数理解析研究所講究録 1202 巻 2001 年 61-65

61NONREFLECTING STATIONARY SETS IN$P_{\kappa}\lambda$

SAHARON SHELAH AND MASAHIRO SHIOYA
is constructed
ABSTRACT. Anonreflecting stationary subset ofcardinal.
e.g. when \kappa is the successor of aregular uncountable$\mathcal{P}_{\kappa}\kappa^{+}$

1. INTRODUCTION
be aregular cardinal. The reflection principle for staLet
is acardinal, was introduced and, wheretionary subsets ofshown consistent relative to asupercompact cardinal in caseby Foreman, Magidor and Shelah [3]. The corresponding principle forwas refuted in ZFC by Feng and Magidor [1] when is asuccessor cardinal, and in general by Foreman and Magidor [2]. Specifically,“combinatorialization” of the latter argument (see Section 4below)
yieldshas a nonreflecting stationary subset when
Theorem 1.
and?In this note, we give aparallel result for
What aboutregular. More generally, we showwith$\kappa$$>\omega$$\lambda>\kappa$$P_{\kappa}\lambda$$\kappa$$=\omega_{1}$$\kappa$$\kappa$$>\omega_{1}$$\kappa$$P_{\kappa}\lambda$$>\omega_{1}$$\lambda\geq 2^{\kappa^{+}}$$\kappa=\nu^{+}$$P_{\kappa}\kappa^{+}$$\nu>\omega$

Theorem 2. Assume. Thenfor all$\gamma<\kappa$$\kappa$, $\omega<\nu<\kappa$ is regular andhas a nonreflecting stationary subset.

$\mathrm{c}\mathrm{f}[\gamma]^{<\nu}<$$\mathrm{c}\mathrm{f}[\lambda]^{\kappa}=\lambda$$P_{\kappa}\lambda$

2. PRELIMINARIESOur terminology generally follows Kanamori [5] with the following
denotes aregular cardinal
exceptions. For the rest of this paper,and aregular,
acardinal from, Aacardinal$[\lambda]^{\mu}=\{x\subset\lambda : |x|=\mu\}$ ,the. We let
cardinal from.
minimal size of its unbounded subsets andforordinals,
anddenotes the set of limit points of aset of
Also
under
the closure of,amap defined on asubset of$\kappa$$>\kappa$$>\omega_{1}$$\lambda-\kappa$

$\nu$

$\mu$

$\mathrm{c}\mathrm{f}[\lambda]^{\mu}$$\kappa-\omega_{1}$$S_{\kappa}^{\nu}=\{\gamma<\kappa :$A$$\lim$ $A$$f$\mathrm{c}\mathrm{f}\gamma=\nu\}$$\lambda^{<\omega}$$\mathrm{c}1_{f}x$$x\in P_{\kappa}\lambda$.
1991 Mathematics Subject Classification.
Foundation founded by theScience
Israelthebysupported
The first author was764. The second author
Publication
Humanities.
Israel Academy of Sciences and(No. 12640098),
Researchfor
Scientific
was partially supported by Grant-in-AidJapanofandCulture
Ministry of Education, Science, Sports$03\mathrm{E}05$

62
and $C(f)$ the set. The reflection
principle we consider states that for allandstationary
there issuch thatis stationary. Astationary setwitnessing its failure is called nonreflecting. More generally,
iscalled -nonreflecting if
is nonstationary for all.
A -complete filter onextending
iscalled fine. The specific example relevant to us was introduced in [7]:Lemma 1. The sets {: ::$i<n\rangle)\subset A_{n})\}$ , where:, generate a fine filter on$f$$\{x\in P_{\kappa}\lambda :x\cap\kappa\in\kappa\wedge \mathrm{c}1_{f}x=x\}$$\lambda>\kappa$$S\subset P_{\kappa}\lambda$$S\cap P_{\kappa}A$$\kappa\subset A\in[\lambda]^{\kappa}$$S\subset P_{\kappa}\lambda$$S\cap P_{\kappa}A$$\mu$$\mu^{+}$$\bigcup_{i<\omega}A$$\{A_{i} :\alpha\in x\} :\alpha<\lambda\}$i<\omega\}\subset[\lambda]^{\mu}\wedge\forall n<\omega(\tau(\langle A_{i}$$([\lambda]^{\mu})^{<\omega}arrow[\lambda]^{\mu}$$\tau$$[\lambda]^{\mu}$$\mu\subset A\in[\lambda]^{\mu}$$\{\{x\in[\lambda]^{\mu} :$[\lambda]^{\mu}$.

