数理解析研究所講究録1095 巻 1999 年 42-51

42

THE

STRENGTH IN THE CLUBEXTENT.OF FILTERSDOUGLAS R. BURKE AND YO MATSUBARA$(*^{/}\text{ノ}.

\mathit{4}^{\backslash }’\backslash \Gamma/:\S_{\backslash } \backslash \grave{\mathrm{v}}.\neq^{\backslash \prime})$1. INTRODUCTION

This paper gives a number of partial results towards the following
conjectures. Unless otherwise noted, is a regular, uncountable cardi.
nal and is an infinite cardinal
is not precipitous –unlessConjecture 1. The club filter onis regular.
is not pre-saturated–unlessConjecture 2. The club filter on
is weakly inaccessible.
and is regular or$\kappa$$(\lambda\geq\kappa)$$\lambda$$\lambda$$\mathcal{P}_{\kappa}\lambda$$\mathcal{P}_{\kappa}\lambda$$\kappa=\aleph_{1}$$\kappa=\lambda$$\lambda$

The corresponding conjecture for saturation has been established by
Foreman and Magidor:
is not saturated
Theorem (Foreman-Magidor). The club filter on.
–unless
The results of section 2 of this paper are the authors partial results
towards the above theorem. Shortly after the results of this paper wereannounced, Foreman and Magidor proved the above theorem. Their
proof does not use any of the results of this paper, and in fact in thecase covered by Theorem 2.10, they establish the stronger result that.
the club filter is not even
Remarks. 1. [She87] It is consistent that the club filter on is saturated (assuming the consistency of a Woodin cardinal).
weakly inacces2. [Git95] It is consistent that the club filter onsible, is pre-saturated (assuming the consistency of an up-repeat$\mathcal{P}_{\kappa}\lambda$$\kappa=\lambda=\aleph_{1}$$\lambda^{++_{\mathrm{S}\mathrm{a}}}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}$$\mathrm{N}_{1}$$\kappa,$$\kappa$

point).

3. [Go192] If is Woodin then for every regularis pre-saturated”.
“the club filter on
4. [Gol] If is Woodin then for every regularis precipitous”.
“the club filter on$\lambda(\aleph_{1}\leq\lambda<\delta)$$\delta$$V^{C\Phi}l(\lambda,<\delta)\models$$P_{\aleph_{1}}\lambda$$\kappa<\lambda(\aleph_{1}\leq\kappa\leq\lambda<\delta)$$\delta$$V^{co}l(\lambda,<\delta)\models$,$P_{\kappa}\lambda$

Date: August 12, 1997.
The first author was partially supported by NSF grant DMS-9626212.

,

43

DOUGLAS R. BURKE AND YO MATSUBARA$F$

We now give our basic definitions and conventions.
is a normal filter on
if$P(\lambda)$

1.
2. (fine)3.

$\mathcal{F}\subseteq \mathcal{P}P(\lambda)$

is a filter.

.

$\forall\alpha\in\lambda\{a\subseteq\lambda|\alpha\in a\}\in \mathcal{F}$$\mathcal{F}(\mathrm{n}\mathrm{o}\mathrm{m}\mathrm{a}\mathrm{l})$If$C_{\alpha}\in F(\alpha\in\lambda)$, then$\{a\subseteq\lambda|\forall\alpha\in a(a\in C_{a})\}\in$

Throughout this paper, filter will mean normal filter.
.
has an associatedpartial ordering: $A\leq B$ iff.
A filter on
is saturated if every antichain in
has size.
$F$ is pre-satumted
if
antichainsand, there isa $T\leq S$ such that for all.
Forcing withextends to a -nomal, -ultrafilter –so we
get a generic embedding.
is precipitous if this ultrapower iswell-founded. If is
pre-saturated, thenis precipitous and the ultrapower is closed under sequences in. For more on the basic facts about genericembeddings see [For86].
The club filter on(or just) consists of all
such thatwith.
Sets inare called stationary. CF is the smallest normal filter on$\mathcal{F}^{+}=_{\mathrm{d}\mathrm{e}\mathrm{f}}\{A\subseteq \mathcal{P}(\lambda)|\forall C\in \mathcal{F}(c\cap A\neq\emptyset)\}$$\mathcal{F}^{+}$$A\subseteq B$$\mathcal{F}$$P(\lambda)$$\mathcal{F}^{+}$$A_{a}(\alpha<\lambda)$$\dot{\mathrm{g}}\mathrm{v}\mathrm{e}\mathrm{n}$$\alpha<\lambda,$$F^{+}$$\leq\lambda$$S\in \mathcal{F}^{+}$$|\{A\in A_{a}|A\cap\tau_{\in \mathcal{F}}+\}|\leq\lambda$$V$$\mathcal{F}$$V$$\mathcal{G}$$j:Varrow \mathrm{U}\mathrm{l}\mathrm{t}(V, \mathcal{G})\subseteq V[\mathcal{G}]$$\mathcal{F}$$\mathrm{a}\mathrm{l}\mathrm{w}\mathrm{a}\mathrm{y}_{8}$$\mathcal{F}$$\mathcal{F}$$\lambda$$V[\mathcal{G}]$$P(\lambda)$$\mathrm{C}\mathrm{F}$$\mathrm{C}\mathrm{F}_{\mathcal{P}}(\lambda)$$A\subseteq P(\lambda)$$\exists f:\lambda^{<\omega}arrow\lambda$$\mathrm{c}1_{f}\subseteq A(\mathrm{c}1_{f}=\{a\subseteq\lambda|f’’a^{<\omega}\subseteq a\})$$\mathrm{C}\mathrm{F}^{+}$$P(\lambda)$.

