数理解析研究所講究録1095 巻 1999 年 1-15

1Partition properties on$P_{\kappa}\lambda$

Shizuo Kamo*
University of Osaka Prefecture, Sakai Osaka Japane-mail:kamo@center.osakafu-u. ac.jp(カロ$f”\nwarrow^{\backslash }$$\neq_{\backslash }.)$$\not\leqq \mathrm{a}F$

Abstract

M. Magidor [9] showed that if$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

holds then$\kappa$

is -ineffable.
$\lambda$$\mathrm{I}\mathrm{n}\cdot[6]$,we showed that, under some additional assumption, the reverse implication also
holds. The main purpose of this paper is to improve this result. We will provethat if$\kappa$

is$\lambda^{<\kappa}$-ineffable then$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

holds. Also, using a similar technic in

the proof of the above result, we prove that if
is a normal ultrafilter on$\mathcal{P}_{\kappa}\lambda$

1$\mathcal{P}_{\kappa}\lambda$

is -supercompact then there$\lambda$with the partition property. This result is somevariation of a Menas’s result: If
normal ultrafilters on$\kappa$$\kappa$is$2^{\lambda^{<\kappa}}$-supercompact, then there are$2^{2^{\lambda}}<\kappa$

with the partition property.

Introduction
There are several combinatorial properties related with supercompactness. Partition

properties and ineffabilities are some of these properties. In fact, Menas [11, Theorem
3] proved that ifon$P_{\kappa}\lambda$

then$\kappa$$\kappa$is$2^{\lambda^{<\kappa}}$-supercompact then there are$2^{2^{\lambda^{<\kappa}}}$many normal ultrafilterswith the partition property, and Magidor [9] proved that if

is -ineffable and that if$\lambda$

is -ineffable, for all$\lambda$$\kappa$$\lambda\geq\kappa$then$\kappa$$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

holds

is supercompact.

In the above results, it is natural to ask whether -ineffability imply the partition$\lambda$–*The author is partially supported by-Aid for Scientific Research (No. 09640288), Ministry of Education, Science, Sports and Culture, Japan$\mathrm{G}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{t}-\mathrm{i}\mathrm{n}$2property. In [6], we proved that, under some additional assumption of , if$\kappa$

ineffable then$\kappa$

is$\lambda^{<\hslash_{-}}$

holds. The main purpose of this paper is to improve this$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

result. We prove
If

Theorem 4.3$\kappa$

is$\lambda^{<\hslash}$-ineffable, then$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

holds.

In the proof of Theorem 4.3, we bollow an idea of how to use the notion of presubtlity
which is appeared in Kunen and Pelletier’s paper [8]. (The notion of presubtlity does
not appear in this paper.) Also, using a similar technic in the proof of Theorem 4.3,
we prove

Theorem 5.1

If

is -supercompact then there is a normal ultrafilter on$\lambda$$\kappa$$\mathcal{P}_{\kappa}\lambda$

withthe partition property.
This is a variation of the ab

Menas result.
$\iota_{\mathrm{S}}$$o\mathrm{v}\mathrm{e}$The paper consists five sections. In the next section, we give some notations anddefinitions which we will be used the following sections. Section 3 will be devoted
to give some lemmas which will be used to prove Theorem 4.3. Theorem 4.3 will be
proved in section 4. Theorem 5.1 will be proved in section 5.

Notations and definitions

2

We use standard
paper,$\mathcal{P}_{\kappa}\lambda$-combinatorial terminologies (e.g., see [7]). Throughout thisdenotes a regular uncountable cardinal. Let$\kappa$

denotes the dual filter anddenotes$\mathcal{I}\cap \mathcal{P}(S’)$$f^{-1}X\in \mathcal{I}\}$

Let

Let
a

$Y$

$T$

,

$Q_{x}$$P(S)\backslash \mathcal{I}$be an ideal on a set S.

. For any subset$f_{*}(\mathcal{I})$$S’\subseteq S,$$\mathcal{I}^{*}$$\mathcal{I}\lceil S’$

denotes the ideal $\{X\subset T|$.
$\kappa\subset A$

denotes the set

be a subset of$y\in \mathrm{Y}$denotes the set. For any function $f$ : $Sarrow T,$

be a set such that$A$$x\in \mathcal{P}_{\kappa}A$on$\mathcal{I}^{+}$$\mathcal{I}$such that$\mathcal{P}_{\kappa}A$.

$x\subset y$

$Y$

.

.

$\mathcal{P}_{\kappa}$$\mathcal{P}_{1x\cap\kappa}|x$A denotes the set $\{x\subset A||x|<\kappa\}$ . For each

. For any

$x,$ $y\in \mathcal{P}_{\kappa}A,$$x\prec y$is said to be unbounded, if for any$Y$is called, if$\mathrm{a}\underline{\mathrm{c}\mathrm{l}\mathrm{u}\mathrm{b}}$$Y$means that $x\in Q_{y}$ .
$x\in \mathcal{P}_{\kappa}A$

there exists

is unbounded and closed under3$\subset$-increasing chains with lengthclub$C\subset \mathcal{P}_{\kappa}A$

A function

ideal

on

$\mathcal{I}$. Let us denote by$\mathcal{P}_{\kappa}A$

the set of all non-stationary subsets of$\mathrm{N}\mathrm{S}_{\kappa,A}$

is normal, if it contains all bounded subsets and, for any

is known [11] that

: $Xarrow A$ , there exists an $a\in A$ such that$f$is the smallest normal ideal on$\mathrm{N}\mathrm{S}_{\kappa,A}$

For each function$Q_{x}(\tau(t)\in Q_{x})\}$:$\tau$. For each$p$:$\mathcal{P}_{\kappa}A$$\mathcal{P}_{\kappa}A$.

