数理解析研究所講究録
第 1895 巻 2014 年 103-107

103

THE APPROXIMATION PROPERTY AND THE CHAIN
CONDITION

Toshimichi Usuba (薄葉季路)
Organization of Advanced Science and Technology
Kobe University
1,

Definition 1,1. Let

THE APPROXIMATION PROPERTY

a be poset anda cardinal. We say that the poset has
the -approximation property if for every ordinal and every f ( 2)VlP, if f x V
for every x ([ ] )V, then f V.
$\mathbb{P}$$\mathbb{P}$$\kappa$$\tau$$\kappa$$\in$$\tau$$\in$$|$$<\kappa$

$\in$

$\in$

$\tau$

It is known that for an uncountable , if is an atomless poset of size
andis a -name for a -closed poset, thenhas the -approximation property(e.g., see Mitchell [1]). In this note, we show that the size assumption for a poset
can be relaxed to the chain condition assumption.
$\kappa$$\dot{\mathbb{Q}}$$\mathbb{P}$$<\kappa$$\mathbb{P}$$\mathbb{P}*\dot{\mathbb{Q}}$$\kappa$$\kappa$$\mathbb{P}$Definition 1.2. Let be a regular uncountable cardinal.
poset
strong -chain condition strong $\kappa-c.c., for$ short) if satisfies the
every -Suslin tree $T,$ does not add a cofinal branch of $T.$$A$$\kappa$$($$\kappa$$\mathbb{P}$$\mathbb{P}$satisfies the. and for$\kappa-c.c$$\mathbb{P}$$\kappa$

Note 1.3.
(1) If there is no -Suslin tree, then the. is equivalent to thestrong(2) For a poset , if
, then satisfies the strong
satisfies the$\kappa-c.c$$\kappa$$\kappa-c.c.$$\mathbb{P}\cross \mathbb{P}$$\mathbb{P}$$\kappa-c.c.$$\mathbb{P}$$\kappa-c.c.$Lemma 1.4. If a poset satisfies the $\mu-c.c$ . for some, thensatisfies thestrong. In particular, $ever1/$ poset of $size<\kappa$ satisfies the strong$\mathbb{P}$$\mu<\kappa$$\mathbb{P}$$\kappa-c.c$$\kappa-c.c.$

Proof. Suppose to the contrary that there is a -Suslin tree$\kappa$$T$such that$|\vdash {}_{\mathbb{P}}T$

has a cofinal branch
Let $T’=\{t\in T$ :“for some
. It
$T’$
easy
is
to check thatis a downward closed subtree of of height . Since$\mu<\kappa$
satisfies the c.c. and, each level of $T’$ has size $<\mu$ . Now, by Kurepa’s
theorem, $T’$ has a cofinal branch. Then this branch is a cofinal branch of , this isa contradiction.
口$\dot{B}$“$p|\vdash_{\mathbb{P}}t\in\dot{B}$$p\in \mathbb{P}\}$$T$$\mathbb{P}$$\kappa$$\mu-$$T$

The following is a main result of this note:

Lemma 1,5. Let be a regular uncountable cardinal. Let be an atomless poset
which satisfies the strong. Let
be a -name for a -closed poset (trivial
poset is possible). Then
has the -approximation property.
$\mathbb{P}$$\kappa$$\kappa-c.c$$\dot{\mathbb{Q}}$$\mathbb{P}$$\kappa$$\mathbb{P}*\dot{\mathbb{Q}}$$\kappa$

104
is the set of all -names withbe a term poset of , that is,, define
if
inFor
Since is a namefor a -closed poset, is -closed.
Let be a-name such that
We say that a condition

Proof. Let$|\vdash_{\mathbb{P}}\dot{q}\in\dot{\mathbb{Q}}$$\dot{\mathbb{Q}}$$\tilde{\mathbb{Q}}$“$\dot{q}_{0},\dot{q}_{1}\in\tilde{\mathbb{Q}}$$\tilde{\mathbb{Q}}$$\mathbb{P}$$\dot{\mathbb{Q}}$$|\vdash_{\mathbb{P}}\dot{q}_{0}\leq\dot{q}_{1}$$\dot{q}_{0}\leq\dot{q}_{1}$$\dot{q}$”$\dot{\mathbb{Q}}$$\tilde{\mathbb{Q}}$$\kappa$$\kappa$

decides$|\vdash\dot{x}\in V$$\mathbb{P}*\dot{\mathbb{Q}}$$\dot{x}$$\dot{x}$if there is

