# Another method to add a closed unbounded set of former regulars

Moti Gitik  School of Mathematical Sciences, Tel Aviv University, Tel Aviv-
Yafo, Tel Aviv, Israel, 6997801<EMAIL_ADDRESS>and  Sittinon
Jirattikansakul  School of Mathematical Sciences, Tel Aviv University, Tel
Aviv-Yafo, Tel Aviv, Israel, 6997801<EMAIL_ADDRESS>
###### Abstract.

A club consisting of former regulars is added to an inaccessible cardinal,
without changing cofinalities outside it. The initial assumption is optimal. A
variation of the Radin forcing without a top measurable cardinal is introduced
for this.

The authors were partially supported by ISF grant No. 1216/18. The first
author is grateful to Carmi Merimovich for many helpful discussions on the
subject.

##  1\. introduction

From the assumption $\circ(\kappa)=\kappa^{+}$, by Mitchell, Radin forcing [1]
adds a club which consists of ground model regular cardinals, while
maintaining inaccessibility at $\kappa$. The large cardinal assumption for
that is not optimal. Later, the first author [2] constructed a forcing which
adds a club subset of $\kappa$ containing regular cardinals in $V$, while
maintaining inaccessibility of $\kappa$, from the optimal assumption. The
forcing was done in two steps. The first step is to build a $\kappa$-c.c.
forcing iteration which is a cardinal-preserving forcing $\mathbb{P}_{0}$
ascertaining that in $V^{\mathbb{P}_{0}}$, the set
$E:=\\{\alpha<\kappa\mid\alpha$ is regular in $V\\}$ is a fat set, namely for
every $\delta<\kappa$ and every club subset $C$ of $\kappa$, there is a closed
subset of $C\cap E$ of order-type $\delta$. Then the second step was done by
shooting a club subset of $E$ using closed bounded subsets of $E$. By fatness,
the forcing $\mathbb{P}_{1}$ is $(\kappa,\infty)$-distributive. Hence, the
forcing produces a club subset of $\kappa$, preserves all cardinals, preserves
inaccessibility of $\kappa$. However, $V$-regular cardinals even outside the
club added by the forcing maybe singularized. Here we prove the sharper
result. We state our result as the following.

######  Theorem 1.1.

Let $\kappa$ be a strongly inaccessible cardinal. Assume there are a set
$X\subseteq\kappa$, $\circ:X\to\kappa$ ($\circ$ is the function representing
Mithchell orders), and $\vec{U}$ such that

  1. (1)

for $\alpha\in X$, $\circ(\alpha)<\alpha$.

  2. (2)

for $\alpha\in X$,
$\vec{U}(\alpha)=\\{U(\alpha,\beta)\mid\beta<\circ(\alpha)\\}$ is a Mitchell
increasing sequence of normal measures.

  3. (3)

for $\nu<\kappa$, $\\{\alpha\in X\mid\circ(\alpha)\geq\nu\\}$ is stationary.

Then there is a cardinal-preserving extension of size $\kappa$ which adds a
club subset $C$ of $\kappa$ containing only regular cardinals in $V$, $\kappa$
is strongly inaccessible in the forcing extension. Furthermore, $V$-regular
cardinals in $\kappa\setminus\lim(C)$ remain regular in the extension.

In this paper, we present a different method to add a club subset of $\kappa$
while maintaining inaccessibility of $\kappa$. Although we work in detail
where the assumption (as in Assumption 2.1) is slightly stronger than the
assumption in Theorem 1.1, one can easily modify our detail to obtain the same
result from the optimal assumption, namely the assumption in Theorem 1.1.

The organization of this paper is the followings.

  * •

In Section 2 we give an initial assumption, give some basic facts regarding
large cardinals and forcings.

  * •

In Section 3 and 4 we define the forcing $P_{\alpha}$ for $\alpha<\kappa$. The
forcing will approximate the main forcing $\mathbb{P}_{\kappa}$ in the end. We
show basic properties and discuss on quotients.

  * •

In Section 5 we prove the Prikry property for $P_{\alpha}$.

  * •

In Section 6 we investigate the cardinal behavior after forcing with
$P_{\alpha}$. We also investigate the canonical sets $C_{\alpha}$ and
$C_{\alpha/Q}$ derived from generic objects. These are served as the initial
segments of the final club subset of $\kappa$.

  * •

In Section 7 we define the final forcing $\mathbb{P}_{\kappa}$ and show that
it gives the main result.

  * •

In Section 8 we show that the assumption in Theorem 1.1 is optimal.

We assume the readers are familiar with forcings, Radin-Magidor forcing from a
coherent sequence of measure (for example, the version that appear in Section
5.2 of Gitik’s chapter in [3], iterated forcings, and quotient forcings,
although we attempt to present in significant details.

##  2\. Basic facts

We start by stating our initial assumption, which we use from Section 2 to
Section 7.

######  Assumption 2.1.

Assume GCH and $\kappa$ is inaccessible. Assume there is a set
$X\subseteq\kappa$, $\circ:X\to\kappa$, and $\vec{U}$ such that

  1. (1)

for $\alpha\in X$, $0<\circ(\alpha)<\alpha$.

  2. (2)

$\vec{U}=\\{U(\alpha,\beta)\mid\alpha\in X,\beta<\circ(\alpha)\\}$ is a
coherent sequence of normal measures.

  3. (3)

for $\nu<\kappa$, $\\{\alpha\in X\mid\circ(\alpha)\geq\nu\\}$ is stationary.

We remind readers the definition of a coherent sequence of measures. Define
$(\alpha_{0},\beta_{0})<(\alpha_{1},\beta_{1})$ for each ordered pairs
$(\alpha_{0},\beta_{0})$ and $(\alpha_{1},\beta_{1})$ if
$\alpha_{0}<\alpha_{1}$ or ($\alpha_{0}=\alpha_{1}$ and
$\beta_{0}<\beta_{1}$). Let
$\vec{U}\restriction\gamma=\\{U(\alpha,\beta)\in\vec{U}\mid\alpha<\gamma\\}$
and
$\vec{U}\restriction(\alpha,\beta)=\\{U(\alpha^{\prime},\beta^{\prime})\mid(\alpha^{\prime},\beta^{\prime})<(\alpha,\beta)\\}$.
Coherence states that $j:V\to\operatorname{Ult}(V,U(\alpha,\beta))$, then
$j(\vec{U})\restriction(\alpha+1)=\vec{U}\restriction(\alpha,\beta)$. It is
known in [2] that Assumption 2.1 is sufficient to construct a forcing which
adds a club subset of $\kappa$ of ground model regular cardinals, while
maintaining inaccessibility of $\kappa$. We note again that Assumption 2.1 is
slightly stronger than the assumption in Theorem 1.1, which is known to be
optimal to build a forcing to add a club subset of $\kappa$ containing
$V$-regular cardinals, and $\kappa$ is inaccessible in the extension. We can
construct a forcing that gives the same effect using only the assumption in
Theorem 1.1.

Notations and conventions: From this point to Section 7, we assign some
notations and agreements for our convenience. Some of them are not standard
notations. For each pair of sets $A$ and $B$, we denote $A\sqcup B$ the usual
union of $A$ and $B$, whenever $A\cap B=\emptyset$. We assume that all
measure-one sets concentrate on inaccessible cardinals. Define
$\circ(\gamma)=0$ if and only if $\gamma$ is inaccessible and not measurable.
We also assume that $X$ includes all points $\gamma$ with $\circ(\gamma)=0$,
so that $X=\\{\alpha<\kappa\mid\alpha$ is inaccessible$\\}$. For $\alpha\in X$
with $\circ(\alpha)>0$, let
$\vec{U}(\alpha)=\\{U(\alpha,\beta)\mid\beta<\circ(\alpha)\\}$ and we say that
$A\in\vec{U}(\alpha)$ if and only if
$A\in\cap_{\beta<\circ(\alpha)}U(\alpha,\beta)$. If $\circ(\alpha)=0$,
$\vec{U}(\alpha)=\\{\emptyset\\}$. We assume that each $A\in\vec{U}(\alpha)$
is written as a disjoint union of $\langle
A_{\beta}\mid\beta<\circ(\alpha)\rangle$ where $A_{\beta}\in U(\alpha,\beta)$
and for $\gamma\in A_{\beta}$, $\circ(\gamma)=\beta$. For each
$A\in\vec{U}(\alpha)$ and $\gamma\in A$, we define
$A\restriction\gamma=\\{\xi\in A\mid\xi<\gamma$ and
$\circ(\xi)<\circ(\gamma)\\}$ and $A\setminus\gamma=\\{\xi\in
A\mid\xi>\gamma\\}$. We note that if $\gamma\in A\in\vec{U}(\alpha)$ and
$\circ(\gamma)=0$, then $A\restriction\gamma=\emptyset$. Assume that for
$\gamma\in A$, $A\restriction\gamma\in\vec{U}(\gamma)$. For $P$-name of
ordinal $\dot{\nu}$, define $\bar{\nu}=\sup\\{\nu\mid\exists p\in
P(p\Vdash\dot{\nu}=\check{\nu})\\}+1$. For each function $F$ with domain $A$
of ordinals, and $\gamma$ is an ordinal, let
$F\setminus\gamma=F\restriction(A\setminus\gamma)$. When we force with a
forcing notion $P_{\alpha}$, then the corresponding relations and other
relevant objects may be distinguished by the subscript $\alpha$, e.g.
$\leq_{\alpha}$ and $\leq^{*}_{\alpha}$, $\Vdash_{\alpha}$, and so on. If
$\dot{\nu}$ is a $Q$-name of an ordinal which is at least $\alpha$, we put
subscripts $\dot{\nu}/Q$ for the forcing $P_{\dot{\nu}}/Q$ e.g.
$\leq_{\dot{\nu}/Q}$ and $\leq^{*}_{\dot{\nu}/Q}$. If $Q=P_{\alpha}$, then we
abbreviate the subscript $\dot{\nu}/Q$ by $\dot{\nu}/\alpha$. We remove
subscripts sometimes if they are clear from the context. We often omit the
check symbols for the names representing ground model elements in a forcing
(except in Section 3 where we emphasize the places where names are used). Fix
a limit ordinal $\gamma$. A club in $\gamma$ is a closed unbounded subset of
$\gamma$. Even though $\operatorname{cf}(\gamma)=\omega$, a cofinal subset of
$\gamma$ of order-type $\omega$ is also considered as a club.

We start off in this section by recalling some known facts, which will be used
in further sections.

######  Proposition 2.2.

Let $A\in\vec{U}(\alpha)$. For $\gamma\in A$, let
$B_{\gamma}\in\vec{U}(\gamma)$ and $B_{\gamma}\subseteq A\restriction\gamma$.
Then there is $A^{*}\in\vec{U}(\alpha)$, such that $A^{*}\subseteq A$, and for
$\gamma\in A^{*}$, $A^{*}\restriction\gamma\subseteq B_{\gamma}$.

###### Proof.

For each $\beta<\circ(\alpha)$, we define $X_{1}^{\beta}\cup X_{2}^{\beta}\cup
X_{3}^{\beta}$ as follows. Let
$X_{1}^{\beta}=j_{U(\alpha,\beta)}(\gamma\mapsto B_{\gamma})(\alpha)$. Then
$X_{1}^{\beta}\in
j_{U(\alpha,\beta)}(\vec{U})(\alpha)=\vec{U}(\alpha)\restriction\beta$. Also,
$X_{2}^{\beta}:=\\{\gamma\in A\mid\circ(\gamma)=\beta$ and
$X_{1}^{\beta}\cap\gamma=B_{\gamma}\\}\in U(\alpha,\beta)$. Finally, let
$X_{3}^{\beta}=\\{\gamma\in A\mid\circ(\gamma)>\beta\\}$. Then
$X_{3}^{\beta}\in\vec{U}(\alpha)\setminus(\beta+1)$. Let
$A^{*}=\cap_{\beta<\circ(\alpha)}(X_{1}^{\beta}\cup X_{2}^{\beta}\cup
X_{3}^{\beta})$. By $\alpha$-completeness of measures,
$A^{*}\in\vec{U}(\alpha)$. Furthermore, we have that for $\gamma\in A^{*}$
with $\circ(\gamma)=\beta$, $A^{*}\restriction\gamma\subseteq
X_{1}^{\beta}\cap\gamma\subseteq B_{\gamma}$.

∎

######  Proposition 2.3.

Let $P\in V_{\xi+1}$ be a forcing notion and $G$ be $P$-generic. Then there is
a coherent sequence of measures $\langle
U(\alpha,\beta)\mid\beta<\circ(\alpha),\alpha>\xi,$ and $\alpha\in X\rangle$,
each $U^{\prime}(\alpha,\beta)$ extends $U(\alpha,\beta)$.

###### Proof.

In $V[G]$, for $\rho>\xi$, define
$U^{\prime}(\rho,\gamma)=\\{A\subseteq\rho\mid\exists B\in
U(\rho,\gamma)(B\subseteq A)\\}$. For each ordered pair $(\alpha,\beta)$,
$\alpha>\xi$, let $j:=j_{U(\alpha,\beta)}:V\to M$. Since $P\in V_{\xi+1}$ and
$\alpha>\xi$ is inaccessible, $j$ extends to $j:V[G]\to M[G]$. Since
$\operatorname{dom}(\vec{U}^{\prime})=\\{(\rho,\xi)\in\operatorname{dom}(\vec{U})\mid\rho>\xi\\}$,
$\operatorname{dom}(j(\vec{U}^{\prime}))\restriction\alpha+1=\operatorname{dom}(\vec{U}^{\prime})\restriction(\alpha,\beta)$.
Since $\operatorname{crit}(j)=\alpha$, $M[G]\subseteq V[G]$, and
$V_{\alpha+1}^{M[G]}=V_{\alpha+1}^{V[G]}$, by following the coherence of
$\vec{U}$, we have that
$j(\vec{U}^{\prime})\restriction\alpha+1=\vec{U}^{\prime}\restriction(\alpha,\beta)$.

∎

Observe that in Proposition 2.3, $U(\rho,\xi)$ is dense in
$U^{\prime}(\rho,\xi)$: $U(\rho,\xi)\subseteq U^{\prime}(\rho,\xi)$ and for
$A\in U^{\prime}(\rho,\xi)$, there is $B\in U(\rho,\xi)$ such that $B\subseteq
A$.

######  Proposition 2.4.

Let $P\in V_{\alpha}$ and $\alpha\in X$. If $\dot{A}$ is a $P$-name of an
element in $\vec{U}(\alpha)$. Then there is $B\in\vec{U}(\alpha)$ such that
$\Vdash_{P}\check{B}\subseteq\dot{A}$.

###### Proof.

Take $B=\cap\\{A^{\prime}\mid\exists p\in
P(p\Vdash_{P}\dot{A}=\check{A}^{\prime}\rangle\\}$. Then $B\in\vec{U}(\alpha)$
since $|P|<\alpha$.

∎

We end this section by recalling that if for $\gamma<\alpha$,
$A_{\gamma}\in\vec{U}(\alpha)$, the diagonal intersection of $\langle
A_{\gamma}\mid\gamma<\alpha\rangle$, denoted by
$\triangle_{\gamma<\alpha}A_{\gamma}$, is the set
$\\{\gamma<\alpha\mid\forall\xi<\gamma(\gamma\in A_{\xi})\\}$. It is easy to
check that $\triangle_{\gamma<\alpha}A_{\gamma}\in\vec{U}$. A variation of the
diagonal intersection is also defined: for $\vec{\gamma}\in[\alpha]^{n}$, let
$A_{\vec{\gamma}}\in\vec{U}(\alpha)$, we define
$\triangle_{\vec{\gamma}}A_{\vec{\gamma}}=\triangle_{\xi<\alpha}\cap_{\\{\vec{\gamma}\mid\max(\vec{\gamma})=\xi\\}}A_{\vec{\gamma}}$.

##  3\. The forcing $\mathbb{P}_{\alpha}$: basic cases

In Section 3 and 4 we build forcings $P_{\alpha}$ for $\alpha\in X$ by
recursion. In this section we define $P_{\alpha}$ for the first two ordinals
in $X$. The reason we define forcings at two levels is that we want to
introduce quotient forcings, which are more meaningful when we consider two
levels. All the forcings and the quotients in this section are equivalent to
the trivial forcing. However, the construction of $P_{\alpha}$ for $\alpha\in
X$ and their quotients which appear in both sections will satisfy a long list
of the properties, which will be verified inductively. We also introduce all
necessary notions which will appear in the list of properties, which will be
elaborated later in Proposition 4.1.

