数理解析研究所講究録
第 1595 巻 2008 年 84-88

84

CAN BE CONSIDERED AS AN
ANALOGUE OF SUSLIN’S HYPOTHESISTHE INEQUALITY$\mathfrak{b}>\aleph_{1}$TERUYUKI YORIOKA

ABSTRACT. In [3], the author introduced a chain condition, called the antirectangle reflning property, of forcing notions and the statement $\neg C(arec)$ that
We show that every forcing notion with the anti-rectangle refining property
ha8 an uncountable antichain. Since a typical example of a forcing notion(arec) is awith the anti-rectangle reflning property is an Aronszajn tree,
generalization of Suslin’s Hypothesis. We show that 4(arec) implies that the, that is, this statement can be consideredbounding number is larger thanas an analogue of Suslin’8 Hypothesis.
$\neg C$$\aleph_{1}$1. INTRODUCTION
The author investigated several fragments of Martin’s Axiom in [3]. Fragmentsof Martin’s Axiom were studied mainly by Stevo Todor\v{c}evi\v{c} in $1980’ s$ , and many
applications are discovered (see [2] and his many other articles). In this manuscript,
we give a proof of one question in this area as follows.

We explain some notions in [3]. Aforcing notion has the anti-rectangle refining
prvpeny if for any uncountable subset and of , there exists uncountable subsets
and $J’$ of and respectively such that for every $p\in I’$ and $q\in J’,$ and areincompatible in .
is the statement that every forcing notion with the antirectangle refining property has an uncountable antichain. Since an Aroozajn tree
has the anti-rectangle refining property, $\neg C(arec)c\bm{t}$ be considered ageneralizationthat everyof Suslin’s Hypothesis. In fact, $\neg C(arec)$ implies Suslin’s Hypothesis-gaps are indestructible. The author would like to find other examples of ageneralization of Suslin’s Hypothesis, that is, other statements about combinatorics
onwhich is deduced $hom\neg C(arec)$ . One candidate is the statement that thebounding number is larger than .
, which is aweak hagments of Martin’s
We had already known that. So it is naturaUy arisen aAxiom and implies $\neg C(arec),$ implies that
. In this manuscript, we show apositive
question that $\neg C(arec)$ implies
in section 3.
answer of this question, that $i_{8\urcorner}C(arec)implie8$ that
Aproof of the theorem is self contained in this manuscript, however Iomit someproofs of $weU$ known results in section 2. All of them are written in [3] or [1].
$\mathbb{P}$

$I$

$I’$

$J$

$\mathbb{P}$

$J$

$I$

$\mathbb{P}$

$p$

$q$

$\urcorner C(arec)$$\bm{t}d$$(\omega_{1},\omega_{1})$$w_{1}$$\mathfrak{b}$$\aleph_{1}$$\mathcal{K}_{2}(rec)$$\mathfrak{b}>\aleph_{1}$$\mathfrak{b}>\aleph_{1}$$\mathfrak{b}>\aleph_{1}$2. A REASON WHY WE WILL PROVE AS BELOW(rec) implies. A partition
At first, we will see a proof thathas the rectangle refining property if for any uncountable subset$\mathcal{K}_{2}$$K_{0}\cup K_{1}$$\mathfrak{b}>\aleph_{1}$$[w_{1}]^{2}=$
$I$

and

Supported by $Grant-in$-Aid for JSPS FeUow, No. 18840022, Ministry of Education, Culture,Sports, Science and Technology.