If, thena normal filter. If
smallest nomal filter on$S\in_{-}\mathcal{F}^{+}$

is, then the club filter on , CF
, is thecontaining .
. This definition is slightly”
non-standard: usually the condition “is dropped. The set
is stationary in. If is a filter onand, thenis -complete, and so.
If
Ord, thenis the cofinality of the order type of . A
sequence is a set
such that for all,$\{a\in p_{\kappa}\lambda|a\cap A=s_{a}\}$is stationary.
The following fact was proved in [BTW77] for filters on cardinals. A
similar proof works here.
$\mathcal{F}\square S=_{\mathrm{d}\mathrm{e}\mathrm{f}}\{A\subseteq \mathcal{P}(\lambda)|(\exists C\in \mathcal{F})c\cap s\subseteq A\}$

$S\in \mathrm{C}\mathrm{F}^{+}$$S$$\mathrm{r}S$$S$$\mathcal{P}(\lambda)$$P_{\kappa}\lambda=_{\mathrm{d}_{\mathrm{G}\mathrm{f}}}\{a\subseteq\lambda||a|<\kappa\ a\cap\kappa\in\kappa\}$$a\cap\kappa\in\kappa$$\mathcal{P}_{\kappa}\lambda$$P(\lambda)$$\mathcal{F}$$\mathcal{P}(\lambda)$$\mathcal{P}_{\kappa}\lambda\in \mathcal{F}$$\mathcal{F}$$\kappa$$\forall s\in P_{\kappa}\lambda,$$a\subseteq$$\{a\in \mathcal{P}_{\kappa}\lambda|s\subseteq a\}\in \mathcal{F}$$\mathrm{c}\mathrm{o}\mathrm{f}(a)$$a$$0_{\kappa,\lambda}$$\langle s_{a}\subseteq a :Ehct 1.1. Assume

filters$\mathcal{G}\supseteq \mathcal{F},$$\mathcal{F}$a\in P_{\kappa}\lambda\rangle$

is a filter onsuch that$\exists S\in F^{+}$$P(\lambda)$$\mathcal{G}=\mathcal{F}\mathrm{r}S$

Corollary 1.2. Suppose the club filter
filter on is saturated.
$S$.

$\mathit{0}’ nS$$A\subseteq\lambda$$\mathcal{F}$.

is saturated iff for all

is saturated. Then every

44

THE EXTENT OF STRENGTH IN THE CLUB FILTERS

2. SATURATION
One of the first results about the failure of saturation is a theoremof Shelah (, p. 440) that says, for example, if is a saturated
filter on , then {a. The proof of this}$[\mathrm{S}\mathrm{h}\mathrm{e}821$$\mathcal{F}$$<\omega_{2}|\mathrm{c}\mathrm{o}\mathrm{f}(\alpha)=\omega_{1}$$\omega_{2}$uses the following result (with$\in \mathcal{F}$). We also use this result to getsimilar facts about saturated filters on.
$\lambda=\omega_{2}$$P_{\kappa}\lambda$

Theorem 2.1. (of $ZFC,$ is a cardinal of

Assumeare inner models
is a cardinal of $W$ , and.
$V\subseteq W$$\mathrm{l}\mathrm{S}\mathrm{h}\mathrm{e}82],[\mathrm{C}\mathrm{u}\mathrm{m}97]\mathit{1}$$\lambda$$A_{\mathit{8}Su}ming(\#),$$V_{f}\rho$$W\models \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)=\mathrm{C}\mathrm{o}\mathrm{f}(\rho)$$\lambda_{V}^{+}=\rho_{W}^{+}$.

is regular, or ( is singular and) there is a good scale on , or
( is singular and) $W$ is a
forcing extension of .

$(^{*})\lambda$$\lambda$$\lambda$$\lambda$$\lambda^{+}-_{CC}$$V$

See the next section for the definition of good scale.

Definition 2.2.
Theorem 2.3. Assume.

$S_{\lambda}=_{def}\{a\subseteq\lambda|\mathrm{c}\mathrm{o}\mathrm{f}(a)=\mathrm{c}\mathrm{o}\mathrm{f}(|a|)\}$$\mathcal{F}$is a saturated filter on$\mathcal{P}(\lambda)$. Then$S_{\lambda}\in \mathcal{F}$.

Proof. Suppose not. So we get $j:Varrow M\subseteq V[G]$ with.
. Sincein $V[G]$ ,. This contradicts Theorem 2.1 since $V[G]$ is$P(\lambda)\backslash S_{\lambda}\in G$

Since$P(\lambda)\backslash S_{\lambda}\in G,$$M^{\lambda}\subseteq M$$M\models \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\neq \mathrm{c}\mathrm{o}\mathrm{f}(|\lambda|)$$V[G]\models \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\neq \mathrm{c}\mathrm{o}\mathrm{f}(|\lambda|)$a$\lambda^{+}- \mathrm{C}\mathrm{C}$generic extension of V.

Lemma 2.4. Assume
is non-stationary.

$\kappa=\rho^{+},$$\square$, and$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)<\kappa$. Then$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\neq \mathrm{c}\mathrm{o}\mathrm{f}(\rho)$$S_{\lambda}\cap P_{\kappa}\lambda$

Proof. Let$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)<\kappa$. On a club, $|a|=\rho$ and so, on a club $\sup(a)=\lambda$ and so$a\in P_{\kappa}\lambda$$\mathrm{c}\mathrm{o}\mathrm{f}(|a|)=\mathrm{c}\mathrm{o}\mathrm{f}(\rho)$$\mathrm{c}\mathrm{o}\mathrm{f}(a)=\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$S_{\lambda}\cap \mathcal{P}_{\kappa}\lambda$

is non-stationary.

Corollary 2.5. For
Remark. If$\kappa=\rho^{+}$$\lambda$$\kappa,$andLemma 2.6. Assume
stationary,$co$

Proof. Let$f$. Since. Therefore$\square$as above, there is no saturated filter on$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)=\mathrm{c}\mathrm{o}\mathrm{f}(\rho)$$\kappa=\rho^{+}\geq\aleph_{2}$-stationary in$\mathcal{P}_{\kappa}\lambda$.

