# The gluing property

Yair Hayut  Einstein Institute of Mathematics, Hebrew University of Jerusalem,
Givat-Ram, 91904, Israel<EMAIL_ADDRESS>and  Alejandro Poveda
Center of Mathematical Sciences and Applications, Harvard University,
Cambridge MA, 02138, USA<EMAIL_ADDRESS>
###### Abstract.

We introduce a new compactness principle which we call the gluing property.
For a measurable cardinal $\kappa$ and a cardinal $\lambda$, we say that
$\kappa$ has the $\lambda$-gluing property if every sequence of $\lambda$-many
$\kappa$-complete ultrafilters on $\kappa$ can be glued into an extender. We
show that every $\kappa$-compact cardinal has the $2^{\kappa}$-gluing
property, yet non-necessarily the $(2^{\kappa})^{+}$-gluing property. Finally,
we compute the exact consistency-strength for $\kappa$ to have the
$\omega$-gluing property; this being $o(\kappa)=\omega_{1}$.

###### Key words and phrases:

Gluing property, Prikry-type forcings.

###### 2020 Mathematics Subject Classification:

03E35, 03E55

Hayut’s research was supported by the ISF grant 1967/21.

Poveda wishes to thank the Einstein Institute of Mathematics at the Hebrew
University and the Center of Mathematical Sciences and Applications at Harvard
University for their generous support during the entire duration of this
project.

##  1\. Introduction

Suppose that $\mathfrak{M}$ is a mathematical structure all of whose _small_
substructures $\mathfrak{N}\subseteq\mathfrak{M}$ have a property $\Phi$. Is
it true that $\mathfrak{M}$ itself satisfies $\Phi$?

An affirmative answer to the above is an instance of a phenomenon called
_compactness._ Historical experience in Mathematics indicates that there is a
natural expectation for compactness. Indeed, several classical results, in
various areas, indicate that this is the case when _small_ means finite.

For instance, in Graph Theory, a classical theorem of De Brujin and Erdős
[BE51] establishes that if $\mathcal{G}$ is an infinite graph such that all of
its finite subgraphs $\mathcal{H}\subseteq\mathcal{G}$ have chromatic number
${\leq}n$ then $\mathcal{G}$ itself has chromatic number ${\leq}n$. Another
example in this vein is _Gődel’s Compactness theorem_ for First Order Logic
(FOL). This latter says that an arbitrary set of sentences $T$ is satisfiable
provided all of its finite subsets $S\subseteq T$ are satisfiable. This
result, in turn, has several applications in other areas of Mathematics, such
as Ramsey Theory, Algebra or Topology. To mention just one of these, it is a
consequence of the compactness theorem for FOL that every one-to-one
polynomial map $f\colon\mathbb{C}^{n}\rightarrow\mathbb{C}^{n}$ is in fact a
bijection.

The above results witness that the first infinite cardinal $\aleph_{0}$ is
compact. This invites to the following natural question: How about if small
means, e.g., countable (i.e., cardinality ${<}\aleph_{1}$)? Should we still
expect such a garden-variety of examples in favor of compactness of
$\aleph_{1}$? The truth is that this is not the case. To mention one of the
various existing counter-examples, the De Brujin-Erdős’ theorem does not apply
in this context. Indeed, a theorem of Erdős and Hajnal [EH66] shows that, if
the _Continuum Hypothesis_ holds, there is a graph $\mathcal{G}$ of size
$\aleph_{2}$ whose chromatic number is $\aleph_{1}$ but all of its subgraphs
$\mathcal{H}$ with $|\mathcal{H}|<\aleph_{1}$ do have chromatic number
$\leq\aleph_{0}$.

The above counter-example is in fact an evidence that “finite” cannot be
replaced by “countable” in Gődel’s compactness theorem for FOL. Inspired by
this, Keisler and Tarski investigated in [KT64] under what assumptions the
infinitary logic $\mathcal{L}_{\kappa,\kappa}$ does satisfy an analogue of
Gődel’s theorem, provided $\kappa$ is an uncountable cardinal. This lead to
the discovery of _strongly compact cardinals_ , a cornerstone notion in the
theory of large cardinals.

Even though our understanding of the large cardinal hierarchy has vastly
improved since the 60’s [Kan09], some natural problems regarding strongly
compact cardinals are among the most stubborn open questions in set theory.

Mimicking the tight connection between Gődel’s compactness theorem and the
_Boolean Prime Ideal theorem_ , one can characterize the strong compactness of
a cardinal $\kappa$ as follows: every $\kappa$-complete filter $\mathscr{F}$
extends to a $\kappa$-complete ultrafilter $\mathscr{U}$. In [Mit79] Mitchell
asked whether the restricted version of this statement —“every
$\kappa$-complete filter _on $\kappa$_ can be extended to a $\kappa$-complete
ultrafilter”, is witnessed in a model of the form $L[\vec{\mathcal{U}}]$, the
canonical inner model for the axiom
$``\forall\kappa\,(o(\kappa)\leq\kappa^{++})$”.

A negative answer to this question was quickly provided by Gitik in [Git93,
§2], showing that the _consistency strength_ of this statement is at least
that of the existence of a strong cardinal. In his proof, Gitik constructs an
extender of arbitrary length using the filter extension property of a cardinal
$\kappa$ that is $\kappa$-compact. Unfortunately, Gitik’s argument contains a
subtle gap. More specifically, in order to get past the limit steps in his
construction of [Git93, Theorem 2.1] a new compactness property is required –
we have coined this principle the name of the _Gluing property_ :

######  Definition 1.1 (The Gluing Property).

Let $\kappa$ be a measurable cardinal. We say that $\kappa$ has the _$\lambda$
-gluing property_ if for every sequence of $\kappa$-complete ultrafilters on
$\kappa$, $\langle U_{\gamma}\mathrel{|}\allowbreak\gamma<\lambda\rangle$,
there is a $\kappa$-complete extender $E$ and an increasing sequence of
ordinals $\langle\eta_{\gamma}\mathrel{|}\allowbreak\gamma<\lambda\rangle$
such that

$U_{\gamma}=\\{X\subseteq\kappa^{\\{\gamma\\}}\mathrel{|}\allowbreak\\{\langle
j_{E}(\gamma),\eta_{\gamma}\rangle\\}\in j_{E}(X)\\}.$

One may speculate whether it is also possible to glue together various
_extenders_. We will not deal with this issue here.

The gluing property is catching a weakening of the filter extension property.
Instead of trying to extend an arbitrary $\kappa$-complete filter, we are
extending a filter on increasing sequences which has the property that each
projection is already a $\kappa$-complete ultrafilter.

The adoption of the gluing property is inspired by Gitik’s argument in [Git93,
Theorem 2.1]. In fact, Gitik’s argument immediately shows:

######  Theorem (Gitik).

If $\kappa$ has the $\lambda$-gluing property for every cardinal $\lambda$
then there is an inner model with a strong cardinal.

Namely, the assumption “$\kappa$ is a $\kappa$-compact cardinal” in [Git93,
§2] should be replaced by “$\kappa$ has the $\lambda$-gluing property for all
cardinals $\lambda$”.

As our title announces, in this paper we investigate the gluing property. In
§3 we show that $\kappa$-compact cardinals enjoy the $2^{\kappa}$-gluing
property, yet not more than that. More precisely, we prove the following
result:

######  Theorem.

If $\kappa$ has the $\lambda$-filter extension property then $\kappa$ has the
$2^{\lambda}$-gluing property. However, it is consistent that $\kappa$ is a
$\kappa$-compact cardinal which does not have the $(2^{\kappa})^{+}$-gluing
property.

Nevertheless we would like to stress that the $2^{\kappa}$-gluing property
(and hence $\kappa$-compactness) is sufficiently strong to derive another
striking result due to Gitik – the existence of a Prikry type forcing that is
universal for all $\kappa$-distributive forcings of size $\kappa$ (see
[Git20]). In §2 we provide the reader with a self-contained and complete
exposition of this extender-based poset.

Continuing with our study of the gluing property, in §4 we derive the
consistency of the $\omega$-gluing property from the existence of a strong
cardinal:

######  Theorem.

Assume that $\kappa$ is a strong cardinal. Then, there is a cardinal-
preserving generic extension in which $\kappa$ has the $\omega$-gluing
property.

Thus, strongness yields an upper bound for the consistency strength of the
$\omega$-gluing property. Following this vein, we conclude the paper with §§5
and 6 where we compute the consistency strength of the $\omega$-gluing
property:

######  Theorem.

The following statements are equiconsistent:

  1. (1)

$\kappa$ has the $\omega$-gluing property.

  2. (2)

$\kappa$ is a measurable cardinal with $o(\kappa)=\omega_{1}$.

The difference between those two results is that in order to prove the second
one, we work over a suitable core model, with some anti-large cardinal
hypothesis, while the first one is a straight-forward forcing argument.

The set-theoretic notations of this paper are mostly standard. All the new
concepts and terminologies will be timely introduced. Nevertheless, we assume
our readers to be fluent with the theory of elementary embeddings and Prikry-
type forcings. For a comprehensive account in these matters we refer the
reader to [Kan09] and [Git10], respectively.

##  2\. Gitik’s universal extender-based Prikry forcing

An uncountable cardinal $\kappa$ is said to be _$\kappa$ -compact_ if every
$\kappa$-complete filter on $\kappa$ can be extended to a $\kappa$-complete
ultrafilter. In this section we would like to provide a full-detailed account
of the following theorem:

######  Theorem 2.1 (Gitik).

Assume that $\kappa$ is a $\kappa$-compact cardinal. Then, there is a Prikry-
type poset $\mathbb{P}$ forcing the following: for every poset $\mathbb{Q}$
with $|\mathbb{Q}|=\kappa$ and $V\models``\text{$\mathbb{Q}$ is
$\kappa$-distributive}$” there is a $\mathbb{Q}$-generic filter over $V$.

Gitik’s result provides a substantial improvement of other _universality
theorems_ concerning Prikry-type forcings. For instance, it was well known
that if $\kappa$ is ${2}^{\kappa}$-supercompact then the usual _Supercompact
Prikry forcing_ with respect a measure on $\mathcal{P}_{\kappa}(2^{\kappa})$
projects onto any $\kappa$-distributive forcing of size $\kappa$ (see [Git10,
§6.4]). Gitik improved that to a $\kappa$-compact cardinal, while the exact
consistency strength is still unclear. More recent developments in this vein
have been obtained by Benhamou, Hayut and Gitik in [BGH21].

In this paper, however, our interest in Theorem 2.1 relies on the fact that it
hints the gluing property. Succinctly speaking the idea is the following.

Firstly, it suffices to establish the theorem for $\kappa$-distributive
forcing posets (in $V$) whose underlying set is $\kappa$. So, let
$\langle\mathbb{Q}_{\alpha}\mathrel{|}\allowbreak\kappa\leq\alpha<2^{\kappa}\rangle$
be an injective enumeration of all such forcings and
$\langle\mathcal{F}_{\alpha}\mathrel{|}\allowbreak\kappa\leq\alpha<2^{\kappa}\rangle$
be the corresponding filters of dense open sets. All of these are
$\kappa$-complete filters on $\kappa$. Hence, by $\kappa$-compactness of
$\kappa$, each $\mathcal{F}_{\alpha}$ can be extended to a $\kappa$-complete
ultrafilter $U_{\alpha}$. Since $\kappa$-compact cardinals have the
$2^{\kappa}$-gluing property (see Theorem 3.2) we can glue all these measures
into a $(\kappa,2^{\kappa})$-extender $E$. Once this is done we shall define
an extender-based forcing (_Gitik’s Universal Extender-based Prikry forcing_)
that will simultaneously project onto every poset $\mathbb{Q}_{\alpha}.$

We wish to provide a full-detailed exposition of Gitik’s theorem, which first
appeared with fewer details in [Git20]. For this we will use Merimovich’s
framework for extender-based forcings. The core concept in Merimovich setup is
the the notion of object (see e.g., [Mer03, Mer07, Mer11]).

So, put $\mathcal{I}:=[2^{\kappa}\setminus\kappa]^{\leq\kappa}$ and invoke the
$2^{\kappa}$-gluing property of $\kappa$ to produce an extender $E=\langle
E_{a}\mathrel{|}\allowbreak a\in\mathcal{I}\rangle$ such that, for each
$a\in\mathcal{I}$, $E_{a}$ concentrates on

$X_{a}:=\\{\nu\colon a\rightarrow\kappa\mathrel{|}\allowbreak\text{$\nu$ is
order-preserving and $|\nu|<\kappa$}\\}$

and $\\{e_{\alpha}``X\mathrel{|}\allowbreak X\in
E_{\\{\alpha\\}}\\}=U_{\alpha}$, where $e_{\alpha}$ is the map
$e_{\alpha}(\nu):=\nu(\alpha).$

For each $\alpha\in[\kappa,2^{\kappa})$, the seed of $U_{\alpha}$ is denoted
by $\sigma_{\alpha}$; to wit,
$\sigma_{\alpha}:=[\operatorname{id}]_{U_{\alpha}}$.

######  Definition 2.2.

Let $\mathbb{P}^{*}$ be the poset consisting of maps
$f\colon\operatorname{dom}(f)\rightarrow[\kappa]^{<\omega}$, where
$\operatorname{dom}(f)\in\mathcal{I}$, $\kappa\in\operatorname{dom}(f)$ and
for each $\alpha\in\operatorname{dom}(f)$,

$f(\alpha)$ is both $<$-increasing and
$\leq_{\mathbb{Q}_{\alpha}}$-decreasing.

Given $f,g\in\mathbb{P}^{*}$ we write $f\leq g$ if $f\supseteq g.$

For each $f\in\mathbb{P}^{*}$ and $\alpha\in\operatorname{dom}(f)$, denote
$\mu^{f}_{\alpha}:=\max(f(\alpha))$.111When $f$ is clear we shall suppress the
superscript $f$ and just write $\mu_{\alpha}.$

It is clear that $\mathbb{P}^{*}$ is a $\kappa^{+}$-directed-closed forcing.

######  Definition 2.3 (Objects).

Fix $f\in\mathbb{P}^{*}$. A map
$\nu\colon\operatorname{dom}(\nu)\rightarrow\kappa$ is called an _$f$ -object_
if it has the following properties:

  1. (1)

$|\nu|\leq\nu(\kappa)$;

  2. (2)

$\operatorname{dom}(\nu)\in[\operatorname{dom}(f)]^{<\kappa}$ and
$\kappa\in\operatorname{dom}(\nu)$;

  3. (3)

for all $\alpha\in\operatorname{dom}(\nu)$, $\mu_{\alpha}<\nu(\alpha)$ and
$\nu(\alpha)\leq_{\mathbb{Q}_{\alpha}}\mu_{\alpha}.$

The set of $f$-objects is denoted by $\operatorname{OB}(f)$.

Given $\nu,\eta\in\operatorname{OB}(f)$, write $\nu\prec\eta$ iff
$\operatorname{dom}(\nu)\subseteq\operatorname{dom}(\eta)$ and for every
$\alpha\in\operatorname{dom}(\nu)$, $\nu(\alpha)<\eta(\kappa)$ and
$\eta(\alpha)\leq_{\mathbb{Q}_{\alpha}}\nu(\alpha)$.

######  Definition 2.4 (Measures).

Given $f\in\mathbb{P}^{*}$ define $E(f)$ as follows:

$X\in
E(f)\;\;\Longleftrightarrow\;\;X\subseteq\operatorname{OB}(f)\;\wedge\;\sigma_{f}\in
j_{E}(X),$

where $\sigma_{f}:=\\{\langle
j_{E}(\alpha),\sigma_{\eta}\rangle\mathrel{|}\allowbreak\alpha\in\operatorname{dom}(f)\\}$
and $\sigma_{\eta}$ denotes the seed of the measure $U_{\eta}$, where $\eta$
is the unique ordinal such that
$\mathbb{Q}_{\alpha}/\mu_{\alpha}=\mathbb{Q}_{\eta}$, where
$\mathbb{Q}_{\alpha}/\mu_{\alpha}$ is the forcing that consists of all
conditions stronger than $\mu_{\alpha}$ in $\mathbb{Q}_{\alpha}$.222This make
sense because $\mathbb{Q}_{\alpha}/\mu_{\alpha}$ is $\kappa$-distributive of
size $\kappa$ and our enumeration is injective.

It is automatic that $E(f)$ is a $\kappa$-complete measure on
$\operatorname{OB}(f)$.

######  Definition 2.5.

Let $f\in\mathbb{P}^{*}$ and $\nu\in\operatorname{OB}(f)$. Then,
$f_{\langle\nu\rangle}$ denotes the function with domain
$\operatorname{dom}(f)$ such that for all $\alpha\in\operatorname{dom}(f)$,

$f_{\langle\nu\rangle}(\alpha):=\begin{cases}f(\alpha)^{\smallfrown}\langle\nu(\alpha)\rangle,&\text{if
$\alpha\in\operatorname{dom}(\nu)$;}\\\
f(\alpha),&\text{otherwise.}\end{cases}$

In general, if $\vec{\nu}$ is a finite $\prec$-increasing sequence in
$\operatorname{OB}(f)$, the function $f_{\vec{\nu}}$ is recursively defined as
$f_{\vec{\nu}}:=(f_{\vec{\nu}\mathbin{\upharpoonright}|\vec{\nu}|})_{\langle\vec{\nu}_{|\vec{\nu}|-1}\rangle}.$

######  Definition 2.6 ($f$-Trees).

A set $T$ of finite $\prec$-increasing sequences in $\operatorname{OB}(f)$ is
called an _$f$ -tree_ if it is a tree with respect to end-extensions333I.e.,
if $\vec{\eta}\in T$ then $\vec{\eta}\mathbin{\upharpoonright}n\in T$ for all
$n\leq|\vec{\eta}|.$ and

$\mathrm{Succ_{T}}(\vec{\nu}):=\\{\eta\in\operatorname{OB}(f)\mathrel{|}\allowbreak\max_{\prec}\vec{\nu}\prec\eta\;\wedge\;\vec{\nu}\,^{\smallfrown}\langle\eta\rangle\in
T\\}\in E(f_{\vec{\nu}}),$

for all $\vec{\nu}\in T\cup\\{\langle\rangle\\}$.

For an $f$-tree $T$ and $\vec{\nu}\in T$, denote
$T_{\vec{\nu}}:=\\{\vec{\eta}\in
T\mathrel{|}\allowbreak\vec{\nu}{}^{\smallfrown}\vec{\eta}\in T\\}$.

######  Remark 2.7.

For every $f$-tree $T$ and every $\vec{\nu}\in T$, $T_{\vec{\nu}}$ is again an
$f_{\vec{\nu}}\,$-tree.

Given $f,g\in\mathbb{P}^{*}$ with $f\leq g$, denote by
$\pi_{\operatorname{dom}(f^{p}),\operatorname{dom}(f^{q})}$ the restriction
map from $\operatorname{OB}(f^{p})$ to $\operatorname{OB}(f^{q})$; namely, the
map $\nu\mapsto\nu\mathbin{\upharpoonright}\operatorname{dom}(f^{q})$.

