreas, if you are suggesting that the maximality of $\mathcal{S} $ is irrelevant, I agree! –  Isaac Gorelic Feb 16 '12 at 19:36
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Just a brief remark that the next to last solution is false.

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