$\mathrm{c}1(\tau)$:$P_{\kappa}Barrow \mathcal{P}_{\kappa}A$$A\cross Aarrow 7^{\mathit{2}_{\kappa}}A$, and$f^{-1}\{a\}\in \mathcal{I}^{+}$. It$\{x\in \mathcal{P}_{\kappa}A|\forall t\in$denotes the set $\{x\in P_{\kappa}A|$. It is known [10] that, for any$\tau$$X\in \mathcal{I}^{+}$.

denotes the set$\mathrm{c}1(\tau)$: $A\cross Aarrow P_{\kappa}A,$$\tau$

a club if and only if there exists a, the function,$\mathcal{P}_{\kappa}Aarrow \mathcal{P}_{\kappa}A$$\beta\in x(\tau(\alpha, \beta)\subset x)\}$$B\supset A$. $Y$ is said to be stationary, if $X\cap C\neq\phi$ , for any: $Yarrow A$ is said to be regressive, if $f(x)\in x$ holds, for all $x\in Y$ . An$f$

any regressive function$\forall\alpha,$$<\kappa$$X$$X\subset \mathcal{P}_{\kappa}A,$such that$\mathrm{c}1(\tau)\subset X$

contains. For any

which is defined by $p(x)=x\cap A$ is called the

projection.
Let$Y\subset \mathcal{P}_{\kappa}A$

any function$|f$“$[H]^{2}|=1$.

$[Y]^{2}$denotes the set { $(x,y)\in Y\cross Y|x\subset y$ and $x\neq y$ }. For: $[Y]^{2}arrow 2$ , a subset $H$ of$f$. We say that$Y$

is said to be homogeneous forhas the partition property, if for any

exists a stationary subset $H$ ofineffable (

$Y$

$Y$

$Y$

$f$

ineffable), if for any$s_{x}\subset x$, if: $[Y]^{2}arrow 2$ , there$f$such that $H$ is homogeneous for .

$\underline{\mathrm{a}\mathrm{l}\mathrm{m}\mathrm{o}\mathrm{s}\mathrm{t}}$$f$is said to be(for $x\in Y$ ), there exists an $S\subset A$ suchthat $\{x\in Y|s_{x}=S\cap x\}$ is stationary (unbounded). Set$\mathrm{N}\mathrm{P}_{\kappa,A}=$$\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,A}=${$X\subset \mathcal{P}_{\kappa}A|X$

does not have the partition property},{$X\subset \mathcal{P}_{\kappa}A|X$

is not ineffable}, and$\mathrm{N}\mathrm{A}\mathrm{I}\mathrm{n}_{\kappa,A}=${

It is known [3] that

We say that$\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,A}$$X\subset \mathcal{P}_{\kappa}A|X$$\mathrm{N}\mathrm{P}_{\kappa,A},$$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, A)$$\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,A}$, andholds, if

is a proper ideal, and that

It is known that, for any

is not almost ineffable}.

$B\supset A$,$\kappa$$\mathrm{N}\mathrm{A}\mathrm{I}\mathrm{n}_{\kappa,A}$$\mathrm{N}\mathrm{P}_{\kappa,A}$

are normal ideals on

is a proper ideal and thatis almost$A$-ineffable, if NAIn$\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,A}\subset p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,B})$, where$p$$\kappa,A$$\kappa$$\mathcal{P}_{\kappa}A$

is

$A$

.

-ineffable, if

is a proper ideal.

denotes the projection

4

from

to

$\mathcal{P}_{\kappa}B$Let$P_{\kappa}A$.

be an ultrafilter on$U$a nornal ideal.

$U$$\mathcal{P}_{\kappa}A$. We say that$U$

is normal, if the dual ideal of

has the partition property, if for any $X\in U$ and any

there exists $Y\in U$ such that $Y\subset X$ and

$Y$

$U$

is

: $[X]^{2}arrow 2$ ,$f$is homogeneous for .
$f$

Several Lemmas

3

In this section, we will state some lemmas which will be used in the next section.
From now on,$\lambda$

denotes a cardinal greater than or equal to .
$\kappa$

The -Shelah property

3.1$\lambda$

The -Shelah property was first introduced by Carr [3]. A subset$\lambda$

Shelah property, if for any$f_{x}$

has the$X\subset \mathcal{P}_{\kappa}\lambda$: $xarrow x$ (for $x\in X$ ), there exists a function$f$:$\lambdaarrow\lambda$

such that$\forall x\in \mathcal{P}_{\kappa}\lambda\exists y\in X$

Set$\mathrm{N}\mathrm{S}\mathrm{h}_{\kappa,\lambda}=$$\mathrm{N}\mathrm{S}\mathrm{h}_{\kappa,\lambda}$$\mathrm{N}\mathrm{S}\mathrm{h}_{\kappa,\lambda}${($x\subset y$$X\subset \mathcal{P}_{\kappa}\lambda|X$

is a normal ideal onand$f_{x}=f\square x$).

does not have the Shelah property}. It is known that$\mathcal{P}_{\kappa}\lambda$

and$\mathrm{N}\mathrm{S}\mathrm{h}_{\kappa,\lambda}\subset \mathrm{N}\mathrm{A}\mathrm{I}\mathrm{n}_{\kappa,\lambda}$. We say thatis -Shelah, if$\lambda$$\kappa$

is a proper ideal.