$y$

with

$\langle p,\dot{q}\rangle|\vdash\dot{x}=y$$f$$x\in([\tau]^{<\kappa})^{V}$$(\wp,\dot{q}\rangle\in \mathbb{P}*\dot{\mathbb{Q}}$”Claim 1.6. Let be an ordinal and be a
$f|x\in V$ for every”. Letandsuch that:$\tau$”$\mathbb{P}*\dot{\mathbb{Q}}$-name suchand$\langle p,\dot{q}\rangle\in \mathbb{P}*\dot{\mathbb{Q}}$:and. Then there are$that|\vdash\dot{f}$$x\in[\tau]^{<\kappa}$$\tauarrow 2$$F\subseteq x2$$\dot{q}^{*}\leq\dot{q}$(1) $|F|<\kappa.$(2) For every $g\in F$ , there is $p’\leq p$ such that
(3) For every $p’\leq p$ , there are $p”\leq p’$ and $g\in F$ such that$\langle p’,\dot{q}^{*}\rangle|\vdash\dot{f}|x=g$

Proof. It is easy to check that the set”.

$\langle p",\dot{q}^{*}\rangle|\vdash\dot{f}|x=g$”.

anddecides $f|x)\}$ is predense below . Take a maximal antichain which is contained, we know that $|A|<\kappa$ . Then for eachin this set. Since satisfies the”$r\in A$ , there are
andLetsuch that
and”$F=\{g_{r}:r\in A\}$ and one can takesuch thatandfor every$F$$r\in A$ . Thenand work.
$\{p’\leq p:\exists\dot{q}’(\langle p’,\dot{q}’\rangle\leq\langle p,\dot{q}\rangle$$\langle p’,\dot{q}’\rangle$

$A$

$p$

$\kappa-c.c.$$\mathbb{P}$$\dot{q}_{r}$$\langle r,\dot{q}_{r}\rangle|\vdash f|x=g_{r}$$\langle r,\dot{q}_{r}\rangle\leq\langle p,\dot{q}\rangle$$g_{r}$$r|\vdash\dot{q}^{*}=\dot{q}_{r}$$\dot{q}^{*}\leq\dot{q}$$\dot{q}^{*}$$\square [Claim]$$\dot{q}^{*}$