We need an analogue of Ulam’s result (see [4] or [6] for aproof):

Lemma 2. Let

be a fine filter on
joint $F$ -positive sets exist.
$F$$[\lambda]^{\mu}$. Then Amany mutually dis-3. MAIN THEOREM
In this section we prove Theorem 2in an even more general form:
Theorem 3. Let

be as in Section 2. Assume
is unbounded infor someand {::
is unbounded in $[z]$ ”’} has a stationary
subset of size for some
. Thenhas anonrefiecting stationary subset.
$\{\bigcup_{\alpha\in a}E_{\alpha} :$[\lambda]^{\mu}$$z$$\in P_{\kappa}\mu$$\{\mathrm{C}\xi$$e$$\{E_{\alpha} :$[\lambda]^{\mu}$\alpha<\lambda\}\subset$$\xi\in z\}$$\{c_{\xi} :$\mu$Proof. Define :Aon$<\nu<\kappa\leq\mu<\lambda$

a\in[\lambda]^{<\nu}\}$$T$$[\lambda]^{\mu}$$\omega$$\cross\muarrow\lambda$

so that\xi<\mu\}\subset[\mu]^{<\nu}$$e$$P_{\kappa}\lambda$“ $\{\alpha\}\cross\mu=\mathrm{E}\mathrm{a}$as in Lemma 1. Fix amutually disjoint. Let$\{X_{z} :$F$$\mu-$be the filter
byz\in T\}\subset F^{+}$

Lemma 2. We show that $S=\{x\in C(e)$ :is as desired.
To show stationary, fix :. Consider the following
gamefor $z\in T$ :Iand II take in turnand atriple of, abijection:
and $x_{n}\in C(f)$respectively so that,,“$(x_{n}\cap\mu)=x_{n}$ andandare increasing. II
wins iff $x_{n}\cap\mu=z$ for all $n<\omega$ . We first claim $T\cap D\subset\{z\in T$ : II
has awinning strategy in.
} for some club
Suppose to the contrary $T’=\{z\in T$ :II has no winning strategyin} is stationary. For $z\in T’$ , we have awinning strategy forIin, since the game is closed for II and hence determined. By
induction on $n<\omega$ , build
for $z\in T’$ as follows: First takewith$C(f)$ . Next take abijection:. For $z\in T’$ , take“with, and if possible$x\cap\mu\in T\wedge\exists b\in[x]^{<\nu}(x\subset$$\mathrm{c}1_{e}(b\cup\mu)\in X_{x\cap\mu})\}$

$S$

$f$

$\lambda^{<\omega}arrow P_{\kappa}\lambda$$G_{f,z}$$\mu\subset A_{n}\in[\lambda]^{\mu}$$b_{n}\in[\lambda]^{<\nu}$$\muarrow \mathrm{c}1_{e}(b_{n}\cup\mu)$$\pi_{n}$$A_{n}\subset \mathrm{c}1_{e}(b_{n}\cup\mu)\subset A_{n+1}$$\pi_{n}$$\langle b_{\dot{1}} :$\langle x:i<\omega\rangle$$G_{f,z}$:$b_{n}\subset x_{n}\subset \mathrm{c}1_{e}(b_{n}\cup\mu)$i<\omega\rangle$$D\subset P_{\kappa}\mu$$G_{f,z}$$\sigma_{z}$$G_{f,z}$$(b_{n}, \pi_{n}, x_{n}^{z})$$b_{n-1}\subset b_{n}\in[\lambda]^{<\nu}$$\bigcup_{z\in T’}\sigma_{z}(\langle(b:, \pi:,x_{\dot{1}}^{z}) :$\pi_{n}$$b_{n}\subset x_{n}^{z}\in C(f)\cap P_{\kappa}\mathrm{c}1_{e}(b_{n}\cup\mu)$i<n\rangle)\subset \mathrm{c}1_{e}(b_{n}\cup\mu)\in$$\muarrow \mathrm{c}1_{e}(b_{n}\cup\mu)$$\pi_{n}$$(x_{n}^{z}\cap\mu)=x_{n}^{z}$63$x_{n}^{z}\cap\mu=z$. Now set

$b= \bigcup_{n<\omega}b_{n}\in[\lambda]$”’and$A=\mathrm{c}1_{e}(b\cup\mu)\in[\lambda]^{\mu}$.