85
TERUYUKI YORIOKAof , there exist uncountable subsets $I’$ and $J’$ of and respectIvely suchthat for every, ifand, then. We note that therectangle refining property is a strong property than the countable chain condition.
(rec) is the statement that every partition
with the rectangle
refining property has an uncountable-homogeneous set. We note that(rec)is deduced from Martin’s Axiom for -dense sets, and(rec) implies $\neg C(arec)$ .
Let
be a set of strictly increasing functions from into suchthat for every and in, if $\xi<\eta$ , then, i.e. there exists $m\in\omega$such that for all $n\geq m,$. For this family, we define a partitionby letting
iff there exists $m$ and in such that$f_{\xi}(m)<f_{\eta}(m)$ and $f_{\eta}(n)<f_{\xi}(n)$ . We call that $F$ is unbounded when for every, there exists $f\in F$ such thatfunction in. We note that if $F$ isunbounded, then this partition has the rectangle refining property. (This follows
from Lemma 3.2 below.) However, in [1, Lemma 16], if $F$ is unbounded, since an
uncountable subset of $F$ is also unbounded, for every uncountable subset $F’$ of $F$ ,there are two functions and in $F$ such that dominates everywhere, i.e., for$f(n)\leq g(n)$ . Therefore,
every(rec) implies
.
$J$

$I$

$\omega_{1}$$\alpha\in I’$$\beta\in J’$$\alpha<\beta$$J$$\{\alpha,\beta\}\in K_{0}$$[\omega_{1}]^{2}=K_{0}\cup K_{1}$$\mathcal{K}_{2}$$K_{0}$$\mathcal{K}_{2}$$\aleph_{1}$$\mathcal{K}_{2}$$F=\{f_{\xi};\xi\in\omega_{1}\}$$\xi$$\omega$$w$$f_{\xi}\leq*f_{\eta}$$w_{1}$$\eta$$f_{\xi}(n)\leq f_{\eta}(n)$$[\omega_{1}]^{2}=K_{0}\cup K_{1}$$\{\xi,\eta\}\in K_{0}$$n$$\omega^{w}$$g$$\omega$$f\not\leq^{*}9$

$f$

$g$

$f$

$g$

$n\in\omega,$$b>\aleph_{1}$$\mathcal{K}_{2}$So to try to prove that $\neg C(arec)$ implies, it seems to be natural to modify
the argument above. Let
be a forcing notion which consists of finite subsets
ofsuch that the set
is totally ordered by the dominance everywhere,i.e., for every$n\inw$, maxand, ordered by thereverse inclusion. As the above partition has the rectangle refining property, wenote thathas the anti-rectangle refining property if $F$ is unbounded. So if weshow thatis ccc whenever $F$ is unbounded, we conclude that doesn’t have tobe unbounded. However, unfortunately, in general,
does not have the ccc even
if is unbounded. For example, if the set
is a subset of suchthatfor anyin, and$f_{\xi}(0)=0$ andfor any,
then it is an uncountable antichain in P.
$b>\aleph_{1}$$\mathbb{P}’$$\sigma$$\{f_{\xi};\xi\in\sigma\}$$\omega_{1}$$\xi\in\sigma$$\{f_{\zeta}(n);\zeta\in\sigma\cap\xi\}\leq f_{\xi}(n)$$\mathbb{P}’$$F$$\mathbb{P}’$$\mathbb{P}’$$F$$\mathbb{P}’$$\{\{\xi_{\zeta},\eta_{\zeta}\};\zeta\in\omega_{1}\}$$\bullet$$\zeta<\zeta’$$\bullet$$\zeta\in\omega_{1},$$w_{1},$$\xi_{\zeta}<\eta_{\zeta}<\xi_{\zeta’}$$f_{\eta_{\zeta}}(1)=1$

In section 3, we define a forcing notion
which is a modification ofandshow that (Lemma 3.2) has the anti-rectangle refining property whenever $F$ isunbounded, and (Lemma 3.3) has the countable chain condition whenever $F$ isunbounded. This completes the proof of our theorem.
$\Psi$$\mathbb{P}$$\mathbb{P}$$P$

3. A PROOF
Throughout this section, let $F=\{f_{\xi}; \xi\in w_{1}\}$ be a set of strictly increasing
functioms $kom\omega$ into such that for every and in , if $\xi<\eta$ , then.
We define a forcing notion which consists of finite subsets ofsuch thatfor every
and $n\in\omega$ , either max
or $f_{\xi}(n)\in$, ordered by the reverse inclusion.
$w$