has the -approximation property, takeIn order to show that:an ordinal , and a name such thatand $;|x\in V$ for every””
Suppose to the contrary that
, we would find
By induction onsuch that:(1):
is -increasing.
and
(2)
:
is decreasing in and
(3)
and
”
, there is $p’\leq p$ such that
(4) For every(5) For every $p’\leq p$ there are $p”\leq p’$ andsuch that””, i.e., the set
for someis predense
below, there are(6) For every
andsuch that. Then we can find required
When $\alpha=0$ , pick an arbitraryby Claim 1.6.
andare defined for all
Let $\alpha>0$ and suppose. Sinceforsuch that
Case 1: is limit. We can findby
. Then takeandfor everyis -closed, we can findClaim 1.6.
below
is successor, say $\alpha=\beta+1$ . Pick a maximal antichain
Case 2:”$p’\inA$
Notesuch that
there issuch that for every$\mathbb{P}*\dot{\mathbb{Q}}$$\kappa$$f$$\tau$$\langle p,\dot{q}\rangle|\vdash$;$\tauarrow 2$$\langle p,\dot{q}\rangle|\vdash f\not\in V$$x\in([\tau]^{<\kappa})^{V}$$\alpha<\kappa$$x_{\alpha}\in[\tau]^{<\kappa}$$x_{\alpha},\dot{q}_{\alpha},$$\langle x_{\alpha}$$\alpha<\kappa\rangle$$F_{\alpha}\subseteq x_{\alpha}2$$F_{\alpha}(\alpha<\kappa)$$\subseteq$$\tilde{\mathbb{Q}}$$\alpha<\kappa\rangle$$\langle\dot{q}_{\alpha}$$\langle p,\dot{q}\rangle\in \mathbb{P}*\dot{\mathbb{Q}},$$\dot{q}_{0}\leq\dot{q}.$$|F_{\alpha}|<\kappa.$$l;p’,\dot{q}_{\alpha}\rangle|\vdash;|x_{\alpha}=g$$g\in F_{\alpha}$$\langle p",\dot{q}_{\alpha}\rangle|\vdash f|x_{\alpha}=$$g\in F_{\alpha}$$\{p’\leq p:\langle p’,\dot{q}_{\alpha}\rangle|\vdash;|x_{\alpha}=g$$g$$g\in F_{\alpha}\}$$p.$$g\in F_{\alpha}$$g_{0},$$g_{1}\in F_{\alpha+1}$$g\subseteq g_{0},$$g_{0}\neq g_{1}.$$g_{1}$$x_{0}\in[\tau]^{<\kappa}$$\dot{q}_{0}\leq\dot{q}$$F_{0}$$x_{\beta},\dot{q}_{\beta},$$x_{\alpha}\in[\tau]^{<\kappa}$$\alpha$$\tilde{\mathbb{Q}}$$\beta<\alpha.$$F_{\beta}$$\beta<\alpha$$\dot{q}^{*}\leq\dot{q}_{\beta}$$\kappa$$\beta<\alpha$$\dot{q}_{\alpha}\leq\dot{q}^{*}$$A\subseteq \mathbb{P}$$\alpha$$p$$x_{\beta}\subseteq x_{\alpha}$$g\in F_{\beta}$$\langle p’,\dot{q}_{\beta}\rangle|\vdash^{(}f|x_{\beta}=g$$F_{\alpha}$105that $|A|<\kappa$ , and, for every, there is $p’\in A$ with$|A|<\kappa$”, we can find
Since
and
such that, butdoes not decide
for every $p’\in A.$$\langle p’,\dot{q}_{\beta}\rangle|\vdash f|x_{\beta}=g$$g\in F_{\beta}$$\langle p,\dot{q}_{\beta}\rangle|\vdash f\not\in V$$\beta<\alpha$$x_{\alpha}\in[\tau]^{<\kappa}$$x_{\beta}\subseteq x_{\alpha}$”for$f|x_{\alpha}$$\langle p’,\dot{q}_{\beta}\rangle$

Claim 1.7. For each $p’\in A$ , there are
that
and
”.

$p_{0}’,p_{1}’\leq p’$, g\’o,:$9_{1}’$$x_{\alpha}arrow 2$, and$\dot{r}\leq\dot{q}_{\beta}$such$\langle p_{i}’,\dot{r}\rangle|\vdash\dot{f}|x_{\alpha}=g_{i}$$g_{0}’\neq g_{1}’$Proof. Since

does not decide, we can takeand g\’o, :such thatandWe may assume
thatis incompatible with ; ifandhave a common extension , takesuch thatand replace by”$Now$ take$\square [Claim]$such that
Clearly
and work.
$f|x_{\alpha}$$\langle p’,\dot{q}_{\beta}\rangle$$x_{\alpha}arrow 2$$g_{1}’$$\langle p_{0}’,\dot{q}_{0}\rangle,$$\langle p_{i}’,\dot{q}_{i}\rangle I\vdash;|x_{\alpha}=g_{i}’$$g_{0}’\neq g_{1}’$$p_{0}’$$p_{1}’$$p_{0}",p_{1}"\leq p_{2}$$p_{0}’$$p_{0}"\perp p_{1}"$$p_{1}’$$p_{2}$$p_{i}".$$p_{i}’$$p_{i}"|\vdash\dot{r}=\dot{q}_{i}$$\dot{r}\leq\dot{q}_{\beta}$$\langle p_{1}’,\dot{q}_{1}\rangle\leq\langle p’,\dot{q}_{\beta}\rangle,$”$p_{i}’,$$\dot{r}$$g_{i}’$For each $p’\in A$ , picksuch that there are
:
”
with
and
”
Then picksuch thatfor every $p’\in A$ . Finally, take
and
, and
as in Claim 1.6, The following claim shows thatwork well:$p_{0}’,$$\dot{r}_{p’}\leq\dot{q}_{\beta}$$g_{0}’\neq g_{1}’$$p_{1}’\leq p’,$$g_{0}’,$$g_{1}’$$x_{\alpha}arrow 2$$\langle p_{i}’,\dot{r}_{p’}\rangle|\vdash f|x_{\alpha}=g_{i}’$$p’|\vdash\dot{q}^{*}=\dot{r}_{p’}$$q^{*}\leq q_{\beta}$$F_{\alpha}\subseteq x_{\alpha}2$$\dot{q}_{\alpha}\leq\dot{q}^{*}$$x_{\alpha},\dot{q}_{\alpha}$$F_{\alpha}$