######  Definition 3.1.

(The first basic case)

Let $\alpha_{0}=\min(X)$.

  1. (1)

Define $P_{\alpha_{0}}$ as $\\{1_{\alpha_{0}}$, $p_{0}\\}$ where
$p_{0}=\langle\alpha_{0},\emptyset,\emptyset\rangle$. Define the forcing
relation $\leq$ on $P_{\alpha_{0}}$ so that $p_{0}$ is stronger than
$1_{\alpha_{0}}$. Let $\leq^{*}=\leq$. We call $p_{0}$ a pure condition. There
are no impure conditions. The stems for both conditions are empty. Suppose $G$
is $P_{\alpha_{0}}$-generic, then $G=P_{\alpha_{0}}$. In $V[G]$, define
$C_{\alpha_{0}}=\\{\alpha_{0}\\}$, and let $\dot{C}_{\alpha_{0}}$ be the
$P_{\alpha_{0}}$-name of such $C_{\alpha_{0}}$.

  2. (2)

Assume $G$ is $P_{\alpha_{0}}$-generic, define
$(P_{\alpha_{0}}[G],\leq,\leq^{*})$ as the trivial forcing
$(\\{\emptyset\\},\leq,\leq^{*})$ in $V[G]$. Let
$P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$ be a $P_{\alpha_{0}}$-name of the
forcing $P_{\alpha_{0}}[G]$, where $\check{\alpha}_{0}$ is the
$P_{\alpha_{0}}$-name. If $H$ is $P_{\alpha_{0}}[G]$-generic, in $V[G][H]$,
define $C_{\alpha_{0}/\alpha_{0}}=\emptyset$, and let
$\dot{C}_{\check{\alpha}_{0}/\alpha_{0}}$ be a
$P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$-name for such
$C_{\alpha_{0}/\alpha_{0}}$.

  3. (3)

If $G$ is $P_{0}$-generic, define $P_{\check{\alpha}_{0}}/P_{\alpha_{0}}[G]$
as $P_{\alpha_{0}}[G]$.

######  Definition 3.2.

(The second basic case)

Let $\alpha_{0}=\min(X)$ and $\alpha_{1}=\min(X\setminus(\alpha_{0}+1))$.

  1. (1)

Define $P_{\alpha_{1}}$ as $\\{1_{\alpha_{1}},p_{1},p_{01}\\}$,
$p_{1}=\langle\alpha_{1},\emptyset,\emptyset\rangle$,
$p_{01}=\langle\alpha_{0},\emptyset,\emptyset,\check{\alpha}_{0},\check{\emptyset}\rangle^{\frown}\langle\alpha_{1},\emptyset,\emptyset\rangle$
where the checks are considered as a $P_{\alpha_{0}}$-name. Define the forcing
relation $\leq$ on $P_{\alpha_{1}}$ so that $1_{\alpha_{1}}$ is the weakest
condition, and $p_{1}$ and $p_{01}$ are not comparable. Let $\leq^{*}=\leq$.
We call $p_{1}$ a pure condition and $p_{01}$ an impure condition. The stems
for $1_{\alpha_{1}}$ and $p_{1}$ are empty, and the stem for $p_{01}$ is
$\langle\alpha_{0},\emptyset,\emptyset,\check{\alpha}_{0},\check{\emptyset}\rangle$.
If $G$ is $P_{\alpha_{0}}$-generic, then either
$G=\\{1_{\alpha_{1}},p_{1}\\}$, which we define the corresponding
$C_{\alpha_{1}}=\\{\alpha_{1}\\}$, otherwise, $G=\\{1_{\alpha_{1}},p_{01}\\}$,
in which case, we define $C_{\alpha_{1}}=\\{\alpha_{0},\alpha_{1}\\}$. Let
$\dot{C}_{\alpha_{1}}$ be a $P_{\alpha_{1}}$-name for such $C_{\alpha_{1}}$.

  2. (2)

Assume $G$ is $P_{\alpha_{0}}$-generic, define $P_{\alpha_{1}}[G]$ as
$\\{1_{\alpha_{1}},\langle\alpha_{0},\emptyset\rangle^{\frown}\langle\alpha_{1},\emptyset,\emptyset\rangle\\}$.
We consider $1_{\alpha_{1}}$ as the weakest condition and $\leq^{*}=\leq$. If
$H$ is $P_{\alpha_{1}}[G]$-generic, define
$C_{\alpha_{1}/\alpha_{0}}=\\{\alpha_{1}\\}$, and let
$\dot{C}_{\check{\alpha}_{1}/\alpha_{0}}$ be a
$P_{\alpha_{0}}*P_{\check{\alpha}_{1}/\alpha_{0}}$-name for such
$C_{\alpha_{1}/\alpha_{0}}$. We do not define pure, impure conditions, and
stems in this case. Define $P_{\check{\alpha}_{1}}/P_{\alpha_{0}}$ as the
$P_{\alpha_{0}}$-name of the forcing $P_{\alpha_{1}}[\dot{G}]$ where $\dot{G}$
is the canonical name for generic for $P_{\alpha_{0}}$, where the check is the
$P_{\alpha_{0}}$-name.

  3. (3)

Assume $G$ is
$Q:=P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$-generic where the
check is the $P_{\alpha_{0}}$-name, define
$P_{1}[G]=\\{1_{\alpha_{1}},\langle\alpha_{1},\emptyset,\emptyset\rangle\\}$
(in $V[G]$), with the induced ordering in $P_{1}$, i.e., $1_{\alpha_{1}}$ is
the weakest element. Define
$\langle\alpha_{1},\emptyset,\emptyset\rangle[G]=\langle\alpha_{1},\emptyset,\emptyset\rangle$.
The condition $\langle\alpha_{1},\emptyset,\emptyset\rangle$ is a pure
condition. The stems of both conditions are empty. There are no impure
conditions. Define $P_{\alpha_{1}}/Q$ as the $Q$-name of the forcing
$P_{\alpha_{1}}[\dot{G}]$ where $\dot{G}$ is the canonical name for generic
for $Q$. Define $C_{\alpha_{1}/Q}=\\{\alpha_{1}\\}$. Let
$\dot{C}_{\alpha_{1}/Q}$ be a $P_{Q}*P_{\check{\alpha}_{1}/Q}$ be a name for
such $C_{\alpha_{1}/Q}$.

  4. (4)

Assume $G$ is $P_{\alpha_{1}}$-generic, let $P_{\alpha_{1}}[G]$ as the trivial
forcing $(\\{\emptyset\\},\leq,\leq^{*})$ in $V[G]$. Define
$P_{\check{\alpha}_{1}}/P_{\alpha_{1}}$ (where the check is the
$P_{\alpha_{1}}$-name) as the $P_{\alpha_{1}}$-name of $P_{\alpha_{1}}[G]$. If
$H$ is $P_{\alpha_{1}}[G]$-generic, in $V[G][H]$, define
$C_{\alpha_{1}/\alpha_{1}}=\emptyset$. Let
$\dot{C}_{\check{\alpha_{1}}/{\alpha_{1}}}$ be a
$P_{\alpha_{1}}*P_{\check{\alpha}_{1}}/P_{\alpha_{1}}$-name for such
$C_{\alpha_{1}/\alpha_{1}}$.

  5. (5)

For $i=0,1$, if $G$ is $P_{i}$-generic, define
$P_{\check{\alpha}_{1}}/P_{\alpha_{i}}[G]=P_{\alpha_{1}}[G]$.

Note that $p_{01}$ in Definition 3.2 is considered in the following manner.
First we consider the first tuple of $p_{01}$, which is
$\langle\alpha_{0},\emptyset,\emptyset,\check{\alpha}_{0},\check{\emptyset}\rangle$.
The first three entries $\langle\alpha_{0},\emptyset,\emptyset\rangle$ is an
element in $P_{\alpha_{0}}$. The fourth entry $\check{\alpha}_{0}$ is
considered as a $P_{\alpha_{0}}$-name of an ordinal in the interval
$[\alpha_{0},\alpha_{1})\cap X$, which is just $\\{\alpha_{0}\\}$. It serves
as a parameter in the quotient forcing the fifth entry is living in. In this
case, it is $P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$, where the denominator is
the first entry of the tuple. The last entry $\check{\emptyset}$ is a
$P_{\alpha_{0}}$-name of the element in
$P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$. Hence,
$\langle\alpha_{0},\emptyset,\emptyset,\check{\alpha}_{0},\check{\emptyset}\rangle$
is regarded as a condition in
$P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$. When $G$ is
$P_{\alpha_{0}}$-generic, we consider $P_{\alpha_{1}}[G]$ as the “leftover” of
the conditions in $P_{\alpha_{1}}$ whose initial segments are in $G$. For
example, the condition
$p_{01}=\langle\alpha_{0},\emptyset,\emptyset,\check{\alpha}_{0},\check{\emptyset}\rangle^{\frown}\langle\alpha_{1},\emptyset,\emptyset\rangle$
can be factored to $\langle\alpha_{0},\emptyset,\emptyset\rangle\in
P_{\alpha_{0}}$, and the rest:
$\langle\check{\alpha}_{0},\check{\emptyset}\rangle^{\frown}\langle\alpha_{1},\emptyset,\emptyset\rangle$.
If $G$ is $P_{\alpha_{0}}$-generic, then the non-trivial element in
$P_{\alpha_{1}}[G]$ is
$\langle\check{\alpha}_{0},\check{\emptyset}\rangle^{\frown}\langle\alpha_{1},\emptyset,\emptyset\rangle[G]=\langle\alpha_{0},\emptyset\rangle^{\frown}\langle\alpha_{1},\emptyset,\emptyset\rangle$
which is considered as a condition in $P_{\alpha_{0}}[G]\times
P_{\alpha_{1}}[G]$.

We now list all the properties which we need to be true in the more general
settings. Stronger versions of some properties are true, but we state in a way
that is aligned together with properties in general cases, i.e. Proposition
4.1. Let $\alpha_{0}$ and $\alpha_{1}$ be as in Definition 3.2. Let $0\leq
i\leq j\leq 1$.

  * •

$(P_{\alpha_{0}},\leq,\leq^{*})$ and $(P_{\alpha_{1}},\leq,\leq^{*})$ have the
Prikry property , preserve all cardinals, and GCH.
$\Vdash_{\alpha_{i}}(P_{\check{\alpha}_{j}}/P_{\alpha_{i}},\leq,\leq^{*})$ has
the Prikry property, preserves all cardinals, and GCH, and
$\Vdash_{\alpha_{i}}P_{\check{\alpha}_{i}}/P_{\alpha_{i}}$ is the trivial
forcing.

  * •

$|P_{\alpha_{0}}|\leq\alpha_{0}^{+}$, $|P_{\alpha_{1}}|\leq\alpha_{1}^{+}$
(they are actually finite). Hence, they are $\alpha_{0}^{++}$-c.c., and
$\alpha_{1}^{++}$-c.c., respectively. $\leq_{\alpha_{0}}^{*}$ is
$\alpha_{0}$-closed and $\leq_{\alpha_{1}}^{*}$ is $\alpha_{1}$-closed (both
are $\infty$-closed). Furthermore,
$\Vdash_{\alpha_{i}}(P_{\check{\alpha}_{j}}/P_{\alpha_{i}},\leq^{*})$ is
$\min(X\setminus(\alpha_{i}+1))-$closed (it is $\infty$-closed). It is easy to
check that
$\Vdash_{\alpha_{i}}|P_{\check{\alpha}_{j}}/P_{\alpha_{i}}|\leq\alpha_{j}^{+}$
(it is finite), hence $\alpha_{j}^{++}$-c.c., and $\leq^{*}$ is
$\alpha_{j}$-closed (it is $\infty$-closed). Hence, by counting nice names,
the two-step iteration $P_{\alpha_{i}}*P_{\check{\alpha}_{j}}/P_{\alpha_{i}}$
has size $\alpha_{j}^{+}$ in $V$, hence $\alpha_{j}^{++}$-c.c., has the Prikry
property, preserves all cardinals, and GCH.

  * •

Consider the condition $p_{01}\in P_{\alpha_{1}}$ as in Definition 3.2. Set
$P^{s}=P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$. Let $G$ be
$P^{s}$-generic (it is unique). Then, in $V[G]$, $P_{\alpha_{1}}[G]$ has the
Prikry property, preserves all cardinals, GCH,
$|P_{\alpha_{1}}[G]|\leq\alpha_{1}^{+}$, so $\alpha_{1}^{++}$-c.c. Notice that
$p_{01}=s^{\frown}p_{1}$, where $s\in G$, and $p_{1}=p_{1}[G]\in
P_{\alpha_{1}}[G]$.

  * •

If $p\in P_{\alpha_{i}}$, $\gamma<\alpha_{i}$, $\langle
p_{\beta}\mid\beta<\gamma\rangle$ is a sequence of conditions such that
$p_{\beta}\leq^{*}_{\alpha_{i}}p$ for all $\beta$, then there is
$p^{*}\leq^{*}_{\alpha_{i}}p$ such that $p^{*}\leq^{*}_{\alpha_{i}}p_{\beta}$
for all $\beta$. If $\Vdash_{i}\dot{q}\in
P_{\check{\alpha}_{j}}/P_{\alpha_{i}}$,
$\gamma<\min(X\setminus(\alpha_{i}+1))$, and for $\beta<\gamma$, there are
$p_{\beta},\dot{q}_{\beta}$ such that $p_{\beta}\in P_{\alpha_{i}}$,
$p_{\beta}\Vdash_{i}\dot{q}_{\beta}\leq^{*}\dot{q}$, then there is
$\dot{q}^{*}$ such that $\Vdash\dot{q}^{*}\leq^{*}\dot{q}$ and for all
$\beta$, $p_{\beta}\Vdash\dot{q}^{*}\leq^{*}q_{\beta}$.

  * •

Let $p_{01}\in P_{\alpha_{1}}$ be as before. Set
$Q=P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$ and
$s:=\operatorname{stem}(p_{01})=\langle\alpha_{0},\emptyset,\emptyset,\check{\alpha}_{0},\check{\emptyset}\rangle\in
Q$. Set $P^{s}=P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$. Let $G$
be $P^{s}$-generic, and $H$ be $P_{\alpha_{1}}[G]$-generic containing
$\langle\alpha_{1},\emptyset,\emptyset\rangle$. By setting
$I=\\{1_{\alpha_{1}},p_{01}\\}$, we have that $I$ is $P_{\alpha_{1}}$-generic,
$V[G][H]=V[I]$. Conversely, assume $I$ is $P_{\alpha_{1}}$-generic containing
$p_{01}$. Let $G$ be $P^{s}$-generic, and $H$ be $P_{1}[G]$-generic (these are
unique), then $V[G][H]=V[I]$. Notice that $p_{01}=p_{01}\restriction
Q{}^{\frown}p_{1}$ where $p_{01}\restriction Q=\operatorname{stem}(p_{01})\in
Q$, and $p_{1}=p_{1}[G]\in P_{\alpha_{1}}[G]$.

######  Definition 3.3.

Let $Q=P_{\alpha_{0}}$ or
$P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$ and $G$ be $Q$-generic.
Define $\Vdash_{P_{\alpha_{1}}}[G]\dot{q}\in
P_{\check{\alpha}_{1}}/P_{\alpha_{1}}[G]$ iff
$\dot{q}\in(P_{\check{\alpha}_{1}})[\dot{I}]$ where $\dot{H}$ is the name for
$P_{\alpha_{1}}[G]$-generic, and $\dot{I}$ is the corresponding canonical name
for $P_{\alpha_{1}}$-generic, i.e.
$\Vdash_{P_{\alpha_{1}}[G]}\dot{I}=G*\dot{H}$.

  * •

If $G$ is $P_{\alpha_{0}}$-generic, then
$C_{\alpha_{0}}=\\{\alpha_{0}\\}\subseteq\alpha_{0}+1$,
$C_{\alpha_{0}}\setminus\\{\alpha_{0}\\}=\emptyset$ is a closed bounded subset
of $\alpha_{0}$. If $G$ is $P_{\alpha_{1}}$-generic, then $C_{\alpha_{1}}$ is
either $\\{\alpha_{1}\\}$ or $\\{\alpha_{0},\alpha_{1}\\}$, both are subsets
of $\alpha_{1}+1$. $C_{\alpha_{1}}\setminus\\{\alpha_{0}\\}$ is either
$\emptyset$ or $\\{\alpha_{0}\\}$ which are closed bounded subsets of
$\alpha_{1}$. $C_{\alpha_{i}/\alpha_{i}}=\emptyset$.
$C_{\alpha_{1}/\alpha_{0}}=\\{\alpha_{1}\\}\subseteq\alpha_{1}+1$, and
$C_{\alpha_{1}/\alpha_{0}}\setminus\\{\alpha_{1}\\}=\emptyset$ is a closed
bounded subset of $\alpha_{1}$. All regular cardinals are regular in these
extensions.

