# Prikry type forcings and the Bukovský-Dehornoy phenomena

Yair Hayut  Einstein Institute of Mathematics
Edmond J. Safra Campus
The Hebrew University of Jerusalem
Givat Ram. Jerusalem, 9190401, Israel<EMAIL_ADDRESS>
###### Abstract.

This paper is meant to present in a coherent way several instances of quite
common phenomena that was first identified (independently) by Bukovský and
Dehornoy. We present the basic result for Prikry type forcing and show how to
extend it to the Gitik-Shraon forcing, the Extender Based Prikry forcing,
Prikry forcings with interleaved collapses and Radin forcing for
$o(\kappa)<\kappa^{+}$.

This document is based on a tutorial lectures that were given in Torino, in
the 8th European Set Theory Conference. This research was supported by the
Israel Science Foundation 1967/21.

In 1968, Karl Prikry introduces the Prikry forcing, a forcing notion that
changes the cofinality of a regular cardinal without collapsing cardinals,
[24]. Almost ten years later Bukovský and Dehornoy published independently a
theorem showing that a generic for the Prikry forcing can be obtained in an
iterated ultrapower (see Theorem 14), [6, 8], and moreover — the generic
extension over the iterated ultrapower is exactly the intersection of the
intermediate steps in the iteration.

This theorem did not get a lot of attention and remained mostly unknown.
Dehornoy found a generalizations of this theorem to Magidor forcing, [9] and
later Hamkins generalized it to tree forcing [17]. Woodin and Cummings develop
a generalization of the theorem for the Radin forcing, which remained
unpublished.

In [25], Sakai derived a version of this theorem for Prikry-like forcings
based on ideals but more importantly gave a way to construct a Prikry type
forcing from an iteration of elementary emebedding. This method was used
recently in [2] in order to construct a sophisticated forcing notion for
controlling the intersection of set of indiscernibles for a set of reals $A$.

In [11], Fuchs and Hamkins analysed the possible Bukovský-Dehornoy phenomena
for various forcing notions (both the existence of generic over iterated
ultrapower and the intersection model theorem).

In this paper we would like to continue the line of investigation of Sakai,
but going even further: we would like to show that almost all properties of
the generic extension by a Prikry type forcings can be derived naturally by
considering a variant of the Bukovský-Dehornoy theorem and analysing the
properties of the intersection model.

In Section 1, I collected some facts which are partially folklore but not
known to the wide audience regarding the Bukovský-Dehornoy Theorem. In
particular, I illuminate the potential shift in view, focusing on the generic
extension as the intersection of models instead, and deriving the properties
of the Prikry forcing from it. The results in this section are not due to me,
and I put the right citations when possible.

In Section 2, I derive the results of Gitik-Sharon [16], using the Bukovský-
Dehornoy approach. In Section 3 I derive a Bukovský-Dehornoy Theorem for the
Extender Based Prikry forcing and for Prikry forcing with interleaved
collapses.

In Section 4 we derive a variant of the Bukovský-Dehornoy Theorem for the
Magidor and Radin forcing. This theorem seems to be folklore as well, but I
show there how to derive the main properties of Radin forcing using this
approach. The main result of the last section is the _failure_ of the
Bukovský-Dehornoy Theorem for Radin forcing with $o(\kappa)=\kappa^{+}$.

##  1\. Prikry forcing and the Bukovský-Dehornoy Theorem

In this section we will present the Prikry forcing from the perspective of the
Bokuvský-Dehornoy theorem. Instead of designing a forcing notion and analysing
its properties, our goal is to construct a pair of models of $\mathrm{ZFC}$,
with the same cardinals $N_{0}\subseteq N_{1}$ such that there is an
$N_{0}$-regular cardinal which is singular in $N_{1}$.

###  1.1. Iterated ultrapowers

Recall the definition of a measure on $\kappa$.

######  Definition 1.

Let $\kappa$ be a regular uncountable cardinal.

  * •

An ultrafilter $U\subseteq\mathcal{P}(\kappa)$ is $\kappa$-complete if every
intersection of $<\kappa$ sets from $U$ is in $U$.

  * •

We say that $U\subseteq\mathcal{P}(\kappa)$ is a _normal measure_ if $U$ is an
ultrafilter that contains all co-bounded sets and for every $f\colon
A\to\kappa$ such that $A\in U$ and $f(\alpha)<\alpha$ for all $\alpha\in A$,
there is $\gamma\in\operatorname{range}f$, such that $f^{-1}(\\{\gamma\\})\in
U$.

Note that every normal measure is $\kappa$-complete.

######  Lemma 2 (Scott, [26]).

An ultrafilter $U$ on $\kappa$ is normal if and only if there is an elementary
embedding $j\colon V\to M$, such that $M$ is well founded and
$\operatorname{crit}j=\kappa$, and $U=\\{X\subseteq\kappa\mid\kappa\in
j(X)\\}$.

Throughout this paper, I will try to replace (as much as possible) the
combinatorial definitions of the objects involved, such as measures and
extenders, with properties of elementary embeddings. This fits better with the
Bukovaký-Dehornoy theorem as well as will some recent works of Merimovich, for
example [22]. Nevertheless, we need to know that all our elementary embeddings
have combinatorial definition. This is important in order to be able to
iterated them as well as to control their continuity points.

######  Definition 3 (Kunen, [19]).

Let $U$ be a $\kappa$-complete measure in a model $V$. Let us define by
induction an iteration of the ultrapower embedding.

$\begin{matrix}V&=&M_{0}&j_{0,0}=id\\\
M_{n+1}&=&\operatorname{Ult}(M_{n},j_{n}(U)),&j_{n,n+1}\colon M_{n}\to
M_{n+1}\\\ &\vdots&\\\ M_{\omega}&=&\lim\langle M_{n},j_{n,m}\mid n\leq
m<\omega\rangle&j_{n,\omega}\colon M_{n}\to M_{\omega}\end{matrix}$

Where $\lim$ indicates the direct limit of the directed system of embeddings.

######  Theorem 4 (Gaifman, [13]).

$M_{\omega}$ is well founded.

Let us assume that $U$ is normal, for simplicity. Let
$\kappa_{n}=j_{n}(\kappa)$ for $n\leq\omega$. Let $P=\langle\kappa_{n}\mid
n<\omega\rangle$.

Let us look at $M_{\omega}[P]$ — the least $\mathrm{ZFC}$-model containing
$M_{\omega}$ and $P$. This model can be defined by
$\bigcup_{\alpha\in\beta\in\mathrm{Ord}}L_{\beta}(M_{\alpha}\cup\\{P\\})$, but
as the following theorems will show us, it can also be defined without
referring to the $L(A)$ construction. We will show that $M_{\omega}$ and
$M_{\omega}[P]$ have the same cardinals and $\kappa_{\omega}$ is regular in
$M_{\omega}$ and singular in $M_{\omega}[P]$.

Let us start with a few simple facts about iterated ultrapowers by a measure
on $\kappa$.

######  Claim 5.

  1. (1)

For every element $x\in M_{n}$, there is $f\in V$, $f\colon\kappa^{n}\to V$
such that $x=j_{n}(f)(\kappa_{0},\dots,\kappa_{n-1})$.

  2. (2)

For every $x\in M_{\omega}$ there is $n<\omega$ and $\bar{x}\in M_{n}$ such
that $x=j_{n,\omega}(\bar{x})$. In particular, there is $f\colon\kappa^{n}\to
V$ such that $x=j_{\omega}(f)(\kappa_{0},\dots,\kappa_{n-1})$.

  3. (3)

$\operatorname{crit}j_{n,m}=\kappa_{n}$ for all $n<m\leq\omega$.

  4. (4)

$\kappa_{\omega}=\sup\kappa_{n}$.

######  Definition 6.

An ordinal $\alpha$ is a _continuity point_ of an elementary embedding $j$, if
$j(\alpha)=\sup j\operatorname{{}^{\prime\prime}}\alpha$. An ordinal $\alpha$
is a _fixed point_ of $j$, if $j(\alpha)=\alpha$.

Since $j(\beta)\geq\beta$ for all $\beta$, every fixed point is a continuity
point.

######  Claim 7.

Let $n<\omega$. Every regular cardinal in $M_{n}$ which is not $\kappa_{n}$ is
a continuity point of $j_{n,m}$ for all $n<m\leq\omega$.

###### Proof.

Let $j=j_{n,m}$. If $\lambda<\kappa_{n}=\operatorname{crit}j$, then it is a
fixed point. Otherwise, let $\alpha<j(\lambda)$. Then, there is a function
$f\in M_{n}$ from some finite power of $\kappa_{n}$ to $\lambda$ representing
$\alpha$. But then, taking $\gamma=\sup\operatorname{range}f<\lambda$,
$\alpha<j(\gamma)$ by Łoš theorem. ∎

The embeddings $j_{n,\omega}$ as well as the models $M_{m}$ for $m\geq n$ are
definable in $M_{n}$. In particular, $M_{\omega}\subseteq M_{n}$ as a
definable subclass and $P\in M_{n}$. Therefore, $M_{\omega}[P]\subseteq
M_{n}$. We conclude that:

######  Lemma 8.

$M_{\omega}[P]\subseteq\bigcap M_{n}$.

From this lemma, let us conclude that $M_{\omega}$ and $M_{\omega}[P]$ have
the same cardinals.

######  Lemma 9.

$M_{\omega}[P]\cap V_{\kappa_{\omega}}\subseteq M_{\omega}$.

###### Proof.

Pick $X\in M_{\omega}[P]\cap V_{\kappa_{\omega}}$ and let $n$ be large enough
so that $\kappa_{n}>\operatorname{rank}X$. Then, $X\in M_{n}$ and therefore
$X=j_{n,\omega}(X)\in M_{\omega}$. ∎

######  Lemma 10.

If $\lambda\neq\kappa_{\omega}$ is a regular cardinal in $M_{\omega}$, then it
remains regular in $M_{\omega}[P]$.

###### Proof.

Let $n<\omega$ be large enough and let $\bar{\lambda}$ be the pre-image of
$\lambda$ under $j_{n,\omega}$ and let $\bar{\rho}$ be the pre-image of
$\rho=\operatorname{cf}^{M_{\omega}[P]}(\lambda)$. So $\bar{\lambda}$ is
regular in $M_{n}$ and $\bar{\rho}\leq\bar{\lambda}$. Note that
$\rho\neq\kappa_{\omega}$, which is singular in $M_{\omega}[P]$ and therefore
$\bar{\rho}\neq\kappa_{n}$.

Since $\bar{\lambda},\bar{\rho}$ are regular cardinals in $M_{n}$, and
different than $\kappa_{n}$, by Lemma 7 they are continuity points of the
embedding $j_{n,\omega}$:

$\lambda=j_{n,\omega}(\bar{\lambda})=\sup\\{j_{n,\omega}(\alpha)\mid\alpha<\bar{\lambda}\\},$
$\rho=j_{n,\omega}(\bar{\rho})=\sup\\{j_{n,\omega}(\alpha)\mid\alpha<\bar{\rho}\\}.$

Thus,

$\bar{\rho}=\operatorname{cf}^{M_{n}}(\rho)=\operatorname{cf}^{M_{n}}(\operatorname{cf}^{M_{\omega}[P]}\lambda)=\operatorname{cf}^{M_{n}}\lambda=\bar{\lambda}$

∎

So, we conclude that $M_{\omega}\subseteq M_{\omega}[P]$ have the same
cardinals and $P$ witness the singularity of $\kappa_{\omega}$ in
$M_{\omega}[P]$.

We would like to show that $M_{\omega}[P]$ is a generic extension of
$M_{\omega}$ (without pointing on an explicit forcing notion), and get some
general information about the properties of the forcing.

######  Lemma 11.

$M_{\omega}[P]$ is a generic extension of $M_{\omega}$ using a
$\kappa_{\omega}^{+}$-c.c. forcing notion.

###### Proof.

By Bukovský Theorem, [5], this is equivalent to the following statement: For
every $f\colon\alpha\to\lambda$, $f\in M_{\omega}[P]$, there is
$g\colon\alpha\to\operatorname{\mathcal{P}}(\lambda)$ in $M_{\omega}$ such
that $\forall\zeta<\alpha,f(\zeta)\in g(\zeta)$ and
$|g(\zeta)|\leq\kappa_{\omega}$.

Using Lemma 8, we know that $M_{\omega}[P]\subseteq\bigcap M_{n}$. Let
$f\in\bigcap M_{n}$ be a function from $\alpha$ to $\lambda$ for some ordinals
$\alpha,\lambda$. Let us define for each $n$ a function $f_{n}$ such that if
$\zeta=j_{n,\omega}(\bar{\zeta}),f(\zeta)=j_{n,\omega}(\bar{\xi})$ we let
$f_{n}(\bar{\zeta})=\bar{\xi}$ (so $f_{n}$ is a partial function).