######  Definition 2.8.

Denote by $\mathbb{P}$ the set of all pairs $p=\langle f^{p},T^{p}\rangle$
such that $f^{p}\in\mathbb{P}^{*}$ and $T^{p}$ is an $f^{p}$-tree. Given
$p,q\in\mathbb{P}$, we write $p\leq^{*}q$ iff $f^{p}\supseteq f^{q}$ and
$\pi_{\operatorname{dom}(f^{p}),\operatorname{dom}(f^{q})}``T^{p}\subseteq
T^{q}$.

######  Lemma 2.9.

Suppose $f,g\in\mathbb{P}^{*}$ are such $f\leq g$ and $T$ is a $g$-tree. Then,

$T^{*}:=\\{\langle\nu_{0},\dots,\nu_{n-1}\rangle\in\operatorname{OB}(f)^{<\omega}\mathrel{|}\allowbreak\langle\nu_{0}\mathbin{\upharpoonright}\operatorname{dom}(g),\dots,\nu_{n-1}\mathbin{\upharpoonright}\operatorname{dom}(g)\rangle\in
T\\}$

is an $f$-tree such that
$\pi_{\operatorname{dom}(f),\operatorname{dom}(g)}``T^{*}\subseteq T$.

###### Proof.

Clearly, $T^{*}$ is a tree under end-extension. Fix, $\vec{\nu}\in
T\cup\\{\langle\rangle\\}$.

$\blacktriangleright$ If $\vec{\nu}=\langle\rangle$ then
$``\mathrm{Succ}_{T^{*}}(\langle\rangle)\in E(f)$” is equivalent to
$\sigma_{f}\in j(T^{*})$. Observe that
$\sigma_{f}\mathbin{\upharpoonright}j(\operatorname{dom}(g))=\sigma_{f}\mathbin{\upharpoonright}j``\operatorname{dom}(g)=\sigma_{g}\in
j(T)$, hence $\sigma_{f}\in j(T^{*})$.

$\blacktriangleright$ Let $\vec{\nu}=\langle\nu_{0},\dots,\nu_{n-1}\rangle$.
We need to check that $\max_{\prec}j(\vec{\nu})\prec\sigma_{f}$ and
$\sigma_{f}\in j(T^{*})$. The latter has been proved before. About the former,
$j(\vec{\nu})=\langle j(\nu_{0}),\dots,j(\nu_{n-1})\rangle$, where
$\operatorname{dom}(j(\nu_{n-1}))=j``\operatorname{dom}(\nu_{n-1})$ and
$j(\nu_{n-1})(j(\alpha))=\nu_{n-1}(\alpha)<\kappa\leq\sigma_{f}(j(\kappa))$.
Thus, $j(\nu_{n-1})\prec\sigma_{f}$. ∎

######  Definition 2.10 (One point extensions).

Let $p\in P$ and $\langle\nu\rangle\in T^{p}$. We denote by
$p{}^{\curvearrowright}\langle\nu\rangle$ the pair $\langle
f^{p}_{\langle\nu\rangle},T^{p}_{\langle\nu\rangle}\rangle$. For every
$\vec{\nu}\in T^{p}$, $p{}^{\curvearrowright}\vec{\nu}$ is defined recursively
in the natural way. Also, by convetion,
$p{}^{\curvearrowright}\langle\rangle:=p$.

######  Lemma 2.11.

For each $p\in\mathbb{P}$ and $\vec{\nu}\in T^{p}\cup\\{\langle\rangle\\}$,
$p{}^{\curvearrowright}\vec{\nu}\in\mathbb{P}$.

###### Proof.

Let us prove the lemma by induction on $n$, the length of $\vec{\nu}$.

If $n=0$ the result follows from the very definition of
$p{}^{\curvearrowright}\langle\rangle$. So, suppose the above property holds
for all $\vec{\nu}\in T^{p}$ with $|\vec{\nu}|=n$. Choose
$\langle\nu_{0},\dots,\nu_{n}\rangle\in{}^{n+1}T$ and set
$\vec{\eta}:=\langle\nu_{0},\dots,\nu_{n-1}\rangle$. By the induction
hypothesis, $q:=\langle
f^{p}_{\vec{\eta}},T^{p}_{\vec{\eta}}\rangle\in\mathbb{P}$. Also, by
definition, $\langle\nu_{n}\rangle\in T^{q}$. We thus need to show that
$q{}^{\curvearrowright}\langle\nu_{n}\rangle\in\mathbb{P}$.

On one hand, it is clear that $T^{q}_{\langle\nu_{n}\rangle}$ is an
$f^{q}_{\langle\nu_{n}\rangle}$-tree. On the other hand,
$f^{q}_{\langle\nu_{n}\rangle}\in\mathbb{P}^{*}$: Indeed, $T^{q}$ is a
$f^{p}_{\vec{\eta}}\,$-tree, hence
$\nu_{n}\in\operatorname{OB}(f^{p}_{\vec{\eta}})$, and thus

$\nu_{n}(\alpha)>\max(f^{p}_{\vec{\eta}}(\alpha))=\mu^{q}_{\alpha}\;\;\text{and}\;\;\nu_{n}(\alpha)\leq_{\mathbb{Q}_{\alpha}}\mu^{q}_{\alpha},$

for $\alpha\in\operatorname{dom}(\nu)$. Thereby,
$f^{q}_{\langle\nu_{n}\rangle}$ witnesses the clauses of Definition 2.2. ∎

######  Definition 2.12 (The main poset).

_Gitik’s universal Extender based forcing_ is the set $\mathbb{P}$ endowed
with the following order: for two conditions $p,q\in\mathbb{P}$, we write
$p\leq q$ iff there is $\vec{\nu}\in T^{q}\cup\\{\langle\rangle\\}$ such that
$p\leq^{*}q{}^{\curvearrowright}\vec{\nu}$.

######  Notation 2.13.

Let $m<\omega$ and $p,q\in\mathbb{P}$. We write $p\leq^{m}q$ as a shorthand
for $p\leq q$ and $|\vec{\nu}|=m$, where $\vec{\nu}$ is some sequence
witnessing $p\leq^{*}q{}^{\curvearrowright}\vec{\nu}.$

Note that, under this convention, $\leq^{0}=\leq^{*}.$

######  Lemma 2.14.

$\langle\mathbb{P},\leq^{*}\rangle$ is $\kappa$-directed-closed.

###### Proof.

Let $D=\\{p_{\alpha}\mathrel{|}\allowbreak\alpha<\theta\\}$ be a
$\leq^{*}$-directed set of conditions in $P$ with $\theta<\kappa$. Set
$f:=\bigcup_{\alpha<\theta}f^{p_{\alpha}}$. Since $D$ is $\leq^{*}$-directed
it is immediate that $f\in P^{*}$. Put,
$T:=\bigcup_{\alpha<\theta}\pi_{\operatorname{dom}(f),\operatorname{dom}(f^{p_{\alpha}})}^{-1}T^{p_{\alpha}}$.
Arguing as in Lemma 2.9 one can check that $T$ is a
$\operatorname{dom}(f)$-tree. Altogether, $p:=\langle f,T\rangle\in P$ yields
a $\leq^{*}$-lower bound for the set $D$. ∎

######  Lemma 2.15.

$\mathbb{P}$ has the $\kappa^{++}$-cc.

###### Proof.

This is as in [Mer07, Proposition 3.17]. ∎

The proof of the _Strong Prikry Property_ is essentially the same as that from
[Mer03]. We reproduce it just for completeness and for the reader’s benefit.
Let us begin with an useful lemma about _integration_ of conditions:

######  Lemma 2.16.

Let $p\in\mathbb{P}$ and $D\colon T^{p}\rightarrow V$ be a function such that,
for each $\vec{\nu}\in T$, $D(\vec{\nu})$ is a $\leq^{*}$-dense open set below
$p{}^{\curvearrowright}\vec{\nu}$. Then, there is $p^{*}\leq^{*}p$ such that
for each $\vec{\mu}\in T^{p^{*}}$, $p^{*}{}^{\curvearrowright}\vec{\mu}\in
D(\vec{\mu}\mathbin{\upharpoonright}\operatorname{dom}(f^{p})).$

###### Proof.

Fix $\chi$ a big enough regular cardinal and let $N\prec H_{\chi}$ be such
that $|N|=\kappa$, $N^{<\kappa}\subseteq N$ and
$\\{p,\mathbb{P},\mathbb{P}^{*},\\{D(\vec{\nu})\mathrel{|}\allowbreak\vec{\nu}\in
T^{p}\\}\\}\subseteq N$. Using a properness argument one can show that there
is $f^{*}\leq f^{p}$ that is totally $\langle
N,\mathbb{P}^{*}\rangle$-generic; namely, for every $E\subseteq\mathbb{P}^{*}$
dense open set, $E\in N$, there is $h\in N\cap E$ with $f^{*}\leq h$. For each
$\vec{\mu}\in\operatorname{OB}(\operatorname{dom}(f^{*}))$ define
$D^{*}(\vec{\mu})$ as

$\\{f\leq f^{p}\mathrel{|}\allowbreak\exists
q\in\mathbb{P}\,(f^{q}=f_{\vec{\mu}}\,\wedge\,q\leq^{*}p{}^{\curvearrowright}\langle\vec{\mu}\mathbin{\upharpoonright}\operatorname{dom}(f^{p})\rangle\,\wedge\,q\in
D(\vec{\mu}\mathbin{\upharpoonright}\operatorname{dom}(f^{p})))\\}.$

Clearly, $D^{*}(\vec{\mu})\in N$ (because $\vec{\mu}\in N$) and it is an open
subset of $\mathbb{P}^{*}$. In addition, it is not hard to check that
$D^{*}(\vec{\mu})$ is dense below $f^{p}$. Thereby we have that
$D^{*}(\vec{\mu})\cup\\{f\in\mathbb{P}^{*}\mathrel{|}\allowbreak f\perp
f^{p}\\}$ is a dense subset of $\mathbb{P}^{*}$ in $N$ and so, by properness,
there is $h\in N$ in this set with $f^{*}\leq h$, Thus, $h\in
D^{*}(\vec{\mu})$.

The above shows that, for each
$\vec{\mu}\in\operatorname{OB}(\operatorname{dom}(f^{*}))$, there is a map
$h_{\vec{\mu}}$ in $D^{*}(\vec{\mu})\cap N$, which amounts to saying that
there is a condition $\langle f_{\vec{\mu}},T^{\vec{\mu}}\rangle$ in the dense
open set $D(\vec{\mu}\mathbin{\upharpoonright}\operatorname{dom}(f^{p}))$ such
that $\langle
f_{\vec{\mu}},T^{\vec{\mu}}\rangle\leq^{*}p{}^{\curvearrowright}\langle\vec{\mu}\mathbin{\upharpoonright}\operatorname{dom}(f^{p})\rangle$.

We would like to integrate all these trees $T^{\vec{\mu}}$ into a single one
$T^{*}$ for which $\langle f^{*},T^{*}\rangle\leq^{*}\langle
f^{p},T^{p}\rangle$ and, for each $\vec{\mu}\in T^{*}$, $\langle
f^{*},T^{*}\rangle{}^{\curvearrowright}\vec{\mu}\leq^{*}\langle
f_{\vec{\mu}},T^{\vec{\mu}}\rangle$.

Let
$T^{*}:=\pi_{\operatorname{dom}(f^{*}),\operatorname{dom}(f^{p})}^{-1}T^{p}\cap\bigtriangleup_{\vec{\mu}\in\operatorname{OB}(\operatorname{dom}(f^{*}))}T^{\vec{\mu}}$,
where this latter tree (call it $\bigtriangleup T$) is defined by induction so
as to ensure that, for each $\vec{\mu}\in\bigtriangleup T$, $(\bigtriangleup
T)_{\vec{\mu}}=T^{\vec{\mu}}$. It should be clear that $T^{*}$ is the sought
tree. ∎

######  Lemma 2.17 (Strong Prikry property).

For each $p\in\mathbb{P}$ and $D\subseteq\mathbb{P}$ dense open there is a
condition $p^{*}\leq^{*}p$ and an integer $n<\omega$ such that, for each
$m\geq n$, if $q\leq^{m}p^{*}$ then $q\in D$.

###### Proof.

Let $p\in\mathbb{P}$ and $D\subseteq\mathbb{P}$ be as above. The next is the
crux of the matter:

######  Claim 2.17.1.

There is $p^{*}\leq^{*}p$ such that

  1. (1)

either $``p^{*}{}^{\curvearrowright}\vec{\mu}\in D$” for all $\vec{\mu}\in
T^{p^{*}}$,

  2. (2)

or $``\forall q\leq^{*}p{}^{\curvearrowright}\vec{\mu}\,(q\notin D)$” for all
$\vec{\mu}\in T^{p^{*}}$.

###### Proof of claim.

For each $\vec{\nu}\in T^{p}$ put

$D^{0}(\vec{\nu}):=\\{q\leq^{*}p{}^{\curvearrowright}\vec{\nu}\mathrel{|}\allowbreak
q\in D\\},$

and

$D^{1}(\vec{\nu}):=\\{r\leq^{*}p{}^{\curvearrowright}\vec{\nu}\mathrel{|}\allowbreak\forall
q\leq^{*}r\,(q\notin D)\\}.$

For each $\vec{\nu}\in T^{p}$, $D(\vec{\nu}):=D^{0}(\vec{\nu})\cup
D^{1}(\vec{\nu})$ is $\leq^{*}$-dense open below
$p{}^{\curvearrowright}\vec{\nu}$.

Applying the previous claim we find $p^{*}\leq^{*}p$ such that, for each
$\vec{\mu}\in T^{p^{*}}$, $p^{*}{}^{\curvearrowright}\vec{\mu}\in
D(\vec{\mu}\mathbin{\upharpoonright}\operatorname{dom}(f^{p}))$. Let us
define, by induction, a tree $T^{*}\subseteq T^{p^{*}}$ such that $\langle
f^{p^{*}},T^{*}\rangle$ is a condition $\leq^{*}$-below $p$ witnessing the
claim.

Firstly, either the set

$\\{\mu\in\operatorname{OB}(\operatorname{dom}(f^{p^{*}}))\mathrel{|}\allowbreak
p^{*}{}^{\curvearrowright}\langle\mu\rangle\in
D^{0}(\mu\mathbin{\upharpoonright}\operatorname{dom}(f))\\}$

is $E(\operatorname{dom}(f^{p^{*}}))$-large or so is its complement,

$\\{\mu\in\operatorname{OB}(\operatorname{dom}(f^{*}))\mathrel{|}\allowbreak
p^{*}{}^{\curvearrowright}\langle\mu\rangle\in
D^{1}(\mu\mathbin{\upharpoonright}\operatorname{dom}(f))\\}.$

Shrink the first level of $T^{p^{*}}$ to such
$E(\operatorname{dom}(f^{p^{*}}))$-large set and for each $\mu$ in there let
$(T^{1})_{\mu}:=T^{p^{*}}_{\mu}$. This defines another
$E(\operatorname{dom}(f^{p^{*}}))$-tree $T^{1}\subseteq T^{p^{*}}$ which is
our first approximation to $T^{*}$.

The second approximation to $T^{*}$ is defined as follows: for each object
$\mu\in\mathrm{Lev}_{1}(T^{1})$ look at

$X^{0}_{\mu}:=\\{\nu\in\operatorname{OB}(\operatorname{dom}(f^{*}))\mathrel{|}\allowbreak\mu\prec\nu\,\wedge\,p^{*}{}^{\curvearrowright}\langle\mu,\nu\rangle\in
D^{0}(\langle\mu,\nu\rangle\mathbin{\upharpoonright}\operatorname{dom}(f))\\}$

and at its complement

$X^{1}_{\mu}:=\\{\nu\in\operatorname{OB}(\operatorname{dom}(f^{*}))\mathrel{|}\allowbreak\mu\prec\nu\,\wedge\,p^{*}{}^{\curvearrowright}\langle\mu,\nu\rangle\in
D^{1}(\langle\mu,\nu\rangle\mathbin{\upharpoonright}\operatorname{dom}(f))\\}$

Denote by $X_{\mu}$ the unique of the above sets that is a
$E(\operatorname{dom}(f^{*}))$-large. Set
$\\{\mu\in\mathrm{Lev}_{1}(T^{1})\mathrel{|}\allowbreak
X_{\mu}=X^{0}_{\mu}\\}$ and
$\\{\mu\in\mathrm{Lev}_{1}(T^{1})\mathrel{|}\allowbreak
X_{\mu}=X^{1}_{\mu}\\}$. At least one of them is large, call it $X^{i}$, and
stipulate that the first level of $T^{2}$ to be $X$. For each $\mu\in X$,
$(T^{2})_{\mu}:=X^{i}_{\mu}$ and for each $\langle\mu,\nu\rangle\in
X^{i}\times X^{i}_{\mu}$,
$(T^{2})_{\langle\mu,\nu\rangle}:=(T^{1})_{{\langle\mu,\nu\rangle}}$. This way
we define our second approximation, $T^{2}$.

Proceeding in this vein we generate an inclusion-decreasing sequence of
$E(\operatorname{dom}(f^{p^{*}}))$-trees, $\langle T^{n}\mathrel{|}\allowbreak
n\geq 1\rangle$, so that $T^{*}:=\bigcap_{1\leq n<\omega}T^{n}$ is the sought
tree. ∎

We are now ready to prove the lemma. Since $D$ is dense there must be some
$q\leq p^{*}$ in $D$. Then Clause (2) above fails and so for every
$\vec{\mu}\in[T^{*}]^{n}$, $p^{*}{}^{\curvearrowright}\vec{\mu}\in D$. The
claim follows right away from this and openess of $D$. ∎

######  Corollary 2.18 (Prikry Property).

For each $p\in\mathbb{P}$ and every sentence $\varphi$ in the language of
forcing there is $p^{*}\leq^{*}p$ such that $p^{*}$ decides $\varphi$.

Let $G$ a $\mathbb{P}$-generic filter over $V$. For each
$\alpha\in[\kappa,2^{\kappa})$ define

$G_{\alpha}:=\bigcup\\{f^{p}(\alpha)\mathrel{|}\allowbreak p\in
G\,\wedge\,\alpha\in\operatorname{dom}(f^{p})\\}.$

For each $\alpha\in[\kappa,2^{\kappa})$, $\sup(G_{\alpha})=\kappa$ and
$\operatorname{otp}(G_{\alpha})=\omega$. Also, $G_{\alpha}\neq G_{\beta}$, for
all $\alpha,\beta\in[\kappa,2^{\kappa})$ with $\beta\neq\alpha$. The following
lemma shows that, indeed, $\mathbb{P}$ produces the configuration described in
Theorem 2.1.

######  Lemma 2.19 (Universality).

For each $\alpha\in[\kappa,2^{\kappa})$, $G_{\alpha}$ is eventually contained
in every dense open subset of $\mathbb{Q}_{\alpha}$ lying in $V$. In
particular,

$\\{q\in\mathbb{Q}_{\alpha}\mathrel{|}\allowbreak\exists p\in
G_{\alpha}\,(p\leq_{\mathbb{Q}_{\alpha}}q)\\}$

defines a $\mathbb{Q}_{\alpha}$-generic filter over $V$.