The following two lemmas are due to Carr $[2, 3]$ .

Lemma 3.1 (Carr [2]){Lemma 3.2 (Carr [3])

If

Let
for

$S$

$S$

$x\in P_{\kappa}\lambda|x\cap\kappa$$\kappa$is$2^{\lambda^{<\kappa}}$

is an inaccesible $cardinal$ }-Shelah, thenbe an infinite set. A function $F$ from, if for any $T\subset S$ with cardinality$|S|$$S^{\omega}$$\in \mathrm{N}\mathrm{S}\mathrm{h}_{\kappa,\lambda}^{*}$

is -supercompact.
$\lambda$$\kappa$

to

$S$

.

$\square$$\square$is called an -Jonsson function$\omega$, it holds that $F$ “ $T\{v=S$ . Concerning this,Erd\"os-Hajnal (e.g., see [7, Theorem 23.13]) proved

Lemma 3.3 (Erd\"os-Hajnal) For any infinite set , there exists an -Jonsson func$S$

tion for$S$.

$\omega$$\square$5

Furthermore, we need

Lemma 3.4 (Johnson [4]) Let

and$\delta\leq\lambda$

$F$

an

$\omega$-Jonsson fun ction for .

$Then_{f}$$\delta$itholds that{$x \in \mathcal{P}_{\kappa}\lambda|F\int(x\cap\delta)^{\mathrm{t}v}$

is an$\omega$-Jonsson function for$x\cap\delta$}

How reflects cardinals on a certain set in

3.2$\in \mathrm{N}\mathrm{S}\mathrm{h}_{\kappa,\lambda}^{*}$.

$\square$$\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}$

In this subsection, we begin with an easy lemma. We left a proof to the reader.
Lemma 3.5{$\delta$

is a cardinal$x\in \mathcal{P}_{\kappa}\lambda|\mathrm{o}\mathrm{t}(x\cap\delta)i_{\mathit{8}}$Lemma 3.6

{

If

If

$\gamma$a cardinal}$\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}^{*}$

is a strong limit cardinal$x\in P_{\kappa}\lambda||x\cap\gamma|$

Proofthen it holds that$\leq\lambda_{J}$$\leq\lambda$.

$\square$, then it holds thatis a strong limit cardinal}$\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}^{*}$.

To get a contradiction, assume that$X=${

$x\in \mathcal{P}_{\kappa}\lambda||x\cap\gamma|$

For each $x\in X$ , takethere is an$\alpha<\gamma$

is not a limit$\alpha_{x}\in x\cap\gamma$such that$\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{a}1$}$\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}^{+}$$|x\cap\gamma|\leq 2^{|x\cap\alpha_{x}|}$.

. Since$\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}$

is normal,

such that.

$X’=\{x\in X|\alpha_{x}=\alpha\}\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}^{+}$For each $x\in X’$ , take an injection$a_{x}=$

Since{$(\xi,$$f_{x}$, there exists an$x\cap\gammaarrow \mathcal{P}(x\cap\alpha)$$A\subset\gamma\cross\alpha$such that$Y=\{x\in X’|a_{x}=A\cap X\cross x\}\in \mathrm{N}\mathrm{S}_{\kappa,\lambda}^{+}$

Define$f$:$\gammaarrow \mathcal{P}(\alpha)$

$Y$

.

by

$f(\xi)=\{\eta<\alpha|(\xi,\eta)\in A\}$Since

and setand $f_{x}(\xi)=\eta$ }.

$\eta)\in x\cross x|\xi<\gamma$$X’\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}^{+}$:.

is unbounded, it holds that$f$is an injection. This contradicts thatstrong limit cardinal.
The following lemma is due to Abe [1].

$\gamma$is a$\square$6

Lemma 3.7 (Abe [1]) Let$\delta$$\gamma,$be cardinals such that$2^{\gamma}=\delta\leq\lambda.$

Then, it hol& $that$.

$\{x\in \mathcal{P}_{\kappa}\lambda|2^{|x\mathrm{n}|}\gamma=|x\cap\delta|\}\in \mathrm{N}\mathrm{S}\mathrm{h}_{\kappa,\lambda}^{*}$

A similar argment gives a proof of the next lemma. We left a proof to the reader.

If

Lemma 3.8$\gamma\leq\lambda\leq 2^{\gamma}$: then it holds that.

$\square$$\{x\in P_{\kappa}\lambda||x|\leq 2^{1x\cap}\gamma|\}\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\lambda}^{*}$

3.3$P_{\kappa}$Let

A vasus$\theta=\lambda^{<\kappa}$

defineand$\mathcal{P}_{\kappa}\lambda^{<\kappa}$$p:\mathcal{P}_{\kappa}\thetaarrow \mathcal{P}_{\kappa}\lambda$$q:\mathcal{P}_{\kappa}\thetaarrow \mathcal{P}_{\kappa}\lambda$$q(y)=\cup h^{-1}y$the projection. Take a bijection$h:\mathcal{P}_{\hslash}\lambdaarrow\theta$

and

by

, for all$y\in \mathcal{P}_{\kappa}\theta$.

The lemmas of this subsection were appeared in $[5, 6]$ implicitly. Set$Y_{0}=${

Lemma 3.9
Proof$y\in \mathcal{P}_{\kappa}\theta|p(y)=q(y)$$Y_{0}\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta}^{*}$

and$Q_{p(y)}=y$}.

