数理解析研究所講究録1095 巻 1999 年 28-41

28Splitting$P_{\kappa}\lambda$

into maximally many stationary sets
MASAHIRO SHIOYAA BSTRACT. Let$P_{\hslash}\lambda$

splits into

be a regular cardinal andstationary sets.

$\kappa>\omega$$\lambda^{\omega}$

$0$

Let

for

$\lambda>\kappa$

a cardinal. We show that

. INTRODUCTION

be a regular cardinal and$\lambda>\kappa$

a cardinal. Solovay’s classical result[So] led Menas [Me] to conjecture that a stationary subset of$\kappa$

into$\kappa>\omega$$r \overline{\dot{\mathbb{R}}}\iota\wedge 7-,\frac{\mathrm{H}}{\backslash }\overline{2}(\sim\backslash \backslash$$\lambda^{<\kappa}$$P_{\kappa}\kappa^{+}$

stationary sets. Unfortunately his conjecture fails when

carries a stationary set of size

size is$(\kappa^{+})^{<\kappa}$$\kappa^{+}$$P_{\kappa}\lambda$

would split

While$2^{<\kappa}>\kappa^{+}:$(see [BT]), the conjecture implies that theas well.

What about splitting a stationary set$S$into $\min$ { $|S\cap C|$ : $C$ is club} many

sets? Gitik’s answer [G] was again negative: Relative to supercompactness, it isconsistent that some stationary subset of$P_{\kappa}\kappa^{+}$

splits into at most$\kappa$

stationarysets.
Now it seems natural to ask the same question as above for a canonical stationary

set. Let us concentrate on the case where the canonical set is$\kappa=\omega_{1}$case$P_{\kappa}\lambda$

itself. When

, we have a satisfactory answer by the works of Baumgartner-Taylor [BT] (the$\lambda\leq 2^{\omega}$) and Donder-Matet [DM] (otherwise):$P_{\omega_{1}}\lambda$

sets. In fact the latter proved the diamond principle forsplits into

$P_{\kappa}\lambda$

when$\lambda^{\omega}$stationary$\lambda>2^{<\hslash}$.

Part of this work was done during the author’s stay at Boston University as one of the JapaneseOverseas Research Fellows. He gratefully acknowledges Professor Akihiro Kanamori’s hospitality.
He also wishes to thank members of the set theory seminar at Waseda University for their interestat the early stage.
Typeset by$A_{\mathcal{M}}s_{-}\mathrm{I}\mathrm{F}\mathrm{x}$29
In this paper we are mainly concerned with the general type of result as follows(see [Ka]):$P_{\kappa}\lambda$

splits into$\lambda$

stationary sets. As suggested above, we should firstmeasure the minimum size of a club subset of$\mathcal{P}_{\kappa}\lambda$. Elaborating his earlier result[BT], Baumgartner [B] has already shown that it is at least

following result of Magidor [Mag] imply thatspecific splitting problem: If there is nomodel,$P_{\kappa}\lambda$

carries a club set of size$\lambda^{\omega}$$\omega_{1^{-}}\mathrm{E}\mathrm{r}\mathrm{d}_{\acute{\acute{\mathrm{O}}}}\mathrm{s}$$\lambda^{\omega}$

when cf$\lambda^{\omega}$. This and theis the critical number for our

cardinal in the Dodd-Jensen core$\lambda\geq\kappa$, and of size$\max\{\lambda^{\omega}, \lambda^{+}\}$

otherwise.
Unifying three of the results above, we establish the desired splitting:
Theorem 1.

$P_{\kappa}\lambda$

splits into$\lambda^{\omega}stati_{ona}w$

sets.

We also reahize the splitting suggested in the latter case of Magidor’s theorem:
Theorem 2.

$P_{\kappa}\lambda$

splits into

stationary sets when$\lambda^{+}$$\mathrm{c}\mathrm{f}\lambda<\kappa$.

1. PRELIMINARIESOur notation should be standard. Kanamori’s book [Ka] is an excellent sourcefor background material. We reserve$>\kappa$and$\mu,$$\nu$

for a cardinal$\geq\omega$

denotes the set of limit ordinals$\kappa$

for a regular cardinal. When$<\kappa$$\mu<\kappa$of cofinality

is regular,$\mu$$>\omega,$$S_{\kappa}^{\mu}$(resp. $<\mu,$$\lambda$

for a cardinal(resp.
$\geq\mu$$S_{\kappa}^{<\mu},$$S_{\kappa}^{\geq\mu}$)). For a set $X$of ordinals let $\lim X$ be the set $\{\gamma<\sup X : \sup(X\cap\gamma)=\gamma>0\}$ of limit points
of $X$ and cl$\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\cup f$$f^{X}$

the closure of $X$ under“ $\mathrm{Y}^{<\omega}\subset \mathrm{Y}$$f$:$\lambda^{<\omega}arrow P_{\kappa}\lambda,$$\mathrm{i}.\mathrm{e}$. the minimal set$\mathrm{Y}\supset X$. Unless otherwise stated, we understand that a set of ordinals

is listed in increasing order and a splitting of a stationary set is mutually disjoint.
Thoughout the paper we freely use Solovay’s theorem [So] mentioned earlier:

30

Theorem. A$stati_{\mathit{0}}naw$

subset ofsplits into$\kappa$$\kappa$

stationary sets.

We need of a version of Shelah’s club guessing sequence (see [Ko]). Let us sketcha proof due to Hirata [H]:
Theorem. Let $\mu<\kappa<\lambda$ be all regular andthere is a sequence
type

$\mu$

for any$\gamma\in S$

Proof. First for$\gamma\in S$$\langle c_{\gamma} :such thatis$D \subset\lim\lambda$

set. Note that$x_{\gamma}^{D}\subset D$

induction on $n<\omega$ . First we find a club set
stationary for any club set$C\subset\lambda$$d_{\beta}\subset\beta$.and, where$\lim\lambda$$\gamma\in S\cap \mathrm{n}_{\xi}<\kappa C\xi$

for any

such thatsuch that$|x_{\gamma}^{C_{\xi n}},n|<\kappa$$\beta\in x_{\gamma,n}^{D}\mathrm{n}S\lambda<\kappa$$x_{\gamma,0}^{D}$

impliesis defined:$\exists\beta\in$$|x_{\gamma,n}^{D}|<\kappa$$\langle C_{\xi} :\xi<\kappa\rangle$

for any

such thatbyisof club$\xi<\kappa$. Fix$x_{\gamma^{\epsilon_{n}}\gamma,n}^{cC_{\xi_{n}}},=X$

is decreasing for any$\alpha<\lambda$

and$x_{\gamma^{\xi}}^{C}=x_{\gamma}^{C_{\xi+1}}\subset C_{\xi+1}$$C_{\xi+1}\cap\{\gamma\in S:x_{\gamma}^{c}\xi\subset C_{\xi+1}\}=\emptyset$.

as above. Then $S^{*}= \{\gamma\in S\cap\lim D : x_{\gamma}^{D}\subset\lim D\}$is stationary by the claim above. Fix$\gamma$n<\omega\}\subset\kappa$$\xi \text{ト}arrow\sup(C_{\xi}\cap\alpha)$by the note above. This contradicts

in , since$\{\xi_{n} :by the note above. Set $\xi=\sup_{n<\omega}\xi_{n}<\kappa$ . Then

$D\subset\lambda$. For$\{\gamma\in S:x_{\gamma}^{D}\subset C\}$

$C_{\xi+1}\cap\{\gamma\in S : x_{\gamma^{\xi}}^{C}\subset C_{\xi+1}\}=\emptyset$, since the map

Now fix a club set$\beta$.