Let $g_{n}\colon\kappa^{n}\to V$ represent $f_{n}$, so
$j_{n}(g_{n})(\kappa_{0},\dots,\kappa_{n-1})=f_{n}$. Finally, let:

$\bar{g}(\zeta)=\\{g_{n}(\eta)(\zeta)\mid
n<\omega,\eta\in\kappa^{n},\zeta\in\operatorname{dom}g(\eta)\\},$

and let $g=j_{\omega}(\bar{g})$. Let us verify that for every $\zeta<\alpha$,
$f(\zeta)\in g(\zeta)$.

Let $n$ be large enough to that $\zeta,f(\zeta)$ are in the range of
$j_{n,\omega}$ and in particular the pre-image of $\zeta$ under $j_{n,\omega}$
is in the domain of $f_{n}$. In particular,

$f(\zeta)=j_{n,\omega}(f_{n}(\bar{\zeta}))\in
j_{n,\omega}(\\{g_{n}(\eta)(\zeta)\mid\eta\in j_{n}(\kappa)^{n}\\})\subseteq
j_{\omega}(\bar{g})(\zeta).$

Since $|\bar{g}(\zeta)|\leq\kappa$ for all $\zeta$, the result follows. ∎

######  Remark 12.

Let us consider an arbitrary iteration of measures, so we let $V=M_{0}$ and
$M_{n+1}=\operatorname{Ult}(M_{n},\mathcal{V}_{n})$, with a direct limit
$M_{\omega}$. Any model of $\mathrm{ZFC}$ between $M_{\omega}$ and $\bigcap
M_{n}$ is a $j_{\omega}(\lambda)$-c.c. extension of $M_{\omega}$, where
$\lambda$ bounds the size of the sets of the measures. More precisely, if the
_width_ of the embedding $j_{n}$ is $<j_{n}(\lambda)$, then the forcing is
$j_{\omega}(\lambda)$-c.c.111Recall that the width of an elementary embedding
$j\colon M\to N$ between two models of $\mathrm{ZFC}$ is $\leq\mu$ iff for
every $x\in N$ there is a set $a\in M$ with $|a|\leq\mu$ such that $x\in
j(a)$. Equivalently, if for every $x\in N$ there is $f\in M$,
$|\operatorname{dom}f|\leq\mu$ and $a\in N$ such that $j(f)(a)=x$.

Note that while this approach allows us to compute the chain condition of the
forcing quite easily, it is still unclear whether one can deduce that the size
of the corresponding forcing is $\leq 2^{\kappa}$.

######  Question 13.

Can we show abstractly, without referring to the Prikry forcing, that
$M_{\omega}[P]$ is a generic extension of $M_{\omega}$ using a forcing notion
of cardinality $\leq 2^{\kappa_{\omega}}$?

So far our arguments only used the inclusion $M_{\omega}[P]\subseteq\bigcap
M_{n}$, and applied absoluteness. In general, the intersection of models of
$\mathrm{ZFC}$ does not have to satisfy even $\mathrm{ZF}$. In this case, the
model $\bigcap M_{n}$ is definable in every $M_{n}$, so we can abstractly get
more information, using the methods of Set Theoretic Geology, [12, Lemma 21].

Fortunately, we do not need to analyse this model as, surprisingly, the
Bukovský-Dehornoy Theorem shows that the intersection model is the minimal
possible model:

######  Theorem 14 (Bukovský-Dehornoy, [6, 8]).

$\bigcap M_{n}=M_{\omega}[P]$.

We include the proof as its main ideas are going to repeat throughout the
other, more complicated, cases.

###### Proof.

The proof consists of two steps. In the first step we show that the model
$M_{\omega}[P]$ is closed under countable sequences of ordinals. In the second
step we use the elementary embeddings in order to approximate an arbitrary set
of ordinals from $\bigcap M_{n}$ using a countable sequence of sets in
$M_{\omega}$. From this approximation, $M_{\omega}[P]$ can compute $X$.

######  Lemma 15.

$M_{\omega}[P]$ is closed under countable sequences of ordinals.

###### Proof.

Let $\langle\alpha_{n}\mid n<\omega\rangle$ be a sequence of ordinals. Fix, in
$V$, a countable sequence of functions
$f_{n}\colon\kappa^{m_{n}}\to\mathrm{Ord}$, such that
$j_{\omega}(f_{n})(\kappa_{0},\dots,\kappa_{m_{n}-1})=\alpha_{n}$. Let
$\vec{f}=\langle f_{n}\mid n<\omega\rangle$.

Then $j_{\omega}(\vec{f})\in M_{\omega}$. Therefore,

$\langle\alpha_{n}\mid n<\omega\rangle=\langle j_{\omega}(f_{n})(P\restriction
m_{n})\mid n<\omega\rangle\in M_{\omega}.$

∎

######  Lemma 16.

Let $X\in\bigcap M_{n}$ be a set of ordinals. Then $X\in M_{\omega}[P]$.

###### Proof.

Let

$Y_{n}=\\{\alpha\mid j_{n,\omega}(\alpha)\in X\\},\quad
Z_{n}=j_{n,\omega}(Y_{n}).$

By the $\sigma$-closure, $\langle Z_{n}\mid n<\omega\rangle\in M_{\omega}$.
Now,

$\beta\in X\iff\forall^{*}n\beta\in Z_{n},$

where $\forall^{*}$ means for all large $n$. Indeed, if
$\beta\in\operatorname{range}j_{n,\omega}$ then $\beta\in X\iff\beta\in
Z_{n}$. ∎

∎

From the proof one can extract a more concrete definition for the model
$M_{\omega}[P]$:

######  Proposition 17.

$M_{\omega}[P]$ is the least transitive class $C$ that contains
$M_{\omega}\cup\\{P\\}$ and closed under the operations:

  1. (1)

If $\vec{g}=\langle g_{n}\mid n<\omega\rangle\in C$ such that
$\operatorname{dom}g_{n}=\kappa_{\omega}^{n}$ then $\langle
g_{n}(P\restriction n)\mid n<\omega\rangle$ is in $C$.

  2. (2)

If $\langle X_{n}\mid n<\omega\rangle\in C$ then $X=\\{z\mid\forall^{*}n,z\in
X_{n}\\}$.

  3. (3)

If $X=\langle A,E\rangle\in C$ and $E$ is well founded and extensional, then
the Mostowski collapse of $X$ is in $C$.

##  2\. Gitik-Sharon forcing

Let us consider now the diagonal supercompact forcing, introduced by Gitik and
Sharon in [16]. This forcing is using a stronger type of large cardinal axioms
— supercompactness.

######  Definition 18.

Let $\kappa\leq\lambda$ be cardinals. Let $P_{\kappa}\lambda$ be the set of
all subsets of $\lambda$ of cardinality $<\kappa$.

A $\kappa$-complete ultrafilter $\mathcal{U}$ on $P_{\kappa}\lambda$ is _fine_
if for every $\alpha\in\lambda$, the set $\\{x\in
P_{\kappa}\lambda\mid\alpha\in x\\}\in\mathcal{U}$. $\mathcal{U}$ is _normal_
if it is fine and for every choice function $f\colon A\to\lambda$, where
$A\in\mathcal{U}$, there is an $\mathcal{U}$-large set $B\subseteq A$, such
that $f\restriction B$ is constant.

######  Lemma 19 (Reinhardt, Magidor).

There is a normal measure on $P_{\kappa}\lambda$ if and only if there is an
elementary embedding $j\colon V\to M$, with critical point $\kappa$ such that
$M$ is closed under $\lambda$-sequences and $j(\kappa)>\lambda$.

Let us recall that given an elementary embedding $j\colon V\to M$, such that
$\operatorname{crit}j=\kappa$, $j(\kappa)>\lambda$ and ${}^{\lambda}M\subseteq
M$, the ultrafilter $\mathcal{U}=\\{X\subseteq P_{\kappa}\lambda\mid
j\operatorname{{}^{\prime\prime}}\lambda\in j(X)\\}$ is a normal measure on
$P_{\kappa}\lambda$. In other words, the _seed_ of $\mathcal{U}$ is
$j\operatorname{{}^{\prime\prime}}\lambda$.

######  Remark 20.

Throughout this section we will always assume that $\kappa$ carries a sequence
of normal measures $\mathcal{V}_{n}$ on $P_{\kappa}\kappa^{+n}$. Except for a
couple of changes in the notations, there is no harm in reducing the
hypothesis to $\mathcal{V}_{n}$ being $\kappa$-complete fine measure on
$P_{\kappa}\kappa^{+n}$.

We would like to define an iteration using this sequence of supercompact
measures.

$\begin{matrix}V&=&M_{0}&j_{0,0}=id\\\
M_{n+1}&=&\operatorname{Ult}(M_{n},j_{n}(\mathcal{V}_{n})),&j_{n,n+1}\colon
M_{n}\to M_{n+1}\\\ &\vdots&\\\ M_{\omega}&=&\lim\langle M_{n},j_{n,m}\mid
n\leq m<\omega\rangle&j_{n,\omega}\colon M_{n}\to M_{\omega}\end{matrix}$

Either Gaifman’s arguments, or the modern version using the completeness of
the measures show that:

######  Lemma 21.

$M_{\omega}$ is well founded.

As usual, we let $j_{n}=j_{0,n}$ for $n\leq\omega$.

Unlike the case of the Prikry forcing, this time the seeds of the embeddings
are moved by the later steps of the iterations. Thus, we need to define the
sequence $P$ in a more precise way.

Let

$p_{n}=j_{n+1,\omega}\left(j_{n,n+1}\operatorname{{}^{\prime\prime}}j_{n}(\kappa^{+n})\right)=j_{n,\omega}\operatorname{{}^{\prime\prime}}j_{n}(\kappa^{+n})$

and let $P=\langle p_{n}\mid n<\omega\rangle$222Here, if we assume that
$\mathcal{V}_{n}$ are merely $\kappa$-complete fine ultrafilters, instead of
normal ones, we will need to change the definition of $p_{n}$ to be
$j_{n+1,\omega}\left([id]_{j_{n}(\mathcal{V}_{n})}\right)$..

Let us collect a couple of useful properties of the iteration.

######  Lemma 22.

For every $n<m\leq\omega$, $\operatorname{crit}j_{n,m}=j_{n}(\kappa)$.

######  Lemma 23.

For every $x\in M_{n}$ there is a function
$f\colon\prod_{m<n}P_{\kappa}\kappa^{+m}\to V$ such that

$x=j_{n}(f)(\bar{p}_{0},\dots,\bar{p}_{n-1}),$

where $\bar{p}_{m}=j_{n,m}\operatorname{{}^{\prime\prime}}j_{m}(\kappa^{+m})$.
In particular, for every $x\in M_{\omega}$ there is a natural number $n$ and
$f\colon\prod_{m<n}P_{\kappa}\kappa^{+m}\to V$ such that
$x=j_{\omega}(f)(p_{0},\dots,p_{n-1})$.

######  Lemma 24.

Let $\lambda$ be a regular cardinal in $M_{n}$,
$\lambda\notin[j_{n}(\kappa),j_{n}(\kappa^{+\omega}))$. Then, $\lambda$ is a
continuity point of $j_{n,\omega}$.

###### Proof.

This is a corollary of Lemma 22 and Lemma 23. Let us prove the lemma for
$n=0$. The general case is similar, with slightly more complicated notations.
If $\lambda<\kappa$ then $\lambda$ is a fixed point of $j_{\omega}$. If
$\lambda>\kappa^{\omega}$, then let $\alpha<j_{\omega}(\lambda)$. So, $\alpha$
is represented by some function $f$ as above. But,
$|\operatorname{dom}f|\leq\kappa^{+n}<\lambda$, so
$\sup\operatorname{range}f=\gamma<\lambda$, and therefore
$\alpha<j_{\omega}(\gamma)<j_{\omega}(\lambda)$. ∎

So, the proof of Lemma 10 goes throughout unchanged and we conclude that every
$M_{\omega}$-regular cardinal $\lambda$ which is not in the interval
$[j_{\omega}(\kappa),j_{\omega}(\kappa^{+\omega}))$ remains regular in
$M_{\omega}[P]$.

######  Lemma 25.

In $M_{\omega}[P]$, $\forall
n<\omega,\,\operatorname{cf}j_{\omega}(\kappa^{+n})=\omega$.

###### Proof.

Look at $\alpha_{m}=\sup
j_{m,\omega}\operatorname{{}^{\prime\prime}}j_{m}(\kappa^{+n})$. For every
$m\geq n$, this is an ordinal below $j_{\omega}(\kappa^{n})$, as
$j_{m}(\kappa^{+n})$ is a discontinuity point of $j_{m,m+1}$. But, for every
ordinal $\gamma<j_{\omega}(\kappa^{+n})$ there is some large enough $m$ such
that $\gamma\in\operatorname{range}j_{m,\omega}$ and thus $\gamma<\alpha_{m}$.
Therefore, $\sup_{m\geq n}\alpha_{m}=j_{\omega}(\kappa^{+n})$.