$\xi$

$\eta$$f_{\xi}\leq^{*}f_{\eta}$$\omega_{1}$$\mathbb{P}$$\sigma$$\xi\in\sigma$$\omega_{1}$$\{f_{\zeta}(n);\zeta\in\sigma\cap\xi\}\leq f_{\xi}(n)$$\{f_{\zeta}(n);\zeta\in\sigma\cap\xi\}$

Proposition 3.1. Suppose that$e\in w$ such that for every
uncountable.

is unbounded. Then there enists
and $k\in w$ , the set $\{\xi\in w_{1}; f_{\xi}(n)\geq k\}$ is$F=\{f_{\xi}; \xi\in\omega_{1}\}$$n\in w\backslash e$

86$\neg C(arec)\Rightarrow b>\aleph_{1}$Proof. Assume not, i.e. there exists an infinite setof natural numbers such thatfor every $n\in Z$ , there exists $k_{n}\in w$ such that the set
isbe such that for all $n\in Z,$
countable. Let
is a subset
of . Let $\{n_{i};i\in w\}$ be an increasing enumeration of , and we define a function
on by
$Z$

$\{\xi\in\omega_{1}; f_{\xi}(n)\geq k_{n}\}$$\{\xi\in w_{1}; f_{\xi}(n)\geq k_{n}\}$$\delta\in\omega_{1}$$Z$$\delta$$g$$w$$g(m)$ $:= \max(\{f_{\delta}(m)\}\cup\{k_{n} ; i\in m+1\}\cup\{g(i)+1;i\in m\})$for each $m\in w$ . We notice that for each, since $m\leq n_{m}$ ,and$\xi\in\delta,$$f_{\xi}\leq g$. Moreover for each$\xi\in w_{1}\backslash \delta$$m\in\omega$$f_{\xi}(m)\leq f_{\xi}(n_{m})<k_{n_{m}}\leq g(m)$.

So $F$ is bounded by 9, which is a contradiction.

$\square$Lemma 3.2. If $F=\{f_{\xi};\xi\in w_{1}\}$ is unbounded, thenfining property.

Pmof Lethas the anti-rectangle oe-$\mathbb{P}$and be uncountable $8ubsets$ of P. By shrinking and if necessary,
we may assume thatI $form8$ a -system with a root , and also forms a -system with a root
$I$

$J$

$I$

$J$

$\Delta$$\bullet$

$\mu$

$\nu$

$\Delta$,

all members of has the same size, and all members ofsize,,for any
and$I$$\bullet$$\max(\mu\cup\nu)<\min(\sigma\backslash \mu)$there exists$\bullet$$J$

also has the same$\tau\in J$$\sigma\in I$$\bullet$$J$$e\in\omega$,$\max(\mu\cup\nu)<\min(\tau\backslash \nu)$, such that for every$\sigma\in I$,

and$(\sigma\backslash \mu)\cap(\tau\backslash \nu)=\emptyset$$\tau\in J$,

and $n\geq e$ ,max$( \{f_{\zeta}(n);\zeta\in\mu\cup\nu\})<\min(\{f_{\xi}(n);\xi\in\sigma\backslash \mu\})$max$( \{f_{(}(n);\zeta\in\mu\cup\nu\})<\min(\{f_{\eta}(n);\eta\in\tau\backslash \nu\})$andWe notice that for every
previous lemma, there exists{$\sigma\in I$; mIn, the set

is unbounded. So by thesuch that for every $k\in w$ , the set$A\in[\omega_{1}]^{\aleph_{1}}$$e_{0}\geq e$$\{f_{\xi};\xi\in A\}$}$(\{f_{\xi}(e_{0});\xi\in\sigma\backslash \mu\})\geq k$

is uncountable. Let $J’$ be uncountable subset of$\tau\in J’$

,

max

.

$J$

and

$k_{0}\in\omega$such that for every,of such that for every$(\{f_{\eta}(e_{0});\eta\in\tau\})\leq k_{0}$and then we take an uncountable subset

$I’$

$I$

$\min(\{f_{\xi}(e_{0});\xi\in\sigma\backslash \mu\})>k_{0}$Then we notice that for any
and $\tau\in J’$ , sincethen and are incompatible in P.
$\sigma\in I’$$\sigma\in I’$,.