###### Proof.

Fix a dense open set $D\subseteq\mathbb{Q}_{\alpha}$ and set

$E:=\\{p\in\mathbb{P}\mathrel{|}\allowbreak\alpha\in\operatorname{dom}(f^{p})\,\wedge\,\forall\langle\nu\rangle\in\mathrm{Succ}_{T^{p}}(\langle\rangle)\;(\nu(\alpha)\in
D)\\}.$

We claim that $E$ is dense: Let $p\in\mathbb{P}$ and assume that
$\alpha\in\operatorname{dom}(f^{p})$. Since $T^{p}$ in an $f^{p}$-tree,
$\mathrm{Succ}_{T^{p}}(\langle\rangle)\in E(f^{p})$. Let
$\eta\in[\kappa,2^{\kappa})$ be the unique ordinal such that
$\mathbb{Q}_{\eta}=\mathbb{Q}_{\alpha}/\mu_{\alpha}$. Note that
$U_{\eta}\leq_{\mathrm{RK}}E(f^{p})$, as witnessed by the map
$e_{\alpha}\colon\langle\nu\rangle\mapsto\nu(\alpha)$. In particular,
$e_{\alpha}``\mathrm{Succ}_{T^{p}}(\langle\rangle)\in U_{\eta}$ and thus the
set $A:=e_{\alpha}``\mathrm{Succ}_{T^{p}}(\langle\rangle)\cap D$ is
$U_{\eta}$-large. Pulling $A$ back with $e_{\alpha}$ one obtains a set
$A^{\prime}\subseteq\mathrm{Succ}_{T^{p}}(\langle\rangle)$ with $A^{\prime}\in
E(f^{p})$ such that for all $\langle\nu\rangle\in A$, $\nu(\alpha)\in D$.
Repeating this process by induction on the height of the tree $T^{p}$ one gets
an $f^{p}$-tree $T^{\prime}$ such that $T^{\prime}\subseteq T^{p}$ and
$\mathrm{Succ}_{T^{\prime}}(\langle\rangle)=A^{\prime}$. Clearly,
$p^{\prime}:=\langle f^{p},T^{\prime}\rangle$ is a condition in $\mathbb{P}$
with $p^{\prime}\leq^{0}p$ and $p^{\prime}\in E$.

Let $p\in G\cap E$. Since
$\\{p{}^{\curvearrowright}\langle\nu\rangle\mathrel{|}\allowbreak\langle\nu\rangle\in
T^{p}\\}$ forms a maximal antichain below $p$ there is some $f^{p}$-object
$\nu$ such that $p{}^{\curvearrowright}\langle\nu\rangle\in G$, hence

$\nu(\alpha)=f^{p{}^{\curvearrowright}\langle\nu\rangle}(\alpha)\in
G_{\alpha}\cap D.$

Say $\nu(\alpha)$ is the $n^{\mathrm{th}}$-member of $G_{\alpha}$. By the very
definition of the forcing it turns out that all the members of $G_{\alpha}$
past stage $n$ are $\leq_{\mathbb{Q}_{\alpha}}$-stronger than $\nu(\alpha)$
and thus, by openess of $D$, $G_{\alpha}\subseteq^{*}D$. ∎

The above concludes the proof of Gitik’s theorem (Theorem 2.1).

##  3\. (Sub)compactness and the gluing property

###  3.1. The filter extension property and the gluing property

In this section we analyze the implications between the $\lambda$-filter
extension property and the gluing property. In particular, we show that if
$\kappa$ is $\kappa$-compact then $\kappa$ has the $2^{\kappa}$-gluing
property. As we will demonstrate this implication is optimal, for there are
$\kappa$-compact cardinals without the $(2^{\kappa})^{+}$-gluing property.
Recall that a cardinal $\kappa$ is said to have the _$\lambda$ -filter
extension property_ (for $\lambda\geq\kappa$) if every $\kappa$-complete
filter on $\lambda$ extends to a $\kappa$-complete ultrafilter.

We commence with a strengthening of [Hay19, Corollary 6]:

######  Lemma 3.1.

Let $\kappa\leq\lambda$ be uncountable cardinals with
$\lambda^{<\kappa}=\lambda$. Then the following two are equivalent:

  1. (1)

$\kappa$ has the $\lambda$-filter-extension property;

  2. (2)

for every transitive model $M$ with $2^{\lambda}\subseteq M$,
$|M|=2^{\lambda}$ and ${}^{\kappa}M\subseteq M$ there is $N$ and an elementary
embedding $j\colon M\rightarrow N$ such that:

    1. (a)

$\operatorname{crit}(j)=\kappa$;

    2. (b)

${}^{\kappa}N\subseteq N$;

    3. (c)

$s\in N$, $j``2^{\lambda}\subseteq s$ and $N\models|s|<j(\kappa)$.

###### Proof.

The implication $(2)\Rightarrow(1)$ follows from [Hay19, Corollary 6]. Assume
that $\kappa$ has the $\lambda$-filter-extension property and let $M$ be as
above. In [Hay19, Corollary 6] it is shown that there is a transitive model
$\mathcal{N}$, $\bar{s}\in\mathcal{N}$ and an elementary embedding $i\colon
M\rightarrow\mathcal{N}$ witnessing (a) and (c) above. We shall now slightly
modify both $\mathcal{N}$ and $i$ in order to obtain the desired outcome.

Define $\mathcal{X}:=\\{i(f)(\kappa,\bar{s})\mathrel{|}\allowbreak f\in M\\}$.
Following Hamkins, [Ham97], $\mathcal{X}\prec\mathcal{N}$. In particular,
$\mathcal{X}$ is well-founded. Let
$\pi\colon\mathcal{X}\rightarrow\mathcal{N}$ be the Mostowski collapse, and
put $j:=\pi\circ i$, $s:=\pi(\bar{s})$ and $N:=\pi``\mathcal{X}$. Note that
$\pi(\kappa)=\kappa$ and

$N=\pi``\\{i(f)(\kappa,\bar{s})\mathrel{|}\allowbreak f\in
M\\}=\\{j(f)(\kappa,s)\mathrel{|}\allowbreak f\in M\\}.$

Clearly, $s\in N$, $j``2^{\lambda}\subseteq s$ and $N\models|s|<j(\kappa)$.
Next, we check Clause (b).

Let $\vec{x}=\langle
x_{\alpha}\mathrel{|}\allowbreak\alpha<\kappa\rangle\subseteq N$. For each
$\alpha<\kappa$, $x_{\alpha}=j(f_{\alpha})(\kappa,s)$ for some $f_{\alpha}\in
M$. Observe that establishing $\vec{x}\in N$ amounts to verify that $\langle
j(f_{\alpha})\mathrel{|}\allowbreak\alpha<\kappa\rangle\in N$. This is indeed
the case: let $h\colon\kappa\times M\rightarrow M$ be defined as
$(\beta,\bar{s})\mapsto\langle
f_{\alpha}(\beta)\mathrel{|}\allowbreak\alpha<\beta\rangle$. Since
${}^{\kappa}M\subseteq M$, the sequence $\langle
f_{\alpha}\mathrel{|}\allowbreak\alpha<\kappa\rangle\in M$ and therefore $h\in
M$. So,

$j(h)(\kappa,s)=\langle
j(f_{\alpha})(\kappa)\mathrel{|}\allowbreak\alpha<\kappa\rangle=\langle
x_{\alpha}\mathrel{|}\allowbreak\alpha<\kappa\rangle\in N.\qed$

######  Theorem 3.2.

Let $\kappa\leq\lambda$ be uncountable cardinals with
$\lambda^{<\kappa}=\lambda$.

If $\kappa$ has the $\lambda$-filter-extension property then it has the
$2^{\lambda}$-gluing property.

###### Proof.

Let $\langle U_{\alpha}\mathrel{|}\allowbreak\alpha<2^{\lambda}\rangle$ be a
sequence of measures on $\kappa$ and consider $\mathcal{X}:=\langle
U_{\alpha}\mathrel{|}\allowbreak\alpha<2^{\lambda}\rangle\cup\\{A\mathrel{|}\allowbreak
A\in U_{\alpha},\,\alpha<2^{\lambda}\\}$. By coding $\mathcal{X}$
appropriately we may assume that $\mathcal{X}\subseteq 2^{\lambda}$. Let $M$
be a transitive model as in the statement of Lemma 3.1. Using the
$\lambda$-filter-extension property of $\kappa$ and Lemma 3.1 we find a
transitive model $N$ with ${}^{\kappa}N\subseteq N$ and $s\in N$, and an
elementary embedding $j\colon M\rightarrow N$ with
$\operatorname{crit}(j)=\kappa$, $j``2^{\lambda}\subseteq N$ and
$N\models|s|<j(\kappa)$. Note that $j``\mathcal{X}\subseteq
j``2^{\lambda}\subseteq s$ hence $j``U_{\alpha}\subseteq s\cap j(U_{\alpha})$
for all $\alpha<2^{\lambda}$. Also, $j(U_{\alpha})\in N$, so that $s\cap
j(U_{\alpha})\in N$ is a subset of $j(U_{\alpha})$ of cardinality
${<}j(\kappa)$. Since $j(U_{\alpha})$ is a $j(\kappa)$-complete measure in $N$
we conclude that $\bigcap(j(U_{\alpha})\cap s)\neq\emptyset$. For each
$\alpha<2^{\lambda}$ pick $s_{\alpha}$ be a witness for it. Note that
$s_{\alpha}\in\bigcap j``U_{\alpha}$.

Let us show that we can pick $s_{\alpha}$ to be increasing. Indeed, in $N$, we
can apply this procedure by recursion for each measure of the form
$j(U)_{\zeta}$, $\zeta\in s$. Namely, for each $\zeta\in s\cap
j(2^{\lambda})$, pick $\tilde{s}_{\alpha}\in\bigcup(j(U)_{\zeta}\cap s)$,
larger than $\sup\\{\tilde{s}_{\eta}\mathrel{|}\allowbreak\eta\in
s\cap\zeta\\}$. Since $N\models|s|<j(\kappa)$, this is possible. Finally, we
let $s_{\alpha}=\tilde{s}_{j(\alpha)}$.

Next, let us form the corresponding gluing extender. For each
$a\in[2^{\lambda}]^{\leq\kappa}$, let $\nu_{a}:=\\{\langle
j(\alpha),s_{\alpha}\rangle\mathrel{|}\allowbreak\alpha\in a\\}$. Since
${}^{\kappa}N\subseteq N$, $\nu_{a}\in N$. Define

$X\in E_{a}\;\Longleftrightarrow\;X\subseteq X_{a}\;\wedge\,\nu_{a}\in j(X),$

where $X_{a}:=\\{f\colon a\rightarrow\kappa\mathrel{|}\allowbreak\text{$f$
increasing}\,\wedge\,|f|<\kappa\\}$.

This yields a $\kappa$-complete ultrafilter on $X_{a}$ such that $E_{a}\in
M$.444Note that here we need to use that $2^{\lambda}\subseteq M$ and
${}^{\kappa}M\subseteq M$. Put $E:=\langle E_{a}\mathrel{|}\allowbreak
a\in[2^{\lambda}]^{\leq\kappa}\rangle$ and observe that for each
$\alpha<2^{\lambda}$ and $A\in U_{\alpha}$,
$\hat{A}:=\\{\\{\langle\alpha,\gamma\rangle\\}\mathrel{|}\allowbreak\gamma\in
A\\}$ is a member of $E_{\\{\alpha\\}}$ as needed. ∎

Combining Theorem 3.2 with [Hay19, Theorem 11] we get:

######  Corollary 3.3.

Let $\kappa\leq\lambda$ be uncountable cardinals with
$\lambda^{<\kappa}=\lambda$.

If $\kappa$ is $\lambda^{+}$-$\Pi^{1}_{1}$-subcompact then it has the
$2^{\lambda}$-gluing property.

In the next subsection we show that the above result is (consistently)
optimal, by exhibiting a generic extension where $\kappa$ is
$\kappa^{+}$-$\Pi^{1}_{1}$-subcompact but it does not have the
$(2^{\kappa})^{+}$-gluing property.

###  3.2. $\Pi^{1}_{1}$-subcompactness and the gluing property

The next auxiliary result will be instrumental later on:

######  Lemma 3.4.

If $\kappa$ is a $\kappa^{+}$-$\Pi^{1}_{1}$-subcompact cardinal then
$o(\kappa)=(2^{\kappa})^{+}$.

###### Proof.

The proof mimics Solovay’s original argument proving the analogous property
for supercompact cardinals. The following is the key claim:

######  Claim 3.4.1.

For $X\subseteq\mathcal{P}(\kappa)$ there is a normal measure $U$ on $\kappa$
such that

$X\in\mathrm{Ult}(V,U).$

###### Proof of claim.

Assume towards contradiction that for some $X\subseteq\mathcal{P}(\kappa)$ the
above statement is false. Then the formula defined by

$\Phi(X,\kappa)\equiv\forall\,U\,(\text{$U$ normal measure on
$\kappa\,\Longrightarrow\,X\notin\mathrm{Ult}(H(\kappa^{+}),U)$)}$

is a true $\Pi^{1}_{1}$-sentence in $\langle H(\kappa^{+}),\in,X\rangle$.
Indeed, one can formulate the statement $X\notin\mathrm{Ult}(H(\kappa^{+}),U)$
as a first order statement in the structure $\langle
H(\kappa^{+}),\in,X,U\rangle$ as follows: for every function $f$ with domain
$\kappa$ such that $f(\alpha)\subseteq\mathcal{P}(\alpha)$ and for every
$y\subseteq\kappa$, $y\in X$ if and only if
$\\{\alpha<\kappa\mathrel{|}\allowbreak y\cap\alpha\in f(\alpha)\\}\in U$.
Using that $\kappa$ is $\kappa^{+}$-$\Pi^{1}_{1}$-subcompact we can find an
ordinal $\varrho<\kappa$, $\bar{X}\subseteq H(\varrho^{+})$ and an elementary
embedding $j\colon\langle H(\varrho^{+}),\in,\bar{X}\rangle\rightarrow\langle
H(\kappa^{+}),\in,X\rangle$ with $\operatorname{crit}(j)=\varrho$,
$j(\varrho)=\kappa$ and such that

$(\star)\;\;\langle
H(\varrho^{+}),\in,\bar{X}\rangle\models\Phi(\bar{X},\varrho).$

Note that $(\star)$ entails $H(\kappa^{+})\models\exists Y\Phi(Y,\varrho)$,
and this is a first order statement.

Let $U_{\varrho}$ be the measure on $\varrho$ induced by $j$: that is,
$U_{\varrho}:=\\{X\subseteq\varrho\mathrel{|}\allowbreak\varrho\in j(X)\\}.$
Put $i:=j\mathbin{\upharpoonright}\langle H(\varrho^{+}),\in\rangle$ and let
$\ell$ be the ultrapower map induced by $U_{\varrho}$. Put
$M:=\mathrm{Ult}(H(\varrho^{+}),U_{\varrho})$ and let $k$ be the elementary
embedding factoring $i$ through $\ell$. Note that
$\operatorname{crit}(k)>|\mathcal{P}(\varrho)^{M}|$. By elementarity of $k$,
$M\models\exists Y\Phi(Y,\varrho)$, hence there is $\mathcal{Y}\in M$ such
that $M\models\Phi(\mathcal{Y},\varrho)$. Since
$\mathcal{Y}\subseteq\mathcal{P}(\varrho)^{M}$ and
$\operatorname{crit}(k)>|\mathcal{P}(\varrho)^{M}|$ we have
$k(\mathcal{Y})=k``\mathcal{Y}=\mathcal{Y}$. Thus, $\Phi(\mathcal{Y},\varrho)$
holds in $H(\kappa^{+})$. However note that $\mathcal{Y}\in
M=\mathrm{Ult}(H(\varrho^{+}),U_{\varrho})^{H(\kappa^{+})}$, which yields the
desired contradiction. ∎

Let $\mathcal{U}=\langle
U_{\alpha}\mathrel{|}\allowbreak\alpha<2^{\kappa}\rangle$ be a sequence of
measures on $\kappa$. Note that $\mathcal{U}$ can be naturally coded as a
subset of $\mathcal{P}(\kappa)$ and hence the above claim yields a measure $U$
such that
$\mathcal{U}\in\mathrm{Ult}(H(\kappa^{+}),U)\subseteq\mathrm{Ult}(V,U)$. Since
we can repeat this argument $(2^{\kappa})^{+}$-many times we infer that
$o(\kappa)=(2^{\kappa})^{+}$. ∎

######  Lemma 3.5.

Assume that the $\mathrm{GCH}$ holds. Suppose that $\kappa$ is a cardinal with
the $\kappa^{++}$-gluing property and $o(\kappa)=\kappa^{++}$. Then, there is
an embedding $j\colon V\rightarrow M$ such that
$\operatorname{crit}(j)=\kappa$, ${}^{\kappa}M\subseteq M$ and
$\kappa^{++}_{M}=\kappa^{++}$.

###### Proof.

Let $\mathcal{U}=\langle
U_{\alpha}\mathrel{|}\allowbreak\alpha<\kappa^{++}\rangle$ be a
$\lhd$-increasing sequence of measures on $\kappa$. Using the
$\kappa^{++}$-gluing property of $\kappa$ we can find a
$(\kappa,\kappa^{++})$-extender $E=\langle E_{a}\mathrel{|}\allowbreak
a\in[\kappa^{++}]^{\leq\kappa}\rangle$ such that for all $\alpha<\kappa^{++}$
and $A\in U_{\alpha}$ then

$\\{\langle\alpha,x\rangle\mathrel{|}\allowbreak x\in A\\}\in
E_{\\{\alpha\\}}.$

Let $i_{\alpha}\colon V\rightarrow
N_{\alpha}\simeq\mathrm{Ult}(V,U_{\alpha})$.

######  Claim 3.5.1.

For every $\alpha<\kappa^{++}$, $\kappa^{++}_{N_{\alpha}}\geq\alpha$.

###### Proof of claim.

Suppose towards a contradiction that this is false, and let
$\alpha<\kappa^{++}$ be minimal such that $\kappa^{++}_{N_{\alpha}}<\alpha$.
Fix $\beta<\alpha$ arbitrary, and let $i_{\alpha,\beta}\colon
N_{\alpha}\rightarrow\mathrm{Ult}(N_{\alpha},U_{\beta})\simeq
N_{\alpha,\beta}$. Similarly, let

$i_{\beta,\alpha}\colon
N_{\beta}\rightarrow\mathrm{Ult}(N_{\beta},i_{\beta}(U_{\alpha}))\simeq
N_{\beta,\alpha}.$

Standard arguments show that $N_{\alpha,\beta}=N_{\beta,\alpha}$ and that
$i_{\alpha,\beta}\circ i_{\alpha}=i_{\beta,\alpha}\circ i_{\beta}.$ Letting
$N:=N_{\alpha,\beta}$ one has in particular that
$i_{\beta,\alpha}(\kappa^{++}_{N_{\beta}})=\kappa^{++}_{N}$, and since
$\operatorname{crit}(i_{\beta,\alpha})=i_{\beta}(\kappa)>\kappa^{++}_{N_{\beta}}$
actually $\kappa^{++}_{N_{\beta}}=\kappa^{++}_{N}.$ By the minimality
assumption, $\kappa^{++}_{N_{\beta}}\geq\beta$ and hence
$\kappa^{++}_{N}\geq\beta$. All in all, we arrive at

$\beta\leq\kappa^{++}_{N}<i_{\alpha,\beta}(\kappa)<\kappa^{++}_{N_{\alpha}}.$

From all of this we conclude that $\kappa^{++}_{N_{\alpha}}\geq\alpha$, which
is impossible.∎

Let $j\colon V\rightarrow M$ be the elementary embedding induced by the
extender $E$. By $\kappa$-completeness of the extender $E$ we have that
${}^{\kappa}M\subseteq M$.