The sequence $\langle\alpha_{m}\mid m<\omega\rangle$ can be computed from $P$:
$\alpha_{m}=\sup\left(p_{m}\cap j_{\omega}(\kappa^{+n})\right)$. ∎

We conclude that the successor of $j_{\omega}(\kappa)$ in $M_{\omega}[P]$ is
$j_{\omega}(\kappa^{+\omega+1})$.

By Remark 12, $M_{\omega}[P]$ is $j_{\omega}(\kappa^{+\omega+1})$-c.c. generic
extension of $M_{\omega}$.

We are now ready for the proof of the Bukovský-Dehornoy Theorem for the Gitik-
Sharon forcing. The proof is essentially the same as the one for the Prikry
forcing, so we will sketch it.

######  Theorem 26.

$M_{\omega}[P]=\bigcap M_{n}$.

###### Proof.

First, we need to show that $M_{\omega}[P]$ is closed under countable
sequences of ordinals. Indeed, if $\langle\alpha_{n}\mid n<\omega\rangle$ is a
countable sequence of ordinals, then for each $n$, pick a function
$g_{n}\colon\prod_{m<m_{n}}P_{\kappa}\kappa^{+m}$ such that

$\alpha_{n}=j_{\omega}(g_{n})(p_{0},\dots,p_{m_{n}-1}),$

using Lemma 23. Then, since $j_{\omega}(\langle g_{n}\mid
n<\omega\rangle)=\langle j_{\omega}(g_{n})\mid n<\omega\rangle\in M_{\omega}$,
we get that:

$\langle\alpha_{n}\mid n<\omega\rangle=\langle j_{\omega}(g_{n})(P\restriction
m_{n})\mid n<\omega\rangle\in M_{\omega}[P].$

Next, given a set of ordinals $X\in\bigcap M_{n}$, let us define the sets:

$Y_{n}=\\{\alpha\mid j_{n,\omega}(\alpha)\in
X\\}\quad,Z_{n}=j_{n,\omega}(Y_{n})$

and verify that $\alpha\in X$ iff for all large $n$, $\alpha\in Z_{n}$. ∎

One of the interesting features of the Gitik-Sharon forcing relates to the
behavior of $\kappa^{+\omega+1}$. In particular, if $\kappa^{+\omega}$ is
strong limit cardinal then it is a fixed point of $j_{\omega}$. Thus, under
suitable cardinal arithmetic assumption, the Gitik-Sharon forcing cannot
introduce certain sufficiently absolute objects.

Let us start with a simple observation.

######  Claim 27.

Let us assume that $\kappa^{+\omega}$ is a strong limit cardinal.

There is no special Aronszajn tree on $\kappa^{+\omega+1}$ in $V$ if and only
if there is no special Aronszajn tree on the successor of $j_{\omega}(\kappa)$
in $M_{\omega}[P]$.

###### Proof.

Let us assume that there is such a tree $T$ in $M_{\omega}[P]$. So, this is a
tree of height
$\left(j_{\omega}(\kappa)^{+}\right)^{M_{\omega}[P]}=\kappa^{+\omega+1}$ such
that there is a function $f\colon T\to j_{\omega}(\kappa)$ which is injective
on chains. Since $M_{\omega}[P]\subseteq V$, $T,f\in V$ and since
$|j_{\omega}(\kappa)|^{V}=\kappa^{+\omega}$, $T$ is a special Aronszajn tree
in $V$.

The other direction is similar: if $T$ is a special Aronszajn tree in $V$,
then $j_{\omega}(T)$ is a special Aronszajn tree in $M_{\omega}$. But being
special is upwards absolute, so it is special in $M_{\omega}[P]$ as well. ∎

Theorem 26 implies that $M_{\omega}[P]$ is closed under $\kappa$-sequences
with respect to $V$, since it is an intersection of models which are closed
under $\kappa$-sequences of ordinals. The following claim (and its variant)
was used in [10] in order to derive instances of Chang’s Conjecture in the
Gitik-Sharon extension.

######  Claim 28.

Let $\kappa^{+\omega}$ be a strong limit. Let us assume that Chang’s
Conjecture
$(\kappa^{+\omega+1},\kappa^{+\omega})\twoheadrightarrow(\rho^{+},\rho)$ holds
for some $\rho<\kappa$.

Then
$M_{\omega}[P]\models(j_{\omega}(\kappa)^{+},j_{\omega}(\kappa))\twoheadrightarrow(\rho^{+},\rho)$.

###### Proof.

Let $\mathcal{A}$ be an algebra in $M_{\omega}[P]$ on
$j_{\omega}(\kappa)^{+}=\kappa^{+\omega+1}$. Then, by Chang’s Conjecture in
$V$, there is $\mathcal{B}\prec\mathcal{A}$ of order type $\rho^{+}$. But
$\mathcal{B}\in M_{\omega}[P]$, by the closure of the model. ∎

Let us remark that the hypothesis of the claim follows from the assumption
that $\kappa$ is $\kappa^{+\omega+1}$-supercompact.

Claim 28 fails if $\kappa^{+\omega}$ is not a strong limit. In this case, the
Gitik-Sharon forcing adds a good scale (see Definition 32), which implies that
every instance of Chang’s Conjecture as in the claim does not hold.

###  2.1. Scales

This subsection deals with the applications of the Bukovský-Dehornoy Theorem
to the behavior of scales at a Prikry type extension. This type of application
was used in [4], and the basic ideas are taken from there. Here we apply it
for the Gitik-Sharon forcing, in order to derive the preservation of a bad
scale in the generic extension by the Gitik-Sharon forcing.

The applications in this subsection used only the fact that the generic
extension $M_{\omega}[P]$ is closed under countable sequences.

Scales are one of the basic objects in the Shelah’s PCF theory, see [28, 1].
Let us present a special case.

######  Definition 29 (Shelah).

Let $\lambda$ be a singular cardinal of countable cofinality and let
$\langle\lambda_{n}\mid n<\omega\rangle$ be a sequence of regular cardinals
converging to $\lambda$. A _scale_ on $\prod\lambda_{n}$ is a sequence of
functions $\langle f_{\alpha}\mid\alpha<\mu\rangle$,
$f_{\alpha}\in\prod\lambda_{n}$ such that:

  * •

For all $\alpha<\beta$, $\\{n\mid f_{\alpha}(n)\leq f_{\beta}(n)\\}$ is co-
finite.

  * •

For all $g\in\prod\lambda_{n}$, there is $\alpha<\mu$ such that $\\{n\mid
g(n)\leq f_{\alpha}(n)\\}$ is co-finite.

We will denote the assertion ”$\\{n\mid f(n)\leq g(n)\\}$ is co-finite” by
$f\leq^{*}g$ and we call the set $n$ of values such that $f(n)>g(n)$, the set
of violations of the inequality.

It is clear that every product carries a scale of some length.

######  Lemma 30 (Shelah).

The lengths of any two scales on the same product have the same cofinality.

In particular, it makes sense to assume that the length of the scale is always
a regular cardinal.

######  Theorem 31 (Shelah).

For every singular cardinal $\lambda$ of countable cofinality, there is a
sequence $\langle\lambda_{n}\mid n<\omega\rangle$ of regular cardinals and a
scale $\langle f_{\alpha}\mid\alpha<\lambda^{+}\rangle$ on $\prod\lambda_{n}$.

The notion of scales is much wider than this limited definition. For our
purposes, we would like to focus on scales of minimal length, and consider
ones with a better behaviour.

######  Definition 32.

Let $\mathcal{S}=\langle f_{\alpha}\mid\alpha<\lambda^{+}\rangle$ be scale. An
ordinal $\beta<\lambda^{+}$ of uncountable cofinality is _good_ if there is
$n<\omega$ and $A\subseteq\beta$ cofinal such that for every
$\alpha_{0}<\alpha_{1}$ in $A$, $\\{m\mid
f_{\alpha_{0}}(m)>f_{\alpha_{1}}(m)\\}\subseteq n$.

An ordinal $\beta$ is _bad_ , if it is not good.

A scale $\mathcal{S}$ is a _good_ if there are club many good points. A scale
is _bad_ if it has stationarily many bad ordinals.

The notion of bad (and good) points in a scale can be traced back to [27]. The
connections between the notions of good scales to other anti-compactness
principles such as square principles were summarized and investigated, for
example, in the seminal paper [7].

######  Remark 33 (Shelah).

If $\mathcal{S},\mathcal{S}^{\prime}$ are both scales on the same product of
length $\lambda^{+}$, then their sets of good ordinals agree up to a non-
stationary error.

###### Proof.

Let $\mathcal{S}=\langle
f_{\alpha}\mid\alpha<\lambda^{+}\rangle,\mathcal{S}^{\prime}=\langle
f^{\prime}_{\alpha}\mid\alpha<\lambda^{+}\rangle$. Let $C$ be the club of all
ordinals $\delta$ such that for all $\alpha<\delta$, there is
$\alpha^{\prime}<\delta$ such that
$f_{\alpha}\leq^{*}f^{\prime}_{\alpha^{\prime}}$ and vise verse. Then, an
ordinal $\delta\in C$ is good in $\mathcal{S}$ if and only if it is good in
$\mathcal{S}^{\prime}$: Take $A\subseteq\delta$ witnessing $\delta$ being good
in $\mathcal{S}$. Then, by induction, pick a sequence of ordinals
$\alpha_{i}<\alpha_{i}^{\prime}<\alpha_{i+1}<\cdots$ such that $\alpha_{i}\in
A$ for all $i$ and
$f_{\alpha_{i}}\leq^{*}f^{\prime}_{\alpha^{\prime}_{i}}\leq^{*}f_{\alpha_{i+1}}$.

For every $i$, there is a bound for the error in the inequality
$f_{\alpha_{i}}\leq^{*}f_{\alpha_{i}^{\prime}}$ and
$f^{\prime}_{\alpha_{i}^{\prime}}\leq^{*}f_{\alpha_{i+1}}$, $n_{i}$. As
$\operatorname{cf}\delta>\omega$, there is some $n_{*}$ such that
$A^{\prime}=\\{\alpha_{i}^{\prime}\mid n_{i}=n_{*}\\}$ is unbounded. So,
$A^{\prime}$ witnesses $\delta$ being good from $\mathcal{S}^{\prime}$. ∎

In the paper [16], Gitik and Sharon solved Woodin’s question whether it is
consistent that in a successor of a singular, $\mathrm{SCH}$ fails but there
is no weak square. They achieved that by showing that in the model obtained by
forcing with the Gitik-Sharon forcing, starting with a supercompact cardinal
$\kappa$ such that $2^{\kappa}>\kappa^{+\omega+1}$, there are bad scales on
the successor of $\kappa$. In our framework, this theorem translates to the
following statement about $M_{\omega}[P]$.

######  Theorem 34 (Gitik-Sharon).

Assume that in $V$ there is a scale on $\prod\kappa^{+n}$ in which the set of
bad points $S$ contains stationarily many ordinals of cofinality $<\kappa$.
Then, there is a bad scale in the successor of $j_{\omega}(\kappa)$ in
$M_{\omega}[P]$.

###### Proof.

Following the ideas from Claims 27 and 28, we would like to somehow use the
scale from $V$ in order to verify that a corresponding scale from
$M_{\omega}[P]$ cannot be good.

Let us note that if $\kappa^{+\omega}$ is a strong limit, this is rather
trivial, but we are mostly interested in the case that $2^{\kappa}$ is large.
In this case, $\langle\kappa^{+n}\mid n<\omega\rangle$ is not necessarily
cofinal at $j_{\omega}(\kappa)$, as even $j_{1}(\kappa)$ might be larger than
all of those cardinals. Thus, we study the product of
$\mu_{n}=\left(j_{n}(\kappa^{+n+1})\right)^{M_{\omega}}$.

######  Claim 35.

For every $n<\omega$, $\kappa^{+n+1}$ is a continuity point of $j_{n}$.

Note that $\mu_{n}$ is a regular cardinal in $M_{n}$, and by absoluteness, it
is regular in $M_{\omega}[P]$ as well. Moreover, since
$j_{n+1}(\kappa)>j_{n}(\kappa^{+n+1})$, $\sup\mu_{n}=j_{\omega}(\kappa)$.

Pick a scale on $\prod\mu_{n}$, $\mathcal{T}=\langle
g_{\alpha}\mid\alpha<\mu_{*}\rangle$. Even though $\mu_{n}$ might not be
regular in $V$, the two defining properties of a scale (weakly increasing and
cofinal) holds in $V$ for $\mathcal{T}$ since $M_{\omega}[P]$ is closed under
$\omega$-sequences or ordinals.

We would like to collapse $\mathcal{T}$ to a scale in $V$. Recall that for all
$n$, $\sup j_{n}\operatorname{{}^{\prime\prime}}\kappa^{+n+1}=\mu_{n}$. Let us
define for every $\alpha$,

$h_{\alpha}(n)=\min\\{\beta<\kappa^{+n+1}\mid g_{\alpha}(n)<j_{n}(\beta)\\},$

and let $\bar{\mathcal{T}}$ be $\langle h_{\alpha}\mid\alpha<\mu_{*}\rangle$.