  * •

Consider $p_{01}\in P_{\alpha_{1}}$ as before, $G$ is
$Q:=P_{\alpha_{0}}*P_{\check{\alpha}_{0}}/P_{\alpha_{0}}$-generic, $H$ is
$P_{\alpha_{1}}[G]$-generic, and $I$ be the canonical $P_{\alpha_{1}}$-generic
such that $V[G][H]=V[I]$. We see that in $V[I]$,
$C_{\alpha_{0}}=\\{\alpha_{0}\\}$,
$C_{\alpha_{0}/\alpha_{0}}=\\{\emptyset\\}$,
$C_{\alpha_{1}/Q}=\\{\alpha_{1}\\}$. Note that $I$ is $P_{\alpha_{1}}$-generic
containing $p_{01}$ so that
$C_{\alpha_{1}}=\\{\alpha_{0},\alpha_{1}\\}=C_{\alpha_{0}}\sqcup
C_{\alpha_{0}/\alpha_{0}}\sqcup C_{\alpha_{1}/Q}$.

##  4\. The forcings $P_{\alpha}$: the general cases

We now state a list of properties that we wish to be true in general. Note
that all the properties are true in our basic cases (we did not prove the
properties for every possible case, but one can simply apply the arguments
from Section 3 to check all the cases).

######  Proposition 4.1.

(The inductive assumptions)

Fix $\alpha\in X$, and $P_{\alpha}$-name $\dot{\nu}$ with
$\Vdash_{\alpha}\dot{\nu}\in X$ and $\dot{\nu}\geq\alpha$. The following
properties hold:

  1. (1)

$(P_{\alpha},\leq_{\alpha},\leq^{*}_{\alpha})$ has the Prikry property,
preserves all cardinals, and GCH.
$(P_{\dot{\nu}}/P_{\alpha},\leq_{\dot{\nu}/\alpha},\leq^{*}_{\dot{\nu}/\alpha})$
is forced by $P_{\alpha}$ to have the Prikry property, preserve all cardinals,
and GCH. $\Vdash_{\alpha}\dot{\nu}=\alpha$ implies
$(P_{\dot{\nu}}/P_{\alpha},\leq,\leq^{*})$ is the trivial forcing.

  2. (2)

$|P_{\alpha}|\leq\alpha^{+}$, hence $\alpha^{++}$-c.c. $(\\{p\in
P_{\alpha}\mid p$ is pure$\\},\leq^{*})$ is $\alpha$-closed
$\Vdash_{\alpha}|P_{\dot{\nu}}/P_{\alpha}|\leq\dot{\nu}^{+}$, hence
$\dot{\nu}^{++}$-c.c., $(\\{p\in P_{\dot{\nu}}/P_{\alpha}\mid p$ is
pure$\\},\leq^{*})$ is $\dot{\nu}$-closed.

Recall from our notation that $\bar{\nu}=\sup\\{\nu\mid\exists p\in
P_{\alpha}(p\Vdash_{\alpha}\dot{\nu}=\check{\nu})\\}+1$. Note that
$\bar{\nu}<\kappa$ since $|P_{\alpha}|<\kappa$ and $\dot{\nu}$ is forced to be
less than $\kappa$. By the basic facts on two-step iteration, and GCH, the
forcing $P_{\alpha}*P_{\dot{\nu}}/P_{\alpha}$ has size at most
$\bar{\nu}^{+}$, has $\bar{\nu}^{++}$-c.c., preserves cardinals and GCH.
Furthermore, $(\\{p\in
P_{\alpha}\mid((\operatorname{stem}(p)_{0})_{0})=\gamma\\},\leq^{*})$ is
$\gamma$-closed and $\Vdash_{\gamma}(P_{\dot{\nu}}/P_{\alpha},\leq^{*})$ is
$\min(X\setminus(\alpha+1))$-closed.

  3. (3)

Let $Q$ be $P_{\beta}$ or $P_{\beta}*P_{\dot{\gamma}}/P_{\beta}$ where
$\beta<\alpha$, $\Vdash_{\beta}\beta\leq\dot{\gamma}<\alpha$. We define

$\displaystyle\Vdash_{Q}(\dot{q}\in P_{\alpha}/Q\Leftrightarrow\exists
p_{0}^{\frown}p_{1}\in Q(p_{0}\in Q\text{ and }p_{1}[\dot{G}]=\dot{q}))$ and
$\displaystyle(\Vdash_{Q}\dot{q}\in P_{\alpha}/Q)\text{ iff for every
}Q\text{-generic }G,\dot{q}[G]\in P_{\alpha}[G].$

(the notion $P_{\alpha}[G]$ and $p_{1}[G]$ will be explained further in
Definition 4.10). Then $\Vdash_{Q}(P_{\alpha}/Q,\leq,\leq^{*})$ has the Prikry
property, preserves all cardinals, GCH. With nice names,
$|P_{\alpha}/Q|^{V}\leq\alpha^{+}$, and so $(P_{\alpha}[G],\leq)$ is
$\alpha^{++}$-c.c. If pure conditions are defined in $P_{\alpha}[G]$, then
$(\\{p\in P_{\alpha}[G]\mid p$ is pure$\\},\leq^{*})$ is $\alpha$-closed. If
$Q$ is $P_{\alpha}$, then $P_{\alpha}/P_{\alpha}$ is the trivial forcing.

  4. (4)

Let $p\in P_{\alpha}$, $\operatorname{stem}(p_{0})_{0}=\gamma$. Suppose that
$\beta<\gamma$ and for all $\xi<\beta$, $p_{\xi}\leq^{*}_{\alpha}p$, then
there is $p^{*}\leq^{*}_{\alpha}p$ such that for all $\xi<\beta$,
$p^{*}\leq^{*}_{\alpha}p_{\xi}$. Furthermore, let $\Vdash_{\alpha}\dot{q}\in
P_{\dot{\nu}}/P_{\alpha}$. Let $\beta^{\prime}<\min(X\setminus(\alpha+1))$.
Suppose that for $\xi<\beta$, there are $p_{\xi}\in P_{\alpha}$ and
$\dot{q}_{\xi}$ such that
$p_{\xi}\Vdash_{\alpha}\dot{q}_{\xi}\leq^{*}\dot{q}$. Then there is
$\dot{q}^{*}$ such that $\Vdash_{\alpha}\dot{q}^{*}\leq^{*}\dot{q}$ and for
all $\xi$, $p_{\xi}\Vdash_{\alpha}\dot{q}^{*}\leq^{*}\dot{q}_{\xi}$.

  5. (5)

With the notions in (3), let $H$ be $V[G]$-generic over $P_{\alpha}[G]$. Then
there is a canonical generic $I$ over $P_{\alpha}$ such that $V[G][H]=V[I]$.
$I$ is defined as $\\{p\in P_{\alpha}\mid p$ is of the form
$p_{0}^{\frown}p_{1}$, $p_{0}\in G$, $p_{1}[G]\in H\\}$. Conversely, let $Q$
be the forcing $P_{\beta}$ or $P_{\dot{\gamma}}/P_{\beta}$ where
$\Vdash_{\beta}\beta\leq\dot{\gamma}<\alpha$. Let $I$ be $P_{\alpha}$-generic,
$p\in I$, and $p\restriction Q\in Q$ exists. Define

$G:=\\{p^{\prime}\restriction Q\mid p^{\prime}\in I$ and
$p^{\prime}\restriction Q$ exists$\\}$

is $Q$-generic,

$H:=\\{(p^{\prime}[G]\mid\exists t\in G(t^{\frown}p^{\prime}\in I)\\}$

is $V[G]$-generic over $P_{\alpha}[G]$, and $V[G][H]=V[I]$. This makes
$P_{\alpha}$ to be equivalent to $Q*P_{\alpha}/Q$.

  6. (6)

With the notions in (3),
$(P_{\dot{\nu}}/P_{\alpha})[G]:=P_{\dot{\nu}[G]}[G]/P_{\alpha}[G]$ is also
defined ($\dot{\nu}[G]$ is a $P_{\alpha}[G]$-name) as follows:
$\Vdash_{P_{\alpha}[G]}\dot{q}\in(P_{\dot{\nu}}/P_{\alpha})[G]$ iff
$\dot{q}\in(P_{\dot{\nu}})[G][\dot{H}]$ where $\dot{H}$ is the canonical name
for $P_{\alpha}[G]$-generic. If $I$ is $P_{\alpha}$-generic, we define
$P_{\dot{\nu}}[I]/P_{\alpha}[I]$ as $P_{\dot{\nu}[I]}[I]$, which was defined
in (5).

  7. (7)

The notions $\dot{C}_{\alpha}$ and $\dot{C}_{\dot{\nu}/\alpha}$ are sets
defined from generics over $P_{\alpha}$ and $P_{\dot{\nu}}/P_{\alpha}$,
respectively. Let $G$ be $P_{\alpha}$-generic and
$C_{\alpha}=\dot{C}_{\alpha}[G]$. Then $C_{\alpha}\subseteq\alpha+1$,
$\max(C_{\alpha})=\alpha$. If $\circ(\alpha)=0$, then
$C_{\alpha}\setminus\\{\alpha\\}$ is a closed unbounded subset of $\alpha$.
Otherwise, $C_{\alpha}\setminus\\{\alpha\\}$ is a club subset of $\alpha$ and
singularizes $\alpha$ to have cofinality
$\operatorname{cf}(\omega^{\circ(\alpha)})$ (where the exponentiation is the
ordinal one). If $\lambda$ is regular in $V$ and
$\lambda\not\in\lim(C_{\alpha}\setminus\\{\alpha\\})$, then $\circ(\lambda)=0$
and $\lambda$ is regular in $V[G]$. Now, let $H$ be
$(P_{\dot{\nu}})[G]$-generic, $\nu=\dot{\nu}[G]$, and
$C_{\nu/\alpha}=\dot{C}_{\nu/\alpha}[G][H]$. Then,
$C_{\nu/\alpha}\subseteq(\alpha,\nu]$ so that if $\nu=\alpha$,
$C_{\nu/\alpha}=\emptyset$, otherwise $\max(C_{\nu/\alpha})=\nu$. Assume
$\nu\neq\alpha$, if $\circ(\nu)=0$, $C_{\nu/\alpha}\setminus\\{\nu\\}$ is a
closed bounded subset of $\nu$, otherwise, $C_{\nu/\alpha}\setminus\\{\nu\\}$
is a club subset of $\nu$, which singularizes $\nu$ to have cofinality
$\operatorname{cf}(\omega^{\circ(\nu)})$ (where the exponentiation is the
ordinal one). If $\lambda$ is regular in $V[G]$ and
$\lambda\not\in\lim(C_{\nu/\alpha})$, then $\circ(\lambda)=0$ and $\lambda$ is
regular in $V[G][H]$.

  8. (8)

Fix the notions as in (3) and (5), i.e. $Q$ is either $P_{\beta}$ or
$P_{\dot{\gamma}}/P_{\beta}$ where
$\Vdash_{\beta}\beta\leq\dot{\gamma}<\alpha$, $G$ is $Q$-generic, $H$ is
$P_{\alpha}[G]$-generic, and $I$ is the canonical generic such that
$V[I]=V[G][H]$. Consider the model $V[I]$. If $Q=P_{\beta}$, let $C_{\beta}$
be derived from $G$ and let $C_{Q}=C_{\beta}$. If
$Q=P_{\beta}*P_{\dot{\gamma}}/P_{\beta}$, let $C_{\beta}$ be derived from
$G\restriction P_{\beta}$, $\gamma=\dot{\gamma}[G\restriction P_{\beta}]$,
$C_{\gamma/\beta}$ be derived from
$G\restriction(P_{\dot{\gamma}}[G\restriction\beta])$, and
$C_{Q}=C_{\beta}\sqcup C_{\gamma/\beta}$. Then there is a set $C_{\alpha/Q}$
derived from $H$, the $P_{\alpha}[G]$-generic. $C_{\alpha/Q}$ is of the form
$C_{\alpha/\beta}$ if $Q=P_{\beta}$, then
$C_{\alpha/Q}\subseteq(\beta,\alpha]$, if $Q=P_{\beta}*P_{\dot{\gamma}/\beta}$
and $\gamma$ is as above, then $C_{\alpha/Q}\subseteq(\gamma,\alpha]$. In both
cases, $\max(C_{\alpha/Q})=\alpha$. If $\circ(\alpha)=0$, then
$C_{\alpha/Q}\setminus\\{\alpha\\}$ is bounded in $\alpha$, otherwise,
$C_{\alpha/Q}\setminus\\{\alpha\\}$ adds a club subset of $\alpha$, and
singularizes $\alpha$ to have cofinality
$\operatorname{cf}(\omega^{\circ(\alpha)})$ (the exponentiation is the ordinal
one). Furthermore, $C_{\alpha}=C_{Q}\sqcup C_{\alpha/Q}$.

Fix $\alpha$. We assume that for $\beta<\alpha$ and $P_{\beta}$-name
$\dot{\nu}$ with $\Vdash_{\beta}\beta\leq\dot{\nu}<\alpha$,
$P_{\dot{\nu}}/P_{\beta}$ are defined and satisfy Proposition 4.1. Our main
task is to define $P_{\alpha}$ and $P_{\alpha}/P_{\beta}$ (or more precisely,
$P_{\dot{\nu}}/P_{\beta}$ when $\dot{\nu}$ is forced in $P_{\beta}$ to be
$\alpha$) for $\beta\leq\alpha$, and show that all the relevant forcings at
level $\alpha$ satisfy Propsition 4.1.

######  Definition 4.2.

(The forcing: general cases) A condition in $P_{\alpha}$ is either the weakest
condition $1_{\alpha}$, or of the form $p=s^{\frown}\langle\alpha,A,F\rangle$,
where $s=\langle s_{i}\rangle_{i<n}$ for some $n$, $s_{i}$ is either
$\langle\alpha_{i},A_{i},F_{i}\rangle$ or
$\langle\alpha_{i},A_{i},F_{i},\dot{\nu}_{i},\dot{q}_{i}\rangle$ such that

  1. (1)

for $i<n$, if $\circ(\alpha_{i})\leq\max_{j<i}(\circ(\alpha_{j}))$, then
$s_{i}$ is a triple, otherwise $s_{i}$ is a $5$-tuple.

  2. (2)

$\alpha_{0}<\cdots<\alpha_{n-1}<\alpha_{n}=:\alpha$ are inaccessible.

  3. (3)

for $i\leq n$,

     * •

$A_{i}\in\vec{U}(\alpha_{i})$ (note that $A_{i}$ is empty if and only if
$\circ(\alpha_{i})=0$).

     * •

$\operatorname{dom}(F_{i})=\\{\xi\in
A_{i}\mid\circ(\xi)>\max_{j<i}(\circ(\alpha_{j}))\\}$.

     * •

for $\gamma\in\operatorname{dom}(F_{i})$,
$F_{i}(\gamma)=\langle\dot{\nu},\dot{q}\rangle$,
$\Vdash_{\gamma}\dot{\nu}\in[\gamma,\alpha_{i})$, $\dot{q}\in
P_{\dot{\nu}}/P_{\gamma}$.

     * •

if $i\neq n$, $\dot{\nu}_{i}$ and $\dot{q}_{i}$ exist, then
$\Vdash_{\alpha_{i}}\dot{\nu}_{i}\in[\alpha_{i},\alpha_{i+1})$ and
$\dot{q}_{i}\in P_{\dot{\nu}_{i}}/P_{\alpha_{i}}$.

     * •

if $i\neq n$, recall that $\bar{\nu}_{i}=\sup\\{\nu\mid\exists r\in
P_{\beta}(r\Vdash_{\beta}\dot{\nu}_{i}=\check{\nu})\\}+1$. Note that
$|P_{\alpha_{i}}|\leq\alpha_{i}^{+}$, so $\bar{\nu}_{i}<\alpha_{i+1}$. We
assume that $\min(A_{i+1})>\bar{\nu}_{i}$.