Claim 1.8. For each$g\in F_{\beta}$

Proof. Take $p’\in A$ so thatg\’o,the choice of$g_{1}’$:$x_{\alpha}arrow 2$

and$\langle p’,\dot{q}_{\beta}\rangle|\vdash(;|x_{\beta}=g$means thatsuch that$g_{0}\neq g_{1}$Then we can take”$\langle p_{i}’,\dot{q}^{*}\rangle|\vdash f|x_{\alpha}=g_{i}’$ ”$g\’{o}\neq g_{1}’$and$\dot{q}_{\alpha}\leq\dot{q}^{*}$$g\subseteq g_{0},$ $g_{1}.$$p_{0}’,p_{1}’\leq p’$$g_{i}\in F_{\alpha}$$\langle p_{i},\dot{q}_{\alpha}\rangle$

and

. By
such that. This$g\subseteq g_{0}’,$$p_{i}\leq p_{i}’$$\dot{q}_{\alpha}$$\langle p_{i},\dot{q}_{\alpha}\rangle|\vdash f|x_{\alpha}=g_{i}$$g_{i}’=g_{i}$$g_{0},$ $g_{1}\in F_{\alpha}$

andClearly, for each $i<2$ , one can takeand”, eachSince
is compatible with, so $g_{0}\neq 91$ andsuch that$F_{\alpha}$, there are$g_{1}’$$\langle p_{i}’,\dot{q}^{*}\rangle$$\square [Claim]$$g\subseteq g_{0},$ $g_{1}.$Supposeare defined for. Note that, for every
and”, we have; take $p’\leq p$ such that
Then one”$p”\leqp’$
can pickandsuch thatis compatible
with
. So
Let
with the inclusion forms a -tree, and each node of has
at least two immediate successors.
$\dot{q}_{\alpha},$$x_{\alpha},$$g\in F_{\beta}$$\alpha<\beta<\kappa$$\alpha<\kappa$$F_{\alpha}$$\langle p’,\dot{q}_{\beta}\rangle|\vdash f|x_{\beta}=g$$g|x_{\alpha}\in F_{\alpha}$$\langle p",\dot{q}_{\alpha}\rangle|\vdash f|x_{\alpha}=h$$h\in F_{\alpha}$$h=g|x_{\alpha}.$$\langle p",\dot{q}_{\alpha}\rangle$$T= \bigcup_{\alpha<\kappa}F_{\alpha}.$Claim 1.9.

$\langle p’,\dot{q}_{\beta}\rangle$

$T$

$T$

$T$

$\kappa$

has no antichain of size

Proof. For each $g\in T$ , there are$\kappa.$”andsuch thatFor
in , if and are incompatible in , thenis incompatible within
. This means that if has an antichain of size , then also has an antichain of$\square [Claim]$
size . This is impossible, hence does not have an antichain of size$T$$g,$ $g’$

$g$

$T$

$g’$

$T$

$\mathbb{P}$$\langle p_{g},\dot{q}_{\alpha_{g}}\rangle|\vdash;|x_{\alpha_{g}}=g$$\alpha_{g}<\kappa$$p_{g}$$p_{g’}$$p_{g}$$\mathbb{P}$$\kappa$$T$$\kappa$$\kappa.$

Hence is a -Suslin tree. We finish the proof by showing the following claim,
which contradicts the strong c.c. of :$T$$\kappa$$\kappa-$$\mathbb{P}$106
Claim 1.10.