It suffices to show that$\{y\in \mathcal{P}_{\kappa}\theta|p(y)\subset q(y)\}\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta}^{*}$(2)$\{y\in \mathcal{P}_{\kappa}\theta|q(y)\subset p(y)\}\in \mathrm{N}\mathrm{h}_{\kappa}^{*},\theta$

(4)

“

.

(1)

(3)

$h$

,
’

$\{y\in \mathcal{P}_{\kappa}\theta|h" Q_{p}(y)\subset y\}\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa}^{*},\theta$$\{y\in \mathcal{P}_{\hslash}\theta|y\subset h" Q_{p}(y)\}\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta}^{*}$’.

These are proved by similar argments. We only deal (3). To get a contradiction,

assume that
$Y=\{y\in p_{\kappa}\theta|h" Q_{p(y)}\not\subset y\}\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa}^{+},\theta$

For each $y\in Y$ , Take
$T\subset\lambda$$t_{y}\in Q_{p(y)}$such thatsuch that$Z=\{y\in Y|T\cap y=t_{y}\}\in \mathrm{N}\mathrm{S}_{\kappa,\theta}^{+}$.

.

$h(t_{y})\not\in y$. Since$Y\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta}^{+}$, there exists

7

Claim 1$|T|<\kappa$.

Proof of Claim 1

Suppose not. Take an injection$C=\{y\in \mathcal{P}\kappa\theta|f"(y\cap\kappa)\subset y\}$

Since

$C$

is a club of$|y\cap\kappa|\leq|y\cap T|$$\mathcal{P}_{\kappa}\theta$:$f$$\kappaarrow T$

and set.

, there exists a $y\in C\cap Z$ . Since $y\in C$ , we have that

. Since $y\in Z$ , we have that $|y\cap T|=|t_{y}|<|y\cap\kappa|$ . This is a desired

contradiction.

QED of Claim 1

By Claim 1, set $\alpha=h(T)$ . Since$T\cup\{\alpha\}\subset y$$A\subset \mathcal{P}_{\kappa}\lambda$

V :$\tau$Let$X\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta})^{+}$“$a_{p}(y)$.

$\square$such that.

$\mathcal{P}_{\kappa}\lambdaarrow \mathcal{P}_{\kappa}\lambda\exists x\in X\cap \mathrm{c}1(\tau)(a_{x}=A\cap Qx)$

. Then, since$b_{y}\subset y$$y\in Y_{0}$$Y\in \mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta}^{+}$. For each $y\in Y$ , set, it holds that, for all $y\in Y$ ,So, there exists a$B\subset\theta$such that$Y’=\{y\in Y|b_{y}=B\cap y\}\in \mathrm{N}\mathrm{S}_{\kappa,\theta}^{+}$

Set $A=h^{-1}B$ . We claim that

Define $\tau’=h\tau h^{-1}$ :is easy to check that
Lemma 3.11$\thetaarrow\theta$$A$. Sinceis as required. To show this, let$Y’\in \mathrm{N}\mathrm{S}_{\kappa,\theta}^{+}$$p(y)\in X\cap \mathrm{c}1(\tau)$

Suppose that.

, there exists a$\tau$:$\mathcal{P}_{\kappa}\lambdaarrow \mathcal{P}_{\kappa}\lambda$$y\in Y’\cap \mathrm{c}1(\tau’)$$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$.

. Then, it

and $a_{p(y)}=A\cap Q_{p(y)}$ .

$\square$fails. Then, it holds that$\{x\in \mathcal{P}_{\kappa}\lambda|\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}*(x\cap\kappa, x)fails\}\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta})^{*}$

Proof$t_{y}$. Then, for any $a_{x}\subset Q_{x}(x\in X)$ , there exists

Set $Y=p^{-1}X\cap Y_{0}$ . By Lemma 3.9,Proof$b_{y}=h$is unbounded, we can take a $y\in Z$ such that

. Then, $t_{y}=T$ and $h(t_{y})=\alpha\in y$ . This contradicts the choise ofLemma 3.10

an

$Z$

.

To get a contradiction, assume that$X=$$/\mathrm{L}\mathrm{e}\mathrm{t}X’=${$x\in \mathcal{P}_{\kappa}\lambda|\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}*(x\cap\kappa,$$x)$

holds}$\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\hslash,\theta})^{+}${ $x\in X|x\cap\kappa$ is inaccesible}. By Lemma 3.1,.
$X’\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\hslash,\theta})^{+}$

.

Since

8

$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

fails, there exists a function$\forall H\in \mathrm{N}\mathrm{S}_{\kappa,\lambda}^{+}$

3.10, there exists an$\forall\tau$

:

$f$

).

and $e_{x}<2$ such that

$H_{x}\in \mathrm{N}\mathrm{S}_{\hslash\cap x}^{+},x$$H\subset \mathcal{P}_{\kappa}\lambda$

such that$[P_{\kappa}\lambda]^{2}arrow 2$( $H$ is not homogeneous for

For each $x\in X’$ , take$(^{*})$

:

$f$

$f$

“

$[H_{x}]^{2}=\{e_{x}\}$. By Lemma

and $e<2$ such that$\mathcal{P}_{\kappa}\lambdaarrow \mathcal{P}_{\kappa}\lambda\exists x\in X’\cap \mathrm{c}1(\tau)$( $H_{x}=H\cap Q_{x}$ and $e_{x}=e$ ).