, thenand$S_{\lambda}\cap \mathcal{P}_{\kappa}\lambda$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\geq\kappa$

is club in. Then$\prime P_{\mathfrak{k}},\lambda$$\mathcal{P}_{\kappa}\lambda$$S_{\lambda}\cap \mathcal{P}_{\kappa}\lambda$

.
.

is

. We may assume
implies. For any regular
we can build aContinuous increasing chain of length to findclosed under with
. Takingshows thatis stationary.
:$\lambda^{<\omega}arrow\lambda$$a\in \mathcal{P}_{\kappa}\lambda\cap \mathrm{c}1_{f}$$\delta<\kappa$$\mathrm{c}\mathrm{o}\mathrm{f}(|a|)=\mathrm{c}\mathrm{o}\mathrm{f}(\rho)$$\delta$$\mathrm{c}\mathrm{o}\mathrm{f}(a)=\delta$

Taking$\delta\neq \mathrm{c}o\mathrm{f}(\rho)$shows that

Corollary 2.7. For
rated.

Remark. If$a\in P_{\kappa}\lambda$$\delta=\mathrm{c}\mathrm{o}\mathrm{f}(\rho)$$\kappa=\aleph_{1}$$\lambda$$\kappa,$$S_{\lambda}\cap \mathcal{P}_{\kappa}\lambda$

is$\mathrm{C}\mathrm{O}$-stationary inas above, the club filter on, then for all$\lambda\geq\kappa,$$f$$S_{\lambda}\cap \mathcal{P}_{\kappa}\lambda$$S_{\lambda}\cap P_{\kappa}\lambda$$P_{\kappa}\lambda$

is club in$\mathcal{P}_{\kappa}\lambda$.

$\square$is not satu$P_{\kappa}\lambda$.

45

DOUGLAS R. BURKE AND YO MATSUBARA

Lemma 2.8. AssumeThenis$S_{\lambda}\cap P_{\hslash}\lambda$

Proof. Let

to findand, if

$f$

:

$a$

$stationa7\mathrm{Y}/,$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)>\kappa,$

oegular limit cardinal and$CO-Stati_{on}aw$ in.
$\rho<\kappa$

(if
Hence

.

.

a regular cardinal. It is easysuch that $|a|=|a\cap\kappa|$ and$\mathrm{c}\mathrm{o}\mathrm{f}(a)=\rho$$\mathrm{c}\mathrm{o}\mathrm{f}(a)=\mathrm{c}\mathrm{o}\mathrm{f}(\lambda))$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\neq\kappa$$\mathcal{P}_{\kappa}\lambda$

and$\lambda^{<\omega}arrow\lambda$$a\in P_{\kappa}\lambda\cap \mathrm{c}1_{f}$$a\in P_{\kappa}\lambda,$

is$\kappa$$\mathrm{c}\mathrm{o}\mathrm{f}(|a\cap\kappa|)=\rho$, then for club manyis stationary (take $\rho=$
is co-stationary. The idea for$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)<\kappa$$S_{\lambda}\cap P_{\kappa}\lambda$

if
. If
then
in
. Finally,
the following argument is[Bau91]. Let. Note thatis club, so we may assume witnesses
this. Letbethat
implies. Defineby $g( \alpha)=\sup(\mathrm{c}1;(\alpha+1))$ . Now choosesuch that$|a|=|a\cap\kappa|,$
, and
.
Let
. Given , let
, and
.
Let
. Thenand. Let. Thenandand. Henceis -stationary in.
Corollary 2.9. For
as above, the club filter on
is not satu$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)<\kappa)$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)<\kappa$$S_{\lambda}\cap P_{\kappa}\lambda$$P_{\kappa}\lambda-\mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}\rho\neq \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\mathrm{a}8\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)>\kappa$$\mathrm{h}\mathrm{o}\mathrm{m}$$\delta=\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\{a\in \mathcal{P}_{\kappa}\lambda\rfloor \mathrm{c}\mathrm{o}\mathrm{f}(a)=\mathrm{C}\mathrm{o}\mathrm{f}(a\mathrm{n}\delta)\}$$f:\delta^{<\omega}arrow\delta$$f$$8\mathrm{u}\mathrm{c}\mathrm{h}$$\mathrm{c}1_{f}(a)\cap\delta=a$$a\in \mathrm{c}1_{\overline{f}}$$g:\deltaarrow\delta$$a\in P_{\kappa}\delta$$a$$\in \mathrm{c}1_{\overline{f}},$$a\in \mathrm{c}1_{g},$$\mathbb{C}\mathrm{o}\mathrm{f}(|a\cap\kappa \mathrm{I})=\aleph_{1},$ $\mathrm{c}\mathrm{o}\mathrm{f}(a)=\aleph_{1}$$a_{0}=a\cap\kappa$$a_{\omega}= \bigcup_{n\in\omega}a_{n}$$a_{\omega}\cap\kappa=a\cap\kappa,$$b=\mathrm{c}1_{f(}a_{\omega})$$S_{\lambda}\cap P_{\hslash}\lambda$$\beta\in a\backslash \sup(a_{n})$$a_{n}$$b\in \mathrm{c}1_{f}$$\mathrm{c}\mathrm{o}$$\kappa\in a$$a_{n+f^{(a_{n}}}1=\mathrm{C}1\cup\{\beta\})$$\mathrm{C}\mathrm{o}\mathrm{f}(a_{\omega})=\omega$$a_{\omega}\in \mathrm{c}1_{\overline{f}}$$\mathrm{c}\mathrm{o}\mathrm{f}(|b|)=\aleph_{1}$$\mathrm{c}\mathrm{o}\mathrm{f}(b)=\omega$$\mathcal{P}_{\kappa}\lambda$$\square$$\lambda$$\kappa,$$P_{\kappa}\lambda$

rated.