######  Claim 3.5.2.

$\kappa^{++}_{M}=\kappa^{++}$.

###### Proof of claim.

Let $\alpha<\kappa^{++}$. By Claim 3.5.1,
$\kappa^{++}_{N_{\alpha}}\geq\alpha$, hence
$\kappa^{++}_{M_{\alpha}}\geq\alpha$. Let $k_{\alpha}\colon
M_{\alpha}\rightarrow M$ be the canonical factor map. Then,
$\kappa^{++}_{M}\geq k_{\alpha}(\alpha)\geq\alpha$. Since the above is valid
for every $\alpha<\kappa^{++}$ we have $\kappa^{++}_{M}\geq\kappa^{++}$. ∎

At this stage we have accomplished the proof of the lemma. ∎

We now use the previous lemma to show that (consistently) there are
$\kappa^{+}$-$\Pi^{1}_{1}$-subcompact cardinals without the
$(2^{\kappa})^{+}$-gluing property.

Let $\mathbb{S}$ be the Easton-support iteration
$\langle\mathbb{P}_{\alpha};\dot{\mathbb{Q}}_{\beta}\mathrel{|}\allowbreak\beta\leq\alpha<\kappa\rangle$
defined as follows: If $\alpha<\kappa$ is an inaccessible cardinal then we
stipulate that

$\mathds{1}_{\mathbb{P}_{\alpha}}\Vdash_{\mathbb{P}_{\alpha}}\text{$``\dot{\mathbb{Q}}_{\alpha}$
is the forcing to add a $\square_{\alpha^{+}}$-sequence''}.$

Otherwise, $\mathop{1\hskip-3.0pt{\rm
l}}_{\mathbb{P}_{\alpha}}\Vdash_{\mathbb{P}_{\alpha}}\text{$``\dot{\mathbb{Q}}_{\alpha}$
is trivial''}.$ Let $G$ a $V$-generic filter for $\mathbb{S}$.

######  Theorem 3.6.

Assume the $\mathrm{GCH}$ holds and that $\kappa$ is a
$\kappa^{+}$-$\Pi^{1}_{1}$-subcompact cardinal such that
$\operatorname{Refl}(E^{\kappa^{++}}_{\omega})$ holds. Then, the following
hold in $V[G]$:

  1. (1)

$\kappa$ is $\kappa^{+}$-$\Pi^{1}_{1}$-subcompact;

  2. (2)

There is no embedding $j\colon V\rightarrow M$ with
$\operatorname{crit}(j)=\kappa$ and $\kappa^{++}_{M}=\kappa^{++}$.

In particular, in $V[G]$ the cardinal $\kappa$ does not have the
$\kappa^{++}$-gluing property.

###### Proof.

We begin proving the following claim:

######  Claim 3.6.1.

In $V[G]$ the principle $\operatorname{Refl}(E^{\kappa^{++}}_{\omega})$ holds.

###### Proof of claim.

Let $\dot{S}$ be a $\mathbb{S}$-name and $p\in G$ a condition such that

$p\Vdash_{\mathbb{S}}\text{$``\dot{S}$ is stationary in
$\dot{E}^{\kappa^{++}}_{\omega}$''.}$

For every $q\leq p$, define
$S_{q}:=\\{\alpha<\kappa^{++}\mathrel{|}\allowbreak
q\Vdash_{\mathbb{S}}\check{\alpha}\in\dot{S}\\}$. Since the condition $p$
forces $``\bigcup_{q\in\mathbb{P}/p}\check{S}_{q}=\dot{S}$” it follows that
for some $q\leq p$ the set $S_{q}$ is stationary.

We may assume without loss of generality that $q\in G$. By our assumption
there is $\delta$ with $\theta:=\operatorname{cf}^{V}(\delta)>\omega$ such
that $S_{q}\cap\delta$ is stationary. Put $S^{*}:=S_{q}\cap\delta.$ Clearly,
$\check{S}^{*}_{G}\subseteq\dot{S}_{G}$, so it suffices to check that
$q\Vdash_{\mathbb{S}}\text{$``\check{S}^{*}$ is stationary''}$.

$\blacktriangleright$ If $\theta$ is a Mahlo cardinal then
$\mathbb{P}_{\theta}$ is $\theta$-cc, hence it preserves the stationarity of
$S^{*}$. Also, observe that the tail of the iteration is
$(\theta+1)$-strategically closed, hence it preserves stationary subsets in
cofinality ${<\kappa}$.666See e.g. [Cum10, Lemma 10.6].

$\blacktriangleright$ If $\theta$ is a non-Mahlo limit of inaccessibles then
$\mathbb{P}_{\theta}$ is the Easton-support iteration of the previous stages.
Assume towards a contradiction that for a $\mathbb{P}_{\theta}$-name $\dot{C}$
and $r\in G_{\theta}$ we have $r\Vdash_{\mathbb{P}_{\theta}}\text{$``\dot{C}$
is a club $\wedge$ $\dot{C}\cap\check{S}^{*}=\emptyset$''}$. Since
$r=r_{\alpha}\ast\mathop{1\hskip-3.0pt{\rm l}}_{[\alpha,\theta)}$ we can
construct a $\mathbb{P}_{\alpha}$-name $\dot{C}^{*}$ such that
$r\Vdash_{\mathbb{P}_{\theta}}\dot{C}=\dot{C}^{*}$. This in particular implies
that $r_{\alpha}\Vdash_{\mathbb{P}_{\alpha}}\text{``$\check{S}^{*}$ is not
stationary''}$. However, this yields a contradiction because
$\mathbb{P}_{\alpha}$ is a small forcing. Thus, $S^{*}$ remains stationary
after forcing with $\mathbb{P}_{\theta}$. Yet again, the rest of the forcing
is distributive enough to preserve the stationarity of $S^{*}$ in $V[G]$.

$\blacktriangleright$ In any other case the argument is easy. In effect,
$\mathbb{S}$ factors as a two-step iteration of a small forcing and a
sufficiently distributive one. ∎

The next claim yields Clause (1) of the theorem:

######  Claim 3.6.2.

The cardinal $\kappa$ is $\kappa^{+}$-$\Pi^{1}_{1}$-subcompact in $V[G]$.

###### Proof of claim.

Working in $V[G]$, assume that $\Phi$ is the $\Pi^{1}_{1}$-sentence of the
form $\forall X,\varphi(X,A)$, where $A\subseteq H(\kappa^{+})$. Let us assume
that $\langle H(\kappa^{+}),\in,A\rangle\models\Phi$. Pick $\dot{A}$ be a
$\mathbb{S}$-name such that $\dot{A}_{G}=A$. Since $\mathrm{GCH}$ holds then
$\mathbb{S}\subseteq H(\kappa^{+})$, hence $\dot{A}\subseteq
H(\kappa^{+})^{V}$. Moreover, every subset $X$ of $H(\kappa^{+})$ in the
generic extension, has a name $\dot{X}\subseteq H(\kappa^{+})$ and every
element $y$ of $H(\kappa^{+})^{V[G]}$ has a name $\dot{y}\in
H(\kappa^{+})^{V}$ such that $\dot{y}^{G}=y$.

Let us assume that in $V[G]$,

$\langle H(\kappa^{+}),\in,A\rangle\models\forall X\subseteq
H(\kappa^{+}),\varphi(X,A),$

So the _Forcing Theorem_ yields that for some condition $p$,

$p\Vdash_{\mathbb{S}}\langle\dot{H}(\kappa^{+}),\in,\dot{A}\rangle\models\forall
X\subseteq\dot{H}(\kappa^{+}),\varphi(X,\dot{A})$

which means by the above discussion:

$\langle
H(\kappa^{+})^{V},\in,\dot{A},\Vdash_{\mathbb{S}},p\rangle\models\forall\dot{X},(\dot{X}\text{
is an
}\mathbb{S}\text{-name}\implies\text{$``p\Vdash_{\mathbb{S}}\varphi(\dot{X},\dot{A})$''}).$

Besides, by virtue of the _Forcing Definability Theorem_ ,
$``p\Vdash_{\mathbb{S}}\varphi(\dot{X},\dot{A})$” is a first order sentence in
the above-displayed structure. Using that $\kappa$ is
$\kappa^{+}$-$\Pi^{1}_{1}$-subcompact in $V$ we find $\varrho<\kappa$,
$\dot{B}\subseteq H(\varrho^{+})^{\mathbb{S}_{\varrho}}$ and
$q\in\mathbb{S}_{\varrho}$, along with an elementary embedding

$j\colon\langle
H(\varrho^{+}),\in,\dot{B},\Vdash_{\mathbb{S}_{\varrho}},q\rangle\rightarrow\langle
H(\kappa^{+}),\in,\dot{A},\Vdash_{\mathbb{S}},p\rangle$

with $\operatorname{crit}(j)=\varrho$, $j(\varrho)=\kappa$ and such that
$\langle
H(\varrho^{+}),\in,\dot{B},\Vdash_{\mathbb{S}_{\varrho}},q\rangle\models\text{$``q\Vdash_{\mathbb{S}}\Phi$''}$.
Since $\operatorname{crit}(j)=\varrho$ and $\mathbb{S}_{\varrho}$ is an
Easton-support iteration of forcings in $H(\varrho^{+})$, it follows that
$j\mathbin{\upharpoonright}\mathbb{S}_{\varrho}=\mathrm{id}$. In particular,
putting $B:=\dot{B}_{G_{\varrho}}$, Silver’s lifting criterion implies that
$j$ lifts to

$j^{*}\colon\langle
H(\varrho^{+})^{V[G_{\varrho}]},\in,B\rangle\rightarrow\langle
H(\kappa^{+})^{V[G]},\in,A\rangle,$

with $\operatorname{crit}(j^{*})=\varrho$ and $j^{*}(\varrho)=\kappa$. Also,
note that since $j(q)=p\in G$ and
$j\mathbin{\upharpoonright}\mathbb{S}_{\varrho}=\mathrm{id}$ then $q\in
G\cap\mathbb{S}_{\varrho}=G_{\varrho}$.

Finally, since the forcing $\mathbb{S}$ splits into
$\mathbb{S}\mathbin{\upharpoonright}\varrho$ iterated by a
$\varrho^{++}$-distributive forcing, since

$\langle
H(\varrho^{+})^{V},\in,\dot{B},\Vdash_{\mathbb{S}\mathbin{\upharpoonright}\varrho},q\rangle\models\forall\dot{X},(\dot{X}\text{
is an
}\mathbb{S}\mathbin{\upharpoonright}\varrho\text{-name}\implies\text{$``q\Vdash_{\mathbb{S}\mathbin{\upharpoonright}\varrho}\varphi(\dot{X},\dot{B})$''}),$

we conclude that $q$ actually forces over $\mathbb{S}$ the same statement.

Hence, $V[G]\models\langle
H(\varrho^{+})^{V[G_{\varrho}]},\in,B\rangle\models\Phi$. ∎

Let us now prove Clause (2). Assume towards contradiction that in $V[G]$ there
is an elementary embedding $j\colon V[G]\rightarrow M$ such that
$\operatorname{crit}(j)=\kappa$ and $(\kappa^{++})^{M}=(\kappa^{++})$. By
elementarity, $\square_{\kappa^{+}}$ holds in $M$. Now, since $M$ computes
correctly $\kappa^{++}$ we have that $\square_{\kappa^{+}}$ holds in $V[G]$,
as well. This indeed forms a contradiction with Claim 3.6.1 above, and thereby
Clause (2) is established. Finally, $\kappa$ cannot have the
$\kappa^{++}$-gluing property, for otherwise lemmas 3.4 and 3.5 would together
imply that Clause (2) fails. ∎

Using that $\operatorname{Refl}(E^{\kappa^{++}}_{\omega})$ holds for
$\kappa^{++}$-$\Pi^{1}_{1}$-subcompact cardinals one infers the following
corollary:

######  Corollary 3.7.

Assume that there is a $\kappa^{++}$-$\Pi^{1}_{1}$-subcompact cardinal.

Then, there is a generic extension where $\kappa$ is $\kappa$-compact but does
not have the $(2^{\kappa})^{+}$-gluing property.

##  4\. The $\omega$-gluing property from a strong cardinal

In the next sections we deal with the consistency of the $\omega$-gluing
property, starting with a large cardinal hypothesis weaker than
subcompactness. Let us observe first that the $\omega$-gluing property is in
fact a property of measures and not of extenders.

######  Lemma 4.1.

A cardinal $\kappa$ has the $\omega$-gluing property if and only if for every
$\omega$-sequence of $\kappa$-complete ultrafilters $\langle
U_{n}\mathrel{|}\allowbreak n<\omega\rangle$ there is a $\kappa$-complete
ultrafilter $W$ on $\kappa^{\omega}$, concentrating on increasing sequences,
such that the projection map $e_{n}$ sending $\eta\in\kappa^{\omega}$ to
$\eta(n)$ is a Rudin-Keisler projection from $W$ to $U_{n}$.

###### Proof.

The backwards direction is clear, as $W$ itself can be represented as an
extender. For the forwards direction, let $E$ be a $\kappa$-complete extender
such that there is a sequence $\langle\eta_{n}\mathrel{|}\allowbreak
n<\omega\rangle$,
$U_{n}=\\{X\subseteq\kappa^{\\{n\\}}\mathrel{|}\allowbreak\\{\langle
n,\eta_{n}\rangle\\}\in j_{E}(X)\\}$.

Let
$W:=\\{X\subseteq\kappa^{\omega}\mathrel{|}\allowbreak\langle\eta_{n}\mathrel{|}\allowbreak
n<\omega\rangle\in j_{E}(X)\\}$. Then, $W$ witnesses the validity of the
lemma. ∎

Assume the GCH holds and let $\kappa$ be a strong cardinal. Let
$\ell\colon\kappa\rightarrow V_{\kappa}$ be a _Laver function_ in the sense of
[GS89, §2]. In this section we define a Gitik iteration $\mathbb{P}_{\kappa}$
yielding a generic extension where $\kappa$ has the $\omega$-gluing property.

Let $A\subseteq\kappa$ be the collection of all inaccessible cardinals
$\alpha<\kappa$ such that $\ell``\alpha\subseteq V_{\alpha}.$ Let
$A^{\ell}:=\overline{A\cup\\{\beta+1\mathrel{|}\allowbreak\beta\in A\\}}$. We
define $\mathbb{P}_{\kappa}$ by induction on $\alpha\in A^{\ell}$ as follows.
Suppose that $\mathbb{P}_{\beta}$ has been defined for all $\beta\in
A^{\ell}\cap\alpha$. If $\alpha$ is a limit point of $A^{\ell}$ then
$\mathbb{P}_{\alpha}$ is simply the Gitik iteration of the previous stages.
Otherwise, suppose that $\alpha$ is a successor point of $\beta$ in
$A^{\ell}$. If $\alpha\neq\beta+1$ then we stipulate
$\mathbb{P}_{\alpha}:=\mathbb{P}_{\beta}\ast\dot{\mathbb{Q}}_{\beta}$, where
$\mathop{1\hskip-3.0pt{\rm
l}}_{\mathbb{P}_{\beta}}\Vdash_{\mathbb{P}_{\beta}}\text{$``\dot{\mathbb{Q}}_{\beta}$
is trivial''}$. Otherwise we shall distinguish two cases:

  * $(a)$

If
$p\Vdash_{\mathbb{P}_{\beta}}$“$\ell(\beta)=\langle\dot{U}^{\beta}_{n}\mathrel{|}\allowbreak
n<\omega\rangle$ are $\beta$-complete uniform measures on $\beta$”, then
$p\Vdash_{\mathbb{P}_{\beta}}\text{$``\dot{\mathbb{Q}}_{\beta}=\dot{\mathbb{T}}(\langle\dot{U}^{\beta}_{n}\mathrel{|}\allowbreak
n<\omega\rangle)$''}$, where this denotes a name for the tree Prikry forcing
relative to $\langle U^{\beta}_{n}\mathrel{|}\allowbreak n<\omega\rangle$ (see
[Ben19, §4]).

  * $(b)$

Otherwise, $\mathop{1\hskip-3.0pt{\rm
l}}_{\mathbb{P}_{\beta}}\Vdash_{\mathbb{P}_{\beta}}\text{$``\dot{\mathbb{Q}}_{\beta}$
is trivial''}.$

######  Theorem 4.2.

Assume the $\operatorname{GCH}$ holds that there is a strong cardinal
$\kappa$.