$e_{0}\geq e$, if$\tau\not\subset m\alpha(\sigma)+1$,$\tau$$\sigma$Conversely, by the previous lemma, there exists $e_{1}>e_{0}$ such that for every, the set
{ $\tau\in J’$ ; min}$I’$
$k_{1}\in
w$ such that for every
is uncountable. Let
be uncountable subset of and$k\in\omega$$(\{f_{\eta}(e_{1});\eta\in\tau\backslash \nu\})\geq k$$I^{u}$$\sigma\in I’’$,max$(\{f_{\xi}(e_{1});\xi\in\sigma\})\leq k_{1}$,87
TERUYUKI YORIOKA

and then we take an uncountable subset $J”$ of $J’$ such that for everymin$(\{f_{\eta}(e_{1});\eta\in\tau\backslash \nu\})>k_{1}$$\tau\in J’’$.

, for any
Then we notice that, sinceand $\tau\in J’’$ , ifthen and are incompatible in P.
By shrinking $I”$ and $J”$ if necessary, we may $w$ sume that for any$\tau\in J’’$ ,either $\tau\not\subset\max(\sigma)+1$ or $\sigma\not\subset\max(\tau)+1$ . Then for every$\tau\in J’’,$
andr are incompatible in P.
$\sigma\in I’’$$e_{1}\geq e$,$\sigma\not\subset\max(\tau)+1$,$\tau$$\sigma$$\sigma\in I’’$$\sigma\in I’’$

andand$\square$$\sigma$Lemma 3.3. If $F=\{f_{\xi};\xi\in w_{1}\}$ is unbounded, thencondition.

Proof. Here, for each$\sigma\in \mathbb{P}$

we denote, letting$\langle\xi_{i};i\in|\sigma|\rangle$;$\vec{\sigma}:=(f_{\xi}$$\mathbb{P}$

has the countable chain

be an increasing enumeration of ,$i\in|\sigma|\rangle$$\sigma$,. Let be an uncountable subset of P. Without
which is a member of the set
may
generality,
loss ofassume that
we
I forms a -system with a root ,for every and in , either $\max(\sigma)<\min(\tau\backslash \mu)$ or $\max(\tau)<\min(\sigma\backslash \mu)$ ,,and
there exists $n_{0}\in w$ such that for every $n\geq n_{0},$$(w^{w})^{|\sigma|}$$I$$\Delta$$\bullet$$\mu$$\bullet$$I$$\tau$$\sigma$$\sigma\in I$$\bullet$max$\bullet$$\bullet$$\{f_{\langle}(n);\zeta\in\mu\}<f_{\xi}(n)$$\xi\in\sigma\backslash \mu$,$|\sigma|=k$ ,there exists $k\in w$ such that for every, i.e. for each $j\in k$ , the initial segment
for every and inis equal to the initial segment of theof the j-th element of a of lengthj-th element of of length .
$\sigma\in I,$$I,\vec{\sigma}\square n_{0}=\tilde{\tau}|n_{0}$$\tau$$\sigma$$n_{0}$$\vec{\tau}$

Then there exists$\{\tilde{\sigma};\sigma\in I\}$each$\sigma\in I$$n_{0}$

such that the set$\gamma\in w_{1}$$\{\tilde{\sigma};\sigma\in I\cap[\gamma]<\aleph_{0}\}$is dense in the set. For. We fix some (any)as a subspace of the space$n\inw$,, we define two functions
on as follows: For eachand$\nu\in I\backslash [\gamma]^{<\aleph_{0}}$$(w^{w})^{k}$$h_{\sigma}$$g_{\sigma}$$w$$g_{\sigma}(n):=mu\{f_{\xi}(n);\xi\in\sigma\}(=m\alpha\{f_{\xi}(n);\xi\in\sigma\backslash \mu\})$and$h_{\sigma}(n)$We notice that for