Remark. Assume( regular limit). Then for club many. So
is club in the (stationary) set$\{a\in P_{\kappa}\lambda||a|=|a\cap\kappa|\}$and is non-stationary in the (possibly
non-stationary) set.
$a\in P_{\hslash}\lambda,$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)=\kappa$$\kappa$$\mathrm{c}\mathrm{o}\mathrm{f}(a)=\mathrm{c}\mathrm{o}\mathrm{f}(a\cap\kappa)$$S_{\lambda}$$\{a\in \mathcal{P}_{\kappa}\lambda||a|=|a\cap\kappa|^{+}\}$

The above method does not handle the cases: (i)and, and (iii) regular limit andis handled in the following:$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)=\mathrm{c}\mathrm{o}\mathrm{f}(\rho)$$\kappa$

Theorem 2.10. Assumeon
is not saturated.

$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)<\kappa$

and$\kappa\geq \mathrm{N}_{2}$$P_{\kappa}\lambda$

Proof. Let$\langle f_{\alpha} :be a scale on\alpha\in\lambda^{+}\rangle$, where thecardinals cofinal in. Case (ii). Then the club filter(see definition 3.3, so each$\lambda$$\beta_{\xi}’ 8$$\lambda$$\kappa<\rho_{0}$$g_{a}( \xi)=\sup(a\cap\rho\epsilon)$$a\in P_{\kappa}\lambda$$g_{a}\in\Pi p_{\xi}$$\alpha\in\lambda^{+}$$\mathcal{F}$$\mathcal{P}_{\kappa}H_{\theta}$$(\mathrm{i}\mathrm{i})\kappa=\rho+$

are an increasing sequence of regularwith). Given
defineby$\pi(a)=$and letleast
such that.
be a filter on. Let, an assume
is a filter on
projecting to . Let$f_{\alpha}\in\square \epsilon<\infty \mathrm{f}(\lambda)\rho\xi$Let$\kappa=\mathrm{N}_{1},$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)=\kappa$$P_{\kappa}\lambda$$\theta>>\lambda$$g_{a}\leq*f_{\alpha}$$\mathrm{d}$$F_{\theta}$$\mathcal{F}$$E=\{b\prec H\theta|b\in p_{\kappa}H_{\theta}\ \langle f_{\alpha} :\alpha\in\lambda^{+}\rangle\in b\$cof(\mbox{\boldmath $\lambda$})b&(p\xi :Claim 1. If $b\in E$ then$\xi<\mathrm{C}\mathrm{o}\mathrm{f}(\lambda)\rangle\in b\}$$\sup(b\cap\lambda^{+})\leq\pi(b\cap\lambda)$

.

.

46

THE EXTENT OF STRENGTH IN THE CLUB FILTERS

sSuppose not. Let $b\in E$ with
Say

with

$\beta\in b\cap\lambda^{+}$

such that$\exists\eta<\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\beta\in b$, and$\langle f_{\alpha} :.
. Therefore
. But
,

. But

$f\rho^{*}>f_{\xi^{*}}\geq g\iota\cap\lambda$$f_{\beta}(\eta)\succ f_{\alpha}(\eta)\geq g_{b\cap}\lambda(\eta)$\alpha\in\lambda^{+}\rangle\in b$$gb \cap\lambda(\eta)=\sup(b\cap p_{\eta})$Let$\beta>\xi$$\sup(b\cap\lambda^{+})>\pi(b\cap\lambda)=_{\mathrm{d}\mathrm{e}\mathrm{f}}\xi$. Therefore$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\subseteq b$$f_{\beta}(\eta)\in b\cap p_{\eta}$. Contradiction. But

and define $S_{T}(a)= \sup\{\sup(b\cap\lambda^{+})| b\cap\lambda=a\ b \in T\}_{j}$Note thatis defined on a set in (the projection of ), if
then $S_{T}(a)\leq s_{\tau’}(a)$ , and if
then $S_{T}(a)\leq\pi(a)$ .
$T\in \mathcal{F}_{\theta}$$S_{T}$$T$$\mathcal{F}$$T\subseteq T$$T\subseteq E$

Claim 2. Givenandwith
set we have $\beta<S_{T}(a)\leq\pi(a)<\lambda^{+}$ .
$\beta\in\lambda^{+}$$T\subseteq E$

We already have that$\tau’\in F_{\theta}$$S_{T’}(a)\leq S_{T}(a)$Claim 3. AssumeThen there is a$f$$\mathcal{F}$$\beta\in\lambda^{+}$$T^{J}$.

:

is such that$\mathcal{P}_{\kappa}\lambdaarrow\lambda^{+}$$T\in F_{\theta}$$S_{T}’(a)=f(a)$, on an $F$ measure one

is defined on an measure one set and. Letand let $\tau’=\{b\in T|\beta\in b\}$ .
and (on the projection of ) $s_{\tau’}(a)>\beta$ and$S_{T}$$S_{T}(a)\leq\pi(a)<\lambda^{+}$

Then$T\in F_{\theta}$.

such that.

$1 \vdash_{f+}[f]=\sup j’’\lambda^{+}$$(\forall T’\subseteq T)T^{J}\in \mathcal{F}_{\theta}$