It is easy to check that $H$ is homogeneous for . It suffices to complete the proof to show$f$that $H$ is stationary. So, let$x\subset\tau(x)$. Then, by$x\in \mathrm{c}1(\tau)$, it holds that$(^{*})$$C$

be a club of, there exists an$C\cap Q_{x}$is club in$\mathcal{P}_{\kappa}\lambda$. Take a function :$\tau$

such thatsuch that $H_{x}=H\cap Q_{x}$ . Since$x\in X’\cap \mathrm{c}1(\tau)$$Q_{x}$$\mathcal{P}_{\kappa}\lambdaarrow C$. So, it holds that$\phi\neq H_{x}\mathrm{n}c\cap Qx\subset H\cap C$.

$\square$

Proof of Main Theorem

4

In this section, we prove the main theorem. In the proof, we will use some known
results. The next lemma is well-known. But, I don’t know who established it.

Lemma 4.1 (folklore)$\theta$If$\kappa$$is<\lambda$-supercompact and$\lambda$

is$\theta_{- \mathit{8}u}percompact$, then-supercompact.

$\kappa$is$\square$

The next lemma was appeared in [5].

{Lemma 4.2
where$pdenote\mathit{8}$Theorem 4.3$\theta=\lambda^{<\kappa},$

DefineIf$\kappa$

the projection fromis$\lambda^{<\kappa}$$\delta_{i}$$p:P_{\kappa}\thetaarrow \mathcal{P}_{\kappa}\lambda$(for $i<\omega$ ) by$\mathcal{P}_{\kappa}\lambda^{<\kappa}$-ineffable, then

$F$

to

$\mathcal{P}_{\kappa}\lambda$$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

is$\kappa$$\lambda^{<\kappa}$$\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa},\lambda<\kappa)^{*}\mathrm{Z}$.

$\square$holds.

-ineffable and$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

the projection, and the largest strong limit cardinal$\delta$$\delta_{0}=\delta$

Take -Jonsson functions$\omega$is almost $x- ineffable$ }To get a contradiction, assume that

Proof

Let

$x\in \mathcal{P}_{\kappa}\lambda|x\cap\kappa$

and

and

$F_{i}$$\delta_{i+1}=2^{S_{i}}$. Let $n<\omega$ be such that(for $i\leq n$ ) for$\lambda$

and$\delta_{i}$fails.
$\leq\lambda$$\delta_{n}\leq\lambda<\delta_{n+1}$.
.

(for $i\leq n$ ), respectively.

9

Let $X$ be the set of all$x\in \mathcal{P}_{\kappa}\lambda$

which satisfies(1)$x\cap\kappa$

is an inaccessible cardinal,(2)$x\cap\delta$

is a strong limit cardinal cardinal,(3)$\mathrm{o}\mathrm{t}(x\cap\delta_{i+1})=2^{\mathrm{o}\mathrm{t}(x\cap}\delta_{i})$(4)$F\lceil x$and$F_{i}\square (x\cap\delta_{i})^{\omega}$, for all $i<n$ and$\mathrm{o}\mathrm{t}(x)\leq 2^{\mathrm{o}\mathrm{t}(x\cap s}n)$are -Jonsson functions for$\omega$

$x$

and

,

$x\cap\delta_{i}$, for $i\leq n$ , respec-tively,is almost x-ineffable,(5)$x\cap\kappa$(6)$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(x\cap\kappa, x)$fails.

By Lemmas 3.1, 3.4, 3.6, 3.7, 3.8, 3.11, and 4.2, it holds that
this, since

is -ineffable, we have that$\theta$$\kappa$$X\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta})^{+}$.

$X\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\kappa,\theta})^{*}$. By

Note that, by Lemma 3.2

and (5), it holds that(7)$x\cap\kappa$isClaim 2$x\cap\alpha$-supercompact, for all$\forall(x, y)\in[X]^{2}$

Proof of Claim 2(if$\alpha\in x\cap[\kappa, \delta)$, for all $x\in X$ .

then $x\prec y$ ).

$x\cap\delta_{n}\neq y\cap\delta_{n}$To get a contradiction, assume that there exists $(x, y)\in[X]^{2}$such that$x\cap \mathit{5}_{n}\neq y\cap\delta_{n}$and not $x\prec y$ .

By (3), it holds that $x\cap\delta\neq y\cap\delta$ . Since$|x\cap\delta|<|y\cap\delta|$(8). Since$2^{|x|}<|y\mathrm{n}\delta|$

and$|y\cap\kappa|$

and$|y\cap\delta|$$y\mathrm{n}\kappa\leq|x\mathrm{n}\delta|$$F\mathrm{r}(y\cap\delta)^{\omega}$is$\omega$-Jonsson, it holds thatis strong limit cardinals, we have that.

By this, and (7), and Lemma 4.1, it holds that$x\cap\kappa$

is x-supercompact.

But this contradicts that$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(x\cap\kappa, x)$fails.

We complete the proof by proving thatcases.
Case 1.

$\lambda=\delta_{n}$.

By Claim 2, it holds that

$X\in \mathrm{N}\mathrm{P}_{\kappa,\lambda}^{+}$

QED of Claim 2. The proof is divided into two10$\forall(x, y)\in[x]^{2}(x\prec y)$

In order to show that$a_{x}=$.