for allwith
Take$n<\omega$ and $z=x\cap\mu\in T’$ . Thenfor all $n<\omega$ ,is the unique set satisfying all the requirements forsincewith the moves
including the extra one. Thus II wins the game
, yet Iplays according to the winning strategy .
:Contradiction.
. Define
Now fix $z\in T$ with awinning strategy for II in$\sigma(t)=(b,\pi,x)$,. Since
bywhere:$X_{z}\in F^{+}$ , we haveandsuch that$n<\omega$. Set
for all
for $n<\omega$ . Then $x= \bigcup_{n<\omega}x_{n}\in S\cap C(f)$ as desired, since $x\cap\mu=$$z\in T$ and, where“$b\subset x\in C(f)\cap P_{\kappa}A$$\pi_{n}$$(x\cap\mu)=x\cap \mathrm{c}1_{e}(b_{n}\cup\mu)$$x_{n}^{z}=x\cap \mathrm{c}1_{e}(b_{n}\cup\mu)$$x\cap \mathrm{c}1_{e}(b_{n}\cup\mu)$$G_{f,z}$$x_{n}^{z}$$\mathrm{C}(\mathrm{h})\pi_{n},$$x_{n}^{z})$$n<\omega\rangle$$\sigma_{z}$$G_{f,z}$$\sigma$$\tau$$([\lambda]^{\mu})"’arrow[\lambda]^{\mu}$$\mathrm{r}(\mathrm{t})=\mathrm{c}\mathrm{l}_{e}(b\cup\mu)$$\{A_{i} :$\tau(\langle A_{i} :i<\omega\}\subset[\lambda]^{\mu}$$\bigcup_{i<\omega}A:\in X_{z}$$(b_{n}, \pi_{n},x_{n})=\sigma(\langle A_{i} :i<n\rangle)\subset A_{n}$i\leq n\rangle)$$x \subset \mathrm{c}1_{e}(b\cup\mu)=\bigcup_{n<\omega}\mathrm{c}1_{e}(b_{n}\cup\mu)=\bigcup_{i<\omega}A_{i}\in X_{z}$$b= \bigcup_{n<\omega}b_{n}\in[x]^{<\nu}$.

is staTo show $S\mu$-nonreflecting, suppose to the contrary. Then “ $(A\cross\mu)\subset A$ , since $C(e)\cap P_{\kappa}A$
tionary for some.
. We next give $a\in[A]^{<\nu}$ withis unbounded in$B=\pi^{-1}\mu\in[\mu]^{\mu}$$\muarrow
A$.
:.
Define:abijection
Set
Fix
$S’$
$=e(\pi(\xi),\pi(\zeta))$. Let
be the stationary
by $\pi(h(\xi, ())$$\{x\cap\mu : \pi"(x\cap\mu)=x\in S\cap P_{\kappa}A\}\subset T$ . For $z\in S’$ , takewith“by $z\in S$ . Take $\xi<\mu$ so that $S^{*}=\{z\in S’ : \xi_{z}=\xi\}$, sincefor all $z\in S^{*}$ .
is stationary. Then, as desired.
HenceofNow we have the desired contradiction to the mutual disjoint$\{X_{z} : z\in T\}:A\in X_{x\cap\mu}$ for all $x\in S\cap P_{\kappa}A$ with, since for some$b\in[x]^{<\nu}$ ,.
by$S\cap P_{\kappa}A$$\mu\subset A\in[\lambda]^{\mu}$$e$$\mathrm{c}1_{e}(a\cup\mu)=A$$P_{\kappa}A$

$h$

$\pi$

$\mu \mathrm{x}Barrow\mu$$\xi_{z}\in z$$z\subset \mathrm{c}1_{h}(c_{\xi_{z}}\cup B)$$\pi$$\mu=\mathrm{c}1_{h}(c_{\xi}\cup B)$$z\subset \mathrm{c}1_{h}(c_{\xi}\cup B)$$A=\mathrm{c}1_{e}(\pi" c_{\xi}\cup\mu)$$a\subset x$$A=\mathrm{c}1_{e}(a\cup\mu)=\mathrm{c}1_{e}(b\cup\mu)$$a\subset \mathrm{c}1_{e}(b\cup\mu)$$\square$