.

. Then we have inductively$\xi_{n}\leq\xi<\kappa$$x_{\gamma,n}^{D}$$x_{\gamma,n+1}^{D}= \{\sup(D\cap\alpha)$

Otherwise we would have inductively a descending sequencesubsets of$C\subset\lambda$

of order type cf

since $D$ is closed, and$D\subset\lambda$

and of order$\gamma$for any club set$x_{\gamma}^{D}= \bigcup_{n<\omega}x_{\gamma},-Dn\{0\}$\alpha\in d_{\gamma}\}$stationary. Thenis unbounded in$stati_{\mathit{0}}naw$

fix an unbounded set$x_{\gamma,0}^{D}= \{\sup(D\cap\alpha) :$x_{\gamma,n}^{D}\cap S_{\lambda}^{<\kappa}(\alpha\in d_{\beta})\}$$c_{\gamma}\subset S_{\lambda}^{\geq\kappa}$$\{\gamma\in S:c_{\gamma}\subset C\}$$\beta\in\lim\lambda$

and a club set

inductively by

and\gamma\in S\rangle$$S \subset S_{\lambda}^{\mu}\cap\lim S_{\lambda}^{\geq\kappa}$$\beta\in\lim x_{\gamma,n+1}^{D}$is unbounded in$\gamma$$\gamma\in S^{*}$. We have

by $\beta\in\lim D$ . Also$x_{\gamma\gamma}^{D}- \lim x^{D}\subset S_{\lambda}^{\geq\kappa}$$x_{\gamma}^{D}- \lim x^{D}\gamma$, sinceis unbounded

by $\gamma\in\lim D$ .

Finally we get the desired sequence by taking an unbounded subset of$x_{\gamma\gamma}^{D}- \lim x^{D}$31

of order type

$\mu$

as

$c_{\gamma}$

for$\gamma\in S^{*}$.

$\square$

In fact we use only the sequence of the form
to the clause$c_{\gamma}\subset S_{\lambda}^{\geq\kappa}$$\langle c_{\gamma} :\gamma\in S_{\lambda}^{\omega}\rangle$

and do not appeal. The second result we quote from Shelah’s pcf theory is ascale on a singu.lar cardinal [Sh] (see also [BMag]):

Theorem. Let$\lambda$

be singular. Then there are an unbounded setof regular cardinals and$\beta<\gamma<\lambda^{+}$

Hereand for any$\leq*\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{S}$$\{f_{\gamma} :\gamma<\lambda^{+}\}\subset\prod_{\xi<\mathrm{C}}\mathrm{f}\lambda\lambda_{\xi}$$g \in\prod_{\xi<\mathrm{C}}\mathrm{f}\lambda\lambda_{\xi}$there issuch that$\gamma<\lambda^{+}$

the eventual dominance: $f\leq^{*}g$ iff

with$\{\lambda_{\xi} :\xi<\mathrm{c}\mathrm{f}\lambda\}\subset\lambda$$f_{\beta}\leq*f_{\gamma}$$g\leq*f_{\gamma}$

for any.

$\{\xi<\mathrm{c}\mathrm{f}\lambda.

f(\xi)\leq g(\xi)\}$is

cobounded. The later development of the theory as presented in [Ko] yields a moretransparent proof of this deep result.

2. MAIN THEOREMSThis section is devoted to establishing Theorems 1 and 2.

Our proof of Theorem 1 consists of two major parts. For the first part we arestrongly indebted to$\mathrm{T}\mathrm{o}\mathrm{d}\mathrm{o}\mathrm{r}\check{\mathrm{c}}\mathrm{e}\mathrm{v}\mathrm{i}\acute{\mathrm{c}}$[T2], who reproved Gitik’s answer [G] to Abraham’squestion [AS] and claimed that his method would yield the Baumgartner-Taylor

result as wel via the following: Letwith $c_{\gamma}=\{\gamma_{n} : n<\omega\}$ . Then {$\lambda$

be regular. We endow

initial segment of$b$}. Let\gamma\in S_{\omega_{2}}^{\omega}\rangle$$x\in P_{\omega_{1}}\omega_{2}$:is stationary for any$x\cap(\gamma_{n+1}-\gamma_{n})\neq\emptyset\}=r)\}$Let$\langle c_{\gamma} :$T$$[\lambda]^{<\omega}$be a club guessing sequence$\exists\gamma\in S_{\omega_{2}}^{\omega}(\sup x=\gamma$$r\in[\omega]^{\omega}$A { $n<\omega$ :.

with the tree ordering $\leq=\{(a, b)$ :be a subtree of$[\lambda]^{<\omega}$, i.e. a subset of$[\lambda]^{<\omega}$

$a$

is an

closedunder initial segments. Set $[T]=\{B\in[\lambda]^{\omega} : \forall\beta\in B(\dot{B}\cap\beta\in T)\}$ , the set of

infinite branches through$T$, and $T^{a}=\{b\in[\lambda]^{<\omega} : a\leq a\cup b\in T\}$ , the tree above

32$a\in[\lambda]^{<\omega}$suc. We call$\tau(a)=\{\alpha<\lambda :$T\neq\emptyset$

stationary if the set of immediate successors ofa\leq a\cup\{\alpha\}\in T\}$$a$$\in T$is always stationary, and $g:Tarrow\lambda$ regressive

when $g(a) \leq g(b)\in\min b\cup\{0\}$ for any $a\leq b\in T$ .
Let us start with a tree version of the regressive function lemma:
Lemma. Let $g:Tarrow\lambda$ be regressive withfor some$stati_{\mathit{0}}naw$

Proof. Forof$g$$\gamma<\lambda$

subtreeof $Tg$$\tau*$

with$\gamma<\lambda$. Take inductively
with$B\in[C_{\alpha}]^{\omega}$$C_{\alpha}$

bysubtree of$stati_{\mathit{0}}naw$$B\in[T_{\alpha}]$$[S_{\lambda}^{\kappa}]^{<\omega}$. Thenis bounded in .