$X\in \mathrm{N}\mathrm{P}_{\kappa,\lambda}^{+}$

, let

$f$

:

$A\subset \mathcal{P}_{\kappa}\lambda$$\forall(x, y)\in[X’\cap A]^{2}(f(x, y)=0)$

So, $X’\cap A$ or:$g(\alpha)=\{$$\kappaarrow\kappa$$f_{0}(\alpha)=\mathrm{t}\mathrm{h}\mathrm{e}$, for all, take$\alpha<\kappa$

and

:

and

$\mathrm{o}\mathrm{t}(x\cap\delta_{n})arrow x\cap\delta_{n}$$s_{x}=\pi_{x}$To show that$a_{x}=$$s_{x}$$w^{x\cap\hslash}\circ\iota(x)(\subset x\cap\delta_{n})$$X\in \mathrm{N}\mathrm{P}_{\kappa,\lambda}^{+}$,, it holds that(for$\xi\leq f_{n+1}(\alpha)$) such that.

byis the order isomorphism, and“.

, let $f$ : $[X]^{2}arrow 2$ . As in the case 1, set{ $t\in Q_{x}|t\in X$ and $f(t,x)=0$ }, for $x\in X$ .

$X\in p_{*}(\mathrm{N}\mathrm{I}\mathrm{n}_{\hslash,\theta})^{+}$$X’=$$\leq g(\alpha)$$\beta$and $i\leq n$ .

$\xi<\forall\eta\leq f_{n}+1(\alpha)(w_{\xi}^{\alpha}\neq w_{\eta}^{\alpha})$$\pi_{x}$$\beta$$\alpha$$f_{n}(x\cap\kappa)=\mathrm{o}\mathrm{t}(x\cap\delta_{n})\leq \mathrm{o}\mathrm{t}(x)\leq f_{n+1}(x\cap\kappa)$$w_{\xi}^{\alpha}\subset f_{n}(\alpha)$

For each $x\in X$ , defineSincesuch that . is not -supercompact, if such exists
otherwise,$\mathrm{o}\mathrm{t}(x\cap\delta)\leq g(x\cap\kappa)\leq \mathrm{o}\mathrm{t}(x)$$f_{0}(x\cap\kappa)=\mathrm{o}\mathrm{t}(x\cap\delta)$$\pi_{x}$and $\forall(x, y)\in[X’\backslash A]^{2}(f(x,y)=1)$ .

largest strong limit cardinal

For any $x\in X$ , sinceV$\beta\geq\alpha$,$f_{i+1}(\alpha)=2^{f(}:\alpha)$$\alpha<\kappa$. It is easy to check that(for $i\leq n+1$ ) by

the smallest$0$

For each$X’\in \mathrm{N}\mathrm{S}_{\kappa,\lambda}^{+}$.

$f_{i}$$g,$.

is as required.

$X’\backslash A$$\delta_{n}<\lambda$

Define(for $x\in X$ ) bysuch that$\mathcal{P}_{\kappa}\lambdaarrow \mathcal{P}_{\kappa}\lambda\exists x\in X\cap \mathrm{C}1(_{\mathcal{T}})(a=A\cap Qxx)$

Set $X’–\{x\in X|a_{x}=A\cap Q_{x}\}$ . Note that

Case 2.

$a_{x}\subset Q_{x}${ $t\in Q_{x}|t\in X$ and $f(t,$ $x)=0$ }.

Then, by Lemma 3.10, there exists an$\forall\tau$: $[X]^{2}arrow 2$ . Define, there exist$S\subset\delta_{n}${ $x\in X|s_{x}=S\cap x$ andand$A\subset \mathcal{P}_{\kappa}\lambda$$a_{x}=A\cap Q_{x}$}such that$\in \mathrm{N}\mathrm{S}_{\kappa,\lambda}^{+}$

.

.

11

Claim 3$\forall(x, y)\in[X’]^{2}(x\cap\delta_{n}\neq y\cap\delta_{n})$Proof of Claim 3$(x, y)\in[X’]^{2}$.

To get a contradiction, assume thatand$x\cap\delta_{n}=y\cap\delta_{n}$.

Note that $s_{x}=s_{y}$ . Set $\alpha=x\cap\kappa(=y\cap\kappa),$
holds that $\xi<\eta$ . Since$x\cap\delta_{n}=y\cap\delta_{n}$$\xi=\mathrm{o}\mathrm{t}(x),$, it holds that$\eta=\mathrm{o}\mathrm{t}(y)$$\pi_{x}=\pi_{y}$. Since $|x|<|y|$ , it

. By this, since$w_{\xi}^{\alpha}\neq w_{\eta}^{\alpha}$,

we have that$s_{x}=\pi_{x}$“$w_{\xi}^{\alpha}\neq\pi_{y}$“$w=s_{y}\eta\alpha$.

This is a contradiction.

QED of Claim 3

By Claims 2 and 3, it holds that$\forall(x, y)\in[X’]^{2}(x\prec y)$

So, $X’\cap A$ or$X’\backslash A$

Corollary 4.4
Proof$\lambda<\mu$.

is as a desired homogeneous set for .