4. REMARKSLet us first deduce Theorem 2from Theorem 3: Assume. Then we haveandfor all. Thenis unbounded insuch that“:stationaryofsubset
isthedesired{}.
unbounded in
The rest of the section is devoted to the$\{c_{\xi} :$\gamma<\kappa$$\{c_{\xi} :$f$\xi<\kappa\}\subset[\kappa]^{<\nu}$\xi<f(\gamma)\}$$f$:$\kappaarrow\kappa$:is$T–\{\gamma\in S_{\kappa}^{\nu}$$[\gamma]^{<\nu}$$\{c_{\xi} :$\gamma<\kappa$$\gamma\subset\gamma$$\mathrm{c}\mathrm{f}[\gamma]^{<\nu}<\kappa$\xi<\gamma\}$$[\gamma]^{<\nu}\}$Proof of Theorem 1. Fix abijection

asurjection$\lim\pi_{\beta}$$g$“ $\pi_{\beta}^{-1}(\alpha)$$\pi_{\gamma}$:$\kappa$$arrow\gamma$for$\in\kappa^{+}-\kappa$$\gamma$andby $h(\alpha, \beta)=$. Define:: A$D$. Let
be the club { $x\in C(h)$ : Vy $\in x\cap(\kappa^{+}-$
V4 $\in x(x\in C(g(\xi)))\}$ . We show that $S=$is as$arrow’+\kappa$$h$$[\kappa^{+}]^{2}arrow P_{\kappa}\kappa^{+}$$\kappa)(\pi_{\gamma}"(x\cap\kappa)=x\cap\gamma)\wedge$$\{x*\cdot\in P_{\kappa}\lambda\tau :\sup\{\sup(y\cap\kappa^{+}) :x\subset y\in D\wedge y\cap\kappa=x\cap\kappa\}<\kappa^{+}\}$

64
To show stationary, suppose otherwise. By induction on $n<\omega$ ,:closed under composition so that $C(f_{0})\subset D-S$
build
for alland for all $m<\omega$ there is $n<\omega$ such that. Define :byandwith$f(t)=\{f_{n}(t) : n<\omega\}$ . Fix $x\in C(f)$ . We claim that $\sup\{\sup(z\cap\kappa^{+})$ :$S$

$f_{n}$$\lambda^{<\omega}arrow\lambda$$f_{m}(t*\langle\gamma\rangle)=g_{f_{n}(t)}(\gamma)$$t\in\lambda^{<\omega}$$\gamma<\kappa^{+}$$x\subset z\in C(f)\wedge z\cap\kappa=x\cap\kappa\}=\kappa^{+}$. By. Then. Then$f$$f_{m}(t*\langle\gamma\rangle)<\kappa$$\lambda^{<\omega}arrow P_{\omega_{1}}\lambda$.

, we have $x\subset y\in D$ with $y\cap\kappa=x\cap\kappa$ andFix,
is as desired: To see$m<\omega$.
for somefixandfor some $n<\omega$ , since $\{f_{n}(t),\gamma\}\subset y\in D$ .
Hence
For $i<2$ , build an increasing and continuous,
so thatandas follows: First fix$C(f)$ with. Take
with$\sup(z\cap\kappa^{+})$:
is the -th element of {so that. Takewithso. Thethat
rest of the construction is routine.
, since. Then
Now setis an. Next to showcountably closed ininitial segment of$\sup(x^{0}\cap\kappa^{+})=\sup(x^{1}\cap\kappa^{+})$ , fixof order type . We have“, andby, as. Thenbywith“desired. Now we have the desired contradiction“, whereis unbounded in$\alpha<\kappa^{+}$$x\not\in S$$\alpha<\gamma\in y\cap\kappa^{+}$$\beta\in z\cap\kappa$$z=\mathrm{c}1_{f}(x\cup\{\gamma\})$$z\cap\kappa\subset y\cap\kappa$$t\in x^{<\omega}$$\beta=f_{m}(t*\langle\gamma\rangle)$$\beta=g_{f_{n}(t)}(\gamma)\in y$$\{x_{\xi}^{i} :\xi<\omega_{1}\}\subset C(f)$$\sup(x_{\xi}^{0}\cap\kappa^{+})<\sup(x_{\xi}^{1}\cap\kappa^{+})<\sup(x_{\xi+1}^{0}\cap$$x_{\xi}^{\dot{l}}\cap\kappa=x_{0}^{0}\cap\kappa\in S_{\kappa}^{\omega_{1}}$$x_{1}^{0} \cap\sup(x_{0}^{1}\cap\kappa^{+})\neq x_{0}^{1}\cap\sup(x_{0}^{1}\cap\kappa^{+})$$\kappa^{+})$$\sup(x_{1}^{0}\cap\kappa^{+})$$x_{0}^{0}\in$$x_{0}^{0}\subset x_{1}^{0}\in C(f)$$\mathrm{c}\mathrm{f}(x_{0}^{0}\cap\kappa)=\omega_{1}$$x_{1}^{0}\cap\kappa$$C(f)\wedge z\cap\kappa=x_{0}^{0}\cap\kappa\}$$=x_{0}^{0}\cap\kappa$$x_{0}^{0}\subset z\in$$\kappa$$x_{0}^{0}\subset x_{0}^{1}\in C(f)$$x_{0}^{1}\cap\kappa=x_{0}^{0}\cap\kappa$$\sup(x_{0}^{0}\cap\kappa^{+})<\sup(x_{0}^{1}\cap\kappa^{+})\in\sup(x_{1}^{0}\cap\kappa^{+})-\lim(x_{1}^{0}\cap\kappa^{+})$$x^{0}\cap\kappa^{+}\neq x^{1}\cap\kappa^{+}$$x^{:}= \bigcup_{\xi<\omega_{1}}x_{\xi}^{\dot{1}}$$x_{\xi}^{i}\cap\kappa^{+}$$x^{:}\cap\kappa^{+}$$x^{:}\cap\kappa^{+}$$a\subset x^{:}\cap\kappa^{+}$$a\subset\beta\in x^{:}\cap\kappa^{+}$$\mathrm{c}\mathrm{f}\sup(x^{:}\cap\kappa^{+})=\omega_{1}$$\pi\rho^{-1}a\subset\pi_{\beta^{-1}}(\alpha)$$\omega$$at\in x^{i}\cap\beta=\pi_{\beta}$$(x^{:}\cap\kappa)$$\sup a\in h(\alpha, \beta)\subset x^{:}$$x^{:}\cap\kappa\in S_{\kappa}^{\omega_{1}}$$x^{:} \cap\kappa^{+}=\bigcup_{\gamma\in C}\pi_{\gamma}$$\kappa)=\bigcup_{\gamma\in C}\pi_{\gamma}$$(x^{:}\cap$$C\subset x^{0}\cap x^{1}\cap\kappa^{+}$$(x_{0}^{0}\cap\kappa)$$\sup(x^{0}\cap\kappa^{+})=\sup(x^{1}\cap\kappa^{+})$.