on a set in$\mathcal{F}$

,

On an
measure one set(If not, thenthere is asuch that $f(b \cap\lambda)<\sup(b\cap\lambda^{+})$ , so on some$f(b\cap\lambda)<\eta$ (is fixed). Projecting to we get$f(b)<\eta$such thaton . Contradiction.)
So letsuch thatand $( \forall b\in T)f(b\cap\lambda)\geq\sup(b\cap$. Therefore, on an measure one set, $f(a)\geq S_{T}(a)$ . Suppose$f(b \cap\lambda)\geq\sup(b\cap\lambda^{+})$$\mathcal{F}_{\theta}$$S\in \mathcal{F}_{\theta}^{+}$$s’\in \mathcal{F}_{\theta^{+}},$$\eta\in\lambda^{+}$$\mathcal{F}$$\overline{S}\in \mathcal{F}^{+}$$\overline{S}$$T\subseteq E$$T\in \mathcal{F}_{\theta}$$\lambda^{+})$$\mathcal{F}$’$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{X}\mathrm{i}_{\mathrm{S}}\mathrm{o}\mathrm{n}S\in \mathcal{F}+_{\mathrm{w}\mathrm{e}}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{S}\overline{S}\subseteq s\mathrm{a}\mathrm{n}\mathrm{d}\eta<\lambda^{+}\mathrm{S}\mathrm{u}\mathrm{C}\mathrm{h}\mathrm{V}\mathrm{e}f(a)>ST(a).\mathrm{T}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{n},S\frac{\mathrm{e}}{S}j’\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}\mathrm{I}\vdash[Tf(a])\leq\eta.\mathrm{h}\mathrm{i}=\sup\lambda^{+}\prime \mathrm{s}$contradicts Claim 2. Finally, assume. Then on $F$ measure$S_{T’}(a)\leq
S_{T}(a)=f(a)$one set. Again by Claim 2, $S_{T’}(a)=f(a)$on an measure one set.
$\mathcal{T}^{f}\subseteq T$$\mathcal{F}$Claim 4. Assume
is regular,
is stationary,, andis IA (intemally approachable) of length (thismeans there is an increasing, continuous sequence
where
each
, and
–see [FMS88] .
Let be the projection of to. Then for all $a\in\overline{T}S_{T}(a)=\pi(a)$$f(\pi(a))=\rho$
and co.
$p<\kappa$$T\subseteq E$$\forall a\in T,$$\rho\neq \mathrm{c}\mathrm{o}\mathrm{f}(\lambda),$$\tau\subseteq \mathcal{P}_{\kappa}H_{\theta}$$a$$\rho$$\langle a_{\xi} :$a_{\xi}\in E,$ $\forall\rho’<\rho\langle a_{\xi} : \xi<\rho’\rangle\in a$$\overline{T}$$T$$a= \bigcup_{\xi<\rho}a_{\xi}$\xi<\rho\rangle$$)$$P_{\kappa}\lambda$

The idea for the proof of Claim 4 comes from [FM97]. Let $b\in T$ ,and
be a witness to IA of length . We may assume. Let $a=b\cap\lambda$ . It is enough to see that{$\sup(b\cap\lambda^{+})=\pi(a)$ . (Notethat) Given$\langle b_{\alpha} :\alpha<\rho\rangle$$\rho$$\forall\alpha\in\rho)b_{\alpha}\in b_{\alpha+1}$$\mathrm{c}\mathrm{o}\mathrm{f}(\sup(b\mathrm{n}\lambda^{+}))=\rho.$$\alpha<\rho$

47

DOUGLAS R. BURKE AND YO MATSUBARA, there(everywhere) and.
. By Claim 1,
such thatis,. For alland we will show
So letunbounded and. Since)( , co.
such thatandBut.
and soLet $p<\kappa$ be regular,.
we have$(\forall f\in b_{\alpha})f<g_{t_{\alpha}}$$\gamma_{\alpha}\in b_{a+1}$$8\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}g\iota\alpha\in b_{\alpha+1}$$\pi(a)\geq\sup(b\cap\lambda^{+})$$g_{b_{\alpha}}\leq^{*}f_{\gamma_{\alpha}}$$\delta=\sup(b\cap\lambda^{+})$$g_{b}\leq^{*}f_{\delta}$$g_{b_{\alpha}}\leq^{*}f_{\gamma_{a}}<*f_{\delta}$$\forall\alpha\in A$$\nu<\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\forall\xi\in$$g_{b}( \xi)=\sup_{\alpha\in A}g_{b}\alpha(\xi)$$\alpha<\rho$$\exists A\subseteq\rho$$\rho\neq \mathrm{c}\mathrm{o}\mathrm{f}(\lambda),$$g_{b_{a}}(\xi)<f_{\delta}(\xi)$$f(\lambda)$$\nu$$g_{b}(\xi)\leq f_{\delta}(\xi)$$\rho\neq \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$

Let $T=$ { $b\in E|b$ is IA of length

Claim 5.
Let

$T$

is stationary.

$f:H_{\theta}^{<\omega}arrow H_{\theta}$$\bigcup_{\epsilon<\xi}b_{\epsilon}$}.

$\rho$. Given. So$b_{\xi}$

. Let

, let

$b_{0}\in E\cap d_{f}$$b_{\xi+1}\in E\cap d_{f}$. If$\xi<\rho$is limit letsuch thatis IA of length$b_{\xi}\cup\{(b_{\epsilon} :. To see$b_{\xi}=$\epsilon\leq\xi\rangle\}\in$