Then, forcing with $\mathbb{P}_{\kappa}$ yields a model for
$\text{$``{\kappa}$ has the $\omega$-gluing property''}.$

###### Proof.

Fix $G\subseteq\mathbb{P}_{\kappa}$ a $V$-generic filter. Working in $V[G]$,
let $\langle U_{n}\mathrel{|}\allowbreak n<\omega\rangle$ be an
$\omega$-sequence of $\kappa$-complete uniform measures on $\kappa$. Let $p\in
G$, and $\langle\dot{U}_{n}\mathrel{|}\allowbreak n<\omega\rangle$ be an
$\omega$-sequence of $\mathbb{P}_{\kappa}$-nice-names such that $p$ forces
these to have the above property. Note that since $\mathbb{P}_{\kappa}$ has
the $\kappa$-cc, and $\mathbb{P}_{\kappa}\subseteq V_{\kappa}$, we have
$\dot{U}_{n}\in V_{\kappa+2}$ for all $n<\omega.$ By [GS89, Lemma 2.1] we can
find a $(\kappa,\kappa+2)$-strong embedding $j\colon V\rightarrow M$ such that
$j(\ell)(\kappa)=\langle\dot{U}_{n}\mathrel{|}\allowbreak n<\omega\rangle.$ We
will use this to form a measure $\mathcal{V}$ on $\kappa^{\omega}$ with
$U_{n}\leq_{\mathrm{RK}}\mathcal{V}$ for all $n<\omega.$

By our $\mathrm{GCH}$ assumption, we can enumerate all the
$\mathbb{P}_{\kappa}$-nice-names for subsets of $\kappa^{\omega}$ as
$\langle\dot{X}_{\alpha}\mathrel{|}\allowbreak\alpha<\kappa^{+}\rangle$. Look
now at $j(\mathbb{P}_{\kappa})$. As usual, this decomposes as an iteration of
the form
$\mathbb{P}_{\kappa}\ast\dot{\mathbb{Q}}_{\kappa}\ast\dot{\mathbb{R}}$, where

${\mathop{1\hskip-3.0pt{\rm
l}}}_{\mathbb{P}_{\kappa}\ast\dot{\mathbb{Q}}_{\kappa}}\Vdash_{\mathbb{P}_{\kappa}\ast\dot{\mathbb{Q}}_{\kappa}}\text{$``\langle\dot{\mathbb{R}},\dot{\leq}^{*}\rangle$
is $\kappa^{+}$-closed''}.$

In addition, since $p\Vdash_{\mathbb{P}_{\kappa}}^{N}\text{$``j(\ell)(\kappa)$
is an $\omega$-sequence of measures''}$, this latter condition forces
$\dot{\mathbb{Q}}_{\kappa}$ to be the tree Prikry forcing with respect to the
sequence $\langle\dot{U}_{n}\mathrel{|}\allowbreak n<\omega\rangle.$ Note that
this definition is absolute between $M[G]$ and $V[G].$

Define by induction on $\alpha<\kappa^{+}$ a sequence of
$\mathbb{P}_{\kappa}\ast\dot{\mathbb{Q}}_{\kappa}$-names
$\langle\dot{r}_{\alpha}\mathrel{|}\allowbreak\alpha<\kappa^{+}\rangle$ for
conditions in $\mathbb{R}$ such that:

  * •

$\langle\dot{r}_{\alpha}\mathrel{|}\allowbreak\alpha<\kappa^{+}\rangle$ is
$\leq^{*}$-decreasing;

  * •

for all $\alpha<\kappa^{+}$,
$\dot{r}_{\alpha}\parallel_{\dot{\mathbb{R}}}\dot{b}_{\kappa}\in
j(\dot{X}_{\alpha}),$ where $\dot{b}_{\kappa}$ denotes the standard name for
the generic object added by $\dot{\mathbb{Q}}_{\kappa}.$

Define $\mathcal{U}$ as follows: $X\in\mathcal{U}$ iff there is $p\in G$ and
$\alpha<\kappa^{+}$ such that

$p\cup\\{\langle\kappa,(\emptyset,\dot{T})\rangle\\}\cup
r_{\alpha}\Vdash^{M}_{j(\mathbb{P}_{\kappa})}\dot{b}_{\kappa}\in
j(\dot{X}_{\alpha}),$

where $\dot{T}$ is a $\mathbb{P}_{\kappa}$-name for a
$\langle\dot{U}_{n}\mathrel{|}\allowbreak n<\omega\rangle$-tree.

Standard arguments prove that this is a $\kappa$-complete measure on (the
increasing sequences in) $\kappa^{\omega}$, in $V[G]$. For more details, see
[Git10, p. 1439].

######  Claim 4.2.1.

$V[G]\models``\forall n<\omega\,(U_{n}\leq_{\mathrm{RK}}\mathcal{U})$”.
Moreover, for each $n<\omega$, this is witnessed by
$e_{n}\colon\kappa^{\omega}\rightarrow\kappa$, the evaluation map $f\mapsto
f(n).$

###### Proof of claim.

We claim that $e_{n}``X\in U_{n}$ for all $X\in\mathcal{U}$. Suppose
otherwise, and let $p\in G$ be such that
$p\Vdash_{\mathbb{P}_{\kappa}}``\dot{X}\in\mathcal{U}\,\wedge\,\dot{Y}\notin
U_{n}$”, where $\dot{Y}$ is a $\mathbb{P}_{\kappa}$-nice name for a subset of
$\kappa$ such that $\dot{Y}_{G}=e_{n}``X.$ Since $\dot{Y}$ is
$\mathbb{P}_{\kappa}$-nice, $\dot{Y}\subseteq V_{\kappa}$.

By definition of $\mathcal{U}$ there is $\alpha<\kappa^{+}$ and a
$\mathbb{P}_{\kappa}$-name $\dot{T}$ such that

$p\cup\\{\langle\kappa,(\emptyset,\dot{T})\rangle\\}\cup
r_{\alpha}\Vdash^{M}_{j(\mathbb{P}_{\kappa})}\dot{b}_{\kappa}\in j(\dot{X}).$

In particular,

$p\cup\\{\langle\kappa,(\emptyset,\dot{T})\rangle\\}\cup
r_{\alpha}\Vdash^{M}_{j(\mathbb{P}_{\kappa})}\dot{b}_{\kappa}(n)\in
j(\dot{Y})\cap\kappa.$

By our choice of $\dot{Y}$, it follows that $j(\dot{Y})\cap\kappa=\dot{Y}.$
So, all in all, the above condition forces $``\dot{b}_{\kappa}(n)\in\dot{Y}$”.
On another front, since $p\Vdash_{\mathbb{P}_{\kappa}}\dot{Y}\notin U_{n}$,
there is $q\leq p$ in $G$ such that
$q\Vdash_{\mathbb{P}_{\kappa}}(\emptyset,\dot{T}^{\prime})\leq^{*}(\emptyset,\dot{T})$,
where $\dot{T}^{\prime}$ is a $\mathbb{P}_{\kappa}$-name for a tree whose
$n^{\mathrm{th}}$-level is disjoint from $\dot{Y}$. Then,

$q\cup\\{\langle\kappa,(\emptyset,\dot{T}^{\prime})\rangle\\}\cup
r_{\alpha}\leq p\cup\\{\langle\kappa,(\emptyset,\dot{T})\rangle\\}\cup
r_{\alpha}$

and, clearly, the stronger forces $``\dot{b}_{\kappa}(n)\notin\dot{Y}$”. Note
that this conflicts with our previous comments, hence producing the sought
contradiction. ∎

By Lemma 4.1, this completes the proof of the theorem. ∎

##  5\. Improving the upper bound for the $\omega$-gluing property

In this section we would like to improve the large cardinal hypothesis
employed in Theorem 4.2. In particular we prove the following result:

######  Theorem.

It is consistent, relative to the existence of a measurable with Mitchell’s
order $\omega_{1}$, that there is a cardinal $\kappa$ with the $\omega$-gluing
property.

Our final verification of the $\omega$-gluing property will follow the steps
of §4. However, we will need to implement some non-trivial modifications to
the above argument in order to have a better control upon the
$\kappa$-complete ultrafilters that will appear in the intermediate models of
the final generic extension (see Lemma 5.8 and Theorem 5.19). To be able to
identify these $\kappa$-complete ultrafilters we will rely on the assumption
that $``\rm{V}=\mathcal{K}$” and suppose that there is no inner model of
$\exists\alpha,\,o(\alpha)=\alpha^{++}$.

On one hand, the first of these assumptions (yet natural) is probably
unnecessary, as demonstrated by Gitik and Kaplan in [GK23]. On the other hand,
our second demand will ensure, by a result of Mitchell [Mit84], (which was
extended by Schindler [Sch06]), that every iteration of ultrapowers is normal
and that it uses only normal measures.

###  5.1. Coding $\kappa$-complete measures after adding a fast function

The first technical problem that we must address is the lack of a Laver
function. Formerly, the existence of such object was granted by our departing
large-cardinal assumption; namely, strongness. Instead, this time we would
like to force some sort of fast function in a way that we can still keep track
over the amount of possible measures. For this, we will consider the following
_non-stationarily-supported_ varation of _Woodin’s fast function_ forcing.

Let $\mathbb{S}$ be the poset consisting on partial functions
$s\colon\kappa\to H_{\kappa}$ such that
$\operatorname{dom}s\subseteq\mathrm{Inacc}$ and the following two hold:

  * •

$(\operatorname{dom}s)\cap\beta\in\mathrm{NS}_{\beta}$ for all
$\beta\in\mathrm{Inacc}\cap\kappa$,

  * •

and $s(\alpha)\in H(\alpha^{+})$ for all $\alpha\in\operatorname{dom}s$.777The
restriction of $s(\alpha)$ to $H(\alpha^{+})$ is inessential at this part.

The order of $\mathbb{S}$ is reverse-inclusion. It is easy to show that
forcing with $\mathbb{S}$ does preserve cardinals (see [Ham00]) and that the
generic function introduced generates the entire generic extension.

Normal iterations are going to play a significant role in our arguments so we
begin recalling the definition and some basic facts about them.

######  Definition 5.1.

A system of elementary embeddings
$\langle\iota_{\alpha,\beta}\mathrel{|}\allowbreak\alpha\leq\beta\leq\delta\rangle$
between transitive models is called a _linear iteration_ if the following
hold:

  1. (1)

$\iota_{\alpha,\beta}\colon\mathcal{M}_{\alpha}\to\mathcal{M}_{\beta}$,888As
customary, $\iota_{\alpha}$ will be a shorthand for $\iota_{0,\alpha}$.

  2. (2)

for all $\alpha\leq\delta$, $\iota_{\alpha,\alpha}=\operatorname{id}$,

  3. (3)

for all $\alpha\leq\beta\leq\gamma\leq\delta$,
$\iota_{\beta,\gamma}\circ\iota_{\alpha,\beta}=\iota_{\alpha,\gamma}$, and

  4. (4)

for every limit ordinal $\gamma\leq\delta$,

$\langle\mathcal{M}_{\gamma},\langle\iota_{\alpha,\gamma}\mathrel{|}\allowbreak\alpha<\gamma\rangle\rangle$
is the direct limit of
$\langle\iota_{\alpha,\beta}\mathrel{|}\allowbreak\alpha\leq\beta<\gamma\rangle$.

A linear iteration is a _normal iteration, using normal (internal) measures_
if:

  1. (5)

$\iota_{\alpha,\alpha+1}$ is an ultrapower embedding by a measure
$\mathcal{U}_{\alpha}\in\mathcal{M}_{\alpha}$ on $\mu_{\alpha}$,

  2. (6)

$\langle\mu_{\alpha}\mathrel{|}\allowbreak\alpha<\delta\rangle$ is a strictly
increasing sequence.

######  Lemma 5.2.

Let
$\langle\iota_{\alpha,\beta}\mathrel{|}\allowbreak\alpha\leq\beta\leq\delta\rangle$
be a normal iteration using normal measures, with critical points
$\langle\mu_{\alpha}\mathrel{|}\allowbreak\alpha<\delta\rangle$. Then, for
every set $x\in\mathcal{M}_{\delta}$ there is a finite normal iteration
$\tilde{\iota}\colon\mathcal{M}_{0}\to\tilde{\mathcal{M}}$ together with a
factor embedding $k\colon\tilde{\mathcal{M}}\to\mathcal{M}_{\delta}$ such that
$k\circ\tilde{\iota}=\iota_{\delta}$ and $x\in\operatorname{range}(k)$.

###### Proof.

A straightforward induction on $\delta$ shows that, for every
$x\in\mathcal{M}_{\gamma}$ ($\gamma\leq\delta$) there is $f\in\mathcal{M}_{0}$
and ordinals $\alpha_{0}<\dots<\alpha_{n-1}<\gamma$ such that

$\iota_{\gamma}(f)(\mu_{\alpha_{0}},\dots,\mu_{\alpha_{n-1}})=x.$

For every $\alpha<\delta$, let us pick a representation of $U_{\alpha}$;
namely

$U_{\alpha}=\iota_{\alpha}(g_{\alpha})(\mu_{\beta^{\alpha}_{0}},\dots,\mu_{\beta^{\alpha}_{m_{\alpha}-1}}),$

where each $\beta^{\alpha}_{i}$ is below $\alpha$.

Let us say that an increasing sequence of ordinals
$\bar{a}\in[\delta]^{<\omega}$ is _saturated_ if
$\\{\beta^{\alpha}_{0},\dots\beta^{\alpha}_{m_{\alpha}-1}\\}\subseteq\bar{a}$
for all $\alpha\in\bar{a}$. Clearly, every set $b\in[\delta]^{<\omega}$ is a
subset of a finite saturated set $\bar{a}$; this is because the generators
corresponding to each $\alpha\in b$ are strictly below $\alpha$. In
particular, given $x\in\mathcal{M}_{\delta}$ there is
$\bar{a}=\\{\alpha_{0},\dots,\alpha_{n-1}\\}$ a saturated set and a function
$f\in\mathcal{M}_{0}$ such that

$\iota_{\delta}(f)(\mu_{\alpha_{0}},\dots,\mu_{\alpha_{n-1}})=x.$

Let us construct a finite iteration
$\langle\tilde{\iota}_{k,m}\mathrel{|}\allowbreak k\leq m\leq n\rangle$
witnessing the validity of the lemma. For each relevant $i$, denote by
$\tilde{\mu}_{i}$ the critical point of $\tilde{\iota}_{i,i+1}$. We will
construct by induction elementary embeddings $\tilde{k}_{i}$ such that
$\tilde{k}_{i}\circ\tilde{\iota}_{i}=\iota_{\alpha_{i}}$ and moreover
$\tilde{k}_{i}(\tilde{\mu}_{j})=\mu_{\alpha_{j}}$, for all $j<i$.

Since $\bar{a}$ is saturated, for every $i<n$, the measure $U_{\alpha_{i}}$ is
represented by generators whose indexes belong to $\bar{a}\cap\alpha_{i}$. Let
$b_{i}$ be this set of indexes. Let
$\tilde{\mu}=\langle\tilde{\mu}_{0},\dots,\tilde{\mu}_{n-1}\rangle$ (note that
each time only an initial segment of this sequence is used in order to define
the next step). For simplicity of notations, let us assume that
$\delta=\alpha_{n}$. For $i=0$, let $\tilde{k}_{0}=\iota_{\alpha_{0}}$. Let us
assume that $\tilde{k}_{i},\tilde{\iota}_{i}$ and
$\tilde{\mu}\mathbin{\upharpoonright}i$ are defined. Let
$\tilde{U}_{i}=\tilde{\iota}_{i}(g_{\alpha_{i}})(\tilde{\mu}\mathbin{\upharpoonright}b_{i})$.
By the inductive hypothesis,

$\tilde{k}_{i}(\tilde{U}_{i})=\iota_{\alpha_{i}}(g_{\alpha_{i}})(\mu_{\beta^{\alpha_{i}}_{0}},\dots,\mu_{\beta^{\alpha_{i}}_{m_{\alpha}-1}})=U_{\alpha_{i}}.$

In particular, $\tilde{U}_{i}$ is a normal measure on $\tilde{\mu}_{i}$ such
that $\tilde{k}_{i}(\tilde{\mu}_{i})=\mu_{\alpha_{i}}$.

Let

$\tilde{k}_{i+1}([r]_{\tilde{U}_{i}}):=\iota_{\alpha_{i}+1,\alpha_{i+1}}(\iota_{\alpha_{i},\alpha_{i+1}}(\tilde{k}_{i}(r))(\mu_{\alpha_{i}})).$

Since $\tilde{k}_{i}(\tilde{U}_{i})=U_{\alpha_{i}}$, it is standard to verify
that $\tilde{k}_{i+1}$ is well defined and elementary. Moreover,
$\tilde{k}_{i+1}\circ\tilde{\iota}_{i+1}=\iota_{\alpha_{i+1}}$. We use the
assumption that the iteration is normal, in order to know that
$\iota_{\alpha_{i}+1,\alpha_{i+1}}$ does not move $\mu_{\alpha_{i}}$.

Finally, $\tilde{k}_{n}\circ\tilde{\iota}_{n}=\iota_{\delta}$, so we set
$k=\tilde{k}_{n}$.

Let us show that $x\in\operatorname{range}k$. Indeed, without loss of
generality,

$x=\iota_{\delta}(f)(\mu_{\alpha_{0}},\dots,\mu_{\alpha_{n-1}})=k\circ\tilde{\iota}(f)(k(\tilde{\mu}))=k(\tilde{\iota}(f)(\tilde{\mu})).\qed$

In the forthcoming Lemma 5.6 we will characterize the measures in
$\mathcal{K}[S]$ where $S\subseteq\mathbb{S}$ is a $\mathcal{K}$-generic. The
next general lemma will be instrumental for this purpose:

######  Lemma 5.3.

Suppose that $\iota\colon\mathcal{M}\rightarrow\mathcal{N}$ is the direct
limit of a normal iteration
$\langle\iota_{\alpha,\beta}\mathrel{|}\allowbreak\alpha\leq\beta<\delta\rangle$
of measures with critical points
$\langle\mu_{\alpha}\mathrel{|}\allowbreak\alpha<\delta\rangle$, and let
$\mathbb{P}\in\mathcal{M}$ be a forcing notion with the following properties:

  1. (1)

Every condition $p\in\mathbb{P}$ is a partial function
$p\colon\mu_{0}\rightarrow\mathcal{M}$;

  2. (2)

If $p\in\mathbb{P}$ then for every $\mathcal{M}$-inaccessible
$\beta\leq\mu_{0}$,
$(\operatorname{dom}p)\cap\beta\in\mathrm{NS}^{\mathcal{M}}_{\beta};$

  3. (3)

For a condition $p\in\mathbb{P}$ and a nowhere stationary set
$T\in\mathrm{NS}^{\mathcal{M}}_{\mu_{0}}$ there is $q\leq p$ such that
$T\subseteq\operatorname{dom}q.$

Then, for every $\mathcal{M}$-generic filter $G\subseteq\mathbb{P}$ we have
that

$\bigcup_{p\in
G}\iota(\operatorname{dom}(p))=\iota(\mu_{0})\setminus\langle\mu_{\alpha}\mathrel{|}\allowbreak\alpha<\delta\rangle.$

###### Proof.

Denote by $G_{\iota}$ the left-hand-side of the above-displayed equation. Here
we just deal with the case where $\iota$ is a finite linear iteration with
critical points $\langle\mu_{0},\dots,\mu_{k}\rangle.$ The general case is
handled similarly, using that any linear iteration is a directed limit of
finite iterations, as demonstrated in Lemma 5.2.

The crux of the matter is the following claim:

######  Claim 5.3.1.

For every $\alpha\in\iota(\mu_{0})\setminus\\{\mu_{0},\dots,\mu_{k}\\}$ there
is a nowhere stationary set $T\subseteq\mu_{0}$ with $\alpha\in\iota(T)$.

Using this latter fact it is easy to derive the desired conclusion:

######  Claim 5.3.2.

$\operatorname{dom}G_{\iota}=\iota(\mu_{0})\setminus\\{\mu_{0},\dots,\mu_{k}\\}$.