######  Claim 36.

$\bar{\mathcal{T}}$ is a scale in $V$ on $\prod\kappa^{+n+1}$.

###### Proof.

Let $\alpha<\beta$. Since $g_{\alpha}(n)\leq g_{\beta}(n)$ for almost all $n$,
we conclude that $h_{\alpha}(n)\leq h_{\beta}(\alpha)$ for almost all $n$.

Let $h$ be arbitrary. Then, let us look at the $\tilde{h}(n)=j_{n}(h(n))$.
Since $M_{\omega}[P]$ is closed under countable sequences, $\tilde{h}\in
M_{\omega}[P]$. Therefore, there is $\alpha$ such that $\tilde{h}(n)\leq
g(\alpha)$ for almost all $n$. We conclude that $h(n)\leq h_{\alpha}(n)$ for
almost all $n$.∎

In particular, $\operatorname{cf}^{V}\mu_{*}=\kappa^{+\omega+1}$. Fix in $V$
an arbitrary continuous sequence of ordinals $\vec{\beta}=\langle\beta_{i}\mid
i<\kappa^{+\omega+1}\rangle$ cofinal at $\mu_{*}$, and let
$\bar{\bar{\mathcal{T}}}=\langle h_{\beta_{i}}\mid
i<\kappa^{+\omega+1}\rangle$. Fix a scale $\mathcal{S}$ as in the hypothesis
of the theorem. Let $C$ be the club from the proof of Remark 33 for the scales
$\mathcal{S},\bar{\bar{\mathcal{T}}}$.

######  Claim 37.

For every $\alpha\in C$ of cofinality $<\kappa$, if $\alpha$ is bad for
$\mathcal{S}$ in $V$ then $\beta_{\alpha}$ it is bad for $\mathcal{T}$ in
$M_{\omega}[P]$.

###### Proof.

Assume that this is not the case. Since $\operatorname{cf}\alpha^{V}<\kappa$
and the sequence $\vec{\beta}$ is continuous,
$\operatorname{cf}^{V}\beta_{\alpha}<\kappa$. By the closure of
$M_{\omega}[P]$, $\operatorname{cf}^{M_{\omega}[P]}\beta_{\alpha}<\kappa$ and
moreover, there is $B\subseteq\beta_{\alpha}$ in $M_{\omega}[P]$, cofinal and
contained in $\\{\beta_{i}\mid i<\alpha\\}$. Pick $A\subseteq B$ witnessing
$\beta_{\alpha}$ being good for $\mathcal{T}$.

Then, $A$ witnesses that $\beta_{\alpha}$ is good for $\bar{\mathcal{T}}$,
which means that $\bar{A}=\\{i\mid\beta_{i}\in A\\}$ witness that $\alpha$ is
good for $\bar{\bar{\mathcal{T}}}$ and thus by Remark 33, $\alpha$ is good
from $\mathcal{S}$. ∎

Finally, if $D$ is a club in $M_{\omega}[P]$ then $D\in V$ and it is a closed
and unbounded subset of the ordinal $\mu_{*}$. We conclude that
$\\{\alpha\mid\beta_{\alpha}\in D\\}\in V$ is a club in $\kappa^{+\omega+1}$.
Together with the previous claim, we see that the set of bad points in
$\mathcal{T}$ has to be stationary in $M_{\omega}[P]$. ∎

The standard proofs of properties of scales in Prikry type extensions need
some bounding lemmas. Looking closely into the arguments, one can identify the
parallel parts: the bounding lemmas correspond to the construction of the
scale $\bar{\bar{\mathcal{T}}}$ from $\mathcal{T}$. Nevertheless, using the
Bukovský-Dehornoy method, we do not need to talk about names and use the
strong Prikry Property is order to partially realize them.

Finally, we can prove:

######  Theorem 38 (Gitik-Sharon).

Let $\kappa$ be a supercompact cardinal such that
$2^{\kappa}\geq\kappa^{+\omega+2}$. Then, there is a generic extension in
which $\kappa$ is a strong limit singular, there is a bad scale on
$\kappa^{+}$ and $2^{\kappa}>\kappa^{+}$.

###### Proof.

In $M_{\omega}$, there is a forcing notion adding $P$, by Bukovský Theorem,
and since $M_{\omega}[P]$ satisfies the conclusion of the theorem (the failure
of $\mathrm{SCH}$ and the existence of a bad scale), by the forcing theorem,
there is a condition forcing that.

By elementarity, the same holds in $V$: there is a forcing notion and a
condition in it forcing the failure of $\mathrm{SCH}$ together with the
existence of a bad scale on $\kappa$. ∎

By combining interleaved collapses in the Gitik-Sharon forcing, one can obtain
a model in which $\mathrm{SCH}$ fails at $\aleph_{\omega^{2}}$ and there is a
bad scale on $\aleph_{\omega^{2}}$. In the next section we will address the
issue of adding collapses to a forcing notion in which the Bukovský-Dehornoy
Theorem holds, thus allowing us to obtain the full result.

##  3\. Combining iterated ultrapowers with forcings

In this section, we will give a couple of examples for extensions of an
iterated ultrapower, $M_{\omega}$, using an object which can be obtained only
in a generic extension of $V$. While an additional level of complexity is
added to the whole process, still a few key components are preserved. Our goal
model can be presented as the intersection of a (definable) decreasing
sequence of models in a generic extension, so many of the arguments from the
previous sections will be applicable here as well.

The most notable change is that we are losing the elementary embeddings
between the models in the chain. This makes the proof of the parallel
intersection theorems more involved.

We will deal with two main cases: the Extender Based Prikry Forcing and adding
interleaved collapses.

###  3.1. Extender Based Prikry forcing

In this section, we follow closely results and ideas from Merimovich, [21, 23]
in order to derive a Bukovský-Dehornoy theorem for extender based Prikry
forcing.

There are several definitions for _extenders_ in the literature, see for
example [14, 18]. For our purposes, a $(\kappa,\lambda)$-extender $E$ is a
combinatorial (set) object coding an elementary embedding $j\colon V\to M$
with $\operatorname{crit}j=\kappa$, $j(\kappa)>\lambda$, $M$ is closed under
sequences of length $\kappa$ and for every $x\in M$ there are
$f\colon\kappa\to V$ in $V$ and $\gamma<\lambda$ such that $j(f)(\gamma)=x$.
In particular, the width of the embedding $j$ is $\leq\kappa$.

######  Lemma 39 ($\kappa$-directness).

Let $E$ be a $(\kappa,\lambda)$-extender and $j\colon V\to M$ be the derived
elementary embedding.

For every $A\subseteq\lambda$, $|A|\leq\kappa$ there is $\gamma<\lambda$ such
that for every $\delta\in A$ there is $f_{\delta}\colon\kappa\to\kappa$ such
that $j(f_{\delta})(\gamma)=\delta$.

###### Proof.

Since $M$ is closed under $\kappa$-sequences, $A\in M$ and moreover, some
enumeration $\langle\delta_{i}\mid i<\kappa\rangle\in M$. Thus, there is
$\gamma<\lambda$ and $g\colon\kappa\to V$ such that
$j(g)(\gamma)=\langle\delta_{i}\mid i<\kappa\rangle$.

So, define $f_{\delta_{i}}(\zeta)=g(\zeta)(i)$ and the result follows from
elementarity. ∎

As a definable elementary embedding, the elementary embedding derived from an
extender can be iterated and the direct limit of such an iteration is well
founded.

Fix a $(\kappa,\lambda)$-extender $E$, and let $\langle M_{n},j_{n,m}\mid
n\leq m\leq\omega\rangle$ be the corresponding iteration of $E$.

######  Lemma 40.

For every $x\in M_{\omega}$ there are $n<\omega$, $f\colon\kappa^{n}\to V$ and
$\gamma_{0},\dots,\gamma_{n-1}$, such that
$\gamma_{k}<j_{k}(\lambda)<j_{k+1}(\kappa)$ and
$x=j_{\omega}(f)(\gamma_{0},\dots,\gamma_{n-1})$.

######  Lemma 41.

The width of the embedding $j_{n,m}$ for $n<m\leq\omega$ is $\leq
j_{n}(\kappa)$.

###### Proof.

First, for $m=n+1$, this follows from our initial hypothesis on the extender
$E$, using elementarity.

Let us assume that the width of the embedding $j_{n,m}$ is $\leq
j_{n}(\kappa)$. Let $x\in M_{m+1}$. Since the width of $j_{m,m+1}$ is
$j_{m}(\kappa)$, there is $y\in M_{m}$ such that $|y|\leq j_{m}(\kappa)$ and
$x\in j_{m,m+1}(y)$.

Since the width of $j_{n,m}$ is $j_{n}(\kappa)$, we conclude that there is
$z\in M_{n}$ such that $|z|\leq j_{n}(\kappa)$ and $y\in j_{n,m}(z)$. By
elementarity, taking $z^{\prime}=\\{y^{\prime}\in z\mid|y^{\prime}|\leq
j_{n}(\kappa)\\}$ we conclude that $y\in j_{n,m}(z^{\prime})$. Finally, let
$w=\bigcup z^{\prime}$ — this set is a union of $\leq j_{n}(\kappa)$ many sets
of cardinality $\leq j_{n}(\kappa)$, so $|w|\leq j_{n}(\kappa)$.

So, $x\in j_{n,m+1}(w)$ since $x\in j_{m,m+1}(y)$ and $j_{m,m+1}(y)\in
j_{m,m+1}(j_{n,m}(z))$.

So, by induction, the claim holds for all $n<m<\omega$.

Now, let us deal with the width of $j_{n,\omega}$. If $x\in M_{\omega}$, then
there is $m>n$ (without loss of generality) and $\bar{x}\in M_{m}$ such that
$x=j_{m,\omega}(\bar{x})$. Since $j_{n,m}$ has width $j_{n}(\kappa)$, there is
$y\in M_{n}$ such that $\bar{x}\in j_{n,m}(y)$.

So, $x\in j_{n,\infty}(y)$, as wanted. ∎

Merimovich proved that one can obtain a generic for the extender based Prikry
forcing (defined in a proper way), but forcing with the direct extension order
taking an iterated ultrapower and adding the generator. The following theorem
works the details for the Bukovský-Dehornoy intersection theorem for this
forcing, again without explicitly defining the forcing notion.

######  Theorem 42.

Let $H\subseteq\mathbb{P}^{*}$, be a $V$-generic filter for the forcing
$\mathbb{P}^{*}=\\{f\colon\lambda\to\kappa^{<\omega}\mid|f|\leq\kappa\\}$,
ordered by reverse inclusion.

Let us define inductively

$H_{n+1}=\text{upwards closure of }\\{j_{n,n+1}(p)^{\smallfrown}\\{\langle
j_{n,n+1}(\alpha),\alpha\rangle\mid\alpha\in\operatorname{dom}p\\}\mid p\in
H_{n}\\}$

Let $G\colon j_{\omega}(\lambda)\times\omega\to j_{\omega}(\kappa)$ be defined
by:

$\begin{matrix}G(\alpha,n)=\gamma\iff&&\\\ \exists m<\omega,&p\in
H_{m},&\alpha\in\operatorname{dom}j_{m,\omega}(p),&n\in\operatorname{dom}(j_{m,\omega}(p)(\alpha))\\\
\gamma=j_{m,\omega}(p)(\alpha)(n)\par\end{matrix}$

Then $M_{\omega}[G]=\bigcap M_{n}[H_{n}]$. Moreover, in $M_{\omega}[G]$
$\operatorname{cf}j_{\omega}(\kappa)=\omega$ and
$2^{j_{\omega}(\kappa)}=j_{\omega}(\lambda)$.

###### Proof.

First, the verification that
$\operatorname{cf}^{M_{\omega}[G]}j_{\omega}(\kappa)=\omega$ and $G$ forms a
scale of length $j_{\omega}(\lambda)$ on $j_{\omega}(\kappa)$ is straight-
forward. Let us focus in the proof of the intersection theorem.

As in Theorem 14, we need to show first that the model is closed under
countable sequences.

######  Lemma 43.

$M_{\omega}[G]$ is closed under $\omega$-sequences.

###### Proof.

Since $M_{\omega}[G]$ is a model of $\mathrm{ZFC}$, it is enough to prove the
claim for sequences of ordinals.

The crux of the argument is [21, Corollary 2.6].

Fix $\alpha$ an ordinal in $M_{\omega}[G]$. So, there is a function
$f\colon\kappa^{n}\to\mathrm{Ord}$ and $\gamma_{0},\dots,\gamma_{n-1}$ as in
Lemma 40, such that $j_{\omega}(f)(\gamma_{0},\dots,\gamma_{n-1})=\alpha$.