we just
need
. But
.
[ T. $F$ is gotten by projection. We will show that
Finally, let
is not saturated, and therefore by Corollary (1.2) the club filter on
is not saturated.
For a contradiction, assume $F$ is saturated. So there is ansuch that, and on a set in(otherwise we could force to havein the ultrapower–so
this collapses ).
By Claim 3,
such that for any
on a$S_{R\mathrm{n}\tau}(a)=f(a)$ .
set in$S_{R’}(a)=f(a)$ . So on a set in
But$R\cap T$ is a set as in
Claim 4. Hence on a set in $F$ (the projection of
Therefore on a set in ’
. This contradiction completes the proof.
$b_{\xi+1}$$b= \bigcup_{\epsilon<\rho}b_{\epsilon}\in E\cap d_{f}$$\forall\xi<\rho\langle b_{\alpha} :\alpha\in\xi\rangle\in b$$b$$\langle b_{\alpha} :$\rho$\alpha\in\xi\rangle\in b_{\xi+1}\subseteq b$$\mathcal{F}_{\theta}=\mathrm{C}\mathrm{F}$$\mathcal{F}$$P_{\kappa}\lambda$$f:P_{\kappa}\lambdaarrow$$\lambda^{+}$$1\vdash[f]=\mathrm{s}\mathrm{u}\mathrm{p}j’’\lambda^{+}$$\mathcal{F},$$\mathrm{c}\mathrm{o}\mathrm{f}(f(a))>p$$\mathrm{C}\mathrm{o}\mathrm{f}([f])\leq p$$\lambda^{+}$$R’\subseteq R(R’\in \mathcal{F}_{\theta})$$\exists R\in \mathcal{F}_{\theta}$$\mathcal{F},$$\mathcal{F},$$R\cap T)S_{R\cap T}(a)=\pi(a)\mathrm{s}\mathrm{t}\mathrm{c}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{c}o\mathrm{f}(\pi(a))=\rho \mathrm{t}\mathrm{s}$$\mathcal{F}\square$$\mathrm{c}\mathrm{o}\mathrm{f}(f(a))=\rho$Question. In the above proof, is the projection of CF
club filter restricted to a stationary set?
$\mathcal{F}$$\mathrm{r}T$. Is$\mathcal{F}$the

We conclude this section with three previously known theorems.

Theorem 2.11.
For allurated. In fact, for any regular withis not saturated.
$([\mathrm{G}\mathrm{S}9\eta)$$\kappa>\mathrm{N}_{1}$$\rho$, the club filte on is not sat-$\rho^{+}<\kappa$, CF$(\{\alpha<$.

$\rho\}$

Corollary 2.12. For all regular
is not satumted.
filter on$\kappa$$\kappa$$\mathrm{r}$and all regular$\kappa|\mathrm{c}\mathrm{o}\mathrm{f}(\alpha)$$=$.

$\lambda\geq\aleph_{2\mathrm{z}}$the club$P_{\kappa}\lambda$

Proof. Define$g$:

by $g(a)= \sup(a)$ . Supposeis. Then $g^{-1}(S)$ is stationary (let$P_{\kappa}\lambdaarrow\lambda$$S\subseteq\lambda$

stationary andand choosesuch that is closed under . Now build$\sup(a)=\alpha)$such that. Also, if
is stationary,
thenis stationary (, define.
$(\forall\alpha\in S)\mathrm{c}\mathrm{o}\mathrm{f}(\alpha)<\kappa$$f:\lambda^{<\omega}arrow\lambda$$\alpha\in S$$S\subseteq \mathcal{P}_{\kappa}\lambda$$a\in \mathcal{P}_{\kappa}\lambda\cap \mathrm{c}1_{f}$$g”S\subseteq\lambda$$f$$\alpha$$\mathrm{i}^{}\mathrm{f}f:\lambda^{<\omega}arrow\lambda$$h(\alpha)=\mathrm{c}1_{f}(\alpha+1)$

48

THE EXTENT OF STRBNGTH IN THE CLUB FILTERS, thenIffrom Theorem 2.11.
$a\in S\cap \mathrm{c}1_{h}$$8\mathrm{u}\mathrm{p}(a)$is closed under$f$. The result now follows$\square$

Theorem 2.13. [DM93] If
is not saturated.
filter on
Theorem 2.14. [BT82] For any
many disjoint $stati_{on}aw$ sets.

then$\lambda>2^{<\kappa}$$0_{\kappa,\lambda}$

holds. Hence the club$\mathcal{P}_{\kappa}\lambda$$\lambda>\aleph_{1},$$\mathcal{P}_{\aleph_{1}}\lambda$

can be$\mathit{8}plit$

into$2^{\omega}$

Remark. Piecing everything together, we have the following partial results towards the theorem of Foreman and Magidor: The club filter on
is not saturated unless
1.
(consistent).
2.
is singular.
3. is limit andand.
$P_{\kappa}\lambda$$\kappa=\lambda=\aleph_{1}$$\kappa=\aleph_{1},$$\lambda=2^{\omega}$$\kappa$$2^{<\kappa}\geq\lambda$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)=\kappa$

3.

PRESERVING TO PRE-SATURATION$\mathrm{c}_{\mathrm{A}\mathrm{R}\mathrm{D}\mathrm{I}}\mathrm{N}\mathrm{A}\mathrm{L}$A filter $F$ on
is weakly pre-saturated if is precipitous andis preserved”. The filter $F$ is called $ca\tau dinal$ presenring if
is preserved”.
, then pre-saturated, weakly presaturated and cardinal preserving are all equivalent. It is not known ifthey are equivalent in general.
We use a number of known combinatorial principles to get that theclub filter cannot have these strong properties. For the case regular,
the solution is complete–the club filter on
is not cardinal preserving unlessoris weakly inaccessible (and both these cases
are consistent).
$P(\lambda)$$\mathcal{F}$$\mathrm{t}\vdash_{\mathcal{F}}+‘(\lambda^{+}$$\mathrm{I}\vdash_{\tau+}"$$\lambda^{+}$$\mathrm{I}\mathrm{f}|\mathcal{F}^{+}|=\lambda^{+}$$\lambda$$\mathcal{P}_{\kappa}\lambda$$\kappa=\lambda$$\kappa=\aleph_{1}$Definition 3.1.
$\mathrm{C}\mathrm{o}\mathrm{f}(|\lambda|)_{f}$then$V^{\mathrm{p}}$$Sh(\lambda)mean\mathit{8}$

.
means

collapses

Definition 3.2.
unbounded subset of and$AD(\lambda)$$\lambda$

implies

for any$\exists\langle a_{\alpha} :\alpha\in\lambda^{+}\rangle$$\forall\alpha\in\lambda^{+}\exists f_{\alpha}$$[a_{h}\backslash f_{\alpha}(\beta_{1})]\cap[a_{\beta_{2}}\backslash f_{\alpha}(\beta_{2})]=\emptyset$

Definition 3.3. Supposeif$\mathrm{P}\in V,$$V^{\mathrm{P}}\models \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\neq$$\lambda_{V}^{+}$:

such that eachsuch that$\alphaarrow\lambda$.