###### Proof of claim.

Fix $\alpha\in\iota(\mu_{0})\setminus\\{\mu_{0},\dots,\mu_{k}\\}$ and put

$D_{\alpha}:=\\{p\in\mathbb{P}\mathrel{|}\allowbreak\alpha\in\iota(\operatorname{dom}(p))\\}.$

We show that $D_{\alpha}$ is dense in $\mathbb{P}$. Indeed, let $p\notin
D_{\alpha}$ and use Claim 5.3.1 in order to find a nowhere stationary set
$T\subseteq\mu_{0}$ such that $\alpha\in\iota(T).$ Using Clause (3) above we
find $q\leq p$ with $T\subseteq\operatorname{dom}q$. Thus,
$\iota(\mu_{0})\setminus\\{\mu_{0},\dots,\mu_{k}\\}\subseteq G_{\iota}.$

Conversely, we claim that
$G_{\iota}\cap\\{\mu_{0},\dots,\mu_{k}\\}=\emptyset.$ For this fix $i\leq k$
and suppose towards a contradiction that $\mu_{i}\in G_{\iota}$. Let $p\in G$
such that $\mu_{i}\in\iota(T)$ where $T:=\operatorname{dom}p.$ Notice that
$\mu_{i}\notin\iota_{i+1}(T)$: By elementarity, $\iota_{i}(T)$ is nowhere
stationary, hence
$\iota_{i}(T)\cap\mu_{i}\in\mathrm{NS}_{\mu_{i}}^{\mathcal{N}}$. Also, if
$\mu_{i}\in\iota_{i+1}(T)$ then normality of the iteration would imply that
$\iota_{i}(T)\cap\mu_{i}$ is stationary in $\mu_{i}.$

Therefore $\mu_{i}\notin\iota(T)$, yielding the sought contradiction. ∎

So, let us prove Claim 5.3.1 above. Fix $\alpha<\iota(\mu_{0})$ and let us
find a nowhere stationary set $T\subseteq\mu_{0}$ such that
$\alpha\in\iota(T)$. To avoid trivialities let us further assume that
$\alpha\geq\mu_{0}$. First, $\alpha=\iota(g)(\mu_{0},\dots,\mu_{n-1})$ for
some $g\colon\mu_{0}^{n}\to\mu_{0}$ and $n\leq k$. Assume in addition that $n$
is the minimal such index.999In particular, $\mu_{n-1}<\alpha$.

Define
$B:=\\{\bar{x}\in\mu_{0}^{n}\mathrel{|}\allowbreak\max(\bar{x})<g(\bar{x})\\}$
and $T:=g``B$. Clearly, $(\mu_{0},\dots,\mu_{n-1})\in\iota(B)$. Thus,
$\alpha=\iota(g)(\mu_{0},\dots,\mu_{n-1})\in\iota(T).$

We claim that $T$ is nowhere stationary. Suppose towards a contradiction that
$\rho<\mu_{0}$ is a regular cardinal such that $T\cap\rho$ is stationary. Then
every $\alpha\in T\cap\rho$ is of the form $g(\bar{x})$ for some $\bar{x}\in
B\cap\rho^{n}$. Let $B_{\rho}\subseteq B\cap\rho^{n}$ such that
$T\cap\rho=g``B_{\rho}$. Next, let $h_{\rho}\colon T\cap\rho\rightarrow\rho$
be given by

$\beta\mapsto\min\\{\max(\bar{x})\mathrel{|}\allowbreak g(\bar{x})=\beta\\}.$

This map is well-defined and regressive, by the definition of $B$. Thus, there
is $T_{\rho}\subseteq T\cap\rho$ stationary and $\eta<\rho$ with
$h_{\rho}``T_{\rho}=\\{\eta\\}$. This latter fact outright implies
$T_{\rho}\subseteq g``\eta^{n}$, which is not impossible in that
$|g``\eta^{n}|=|\eta|<\rho$. This completes the proof of Lemma 5.3. ∎

######  Remark 5.4.

The proof of Claim 5.3.1 shows something slightly stronger: for every ordinal
$\alpha$ there is a set $T\subseteq\kappa$ such that $\alpha\in\iota(T)$ and
there is a regressive function $f\colon T\to\kappa$ such that for every
$\rho$, $|f^{-1}(\\{\rho\\})|\leq|\rho|+\aleph_{0}$.

The next strengthening of Claim 5.3.1 comes handy in the proof:

######  Lemma 5.5.

Let $\iota\colon\mathcal{M}\rightarrow N$ be a finite normal iteration with
critical points $\vec{\mu}=\langle\mu_{i}\mathrel{|}\allowbreak i<n\rangle$.
For every nowhere stationary $T\in\mathcal{P}(i(\kappa))^{N}$ with
$T\cap\vec{\mu}=\emptyset$ there is another nowhere stationary
$\bar{T}\in\mathcal{P}(\kappa)^{\mathcal{M}}$ such that
$T\subseteq\iota(\bar{T})$

###### Proof.

Put $\mu_{n}:=\iota(\kappa)$. Since $T\in\mathcal{P}(i(\kappa))^{N}$, $T$
decomposes as $T=\bigcup_{i<n}T_{i}$ where $T_{i}:=T\cap[\mu_{i},\mu_{i+1})$.
Clearly, each $T_{i}$ is nowhere stationary in $\mu_{i+1}$.

To establish the lemma it will suffice to find, for each $i<n$, a nowhere
stationary set $\bar{T}_{i}\in\mathcal{P}(\kappa)^{\mathcal{M}}$ with
$T_{i}\subseteq\iota_{i+1}(\bar{T}_{i})$. So, fix $i<n$.

Let $g\colon\kappa^{i+1}\rightarrow\mathcal{P}(\kappa)^{\mathcal{M}}$ be a map
in $\mathcal{M}$ representing $T_{i}$; namely,
$T_{i}=\iota(g)(\mu_{0},\dots,\mu_{i})$. Without loss of generality,
$g(\bar{x})$ is nowhere stationary for all $\bar{x}\in\kappa^{i+1}$. Define
$B:=\\{\bar{x}\in\kappa^{i+1}\mathrel{|}\allowbreak\max(\bar{x})<g(\bar{x})\\}$
and

$\bar{T}_{i}:=\bigcup\\{g(\bar{x})\setminus\max(\bar{x})+1\mathrel{|}\allowbreak\bar{x}\in
B\\}.$

Clearly, $(\mu_{0},\dots,\mu_{i})\in\iota_{i+1}(B)$ so that, as
$T\cap\vec{\mu}=\emptyset$,

$\iota_{i+1}(\bar{T}_{i})\supseteq\iota(g)(\mu_{0},\dots,\mu_{i})\setminus\mu_{i}+1=T_{i}\setminus(\mu_{i}+1)=T_{i}.$

Also, $\bar{T}_{i}$ is nowhere stationary by the same argument as in Claim
5.3.1. ∎

The following serves as a warm-up for Lemma 5.8 and provides a complete
cartography of the measures in the generic extension $\mathcal{K}[S]$:

######  Lemma 5.6.

Let $S\subseteq\mathbb{S}$ be a $\mathcal{K}$-generic filter and $\mathcal{U}$
a $\kappa$-complete ultrafilter on $\kappa$ in $\mathcal{K}[S]$. Then, there
is a finite iteration $\langle i_{m,n}\mathrel{|}\allowbreak m\leq n\leq
k+1\rangle$ of measures in $\mathcal{K}$ with critical points
$\langle\mu_{0},\dots,\mu_{k}\rangle$, $\kappa=\mu_{0}<\mu_{1}<\dots<\mu_{k}$,
a function $f\colon\kappa^{k+1}\to\kappa$ in $\mathcal{K}$ and $\langle
a_{0},\dots,a_{k}\rangle\in\prod_{i\leq k}H(\mu_{i}^{+})$ such that

$\mathcal{U}=\\{\dot{X}^{S}\subseteq\kappa\mathrel{|}\allowbreak\exists p\in
S\,(\iota(p)\cup\\{\langle\mu_{i},a_{i}\rangle\mathrel{|}\allowbreak i\leq
k\\}\Vdash_{\iota(\mathbb{S})}\check{\epsilon}\in\iota(\dot{X})^{S^{\prime}})\\},$

where $\epsilon=\iota_{k}(f)(\mu_{0},\dots,\mu_{k}).$

###### Proof.

Let $\mathcal{U}\in\mathcal{K}[S]$ be a $\kappa$-complete ultrafilter on
$\kappa$ and $j\colon\mathcal{K}[S]\to M$ be the corresponding ultrapower
embedding. By our anti-large-cardinal hypothesis,
$j\mathbin{\upharpoonright}\mathcal{K}$ is an iteration of normal measures in
$\mathcal{K}$. Let $\epsilon:=[\operatorname{id}]_{\mathcal{U}}$.

######  Claim 5.6.1.

$j\mathbin{\upharpoonright}\mathcal{K}$ is a finite iteration.

###### Proof of claim.

We commence observing that $M=\mathcal{K}^{M}[S^{\prime}]$ for
$S^{\prime}\subseteq j(\mathbb{S})$ a $\mathcal{K}^{M}$-generic filter: First,
by elementarity, $M=N[S^{\prime}]$ for some $N$-generic filter
$S^{\prime}\subseteq j(\mathbb{S})$. Second, by absoluteness of the core
model, $\mathcal{K}^{\mathcal{K}[S]}=\mathcal{K}$, and so
$j(\mathcal{K})=\mathcal{K}^{M}$. Finally, by the Laver-Woodin theorem on
definability of the ground model [Lav07, Woo11], $``\mathcal{K}$ is the ground
model of $\mathcal{K}[S]$” is first-order-expressible in $\mathcal{K}[S]$,
hence $N=\mathcal{K}^{M}$ is the ground model of $M$.

We are now in conditions to argue that $j\mathbin{\upharpoonright}\mathcal{K}$
is a finite iteration. Clearly, $M$ is closed under $\kappa$-sequences in
$\mathcal{K}[S]$, hence also closed under $\omega$-sequences. Since
$j(\mathbb{S})$ is $\omega_{1}$-closed it must be the case that
$\vec{x}\in{}^{\omega}{}M\cap\mathcal{K}[S]\subseteq\mathcal{K}^{M}$. However,
if $j\mathbin{\upharpoonright}\mathcal{K}$ is an infinite iteration then the
sequence of its first $\omega$-many critical points
$\vec{x}=\langle\mu_{n}\mathrel{|}\allowbreak n<\omega\rangle$ belongs to
$M^{\omega}\cap\mathcal{K}[S]$ but not to $\mathcal{K}^{M}$.∎

So, let
$\iota=\langle\iota_{m,n}\colon\mathcal{\mathcal{K}}_{m}\rightarrow\mathcal{K}_{n}\mathrel{|}\allowbreak
m\leq n\leq k+1\rangle$ be this iteration. For each $i\leq k$ put
$\mu_{i}:=\operatorname{crit}(\iota_{i,i+1})$. Without loss of generality
${\iota}$ is normal.

We would like to analyse the $\mathcal{K}^{M}$-generic filter $S^{\prime}$, or
equivalently, the function $\bigcup S^{\prime}$. Let us consider

$G_{\iota_{k+1}}:=\bigcup\\{j(\operatorname{dom}(s))\mathrel{|}\allowbreak
s\in S\\}.$

Clearly, $\mathbb{S}$ satisfies the assumptions of Lemma 5.3 and thus
$G_{\iota_{k+1}}=\iota_{k+1}(\kappa)\setminus\\{\mu_{0},\dots,\mu_{k}\\}.$
Since $\mathcal{U}$ is the measure induced by $j$ we have that

$\mathcal{U}=\\{\dot{X}^{S}\subseteq\kappa\mathrel{|}\allowbreak\exists q\in
S^{\prime}\,(q\Vdash_{j(\mathbb{S})}\check{\epsilon}\in
j(\dot{X})^{S^{\prime}})\\}.$

Let $f\colon\kappa^{k+1}\rightarrow\kappa$ be such that
$\iota_{k+1}(f)(\mu_{0},\dots,\mu_{k})=\epsilon$. Using Lemma 5.3 we find
$\langle a_{0},\dots,a_{k}\rangle\in\left(\prod_{i\leq
k}H(\mu^{+}_{i})\right)^{M}$ such that for every condition $q\in S^{\prime}$
there is $p\in S$, $q\geq
j(p)\cup\\{\langle\mu_{i_{j}},a_{i_{j}}\rangle\mathrel{|}\allowbreak j\leq
k\\}$. Therefore,

$\mathcal{U}=\\{\dot{X}^{S}\subseteq\kappa\mathrel{|}\allowbreak\exists p\in
S\,(j(p)\cup\\{\langle\mu_{i},a_{i}\rangle\mathrel{|}\allowbreak i\leq
k\\}\Vdash_{j(\mathbb{S})}\check{\epsilon}\in j(\dot{X})^{S^{\prime}})\\},$

which completes the proof. ∎

The following useful observation arises from the previous proof:

######  Lemma 5.7.

Suppose $j\colon\mathcal{K}[S]\to M$ is an arbitrary elementary embedding such
that $\iota=j\mathbin{\upharpoonright}\mathcal{K}$ is a normal iteration of
measures in $\mathcal{K}$, say
$\langle\iota_{\alpha,\beta}\mathrel{|}\allowbreak\alpha\leq\beta\leq\delta\rangle$.101010Note
that $M$ might not be closed under countable sequences Then, there is an
iteration $\langle
j_{\alpha,\beta}\mathrel{|}\allowbreak\alpha\leq\beta\leq\gamma\rangle$ such
that each embedding $j_{\alpha,\beta}$ lifts the corresponding embedding
$\iota_{\alpha,\beta}$.

###### Proof.

Let us prove by induction that each embedding $\iota_{\alpha,\alpha+1}\colon
K_{\alpha}\to K_{\alpha+1}$ extends to an embedding $j_{\alpha,\alpha+1}\colon
K_{\alpha}[j(S)\mathbin{\upharpoonright}\mu_{\alpha}]\to
K_{\alpha+1}[j(S)\mathbin{\upharpoonright}\mu_{\alpha+1}]$.

Indeed, by the arguments of Lemma 5.6, in order to lift this embedding, we
must pick a value for the point $\operatorname{crit}(\iota_{\alpha,\alpha+1})$
in the generic filter for $\iota_{\alpha+1}(\mathbb{S})$. This value is fully
determined by the generic $j_{\alpha,\alpha+1}(S)$. ∎

The next lemma characterizes $\kappa$-complete measures in
$\mathcal{K}[S]$-generic extensions by non-stationary support iterations of
Tree Prikry forcings. This will replace the nice-name-argument used in Theorem
4.2 when claiming that the Laver function captures our $\omega$-sequence of
$\mathbb{P}_{\kappa}$-names for ultrafilters.

Specifically, Lemma 5.8 shows that every measure $\mathcal{U}\in K[S][G]$
admits a _code_ in $H(\kappa^{+})$, hence in the range of the generic Laver
function. This is necessary, as in the present context $\mathbb{P}_{\kappa}$
is not $\kappa$-cc nor there are $(\kappa+2)$-strong embeddings capturing the
names for the ultrafilters.

######  Lemma 5.8 (Coding Lemma).

Let
$\mathbb{P}_{\kappa}=\langle\mathbb{P}_{\alpha},\mathbb{Q}_{\beta}\mathrel{|}\allowbreak\alpha<\beta<\kappa\rangle$
be a non-stationary-supported iteration of tree Prikry forcings in
$\mathcal{K}[S]$. Assume that, for each $\alpha<\kappa$, the iteration has the
following properties:

  1. (1)

$|\mathbb{P}_{\alpha}|\leq 2^{\alpha}$ and $\mathop{1\hskip-3.0pt{\rm
l}}\Vdash_{\mathbb{P}_{\alpha}}``\langle\mathbb{Q}_{\alpha},\leq^{*}\rangle\text{
is }\alpha\text{-closed}$”;

  2. (2)

$\mathop{1\hskip-3.0pt{\rm l}}\Vdash_{\mathbb{P}_{\alpha}}``\forall
p,q\in\mathbb{Q}_{\alpha}$ compatible $p\wedge q$ exists”.111111Here $p\wedge
q$ stands for the $\leq$-greatest condition $r\leq p,q,$ if such a condition
exists.

For each generic filter $G\subseteq\mathbb{P}_{\kappa}$ there are
$\kappa^{+}$-many $\kappa$-complete ultrafilters in $\mathcal{K}[S][G]$.
Moreover, for each such ultrafilter $\mathcal{U}\in\mathcal{K}[S][G]$ there
are

  1. $(\alpha)$

a finite sub-iteration $\iota\colon\mathcal{K}[S]\to\mathcal{K}^{M}[\iota(S)]$
of $j_{\mathcal{U}}\mathbin{\upharpoonright}\mathcal{K}[S]$,

  2. $(\beta)$

an ordinal $\bar{\epsilon}<\iota(\kappa)$ with
$\bar{\epsilon}\in\operatorname{range}(k)$,

  3. $(\gamma)$

$r\in\iota(\mathbb{S}\ast\mathbb{P}_{\kappa})$ with finite support compatible
with all conditions in $\iota\operatorname{{}^{\prime\prime}}G$. Moreover, for
every $p\in G$, $\iota(p)\wedge r$ exists, and

for each $p\in G$,
$p\Vdash_{\mathbb{P}_{\kappa}}``\dot{X}\in\dot{\mathcal{U}}$” if and only if

$\exists q\in\iota(\mathbb{P}_{\kappa})(q\leq^{*}\iota_{n}(p)\wedge
r\,\&\,\operatorname{supp}(q)=\operatorname{supp}(\iota(p)\wedge
r)\,\&\,q\Vdash_{\iota(\mathbb{P}_{\kappa})}\bar{\epsilon}\in\iota_{n}(\dot{X})).$

During the proof of the lemma we will be able to describe $r$ explicitly,
based on the finite iteration $\iota$ (see Definition 5.14). Nevertheless, we
do not know whether in general $r$ can be computed from the iteration $\iota$
itself.

We postpone the proof of this lemma to the end of §5.3 (see p. 5.3).

###  5.2. Interlude on non-stationary-supported iterations

In the road to prove Lemma 5.8 we will need some abstract results about non-
stationary-supported iterations. Essentially, we want a lemma saying the
following: Given a finite normal iteration
$\iota\colon\mathcal{M}\rightarrow\mathcal{N}$, a non-stationary supported
iteration $\mathbb{P}\in\mathcal{M}$, and a dense open set $D\in\mathcal{N}$
for $\iota(\mathbb{P})$, then it suffices to perform a finite modification of
the conditions in $\iota``G$ to enter $D$.

To make this precise we introduce the following handy notation:

######  Definition 5.9.

Let
$\mathbb{P}_{\kappa}=\langle\mathbb{P}_{\alpha},\mathbb{Q}_{\beta}\mathrel{|}\allowbreak\beta<\alpha<\kappa\rangle$
be an iteration of forcings. For $p,q\in\mathbb{P}_{\kappa}$ and
$\Gamma\subseteq\kappa$, write $q\leq_{\Gamma}p$ if $q\leq p$ and

$\\{\alpha<\kappa\mathrel{|}\allowbreak
q\mathbin{\upharpoonright}\alpha\not\Vdash_{\mathbb{P}_{\alpha}}p(\alpha)=q(\alpha)\\}\subseteq\Gamma.$

######  Remark 5.10.

Notice that for $p,q\in\mathbb{P}_{\kappa}$, if $q\leq_{\Gamma}p$ then
$\operatorname{supp}q\subseteq\operatorname{supp}p\cup\Gamma$.

We will eventually apply the forthcoming lemma to a Prikry-type iteration of
Tree Prikry forcings endowed with its direct extension order $\leq^{*}$.

For an exposition on non-stationary-supported iterations of Prikry-type
forcings we refer the reader to Ben-Neria and Unger’s paper [BNU17].