can find$n_{1}\geq n_{0}$$g_{\nu}(n)\leq h_{\tau}(n)$and$\sigma$and$\tau$$g_{\sigma}\leq h_{\tau}$$I$$I’\in[I\backslash [\gamma]<\aleph_{0}]^{\aleph_{1}}$, and for every. Since

is unbounded
is unbounded. Hence there exists$\{h_{\tau}(n);\tau\in I’\}$ is infinite. Let$n_{2}:= \min${$\tau$

and$\tau’$

in$I’,\tilde{\tau}\square n_{1}=\tau^{\tilde{\prime}}rn_{1}$$F$$n\geq n_{1}$$\{h_{\tau};\tau\in I’\}$$n\in[n_{1},w);\{h_{\tau}(n);\tau\in I’\}$is infinite}.

such that, we can takeand infinite
an initial, i.e. for every $j\in k$ , the j-th member offor allsegment of the j-th member of ,is infinite.
the setBy the minimality of$\overline{t}\in(\omega^{na})^{k}$$n_{2}$$\tau\in I’’,$$\{h_{\tau}(n);\tau\in I’’\}$$I”\subseteq I’$$t^{\sim}is$$t\subseteq\sim\vec{\tau}$$\tilde{\tau}$$\bullet$.

. So wein , if $\max(\sigma)<\min(\tau\backslash \mu)$ , thensuch that for every $\tau\in I’$ and $n\geq n_{1}$ ,and $I’$ is uncountable, the set
such that the set$\bullet$$;= \min\{f_{\xi}(n);\xi\in\sigma\backslash \mu\}$,88
BIBLIOGRAPHY

By our assumption, there exists
such that for every $n\geq n_{3},$$\sigma\in I\cap[\gamma]<\aleph_{0}$$n_{3}\geq n_{2}$$g_{\nu}(n_{3})<h_{\tau}(n_{2})$.

.

such that
Then there is, and take $\tau\in I’’$ such that$tarrow\subseteq\vec{\sigma}$$g_{\sigma}(n)\leq g_{\nu}(n)$We will show that for every $n\geq n_{2,9\sigma}(n)\leq h_{\tau}(n)$ holds. If $n_{2}\leq n<n_{S}$ , then$g_{\sigma}(n)<g_{\sigma}(n_{3})\leq g_{\nu}(n_{3})<h_{\tau}(n_{2})\leq h_{\tau}(n)$so it is ok. If$n\geq n_{3}$, then since$n\geq n_{3}\geq n_{1}$and$\tau\in I’’\subseteq I’$$g_{\sigma}(n)\leq g_{\nu}(n)\leq h_{\tau}(n)$,,.

We recall thatis an initial segment of both and , for every,, and both and are members of . Therefore
is also a
condition of , I.e. and are compatible in P.
$t\inarrow(w^{n_{2}})^{k}$$g_{\sigma}(n)\leq h_{\tau}(n)$$\sigma$$\check{\sigma}$$\tau$$\mathbb{P}$$n\geq n_{2}$$\tilde{\tau}$$\sigma\cup\tau$$\mathbb{P}$$\sigma$$\tau$$\square$4. ACKNOWLEDGMENT
I would like to thank Stevo Todor\v{c}evi\v{c} for a discussion on the question in thismanuscript during the international conference of Topology in 2007 December atKyoto.

REFERENCES[1] S. Todorfevi6. Remarks on cellularity in products, Compogitio Math. 57 (1986), no. 3, 357-372.

[2] S. Todoreevi\v{c}. Partition Problems in Topology, volume 84 of $Contempo|u\eta$ mathematics,
American Mathematical Society, Providence, Rhode Island, 1989.
[3] T. Yorioka. Some weak frugments of Martm’s Axiom related to the rectangle refining property, to appear in Arch. Math. Logic.

DEPARTMENT OF MATHEMATICS, SHIZUOKA UNIVERSITY, OHYA 836, SHIZUOKA, 422-8529, JAPANE-mail address: styorioeipc. shizuoka. ac. jp