$p|\vdash {}_{\mathbb{P}}T$

has a cofinal branch”.

with $p\in G$ and work in $V[G]$ . Let. Sincefor someis predense below , we can find”andsuch that
Now, for $\alpha<\beta<\kappa,$
is compatible withand. So
is compatible with. This$\square [Claim]$means that, sois a cofinal branch of $T.$

Proof. Take$a(V, \mathbb{P})$-generic$G$$\alpha<\kappa$$\{p’\leq p:\langle p’,\dot{q}_{\alpha}\rangle|\vdash f|x_{\alpha}=g$ ”$p_{\alpha}\in G$$g\in F_{\alpha}\}$$g_{\alpha}\in F_{\alpha}\subseteq T$$p_{\alpha}$$\langle p_{\alpha},\dot{q}_{\alpha}\rangle$$\dot{q}_{\beta}\leq\dot{q}_{\alpha}$$p_{\beta}$$\{g_{\alpha} :$g_{\alpha}\subseteq g_{\beta}$$p$$\langle p_{\alpha},\dot{q}_{\alpha}\rangle|\vdash f|x_{\alpha}=g_{\alpha}$$\langle p_{\beta},$$q_{\beta}\rangle$\alpha<\kappa\}$$\square$Note 1.11, If satisfies the. but does not the strong
have the -approximation property.
$\mathbb{P}$$\kappa-c.c$$\kappa-c.c.$, thencannot$\mathbb{P}$$\kappa$

2. APPLICATIONS
We consider some applications of Lemma 1.5.

Definition 2.1. Let be a regular uncountable cardinal anda cardinal.
set
has the strong tree property if for every
:
, if
with
$|\{d_{x}\cap a:x\in X\}|<\kappa$ for every, then there issuch that for every
the set $\{x\in X:d_{x}\cap a=D\cap a\}$ is unbounded in$A$$\lambda\geq\kappa$$\kappa$$X\subseteq \mathcal{P}_{\kappa}\lambda$$\langle d_{x}$$x\in X\rangle$$d_{x}\subseteq x$$D\subseteq\lambda$$a\in \mathcal{P}_{\kappa}\lambda$$a\in \mathcal{P}_{\kappa}\lambda$$\mathcal{P}_{\kappa}\lambda.$

Fact 2.2 (Viale-Weiss [3]).
(1) The following are equivalent;
has the strong tree property.
$(b)$ There is some unbounded set
such that $X$ has the strong tree$(a)\mathcal{P}_{\kappa}\lambda$$X\subseteq \mathcal{P}_{\kappa}\lambda$

property.
$(c)$ Every unbounded subsetof

has the strong tree property.
(2) has the tree property if and only if
has the strong tree property.
(3) is strongly compact if and only if is inaccessible andhas the strong
tree property for every(4) Suppose Proper Forcing Axiom. Then
has the strong tree property forevery$\mathcal{P}_{\kappa}\lambda$$\mathcal{P}_{\kappa}\kappa$$\kappa$$\kappa$$\mathcal{P}_{\kappa}\lambda$$\kappa$$\lambda\geq\kappa.$$\mathcal{P}_{\omega_{2}}\lambda$$\lambda\geq\omega_{2}.$Viale-Weiss [3] showed that for an inaccessible , if a standard -stage iteration
satisfying the. forces thatand Proper forcing axiom”, then must
be strongly compact in the ground model. The following is a slight improvement$\kappa$$(\kappa=\omega_{2}$$\kappa-c.c$$\kappa$$\kappa$of their result.

Proposition 2.3. Let be a regular uncountable cardinal. Suppose that there is aposet which has the strong. and forces thathas the strong tree property
every
strongproperty”. Thenhas thetree
in theforfor every
ground model.
$\kappa$$\kappa-c.c$$\mathbb{P}$$\lambda\geq\kappa$$\mathcal{P}_{\kappa}\lambda$$\mathcal{P}_{\kappa}\lambda$$\lambda\geq\kappa$