We assume in general that if $\gamma<\gamma^{\prime}$ are both in $A$ or in
$A_{i}$, and $\dot{\nu}^{\gamma}=F(\gamma)_{0}$, then
$\bar{\nu}^{\gamma}<\gamma^{\prime}$. We denote $s$ by
$\operatorname{stem}(p)$, $\alpha_{n_{1}}+1$ by $\operatorname{top}(s)$, and
$n$ by $l(p)$. Define $\circ(s)=\max\\{\circ(\alpha_{i})\mid i<l(p)\\}$ if
exists, otherwise $\circ(s)=-1$. We sometimes refer $\alpha_{i}$ as
$(s_{i})_{0}$. We say that $p$ is pure if $p=\langle\alpha,A,F\rangle$, in
which case, $\operatorname{dom}(F)=A$, otherwise, we say that $p$ is impure.
For each condition $p$, $\operatorname{stem}(p)\in P_{\alpha_{n-1}}\cup
P_{\alpha_{n-1}}*P_{\dot{\nu}_{n-1}}/P_{\alpha_{n-1}}$ for some
$P_{\alpha_{n-1}}$-name of an ordinal $\dot{\nu}_{n-1}$. Also, note that if
$\circ(\alpha_{i})\leq\max_{j<i}\circ(\alpha_{j})+1$, then $F_{i}=\emptyset$.
Thus, we can see that if $i<n$ and $s_{i}$ is a triple, then
$F_{i}=\emptyset$. The reason that we do not remove $F_{i}$ from a triple is
because we always consider $\langle
s_{k}\rangle_{k<i}{}^{\frown}\langle\alpha_{i},A_{i},F_{i}\rangle$ as a
condition in $P_{\alpha_{i}}$. One may wonder if it may be simpler by saying
that $\operatorname{dom}(F_{i})$ is $A_{i}$, however, by the transitivity of
the relations, the restrictions of the domains or ordinals of certain Mitchell
orders are necessary.

######  Definition 4.3.

(The direct extension relation)

First, let $p\leq^{*}_{\alpha}1_{P_{\alpha}}$ for all $p\in P_{\alpha}$. Let
$p=s^{\frown}\langle\alpha,A,F\rangle$ and
$p^{\prime}=s^{\prime}{}^{\frown}\langle\alpha,A^{\prime},F^{\prime}\rangle$.
Write $s=\langle s_{i}\rangle_{i<l(p)}$, and $s^{\prime}=\langle
s_{i}^{\prime}\rangle_{i<l(p^{\prime})}$. Let
$A_{l(p)}=A,F_{l(p)}=F,A_{l(p^{\prime})}=A^{\prime}$, and
$F_{l(p^{\prime})}=F^{\prime}$. We say that $p$ is a direct extension of
$p^{\prime}$, denoted by $p\leq^{*}_{\alpha}p^{\prime}$, if

  1. (1)

$n:=l(p)=l(p^{\prime})$.

  2. (2)

for $i\leq n$,

     * •

$\alpha_{i}=\alpha_{i}^{\prime}$.

     * •

$A_{i}\subseteq A_{i}^{\prime}$.

     * •

for $\gamma\in A_{i}$,

$s\restriction
i^{\frown}\langle\gamma,A_{i}\restriction\gamma,F_{i}\restriction(\operatorname{dom}(F_{i})\cap
A_{i}\restriction\gamma)\rangle\leq^{*}_{\gamma}s^{\prime}\restriction
i^{\frown}\langle\gamma,A_{i}^{\prime}\restriction\gamma,F_{i}^{\prime}\restriction(\operatorname{dom}(F_{i}^{\prime})\cap
A_{i}\restriction\gamma)\rangle$,

and furthermore, if $\gamma\in\operatorname{dom}(F_{i})$,

$s\restriction
i^{\frown}\langle\gamma,A_{i}\restriction\gamma,F_{i}\restriction(\operatorname{dom}(F_{i})\cap
A_{i}\restriction\gamma)\rangle\Vdash_{\gamma}F_{i}(\gamma)_{0}=F_{i}^{\prime}(\gamma)_{0}$
and
$F_{i}(\gamma)_{1}\leq^{*}_{F_{i}(\gamma)_{0}/\gamma}F_{i}^{\prime}(\gamma)_{1}$.

     * •

$s\restriction
i^{\frown}\langle\alpha_{i},A_{i},F_{i}\rangle\leq^{*}_{\alpha_{i}}s^{\prime}\restriction
i^{\frown}\langle\alpha_{i},A_{i},F_{i}\rangle$, and if $s_{i}$ is a $5$-tuple
(note that this is true if and only if $s_{i}^{\prime}$ is a $5$-tuple), then

$s\restriction
i^{\frown}\langle\alpha_{i},A_{i},F_{i}\rangle\Vdash_{\alpha_{i}}\dot{\nu}_{i}=\dot{\nu}_{i}^{\prime}$
and $\dot{q}_{i}\leq^{*}_{\dot{\nu}_{i}/\alpha_{i}}\dot{q}_{i}^{\prime}$.

We now exemplify how non-direct extension works. We first start by describing
an instance of a minimal one-step extension, as it naturally occurs as a part
of an extension relation.

To exemplify, assume $\circ(\alpha)=\omega$ and $p=\langle\alpha,A,F\rangle$
is pure. Let $\gamma\in A$. Assume $\circ(\gamma)=5$,
$\dot{\nu}=F(\gamma)_{0}$, and $\bar{\nu}=\sup\\{\nu\mid\exists p\in
P_{\gamma}(p\Vdash_{\gamma}\dot{\nu}=\check{\nu})\\}+1$ is the corresponding
value. Then the minimal one-step extension of $p$ by $\gamma$, which will be
denoted by $p+\langle\gamma\rangle$, is the condition

$p^{\prime}=\langle\gamma,A\restriction\gamma,F\restriction(A\restriction\gamma),\dot{\nu},\dot{q}\rangle^{\frown}\langle\alpha,A^{\prime},F^{\prime}\rangle$,

where $A\restriction\gamma=\\{\xi<\gamma\mid\xi\in A$ and $\circ(\xi)<5\\}$,
$\dot{q}=F(\gamma)_{1}$, $A^{\prime}=A\setminus(\gamma+1)$ (note that
$\min(A\setminus(\gamma+1))>\bar{\nu}$), and
$F^{\prime}=F\restriction\\{\xi\in A^{\prime}\mid\circ(\xi)>5\\}$. A triple
$\langle\gamma,A\restriction\gamma,F\restriction(A\restriction\gamma)\rangle$
is now considered as a pure condition in $P_{\gamma}$, the $5$-tuple
$\langle\gamma,A\restriction\gamma,F\restriction(A\restriction\gamma),\dot{\nu},\dot{q}\rangle$
is considered as a condition in $P_{\gamma}*P_{\dot{\nu}}/P_{\gamma}$.

A further one-step extension of $p^{\prime}$ can be done by either adding
$\gamma^{\prime}\in A\restriction\gamma$ to extend the stem to the left as
before, or extending by using $\gamma^{\prime}\in A^{\prime}$. Consider the
latter case. Suppose that $\circ(\gamma^{\prime})\leq 5$, then
$\gamma^{\prime}\not\in\operatorname{dom}(F^{\prime})$. The one-step extension
of $p^{\prime}$ by $\gamma^{\prime}$ will be

$\langle\gamma,A\restriction\gamma,F\restriction(A\restriction\gamma),\dot{\nu},\dot{q}\rangle^{\frown}\langle\gamma^{\prime},A^{\prime}\restriction\gamma^{\prime},F^{\prime}\restriction(A^{\prime}\restriction\gamma^{\prime})\rangle^{\frown}\langle\alpha,A^{\prime}\setminus(\gamma^{\prime}+1),F^{\prime}\setminus(\gamma^{\prime}+1)\rangle$.

We see that
$\langle\gamma,A\restriction\gamma,F\restriction(A\restriction\gamma)\rangle^{\frown}\langle\gamma^{\prime},A^{\prime}\restriction\gamma^{\prime},F^{\prime}\restriction(A^{\prime}\restriction\gamma^{\prime})\rangle\in
P_{\gamma^{\prime}}$. Note that
$F^{\prime}\restriction(A^{\prime}\restriction\gamma^{\prime})$ is empty since
ordinals in $\operatorname{dom}(F^{\prime})$ have Mitchell orders at least
$6$, but the ordinals in $A^{\prime}\restriction\gamma^{\prime}$ have Mitchell
order less than $\circ(\gamma^{\prime})$, which is at most $5$. If, however,
$\circ(\gamma^{\prime})>5$, the one-step extension of $p^{\prime}$ by
$\gamma^{\prime}$ will be

$\langle\gamma,A\restriction\gamma,F\restriction(A\restriction\gamma),\dot{\nu},\dot{q}\rangle^{\frown}\langle\gamma^{\prime},A^{\prime}\restriction\gamma^{\prime},F^{\prime}\restriction(\operatorname{dom}(F^{\prime})\cap
A^{\prime}\restriction\gamma^{\prime}),F^{\prime}(\gamma^{\prime})_{0},F(\gamma^{\prime})_{1}\rangle^{\frown}\langle\alpha,A^{\prime\prime},F^{\prime\prime}\rangle$,

for some certain $A^{\prime\prime}$ and $F^{\prime\prime}$. Suppose that
$\circ(\gamma^{\prime})=8$ and
$\dot{\nu}^{\prime}=F^{\prime}(\gamma^{\prime})_{0}$. Then, ordinals in
$\operatorname{dom}(F^{\prime}\restriction(A^{\prime}\restriction\gamma^{\prime}))$
have Mitchell orders $6$ or $7$.
$A^{\prime\prime}=A^{\prime}\setminus(\gamma^{\prime}+1)$ (recall
$\min(A^{\prime}\setminus(\gamma^{\prime}+1))>\bar{\nu}^{\prime}$).
$F^{\prime\prime}=F^{\prime}\restriction\\{\xi\in
A^{\prime\prime}\mid\circ(\xi)>8\\}$. We now formally define a minimal one-
step extension.

######  Definition 4.4.

(A minimal one-step extension) Let $p\in P_{\alpha}$, $n=l(p)$,
$p=s^{\frown}\langle\alpha,A,F\rangle$, for $i<n$, write
$s_{i}=\langle\alpha_{i},A_{i},F_{i}\rangle$ or
$\langle\alpha_{i},A_{i},F_{i},\dot{\nu}_{i},\dot{q}_{i}\rangle$. Let
$A_{n}=A$, $F_{n}=F$, $i\leq n$, and $\gamma\in A_{i}$. The one-step extension
of $p$ by $\gamma$, denoted by $p+\langle\gamma\rangle$, is the condition
$p^{\prime}=s^{\prime}{}^{\frown}\langle\alpha,A^{\prime},F^{\prime}\rangle$
such that

  1. (1)

$l(p^{\prime})=l(p)+1$.

  2. (2)

$s^{\prime}\restriction i=s\restriction i$.

  3. (3)

if $\gamma\not\in\operatorname{dom}(F_{i})$ (i.e.
$\circ(\gamma)\leq\max_{j<i}\circ(\alpha_{j}))$),
$s^{\prime}_{i}=\langle\gamma,A_{i}\restriction\gamma,F_{i}\restriction(\operatorname{dom}(F_{i})\cap
A_{i}\restriction\gamma)\rangle$, otherwise,
$s^{\prime}_{i}=\langle\gamma,A_{i}\restriction\gamma,F_{i}\restriction(\operatorname{dom}(F_{i})\cap
A_{i}\restriction\gamma),F_{i}(\gamma)_{0},F_{i}(\gamma)_{1}\rangle$.

  4. (4)

if $i<n$, then,

     * •

if $s_{i+1}$ is a triple, then
$s_{i+1}^{\prime}=\langle\alpha_{i},A_{i}\setminus(\gamma+1),F_{i}\restriction\\{\xi\in
A_{i}\setminus(\gamma+1)\mid\circ(\xi)>\circ(\gamma)\\}\rangle$ (note that if
$\gamma\not\in(\operatorname{dom}(F_{i}))$, then $F_{i}\restriction\\{\xi\in
A_{i}\setminus(\gamma+1)\mid\circ(\xi)>\circ(\gamma)\\}$ is simply
$F_{i}\restriction(A_{i}\setminus(\gamma+1))$).

     * •

if $s_{i+1}$ is a $5$-tuple, then
$s_{i+1}^{\prime}=\langle\alpha_{i},A_{i}\setminus(\gamma+1),F_{i}\restriction\\{\xi\in
A_{i}\setminus(\gamma+1)\mid\circ(\xi)>\circ(\gamma)\\},\dot{\nu}_{i},\dot{q}_{i}\rangle$
(note that if $\gamma\not\in(\operatorname{dom}(F_{i}))$, then
$F_{i}\restriction\\{\xi\in
A_{i}\setminus(\gamma+1)\mid\circ(\xi)>\circ(\gamma)\\}$ is simply
$F_{i}\restriction(A_{i}\setminus(\gamma+1))$).

     * •

for $j>i+1$, $s^{\prime}_{j}=s_{j}$, $A^{\prime}=A$, $F^{\prime}=F$.

  5. (5)

if $i=n$, then
$A^{\prime}=A\setminus(\gamma+1),F^{\prime}=F\restriction\\{\xi\in
A_{i}\setminus(\gamma+1)\mid\circ(\xi)>\circ(\gamma)\\}$ (note that if
$\gamma\not\in\operatorname{dom}(F)$, then $F\restriction\\{\xi\in
A_{i}\setminus(\gamma+1)\mid\circ(\xi)>\circ(\gamma)\\}$ is simply
$F\restriction(A\setminus(\gamma+1))$).

We define an $n$-step extension recursively by
$p+\langle\vec{\gamma}\rangle=(p+\langle\gamma_{0},\cdots,\gamma_{n-2}\rangle)+\langle\gamma_{n-1}\rangle$,
as long as the extensions are legitimate.

######  Definition 4.5.

(The extension relation) With the setting in Definition 4.3, we say that $p$
is an extension of $p^{\prime}$, denoted by $p\leq_{\alpha}p^{\prime}$, if

  1. (1)

$l(p)\geq l(p^{\prime})$.

  2. (2)

there are $i_{0}<i_{1}<\cdots<i_{l(p^{\prime})-1}<i_{l(p^{\prime})}=l(p)$ such
that

     * •

$\alpha_{i_{j}}=\alpha_{j}^{\prime}$.

     * •

$A_{i_{j}}\subseteq A_{j}^{\prime}$.

     * •

for $\gamma\in A_{i_{j}}$,

$s\restriction
i_{j}^{\frown}\langle\gamma,A_{i_{j}}\restriction\gamma,F_{i_{j}}\restriction(\operatorname{dom}(F_{i_{j}}\cap
A_{i_{j}}\restriction\gamma)\rangle\leq_{\gamma}s^{\prime}\restriction
j^{\frown}\langle\gamma_{,}A_{j}^{\prime}\restriction\gamma,F_{j}^{\prime}\restriction(\operatorname{dom}(F_{j}^{\prime})\cap
A_{j}^{\prime}\restriction\gamma)\rangle$,

and if $\gamma\in\operatorname{dom}(F_{i_{j}})$,

$s\restriction
i_{j}^{\frown}\langle\gamma^{\frown}A_{i_{j}}\restriction\gamma,F_{i_{j}}\restriction(\operatorname{dom}(F_{i_{j}}\cap
A_{i_{j}}\restriction\gamma)\rangle\Vdash_{\gamma}F_{i_{j}}(\gamma)_{0}=F_{j}^{\prime}(\gamma)_{0}$
and
$F_{i_{j}}(\gamma)_{1}\leq^{*}_{F_{i_{j}}(\gamma)_{0}/\gamma}F_{j}^{\prime}(\gamma)_{1}$
(the $\leq^{*}$ relation is intended).

     * •

$s\restriction
i_{j}^{\frown}\langle\alpha_{i_{j}},A_{i_{j}},F_{i_{j}}\rangle\leq_{\alpha_{i_{j}}}s^{\prime}\restriction
j^{\frown}\langle\alpha_{j},A_{j},F_{j}\rangle$, and $s_{i_{j}}$ is a
$5$-tuple, iff $s_{j}^{\prime}$, in which case

$s\restriction
i_{j}^{\frown}\langle\alpha_{i_{j}},A_{i_{j}},F_{i_{j}}\rangle\Vdash_{\alpha_{i_{j}}}\dot{\nu}_{i_{j}}=\dot{\nu}_{j}^{\prime}$
and
$\dot{q}_{i_{j}}\leq_{\dot{\nu}_{i_{j}}/\alpha_{i_{j}}}\dot{q}_{j}^{\prime}$.