Let

Assume that$f$$\kappa\leq\lambda\leq\mu$. If$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \mu)$$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \mu)holds_{f}$holds. The case

. By a result of Magidor [9], it holds thatJohnson [ , it holds that
$4_{\mathrm{J}}^{1}$$\lambda^{<\kappa}\leq\lambda^{+}\leq\mu$. So,$\kappa$$\kappa$isthen$\lambda=\mu$$\square$$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

holds.

is trivial. We assume thatis -ineffable. Then, by a result of$\mu$$\lambda^{<\kappa}$-ineffable. So,$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}^{*}(\kappa, \lambda)$

holds.
$\square$5

Normal ultrafilters with the partition property
Concerning the partition property of a normal ultrafilter on$\mathcal{P}_{\kappa}\lambda$, Solovay (see Menas[11] proved the existence of a normal ultrafilter without the partition property under$)$

the assumption of that the existence of a certain large cardinal greater than . After$\kappa$

Solovay established this result, Kunen (see Kunen-Pelletier [8]) improved his results,
and proved that the existence of a normal ultrafilter without the partition property
implies the existence of a certain large cardinal above . On the other hand, Menas$\kappa$[11] proved that there exist$2^{2^{\lambda}}<\kappa$

normal ultrafilters with the partition property, under

12

the assumption that

If

Theorem 5.1$\kappa$is$2^{\lambda^{<\kappa}}$

supercompact. In this section, we proveis -supercompactf then there exists a normal ultarfilter on$\mathcal{P}_{\kappa}\lambda$$\lambda$$\kappa$with the partition property.
The proof will be done by a similar, but different argment as in the proof of Menas.
We first reduce this theorem to a certain lemma (Lemma 5.4, below). Let$U$

the ultrapower of the universe by$U$

normal ultrafilter on$P_{\kappa}\lambda$. Denote by$M_{U}$

be a

. The

following two lemmas are due to Menas $[10, 11]$ .

IfLemma 5.2$\kappa$

is $\lambda- supercompact$, then there$exist\mathit{8}$a no rmal ultrafilter

$U$

on

$\mathcal{P}_{\kappa}\lambda$

such that$M_{U}\models\kappa$

is not -supercompact.
$\lambda$

Let

Lemma 5.3

$U$

be a normal ultrafilter on$P_{\kappa}\lambda$.

$Then_{J}$the following (a) and (b)are equivalent.
has the partition property.

(a)

$U$

(b)

There exists an $X\in U$ such that $\forall(x, y)\in[X]^{2}(x\prec y)$ .

By these results, the next lemma directly follows Theorem 5.1.
Lemma 5.4

Suppose that(1)

is not -supercompact.

$M_{U}\models\kappa$

there exists an $X\in U$ such that

$Then_{J}$(2)$\lambda$$x\prec y$

Proof, for all $(x, y)\in[X]^{2}$ .

Suppose that$U$

is a normal ultrafilter onlargest strong limit cardinal$n<\omega$for$\lambda$

be such thatand$\delta_{i}’ \mathrm{s}$$\leq\lambda$. Define$\delta_{n}\leq\lambda<\delta_{n+1}$, respectively. Define$x\in X_{0}$if and only if. Let$\delta_{i}$$F$$\mathcal{P}_{\kappa}\lambda$

which satisfies (1). Let(for $i<\omega$ ) byand

$X_{0}\subset P_{\kappa}\lambda$$F_{i}$$\delta_{0}=\delta$, and$\delta$be the$\delta_{i+1}=2^{\mathit{5}:}$. Let(for $i\leq n$ ) be -Jonnson functions

by, for any$\omega$$x\in \mathcal{P}_{\kappa}\lambda$,

13

is inaccesible and(3)$x\cap\kappa$(4)$\mathrm{o}\mathrm{t}(x\cap\delta_{i})$(5)$2^{|x\cap s_{1}}i=|x\cap\delta_{i+1}|$(6)$F_{i}\lceil(X\cap\delta_{i})^{\omega}$(7)$F[X^{\mathrm{t}}v$

is not x-supercompact,$x\cap\kappa$

is a cardinal, for $i\leq n$ ,, for $i<n$ and,$|x|\leq 2^{|x\cap \mathit{5}_{n}|}$is an -Jonnson function for$\omega$$x\cap\delta_{i}$, for $i\leq n$ ,is an -Jonnson function for .
$x$$\omega$

Note that $X_{0}\in U$ .

Claim 4and$(x, y)\in[X_{0}]^{2}$$y\cap\kappa$$x\cap\alpha$$x\cap\delta_{n}\neq y\cap\delta_{n}$then $x\prec y$ ).

and not $x\prec y$ .

is a strong limit cardinal, it holds that-supercompact, for all$x\cap\kappa$$x\cap\delta_{n}\neq y\cap\delta_{n}$To get a contradiction, assume that

Proof of Claim 4
Since

(if

$\forall(x, y)\in[X_{0}]^{2}$is$y\cap\alpha$$\alpha\in x\cap[\kappa, \delta)$$y\cap\kappa$is. Since$x\cap\kappa$is, we have that-supercompact, for all

By this and Lemma 4.1, since$y\cap\kappa\leq|x\cap\delta_{0}|$$\alpha\in[x\cap\kappa, y\cap\kappa)$$y\cap\alpha$.

-supercompact, for all$\alpha\in y\cap[\kappa, \delta)$, wehave that$x\cap\kappa$

is -supercompact, for all$\alpha$By this, since$x\cap\kappa$$\delta_{n}<\lambda$$x\cap\delta_{n}=y\cap\delta_{n}$. Define$g(\alpha)=\{$.

$x$

By Claim 4, in the casethat$\mathrm{o}\mathrm{t}(y\cap\delta))$is not -supercompact, it holds that $|y\cap\delta|\leq|x|$ . Sincea strong limit cardinal, we have that

This implies that$\alpha\in[x\cap\kappa,$. By this and (6), $x\cap\delta=y\cap\delta$ .