“ $T^{*}$$\lambda$$[T_{\gamma}]\cap[C]^{\omega}\neq\emptyset$

Suppose to the contrary that for$\alpha<\min B$a

set $T_{\gamma}=\{a\in T:g(a)<\gamma\}$ , a subtree of. First we find$[C_{\gamma}]^{\omega J}=\emptyset$$T$$\gamma<\lambda$$T$by order preservation

for any club set

we have a club set$B\in[T]\cap[\triangle_{\gamma<\lambda}C_{\gamma}]^{\omega}$$C\subseteq\lambda$.

$C_{\gamma}\subset\lambda$

with$[T_{\gamma}]\cap$by stationarity of $T$ . Takeby cf $\min B=\kappa>\omega$ and regressiveness of . Then$g$

$B\in[\Delta_{\gamma<\lambda}C_{\gamma}]^{\omega}$. This contradicts$[T_{\alpha}]\cap[C_{\alpha}]^{\omega}=\emptyset$

by the choice of.
Fix$\gamma<\lambda$

as above. Set $T^{*}=$ {a subtree of . Note that$T$$\emptyset\in\tau*$$a\in T_{\gamma}$:$\forall b\leq a\forall C\subset\lambda$

by the choice of$\gamma$club$([T_{\gamma}b]\cap[C]^{\omega}\neq\emptyset)$. We claim that$T^{*}$},is stationary

as desired.
Suppose to the contrary$D\subset\lambda$$a\in\tau*$. Then forand$\alpha\in D$$a\cup\{\alpha\}\not\in\tau*$

by $a\in\tau*$ .

$D\cap \mathrm{s}\mathrm{u}\mathrm{c}\tau*(a)=\emptyset$

we have a club set. Thusand the choice ofby$C_{\beta}$$B\in[C]^{\omega}$.

$C_{\alpha}\subset\lambda$$C=D\cap\Delta_{\alpha\in}DC_{\alpha}$

Set $\beta=\min B$ . Then

$B-\{\beta\}\in[C_{\beta}]^{\omega}$

for some $a\in T^{*}$ and some club set
with$[T_{\gamma}^{a\cup\{\alpha\}}]\cap[C_{\alpha}]^{\omega}=\emptyset$

is club in$\lambda$. Take$B-\{\beta\}\in[T_{\gamma}^{a\cup\{\beta}\}]$. This contradictsbyby$B\in[T_{\gamma}^{a}]\cap[C]^{\omega}$$B\in[T_{\gamma}^{a}]$$[T_{\gamma}^{a\cup\{\beta}\}]\cap[C_{\beta}]^{\omega}=\emptyset$, andby $\beta\in D$$\square$

For the following lemma we fix a club guessing sequence$\langle c_{\gamma} :\gamma\in S_{\lambda}^{\omega}\rangle$

with

33

$c_{\gamma}=\{\gamma_{n} :n<\omega\}$.

Main Lemma 1. Let$S_{\lambda}^{\omega}$$S_{n}\subset S_{\lambda}^{\kappa}$

be stationary for $n<\omega$ . Then {:element of$a$$\exists\gamma\in$( $\sup x=\gamma$ A $\forall n<\omega(\min(x-\gamma n)\in S_{n})$ ) is stationary.
$\}$

Proof. Fix$S_{n})\}$:$f$$\lambda^{<\omega}arrow P_{\kappa}\lambda$. Set $T=\{a$ :, a stationary subtree ofof$T_{n}$

and$T$$h_{n}$:$[S_{\lambda}^{\kappa}]^{<\omega}$$T_{n}\cap[\lambda]^{n}arrow\lambda$$g_{a}$:$T_{n+1}\subset T_{n},$$a\in T_{n+1}\cap.[\lambda]^{n}$First set $T_{0}=T$ . Next suppose thatthe map$\forall n<|a|(\mathrm{t}\mathrm{h}\mathrm{e}n\mathrm{t}\mathrm{h}$$T_{n}$$T_{a}\subset h_{n}(a)$. Then$T_{n+1}=(T_{n}\cap[\lambda]^{<n})\cup$

the desired stationary subtree of$T_{n}$

: Fix

and

$B\in[T_{n_{\Gamma}^{\mathrm{I}}}- 1]a$is regressive on$b \mapsto\sup(\mathrm{c}1_{f(}a\mathrm{U}b)\cap\min b)${

and$T_{n+1}\cap[\lambda]^{n}=T_{n}\cap[\lambda]^{n}$is defined. Fix

By the lemma above we have a stationary subtree
“

is in

. We build inductively a stationary subtree

so thatcl $f(a \cup B)\cap\min B\subset h_{n}(a)$ for any$g_{a}$$x\in P_{\kappa}\lambda$

of$T_{a}$$a\cup b$:.

$a\in T_{n}\cap[\lambda]^{n}$$T_{n}a$$T_{n}^{a}$by cf $\min b=\kappa$ .

and $h_{n}(a)<\lambda$ with

$a\in T_{n}\cap[\lambda]^{n}$$a\in T_{n+1}\cap[\lambda]^{n}$. ThenA $b\in T_{a}$ } isand $B\in[T_{n+1^{a}}]$ . Thencl $f(a \cup B)\cap\min B=\bigcup_{\beta\in B}$ cl$f(a \cup(B\cap\beta))\cap\min B\subset\bigcup_{\beta\in B}g_{a}(B\cap\beta)\subset h_{n}(a)$ .

Now setThen $C=$ {$T^{*}= \bigcap_{n<\omega}\tau_{n}$$\gamma<\lambda$

a club set. Fix$n<\omega\}\in[T^{*}]$$\{\beta_{i} :: cl, a stationary subtree of , and$T$:$h= \bigcup_{n<\omega}h_{n}$$f^{\gamma}= \gamma\wedge\forall a\in T^{*}\cap[\gamma]^{<}\omega(h(a)<\gamma\wedge\gamma\in\lim \mathrm{s}\mathrm{u}\mathrm{c}\tau*(a))$$\gamma\in S_{\lambda}^{\omega}\cap C$

so thatwith$\gamma_{n}<\beta_{n}<\gamma_{n+1}$

i<n\}\in T^{*}\cap[\gamma_{n}]^{<\omega}$since $\sup B=\gamma$ and cl$c_{\gamma}=\{\gamma_{n} :. Then cl$f^{B}$$f^{B}\subset \mathrm{c}1_{f^{\gamma}}=\gamma$by. Take inductivelyn<\omega\}\subset C$$\gamma_{n+1}\in C$$T^{*}arrow\lambda$.