To show nonreflecting, suppose to the contrary
is sta. Fix abijection :. Thentionary for some“$S’=\{\gamma<\kappa : \pi"\gamma\in S\wedge\pi"\gamma\cap\kappa=\gamma\}$ and:{$y\in D\wedge y\cap\kappa\in S’\}$ are stationary inrespectively. Henceand$\sup\{\sup(y\cap\kappa^{+}) : \pi"(y\cap\kappa)\subset y\in D\wedge y\cap\kappa\in S’\}=\kappa^{+}$ . Thus we have,such that“.
contradicting$S$$S\cap P_{\kappa}A$$\kappa\subset A\in[\lambda]^{\kappa}$$\pi$$y\cap\kappa^{+}$$\kappa$$\sup\{\sup(y\cap\kappa^{+}) :$\gamma\in S’$$\pi$$\kappaarrow A$$\pi$$(y\cap\kappa)\subset$$P_{\kappa}\kappa^{+}$\pi"(y\cap\kappa)\subset y\in D\wedge y\cap\kappa=\gamma\}=\kappa^{+}$$\gamma\in S$$\square$

REFERENCES[1] Q. Feng and M. Magidor, On reflection of stationary sets, Fund. Math. 140(1992) 175-181.
[2] M. Foreman and M. Magidor, Large cardinals and definable counterexamples
to the continuum hypothesis, Ann. Pure Appl. Logic 76 (1995) 47-97.
[3] M. Foreman, M. Magidor and S. Shelah, Martin’s maximum, saturated ideals,
and nonregular ultrafilters. I, Ann. Math. 127 (1988) 1-47.
, Fund. Math. 129 (1988) 215-221[4] C. Johnson, On saturated ideals and$P_{\kappa}\lambda$

65[5] A. Kanamori, The Higher Infinite, Springer, Berlin, 1994.
[6] Y. Matsubara, Menas’ conjecture and generic ultrapowers, Ann. Pure Appl.
Logic 36 (1987) 225-234.
[7] S. Shelah, A compactness theorem for singular cardinals, free algebras, Whitehead problem and transversals, Israel J. Math. 21 (1975) 319-349.
INSTITUTE OF MATHEMATICS, HEBREW UNIVERSITY, JERUSALEM, 91904 IsRAEL.
$E$-mail address: shelahGmath.

$\mathrm{h}\mathrm{u}\mathrm{j}\mathrm{i}.\mathrm{a}\mathrm{c}.$il

INSTITUTE OF MATHEMATICS, UNIVERSITY OF TSUKUBA, TSUKUBA, 305-8571
JAPAN.
-mail address: shioya(hnath.tsukuba.
$E$$\mathrm{a}\mathrm{c}.\mathrm{j}\mathrm{p}$