  3. (3)

let $i_{0},i_{1},\cdots,i_{l(p^{\prime})}$ be as in (2),
$k\not\in\\{i_{0},\cdots,i_{l(p^{\prime})}\\}$, and $j$ be the least such that
$k<i_{j}$. Then,

     * •

$\alpha_{k}\in A_{j}^{\prime}$.

     * •

$A_{k}\subseteq A_{j}^{\prime}\restriction\alpha_{k}$.

     * •

for $\gamma\in A_{k}$,

$s\restriction
k^{\frown}\langle\gamma,A_{k}\restriction\gamma,F_{k}\restriction(\operatorname{dom}(F_{k})\cap
A_{k}\restriction\gamma)\rangle\leq_{\gamma}s^{\prime}\restriction
j-1^{\frown}\langle\gamma,A_{j}^{\prime}\restriction\gamma,F_{j}^{\prime}\restriction(\operatorname{dom}(F_{j}^{\prime})\cap
A_{j}^{\prime}\restriction\gamma)\rangle$,

and if $\gamma\in\operatorname{dom}(F_{k})$,

$s\restriction
k^{\frown}\langle\gamma,A_{k}\restriction\gamma,F_{k}\restriction(\operatorname{dom}(F_{k})\cap
A_{k}\restriction\gamma)\rangle\Vdash_{\gamma}F_{k}(\gamma)_{0}=F_{j}^{\prime}(\gamma)_{0}$
and
$F_{k}(\gamma)_{1}\leq^{*}_{F_{k}(\gamma)_{0}/\gamma}F_{j}^{\prime}(\gamma)_{1}$
(the $\leq^{*}$ relation is intended).

     * •

$\alpha_{k}\in\operatorname{dom}(F_{j}^{\prime})$ if and only if $s_{k}$ is a
$5$-tuple, in which case,
$\Vdash_{\alpha_{k}}\dot{\nu}_{k}=F_{j}^{\prime}(\alpha_{k})_{0}$, and

$s\restriction
k^{\frown}\langle\alpha_{k},A_{k},F_{k}\rangle\Vdash_{\alpha_{k}}\dot{q}_{k}\leq_{\dot{\nu}_{k}/\alpha_{k}}F_{j}^{\prime}(\alpha_{k})_{1}$.

It is not true that $p\leq_{\alpha}p^{\prime}$ iff
$p\leq^{*}_{\alpha}p^{\prime}+\vec{\gamma}$ for some $\vec{\gamma}$. However,
$p\leq_{\alpha}p^{\prime}$ iff there is
$p^{*}\leq^{*}_{\alpha}p^{\prime}+\vec{\gamma}$ for some $\vec{\gamma}$ such
that $p$ is obtained from $p^{*}$ by extending the names appearing at the
fifth-coordinate of each $5$-tuple sequence in $\operatorname{stem}(p^{*})$.
We now consider some basic properties of $P_{\alpha}$.

######  Proposition 4.6.

$|P_{\alpha}|\leq\alpha^{+}$, and hence $\alpha^{++}$-c.c.

###### Proof.

Each condition $p=s^{\frown}\langle\alpha,A,F\rangle$, we have $s\in
V_{\alpha}$, $A\subseteq\alpha$, and $F$ is a partial function from $\alpha$
to $V_{\alpha}$. Since GCH holds and $\alpha$ is inaccessible,
$|P_{\alpha}|\leq 2^{\alpha}=\alpha^{+}$.

∎

We now prove the strong form of closure under the direct extension relation,
i.e. Proposition 5 (4).

######  Proposition 4.7.

Let $p\in P_{\alpha}$ and $(\operatorname{stem}(p)_{0})_{0}>\gamma$. Let
$\xi<\min(X\setminus(\gamma+1))$. Assume for $\epsilon<\xi$, there is
$p_{\epsilon}$ such that $p_{\epsilon}\leq^{*}_{\alpha}p$. Then there is
$p^{*}\leq^{*}_{\alpha}p$ such that for $\epsilon$,
$p^{*}\leq^{*}_{\alpha}p_{\epsilon}$.

###### Proof.

Write
$p_{\epsilon}=\operatorname{stem}(p_{\epsilon})^{\frown}\langle\alpha,A^{\epsilon},F^{\epsilon}\rangle$.
Let $Q$ be the forcing such that $\operatorname{stem}(p)\in Q$. Then for all
$\epsilon$, $\operatorname{stem}(p_{\epsilon})\in Q$. By proposition 4.1 (4),
and the fact that $(\operatorname{stem}(p)_{0})_{0}>\gamma$, let $s^{*}$ be
such that for all $\epsilon$,
$s^{*}\leq^{*}\operatorname{stem}(p_{\epsilon})$. Take
$A^{*}=\cap_{\epsilon}A^{\epsilon}$ and for $\beta\in A^{*}$, we will take
$F^{*}(\beta)_{1}$ as a name with is forced to be $\leq^{*}$-stronger than
$F^{\epsilon}(\beta)_{1}$ for all $\epsilon$. More explicitly, we do this by
induction on ordinals in $A^{*}$. First, let $\beta=\min(A^{*})$. Consider
$t_{\epsilon}=\operatorname{stem}(p_{\epsilon})^{\frown}\langle\beta,A^{\epsilon}\restriction\beta,F^{\epsilon}\restriction\\{\xi\in\operatorname{dom}(F^{\epsilon})\cap
A^{\epsilon}\restriction\beta\mid\circ(\xi)>\circ(\beta)\\}$. For each $G$
which is $P_{\beta}$-generic, let $F^{*}(\beta)_{1}[G]$ be such that for any
$\epsilon$ with $t_{\epsilon}\in G$,
$F^{*}(\beta)_{1}[G]\leq^{*}F^{\epsilon}(\beta)_{1}[G]$. Inductive steps can
be done similarly. Now, let $F^{*}(\beta)=\langle
F^{p}(\beta)_{0},F^{*}(\beta)_{1}\rangle$. Let
$p^{*}=s^{*}{}^{\frown}\langle\alpha,A^{*},F^{*}\rangle$, then $p^{*}$ is as
required.

∎

In particular, we have that

######  Corollary 4.8.

$(\\{p\in
P_{\alpha}\mid(\operatorname{stem}(p)_{0})_{0}>\gamma\\},\leq^{*}_{\alpha})$
is $\min(X\setminus(\gamma+1))$-closed. In particular, $(\\{p\in
P_{\alpha}\mid p$ is pure$\\},\leq^{*}_{\alpha})$ is $\alpha$-closed.

######  Remark 4.9.

When we restrict $P_{\alpha}$ to the set $\\{p\in P_{\alpha}\mid p$ is
pure$\\}$, $\leq^{*}_{\alpha}$ and $\leq_{\alpha}$ coincide. In addition, if
$\circ(\alpha)=0$, a pure condition in $P_{\alpha}$ is simply a trivial
forcing, which makes $(\\{p\in P_{\alpha}\mid p$ is
pure$\\},\leq^{*}_{\alpha})$ $\infty$-closed.

######  Definition 4.10.

(The quotient forcing) Let $\beta<\alpha$, and $\dot{\nu}$ is a
$P_{\beta}$-name of an ordinal in the interval $[\beta,\alpha)$. Let $Q$ be
either $P_{\beta}$ or $P_{\beta}*P_{\dot{\nu}}/P_{\beta}$-generic. Let $G$ be
$Q$-generic. We define $P_{\alpha}[G]$ as the collection of
$s_{1}[G]^{\frown}\langle\alpha,A,F[G]\rangle$ where there is $s_{0}\in G$,
$s_{0}^{\frown}s_{1}{}^{\frown}\langle\alpha,A,F\rangle\in P_{\alpha}$, and
$F[G](\gamma)=F(\gamma)[G]$ for all $\gamma\in\operatorname{dom}(F)$.

We note that those notions make sense due to Proposition 4.1 (3), (5), and (6)
below $\alpha$, and the fact that $\min(A)>\bar{\nu}>\beta$. We say that
$P_{\alpha}/Q$ is the collection of $Q$-name $\dot{q}$ such that
$\Vdash_{\beta}\dot{q}\in P_{\alpha}[\dot{G}]$ where $\dot{G}$ is the
canonical name for a generic for $Q$. We assume that $\dot{q}$ is also a nice
name. The extension relation, direct extension relation, minimal one-step
extension are those induced from $P_{\alpha}$. Note that it is possible that
$P_{\alpha}[G]$ has no conditions of the form $\langle\alpha,A,F\rangle$ (for
example, item $(2)$ in Definition 3.2). In that case, we do not define pure
conditions, impure conditions, and stems. Assume $P_{\alpha}[G]$ has
conditions of the form $\langle\alpha,A,F\rangle$. $P_{\alpha}[G]$, a pure
condition is of the form $\langle\alpha,A,F\rangle$. Other conditions which
are not $1_{P_{\alpha}[G]}$ are called impure. We will call the corresponding
relations $\leq_{\alpha/Q}$ and $\leq^{*}_{\alpha/Q}$, respectively. If
$Q=P_{\beta}$, we abbreviate $\leq_{\alpha/Q}$ by $\leq_{\alpha/\beta}$, and
so on.

######  Definition 4.11.

Define $P_{\alpha}/P_{\alpha}$ as the trivial forcing (in $V^{P_{\alpha}}$).

######  Proposition 4.12.

Let $\beta$, $\dot{\nu}$, $Q$, and $G$ be as in Definition 4.10. Then,

  1. (1)

$\Vdash_{Q}|P_{\alpha}/Q|\leq\alpha^{+}$, so $P_{\alpha}/Q$ is
$\alpha^{++}$-c.c. Furthermore, $|P_{\alpha}/Q|\leq\alpha^{+}$.

  2. (2)

Suppose that pure conditions in $P_{\alpha}/Q$ can be defined,
$\Vdash_{Q}\dot{q}\in P_{\alpha}/Q$ is pure. Fix $\xi<\alpha$ and for
$\epsilon<\xi$, there are $p_{\epsilon}\in Q$ and $\dot{q}_{\epsilon}$ such
that $p_{\epsilon}\Vdash_{Q}\dot{q}_{\epsilon}\leq^{*}\dot{q}$. Then there is
$\dot{q}^{*}$ such that $\Vdash_{Q}\dot{q}^{*}\leq^{*}\dot{q}$, and for all
$\epsilon$, $p_{\epsilon}\Vdash_{Q}\dot{q}^{*}\leq^{*}\dot{q}_{\epsilon}$.

  3. (3)

Suppose that pure conditions in $P_{\alpha}[G]$ are defined, then $\\{p\in
P_{\alpha}[G]\mid p$ is pure$\\},\leq^{*})$ is $\alpha$-closed.

  4. (4)

Let $\gamma<\min(X\setminus(\beta+1))$. Suppose that $\Vdash_{\beta}\dot{q}\in
P_{\alpha}/P_{\beta}$. For $\epsilon<\gamma$, let $p_{\epsilon}\in P_{\beta}$
and $\dot{q}_{\epsilon}$ be such that
$p_{\epsilon}\Vdash_{\beta}\dot{q}_{\epsilon}\leq^{*}\dot{q}$. Then there is
$\dot{q}^{*}$ such that $\Vdash_{\beta}\dot{q}^{*}\leq^{*}\dot{q}$ and for all
$\epsilon$, $\dot{q}^{*}\leq^{*}\dot{q}_{\epsilon}$. In particular,
$\Vdash_{Q}(P_{\alpha}/P_{\beta},\leq^{*})$ is
$\min(X\setminus(\beta+1))$-closed.

###### Proof.

  1. (1)

By Proposition 4.1 (1) and (2), $Q$ preserves cardinals and GCH, by the same
counting argument as in Proposition 4.6, $|P_{\alpha}[G]|\leq\alpha^{+}$ in
$V[G]$. Now since $|Q|\leq\alpha^{+}$ and
$\Vdash_{Q}|P_{\alpha}/Q|\leq\alpha^{+}$ and GCH holds in $V$, we have
$|P_{\alpha}/Q|\leq\alpha^{+}$.

  2. (2)

Let $G$ be $Q$-generic. In $V[G]$, write $\dot{q}[G]=\langle\alpha,A,F\rangle$
and for $\epsilon$ with $p_{\epsilon}\in G$, set
$\dot{q}_{\epsilon}[G]=\langle\alpha,A^{\epsilon},F^{\epsilon}\rangle$. Take
$A^{*}=\cap_{\\{\epsilon\mid p_{\epsilon}\in G\\}}A^{\epsilon}$ and for
$\beta\in A^{*}$. Set $F^{*}(\gamma)_{1}$ such that for each $\epsilon$
$F^{*}(\gamma)_{1}$ is forced to be $\leq^{*}$-stronger than
$F^{\epsilon}(\gamma)_{1}$, and let $F^{*}(\gamma)=\langle
F^{p}(\gamma)_{0},F^{*}(\gamma)_{1}\rangle$. Let
$q^{*}=\langle\alpha,A^{*},F^{*}\rangle$ and $\dot{q}^{*}$ be a name of such
$q^{*}$. Then we can check that $\dot{q}^{*}$ is as required.

  3. (3)

It is a consequence of (2).

  4. (4)

The proof is similar to $(2)$.

∎

We now discuss on factorizations on $P_{\alpha}$. Let $\beta,\dot{\nu}$, $Q$,
and $G$ be as in Definition 4.10. Let $H$ be $P_{\alpha}[G]$-generic. Then,
let $I$ be the collection of
$s_{0}^{\frown}s_{1}^{\frown}\langle\alpha,A,F\rangle\in P_{\alpha}$ such that
$s_{0}\in G$ and $(s_{1}^{\frown}\langle\alpha,A,F\rangle)[G]\in H$. Then $I$
is $P_{\alpha}$-generic, and $V[I]=V[G][H]$. On the contrary, let $I$ be
$P_{\alpha}$-generic. Let $p\in I$ (say $p\neq 1_{P_{\alpha}})$. Let
$p=s_{0}^{\frown}s_{1}^{\frown}\langle\alpha,A,F\rangle$. Set $Q^{\prime}$ be
the forcing containing $s_{0}$. Let $G=\\{t_{0}\in Q^{\prime}\mid$ there are
$t_{1},A^{\prime},$ and $F^{\prime}$ such that
$t_{0}^{\frown}t_{1}^{\frown}\langle\alpha,A^{\prime},F^{\prime}\rangle\in
I\\}$. Then, $G$ is $Q^{\prime}$-generic. Let
$H=\\{(t_{1}^{\frown}\langle\alpha,A^{\prime},F^{\prime}\rangle)[G]\mid\exists
t_{0}\in
G(t_{0}^{\frown}t_{1}^{\frown}\langle\alpha,A^{\prime},F^{\prime}\rangle)\in
I\\}$. Then, $H$ is $P_{\alpha}/G$-generic over $V[G]$, and $V[I]=V[G][H]$.
Hence, Proposition 4.1 (3) and (5) follow at level $\alpha$.

We further discuss on $P_{\alpha}[G]/P_{\gamma}[G]$. Again, let
$\beta,\dot{\nu},Q,$ and $G$ be as in Definition 4.10. We define
$P_{\alpha}[G]/P_{\gamma}[G]$ as follows. Recall that by Proposition 4.1 (3),
if $H$ is $P_{\gamma}[G]$-generic, then there is the canonical
$P_{\alpha}$-generic $I$ with $V[G][H]=V[I]$. Define
$P_{\alpha}[G]/P_{\gamma}[G]$ as the collection of $P_{\gamma}[G]$-name
$\dot{q}$ such that $\Vdash_{P_{\gamma}[G]}\dot{q}\in P_{\alpha}[G][\dot{I}]$,
where $\dot{H}$ is the name for $P_{\gamma}[G]$-generic, and $\dot{I}$ is the
name for $P_{\gamma}$-canonical generic such that
$\Vdash_{P_{\gamma}[G]}V[G][\dot{H}]=V[\dot{I}]$. Finally, for $\dot{q}$ with
$\Vdash_{\gamma}\dot{q}\in P_{\alpha}/P_{\gamma}$, define $\dot{q}[G]$ as a
$P_{\gamma}[G]$-name of an element in $P_{\alpha}[G]/P_{\gamma}[G]$ as
follows. For each $H$ which is $P_{\gamma}[G]$-generic, and $I=G*H$, we see
that $\dot{q}[I]\in P_{\alpha}[I]$. Find $p\in P_{\alpha}$ such that
$p=s_{0}^{\frown}s_{1}$ where $s_{0}\in I$. Let $\dot{q}[G][H]=s_{1}[I]$. We
need that $\dot{q}[G][H]=s_{1}[I]$ for each such $H$ and the corresponding
generic $I$. One can trace these genericities to show that Proposition 4.1 (6)
holds at level $\alpha$.