$a_{\alpha}$

is an$\beta_{1}<\beta_{2}<\alpha$is singular.
scale on is an increas$ing$ sequence
regularcardinalofcofinal in , and asequence ( :such that for each)
implies, andsuch that $f<*$. We will assume
is discontinuous everywhere and. An ordinal isgood forunboundedandifsuch thatfrom and. The scale is goodsuch that
if club, then is goodifthemeans theoe is a good scale on .
for scale.
$A$$\lambda$$\langle\rho_{\xi} :$f_{a}$$\lambda$\xi\in \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\rangle$$\lambda$$\alpha\in\lambda^{+}\rangle$$\alpha,$$f_{\alpha}<*f_{\alpha’}$$f_{a}$$f_{a}\in\Pi_{\xi \mathrm{f}(\lambda}\in \mathrm{c}\circ$$\rho\xi,$$\alpha<\alpha’$$\forall f\in\Pi_{\xi\lambda}\in \mathrm{c}\circ \mathrm{r}()\rho_{\xi}\exists\alpha\in\lambda^{+}$$\langle\rho_{\alpha} :\alpha\in \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\rangle$$\forall\alpha\in\lambda^{+}\forall\xi\in \mathrm{c}\mathrm{o}\mathrm{f}(\lambda)f_{\alpha}(\xi)>\sup\{\rho_{\xi’}|\xi’<\xi\}$$\gamma$$\langle f_{a} :$\forall\alpha<\alpha’$$\exists$\alpha\in\lambda^{+}\rangle$$A$$C\subseteq\lambda^{+}$$GS(\lambda)$$\exists A\subseteq\gamma$$\sigma<\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\nu\in(\sigma,\mathrm{c}\mathrm{o}\mathrm{f}(\lambda))fa(\nu)<f_{\alpha’}(\nu)$$\forall\alpha\in C$$\mathrm{c}\mathrm{o}\mathrm{f}(\alpha)>\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\lambda$$\alpha$49
DOUGLAS R. BURKE AND YO MATSUBARA

2.

3.

regular implies AD. [She82]
impliesand singular implies

1.

Remarks.
$\mathrm{A}\mathrm{D}(\lambda)$$\lambda$.

$\mathrm{S}\mathrm{h}(\lambda)$$\lambda$$\mathrm{G}\mathrm{S}(\lambda)$$(\lambda)$$\mathrm{S}\mathrm{h}(\lambda)$. [Cum97]. [CFM]
implies
is consistent (it is consistent to have
5. It is not known if
] .
and
[

4.

$\mathrm{A}\mathrm{D}(\lambda)$$\square _{\lambda}^{*}$$\exists\lambda\urcorner \mathrm{S}\mathrm{h}(\lambda)$$\exists\lambda$$\neg \mathrm{G}\mathrm{S}(\lambda)$$\neg \mathrm{A}\mathrm{D}(\lambda)$$)$

6. Shelah has proved that there is a scale for all singularand that$\lambda$

the set of good points is stationary for all scales ([HJS86]; also see[Cum97] . Shelah also gives an example of a model with no good
. Another example of a model with no good scale
scale, where they show
is given by Foreman and Magidor in [, implies
a version of Chang’s Conjecture,
there is no good scale on .
The proof of the following theorem is essentially the same as Theorem
2.3.
Theorem 3.4. Assume.
and is a pre-saturated filter on
Then.
Theorem 3.5. Suppose is a cardinal preserwing filter onand. Then.
$)$$([\mathrm{H}\mathrm{J}\mathrm{S}86])$$\mathrm{F}\mathrm{M}9\eta$$(\aleph_{\omega+1},.\aleph_{\omega})arrow(\aleph_{1}, \aleph_{0})$$\aleph_{\omega}$$Sh(\lambda)$$\mathcal{F}$$\mathcal{P}(\lambda)$$S_{\lambda}\in \mathcal{F}$$P(\lambda)$$\mathcal{F}$$AD(\lambda)$$S_{\lambda}\in \mathcal{F}$

Proof. We will use Shelah’s method of proof of Theorem 2.1 (page 440,

. Suppose. So we get
be generic with
$j:Varrow(M, E)\subseteq V[G]$ with
(we collapse the well-foundedpart of $M$ ), and, and. Work in $M$ :we writewhere theare increasing, continuousand. So ifis unbounded, thensuch thatis unbounded in . Now work in $V[G]$ : we havesuch thatis unbounded in . So there is a fixed
and an unboundedsuch thatis unboundedin . Letbe such thathas order type . Note that, andare all in $M$ . Now work in $M$ : The set
is unbounded in }{is a family
many non-empty pairwise disjoint subsets of. But[She82] . Let (
. Let
$)$

$a_{\alpha}$:$\alpha\in\lambda^{+}\rangle$,$\langle f_{\alpha} :\alpha\in\lambda^{+}\rangle$$G\subseteq \mathcal{F}^{+}$$S_{\lambda}\not\in \mathcal{F}$witness$AD(\lambda)$$P(\lambda)\backslash S_{\lambda}\in G$$\lambda^{+}\subseteq M$$P^{V}(\lambda)\subseteq M$$\lambda=\bigcup_{\alpha\in \mathrm{c}\mathrm{o}\mathrm{f}}(1^{\lambda}|)Aa$$|A_{\alpha}|<|\lambda|$$a\cap A_{\alpha}$$A_{a}’ \mathrm{s}$$a\subseteq\lambda$$\exists\alpha<\mathrm{c}\mathrm{o}\mathrm{f}(|\lambda|)$$\forall\alpha\in\lambda^{+}\exists\beta\in$$\lambda$$\mathrm{C}\mathrm{o}\mathrm{f}(|\lambda|)^{M}$$\lambda$$a_{a}\cap A_{\beta}$$A\subseteq\lambda^{+}$$\alpha_{0}\in A$$\lambda$$\langle a_{\alpha} :$M\models \mathrm{c}\mathrm{o}f(\lambda)\neq \mathrm{c}\mathrm{o}f(|\lambda|)$\alpha\in\alpha_{0}\rangle,A_{h}$$(\forall\alpha\in A)a_{\alpha}\cap A_{h}$$A\cap\alpha_{0}$$\lambda$$f_{\alpha_{0}}$$(a_{a}\cap A_{\beta_{\mathfrak{c}1}})\backslash f_{a_{0}}(\alpha)|\alpha<\alpha_{0}\ a_{a}\cap A_{h}$$\mathrm{o}\mathrm{f}|\lambda|$$|A_{h}|<|\lambda|$$\beta_{0}$$\lambda$$A_{\beta 0}$, contradiction.