} contains$B=\{\beta_{n}$:and the inductive hypothesis

is as desired: First we have $\sup$ cl $f^{B}=\gamma$ ,by$\gamma\in C$

cl $f^{B\cap\beta_{n}}\subset h_{n}(B\cap\beta n)=h(B\cap\beta n)<\gamma_{n}$ by. Next$\gamma_{n}\in C$$\min(\mathrm{c}1_{f^{B-}\gamma_{n})}=\beta_{n}$and, since$B\cap\beta_{n}\in\tau*\cap[\gamma_{n}]^{<\omega}$.

$\square$

The following lemma is due to Foreman-Magidor [FM], who introduce the notionof mutual stationarity and show that the club filter onwhen$\lambda$

is singular.

$P_{\omega_{1}}\lambda$

is not$\lambda^{\mathrm{c}\mathrm{f}\lambda}$-saturated

34

Let

$\mathrm{c}\mathrm{f}\lambda=\omega$

cardinak$>\kappa$

and$\{\lambda_{n} :such that

Let $W$ be the tree$\omega(\sup(x\cap\lambda_{n})\in S_{n})\}$

Proof. Fix:$f$$T_{n+1}\subset T_{n},$is($\mathrm{c}1_{f}$

ran

be

$stati_{\mathit{0}}naw$

$C=$

{

$\gamma<\lambda_{n}$:$\forall U$$\gamma$(resp.

$T$

of $W$ set
$T$

,

for

$n<\omega$. Then {$x\in P_{\kappa}\lambda$:$\forall n<$.
of $W$ so that

$T_{n}$) $\in S_{n-1}$ for any $B\in[T_{n}]$ and for any $s\in T_{n}$, and is unbounded in$\kappa_{|s|}<\lambda_{n}$of $W$ cobounded below

and is cobounded in

is infinite for any $n<\omega$ .

, the set of infinite branches through$stati_{\mathit{0}}naw$

First set $T_{0}=W$ . Next suppose that
$U\neq\emptyset$\kappa_{i}=\lambda_{n}\}$. We build inductively a subtree$B\cap\lambda_{n-1}$

is a singleton if$\mathrm{s}\mathrm{u}\mathrm{c}\tau_{n}(s)$an unbounded set of regular}, the set of immediate successors of $s\in T$ .

$S_{n}\subset S_{\lambda_{n}}^{\omega}$$\lambda^{<\omega}arrow P_{\kappa}\lambda$$\sup$$\{i<\omega :\forall m<\omega(B|m\in T)\}$$s*\langle\alpha\rangle\in T$

Main Lemma 2. Let

i<\omega\}\subset\lambda$

ordered by inclusion. For a subtree$\bigcup_{m<\omega}\prod_{i<m}\kappa i${a :$\mathrm{s}\mathrm{u}\mathrm{c}\tau(S)=$and$\lambda_{n}.<\lambda_{n+1}$$[T]= \{B\in\prod_{i<\omega}\kappa_{i} :andn<\omega\}=\{\kappa_{i} :$\gamma$is defined. For$\gamma<\lambda_{n}$if for any $s\in U$ suc $u(s)$ is) if$\kappa_{|s|}$$T_{n}$

otherwise.

$\kappa_{|s|}$cobounded below$\kappa_{|s|}=\lambda_{n}$(resp.

$\gamma\exists B\in[T_{n}]\cap[U]$($\kappa_{|s|}$$\kappa_{|s|}>\lambda_{n}$$\mathrm{c}1_{f}$ranwe call a subtree
if$\kappa_{|s|}<\lambda_{n}$,). We claim that

$B\cap\lambda_{n}\subset\gamma$)} contains

a club set.
Suppose to the contrary that we have a stationary set

subtree$U_{\gamma}$

of $W$ cobounded below

Build inductively a subtree$\{\alpha\}$with$s*\langle\alpha\rangle\in\cap\{U_{\gamma} :is injective on$T_{n}$

s\in U_{\gamma}\}$($\mathrm{c}1_{f}$ran$\mathrm{s}\mathrm{u}\mathrm{c}\tau(S)\cap$with cl ran$f$so thatsuc$B\cap\lambda_{n}\not\subset\gamma$$\mathrm{s}\mathrm{u}\mathrm{c}\tau(S)$is. HenceA$\mathrm{s}\mathrm{u}\mathrm{c}_{T_{n}}(s)$if$B\in[T_{n}]\cap[U_{\gamma}]$$\kappa_{|e|}\leq\lambda_{n}$

)

$)\}$

, and is

A

$s$

“{ :
$i$

contains a club set. Fix

as follows: Suppose that, since.

$s\mapsto s|\{i:\kappa_{i}=\lambda_{n}\}$.

$\gamma\in\lim_{\mathrm{S}\mathrm{u}\mathrm{c}\tau}(S)$a$\gamma\in S$$D=\{\gamma<\lambda_{n^{*}}$ $\forall s\in T((\kappa_{1s}|=\lambda_{n}$$B\in[T]\cap[U_{\gamma}]$$\sigma_{\gamma}(s)\neq\emptyset$

and forfor any

otherwise. Note that the map$s\cap\lambda_{n}\subset\gamma$. Take inductively

defined. Thenof$\{s\in T:\kappa_{|s|}=\lambda_{n}\}$$\kappa_{i}=\lambda_{n}\}\subset\gamma)\Rightarrow$$\gamma\in S\cap D$$T$$\gamma$$S\subset\lambda$$\mathrm{s}\mathrm{u}\mathrm{c}_{U_{\gamma}}(s)=\kappa_{|s|}$when$s\in T\cap U_{\gamma}$$\kappa_{|s|}<\lambda_{n}$is, since35$\mathrm{s}\mathrm{u}\mathrm{c}_{U\gamma}(S)$$s$“$\{i :

is cobounded inwhen\kappa_{i}=\lambda_{n}\}\subset\gamma$$\kappa_{|s|}>\lambda_{n}$$B|i\in T$. Thus cl ran
$f$

and

$\gamma$is unbounded in$\mathrm{s}\mathrm{u}\mathrm{c}_{T}(S)$$\kappa_{|s|}=\lambda_{n}$, and by. This contradicts cl ran$f$$B\cap\lambda_{n}\not\subset\gamma$

Fix $\gamma\in S_{n}\cap C$ . Set $\tau*=\{s\in T_{n}$ :$[T_{n}]\cap[U]$($t\subset B$

A cl ran$f$$B\cap\lambda_{n}\subset\gamma$

Fix $s\in\tau*$ . We claim thatin$\gamma$case(resp.