##  5\. The Prikry property

In this section we give a proof of the Prikry property at level $\alpha$. If
$\circ(\alpha)=0$, then $P_{\alpha}$ is equivalent to $P_{\beta}$ or
$P_{\beta}*P_{\dot{\nu}}/P_{\beta}$ for some $\beta,\dot{\nu}$ such that
$\Vdash_{\beta}\dot{\nu}\in[\beta,\alpha)$. These are either a Prikry-type
forcing or a two-step iteration of Prikry-type forcings, hence, has the Prikry
property. We assume throughout this section that $\circ(\alpha)>0$. Note that
when $p=\langle\alpha,A,F\rangle$ is pure, then $\operatorname{dom}(F)=A$.
However, for each impure condition $p=s^{\frown}\langle\alpha,A,F\rangle$, it
is possible that $\operatorname{dom}(F)\neq A$. Our proofs in this section
focus on the pure conditions, with some modification, they also work for
impure conditions.

######  Lemma 5.1.

Let $D\subseteq P_{\alpha}$ be a dense open set,
$p=\langle\alpha,A,F\rangle\in P_{\alpha}$. Then there is
$p^{\prime}=\langle\alpha,A^{\prime},F^{\prime}\rangle$ such that for each
stem $s$, if
$s^{\frown}\langle\alpha,B,F^{\prime}\rangle\leq_{\alpha}p^{\prime}$ is in
$D$, then
$s^{\frown}\langle\alpha,A^{\prime}\setminus\operatorname{top}(s),F^{\prime}\restriction\\{\gamma\in
A^{\prime}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\in
D$.

###### Proof.

We build $\langle A_{\gamma},F_{\gamma}\mid\gamma\in A\rangle$ as follows:

  1. (1)

$\langle A_{\gamma}\mid\gamma\in A\rangle$ is $\subseteq$-decreasing,
$A_{\gamma}\subseteq A$, $A_{\gamma}\in\vec{U}(\alpha)$, and
$\min(A_{\gamma})>\gamma$.

  2. (2)

for $\gamma^{\prime}<\gamma<\xi$ and $\xi\in A_{\gamma}$, we have that

     * •

$\langle\alpha,A_{\gamma}\restriction\xi,F_{\gamma}\restriction(A_{\gamma}\restriction\xi)\rangle\leq^{*}_{\xi}\langle\alpha,A_{\gamma^{\prime}}\restriction\xi,F_{\gamma^{\prime}}\restriction(A_{\gamma^{\prime}}\restriction\xi)\rangle$.

     * •

$\langle\alpha,A_{\gamma}\restriction\xi,F_{\gamma}\restriction(A_{\gamma}\restriction\xi)\rangle\Vdash_{\xi}F_{\gamma}(\xi)_{0}=F_{\gamma^{\prime}}(\xi)_{0}$
and $F_{\gamma^{\prime}}(\xi)_{1}\leq^{*}F_{\gamma^{\prime}}(\xi)_{1}$ (we
assume that the name $F_{\gamma}(\xi)_{0}$ is always the same, which is
$F(\xi)_{0}$).

  3. (3)

let $\dot{\nu}_{\gamma}=F(\gamma)_{0}$. For each $s\in P_{\gamma}\cup
P_{\gamma}*P_{\dot{\nu}_{\gamma}}/P_{\gamma}$, if
$s^{\frown}\langle\alpha,B,H\rangle\leq^{*}_{\alpha}s^{\frown}\langle\alpha,A_{\gamma},F_{\gamma}\rangle$
is in $D$, then already
$s^{\frown}\langle\alpha,A_{\gamma}\setminus\operatorname{top}(s),F_{\gamma}\restriction\\{\gamma\in
A^{\gamma}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\in
D$.

Suppose for $\gamma^{\prime}<\gamma$, $A_{\gamma^{\prime}}$ and
$F_{\gamma^{\prime}}$ is built (note that the case $\gamma=\min(A)$ can be
done in a slightly simpler setting). We now build $A_{\gamma}$ and
$F_{\gamma}$. Let
$A_{\gamma}^{\prime}=\cap_{\gamma^{\prime}<\gamma}A_{\gamma^{\prime}}\setminus(\gamma+1)$.
For $\xi\in A_{\gamma}^{\prime}$, let
$F_{\gamma}^{\prime}(\xi)_{0}=F(\xi)_{0}$ and $F_{\gamma}^{\prime}(\xi)_{1}$
be a name such that $F_{\gamma}^{\prime}(\xi)_{1}$ is forced to be
$\leq^{*}$-stronger than $F_{\gamma^{\prime}}(\xi)_{1}$ for all
$\gamma^{\prime}<\gamma$. This is possible since by Proposition 4.1, item (4),
$\Vdash_{\xi}(P_{F(\xi)_{0}}/P_{\xi},\leq^{*})$ is $\xi^{+}$-closed, and
$\xi>\gamma$. Let
$p^{\prime}_{\gamma}=\langle\alpha,A_{\gamma}^{\prime},F_{\gamma}^{\prime}\rangle$.
Let $\dot{\nu}_{\gamma}=F(\gamma)_{0}$ and $\bar{\nu}_{\gamma}$ be the
corresponding value. By Proposition 4.1, item (2), let $\langle
s_{\delta}\mid\delta<\bar{\nu}_{\gamma}^{+}\rangle$ be an enumeration of
conditions in $P_{\gamma}\cup P_{\gamma}*P_{\dot{\nu}}/P_{\gamma}$. For each
$\delta$, if there is
$s_{\gamma,\delta}{}^{\frown}\langle\alpha,B,H\rangle\leq^{*}_{\alpha}s_{\gamma,\delta}{}^{\frown}p_{\gamma}^{\prime}$
being in $D$, we record $A_{\gamma,\delta}=B$, and for
$\xi\in\operatorname{dom}(H)$, let $\dot{q}_{\gamma,\delta}=H(\xi)_{1}$.
Otherwise, let $A_{\gamma,\delta}=A_{\gamma}^{\prime}$ and for
$\xi\in\operatorname{dom}(H)$, let
$\dot{q}_{\gamma,\delta}=F_{\gamma}^{\prime}(\xi)_{1}$. We now let
$A_{\gamma}=\cap_{\delta<\bar{\nu}_{\gamma}^{+}}A_{\gamma,\delta}$, and for
$\xi\in A_{\gamma}$, let $F_{\gamma}(\xi)_{1}$ be a name such that
$F_{\gamma}(\xi)$ is forced to be $\leq^{*}$-stronger than
$\dot{q}_{\gamma,\delta}$ for all $\delta$. Again, this is possible by
Proposition 4.1 (4).

We now let $A^{\prime}=\triangle_{\gamma\in A}A_{\gamma}$. For each $\gamma\in
A^{\prime}$, let $F^{\prime}(\gamma)_{1}$ be a name which is forced to be
$\leq^{*}$-stronger than $F_{\gamma^{\prime}}(\gamma)_{1}$ for all
$\gamma^{\prime}<\gamma$. Again, this is possible by Proposition 4.1 (4). Let
$F^{\prime}(\gamma)=\langle F(\gamma)_{0},F^{\prime}(\gamma)_{1}\rangle$ and
$p^{\prime}=\langle\alpha,A^{\prime},F^{\prime}\rangle$. It remains to show
that $p^{\prime}$ satisfies our criteria. Let
$s^{\frown}\langle\alpha,B,H\rangle\leq_{\alpha}p^{\prime}$ being in $D$. Let
$Q$ be the forcing such that $s\in Q$. Hence, $s=s_{\gamma,\delta}$ for some
$\delta$. Since $B\subseteq A^{\prime}\setminus(\gamma+1)$, we have that
$B\subseteq A_{\gamma}$. For each $\xi\in B$, it is forced that
$H(\xi)_{1}\leq^{*}F_{\gamma}(\xi)_{1}$. By (3),
$s^{\frown}\langle\alpha,A_{\gamma}\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A^{\gamma}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\in
D$. Since $D$ is open,
$s^{\frown}\langle\alpha,A^{\prime}\setminus\operatorname{top}(s),F^{\prime}\restriction\\{\gamma\in
A^{\prime}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\in
D$ as required.

∎

######  Lemma 5.2.

Fix $n>0$. Let $p=\langle\alpha,A,F\rangle\in P_{\alpha}$. For each
$\vec{\gamma}\in[A]^{n}$, let
$s_{\vec{\gamma}}\leq^{*}\operatorname{stem}(p+\vec{\gamma})$. Then there is
$p^{*}=\langle\alpha,A^{*},F^{*}\rangle\leq^{*}_{\alpha}p$ such that for every
$\vec{\gamma}\in[A^{*}]^{n}$,
$\operatorname{stem}(p^{*}+\vec{\gamma})\leq^{*}s_{\vec{\gamma}}$.

###### Proof.

We induct on $n$. For $n=1$, each $s_{\langle\gamma\rangle}$ is of the form
$\langle\gamma,A_{\gamma},F_{\gamma},\dot{\nu}_{\gamma},\dot{q}_{\gamma}\rangle$.
Use Lemma 2.2 to obtain $A^{\prime}$ such that for all $\gamma\in A^{\prime}$,
$A^{\prime}\restriction\gamma\subseteq A_{\gamma}$. For $\gamma\in
A^{\prime}$, let $A^{\gamma}\in\vec{U}(\alpha)$, $A^{\gamma}\subseteq
A^{\prime}$, and $\dot{q}_{\gamma}^{*}$ be such that for $\xi\in
A^{\gamma}\setminus(\gamma+1)$, $F_{\xi}(\gamma)_{1}=\dot{q}_{\gamma}^{*}$
(this is possible because as in Proposition 4.1 (4.8), the number of nice
names is small). Take $A^{*}=\triangle_{\gamma}A^{\gamma}$. For $\gamma\in
A^{*}$, take $F^{*}(\gamma)_{1}$ be a name which is forced to be
$\leq^{*}$-stronger than $\dot{q}_{\gamma}^{*}$ and $\dot{q}_{\gamma}$. Take
$p^{*}=\langle\alpha,A^{*},F^{*}\rangle$. For each $\xi\in A^{*}$, let
$t_{\langle\xi\rangle}=\operatorname{stem}(p^{*}+\langle\xi\rangle)$. Note
that
$t_{\langle\xi\rangle}=\langle\xi,A^{*}\restriction\xi,F^{*}\restriction(A^{*}\restriction\xi),F^{*}(\xi)_{0},F^{*}(\xi)_{1}\rangle$.
Since $A^{*}\restriction\xi\subseteq A_{\xi}$, and for $\gamma<\xi$, $\xi\in
A^{\gamma}$, we have that $F^{*}_{\xi}(\gamma)_{1}$ is forced to be
$\leq^{*}$-stronger than $\dot{q}_{\gamma}^{*}=F_{\xi}(\gamma)_{1}$. Finally,
$F^{*}(\xi)_{1}$ is forced to be $\leq^{*}$-stronger than $\dot{q}_{\xi}$.
Hence, $t_{\langle\gamma\rangle}\leq^{*}s_{\langle\gamma\rangle}$.

We now assume $n=m+1$ for some $m>0$. For each
$\vec{\gamma}^{\frown}\langle\xi\rangle\in[A]^{n}$, let

$\displaystyle s_{\vec{\gamma}^{\frown}\langle\xi\rangle}^{0}$
$\displaystyle=t_{\vec{\gamma}^{\frown}\langle\xi\rangle}\restriction m,$
$\displaystyle s_{\vec{\gamma}^{\frown}\langle\xi\rangle}^{1}$
$\displaystyle=(t_{\vec{\gamma}^{\frown}\langle\xi\rangle})_{m}.$

Note that for $\vec{\gamma}$, there is $\tau$ such that for each $\xi$,
$s_{\vec{\gamma}^{\frown}\langle\xi\rangle}^{0}\in V_{\tau}$ for some
$\tau<\alpha$. Hence, for $\beta<\circ(\alpha)$ and $\vec{\gamma}\in[A]^{m}$,
there are $B_{\vec{\gamma}}(\beta)\in U(\alpha,\beta)$ and
$s_{\langle\gamma_{0},\cdots,\gamma_{m-1}\rangle}(\beta)$ such that for
$\xi\in B_{\vec{\gamma}}(\beta)$,
$s_{\langle\gamma_{0},\cdots,\gamma_{m-1},\xi\rangle}^{0}=s_{\langle\gamma_{0},\cdots,\gamma_{m-1}\rangle}(\beta)$.
Take $A^{\prime}=\triangle_{\vec{\gamma}}\cap_{\beta}B_{\vec{\gamma}}(\beta)$
and let $s_{\langle\gamma_{0},\cdots,\gamma_{m-1}\rangle}$ be such that for
all $\beta<\circ(\alpha)$,
$s_{\langle\gamma_{0},\cdots,\gamma_{m-1}\rangle}\leq^{*}s_{\langle\gamma_{0},\cdots,\gamma_{m-1}\rangle}(\beta)$
and intersect all measure-one sets with $A^{\prime}$. Set
$p^{\prime}=\langle\alpha,A^{\prime},F\rangle$. We see that for each
$\vec{\gamma}=\langle\gamma_{0},\cdots,\gamma_{m-1}\rangle$,
$s_{\vec{\gamma}}$ is $\leq^{*}$-stronger than
$\operatorname{stem}(p^{\prime}+\vec{\gamma})$. Apply our induction hypothesis
with $p^{\prime}$ and $s_{\gamma_{0},\cdots,\gamma_{m-1}}$ for all
$\vec{\gamma}$ to get $p^{\prime\prime}\leq^{*}_{\alpha}p^{\prime}$ such that
$\operatorname{stem}(p^{\prime\prime}+\vec{\gamma})\leq^{*}s_{\vec{\gamma}}$
for $\vec{\gamma}$ of length $m$. Let $A^{*}=A^{\prime\prime}$ and for $\xi\in
A^{*}$, let $F^{*}(\xi)_{1}$ be a name which is forced to be
$\leq^{*}$-stronger than $F^{\prime\prime}(\xi)_{1}$ and
$s_{\vec{\gamma}^{\frown}\langle\xi\rangle}^{1}$ for all $\vec{\gamma}$ in
$\\{\vec{\gamma}\mid|\vec{\gamma}|=m$ and $\max(\vec{\gamma})<\xi\\}$. This is
possible since the number of such $\vec{\gamma}$ is $\xi$ and by Proposition
4.1 (4). Let $p^{*}=\langle\alpha,A^{*},F^{*}\rangle$. One can then check that
$p^{*}$ is as required.

∎

######  Proposition 5.3.

$P_{\alpha}$ has the Prikry property.

###### Proof.

We first deal with the case $p$ is pure. The impure case will be done
similarly, with just a small additional work in the beginning.

Let $p=\langle\alpha,A,F\rangle$. Assume that $\min(A)>\circ(\alpha)$. By
Lemma 5.1, we may extend $p$ so that if $s$ is a stem and
$s^{\frown}\langle\alpha,B,H\rangle\leq p$ decides $\sigma$, then
$s^{\frown}\langle\alpha,A\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle$ decides
$\sigma$. Fix a finite increasing sequence
$\langle\gamma_{0},\cdots,\gamma_{n-1}\rangle\in[A]^{n}$. Let
$\dot{\nu}_{\vec{\gamma}}=F(\gamma_{n-1})_{0}$ (Recall that
$F(\gamma_{n-1})=(\dot{\nu},\dot{q})$ where $\dot{q}$ is a $P_{\gamma}$-name
of a condition in $P_{\dot{\nu}_{\vec{\gamma}}}/P_{\gamma_{n-1}}$). For
$\xi\in A$, $\xi>\bar{\nu}_{\vec{\gamma}}$, let $\dot{\nu}_{\xi}=F(\xi)_{0}$,
$\dot{G}$ be the canonical name of a generic over $P_{\xi}$ if
$\circ(\xi)\leq\max_{i}\circ(\gamma_{i})$, and generic over
$P_{\xi}*P_{\dot{\nu}_{\xi}}/P_{\xi}$ otherwise. Let
$s_{\vec{\gamma}}=\operatorname{stem}(p+\vec{\gamma})$ and
$s_{\xi}=(\operatorname{stem}(p+(\vec{\gamma}^{\frown}\langle\xi\rangle)))_{n}$.