Work in $V$. Fix an elementary substructure $N\prec H(\chi)$ for some large
$\chi$, such that $\kappa\subseteq N$, $|N|=\kappa$, $E,\alpha\in N$. Let us
apply the $\kappa$-directness of $E$, in the sense of Lemma 39, and obtain an
ordinal $\rho$ such that for every $\beta\in N$ there is $h$ such that
$j(h)(\rho)=\beta$. In particular, this is true for $\gamma_{0}$, so there is
$h_{0}$ such that $j(h_{0})(\rho)=\gamma_{0}$. As $N$ knows that the width of
$j$ is $\kappa$, there is $a\subseteq\lambda$ in $N$ of cardinality $\kappa$
such that $\gamma_{1}\in j(a)$. In particular, since $a\subseteq N$, there is
$h_{1}$ such that $j_{2}(h_{1})(j_{1}(\rho))=\gamma_{1}$. Let us claim that we
can ”trace back” $h_{1}$ to a function in $V$. Indeed, in $N$ there is
$\bar{\rho}<\lambda$ which is Rudin-Keisler above $a$. In particular, there is
$\bar{h}_{2}\in M_{1}^{N}$ such that
$j_{1,2}(\bar{h}_{2})(\bar{\rho})=\gamma_{2}$. But,
$\bar{h}_{2}=j_{1}(\bar{\bar{h}}_{2})(\bar{\gamma}_{2})$ for some
$\bar{\gamma}_{2}\in N$ and thus we conclude that all those computations can
be made using only $\rho$.

Continue this way, we conclude that there is a function
$h\colon\kappa^{n}\to\mathrm{Ord}$ such that
$j_{\omega}(h)(\rho,j_{1}(\rho),j_{2}(\rho),\dots,j_{n-1}(\rho))=\alpha$.

Now, let $\langle\alpha_{n}\mid n<\omega\rangle$ be a sequence of ordinals.
Applying the above arguments with $N$ containing the sequence
$\langle\alpha_{n}\mid n<\omega\rangle$ instead of just a single ordinal
$\alpha$, we obtain $\rho$ and a sequence of functions $\langle h_{n}\mid
n<\omega\rangle$ such that for all $n$,
$j_{n}(h_{n})(\rho,j_{1}(\rho),\dots,j_{m_{n}-1}(\rho))=\alpha_{n}$.

Since the sequence of images of $\rho$ is a final segment of
$G(j_{\omega}(\rho))$, we conclude that $\langle\alpha_{n}\mid
n<\omega\rangle\in M_{\omega}[G]$. ∎

Let $X\in\bigcap M_{n}[H_{n}]$ be a set of ordinals. For each $n$, let us pick
a $j_{n}(\mathbb{P}^{*})$-name $\dot{\tau}_{n}$ for

$Y_{n}=\\{\alpha\mid j_{n,\omega}(\alpha)\in X\\}.$

So, $\dot{\tau}_{n}^{H_{n}}=Y_{n}$.

Let $\sigma_{n}=j_{n,\omega}(\tau_{n})\in M_{\omega}$. Since $M_{\omega}[G]$
is closed under $\omega$-sequences, the sequence of names
$\langle\sigma_{n}\mid n<\omega\rangle$ is a member of $M_{\omega}[G]$.

The problem is the we do not have access to the actual generics $H_{n}$ from
which the names were realized. In order to overcome this, we will need to
isolate a relevant version of the Prikry forcing that hides inside our forcing
and show that its generic is unique (up to shifts). Here we must diverge from
the thesis of the paper and work with forcing notion. As we would like to
avoid cluttering this part of the proof with definitions, we refer the reader
to [14, Section 1.2] for the definition of Prikry forcing on trees.

Let $s\subseteq j_{\omega}(\lambda)$ be a set of size $\leq\kappa_{\omega}$ in
$M_{\omega}$, such that $s$ is the intersection with $\lambda$ of an
elementary submodel of $H^{M_{\omega}}(\chi)$. Let $U(s)$ be the corresponding
measure: $A\in U(s)\iff\\{\langle k(\alpha),\alpha\rangle\mid\alpha\in A\\}\in
k(A)$ for $k=j_{j_{\omega}(E)}$.

Let $\mathbb{Q}_{s}$ be the tree Prikry forcing defined using the measure
$U(s)$. Clearly, $U(s)$ is the image of some measure of the form $U(\bar{s})$
on $M_{n}$ for some $n<\omega$.

######  Lemma 44.

There is a sequence $\vec{t}=\langle t_{n}\mid n<\omega\rangle$ which is
generic for $\mathbb{Q}_{s}$, and a condition $p\in
j_{\omega}(\mathbb{P}^{*})$, such that $p^{\smallfrown}\langle
t_{0},\dots,t_{n-1}\rangle$ are compatible with $G$ for all $n$.

Moreover, the sequence $\vec{t}$ is unique, up to a finite shift.

###### Proof.

This follows from the Merimovich’s criteria, [23]:

First, in order to get such a sequence pick any $p\in\bigcup
j_{n,\omega}\operatorname{{}^{\prime\prime}}H_{n}$ such that
$\operatorname{dom}p\supseteq s$. Let $p\in j_{n,\omega}(H_{n})$. Take,
$t_{m}$ to be the added coordinates in step $n+m$, namely for
$\alpha\in\operatorname{dom}j_{n,m}(\bar{p})$ we let
$t_{m}(j_{m,\omega}(\alpha))=\alpha$.

Let $p,p^{\prime}$ be conditions in $j_{\omega}(\mathbb{P}^{*})$ with domain
$s$, such that there are sequences $\langle t_{n}\mid n<\omega\rangle$,
$\langle t_{n}^{\prime}\mid n<\omega\rangle$ are generic for $\mathbb{Q}_{s}$,
and for all $\alpha\in s$,

$p(\alpha)^{\smallfrown}\langle t_{n}(\alpha)\mid
n<\omega\rangle=p^{\prime}(\alpha)^{\smallfrown}\langle
t^{\prime}_{n}(\alpha)\mid n<\omega\rangle=G(\alpha).$

Let us show that there is $k$ such that for all large $n$,
$t_{k+n}=t^{\prime}_{k^{\prime}+n}$ and

$p^{\smallfrown}\langle
t_{0},\dots,t_{k-1}\rangle={p^{\prime}}^{\smallfrown}\langle
t_{0}^{\prime},\dots,t^{\prime}_{k^{\prime}-1}\rangle.$

Pick in $M_{\omega}$ a function $g$ enumerating $s$. For all large $n$,
$g\operatorname{{}^{\prime\prime}}t_{n}(\kappa_{\omega})=\operatorname{dom}t_{n}$.
In particular, by comparing the values at $\kappa_{\omega}$, we obtain the
possible value of $k^{\prime}-k$. Moreover, for all large $n$,

$\forall\alpha\in\operatorname{dom}t_{n},\,t_{n}(\alpha)>\max(p(\alpha),p^{\prime}(\alpha))\text{
and }t_{n}(\alpha)<t_{n+1}(\kappa_{\omega}).$

This is easily obtained by taking the right $U(s)$-large tree. Therefore, for
such $n$-s, the value of $t_{n}(\alpha)$ is the unique ordinal in $G(\alpha)$
which is between $G(\kappa)_{n}$ and $G(\kappa)_{n+1}$. ∎

Given a condition $p$ and a sequence $\vec{t}$ witnessing the validity of the
lemma for $s$, let $p_{n}=p^{\smallfrown}\langle t_{0},\dots,t_{n-1}\rangle$.
We call $\langle p_{n}\mid n<\omega\rangle$ a Prikry sequence for $s$. The
lemma indicates that the Prikry sequence is unique, up to an initial segment.
We will assume always that $|p_{n}(\kappa_{\omega})|=n$. This makes the Prikry
sequences to agree up to an initial segment, without a shift.

######  Lemma 45.

Let $\langle p_{n}\mid n<\omega\rangle$ be a Prikry sequence for some $s$.
Then, for all large $n$, $p_{n}\in j_{n,\omega}(H_{n})$.

###### Proof.

Pick $n$ large enough so that $s=j_{n,\omega}(\bar{s})$. Let $q\in H_{n}$ with
domain $\bar{s}$. Then, the canonical sequence of generators, starting with
$j_{n,\omega}(q)$ is a Prikry sequence for $s$. In particular, letting
$\langle q_{m}\mid m<\omega\rangle$ be the corresponding sequence (adding
dummy conditions in the beginning, if needed), we know that for all large $n$,
$q_{n}=p_{n}$, as we set the shift using the $\kappa_{\omega}$ coordinate.

Since $q_{n}\in H_{n}$, once we go past the dummy coordinates, the conclusion
follows. ∎

Let us claim that $X$ is the set of all $\alpha$ such that there is a
condition $p$ and $\vec{t}$ compatible with $G$, and for all large $n$,

$p_{n}\Vdash\alpha\in\sigma_{n}$

Indeed, if $\alpha\in X$, the existence of such $p$ is clear. Otherwise, since
for all large $n$, $p_{n}$ comes from $H_{n}$, it cannot force contradictory
information. ∎

It is interesting to try to see where the chain condition proof fails. Indeed,
any ordinal can still be captured by a set of size $\kappa$. The problem is
both the generic for $\mathbb{P}^{*}$ and the fact that there is no elementary
embedding from $V[H]$ to $M_{\omega}[G]$.

###  3.2. Interleaved Collapses

In this section, we will describe a situation in which interleaved collapses
can be incorporated into the BD setting. Let us discuss first the simple
setting of the vanilla Prikry forcing using a normal measure $U$. Let $j\colon
V\to M$ be the ultrapower map using the normal measure.

The following lemma is well known.

######  Lemma 46.

If $2^{\kappa}=\kappa^{+}$, then there is an $M$-generic filter for
$\operatorname{Col}(\kappa^{+},<j(\kappa))$ in $V$.

###### Proof.

Let us count the maximal antichains of the forcing
$\operatorname{Col}(\kappa^{+},<j(\kappa)$ in $M$. By the chain condition and
since $j(\kappa)$ is inaccessible in $M$, there are $j(\kappa)$ such
antichains in $M$. In $V$, $|j(\kappa)|=|\kappa^{\kappa}|=\kappa^{+}$. Let
$\vec{A}=\langle A_{\alpha}\mid\alpha<\kappa^{+}\rangle$ be an enumeration of
all maximal antichains in $M$, $\vec{A}\in V$.

The forcing $\operatorname{Col}(\kappa^{+},<j(\kappa))^{M}$ is
$\kappa^{+}$-closed in $M$, and since $M$ is closed under $\kappa$-sequences,
it is $\kappa^{+}$-closed in $V$ as well. Let us define in $V$ a decreasing
sequence of conditions, $\langle p_{\alpha}\mid\alpha<\kappa^{+}\rangle$ with
the property that $p_{\alpha}\leq q_{\alpha}\in A_{\alpha}$. This can be done,
using the closure of the forcing (from the point of view of $V$) at limit
steps.

Let $K$ be the upwards closure of $\langle
p_{\alpha}\mid\alpha<\kappa^{+}\rangle$. Then $K$ is $M$-generic filter. ∎

While it seems like the argument relies on the chain condition of the forcing,
a similar argument works for the forcing
$\operatorname{Col}(\kappa^{+},j(\kappa))$, as the number of dense open sets
is still $\kappa^{+}$ from the point of view of $V$. It does not work if
$2^{\kappa}>\kappa^{+}$. Yet, by carefully constructing the model, at some
cases one can obtain an $M$-generic filter in those cases as well.

From this point, we will only assume that there is an $M$-generic filter for
the collapse $\operatorname{Col}(\kappa^{+},<j(\kappa))$, but modifying this
to other forcing notions that admit a guiding generic does not change the
argument.

Let us consider the direct system of ultrapower embeddings, $j_{n,m}\colon
M_{n}\to M_{m}$, $m\leq\omega$. Let $K_{n+1}=j_{n}(K)$.

######  Lemma 47.

For each $n\leq m$, $K_{n+1}$ is $M_{m}$ generic for the forcing
$\operatorname{Col}(\kappa_{n}^{+},<\kappa_{n+1})$.

###### Proof.

First, by elementarity, $K_{n+1}$ is $M_{n}$-generic. Moreover, since
$M_{m}\subseteq M_{n}$ and

$\operatorname{Col}^{M_{n}}(\kappa_{n}^{+},<\kappa_{n+1})=\operatorname{Col}^{M_{m}}(\kappa_{n}^{+},<\kappa_{n+1}),$

we conclude that $K_{n+1}$ is $M_{m}$ generic as well. ∎

######  Lemma 48.

Let $K_{0}$ be $V$-generic for $\operatorname{Col}(\omega_{1},<\kappa)$.

For each $n$, $K_{0}\times K_{1}\times\cdots\times K_{n}$ is $M_{n}$-generic
for $\prod_{i\leq n}\operatorname{Col}(\kappa_{i-1}^{+},<\kappa_{i})$.

###### Proof.

Using Easton Lemma: if $H$ is generic for a $\lambda$-c.c. forcing and $G$ is
generic for a $\lambda$-closed forcing then $H$ and $G$ are mutually generic.
∎

Let $\vec{K}=\langle K_{n}\mid n<\omega\rangle$.