$\square$

As in section 2, these two theorems have the following three corollary’s:, andCorollary 3.6. Assume $AD(\lambda)[Sh(\lambda)],$ $\kappa=p^{+},$. Then there is no cardinal preserving [poe-saturated]$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)<\kappa$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\neq \mathrm{c}\mathrm{o}\mathrm{f}(\rho)$

filter on$\mathcal{P}_{\kappa}\lambda$.

50

THE EXTENT OF STRENGTH IN THE CLUB FILTERS

Corollary 3.7. AssumeThen the club filter On

and.
is not cardinal poeseiwing [pm-satumted].

$AD(\lambda)[sh(\lambda)\mathit{1},$$P_{\kappa}\lambda$

Corollary 3.8. Assume$\kappa=\rho^{+}\geq\aleph_{2}$is a regular limit cardinal$AD(\lambda)[sh(\lambda)]\mathrm{Z}\kappa$

and. Then the club filter on[poe-saturated].
$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\neq\kappa$$\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)\geq\kappa$$P_{\kappa}\lambda$

is not cardinal preseming

Theorem 3.9. Assume coand there is a good scale on .
Then there is no weakly pre-saturated filter on.
$f(\lambda)<\kappa$$\lambda$$\mathcal{P}_{\kappa}\lambda$

Proof. Suppose not. So there is $j:Varrow M\subseteq V[G]$ such that isstill a cardinal of $V[G],$ $M$ is well-founded,
, and
with
. Let
be a good scale on . So there is a
club
such thatandimplies is good for. Let
Note that(see [BM97]) and
so $\rho\in j(C)$ . Since.
so in $M$ , there is ansuch that $\sup(A)=\rho$ and
such
that
from and
implies.
Now work in $V[G]$ and repeat an argument from [Cum97]. For eachinchooseffom andsuch that.
Do this so
impliesand.
For eachsuch thatbeyond . Since
is regular there is an unbounded
and
fixed
such that. Let. But then if
are from $B$ , then. Hencemust
be collapsed in $V[G]$ .
$\lambda_{V}^{+}$$\mathcal{P}^{V}(\lambda)\subseteq M$$j(\kappa)>\lambda$$\langle f_{\alpha} :$\langle f_{\alpha} :\alpha\in\lambda^{+}\rangle$\alpha\in\lambda^{+}\rangle$$\alpha\in C$$C\subseteq\lambda^{+}$$\lambda$$\mathrm{c}\mathrm{o}\mathrm{f}(\alpha)>\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$p= \sup j’’\lambda^{+}$$M\models \mathrm{c}\mathrm{o}\mathrm{f}(\rho)\geq\lambda^{+}>\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$A\subseteq p$$A$$\alpha$$\rho<j(\lambda^{+})$$V[G]\models \mathrm{c}\mathrm{o}\mathrm{f}(\beta)=\lambda^{+},$$\alpha_{1}<\alpha_{2}$$\mathrm{c}\mathrm{p}(j)=\kappa$$\exists\sigma<\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\nu\in(\sigma, \mathrm{c}\mathrm{o}f(\lambda))$$j(f)_{\alpha_{1}}(\nu)<j(f)_{a_{2}}(\nu)$$\alpha$$\lambda^{+}$$\beta_{\alpha}<\delta_{a}$$\alpha_{1}<\alpha_{2}$$A$$\gamma_{a}\in\lambda^{+}$$\delta_{\alpha_{1}}<\beta_{a_{2}}$$\alpha\in\lambda^{+}\exists\sigma_{\alpha}<\mathrm{c}\mathrm{o}\mathrm{f}(\lambda)$$\beta_{a}<j(\gamma_{\alpha})<\delta_{a}$$\sup\{\beta_{\alpha}|\alpha\in\lambda^{+}\}=\rho$$j(f)_{\beta_{\alpha}}<j(f)_{j(\gamma_{\alpha}})<j(f)_{\delta_{\alpha}}$$\lambda^{+}$$\sigma_{a}$$\sigma_{1}$$\alpha_{1}<\alpha_{2}$$B\subseteq\lambda^{+}$$\forall\alpha\in B\sigma_{a}=\sigma_{1}$$\overline{\sigma}=\max(\sigma, \sigma_{1})$$f_{\gamma_{\alpha_{1}}}(\overline{\sigma}+1)<f_{\gamma_{a_{2}}}(\overline{\sigma}+1)$$\lambda^{+}$$\square$

Precipitousness is ruled out under certain conditions by the following
theorem of Matsubara and Shioya.

Theorem 3.10. [MS] Ifon
is nowhere precipitous.

$\lambda^{<\hslash}=2^{\lambda}$

and$P_{\kappa}\lambda$$2^{<\kappa}<2^{\lambda}$, then the club filter

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$\omega_{1}$DEPARTMENT OF MATHEMATICS, UNIVERSITY OF NEVADA, LAS VEGAS, NV
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SCHOOL OF INFORMATICS AND SCIENCES, NAGOYA UNIVERSITY, NAGOYA,
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