$\kappa_{|s|}$$\kappa_{|s|}>\lambda_{n}$) if(resp.

(resp.

$\kappa_{|s|}<\lambda_{n}$for any$A=\gamma-\mathrm{s}\mathrm{u}\mathrm{C}_{T^{*}}(s)$$s*\langle\alpha\rangle\subset B\in[T_{n}]\cap[U_{\alpha}]$$\gamma$$\gamma$so that$\mathrm{s}\mathrm{u}\mathrm{c}_{Tn+1}(s)$

is

$U$

and the choice ofNow fix an unbounded set$T^{*}$$B\cap\lambda_{n}\subset\gamma$

by the minimal choice of$s*\langle\alpha\rangle\subset B\in[T_{n}]\cap[U_{\alpha}]$of$\kappa_{|s|}<\lambda_{n}$$\gamma$cl ran$f$$\sup$$B \cap\lambda_{n}=\bigcup_{i<\omega}$

Finally$n<\omega$. Then$\bigcap_{n<\omega}T_{n}$as desired.

(cl$\gamma\in C$$\kappa_{|s|}=\lambda_{n}$$f$$s*\langle\alpha\rangle\not\in T^{*}$

.

. The

$\alpha\in A$

we$B\cap\lambda_{n}\not\subset\gamma$. Fix a subtree$\{\gamma_{i} :. This contradicts cl ran$f$$U_{\alpha}$

i<\omega\}\subset\gamma$$\mathrm{s}\mathrm{u}\mathrm{c}\tau*(s)$ifby $s\in T^{*}$ , and then$U$$\mathrm{c}1_{f}$ran$B\cap\lambda_{n}$$f^{Bi}"\cap\lambda_{n}\subset\gamma$

)

with

$B\cap\lambda_{n}\not\subset\gamma$. Build inductively a subtree$\kappa_{|s|}\neq\lambda_{n}$$=\gamma$$\alpha\in A$.

, and is$\{\alpha\}$$T_{n+1}$$T_{n+1}$

with $\gamma_{m}<\alpha<$
is as desired: Fix, since $\sup\{B(i) : \kappa_{i}=\lambda_{n}\}=\gamma$ andby $B|i\in T^{*}$ .

has a unique branch $B$ and $\sup$ ($\square$by$\gamma\exists B\in$, and is unboundedsuch that cl ran

otherwise, where $m=|\{i<|s| : \kappa_{i}=\lambda_{n}\}|$ . Then

$B\in[T_{n+1}]$$\emptyset\in\tau*$.

with $\{t\in U : s<t\}=\bigcup_{\alpha\in A}\{t\in U_{\alpha} : s*\langle\alpha\rangle\leq t\}$.
$f$by. Note thatis cobounded. Then forby $s\in T^{*}$ andTake $s\subset B\in[T_{n}]\cap[U]$ with cl ran$s*\langle\alpha\rangle\subset B\in[U_{\alpha}]$$U_{\gamma}$

cobounded below). We show the caseof $W$ cobounded below$U_{\alpha}\ni s*\langle\alpha\rangle$

of $W$ cobounded below$T_{n}$

is a singleton if$\kappa_{|s|}>\lambda_{n}$by $\gamma\in D$ and) is given by a similar (resp. simpler) argument.

Suppose to the contrary thathave a subtree

$\}$

when

$\mathrm{s}\mathrm{u}\mathrm{c}\tau(S)$and the choice of$\forall \mathrm{t}\leq s\forall U\ni t$

and$s\in T$$\gamma\in D,$\kappa_{i}=\lambda_{n}\}\subset\gamma$$\gamma\in S$) , a subtree of$\mathrm{s}\mathrm{u}\mathrm{c}\tau*(s)$$\kappa_{|s|}=\lambda_{n}$bybyand the choice of$s\in U_{\gamma}$$B\cap\lambda_{n}=\cup\{\mathrm{c}1_{f}B" i\cap\lambda_{n} :$\gamma$$t$$\mathrm{c}1_{f}$ran$B\cap\lambda_{n}$)$\in S_{n}$for any36

We are ready to prove the main result of this paper:
Theorem 1.

$P_{\kappa}\lambda$

Proof. Whensequence$\{S_{\xi} :with\gamma\in S_{\mu}^{\omega}\rangle$. Then for$p$$\omega(\min(x-\gamma_{n})\in s_{\mathrm{p}(n})))\}$

When cfThen$\lambda=\omega$:$c_{\gamma}=\{\gamma_{n} :sets.

n<\omega\}$:$\omegaarrow\mu\{x\in P_{\kappa}\lambda$$\kappa<\mu\leq\lambda$

and split, fix a club guessing$S_{\mu}^{\kappa}$

into stationary sets$\exists\gamma\in S_{\mu}^{\omega}(\sup(x\cap\mu)=\gamma$A$\forall n<$

is stationary by Main Lemma 1 and mutually disjoint.

, fix an unbounded set$| \prod_{n<\omega}\lambda_{n}|=\lambda^{\omega}$

Then for$\lambda^{\omega}stati_{\mathit{0}}naw$

for some regular cardinal$\lambda\leq\mu^{\omega}$$\langle c_{\gamma} :\xi<\mu\}$

splits into$\{\lambda_{n} :. For $n<\omega$ split$p \in\prod_{n<\omega}\lambda_{n}\{x\in P_{\kappa}\lambda :$S_{\lambda_{n}}^{\omega}$n<\omega\}\subset\lambda$

of regular cardinals

into stationary sets$\{S_{n\xi} :\forall n<\omega(\sup(x\cap\lambda_{n})\in S_{np(n)})\}$$>\kappa$\xi<\lambda_{n}\}$.
.

is stationary byMain Lemma 2 and mutually disjoint.

Otherwise we have$\omega<\mathrm{c}\mathrm{f}\lambda<\lambda$

and$\alpha^{\omega}<\lambda$

for any$\alpha<\lambda$, and hence$\lambda^{\omega}=\lambda$.

For completeness we provide a proof implicit in [T1]. First we claim that $\{x\in$$P_{\kappa}\lambda$: $\sup(x\cap\mu)\in S$ A $\sup(x\cap\nu)\in S’\}$ is stationary for any regular cardinals$\kappa\leq\mu<\nu<\lambda$$\beta\in S’$

clwith cland stationary sets$f^{\beta\cap\nu}=\beta$$f(\alpha\cup b)\cap\mu=\alpha$and, and an unbounded set

into stationary sets$\{S_{\xi} :and, and an unbounded set$\sup(\mathrm{c}1_{f(a}\cup b)\cap\nu)=\beta$$\subset\alpha$$b\subset\beta$

. Fix

$f$

:

$\lambda^{<\omega}arrow P_{\kappa}\lambda$

of size , and then$\omega$of size . Then$\omega${. Also fix an unbounded set$\xi<\mathrm{c}\mathrm{f}\lambda$

split$x\in P_{\kappa}\lambda$$S_{\lambda_{\xi}}^{\omega}$$<\lambda$$\{\lambda_{\xi} :into stationary sets

: $\sup(x\cap\mu)\in S_{\xi}$ A

stationary by the claim above and mutually disjoint.