######  Theorem 49.

$M_{\omega}[\vec{K}]=\bigcap M_{n}[\vec{K}\restriction n+1]$.

###### Proof.

As in the proof of Theorem 14, we need to show first closure under
$\omega$-sequences of ordinals and then to use a similar (but simpler)
argument as in Theorem 42 in order to conclude the full theorem.

Let us begin with the closure under $\omega$-sequences.

######  Lemma 50.

$M_{\omega}[\vec{K}]$ is closed under $\omega$-sequence of ordinals from
$V[K_{0}]$.

###### Proof.

Indeed, it is easy to extract $P=\langle\kappa_{n}\mid n<\omega\rangle$ from
$\vec{K}$. Therefore, $M_{\omega}[\vec{K}]\supseteq M_{\omega}[P]$ which is
closed under countable sequences of ordinals from $V$, by Lemma 15. But $V$
and $V[K_{0}]$ have the same countable sequences, so the conclusion holds. ∎

Let $X\in\bigcap M_{n}[\vec{K}\restriction n+1]$ be a set of ordinals. Let us
define $Y_{n}$ and the corresponding name $\dot{\tau}_{n}$ as before:

$Y_{n}=\\{\alpha\mid j_{n,\omega}(\alpha)\in
X\\},\quad\dot{\tau}_{n}^{\vec{K}\restriction n+1}=Y_{n}.$

Let us look at $j_{n,\omega}(\dot{\tau}_{n})$, this is a name with respect to
the forcing

$\mathbb{Q}_{n}^{j}=\left(\prod_{i<n}\operatorname{Col}(\kappa_{i-1}^{+},\kappa_{i})\right)\times\operatorname{Col}(\kappa_{n-1}^{+},\kappa_{\omega}).$

######  Claim 51.

$\alpha\in X$ if and only if for all large $n$, there is a condition
$p\in\vec{K}\restriction n+1$ such that (viewing $p$ as a condition in
$\mathbb{Q}_{n}^{j}$), $p\Vdash_{\mathbb{Q}^{j}_{n}}\alpha\in
j_{n,\omega}(\dot{\tau}_{n})$.

###### Proof.

Indeed, if $\exists\bar{\alpha},j_{n,\omega}(\bar{\alpha})=\alpha$, then
$\bar{\alpha}\in Y_{n}$ and there is some $p\in\vec{K}\restriction n+1$ that
forces that, namely $\bar{\alpha}\in\dot{\tau}_{n}$. Since
$j_{n,\omega}(p)=p$, we conclude that $p\Vdash_{\mathbb{Q}^{j}_{n}}\alpha\in
j_{n,\omega}(\dot{\tau}_{n})$.

On the other hand, if there is $p\in\vec{K}\restriction n+1$ that forces
$\alpha\in j_{n,\omega}(\dot{\tau}_{n})$ and
$\alpha=j_{n,\omega}(\bar{\alpha})$ then again $p=j_{n,\omega}(p)$ and
therefore $p\Vdash\bar{\alpha}\in\dot{\tau}_{n}$. ∎

In particular, since $\langle\dot{\tau}_{n}\mid n<\omega\rangle\in
M_{\omega}[P]$, we conclude that $X\in M_{\omega}[\vec{K}]$. ∎

From Theorem 49, it is easy to deduce which cardinals are preserved in
$M_{\omega}[\vec{K}]$ and even show using Bukovsky’s Theorem
$M_{\omega}[\vec{K}]$ is a generic extension of $M_{\omega}$ using a
$\kappa_{\omega}^{+}$-c.c. forcing notion.

###  3.3. Extender Based Prikry forcing with interleaved collapses

Let us combine the results of Subsections 3.1 and 3.2.

The following well known lemma, shows that it is possible to obtain a guiding
generic filter for extender ultrapower.

######  Lemma 52.

Let $E$ be a $(\kappa,\kappa^{++})$-extender and let us assume that
$2^{\kappa}=\kappa^{+}$. Let $M$ be the extender ultrapower by $E$.

Then, there is an $M$-generic filter, $K$ for
$\operatorname{Col}(\kappa^{+++},<j(\kappa))$.

###### Proof.

Let $U$ be the derived normal ultrafilter, and let $i\colon V\to N$ be the
normal ultrapower. By Lemma 46, there is an $N$-generic filter $K_{0}$ for the
forcing $\operatorname{Col}((\kappa^{+++})^{N},<i(\kappa))$.

Let $k\colon N\to M$ be the map given by $k([f]_{U})=j(f)(\kappa)$. Let $K$ be
the upwards closure of $k\operatorname{{}^{\prime\prime}}K_{0}$.

######  Claim 53.

$K$ is $M$-generic filter.

###### Proof.

For every $x\in M$, there is $g\colon(\kappa^{++})^{N}\to N$ such that
$x\in\operatorname{range}k(g)$. This is clear, by noting that if $x=j(r)(a)$
for some generator $a$, then without loss of generality, $a\in\kappa^{++}$ and
therefore $x\in\operatorname{range}k(i(r)\restriction(\kappa^{++})^{N})$.

Given a dense open set $D\in M$, let $g\colon(\kappa^{++})^{N}\to N$ cover it,
and let $D^{\prime}=\bigcap\\{g(\alpha)\mid g(\alpha)\text{ dense open}\\}$.
By the distributivity of the forcing in $N$, $D^{\prime}$ is dense open and
therefore there is a condition in $K_{0}$ meeting it. ∎

∎

Let us consider now the iteration $\langle j_{n,m}\colon M_{n}\to M_{m}\mid
n\leq m\leq\omega\rangle$ given by iterating the extender embedding. Let
$K_{0}$ be a $V$-generic for $\operatorname{Col}(\omega_{1},<\kappa)$, and let
$H_{0}\subseteq\mathbb{P}^{*}$ be $V$-generic.

Let us define $H_{n}$ and $G$ in the very same way as in Theorem 42 and
$K_{n}$ in the same way as in Theorem 49.

######  Theorem 54.

$\bigcap M_{n}[\vec{K}\restriction n+1][H_{n}]=M_{\omega}[\vec{K}][G]$.

###### Proof.

As the ideas of the proof are very similar to the proofs of Theorem 3.1 and
Theorem 49 let us sketch the proof.

First, we show that the model $M_{\omega}[\vec{K}][G]$ is closed under
$\omega$-sequences, using Lemma 43. Then, we obtain a sequence of names
$\langle\sigma_{n}\mid n<\omega\rangle$ as above. Using Lemma 44, we obtain
local approximations of the filters
$j_{n,\omega}\operatorname{{}^{\prime\prime}}H_{n}$ and by the same argument
of Claim 51 we obtain the relevant conditions for the $K$-parts. ∎

##  4\. Magidor and Radin forcing

In [20], Magidor introduces a variant of the Prikry forcing that enables one
to change the cofinality of a measurable cardinal to be uncountable. This
forcing was revised by Mitchell and generalized by Radin. The version that we
present here follows Mitchell’s definition.333A variant of this presentation
appeared in the unpublished book of Cummings and Woodin. As in the other parts
of the paper, we are not going to define a forcing notion but rather an
extension of an iterated ultrapower.

######  Definition 55.

Let $U,U^{\prime}$ be two normal measures on $\kappa$. Then $U$ is below
$U^{\prime}$ in the Mitchell order if $U\in\operatorname{Ult}(V,U^{\prime})$.

######  Definition 56 (Mitchell).

Let $o^{\mathcal{U}}$ be a function from $\kappa+1$ to ordinals.

A sequence $\mathcal{U}=\langle
U_{\alpha,\beta}\mid\alpha\leq\kappa,\beta<o^{\mathcal{U}}(\alpha)\rangle$ is
a coherent sequence if for every
$\alpha\leq\kappa,\beta<o^{\mathcal{U}}(\alpha)$,

$j_{U_{\alpha,\beta}}(\mathcal{U})\restriction(\alpha,\beta)=\mathcal{U}\restriction(\alpha,\beta)$

where $\mathcal{U}\restriction(\alpha,\beta)=\langle
U_{\gamma,\delta}\mid(\gamma<\alpha,\beta<o^{\mathcal{U}(\gamma)})\vee(\gamma=\alpha\wedge\delta<\beta)\rangle$.

######  Definition 57.

Let $\vec{U}$ be a Mitchell increasing sequence of normal measures on
$\kappa$. We say that $\vec{U}=\langle U_{\alpha}\mid\alpha<\delta\rangle$ is
pre-coherent if there is a function $c$ such that for every $\alpha<\delta$,
$j_{U_{\alpha}}(c)(\kappa)=\alpha$.

Any coherent sequence of normal measures gives rise to a pre-coherent Mitchell
increasing sequence of normal measures of the same length, but not necessarily
vice-verse, see [3].

######  Remark 58.

Let $\vec{U}$ be a Mitchell increasing sequence of normal measures. If
$\operatorname{len}\vec{U}<\kappa^{+}$ then $\vec{U}$ is pre-coherent.

###### Proof.

If $\operatorname{len}\vec{U}=0$, there is nothing to prove. So let us assume
that it is non-zero.

Since the measures are discrete, there is a sequence of pairwise disjoint sets
$\langle A_{\alpha}\mid\alpha<\operatorname{len}\vec{U}\rangle$ such that
$A_{\alpha}\in U_{\beta}\iff\alpha=\beta$.

Define $g(\zeta)=\alpha$ iff $\zeta\in A_{\alpha}$ and $0$ otherwise. Let also
fix a sujective function $r\colon\kappa\to\operatorname{len}\vec{U}$. Now,
$a=\operatorname{range}j(r)\restriction\kappa=j\operatorname{{}^{\prime\prime}}\operatorname{len}\vec{U}$.
In particular, for $\pi_{a}$ the Mostowski collapse of the set $a$, and it is
easy to verify that $\pi_{a}\circ j(g)\in M$ is the desired function.

Finally, let

$c(\zeta)=\pi_{\operatorname{range}r\restriction\zeta}(g(\zeta)),$

where this application is defined, and zero otherwise. ∎

Let $\vec{U}$ be a sequence of normal measures on $\kappa$, increasing in the
Mitchell order, and let $o^{\vec{U}}(\kappa)=\zeta$. Let us define an
iteration as well as a sequence of ordinals
$\langle\gamma_{\xi}\mid\xi<\alpha_{*}\rangle$ as follows:

$M_{0}=V$, $j_{0,0}=id_{V}$. Given $M_{\alpha}$ and maps
$j_{\beta,\alpha}\colon M_{\beta}\to M_{\alpha}$ for all $\beta<\alpha$, we
pick the least index $\gamma_{\alpha}<\operatorname{len}j_{\alpha}(\vec{U})$
such that $\sup\\{\beta<\alpha\mid
j_{\beta,\alpha}(\gamma_{\beta})=\gamma_{\alpha}\\}$ is bounded (and we let
$\gamma_{0}=0$).

Let
$M_{\alpha+1}=\operatorname{Ult}(M_{\alpha},j_{\alpha}(\vec{U})(\gamma_{\alpha}))$
and let $j_{\alpha,\alpha+1}$ be the ultrapower map. Let
$j_{\beta,\alpha+1}=j_{\alpha,\alpha+1}\circ j_{\beta,\alpha}$ for all
$\beta\leq\alpha$.

If $\gamma_{\alpha}$ is undefined, we halt.

Let $\kappa_{\alpha}$ be the critical point of $j_{\alpha,\alpha+1}$ and let
$\alpha_{*}$ be the length of the process.

The following Lemma is a comparison argument due to Mitchell, in disguise.

######  Lemma 59.

The iteration halts. Moreover, if $\operatorname{len}\vec{U}=\mu<\kappa$, it
halts after $\omega^{\mu}$ many steps (ordinal exponentiation).

###### Proof.

Let us assume towards a contradiction that the iteration continues for
$\lambda=(\operatorname{len}(\vec{U})^{\kappa})^{+}$ many steps (cardinal
exponentiation).

Consider $\gamma_{\alpha}$ for limit ordinals $\alpha<\lambda$. As we take
direct limits at limit steps of the iteration, each such $\gamma_{\alpha}$ is
the image of an ordinal from a previous step in the iteration. More precisely,
there are finitely many ordinals $\beta_{0}<\dots<\beta_{n-1}<\alpha$ and a
function $f\colon\kappa^{n}\to\operatorname{len}\vec{U}$ in $V$, such that
$j_{\alpha}(f)(\kappa_{\beta_{0}},\dots,\kappa_{\beta_{n-1}})=\gamma_{\alpha}$.

By Fodor’s lemma, there is a stationary set $S\subseteq\lambda$, such that $f$
and the finite sequence $\beta_{0},\dots,\beta_{n-1}$ are fixed. Pick
$\alpha\in S\cap\operatorname{acc}S$. Then, for every $\beta\in S\cap\alpha$,
$\gamma_{\alpha}=j_{\beta,\alpha}(\gamma_{\beta})$, a contradiction to our
choice of $\gamma_{\alpha}$.