107
Proof. We check that. Fix
has the strong tree property for every
and takefor everysuch thatand. Take-generic and work in $V[G]$ . In $V[G],$is unboundedinsince satisfies the c.c. By the strong tree property ofin $V[G]$ , wecan findis unbounded inforsuch that
$D\in
V$
every
. We see, this completes the proof. For each, there$D\capa=d_{x}\cap
a\in
V$
is
with
. Thus, by the -approximation property of, we have $D\in V.$$\lambda\geq\kappa$$\mathcal{P}_{\kappa}\lambda$$d_{x}\subseteq x$$\langle d_{x}:x\in \mathcal{P}_{\kappa}\lambda\rangle$$|\{d_{x}\cap a:x\in \mathcal{P}_{\kappa}\lambda\}|<\kappa$$G$$a(V, \mathbb{P})$$a\in \mathcal{P}_{\kappa}\lambda$$\mathcal{P}_{\kappa}^{V}\lambda$$\mathbb{P}$$\mathcal{P}_{\kappa}\lambda$$\mathcal{P}_{\kappa}^{V}\lambda$$\kappa-$$\{x\in \mathcal{P}_{\kappa}^{V}\lambda :$D\subseteq\lambda$$\lambda\geq\kappa$d_{x}\cap a=D\cap a\}$$\mathcal{P}_{\kappa}\lambda$$a\in \mathcal{P}_{\kappa}\lambda$$a\in \mathcal{P}_{\kappa}^{V}\lambda$$x\in \mathcal{P}_{\kappa}^{V}\lambda$$\kappa$$\square$$\mathbb{P}$Next we look at the indestructibility of weak compactness.
Definition 2.4. Let be weakly compact. If every -directed closed forcing preserves the weak compactness of , then is said to be indestructibly weakly compact.
$\kappa$$\kappa$$\kappa$$\kappa$

The existence of an indestructibly weakly compact cardinal is consistent (Laver[2] . The following theorem suggests that the consistency of the existence of anindestructibly weakly compact cardinal might be at least strongly compact cardinal.
$)$

Proposition 2.5. Let be a regular uncountable cardinal. If there is a poset which. and forces thatis indestructibly weakly compact”, thensatisfies the strongis strongly compact.
$\kappa$$\kappa-c.c$$\kappa$$\kappa$

Proof. Take$\lambda\geq\kappa$. We see that$\mathcal{P}_{\kappa}\lambda$

has the strong tree property. Take$\langle d_{x}:x\in \mathcal{P}_{\kappa}\lambda\rangle$

with
and
for everyTake-generic , and $a(V[G], Col(\kappa, \lambda))$ -generic $H$ . We work in $V[G][H].$Fix a bijection :. We know thatis unbounded inSince is weakly compact in $V[G][H]$ , by the tree property of , there issuch that $\{\pi x\in \mathcal{P}_{\kappa}\kappa : \pi(d_{x})\cap a=C\cap a\}$ is unbounded for all a. Put$D=\pi^{-1}C$ . Then for every a, the setisunbounded in. We know $D\in V$ sinceCol
has the -approximation
property by Lemma 1.5.
$d_{x}\subseteq x$$|\{d_{x}\cap a:x\in \mathcal{P}_{\kappa}\lambda\}|<\kappa$$a\in \mathcal{P}_{\kappa}\lambda.$$G$$a(V, \mathbb{P})$$\pi$$\lambdaarrow\kappa$$\{\pi x:x\in \mathcal{P}_{\kappa}^{V}\lambda\}$$\mathcal{P}_{\kappa}\kappa.$$\kappa$$C\subseteq\kappa$$\kappa$$\in \mathcal{P}_{\kappa}\kappa$$\{x\in \mathcal{P}_{\kappa}^{v}\lambda :$\in \mathcal{P}_{\kappa}\lambda$$\mathbb{P}*$$\mathcal{P}_{\kappa}\lambda$$(\kappa, \lambda)$d_{x}\cap a=D\cap a\}$$\kappa$$\square$

REFERENCES[1] W. J. Mitchell, On the Hamkins approximation property, Ann. Pure Appl. Logic 144 (2006),
no. 1-3, 126-129.
[2] R. Laver, Making the supercompactness of indestructible under -directed closed forcing.
Israel J. Math. 29 (1978), no. 4, 385-388.
[3] M. Viale, C. Weiss, On the consistency strength of the proper forcing axiom. Adv. Math.
228 (2011), no. 5, 2672-2687.
$\kappa$$\kappa$(T. Usuba) ORGANIZATION OF ADVANCED SCIENCE AND TECHNOLOGY, KOBE UNIVERSITY,ROKKO-DAI 1-1, NADA, KOBE, 657-8501 JAPAN-mail address: usuba@people. kobe-u. ac. jp$E$