. Take$\alpha\in S$

with$\sup(\mathrm{c}1_{f}(a\cup b)\cap\mu)=\alpha$$\lambda$$\kappa$\xi<\mathrm{c}\mathrm{f}\lambda\}$$( \xi\zeta)\in\sum_{\xi<\mathrm{C}\mathrm{f}}\lambda\lambda\xi$$a$$S’\subset S_{\nu}^{\omega}$

as desired. Now set $\mu=\max${ , cf }regular cardinals $>\mu$ and forThen for$S\subset S_{\mu}^{\omega}$and split$S_{\mu}^{\omega}$\xi<\mathrm{c}\mathrm{f}\lambda\}\subset\lambda$$\{S_{\xi\zeta} :of\zeta<\lambda_{\xi}\}$$\sup(x\cap\lambda_{\xi})\in S_{\xi\zeta}$

.

} is

$\square$Our second result is inspired by Burke’s theorem [BMat] that the club filter on$P_{\kappa}\lambda$

is not$\lambda^{+}$-saturated when$\kappa>\omega_{1}$and cf$\lambda<\kappa$:

37

Theorem 2.

$P_{\kappa}\lambda$

Proof. The casesplits into $\lambda^{+}stati_{ona}w$ sets when$\mathrm{c}\mathrm{f}\lambda=\omega$

Otherwise fix a scaleby$\lambda^{+}$$\rho(x)=\min\{\gamma<\lambda^{+} :stationary in

Fix a club set

so that ran$\rho(x_{a})<\gamma$\langle\sup(x\cap\lambda_{\xi}) :with. Construct$\{x_{a} :$\rho(x)\geq\gamma$$f_{\max}(B\cap\beta)\subset x_{B\cap\beta}$, since$\rho(x_{B\cap\beta})<\gamma$

Now split
stationary in$\rho:P_{\kappa}\lambdaarrow$. We show thatis$\rho^{-1}S$by induction on

a\in[\lambda^{+}]^{<\omega}\}\subset C$, and an unbounded set. We claim that. Define$S\subset S_{\lambda}^{\omega}+\cdot$

for any$a\in[\gamma]^{<\omega}$$\lambda_{0}>\kappa$\xi<\mathrm{C}\mathrm{f}\lambda\rangle\leq^{*}f_{\gamma}\}$for any$\xi<\mathrm{c}\mathrm{f}\lambda\rangle\leq*f_{\gamma}$by\gamma<\lambda^{+}\}\subset\prod_{\xi<\mathrm{C}}\mathrm{f}\lambda\lambda_{\xi}$$f_{\max a}\subset x_{a}\subset x_{b}$First we haveran$\{f_{\gamma} :$C\subset P_{\kappa}\lambda$$x= \bigcup_{\beta\in B\cap\beta}x_{B}\in C$.

follows from Theorem 1.

for any stationary set$P_{\kappa}\lambda$$\mathrm{c}\mathrm{f}\lambda<\kappa$$a\subset b\in[\lambda^{+}]^{<\omega}$$\rho(X)=\gamma$by$\mathrm{c}\mathrm{f}\lambda<\kappa$$B\subset\gamma$. Take$\gamma\in S$

of order type$\omega$

$|a|$

with

. Set

as desired.

, since for any $\beta\in B\rho(x)\geq\rho(x_{B\cap\beta})\geq\max(B\cap\beta)$ by. Next$\langle$$\mathrm{c}\mathrm{f}\lambda>\omega$$\sup(x\cap\lambda_{\xi})$

and for any: $\xi<$ cf$\lambda\rangle$$= \langle\sup_{\beta\in B}\sup(x_{B\cap\beta}\cap\lambda_{\xi})$:$\beta\in B\langle\sup(x_{B\beta}\cap\cap\lambda_{\xi}):\xi<\mathrm{c}\mathrm{f}\lambda\rangle\leq*f_{\gamma}$.

$S_{\lambda}^{\omega}+\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{o}$$P_{\kappa}\lambda$

stationary sets$\{S_{\alpha} :\alpha<\lambda^{+}\}$. Then forby the claim above and mutually disjoint.

$\alpha<\lambda^{+}\rho^{-1}S_{\alpha}$is$\square$3. SOME REMARKSFor the moment let us assume that $\mu<\kappa<\lambda$ are all regular and consider
the stationary set$S_{\kappa\lambda}^{\omega}$

splits into$\lambda^{\omega}$$S_{\kappa\lambda}^{\mu}=${and$\lambda<\kappa^{+\omega}$$S_{\kappa\lambda}^{\mu}$splits into$\lambda$

stationary sets. This is optimal when

, since Baumgartner [B] shows that$\lambda(\mathrm{c}\mathrm{f}\sup(x\cap\nu)>\omega)\}\cap C|=\lambda$$x \mapsto\langle\sup(x\cap\nu) :: cf $\sup x=\mu$ }. Main Lemma 1 implies thatstationary sets. On the other hand Matsubara [Mat] provedthat a stationary subset of$\mu>\omega$$x\in P_{\kappa}\lambda$\kappa\leq\nu\leq\lambda\rangle$

for some club set$|\{x\in P_{\kappa}\lambda$$C\subset P_{\kappa}\lambda$:$\kappa\leq\forall\nu\leq$. In fact the map

is injective on this set. Complementing a result of

38

Abe [A], we remark that the map $x \mapsto\sup x$ is not injective on
: Fix

$f$

an unbounded set

$a$

set

$C\subset P_{\kappa}\lambda$$\sup$:.

$\lambda^{<\omega}arrow$$\subset\gamma$$\mathrm{p}\kappa\lambda$

of size$\mu$

generating $C$ . Takeand$\alpha\in\gamma-\mathrm{c}1fa$$S_{\kappa\lambda}^{\mu}\cap C$$\kappa<\gamma\in S_{\lambda}^{\mu}$. Then cl

for any club

closed under ,
$f$

and

$f^{a}\neq \mathrm{c}1_{f}(a\cup\{\alpha\})$cl $f^{a}= \sup$ cl$f(a\cup\{\alpha\})=\gamma$ as desired.

The rest of the section is devoted to a detailed proof of the Donder-Matet theorem

mentioned earlier.
Let $\mu>\omega$ be regular and$d_{\gamma}=\{\gamma_{n} :n<\omega\}\subset\gamma$

unbounded for$\gamma\in S_{\mu^{d}}($.