By the Prikry property below $\alpha$, let
$s^{0}_{\vec{\gamma},\xi}{}^{\frown}s^{1}_{\vec{\gamma},\xi}\leq^{*}s_{\vec{\gamma}}{}^{\frown}s_{\xi}$
be such that $s^{0}_{\vec{\gamma},\xi}{}^{\frown}s^{1}_{\vec{\gamma},\xi}$
decides

$\exists
s\in\dot{G}(s^{\frown}\langle\alpha,A\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\parallel\sigma$).

For $\beta<\circ(\alpha)$, let $A_{\vec{\gamma}}(\beta)\in U(\alpha,\beta)$ be
such that for each $\xi\in A_{\vec{\gamma}}(\beta)$,
$s^{0}_{\vec{\gamma},\xi}{}^{\frown}s^{1}_{\vec{\gamma},\xi}$ gives the same
decision, i.e. either for all $\xi\in A_{\vec{\gamma}}(\beta)$,

$s^{0}_{\vec{\gamma},\xi}{}^{\frown}s^{1}_{\vec{\gamma},\xi}\Vdash\exists
s\in\dot{G}(s^{\frown}\langle\alpha,A\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\parallel\sigma$),

or for all $\xi\in A_{\vec{\gamma}}(\beta)$

$s^{0}_{\vec{\gamma},\xi}{}^{\frown}s^{1}_{\vec{\gamma},\xi}\Vdash\nexists
s\in\dot{G}(s^{\frown}\langle\alpha,A\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\parallel\sigma$),

If the decision is positive, shrink the set further so that either for all
$\xi$,

$s^{0}_{\vec{\gamma},\xi}{}^{\frown}s^{1}_{\vec{\gamma},\xi}\Vdash\exists
s\in\dot{G}(s^{\frown}\langle\alpha,A\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\Vdash\sigma)$,

or for all $\xi$,

$s^{0}_{\vec{\gamma},\xi}{}^{\frown}s^{1}_{\vec{\gamma},\xi}\Vdash\exists
s\in\dot{G}(s^{\frown}\langle\alpha,A\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\Vdash\neg\sigma)$.

Let $A^{\prime}=\triangle_{\vec{\gamma}}\cap_{\beta}A_{\vec{\gamma}}(\beta)$.
Use Lemma 5.2 with $p=\langle\alpha,A^{\prime},F\rangle$ and
$s_{\vec{\gamma},\xi}^{0}{}^{\frown}s_{\vec{\gamma},\xi}^{1}$ to obtain
$p^{n}\leq^{*}_{\alpha}p$ such that
$\operatorname{stem}(p^{n}+(\vec{\gamma}^{\frown}\langle\xi\rangle))\leq^{*}s_{\vec{\gamma},\xi}^{0}{}^{\frown}s_{\vec{\gamma},\xi}^{1}$.
We now take $p^{*}$ be such that $p^{*}\leq^{*}_{\alpha}p^{n}$ for all $n$.

Suppose for a contradiction that there is no direct extension of $p^{*}$
deciding $\sigma$. Let $\bar{p}\leq p^{*}$ with the minimal number of steps
such that $\bar{p}\parallel\sigma$. Assume $\bar{p}\Vdash\sigma$. Say
$\bar{p}\leq p^{*}$, $l(\bar{p})=n+1$, $\bar{p}\leq
p^{*}+(\vec{\gamma}^{\frown}\langle\xi\rangle)$ where $|\vec{\gamma}|=n$. Let
$s=\operatorname{stem}(\bar{p})$, $s_{0}=s\restriction n$ and $s_{1}=s_{n}$,
so that $s=s_{0}^{\frown}s_{1}$. We note that by maximality,
$s^{\frown}\langle\alpha,A\setminus\operatorname{top}(s),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\Vdash\sigma$.

######  Claim 5.3.1.

$\operatorname{stem}(p^{*}+(\vec{\gamma}^{\frown}\langle\xi\rangle))$ forces
that there is $t\in\dot{G}$,
$t^{\frown}\langle\alpha,A^{*}\setminus\operatorname{top}(t),F^{*}\setminus\operatorname{top}(t)\rangle\Vdash\sigma$.

Proof (Claim 5.3.1)

Suppose not. For simplicity, assume that
$\operatorname{stem}(p^{*}+(\vec{\gamma}^{\frown}\langle\xi\rangle))$ forces
that there is $t\in\dot{G}$,
$t^{\frown}\langle\alpha,A^{*}\setminus\operatorname{top}(t),F^{*}\setminus\operatorname{top}(t)\rangle\Vdash\neg\sigma$
(the other case can be done similarly). Let $G$ be generic containing
$\operatorname{stem}{\bar{p}}$, then it contains
$\operatorname{stem}(p^{*}+(\vec{\gamma}^{\frown}\langle\xi\rangle))$. Let
$s^{\prime}\in G$ be such that
$s^{\prime}{}^{\frown}\langle\alpha,A\setminus\operatorname{top}(s^{\prime}),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s^{\prime})\mid\circ(\gamma)>\circ(s^{\prime})\\}\rangle\Vdash\neg\sigma$
(note that $\operatorname{top}(s^{\prime})=\operatorname{top}(s)$). We may
assume that $s^{\prime}\leq s$, but then
$s^{\prime}{}^{\frown}\langle\alpha,A\setminus\operatorname{top}(s^{\prime}),F\restriction\\{\gamma\in
A\setminus\operatorname{top}(s^{\prime})\mid\circ(\gamma)>\circ(s^{\prime})\\}\rangle\Vdash\sigma$
and $\neg\sigma$, which is a contradiction.

$\blacksquare$(Claim 5.3.1)

With a similar proof as in Claim 5.3.1, we can conclude that
$p^{*}+(\vec{\gamma}^{\frown}\langle\xi\rangle)\Vdash\sigma$, and hence
$s_{0}^{\frown}\langle\xi,A^{*}\restriction\xi,F^{*}\restriction(A^{*}\restriction\xi),F^{*}(\xi)_{0},F^{*}(\xi)_{1}\rangle^{\frown}\langle\alpha,A^{*}\setminus\bar{\nu}_{\xi},F^{*}\restriction\\{\tau\mid\tau>\bar{\nu}_{\xi}$
and $\circ(\tau)>\circ(\xi)\\}\rangle\Vdash\sigma$. This implies that for each
$\xi^{\prime}\in A^{*}$ with $\circ(\xi^{\prime})=\beta$, we have that
$s_{0}^{\frown}\langle\xi^{\prime},A^{*}\restriction\xi^{\prime},F^{*}\restriction(A^{*}\restriction\xi^{\prime}),F^{*}(\xi^{\prime})_{0},F^{*}(\xi^{\prime})_{1}\rangle^{\frown}\langle\alpha,A^{*}\setminus\bar{\nu}_{\xi^{\prime}},F^{*}\restriction\\{\tau\mid\tau>\bar{\nu}$
and $\circ(\tau)>\circ(\xi^{\prime})\\}\rangle\Vdash\sigma$. Since every
extension of
$s_{0}^{\frown}\langle\alpha,A^{*}\setminus\operatorname{top}(s),F^{*}\restriction\\{\gamma\in
A^{*}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle$ can
be extended further to be an extension of
$s_{0}^{\frown}\langle\alpha,A^{*}\setminus\operatorname{top}(s),F^{*}\restriction\\{\gamma\in
A^{*}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle+\langle\xi^{\prime}\rangle$
for some $\xi^{\prime}$ with $\circ(\xi^{\prime})=\beta$, and the further
extension forces $\sigma$. By density,
$s_{0}{}^{\frown}\langle\alpha,A^{*}\setminus\operatorname{top}(s),F^{*}\restriction\\{\gamma\in
A^{*}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\Vdash\sigma$,
contradicting the minimality of the number of steps used in extension from
$p^{*}$ to $\bar{p}$.

The proof for an impure case is the following: let
$s^{\frown}\langle\alpha,A,F\rangle\in P_{\alpha}$. Let $Q$ be the forcing $s$
is living in. Note that cardinal arithmetic in $V$ is the same as in $V[G]$.
Let $G$ be $Q$-generic, then work in $P_{\alpha}[G]$ with
$(\langle\alpha,A,F\rangle)[G]$ in a similarly to before to get
$\langle\alpha,B,H\rangle$ deciding $\sigma[G]$ (note that Proposition 2.3 is
necessary here, since when we split a measure-one set $A\in\vec{U}(\alpha)$
into disjoint sets, and one of them is of measure-one, then that set may have
measure one in the ultrafilters in the extension, however, we can shrink the
set to have measure one with respect to ultrafilters in the ground model).
Back in $V$, by the Prikry property of $Q$, let $s^{*}\leq^{*}s$ such that
$s^{*}\Vdash_{Q}\langle\alpha,\dot{B},\dot{H}\rangle\Vdash_{P_{\alpha}/Q}\sigma^{i}$
for some $i=0,1$, where $\sigma^{0}=\sigma$ and $\sigma^{1}=\neg\sigma$. By
Proposition 2.4, we can shrink $\dot{B}$ to $B\in V$ such that
$\Vdash_{Q}\check{B}\subseteq\dot{B}$. Then find $H$ such that for all
$\gamma$, $H(\gamma)_{1}$ is forced to be $\leq^{*}$-stronger than
$\dot{H}(\gamma)_{1}$, Hence, $s^{*}{}^{\frown}\langle\alpha,B,H\rangle$
decides $\sigma$ as required.

∎

With a similar proof of the proof for the Prikry property, we obtain the
following.

######  Proposition 5.4.

Let $\beta,\dot{\nu}$, and $Q$ be as in Definition 4.10, then
$\Vdash_{Q}(P_{\alpha}/Q,\leq,\leq^{*})$ has the Prikry property.

###### Proof.

First, suppose that pure conditions in $P_{\alpha}/Q$ are defined. Let $G$ be
$Q$-generic and in $V[G]$, let $p\in P_{\alpha}[G]$ is pure. Then proceed as
in Proposition 5.3. Now, suppose either impure conditions in $P_{\alpha}/Q$
are defined, and in $V[G]$, $p\in P_{\alpha}[G]$ is impure, or impure
conditions in $P_{\alpha}/Q$ are not defined and $p\in P_{\alpha}[G]$ be any
conditions. Then $p=s^{\frown}\langle\alpha,A,F\rangle$. Proceed the proof of
the Prikry property as in Proposition 7 for impure conditions.

∎

##  6\. The cardinal behavior in $P_{\alpha}$

By a chain condition of $P_{\alpha}$, cardinals above $\alpha^{+}$,
exclusively, are preserved, and the forcing preserves cofinalities of regular
cardinals above $\alpha^{+}$ We first show that cardinals up to, but not
including $\alpha$, are preserved.

######  Proposition 6.1.

All cardinals up to and including $\alpha$, are preserved.

###### Proof.

We first show that cardinals below $\alpha$ are preserved. Consider each
cardinal $\gamma<\alpha$ and $p\in P_{\alpha}$. Set
$p=s^{\frown}\langle\alpha,A,F\rangle$. Let $G$ be generic containing $s$.
Since it is from the forcing below $\alpha$, by induction, $\gamma$ is a
cardinal in $V[G]$. Let $\bar{p}=(\langle\alpha,A,F\rangle)[G]$. By the Prikry
property and the fact that the direct extension relation for pure conditions
in $P_{\alpha}[G]$ is $\alpha$-closed for arbitrary $G$, we get that
$p\Vdash_{\alpha}\gamma$ is preserved. Finally, note that since $\alpha$ is
strongly inaccessible in $V$, $\alpha$ is preserved in the extension.

∎

######  Proposition 6.2.

GCH below $\alpha$ holds in $V^{P_{\alpha}}$.

###### Proof.

Let $\gamma<\alpha$. Let $p=s^{\frown}\langle\alpha,A,F\rangle$. Forcing with
$p$ is factored into $P^{s}$, the forcing containing $s$, and
$P_{\alpha}/P^{s}$. Since $P^{s}$ is a forcing below
$\alpha$,$\Vdash_{P^{s}}(P_{\alpha}/P^{s},\leq,\leq^{*})$ has the Prikry
property, and $\Vdash(P_{\alpha}/P^{s},\leq^{*})$ is $\alpha$-closed, forcing
with $p$ preserves GCH at $\gamma$.

∎

By Propositions 6.1 and 6.2,

######  Corollary 6.3.

$\alpha$ is preserved (as a cardinal).

If $\circ(\alpha)=0$, $P_{\alpha}$ is equivalent to
$P_{\beta}*P_{\dot{\nu}}/P_{\beta}$ for some $\beta$ and $\dot{\nu}$, which is
forced to be below $\alpha$, hence, $\alpha$ is still inaccessible. If
$\circ(\alpha)>0$, the forcing $P_{\alpha}$ clearly projects down to a Prikry
forcing or a Magidor forcing, adding a club subset of $\alpha$ of order-type
$\omega^{\circ(\alpha)}$ (the ordinal exponentiation). If $\circ(\alpha)=0$,
then obviously $\alpha^{+}$ is still regular in $V^{P_{\alpha}}$.

######  Proposition 6.4.

If $\circ(\alpha)>0$, then $\alpha^{+}$ is preserved and remains regular in
$V[G_{\alpha}]$.

###### Proof.

Since $0<\circ(\alpha)<\alpha$, in $V[G_{\alpha}]$,
$\operatorname{cf}(\alpha)<\alpha$. Suppose $\alpha^{+}$ is collapsed, then
there is $\delta<\alpha$ and a cofinal map $f:\delta\to(\alpha^{+})^{V}$. Let
$\dot{f}$ be a such name, forced by $p$. Assume for simplicity that
$p=\langle\alpha,A,F\rangle$ (if $p$ is not pure, then we may first force over
the stem of $p$ and proceed as usual). Since $\delta<\alpha$, we may find
$p^{\prime}\leq_{\alpha}^{*}p$ such that if
$p^{\prime}=\langle\alpha,A^{\prime},F^{\prime}\rangle$, then for all
$\gamma<\delta$, if
$s^{\frown}\langle\alpha,B,H\rangle\leq_{\alpha}p^{\prime}$ and
$s^{\frown}\langle\alpha,B,H\rangle$ decides $\dot{f}(\gamma)$, then so is
$s^{\frown}\langle\alpha,A^{\prime}\setminus\operatorname{top}(s),F^{\prime}\restriction\\{\gamma\in
A^{\prime}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle$.
Let $X=\\{\xi\mid\exists\gamma<\delta\exists\text{stem
}s(s^{\frown}\langle\alpha,A^{\prime}\setminus\operatorname{top}(s),F^{\prime}\restriction\\{\gamma\in
A^{\prime}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\Vdash_{\alpha}\dot{f}(\check{\gamma})=\check{\xi})\\}$.
Then $p^{\prime}\Vdash_{\alpha}\operatorname{rge}(\dot{f})\subseteq\check{X}$
and $|X|\leq\alpha\otimes\delta=\alpha$. Hence,
$p^{\prime}\Vdash_{\alpha}\operatorname{rge}(\dot{f})$ is bounded in
$(\alpha^{+})^{V}$, which is a contradiction.

A similar proof shows that the cofinality of $\alpha^{+}$ is preserved, so
$\alpha^{+}$ is regular.

∎

Since $|P_{\alpha}|\leq\alpha^{++}$, we have that GCH hold everywhere except
possibly at $\alpha$. The next proposition shows that GCH holds at $\alpha$.

######  Proposition 6.5.

GCH holds at $\alpha$.