######  Lemma 5.11.

Let
$\mathbb{P}_{\kappa}=\langle\mathbb{P}_{\alpha},\mathbb{Q}_{\beta}\mathrel{|}\allowbreak\beta<\alpha<\kappa\rangle$
be a non-stationary-supported iteration satisfying, for each $\alpha<\kappa$,
the following properties:

  1. (1)

$|\mathbb{P}_{\alpha}|\leq 2^{\alpha}$ and $\mathbb{Q}_{\alpha}$ is trivial
unless $\alpha$ is inaccessible;

  2. (2)

$\mathop{1\hskip-3.0pt{\rm
l}}\Vdash_{\mathbb{P}_{\alpha}}``\langle\mathbb{Q}_{\alpha},\leq\rangle\text{
is }\alpha\text{-closed}$”;

  3. (3)

$\mathop{1\hskip-3.0pt{\rm l}}\Vdash_{\mathbb{P}_{\alpha}}``\forall
p,q\in\mathbb{Q}_{\alpha}$ compatible $p\wedge q$ exists”.

Let $\iota\colon\mathcal{M}\to\mathcal{N}$ be a finite iteration of normal
measures and consider $\langle\iota_{k,\ell}\mathrel{|}\allowbreak
k\leq\ell\leq n\rangle$ the collection of all sub-iterations. Put

$\text{$\mu_{i}:=\operatorname{crit}(\iota_{i,i+1})$ and
$\Gamma:=\\{\iota_{k,\ell}(\mu_{k})\mathrel{|}\allowbreak k\leq\ell\leq
n\\}$}.$

Let $D\in\mathcal{N}$ be a dense open subset of $\iota(\mathbb{P}_{\kappa})$
and $r\in\iota(\mathbb{P}_{\kappa})$ a condition with
$\operatorname{supp}r\subseteq\Gamma$. Let us assume, in addition, that
$p\in\mathbb{P}_{\kappa}$ is a condition such that, for each $q\leq p$, $r$ is
compatible with $\iota(q)$.

Then, there is a dense open set $E\subseteq\mathbb{P}_{\kappa}$ below $p$ such
that for every $q\in E$,

  * $(\aleph)$

$\iota(q)\wedge r$ exists, hence $\iota(q)\wedge r\leq_{\Gamma}\iota(q)$;

  * $(\beth)$

there is $q^{\prime}\leq_{\Gamma}\iota(q)\wedge r$ such that $q^{\prime}\in
D$.

###### Proof.

Before addressing the proof of the lemma we need some preliminary
considerations. Given $u,v\in\mathbb{P}_{\kappa}$ with disjoint supports we
write

$(u\cup v)(\alpha):=\begin{cases}u(\alpha),&\alpha\in\operatorname{supp}u;\\\
v(\alpha),&\alpha\in\operatorname{supp}v.\end{cases}$

Clearly, $u\cup v\in\mathbb{P}_{\kappa}.$ Similarly, if
$u\in\mathbb{P}_{\kappa}$ and $X\cup Y\supseteq\operatorname{supp}u$ with
$X,Y$ disjoint, we can define
$u\mathbin{\upharpoonright}X,u\mathbin{\upharpoonright}Y$ in the natural way
so that

$u=(u\mathbin{\upharpoonright}X)\cup(u\mathbin{\upharpoonright}Y).$

######  Claim 5.11.1.

There is a $\leq_{\iota(\kappa)\setminus\Gamma}$-dense open set $D^{\prime}$
such that for every condition $q^{\prime}\in D^{\prime}$, there is yet another
condition $q^{\prime\prime}\leq_{\Gamma}q^{\prime}$ in $D$.

###### Proof.

Enumerate $\Gamma$ as $\\{\gamma_{0},\dots,\gamma_{m-1}\\}$ and put
$\gamma_{-1}:=0$ and $\gamma_{m}:=\iota(\kappa).$

One can regard $\iota(\mathbb{P}_{\kappa})$ as a finite iteration of the
forcings
$\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}[\gamma_{n-1},\gamma_{n})$’s.
By standard facts about the term-space forcing there is a canonical projection
from $\mathbb{A}({\mathop{1\hskip-3.0pt{\rm
l}}},\mathbb{R}_{0})\times\left(\prod_{n=0}^{m-1}\mathbb{A}(\mathbb{R}_{n},\iota(\mathbb{P}_{\kappa})_{\gamma_{n}})\right)\times\mathbb{A}(\mathbb{R}_{m},\\{\mathop{1\hskip-3.0pt{\rm
l}}\\})$ to $\iota(\mathbb{P}_{\kappa})$, where $\mathbb{R}_{n}$ is the
iteration
$\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}[0,\gamma_{n})$.121212See,
e.g., [Cum10, Proposition 2.25]. In particular, there is
$D_{-1}\times(\prod_{n=0}^{m-1}D_{n})\times D_{m}$ which is dense and open in
this product and that it is contained in $D$.131313Formally speaking, $D$
contains the image of $D_{-1}\times(\prod_{n=0}^{m-1}D_{n})\times D_{m}$ under
the canonical projection.

Let $q^{\prime}\leq_{\iota(\kappa)\setminus\Gamma}q$. By induction on
$n\in[-1,m]$ define $r\leq_{\iota(\kappa)\setminus\Gamma}q^{\prime}$ such that
$q^{\prime\prime}:=r\mathbin{\upharpoonright}\gamma_{0}^{\smallfrown}\langle
s(\gamma_{0})\rangle^{\smallfrown}r\mathbin{\upharpoonright}(\gamma_{0},\gamma_{1})^{\smallfrown}\langle
s(\gamma_{1})\rangle^{\smallfrown}\cdots^{\smallfrown}r\mathbin{\upharpoonright}(\gamma_{m-1},\gamma_{m})$
is a condition where $r\mathbin{\upharpoonright}\gamma_{0}\in D_{-1}$,
$\langle s(\gamma_{n})\mathrel{|}\allowbreak
i<m\rangle\in\prod_{n=0}^{m-1}D_{n}$ and
$r\mathbin{\upharpoonright}(\gamma_{m-1},\gamma_{m})\in D_{m}$. This is
possible by density and openess of the $D_{n}$’s. Clearly,
$q^{\prime\prime}\leq_{\Gamma}r$ and $q^{\prime\prime}\in D$. Therefore there
is a $\leq_{\iota(\kappa)\setminus\Gamma}$-dense open set $D^{\prime}$ as in
the claim. ∎

By the claim, it is enough to show that for every
$\leq_{\iota(\kappa)\setminus\Gamma}$-dense open set $D\in\mathcal{N}$ the set
$E\subseteq\mathbb{P}$ of all conditions $q$ such that $\iota(q)\wedge r\in D$
is dense: Indeed, fix $D\in\mathcal{N}$ and $r\in\iota(\mathbb{P}_{\kappa})$
be as in the lemma. Invoking Claim 5.11.1 with respect to $D$ we get in return
a $\leq_{\iota(\kappa)\setminus\Gamma}$-dense open set $D^{\prime}$ with the
above-mentioned property. Using our assumption we have that the set
$E\subseteq\mathbb{P}$ of all $q$’s such that $\iota(q)\wedge r\in D^{\prime}$
is dense. By the properties of $D^{\prime}$ there is a condition
$q^{\prime}\leq_{\Gamma}\iota(q)\wedge r$ in $D$, so we are done.

We will eventually prove the lemma by induction on the length of the
iteration. Let us first deal with the case of a single ultrapower. In order to
make the inductive hypothesis run smoothly we are going to tweak it slightly.
Specifically, the set $\Gamma$ of the forthcoming proof will contain the
critical points of the embedding, their images and other possible cardinals.

So, let $\iota\colon\mathcal{M}\to\mathcal{N}$ be a normal ultrapower
embedding with critical point $\mu$ and let $\Gamma^{\prime}$ be a finite set
of inaccessible cardinals. Put
$\Gamma:=\iota(\Gamma^{\prime})\cup\\{\mu,\iota(\mu)\\}$. Let
$\mathbb{P}_{\kappa}$ be our non-stationary-supported iteration,
$r\in\iota(\mathbb{P}_{\kappa})$ with $\operatorname{supp}r=\Gamma$ and
$D\subseteq\iota(\mathbb{P}_{\kappa})$ a dense open set with respect to
$\leq_{\iota(\kappa)\setminus\Gamma}$.

######  Claim 5.11.2.

Let $E$ be the collection of all conditions $q$ such that

  1. (1)

either there is $\alpha\in\Gamma$ such that

$\iota(q)\mathbin{\upharpoonright}\alpha\Vdash_{\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}\alpha}\text{$``\iota(q)(\alpha)$
is incompatible with $r(\alpha)$''}$

  2. (2)

or $\iota(q)\wedge r\leq_{\Gamma}\iota(q)$ and $\iota(q)\wedge r\in D$.

Then $E$ is $\leq_{\kappa\setminus(\Gamma^{\prime}\cup\\{\mu\\})}$-dense open.

Proving the claim suffices to infer the first step of the induction: In
effect, suppose that $p\in\mathbb{P}_{\kappa}$ is such that $\iota(q)$ is
compatible with $r$ for all conditions $q\leq p$. In that case,
$E\cap(\mathbb{P}_{\kappa}/p)$ is a dense open set $\leq$-below $p$ consisting
of conditions $q$ for which clause (2) above holds.

###### Proof of Claim 5.11.2.

$E$ is open in $\leq_{(\kappa\setminus\Gamma^{\prime}\cup\\{\mu\\})}$ so we
shall show it is dense.

Fix $d$ and $\tilde{r}$ be functions with domain $\mu$ representing $D$ and
$r$, respectively; i.e., $\iota(d)(\mu)=D$ and $\iota(\tilde{r})(\mu)=r$.
Without loss of generality we may assume that, for each $\alpha<\mu$,
$\operatorname{supp}\tilde{r}(\alpha)=\Gamma^{\prime}\cup\\{\alpha,\mu\\}$ and
that $d(\alpha)$ is
$\leq_{\kappa\setminus\Gamma^{\prime}\cup\\{\alpha,\mu\\}}$-dense open. In
what follows $q$ will be a fixed condition in $\mathbb{P}_{\kappa}$.

We inductively define a
$\leq_{\kappa\setminus\Gamma^{\prime}\cup\\{\mu\\}}$-decreasing sequence
$\langle
q_{\alpha}\mathrel{|}\allowbreak\alpha\leq\mu\rangle\subseteq\mathbb{P}_{\kappa}$
below $q$ together with a continuously-decreasing sequence of clubs $\langle
C_{\alpha}\mathrel{|}\allowbreak\alpha<\mu\rangle$ on $\mu$ such that
$(\operatorname{supp}q_{\alpha})\cap C_{\alpha}=\emptyset$. For each
$\alpha<\mu$ we stipulate that

$\gamma_{\alpha}:=\min(C_{\alpha}\setminus(\sup_{\beta<\alpha}\gamma_{\beta})+1).$

Our mission is to find a dense subset of conditions
$e(\gamma_{\alpha})\subseteq\mathbb{P}_{\gamma_{\alpha}+1}$ with the following
property: for all $s\in e(\gamma_{\alpha})$ such that $s\leq
q_{\alpha}\mathbin{\upharpoonright}\gamma_{\alpha}+1$, put
$q^{\prime}_{s}:=s\cup
q_{\alpha}\mathbin{\upharpoonright}[\gamma_{\alpha}+1,\kappa)$. Then,

  1. $(\alpha)$

either $q^{\prime}_{s}$ is incompatible with $\tilde{r}(\gamma_{\alpha})$,

  2. $(\beta)$

or $q^{\prime}_{s}\wedge\tilde{r}(\gamma_{\alpha})$ exists,
$q^{\prime}_{s}\wedge\tilde{r}(\gamma_{\alpha})\leq_{\Gamma^{\prime}\cup\\{\gamma_{\alpha},\mu\\}}q^{\prime}_{s}$
and $q^{\prime}_{s}\wedge\tilde{r}(\gamma_{\alpha})\in d(\gamma_{\alpha})$.

We will inductively maintain that
$q_{\alpha}\mathbin{\upharpoonright}\gamma_{\alpha}+1=q_{\beta}\mathbin{\upharpoonright}\gamma_{\alpha}+1$
for all $\alpha<\beta$. In particular, $\gamma_{\alpha}$ will never be in the
support of $q_{\beta}$, as $(\operatorname{supp}q_{\alpha})\cap
C_{\alpha}=\emptyset$. Moreover, we will assume that $q_{\alpha}$ agrees with
$q$ on $\Gamma^{\prime}\cup\\{\mu\\}$. We proceed with the construction of
$q_{\alpha}$ in a fusion-like fashion.

$\blacktriangleright$ Set $q_{0}:=q$ and let $C_{0}\subseteq\mu$ be any club
disjoint from $\operatorname{supp}q_{0}$.

$\blacktriangleright$ Assume that $\langle
q_{\alpha}\mathrel{|}\allowbreak\alpha<\beta\rangle$ and $\langle
C_{\alpha}\mathrel{|}\allowbreak\alpha<\beta\rangle$ were constructed for some
$\beta>0$. If $\beta$ is limit then, by our inductive assumption, there is a
condition $q^{*}_{\beta}$ that serves as a lower bound for $\langle
q_{\alpha}\mathrel{|}\allowbreak\alpha<\beta\rangle$ and whose support is
disjoint from $\bigcap_{\alpha<\beta}C_{\alpha}$. Note that this latter is a
club for all $\beta<\mu$. Likewise, for $\beta=\mu$, the assumption on
freezing initial segments ensures that this lower bound exists and that its
support is disjoint from $\bigtriangleup_{\beta<\mu}C_{\alpha}.$

Let $\beta<\mu$ be arbitrary. Let us show how to define $q_{\beta}$ together
with the dense set $e(\gamma_{\beta})$. Let us consider $\langle
s_{\zeta}\mathrel{|}\allowbreak\zeta<\zeta_{*}\rangle$, an enumeration of all
the conditions in $\mathbb{P}_{\gamma_{\beta}+1}$ which are below
$q^{*}_{\beta}\mathbin{\upharpoonright}\gamma_{\beta}+1$. By our departing
assumption upon $\mathbb{P}_{\kappa}$, $\zeta_{*}\leq 2^{\gamma_{\beta}}$,
which is below the degree of closure of
$\mathbb{P}_{\kappa}/\mathbb{P}_{\gamma_{\beta}+1}.$ This enable us to define
a decreasing sequence $\langle
q^{\beta}_{\zeta}\mathrel{|}\allowbreak\zeta<\zeta_{*}\rangle$ as follows:

$\blacktriangleright$ If $s_{0}\cup
q^{*}_{\beta}\mathbin{\upharpoonright}[\gamma_{\beta}+1,\kappa)$ is
incompatible with $\tilde{r}(\gamma_{\beta})$ then stipulate
$q^{\beta}_{0}:=q^{*}_{\beta}$. Otherwise, $(s_{0}\cup
q^{*}_{\beta}\mathbin{\upharpoonright}[\gamma_{\beta}+1,\kappa))\wedge\tilde{r}(\gamma_{\beta})$
exists and we can
$\leq_{\kappa\setminus\Gamma^{\prime}\cup\\{\gamma_{\beta},\mu\\}}$-extend it
to a condition $q^{\prime\prime}\in d(\gamma_{\beta}).$ In this case define

$q^{\beta}_{0}:=(q^{*}_{\beta}\mathbin{\upharpoonright}\gamma_{\beta}+1)\cup(q^{\prime\prime}\mathbin{\upharpoonright}[\gamma_{\beta}+1,\kappa)).$

$\blacktriangleright$ In general, let $r_{\eta}$ be a lower bound for $\langle
q^{\beta}_{\zeta}\mathrel{|}\allowbreak\zeta<\eta\rangle$ and proceed exactly
as before with respect to the condition $s_{\eta}\cup
r_{\eta}\mathbin{\upharpoonright}[\gamma_{\beta}+1,\kappa)$.

Finally, let $q_{\beta}$ be the condition

$q^{*}_{\beta}\mathbin{\upharpoonright}\left(\gamma_{\beta+1}\cup\Gamma^{\prime}\cup\\{\mu\\}\right)\cup
q^{\beta}_{\zeta*}\mathbin{\upharpoonright}\left([\gamma_{\beta}+1,\kappa)\setminus\Gamma^{\prime}\right),$

and $e(\gamma_{\beta})$ be the dense open set defined as

$\begin{matrix}\\{s\mathrel{|}\allowbreak
s\perp(q^{*}_{\beta}\mathbin{\upharpoonright}\gamma_{\beta}+1)\\}\cup\\\
\\{s\mathrel{|}\allowbreak\exists\zeta(s=s_{\zeta}\;\wedge\;s\,\cup\,q^{\beta}_{\zeta}\mathbin{\upharpoonright}[\gamma_{\beta}+1,\kappa)\text{
witnesses $(\alpha)$ or $(\beta)$})\\}.\end{matrix}$

Applying $\iota$, for densely many $s\in\iota(e)(\mu)$, which are below
$q_{\mu}\mathbin{\upharpoonright}\mu+1=\iota(q)_{\mu}\mathbin{\upharpoonright}\mu+1$,
the condition
$s\cup\iota(q_{\mu})\mathbin{\upharpoonright}[\mu+1,\iota(\kappa))$ satisfies
the requirements of the claim (here we use the fact that $\iota(q_{\mu})$ is
stronger than $\iota(q)_{\mu}$). But $s\leq
q_{\mu}\mathbin{\upharpoonright}\mu+1$ and therefore $s\cup
q_{\mu}\mathbin{\upharpoonright}[\mu,\kappa)\leq q$ belongs to $E$. ∎

As mentioned earlier we will prove the lemma by induction on the length of the
iteration $\iota\colon\mathcal{M}\rightarrow\mathcal{N}$. Recall that it is
enough to show it for $\leq_{\iota(\kappa)\setminus\Gamma}$-dense open sets.
So assume by induction that the previous claim holds for iterations of length
$n-1$. Let $\iota\colon\mathcal{M}\rightarrow\mathcal{N}_{n}$ be an iteration
of length $n$, $D\in\mathcal{N}$ a $\leq_{\iota(\kappa)\setminus\Gamma}$-dense
open set and $r\in\iota(\mathbb{P}_{\kappa})$ with
$\operatorname{supp}r\subseteq\Gamma$. Let $\mu$ be the last critical point of
$\iota$ and denote by
$\iota_{n-1}\colon\mathcal{M}\rightarrow\mathcal{N}_{n-1}$ and
$\iota_{n-1,n}\colon\mathcal{N}_{n-1}\to\mathcal{N}_{n}$ the first $n-1$
ultrapowers and the last ultrapower of $\iota$, respectively.

Let $\Gamma_{n-1}$ be the collection of critical points of $\iota_{n-1}$
together with their images. Then
$\Gamma=\iota_{n-1,n}(\Gamma_{n-1})\cup\\{\mu,\iota_{n-1,n}(\mu)\\}$. Applying
Claim 5.11.2 to $\iota_{n-1,n}$ we get a
$\leq_{\iota_{n-1}(\kappa)\setminus\Gamma_{n-1}\cup\\{\mu_{n-1}\\}}$-dense
open set $E^{\prime}$ with the above-stated properties. Next we invoke our
induction hypothesis to $\iota_{n-1}$, $E^{\prime}$ and
$r\mathbin{\upharpoonright}\mu_{n-1}$ and find a dense open $E$ satisfying the
conclusion of Claim 5.11.2.