The following lemma ffom [B] (see also [BT]), where it is stated in (harmlessly)

inaccurate form, is implicit in Lemma 9.1 of [DM].
Lemma 1. Let$S\subset S_{\mu}^{\omega}$

be stationaw.

Then { $\alpha<\mu$ :$\{\gamma\in S : \alpha\in d_{\gamma}\}$isstationary} is unbounded.

Proof. Suppose to the contrary that we have
$C_{\alpha}\subset\mu$any$\gamma$.

with$C_{\alpha}\cap\{\gamma\in S : \alpha\in d_{\gamma}\}=\emptyset$

by$\beta<\alpha<\gamma\alpha\not\in d_{\gamma}$$\gamma\in S\cap C_{\alpha}$$\beta<\mu$. Takeand for $\beta<\alpha<\mu$ a club set$\beta<\gamma\in S\cap\Delta_{\beta<\alpha<\mu}c_{\alpha}$. Then for. This contradicts the unboundedness of$d_{\gamma}$

in$\square$We call a subtree
cobounded) ifof$T\neq\emptyset$$\mathrm{s}\mathrm{u}\mathrm{c}\tau(a)$$[\mu]^{<\omega}$in the sense of Section 2 unbounded (resp.

is unbounded (resp. cobounded) in$\mu$

for any $a\in T$ . The

following lemma from [RS] (see also [BMag]) would ensure that the map in Lemma$\xi$

9.2 of [$\mathrm{D}\mathrm{M}_{\rfloor}^{1}$is$\mathrm{w}\mathrm{e}\mathrm{U}$Lemma 2. Let$g$-defined (at least in the case we are interested in).

: $Tarrow\nu$ withfor some unbounded subtree
Proof. For$h$:$\omegaarrow\nu$$\tau*$$T$an unbounded$\mathit{8}ubtree$

of $Tg$ is constant onof$[\mu]^{<\omega}$$T^{*}\cap[\mu]^{n}$and$\nu^{\omega}<\mu$

for any. Then$n<\omega$.

set $T_{h}=\{a\in T : \forall b\leq a(g(b)=h(|b|))\}$, a subtree ofFirst we find $h:\omegaarrow\nu$ with$[T_{h}]\cap[U]\neq\emptyset$

for any cobounded subtree

$U$

of

$T$

$[\mu]^{<\omega}$

.

.

39

Suppose to the contrary that for $h:\omegaarrow\nu$ we have a cobounded subtree$[\mu]^{<\omega}$with$\nu^{\omega}<\mu$$[T_{h}]\cap[U_{h}]=\emptyset$. TakeNow fix$\omegaarrow\nu$

as above. Set $\tau*=\{a\in T_{h}$ :$\exists B\in[T_{h}]\cap[U](b\subset B)\}$

claim that

$B\in[T]\cap[\cap\{U_{h} :with $B\in[T_{h}]$ . This contradicts$h:\omegaarrow\nu$:$h$. Take inductively, a subtree of . Note that$T$

Suppose to the contrary that

$A=\mu-\mathrm{s}\mathrm{u}\mathrm{C}_{T^{*}}(a)$Then for $\alpha\in A$ we have a cobounded subtree

subtreefor anyof$U$$B\in[T_{h}]\cap[U_{\alpha}]$.

$\forall b\leq a\forall U\ni b$$\emptyset\in T^{*}$

bycoboundedby the choice of

$h$

. We

minimal choice of. This contradicts.

is cobounded for some $a\in\tau*$ .

of$a\cup\{\alpha\}\not\in T^{*}$

such that$[\mu]^{<\omega}$. Fix a coboundedwith $\{b\in U : a<b\}=\bigcup_{\alpha\in A}\{b\in U_{\alpha} : a\cup\{\alpha\}\leq b\}$ . Take$[\mu]^{<\omega}$

by $a\in T^{*}$ , and then$U_{\alpha}$$[T_{h}]\cap[U_{h}]=\emptyset$$U_{\alpha}\ni a\cup\{\alpha\}$by $a\in T^{*}$ and$a\subset B\in[T_{h}]\cap[U]$choice of\omegaarrow\nu\}]$is unbounded as desired.

$\tau*$$a\cup\{\alpha\}\not\subset B$

h :

of

$U_{h}$$U$$\alpha\in A$

with$a\cup\{\alpha\}\not\subset B$$a\cup\{\alpha\}\subset B\in[U_{\alpha}]$by theby $B\in[T_{h}]\cap[U_{\alpha}]$ and the$\square$

We are ready to prove the main claim of Proposition 9.6 of [DM]:
Theorem. Let$\{x\in P_{\kappa}\lambda :Proof. Set$x\in P_{\kappa}\lambda$

for

$a$

:

$\in T$.

Then there is a sequenceis stationaw for any

and split$S_{\mu}^{\omega}$$X\subset\lambda$$\langle v_{x} :such thatx\in P_{\kappa}\lambda\rangle$.

into stationary sets$\{S^{w} :w\approx\in P_{\kappa}\kappa\}$. For

with cf $\sup(x\cap\mu)=\omega$ set $v_{x}=\pi(x)^{-1}w$ , where $\sup(x\cap\mu)\in S^{w}$$xarrow \mathrm{o}\mathrm{t}x$$\{x\in P_{\kappa}\lambda :$f$v_{x}=X\cap x\}$$\mu=(2^{<\hslash})^{+}$

and $\pi(x)$ :Fix$\lambda>2^{<\kappa}$

is the increasing bijection. Fixv_{x}=X\cap X\}$$\lambda^{<\omega}arrow P_{\kappa}\lambda$$X\subset\lambda$. We show thatis stationary.

. We build inductively an unbounded subtree

a stationary set$S_{a}\subset S_{\mu}^{\omega}$

and an increasing inj\’ectionthat for any $a\leq b\in TS_{b}\subset S_{a}$ and for any$\gamma\in S_{a\prime}a\subset d_{\gamma}$

and$\chi_{a}$

$T$

of

: cl

$[\mu]^{<\omega}$and$f^{a}arrow\kappa$so$\pi(\mathrm{c}_{l}1_{f}d)\gamma|\mathrm{C}1_{f^{a}}=\chi_{a}$.

40Note that these conditions imply
First set

$S_{\emptyset}=S_{\mu}^{\omega}$are defined. Fix

and$\chi\emptyset=\emptyset$$a\in T\cap[\mu]^{n}$for any $a\leq b\in T$ .