For the moreover part, one can verify by induction that for every non-zero
ordinal $\alpha<\kappa$ of Cantor’s normal form
$\alpha=\omega^{\beta_{0}}\cdot n_{0}+\omega^{\beta_{1}}\cdot
n_{1}+\cdots+\omega^{\beta_{m}}\cdot n_{m}$, where
$\beta_{0}>\beta_{1}>\cdots>\beta_{m}$, and $n_{i}\neq 0$ for all $i$,
$\gamma_{\alpha}=\beta_{m}$. Since the length of $\vec{U}$ in this case is
below the critical point of $j_{\alpha}$, the process terminates at step
$\alpha_{*}=\omega^{\mu}$. ∎

Let $P=\langle\kappa_{\alpha}\mid\alpha<\alpha_{*}\rangle$, be the sequence of
the critical points. In this case, the model $M_{\alpha_{*}}[P]$ clearly
contains sets which are not in $M_{\beta}$ for large $\beta$. The reason is
that for every infinite $\beta$, $M_{\beta}$ is not closed under countable
sequences and in particular will not contain the initial segments of $P$.

Thus, the correct models for this theorem are
$M_{\alpha}[P\restriction\alpha]$. Note that for $\alpha$ limit,
$\kappa_{\alpha}$ is (typically) a singular cardinal in this model. Thus, in
those cases there is no elementary embedding from
$M_{\alpha}[P\restriction\alpha]$ to $M_{\beta}[P\restriction\beta]$.

######  Lemma 60.

$M_{\alpha_{*}}[P]\subseteq\bigcap_{\alpha<\alpha_{*}}M_{\alpha}[P\restriction\alpha]$.

###### Proof.

Working in $M_{\alpha}[P\restriction\alpha]$ one can compute from the
parameter $\gamma_{\alpha}$ the rest of the models $M_{\alpha^{\prime}}$ for
$\alpha^{\prime}>\alpha$ and the iteration, and thus $P\setminus\alpha$. ∎

######  Theorem 61.

Let us assume that $\operatorname{len}\vec{U}\leq\kappa^{+}$ and that it is
pre-coherent then $M_{\alpha_{*}}[P]$ is closed under $\kappa$-sequences.

###### Proof.

Fix a function $c$ witnessing the pre-coherency of $\vec{U}$.

Let $\alpha\in\mathrm{Ord}$. Then, $\alpha$ is represented by some function
$f\colon\kappa^{n}\to\mathrm{Ord}$ and a finite sequence
$\zeta_{0}<\cdots<\zeta_{n-1}$ in $P$, such that:

$\alpha=j_{\alpha_{*}}(f)(\zeta_{0},\dots,\zeta_{n-1}).$

The main challenge is to ”describe” which elements from $P$ we evaluate
$j_{\alpha_{*}}(f)$ in, in order to obtain $\alpha$, in particular where
$\operatorname{len}\vec{U}\geq\kappa$.

######  Lemma 62.

Let $\zeta\in P$, $\beta=\mathrm{otp}(P\cap\zeta)$.

Then, there is a function $g\in M_{\beta}$ and $\zeta^{\prime}<\zeta$ in $P$
such that $\zeta=\min(\\{\rho\in P\mid\rho>\zeta^{\prime}\text{ and
}j_{\beta,\alpha_{*}}(g)(\rho)=j_{\alpha_{*}}(c)(\rho)\\}$.

In particular, there are $\zeta_{0},\dots,\zeta_{n-1}<\zeta$ in $P$ and
$h\colon\kappa^{n}\to o(\kappa)$ in $V$ such that

$\zeta=\min(\\{\rho\in P\mid\rho>\zeta_{n-1}\text{ and
}j_{\alpha_{*}}(h)(\zeta_{0},\dots,\zeta_{n-1},\rho)=j_{\alpha_{*}}(c)(\rho)\\}.$

###### Proof.

For $\zeta\in P$ which is not a limit point, the claim is obvious: its
Mitchell order under $\vec{U}$ is simply $0$ and one can read it from its
predecessor.

For $\zeta$ limit, as $\gamma_{\beta}<\zeta^{+}$, there is a canonical
function representing it, $g\in M_{\beta}$. As $\beta$ is limit,
$g=j_{\bar{\beta},\beta}(\bar{g})$ for some $\bar{g}\in M_{\bar{\beta}}$,
which is the canonical function for some ordinal $\bar{\delta}$. Being
canonical, the ordinal $\bar{\delta}$ is definable from
$\bar{g}=g\restriction\zeta^{\prime}$ for some $\zeta^{\prime}$ and in
general, for every $\zeta^{\prime}\leq\xi<\zeta$ in $P$, $g(\zeta)$ is the
unique ordinal which is the height of the canonical function
$g\restriction\xi$. Since $j_{\delta,\beta}(g\restriction\xi)=g$ for
$\delta=\mathrm{otp}(P\cap\xi)$, $\xi\in P$, we conclude that if equality
holds, then $j_{\delta,\beta}(\gamma_{\xi})=\gamma_{\beta}$, but this can only
hold for boundedly many values in $P$.

The second part follows from the assumption that we are taking direct limits
at limit steps and thus $g$ is the image of a function from a shorted
iteration. Thus, by induction, one can represent $g$ as
$j_{\beta}(h)(\zeta_{0},\dots,\zeta_{n-1})$ (recall that
$\beta=\mathrm{otp}(P\cap\zeta)$), $\zeta_{0},\dots,\zeta_{n-1}<\zeta$ in $P$
and $h\colon\kappa^{n}\to o(\kappa)$. ∎

Let

$\begin{matrix}j_{\alpha_{*}}(h)^{P}(\zeta_{0},\dots,\zeta_{n-1})&=&\min(\\{\rho\in
P\mid&\rho>\zeta_{n-1}\text{ and }\\\
&&&j_{\alpha_{*}}(h)(\zeta_{0},\dots,\zeta_{n-1},\rho)=o^{j_{\alpha_{*}}(\mathcal{U})}(\rho)\\}.\end{matrix}$

Given $\zeta\in P$ we may find an increasing sequence of elements of $P$,
$\zeta_{0},\dots,\zeta_{n}$ and functions $g_{0},\dots,g_{n-1}\in V$ such
that:

$\zeta_{k}=j_{\alpha_{*}}(g_{k})^{P}(\zeta_{0},\dots,\zeta_{k-1})$

This is always possible, using standard arguments, see [15].

So, we conclude that each ordinal $\alpha$ can be represented using a function
$f$ and finitely many functions $g_{0},\dots,g_{n-1}$ that allows us to ”read”
the relevant elements from $P$.

Let $\langle\alpha_{i}\mid i<\kappa\rangle$ be a sequence of ordinals. Let
$\langle f^{i},g_{0}^{i},\dots,g_{n_{i}-1}^{i}\mid i<\kappa\rangle$ be a
choice of representatives, as above. Namely, take
$\beta^{i}_{j}=j_{\alpha_{*}}^{P}(\beta^{i}_{0},\dots,\beta^{i}_{j-1})$ and
$\alpha_{i}=j_{\alpha_{*}}(f)(\beta^{i}_{0},\dots,\beta^{i}_{n_{i}-1})$.

Apply $j_{\alpha_{*}}$, and truncate at $\kappa$, and then compute from $P$
first the indexes $\beta^{0}_{i},\dots,\beta^{n-1}_{i}$ and then the ordinals
$\alpha_{i}$. ∎

Let us analyse $\operatorname{cf}\alpha_{*}$ (which is the cofinality of
$j_{\alpha_{*}}(\kappa)$), under the assumption that
$\operatorname{len}\vec{U}<\kappa^{+}$.

######  Lemma 63.

Let us assume that $\vec{U}$ is weakly-coherent and that
$\operatorname{cf}\operatorname{len}\vec{U}\leq\kappa$.

Let $\lambda=\operatorname{cf}^{M_{\alpha_{*}}[P]}(j_{\alpha_{*}}(\kappa))$.
Then,

$\lambda=\begin{cases}\omega&\operatorname{len}\vec{U}\text{ is a successor
ordinal}\\\
\operatorname{cf}^{V}(\operatorname{len}\vec{U})&\operatorname{cf}^{V}(\operatorname{len}\vec{U})<\kappa\\\
\omega&\operatorname{cf}^{V}(\operatorname{len}\vec{U})=\kappa\end{cases}$

###### Proof.

We split into cases.

Case 0: $\operatorname{len}\vec{U}$ is a successor ordinal. In this case
$\operatorname{cf}\alpha_{*}$ is $\omega$. Indeed, the ordinals $\alpha$ in
which $\gamma_{\alpha}$ is the last measure in $j_{\alpha}(\vec{U})$ form an
$\omega$-sequence, $\alpha_{n}$ cofinal at $\alpha_{*}$. Using the pre-
coherency of the sequence, $\gamma_{\alpha}$ can be read from $P$.

Case 1: $\operatorname{cf}\operatorname{len}\vec{U}=\mu<\kappa$. Pick a
cofinal sequence $\langle\rho_{i}\mid i<\mu\rangle$, cofinal at
$\operatorname{len}U$. Since $\mu<\kappa$ which is the critical point, for all
$\alpha$ (and in particular, for $\alpha_{*}$),
$\operatorname{cf}^{M_{\alpha}}\operatorname{len}j_{\alpha}(\vec{U})=\mu$.

Consider $\alpha_{*}$. For each $i$, there are unboundedly many
$\beta<\alpha_{*}$ such that $\gamma_{\beta}=j_{\beta}(\rho_{i})$. Let
$\beta_{i}$ be the least such ordinal. Then, $\beta_{i}$ are increasing, and
cofinal at $\alpha_{*}$. Indeed, below each $\beta_{i}$, the images of
$\rho_{j}$, for all $j<i$ as well as all lower ordinals appear unboundedly.
Again, using the pre-coherency, this cofinal sequence at
$j_{\alpha_{*}}(\kappa)$ can be computed in $M_{\alpha_{*}}[P]$. The
cofinality cannot be lower than that, as otherwise, the cofinality of $\mu$
must be collapsed, but $M_{\alpha_{*}}[P]\subseteq V$.

Case 2: $\operatorname{cf}^{V}\operatorname{len}\vec{U}=\kappa$. In this case
we need to show that $\operatorname{cf}\alpha_{*}=\omega$.

Fix a sequence $\vec{\delta}=\langle\delta_{\xi}\mid\xi<\kappa\rangle$ cofinal
at $\operatorname{len}\vec{U}$. Let us define by induction a sequence of
ordinals cofinal at $\alpha_{*}$.

Let $\alpha_{0}=0$.

Let $n<\omega$. Denote
$\rho_{n}=j_{\alpha_{n}+1}(\vec{\delta})(j_{\alpha_{n}}(\kappa))$. Since
$\rho_{n}<\operatorname{len}\vec{U}$, we may define $\alpha_{n+1}$ to be the
least ordinal such that
$\gamma_{\alpha_{n+1}}=j_{\alpha_{n}+1,\alpha_{n+1}}(\rho_{n})$.

Let us show that $\alpha_{*}=\sup\alpha_{n}$. Indeed, let
$\alpha_{\omega}=\sup\alpha_{n}$, and let us assume towards a contradiction,
that $\alpha_{\omega}\neq\alpha_{*}$, so $\gamma_{\alpha_{*}}$ is an ordinal
$<j_{\alpha_{\omega}}(\operatorname{len}\vec{U})$.

Since $\alpha_{\omega}$ is a limit ordinal,
$\kappa_{\alpha_{\omega}}=\sup_{n<\omega}\kappa_{\alpha_{n}}$ and thus
$j_{\alpha_{\omega}}(\vec{\delta})$ is cofinal at
$j_{\alpha_{\omega}}(\operatorname{len}\vec{U})$, and there is $n$ such that
the $\kappa_{\alpha_{n}}$-th point of this sequence exceeds
$\gamma_{\alpha_{\omega}}$. Without loss of generality,
$\gamma_{\alpha_{\omega}}=j_{\alpha_{n},\alpha_{\omega}}(\bar{\gamma})$.

Now, as for every $m>n$, $\rho_{m}>j_{\alpha_{n},\alpha_{m}}(\bar{\gamma})$,
there must be an ordinal $\beta$ between $\alpha_{m}$ and $\alpha_{m+1}$ such
that $j_{\beta,\alpha_{\omega}}(\gamma_{\beta})=\gamma$, contradicting the
definition of $\gamma_{\alpha_{\omega}}$. ∎

We conclude that if $\operatorname{len}\vec{U}<\kappa^{+}$ then
$\operatorname{cf}\alpha_{*}<\kappa$. Moreover, this can be decoded from $P$
itself, and thus this cofinality is going to be correctly computed in
$M_{\alpha_{*}}[P]$.

######  Lemma 64.

If $\operatorname{cf}^{V}(\operatorname{len}\vec{U})=\kappa^{+}$ then
$M_{\alpha_{*}}[P]\models j_{\alpha_{*}}(\kappa)$ is regular.