. This contradicts the assumption. QED of Claim 4$X=X_{0}$ satisfies (2). So, henceforth, we assumeand$f_{i}$:$\kappaarrow\kappa$

such that$\beta\geq\alpha$$\alpha$(for $i\leq n+1$ ) by

is not -supercompact,$\beta$$0$$f_{0}(\alpha)=\mathrm{t}\mathrm{h}\mathrm{e}$largest strongly limit cardinal$f_{i+1}(\alpha)=2^{f_{i}(\alpha)}$For each$\alpha<\kappa$$s_{\xi}^{\alpha}\subset f_{n}(\alpha)$

and:$\pi_{x}$$a_{x}\subset x\cap\delta_{n}$

, if

and

$\mathrm{o}\mathrm{t}(x\cap\delta_{n})arrow x\cap\delta_{n}$$a_{x}=\pi_{x}‘ \mathit{8}_{\mathrm{o}}(x\cap \mathrm{t}(\kappa x)$

Since$\langle s_{\xi}^{\alpha}|\xi<f_{n+1}(\alpha)\rangle$$s_{(}^{\alpha}\neq s_{\eta}^{\alpha}$

For each $x\in X_{0}$ , define$\pi_{x}$$\leq g(\alpha)$,, for $i\leq n$ .

, takeis$\lambda=\delta_{n},$$g:\kappaarrow\kappa$

the least,$|y\cap\delta|\leq|x\cap\delta|$$|y\cap\delta|$$\xi\neq\eta$$a_{x}$

such that.

byis the order isomorphism,.

, for all $x\in X_{0}$ , there exists an$X=\{x\in X_{0}|a_{x}=A\cap x\}\in U$ .

$A\subset\delta_{n}$such that

if suchotherwise,$\beta<\kappa$

exists,14

To get a contradiction, assume that there exists

We claim that $X$ satisfies (2).
$(x, y)\in$[X]2 such that not $x\prec y$ . By Claim 4, it holds thatSo, we have that

we have that$\pi_{x}=\pi_{y}$$s_{\xi}^{\alpha}\neq s_{\eta}^{\alpha}$

. Set

. So,

$\alpha=x\cap\kappa(=y\cap\kappa),$$a_{x}=\pi_{x}$“$\xi=\mathrm{o}\mathrm{t}(x),$$\eta=\mathrm{o}\mathrm{t}(y)$““$s_{\xi}^{\alpha}=\pi_{y}$$s_{\xi}^{\alpha}\neq\pi_{y}$$x\cap\delta_{n}=y\cap\delta_{n}$$s_{\eta}^{\alpha}=a_{y}$. Since$\xi\neq\eta$.

,. But, this contradicts

the fact $a_{x}=A\cap x=A\cap y=a_{y}$ .

$\square$byon the set of normal ultrafilters on
Define the Mitchell
if and only if $F\in M_{U}$ .
is well-founded ordering and it
Similar to measurable cardinals (see Mitchell [12]),can be defined$\mathrm{o}(U)=\sup\{\mathrm{o}(F)+1|F\triangleleft U\}$ , for all normal ultrafilter.
on$\mathcal{P}_{\kappa}\lambda$$\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\triangleleft$$F\triangleleft U$$\triangleleft$$U$$\mathcal{P}_{\kappa}\lambda$

Using this, Theorem 5.1 can be restated as:
If $\mathrm{o}(U)=0$ , then$U$

has the partition property.

So, the following question is natural.
Question Suppose that is a normal ultrafilter which does not have the partition$U$

property. How small the value$\mathrm{o}(U)$is?

References[1] Y. Abe, Combinatorial characterization of$\Pi_{1^{-}}^{1}indesC\dot{n}bability$in$P_{\kappa}\lambda$, Archive forMathematical Logic, 37 (1998) pp. 261-272.
[2] D. M. Carr,$P_{\kappa}\lambda$

partition relations, Fundamenta Math., 128 (1987) pp. 181-195.

[3] D. M. Carr, The stmcture of ineffability property of$\mathcal{P}_{\kappa}\lambda$, Acta Math. Hung., 47(1986) pp. 325-332.
[4] C. A. Johnson, Some partition relations for ideal8 on$P_{\kappa}\lambda$, Acta Math. Hung., 56(1990) pp. 269-282.
[5] S. Kamo, Remarks on$\mathcal{P}_{\kappa}\lambda$-combinatorics, Fundamenta Math., 145 (1994) pp.

141-151.
[6] S. Kamo, Ineffability and partition property on$\mathcal{P}_{\kappa}\lambda$, J. Math. Soc. Japan, 49(1997) pp. 125-143.

[7] A. Kanamori, The Higher Infinite, Springer-Verlag (1994).

15$[8],\mathrm{K}$. Kunen and D. H. Pelletier, On a combinatorial property of Menas related tothe partition property for measures on supercompact cardinals, J. Symbolic Logic,
48 (1983) pp. 475-481.
[9] M. Magidor, Combinato$7\dot{\mathrm{v}}al$characterization of supercompact cardinals, Proc.

Amer. Math. Soc., 42 (1974) pp. 279-285.
[10] T. K. Menas, On strong compactness and supercompactness, Annals of Math.

Logic, 7 (1974) pp. 327-359.
[11] T. K. Menas, A combinato r.ial property of$\mathcal{P}_{\kappa}\lambda$, J. Symbolic Logic, 42 (1976) pp.

225-234.
[12] W. J. Mitchell, Set constructible from sequences of ultrafilters, J. Symbolic Logic,
39 (1974) pp. 57-66.