$\chi_{a}\subset\chi_{b}$. Next suppose that

. Let

$S_{a\cup\{\}}\alpha\subset\{\gamma\in S_{a} :\alpha\in d_{\gamma}\}$$\omega\},$$\{z_{n} :$\nu=2^{<\kappa}$

and$xa\cup \mathrm{t}\alpha$$a\in T^{*}\cap[\mu]^{n}$. Then $C=$ {contains a club set. Set$\gamma<\mu$: cl$\mathrm{h}_{\mathfrak{M}}$by$2^{<\kappa}<\mu$$\chi_{a}=y_{n}$

have
Next

$\sup(\mathrm{c}1_{f^{B}}\cap\mu)=\gamma$$\pi(\mathrm{c}1_{f}B)$““

increasing bijection and

“

and

“
$\chi_{a}$$\tau*$. Take a stationary

of

$T$

and {

$y_{n}$$(X\cap \mathrm{c}1_{f}a)=z_{n}$, since. Fix$\gamma_{n}<\beta_{n}<\gamma$. Then cl, since $\sup B=\gamma$ and cl$\chi$\alpha\in d_{\gamma}\}$.

$\bigcup_{n<\omega}z_{n}\in P_{\kappa}\kappa$

i<n\}\in\tau*\cap[\gamma]^{<\omega}$$(x\cap \mathrm{c}1_{f}B)=w$$\{\gamma\in S_{a} :: $n<$

for any$f \gamma\cap\mu=\gamma\wedge\forall a\in T^{*}\cap[\gamma]^{<\omega}(\gamma\in\lim \mathrm{s}\mathrm{u}\mathrm{c}\tau*(a))$$w= \pi(\bigcup_{n<\omega}y_{n})$$\{\beta_{i} :for $a\in T\cap[\mu]^{n}$} : cl$f(a\cup\{\alpha\})arrow\kappa$ so that for any

inductively $B=\{\beta_{n} : n<\omega\}\in[T^{*}]$ so thatinductive . pothesis$\alpha\in \mathrm{s}\mathrm{u}\mathrm{c}_{T}(a)$take an unbounded subtree

so that ran

n<\omega\}\subset P_{\kappa}\kappa$$S_{a}$

A$\gamma\in S_{a\cup\{\alpha\}}\pi(\mathrm{c}1fd)\gamma|\mathrm{c}1f(a\cup\{\alpha\})=\chi_{a\cup}\{\alpha\}$By Lemma 2 withand$\mathrm{s}\mathrm{u}\mathrm{c}\tau(a)=\{\alpha<\mu:\max a<\alpha$is stationary}, which is unbounded by Lemma 1. Fix

set$T\cap[\mu]^{n}$$f^{B}$$f^{B}\cap\mu\subset$$\chi=\bigcup_{\beta\in B}\chi B\cap\beta$$(X \mathrm{n}_{\mathrm{C}}1_{f^{B}})=\bigcup_{n<\omega}z_{n}$$\gamma\in S^{w}\cap C$}. Takeby $\gamma\in C$ and theis as desired: First wecl$f^{\gamma}\cap\mu=\gamma$

: cl

by

$\gamma\in C$$f^{B} arrow\bigcup_{n<\omega}y_{n}$by the note above.

.

is an

$\square$

REFERENCES
[A]
[AS]

[B]
[BT]

Y. Abe, Strongly no rmal ideals on

and the $Sup$ -function, Top. Appl. 74 (1996),
97-107.
U. Abraham and S. Shelah, Forcing closed unbounded sets, J. Symbolic Logic 48 (1983),
643-657.
J. Baumgartner, On the size of the closed unbounded sets, Ann. Pure Appl. Logic 54(1991), 195-227.
J. Baumgartner and A. Taylor, Saturation properties of ideals in generic extensions.
I,. Amer. Math. Soc. 270 (1982), 557-574.
M. Burke and M. Magidor, Shelah’s pcf theory and its applications, Ann. Pure Appl.
Logic 50 (1990), 207-254.
D. Burke and Y. Matsubara, The extent of strength in the dub filters, Israel J. Math.,
To appear.
H.-D. Donder and P. Matet, Two cardinal versions of diamond, Israel J. Math. 83(1993), 1-43.
M. Foreman and M. Magidor, Mutually stationary sequences of sets and the non, Preprint, 1998.
saturation of the non-stationary ideal on, J. Symbolic Logic 50 (1985), 881-894.
M. Gitik, Nonspliuing subset of$\mathcal{P}_{\hslash}\lambda$$?\mathrm{k}\mathrm{a}\mathrm{n}8$[BMag][BMat][DM][FM]$P_{\kappa}(\lambda)$[G]$P_{\kappa}(\kappa^{+})$

41

[H]

Y. Hirata, Nonsaturation of the club fiuer on, Master’s Thesis at University ofTsukuba, 1997.
A. Kanamori, The Higher Infinite, Corrected second printing, Springer, Berlin, 1997.
M. Kojman, The A, B, C of pcf.. a
to pcf theory, part I, PrePrint, 1995.
M. Magidor, Representing sets ordinals as countable unions of sets in the core model,
Trans. Amer. Math. Soc. 317 (1990), 91-126.
Y. Matsubara, Mena8’ conjecture andultrapowers, Ann. Pure Appl. Logic 36(1987), 225-234.
T. Menas, On strong compactness and supercompactness, Ann. Math. Logic 7 (1974),$\mathcal{P}_{\kappa}\lambda$

[Ka]
[Ko]
[Mag]

[Mat]
[Me]

$compani_{\mathit{0}}n$$\mathit{0}\dot{f}$$gene7\dot{\eta}c$327-359.
M. Rubin and S. Shelah, Combinatorial problems on trees: partitions,andlarge free subsets, Ann. Pure Appl. Logic 33 (1987), 43-81.
S. Shelah,
has a Jonsson algebra, Cardinal Arithmetic, Oxford University Press,
Oxford, 1994, pp. 34-116.
R. Solovay, Real-valued measurable cardinals, Axiomatic Set Theory (D. Scott, ed.),Proc. Symp. Pure Math. vol. 13, part 1, Amer. Math. Soc., Providence, 1971, pp. 397428.
S. Todor\v{c}evi\v{c}, Partitioning pairs of countable sets, Proc. Amer. Math. Soc. 111 (1991),
841-844.
–, Coding reals by sets of ordinals, Lectures at Nagoya University, 1994.

[RS]$\Delta_{- sy_{\mathit{8}}em}ts$[Sh]$\aleph_{\omega+1}$

[So]
[T1]
[T2]

IN ST IT UTE OF M AT HEM AT ICS, UN IV ERS IT YOFTSuKuBA, TSUKUB A 305-8571, J A PA

-mail addre8s: shioyamath. tsukuba. ac.jp$E$$\mathrm{N}$