###### Proof.

We will prove the claim for the case $\operatorname{len}\vec{U}=\kappa^{+}$.
The general case is similar.

######  Claim 65.

for every $\alpha$, $M_{\alpha}[P\restriction\alpha]$ is a
$\kappa_{\alpha}^{+}$-c.c. extension of $M_{\alpha}$ and in particular
$\kappa_{\alpha}^{+}$ is preserved.

The proof for this claim is very similar to the proof of Lemma 11, with the
additional complication that every function in $M_{\beta}[P\restriction\beta]$
needs to be bounded (using an inductive hypothesis) by a corresponding
function from $M_{\beta}$. 444There is a different way to show that the
regularity of $\kappa_{\alpha}^{+}$ is preserved, by showing that
$\kappa_{\alpha}$ must be singular in $M_{\alpha}[P\restriction\alpha]$ (using
an inductive hypothesis and the previous theorem) and thus the cofinality of
this cardinal must be below $\kappa_{\beta}$ for some $\beta<\alpha$. Similar
argument appears ahead.

Let us assume now that
$\operatorname{cf}^{M_{\alpha_{*}}[P]}(j_{\alpha_{*}}(\kappa))<j_{\alpha_{*}}(\kappa)$.
So, there is $\alpha<\alpha_{*}$ such that
$\operatorname{cf}^{M_{\alpha_{*}}[P]}(j_{\alpha_{*}}(\kappa))<\kappa_{\alpha}$
and thus
$M_{\alpha}[P\restriction\alpha]\models\operatorname{cf}\alpha_{*}<\kappa_{\alpha}$.

In this model, $M_{\alpha}[P\restriction\alpha]$, one can compute the rest of
the iteration $j_{\alpha,\alpha_{*}}$ and in particular, the sequence
$\langle\gamma_{\alpha}\mid\alpha<\alpha_{*}\rangle$. It is clear that this
sequence is cofinal at
$j_{\alpha_{*}}(\kappa^{+})=\operatorname{len}j_{\alpha_{*}}(\vec{U})$.

As $P$ is cofinal at $j_{\alpha_{*}}(\kappa)$,
$\operatorname{cf}^{M_{\alpha}[P\restriction\alpha]}(\alpha_{*})=\operatorname{cf}^{M_{\alpha}[P\restriction\alpha]}(j_{\alpha_{*}}(\kappa))<\kappa_{\alpha}$.

Since the embedding $j_{\alpha,\alpha_{*}}$ is continuous at
$j_{\alpha_{*}}(\kappa^{+})$, we conclude that the cofinality of
$j_{\alpha_{*}}(\kappa^{+})$ in $M_{\alpha}[P\restriction\alpha]$ must be the
same as the cofinality of $j_{\alpha}(\kappa^{+})$ which is strictly larger
than $\kappa_{\alpha}$, by the chain condition of the forcing.

But this is a contradiction — on the one hand the cofinality of $\alpha_{*}$
must be strictly below $\kappa_{\alpha}$ and on the other hand it must be
$j_{\alpha}(\kappa^{+})$.∎

######  Lemma 66.

Let $\vec{U}$ be a Mitchell increasing sequence of measures and let us
consider the corresponding iteration.

If $\alpha<\alpha_{*}$ satisfies that $\gamma_{\alpha}$ is strictly larger
than $j_{\beta,\alpha}(\gamma_{\beta})$ for all $\beta<\alpha$, then the
embedding $j_{\alpha+1,\alpha_{*}}\colon M_{\alpha+1}\to M_{\alpha_{*}}$ lifts
to an embedding $\tilde{j}_{\alpha+1,\alpha_{*}}\colon
M_{\alpha+1}[P\restriction\alpha]\to M_{\alpha_{*}}[P\restriction\alpha]$.

###### Proof.

Without loss of generality, $\gamma_{\alpha}\in
j_{\alpha}^{\prime\prime}\operatorname{len}\vec{U}$. Otherwise, we will need
to repeat the following process finitely many times.

Let $\bar{\gamma}$ be an ordinal such that
$j_{\alpha}(\bar{\gamma})=\gamma_{\alpha}$.

Let us consider the ultrapower by $U_{\bar{\gamma}}$, $k_{0}\colon V\to N$. By
the definition, in this model, the sequence $\vec{U}\restriction\bar{\gamma}$
exists. By our choice of $\gamma_{\alpha}$, and as
$P(\kappa),\vec{U}\restriction\bar{\gamma}\in N$, if we will start to iterate
$N$ based on $\vec{U}\restriction\bar{\gamma}$ we will obtain exactly the
iteration $j_{\alpha}\restriction N$ and
$j_{\alpha}(\bar{\gamma})=\gamma_{\alpha}$.

Next, the following diagram commutes:

${V}$${N}$${N_{\alpha_{*}}}$${M_{\alpha}}$${M_{\alpha+1}}$${M_{\alpha_{*}}}$$\scriptstyle{k_{0}}$$\scriptstyle{j_{\alpha}}$$\scriptstyle{j_{\alpha}}$$\scriptstyle{k_{\alpha+1,\alpha_{*}}}$$\scriptstyle{j_{\alpha}}$$\scriptstyle{j_{\alpha,\alpha+1}}$$\scriptstyle{j_{\alpha+1,\alpha_{*}}}$

where $k_{\alpha+1,\alpha_{*}}$ is the iteration as defined in $N$ using
$k_{0}(\vec{U})$.

In $N$, one can compute $P\restriction\alpha$ and as $k_{0}(\kappa)$ is an
inaccessible much larger than $\kappa$ in $N$, this set is bounded below
$k_{0}(\kappa)$. Thus, the embedding $k_{\alpha+1,\alpha_{*}}$ can be
restricted to the class $M_{\alpha+1}[P\restriction\alpha]$. Moreover, since
$k_{\alpha+1,\alpha_{*}}\restriction M_{\alpha+1}=j_{\alpha+1,\alpha_{*}}$, we
conclude that the image of this map is going to be contained in
$M_{\alpha_{*}}[P\restriction\alpha]$.

In order to show elementarity, we recall the definition

$M_{\alpha+1}[P\restriction\alpha]=\bigcup_{\zeta\in\mathrm{Ord}}L[M_{\alpha+1}\cap
V_{\zeta},P\restriction\alpha]$
$M_{\alpha_{*}}[P\restriction\alpha]=\bigcup_{\zeta\in\mathrm{Ord}}L[M_{\alpha_{*}}\cap
V_{\zeta},P\restriction\alpha]$

so the restriction of $k_{\alpha+1,\alpha_{*}}$ to each component is
elementary, and thus it is elementary. ∎

######  Theorem 67.

Let $\vec{U}$ be a Mitchell increasing sequence of measures and let us assume
that $\operatorname{len}\vec{U}<\kappa^{+}$. Then $\bigcap
M_{\alpha}[P\restriction\alpha]=M_{\alpha_{*}}[P]$.

###### Proof.

Let us assume by induction on $\kappa$ and $\operatorname{len}\vec{U}$ that
the theorem holds, namely that $\bigcap
M_{\alpha}[P\restriction\alpha]=M_{\alpha_{*}}[P]$. Let $\beta_{i}$ the
cofinal sequences at $\alpha_{*}$ defined in the cases above.

Case 0: $\operatorname{len}\vec{U}$ is a successor ordinal. In this case,
apply the inductive hypothesis for $\operatorname{len}\vec{U}$, (and
elementarity) in the model $M_{\beta_{n}+1}[P\restriction\beta_{n}]$. We get
that
$M_{\beta_{n+1}}[P\restriction\beta_{n+1}]=\bigcap_{\alpha\in[\beta_{n},\beta_{n+1})}M_{\alpha}[P\restriction\beta_{n}][P\restriction[\beta_{n},\alpha)]$.
This is true, by Lemma 66, applied finitely many times. So, in order to show
that the theorem holds, consider
$X\in\bigcap_{n<\omega}M_{\beta_{n}+1}[P\restriction\beta_{n}]$, and use the
elementary emebddings $\tilde{j}_{\beta_{n}+1,\alpha_{*}}\colon
M_{\beta_{n}+1}[P\restriction\beta_{n}]\to
M_{\alpha_{*}}[P\restriction\beta_{n}]$, and repeat the argument of Theorem
14.

Case 1: $\operatorname{cf}\operatorname{len}\vec{U}<\kappa$. This case is the
same, we notational differences, using the closure of the models
$M_{\alpha}[P]$ under $\kappa$-sequences.

Case 2: $\operatorname{cf}\operatorname{len}\vec{U}=\kappa$. In this case,
$\operatorname{cf}\alpha_{*}=\omega$ and we can repeat the argument of Case 0.
∎

It worth mentioning that the intersection theorem is quite weak in this case.
Indeed, if we would apply it for an arbitrary sequence of measures, which
might fail to be Mitchell increasing, it still holds, but it might be quite
degenerated. For example, if we look at an iteration of length $\omega+1$ of
the same measure $U$ and look at the model $M_{\omega+1}[P]$, then since
$P\restriction\omega$ defines the normal measure $j_{\omega}(U)$, this model
is going to be simply $M_{\omega}[P]$ (in particular, class many cardinals of
$M_{\omega+1}$ are collapsed in $M_{\omega+1}[P]$).

######  Claim 68.

Let $\vec{U}$ be a Mitchell increasing sequence of measures of length
$\kappa^{+}$.

Then $\bigcap_{\alpha<\alpha_{*}}M_{\alpha}[P\restriction\alpha]$ is strictly
larger than $M_{\alpha_{*}}[P\restriction\alpha_{*}]$.

###### Proof.

Let us show that $\Gamma=\langle
j_{\alpha,\alpha_{*}}(\gamma_{\alpha})\mid\alpha<\alpha_{*}\rangle$ belongs to
the intersection model, but not to the generic extension.

First, let us show by induction on $\beta$ for every sequence of measures on
$\kappa$ of length $\beta<\kappa^{+}$, $\vec{U}^{\prime}$ the corresponding
sequence $\Gamma^{\prime}$ belongs to $M^{\prime}_{\beta_{\star}}[P^{\prime}]$
(where all those objects are defined using $\vec{U}^{\prime}$.

Let us assume that the claim is proved for every $\beta^{\prime}<\beta$, and
let us consider the case of a sequence of length $\beta$. Since
$M_{\beta_{*}}[P]=\bigcap_{\delta<\beta_{*}}M_{\delta}[P\restriction\delta]$,
it is enough to show that for every $\delta<\beta_{*}$, $\Gamma\in
M_{\delta}[P\restriction\delta]$.

We prove that $\Gamma\in M_{\delta}[P\restriction\delta]$ using a second level
of induction, on $\delta<\beta_{*}$. Note that since
$\Gamma\restriction[\delta,\beta_{*})\in M_{\delta}$, it is enough to show
that $\Gamma\restriction\delta\in M_{\delta}[P\restriction\delta]$.

Let $\rho<\delta$ be the last ordinal such that
$j_{\rho,\delta}(\gamma_{\rho})\geq\gamma_{\delta}$, assuming that there is an
ordinal $\rho$ such that $j_{\rho,\delta}(\gamma_{\rho})\geq\gamma_{\delta}$.
If there is such an ordinal then there is a maximal one, since below every
$\rho$ which is large enough so that $\gamma_{\delta}$ will be in the image of
$j_{\rho,\delta}$, and $j_{\rho,\delta}(\gamma_{\rho})>\gamma_{\delta}$ there
are unboundedly many ordinals such that
$j_{\rho,\delta}(\gamma_{\rho^{\prime}})=\gamma_{\delta}$. Using the
definition of $\gamma_{\delta}$, we know that this set is bounded and using
the definition of $\gamma_{\rho}$, we know that it is closed.

By the inductive hypothesis, $\Gamma\restriction\delta\in
M_{\rho}[P\restriction\rho]$. In $M_{\rho+1}$, the iteration up to $\delta$ is
definable using the measure sequence
$j_{0,\rho+1}(\vec{U}\restriction\gamma_{\rho})$, which by the (external)
induction hypothesis, pushed forward by $j_{0,\rho+1}$, satisfies that
$M_{\delta}[P\restriction[\rho,\delta)$ contains
$\Gamma\restriction[\rho,\delta)$. Combining all together, the result follows.

Next, let us verify that $\Gamma\notin M_{\alpha_{*}}[P]$. The map
$\kappa_{\alpha}\mapsto\gamma_{\alpha}$ is a surjection on
$j_{\alpha_{*}}(\operatorname{len}\vec{U})$ which is
$j_{\alpha_{*}}(\kappa^{+})$, and in particular $j_{\alpha_{*}}(\kappa^{+})$
is collapsed. But, this is impossible, by Claim 65. ∎

######  Question 69.

Is $M_{\alpha_{*}}[P]$ is always closed under $\kappa$-sequences?

######  Question 70.

Is there a parallel for the intersection theorem for Radin forcing with
$o(\kappa)\geq\kappa^{+}$?

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