###### Proof.

Suppose for a contradiction that in $V^{P_{\alpha}}$,
$2^{\alpha}\geq\alpha^{++}$. Let
$p\Vdash_{\alpha}\\{\dot{f}_{\gamma}\mid\dot{f}_{\gamma}:\alpha\to
2,\gamma<\alpha^{++}\\}$ a sequence of different subsets of $\alpha$. Since
forcing with stem of $p$ does not change GCH at $\alpha$, we may assume for
simplicity that $p$ is pure. Let $D_{\gamma,\xi}$ be the collection of
$p^{\prime}$ that decides $\dot{f}_{\gamma}(\xi)$. Use Proposition 5.1 to get
a $\leq^{*}$-decreasing sequence
$\vec{p}_{\gamma}=\\{p_{\gamma,\xi}\mid\xi<\alpha\\}$ each $p_{\gamma,\xi}$
witnesses Proposition 5.1 for $D_{\gamma,\xi}$. By GCH in $V$,
$(\alpha^{+})^{\alpha}=\alpha$, we can shrink the family of functions so that
there is $\vec{p}$ with $\vec{p}_{\gamma}=\vec{p}$ for all $\gamma$. Write
$\vec{p}=\\{p_{\gamma}\mid\gamma<\alpha\\}$,
$p_{\gamma}=\langle\alpha,A_{\gamma},F_{\gamma}\rangle$. Note that the number
of conditions that decide $\dot{f}_{\gamma}(\xi)$ for some $\gamma,\xi$ is at
most $\alpha$: i.e. the collection of
$s^{\frown}\langle\alpha,A_{\gamma}\setminus\operatorname{top}(s),F_{\gamma}\restriction\\{\gamma\in
A_{\gamma}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle$
for some $s,\gamma$. Then, we count the number of nice names for functions
from $\alpha$ to $2$, with respect to conditions in
$\\{s^{\frown}\langle\alpha,A_{\gamma}\setminus\operatorname{top}(s),F_{\gamma}\restriction\\{\gamma\in
A_{\gamma}\setminus\operatorname{top}(s)\mid\circ(\gamma)>\circ(s)\\}\rangle\mid$
for some $s$ and $\gamma\\}$. It is at most
$\alpha^{\alpha\otimes\alpha}=\alpha^{+}$, which is a contradiction.

∎

Let $G$ be $P_{\alpha}$-generic. If $\circ(\alpha)=0$, then $P_{\alpha}$ is
isomorphic to $Q$ for some $\beta,\dot{\nu}$ and $Q$ as in Definition 4.10.
Define $C_{\alpha}$ in $V[G]$ by $C_{\alpha}=C_{Q}\cup\\{\alpha\\}$ where
$C_{Q}$ is the set added by $Q$ (if $Q=P_{\beta}$, then $C_{Q}=C_{\beta}$,
otherwise, $C_{Q}=C_{\beta}\sqcup C_{\nu/\beta}$ where
$\nu=\dot{\nu}[G\restriction P_{\beta}]$. Then $C_{\alpha}\subseteq\alpha+1$,
$\max(C_{\alpha})=\alpha$, and $C_{\alpha}\setminus\\{\alpha\\}$ is a closed
bounded subset of $\alpha$.

Consider the general case, $\circ(\alpha)>0$. Let $G$ be $P_{\alpha}$-generic.
Let $C^{*}=\\{\gamma<\alpha\mid\exists p\in P_{\alpha}\exists
i(\operatorname{stem}(p)_{i})_{0}=\gamma\\}$. The tail of $C^{*}$ is simply a
Prikry sequence if $\circ(\alpha)=1$, or a magidor club set of $\alpha$ if
$\circ(\alpha)>1$. For $\gamma\in C^{*}$, let $G_{\gamma}=G\restriction
P_{\gamma}$. For such $\gamma$. For $p\in G$ with
$(\operatorname{stem}(p)_{i})_{0}=\gamma$ for some $i$, let
$s_{\gamma,p}=\operatorname{stem}(p)_{i}$. If $s_{\gamma,p}$ is a triple, then
$s_{\gamma,p^{\prime}}$ is a triple for all $p^{\prime}\in G$ with
$s_{\gamma,p^{\prime}}$ exists. In this case, let $C^{*}_{\gamma}=\emptyset$.
If, otherwise, $s_{\gamma,p}$ is a $5$-tuple. Set
$R=\\{\dot{q}[G_{\gamma}]\mid\exists p^{\prime}\in
G((s_{\gamma,p^{\prime}})_{4}=\dot{q})\\}$. Then, $R$ is
$P_{\dot{\nu}[G_{\gamma}]}[G_{\gamma}]$-generic over $V[G_{\gamma}]$ for some
$\dot{\nu}$. This produces the set
$C_{\nu_{\gamma}/\gamma}\subseteq(\gamma,\nu_{\gamma}]$ where
$\nu_{\gamma}=\dot{\nu}[G_{\gamma}]$. We have that
$\nu_{\gamma}\in[\gamma,\max(C^{*}\setminus(\gamma+1))$. If
$\nu_{\gamma}=\gamma$, then $C_{\nu_{\gamma}/\gamma}=\emptyset$. Assume
$\nu_{\gamma}>\gamma$. Then $\max(C_{\nu_{\gamma}/\gamma})=\nu_{\gamma}$,
$C_{\nu/\gamma}\setminus\\{\nu\\}$ is a club set of $\nu$ if $\circ(\nu)>0$,
or a closed bounded subset of $\nu$ if $\circ(\nu)=0$. Furthermore,
$\nu<\min(C^{*}\setminus(\nu+1))$. Set
$C_{\alpha}=C^{*}\cup\cup\\{C_{\nu_{\gamma}/\gamma}\mid\gamma\in
C^{*}\\}\cup\\{\alpha\\}$.

Now, if $\beta,\dot{\nu}$ and $Q$ be as in Definition 4.10, and $G$ is
$Q$-generic. In $P_{\alpha}[G]$, we can apply the same process to get the
required set $C_{\alpha}/Q$ as follows. Let $H$ be $P_{\alpha}[G]$-generic,
and $I$ be the canonical generic over $P_{\alpha}$ such that $V[G][H]=V[I]$.
In $V[I]$, we have $C_{\alpha}$. Set $C_{\alpha/Q}=C_{\alpha}\setminus C_{Q}$.

######  Proposition 6.6.

Forcing with $P_{\alpha}$ preserves cofinalities of regular cardinals which
are not in $\lim(C_{\alpha})$.

###### Proof.

Let $\gamma\not\in\lim(C_{\alpha})$. First, suppose that $\circ(\alpha)=0$,
then $P_{\alpha}$ is equivalent to some $P_{\beta}*P_{\dot{\nu}}/P_{\beta}$,
and $C_{\alpha}=C_{\beta}\sqcup C_{\nu/\beta}\cup\\{\alpha\\}$ for some $\nu$.
If $\gamma=\alpha$, then $\gamma$ is regular in $V[G]$ since $P_{\alpha}$ is
equivalent to a small forcing. If $\gamma<\alpha$, then
$\gamma\not\in\lim(C_{\beta}\sqcup C_{\nu/\beta})$, and so $\gamma$ is regular
in $V[G\restriction P_{\beta}*P_{\dot{\nu}}/P_{\beta}]=V[G]$. Now, suppose
that $\circ(\alpha)>0$. Then find $\beta\in(\gamma,\alpha)$ such that
$G\restriction P_{\beta}$ exists. Note that $\gamma\not\in\lim(C_{\beta})$, so
$\gamma$ is regular in $V[G\restriction P_{\beta}]$. The forcing
$(P_{\alpha}[G\restriction P_{\beta}],\leq^{*})$ is $\beta^{+}$-closed, so
$\gamma$ is still regular in $V[G]$.

∎

Similar proof with an easy application on factoring forcings show that

######  Proposition 6.7.

$\Vdash_{\alpha}\gamma\not\in\lim(C_{\alpha/Q})$ and $(\gamma$ is
regular)${}^{V^{Q}}$ implies that $\gamma$ is regular.

##  7\. The main forcing $\mathbb{P}_{\kappa}$

We now define the main forcing $\mathbb{P}_{\kappa}$.

######  Definition 7.1.

Define $\mathbb{P}_{\kappa}=\cup_{\alpha<\kappa}P_{\alpha}$. For
$p,q\in\mathbb{P}_{\kappa}$, define $p\leq q$ if $p,q\in P_{\alpha}$ and
$p\leq_{\alpha}q$, or $p\in P_{\beta}$, $q\in P_{\alpha}$, $\beta>\alpha$, and
$p\restriction P_{\alpha}\leq_{\alpha}q$.

If $G$ is $\mathbb{P}_{\kappa}$-generic, then for $\alpha<\kappa$ with some
$p\in G\cap P_{\alpha}$, $G_{\alpha}:=G\restriction P_{\alpha}$ is
$P_{\alpha}$-generic. $\mathbb{P}_{\kappa}$ has size $\kappa$, so it is
$\kappa^{+}$-c.c.

######  Theorem 7.2.

Let $G$ be $\mathbb{P}_{\kappa}$-generic. In $V[G]$, if $f:\alpha\to On$ for
some $\alpha<\kappa$, then there is $\beta<\kappa$ such that $f\in
V[G\restriction P_{\beta}]$.

###### Proof.

Recall that for $\gamma<\kappa$, $\\{\alpha\in
X\mid\circ(\alpha)\geq\gamma\\}$ is stationary. Let $\dot{f}:\alpha\to On$ and
$p\in\mathbb{P}_{\kappa}$. Let $M\prec H_{\theta}$ such that
$M\cap\kappa\in\kappa$, $M$ is transitive below $M\cap\kappa$,
$\mathbb{P}_{\kappa},p,\kappa,\dot{f}\in M$, and if $\gamma:=M\cap\kappa$,
then $\circ(\gamma)>\alpha$ is a regular cardinal and $\circ(\gamma)$ is
greater than $\circ(\xi)$ any $\xi$ which is equal to
$(\operatorname{stem}(p)_{i})_{0}$ for some $i$, say that $\xi_{0}$ is the
maximum of those $\circ(\xi)$s. Let $A\in\vec{U}(\gamma)$. We build $F$ with
$\operatorname{dom}(F)=\\{\beta\in A\mid\circ(\beta)>\xi_{0}\\}$. We define
the value $F$ inductively on the Mitchell order. For each $\beta$ with
$\circ(\beta)=\xi_{0}+1$, let $G_{\beta}$ be the forcing containing
$p^{\frown}\langle\gamma,A\restriction\beta,\emptyset\rangle$. Then, in
$V[G_{\beta}]$, there is $q_{\beta}\in\mathbb{P}_{\kappa}/G_{\beta}$ such that
$q_{\beta}$ decides $\dot{f}[G_{\beta}](0)$. Since $M\cap\kappa$ is transitive
and $\beta<M\cap\kappa$, we may find $q_{\beta}\in M[G_{\beta}]$. Note that
$\kappa\cap M[G_{\beta}]=\kappa\cap M$. Thus, $q_{\beta}\in P_{\nu}/P_{\beta}$
for some $\nu_{\beta}<\gamma$. Let $\dot{\nu}_{\beta}$ and $\dot{q}_{\beta}$
be names for such $\nu_{\beta}$ and $q_{\beta}$ and define
$F(\beta)=\langle\dot{\nu}_{\beta},\dot{q}_{\beta}\rangle$. Suppose that $F$
is defined on ordinals of Mitchell orders in $(\xi_{0},\xi_{1})$ for some
$\xi_{1}<\xi_{0}+\alpha$. Write $\xi_{1}=\xi_{0}+\tau$ for unique $\tau$. For
$\beta\in A$ with $\circ(\beta)=\xi_{1}$, let $G_{\beta}$ be generic
containing $p^{\frown}\langle\alpha,A\restriction\beta,F\restriction\\{\tau\in
A\restriction\beta\mid\circ(\tau)>\xi_{0}\\}\rangle$. Find
$q_{\beta}\in\mathbb{P}_{\kappa}/G_{\beta}\cap M[G_{\beta}]$ deciding
$\dot{f}[G_{\beta}](\tau)$, so $q_{\beta}\in P_{\nu}[G_{\beta}]$, and let
$\dot{\nu}_{\beta}$, $\dot{q}_{\beta}$ be such names. Set
$F(\beta)=\langle\nu_{\beta},\dot{q}_{\beta}\rangle$. This process continues
up to (but not including) stage $\xi_{0}+\alpha$. For $\beta$ with
$\circ(\beta)\geq\xi_{0}+\alpha$, assign any $F(\beta)$. We can see that by
the choice of $F$, if $G$ is $\mathbb{P}_{\kappa}$-generic containing
$p^{\frown}\langle\alpha,A,F\rangle$, then $\dot{f}\in V[G\restriction
P_{\gamma}]$. Thus,
$p^{\frown}\langle\gamma,A,F\rangle\Vdash_{\kappa}\dot{f}\in
V[\dot{G}\restriction P_{\gamma}]$. ∎

######  Corollary 7.3.

In $V^{\mathbb{P}_{\kappa}}$, $\kappa$ is still inaccessible, and GCH holds
below $\kappa$.

Let $G$ be $\mathbb{P}_{\kappa}$-generic. Set
$C_{\kappa}=\cup_{\alpha<\kappa}C_{\alpha}$ where $C_{\alpha}$ is the set
derived from $V[G\restriction P_{\alpha}]$. For $\alpha<\beta<\kappa$,
$C_{\alpha}\sqsubseteq C_{\beta}$.

######  Proposition 7.4.

$C_{\kappa}$ is a club subset of $\kappa$.

###### Proof.

Since for $p\in P_{\alpha}$ and $\beta\in X\setminus(\alpha+1)$, there is
$p^{\prime}\leq_{\kappa}p$, $p^{\prime}\in P_{\beta}$, $C_{\kappa}$ is
unbounded in $\kappa$. To check the closeness, let $\gamma$ be the limit point
of $C_{\kappa}$, then it is forced by some $p\in P_{\alpha}$ for some
$\alpha>\gamma$. We see that $\gamma$ is a limit point in $C_{\alpha}$. Thus,
$\gamma$ is a limit point of $C_{\kappa}$.

∎

######  Proposition 7.5.

If $\gamma\not\in\lim(C_{\kappa})$ is regular in $V$, then in is regular in
$V^{\mathbb{P}_{\kappa}}$.

###### Proof.

Let $\gamma\not\in\lim(C_{\kappa})$ be regular in $V$. Suppose for a
contradiction that $\beta<\gamma$ and $f:\beta\to\kappa$ be cofinal. Find
$\alpha<\kappa$ large such that $f\in V[G\restriction P_{\alpha}]$. Since
$\gamma\not\in\lim(C_{\kappa})$, $\gamma\not\in\lim(C_{\alpha})$, so $\gamma$
is regular in $V[G\restriction P_{\alpha}]$, which is a contradiction.

∎

##  8\. The optimal assumption

As we noted earlier, the following assumption is also enough to build the
forcing that has the same effect:

######  Assumption 8.1.

$\kappa$ is inaccessible, $X\subseteq\kappa$, each $\alpha\in X$,
$\\{U(\alpha,\beta)\mid\beta<\circ(\alpha)\\}$ is a Mitchell increasing
sequence of normal measures, and for $\nu<\kappa$, $\\{\alpha\in
X\mid\circ(\alpha)\geq\nu\\}$ is stationary.

We complete this section by recalling, with proof, that Assumption 8.1 is
optimal.

######  Proposition 8.2.

Assume there is no inner model of $0^{\P}$. Let $V\subseteq W$. Suppose that
in $W$, there is a club subset $C$ of $\kappa$ containing $V$-regular
cardinals, $\kappa$ is inaccessible in $W$, and all $V$-regular cardinals
which are not limit points of $C$ are still regular in $W$. Then, $V$
satisfies Assumption 8.1.

###### Proof.

Let $\nu<\kappa$ and $C^{\prime}\subseteq\kappa$ be a club in $V$. By
absoluteness, $C^{\prime}$ is a club in $W$, and so $C^{\prime}\cap C$ is a
club. Find $\gamma\in C^{\prime}\cap C$ such that
$\operatorname{cf}^{W}(\gamma)\geq\nu$. Since $\gamma$ is regular in $V$, by
[4], $\circ(\gamma)\geq\nu$ in $V$.

∎

## References

  * [1] W. Mitchell, “How weak is a closed unbounded ultrafilter?,” vol. 108, pp. 209–230, 1982.
  * [2] M. Gitik, “On closed unbounded sets consisting of former regulars,” The Journal of Symbolic Logic, vol. 64, no. 1, pp. 1–12, 1999.
  * [3] M. Foreman and A. Kanamori, Handbook of Set Theory.  Springer Netherlands, 2010.
  * [4] W. J. Mitchell, “Applications of the core model for sequences of measures,” Transactions of the American Mathematicla Society, no. 2, pp. 41–58, 1987\.