Let $q\in E$ be compatible with
$r\mathbin{\upharpoonright}\mu_{n-1}$.141414Recall that we were assuming the
existence of a condition $p\in\mathbb{P}_{\kappa}$ such that $\iota(q)$ is
compatible with $r$ for all $q\leq p.$ Since $E$ is dense, any condition in
$E\cap(\mathbb{P}_{\kappa}/q)$ works. Then, $\iota_{n-1}(q)\wedge
r\mathbin{\upharpoonright}\mu_{n-1}$ is $\leq_{\Gamma_{n-1}}$-below
$\iota_{n-1}(q)$ and there is
$q^{\prime}\leq_{\Gamma_{n-1}}\iota_{n-1}(q)\wedge
r\mathbin{\upharpoonright}\mu_{n-1}$ in $E^{\prime}$.

Note that

$\iota_{n-1,n}(q^{\prime})\leq_{\iota_{n-1}(\Gamma_{n-1})}\iota_{n-1,n}(\iota_{n-1}(q)\wedge
r\mathbin{\upharpoonright}\mu_{n-1})=\iota(q)\wedge
r\mathbin{\upharpoonright}\mu_{n-1},$

hence $\iota_{n-1,n}(q^{\prime})$ is compatible with $r$. Thus, by the
property of $E^{\prime}$, there is
$q^{\prime\prime}\leq_{\Gamma}\iota_{n-1}(q^{\prime})\wedge r$ in $D$. Since
$\iota_{n-1}(q^{\prime})\wedge r\leq_{\Gamma}\iota_{n}(q)\wedge r$ this yields
a condition $q^{\prime\prime}\leq_{\Gamma}\iota(q)\wedge r$, which
accomplishes the proof. ∎

###  5.3. Proof of the coding lemma

After a short digression with non-stationary-supported iterations we focus on
proving Lemma 5.8.

######  Setup.

For the rest of this section our ground model is $\mathcal{K}[S]$ and
$\mathbb{P}_{\kappa}$ is a non-stationary-supported iteration (in
$\mathcal{K}[S]$) as described in Lemma 5.8.

We will fix $G\subseteq\mathbb{P}_{\kappa}$ a $V$-generic filter and consider
$j\colon V[G]\to M[H]$, a $V[G]$-elementary embedding with
$\operatorname{crit}(j)=\kappa$. Notice that, by virtue of our anti-large-
cardinal assumption, $j\mathbin{\upharpoonright}\mathcal{K}$ is a normal
iteration and thus so is $j\mathbin{\upharpoonright}V$ (by Lemma 5.7), but
those iterations can be infinite151515In fact, they must be quite long, as the
forcing $j(\mathbb{P}_{\kappa})$ adds many $\omega$-sequences. Invoking Lemma
5.2, for each $x\in M$, there is a finite iteration $\iota\colon V\rightarrow
N$ such that $k\circ\iota=j\mathbin{\upharpoonright}V$ and
$x\in\mathrm{range}(k)$, where $k\colon N\rightarrow M$ is the factor map
between $\iota$ and $j$ (we will say that _$\iota$ factors
$j\mathbin{\upharpoonright}V$_).

######  Lemma 5.12.

Suppose that $\iota\colon V\to N$ is a finite normal iteration with critical
point at least $\kappa$. Then, $N[G]$ is closed under ${<}\kappa$-sequences in
$V[G]$.

Moreover, for every such sequence $\vec{\alpha}$ there is $\zeta<\kappa$ such
that $\vec{\alpha}\in N[G\mathbin{\upharpoonright}\zeta].$

###### Proof.

As usual, let $\vec{\mu}$ be the sequence of critical points. Since
$\min\vec{\mu}\geq\kappa$, $V_{\kappa+1}\subseteq N$ and thus
$\mathbb{P}_{\kappa}$ is computed correctly in $N$. In particular, the model
$N[G]$ is well-defined and it is the generic extension of $N$ by $G$.

Let $\langle\alpha_{i}\mathrel{|}\allowbreak i<i_{*}\rangle$ be a sequence of
ordinals in $V[G]$ with $i_{*}<\kappa$. For each $i<i_{*}$ fix a function
$V\ni f_{i}\colon\kappa^{n}\to\mathrm{Ord}$ such that
$\alpha_{i}=\iota(f_{i})(\vec{\mu})$. Everything boils down to show that
$\langle f_{i}\mathrel{|}\allowbreak i<i_{\star}\rangle$ belongs to
$V[G\mathbin{\upharpoonright}\gamma]$ for some $\gamma<\kappa$.

Invoking the Axiom of Choice in $V$ we let
$h\colon\operatorname{dom}(h)\rightarrow V$, $h\in V$, with
$\operatorname{dom}(h)\in\operatorname{Ord}$ and
$f_{i}\in\operatorname{range}(h)$ for all $i<i_{\star}$. This choice is
possible in that each $f_{i}$ belongs to the ground model. Note that a priori,
we do not put any bound on $\operatorname{dom}h$. Put
$a_{\star}:=\\{h^{-1}(f_{i})\mathrel{|}\allowbreak i<i_{\star}\\}$ and note
that this is a set of ordinals in $V[G]$ whose order-type is $<\kappa$.

######  Claim 5.12.1.

Let $a$ be a set of ordinals in $V[G]$ with $\operatorname{otp}a<\kappa$.
Then, there is $\zeta<\kappa$ and a function $g\colon\zeta\to\mathrm{Ord}$ in
$V$ such that $a\subseteq\operatorname{range}g$.

###### Proof.

First, we may find a function
$\bar{g}\colon\mathbb{P}_{\kappa}\times\operatorname{otp}a\rightarrow\sup a$
in $V$ such that $a\subseteq\operatorname{range}\bar{g}$. In effect, let for
example $\bar{g}$ be send each pair $\langle
p,\xi\rangle\in\mathbb{P}_{\kappa}\times\operatorname{otp}a$ to the $\xi$-th
element of $a$ decided by $p$, if this exists; otherwise, set
$\bar{g}(p,\xi):=0$. Since $|\mathbb{P}_{\kappa}|=2^{\kappa}$, by well
ordering $\mathbb{P}_{\kappa}$, we may assume that

$\operatorname{dom}(\bar{g})=(2^{\kappa})^{V}=(\kappa^{+})^{V}.$

Since $(\kappa^{+})^{V}=(\kappa^{+})^{V[G]}$ is regular in $V[G]$, $\sup
a<\kappa^{+}$. In particular, there must be some ordinal
$\zeta^{\prime}<\kappa^{+}$ such that
$\operatorname{range}(\bar{g}\mathbin{\upharpoonright}\zeta^{\prime})$ already
covers $a$. Note that this $\zeta^{\prime}$ might be determined only in the
generic extension $V[G]$.

Next, let us pick a surjection $\varphi\colon\kappa\to\zeta^{\prime}$ in $V$.
Then, $\bar{g}\circ\varphi\colon\kappa\to\sup a$ covers $a$. But $\kappa$ is
regular in $V[G]$ as well, so we apply the same argument and obtain some
$\zeta<\kappa$ such that
$g:=\bar{g}\circ\varphi\mathbin{\upharpoonright}\zeta$ already covers $a$. ∎

Using the above claim we have that $a_{\star}\subseteq\operatorname{range}g$
for some $g\colon\zeta\rightarrow V$ in $V$ with $\zeta<\kappa.$ Thus, in
order to decide the value of $a_{*}$, it is enough to decide the value of
$g^{-1}(a_{*})$ which is a subset of $\zeta$.

Using the closure of the $\leq^{*}$-order, we may decide the values of
$a_{\star}$ up to the forcing
$\mathbb{P}_{\kappa}\mathbin{\upharpoonright}\zeta+1$. This is achieved by
repeatedly strengthening the components of the condition above $\zeta$. For
details, see Claim 5.11.2.

Since $a_{\star}\in V[G\mathbin{\upharpoonright}\zeta+1]$ and $h\in V$ it
follows that $h``a_{\star}=\\{f_{i}\mathrel{|}\allowbreak i<i_{\star}\\}$
belongs to $V[G\mathbin{\upharpoonright}\zeta+1]$, hence it also does $\langle
f_{i}\mathrel{|}\allowbreak i<i_{\star}\rangle$. Now, the embedding $\iota$
lifts uniquely (as the forcing
$\mathbb{P}_{\kappa}\mathbin{\upharpoonright}(\zeta+1)$ is small) to an
embedding $\iota_{*}$. Thus, $\iota_{*}(\langle f_{i}\mathrel{|}\allowbreak
i<i_{*}\rangle)=\langle\iota(f_{i})\mathrel{|}\allowbreak i<i_{*}\rangle\in
N[G\mathbin{\upharpoonright}\zeta+1]\subseteq N[G]$. By plugging the
generators $\vec{\mu}$ to this sequence we infer that
$\langle\alpha_{i}\mathrel{|}\allowbreak i<i_{\star}\rangle\in N[G].$ ∎

The next will help us isolate the condition $r$ from Lemma 5.8$(\gamma)$:

######  Lemma 5.13.

Let $\iota\colon V\to N$ be a finite iteration and $k\colon N\rightarrow M$ be
such that $k\circ\iota=j\mathbin{\upharpoonright}V.$ Then, the set $A_{\iota}$
defined as

$\\{\alpha\in[\kappa,\iota(\kappa))\mathrel{|}\allowbreak\exists r\in
k^{-1}(H)\,\forall p\in
G\,(k(r(\alpha))_{H\mathbin{\upharpoonright}k(\alpha)}<j(p)(k(\alpha))_{H\mathbin{\upharpoonright}k(\alpha)})\\}$

is finite. Moreover, for each $\alpha\in A_{\iota}$

$\bar{H}_{\alpha}=\\{\mathrm{stem}(k(r(\alpha)))_{H\mathbin{\upharpoonright}k(\alpha)}\mathrel{|}\allowbreak\text{$r$
witnesses $\alpha\in A_{\iota}$}\\}$

is finite and has $\bigcup\bar{H}_{\alpha}$ as a maximal element.

###### Proof.

Let us show that $\bar{H}:=\bigcup_{\alpha\in A}\bar{H}_{\alpha}$ is finite.
Assume towards a contradiction that $\bar{H}$ is infinite and let $\langle
k(\dot{\eta}_{n})_{H\mathbin{\upharpoonright}k(\alpha_{n})}\mathrel{|}\allowbreak
n<\omega\rangle\in V[G]$ be a $1$-$1$ sequence witnessing this. Put
$\vec{\alpha}:=\langle\alpha_{n}\mathrel{|}\allowbreak n<\omega\rangle\in
V[G]$. Note that we allow $\vec{\alpha}$ to exhibit repetitions, despite
$\bar{H}$ is assumed to be infinite.

For each $n<\omega$, let $r_{n}\in k^{-1}(H)$ be such that

$\mathrm{stem}(k(r(\alpha_{n})))_{H\mathbin{\upharpoonright}k(\alpha_{n})}=k(\dot{\eta}_{n})_{H\mathbin{\upharpoonright}\alpha_{n}}$

and

$k(r_{n}(\alpha_{n}))_{H\mathbin{\upharpoonright}\alpha_{n}}<j(p)(k(\alpha_{n}))_{H\mathbin{\upharpoonright}\alpha_{n}}.$

In particular, as $k(r_{n})\in H$,
$k(\dot{\eta}_{n})_{H\mathbin{\upharpoonright}k(\alpha_{n})}$ is an initial
segment of the $k(\alpha_{n})^{\mathrm{th}}$-Prikry sequence introduced by
$H$.

By the proof of Lemma 5.12, there is $\zeta<\kappa$ such that
$\langle\dot{\alpha}_{n}\mathrel{|}\allowbreak
n<\omega\rangle,\langle\dot{\eta}_{n}\mathrel{|}\allowbreak n<\omega\rangle\in
N[G\mathbin{\upharpoonright}\zeta]$. Clearly, $k$ lifts to $k\colon
N[G\mathbin{\upharpoonright}\zeta]\rightarrow
M[G\mathbin{\upharpoonright}\zeta]$ so that $\langle
k(\alpha_{n})\mathrel{|}\allowbreak n<\omega\rangle,\langle
k(\dot{\eta}_{n})\mathrel{|}\allowbreak n<\omega\rangle\in
M[G\mathbin{\upharpoonright}\zeta]\subseteq M[H].$ Also, $\langle
j(p_{n})\mathrel{|}\allowbreak n<\omega\rangle\in M[H]$.

Combining all of these inputs we find $q\in H$ such that, for each $n<\omega$,

(1)
$q\mathbin{\upharpoonright}k(\alpha_{n})\Vdash_{j(\mathbb{P})\mathbin{\upharpoonright}k(\alpha_{n})}\text{$``k(\dot{\eta}_{n})$
is an initial segment of $\mathrm{stem}(q(k(\alpha_{n})))$''}.$

and

(2)
$q\mathbin{\upharpoonright}k(\alpha_{n})\Vdash_{j(\mathbb{P})\mathbin{\upharpoonright}k(\alpha_{n})}\text{$``\forall
p\in\dot{G}\;(k(\dot{\eta}_{n})\nsubseteq\mathrm{stem}(j(p)(k(\alpha_{n})))$''}.$

We now produce a contradiction by distinguishing two cases: either
$\vec{\alpha}$ is finite or it is not. In the first case there is
$\alpha_{*}\in\vec{\alpha}$ and $I\in[\omega]^{\omega}$ such that
$\alpha_{*}=\alpha_{n}$ for all $n\in I.$ In particular $\langle
k(\dot{\eta}_{n})_{H\mathbin{\upharpoonright}k(\alpha_{*})}\mathrel{|}\allowbreak
n\in I\rangle$ is bounded by
$\dot{\mathrm{stem}}(\dot{q}(k(\alpha_{*})))_{H\mathbin{\upharpoonright}k(\alpha_{*})}$
and thus there are $n,m\in I$ such that
$k(\dot{\eta}_{n})_{H\mathbin{\upharpoonright}k(\alpha_{*})}=k(\dot{\eta}_{m})_{H\mathbin{\upharpoonright}k(\alpha_{*})}$.
This contradicts injectivity of $\langle
k(\dot{\eta}_{n})_{H\mathbin{\upharpoonright}k(\alpha_{n})}\mathrel{|}\allowbreak
n<\omega\rangle.$

Thus, assume that $\vec{\alpha}$ is infinite.

######  Claim 5.13.1.

There is $p_{*}\in G$ such that
$\vec{\alpha}\setminus\vec{\mu}\subseteq\operatorname{dom}(\iota(p_{*}))$.

###### Proof of claim.

Since $\vec{\alpha}\setminus\vec{\mu}\in V[G]$ there is $\zeta<\kappa$ with
$\vec{\alpha}\setminus\vec{\mu}\in N[G\mathbin{\upharpoonright}\zeta]$. A
chain condition argument with $\mathbb{P}_{\zeta}$ allows us to find a set
$B\subseteq[\kappa,\iota(\kappa))$, $|B|<\kappa$ such that
$\mathop{1\hskip-3.0pt{\rm
l}}\Vdash_{\mathbb{P}_{\zeta}}\dot{\vec{\alpha}}\setminus\vec{\mu}\subseteq\check{B}$
and $B\cap\vec{\mu}=\emptyset$. By Lemma 5.5 there must be a condition
$p_{*}\in G$ with $B\subseteq\operatorname{dom}(\iota(p_{*}))$. Thus, $p_{*}$
is as wished. ∎

Let $s\leq q,j(p_{*})$ be in $H$. Then for some $\alpha_{n_{*}}<\omega$

$s\mathbin{\upharpoonright}k(\alpha_{n_{*}})\Vdash_{j(\mathbb{P})\mathbin{\upharpoonright}k(\alpha_{n_{*}})}s(\alpha_{n_{*}})\leq^{*}j(p_{*})(k(\alpha_{n_{*}}))$

and

$s\mathbin{\upharpoonright}k(\alpha_{n_{*}})\Vdash_{j(\mathbb{P})\mathbin{\upharpoonright}k(\alpha_{n_{*}})}s(\alpha_{n_{*}})\leq^{*}q(k(\alpha_{n_{*}})).$

Combining this with (1) and (2) above we have

$s\mathbin{\upharpoonright}k(\alpha_{n_{*}})\Vdash_{j(\mathbb{P})\mathbin{\upharpoonright}k(\alpha_{n_{*}})}k(\dot{\eta}_{n_{*}})\sqsubseteq\dot{\mathrm{stem}}(j(p_{*})(k(\alpha_{n_{*}}))),$

and

$s\mathbin{\upharpoonright}k(\alpha_{n^{*}})\Vdash_{j(\mathbb{P})\mathbin{\upharpoonright}k(\alpha_{n})}\text{$k(\dot{\eta}_{n_{*}})\nsubseteq\mathrm{stem}(j(p_{*})(k(\alpha_{n_{*}})))$},$

which is impossible.

All in all we have shown that $\bar{H}$ is finite and thus so are $A_{\iota}$
and the $\bar{H}_{\alpha}$’s. Finally, we argue that $\bigcup\bar{H}_{\alpha}$
is the maximal element of $\bar{H}_{\alpha}:$

######  Claim 5.13.2.

Every two members of $\bar{H}_{\alpha}$ are $\sqsubseteq$-compatible.

###### Proof of claim.

Suppose
$k(\dot{\eta})_{H\mathbin{\upharpoonright}\alpha},k(\dot{\sigma})_{H\mathbin{\upharpoonright}\alpha}\in\bar{H}_{\alpha}$
and let $r,r^{\prime}\in k^{-1}(H)$ witnessing this. Then there is $H\ni s\leq
k(r),k(r^{\prime})$ such that

$s\mathbin{\upharpoonright}k(\alpha)\Vdash_{j(\mathbb{P})\mathbin{\upharpoonright}k(\alpha)}\text{$``k(r(\alpha))$
and $k(r^{\prime}(\alpha))$ are compatible''},$

which yields $k(\dot{\eta})_{H\mathbin{\upharpoonright}k(\alpha)}\sqsubseteq
k(\dot{\sigma})_{H\mathbin{\upharpoonright}k(\alpha)}$ or the other way
around. ∎

This is the end of the proof. ∎

By the previous lemma we can associate to each finite iteration $\iota\colon
V\rightarrow N$ (factoring $j\mathbin{\upharpoonright}V$) a finite list of
names for stems
$\vec{\eta}:=\langle\dot{\eta}_{\alpha}\mathrel{|}\allowbreak\alpha\in
A_{\iota}\rangle$ where
$k(\dot{\eta}_{\alpha})_{H\mathbin{\upharpoonright}k(\alpha)}$ is the maximal
element of $\bar{H}_{\alpha}$. We may assume, in addition, that
$\eta_{\alpha}$ is forced by the weakest condition of
$\iota(\mathbb{P}_{\kappa})_{\alpha}$ to be a potential stem at coordinate
$\alpha$. This makes the following to be a sound definition:

######  Definition 5.14.

Let $\iota\colon V\rightarrow N$ be a finite iteration factoring
$j\mathbin{\upharpoonright}V.$

We will denote by $r_{\iota}$ the condition with
$\operatorname{supp}r_{\iota}=A_{\iota}$ such that $r(\alpha)$ is a
$\iota(\mathbb{P}_{\kappa})_{\alpha}$-name for a maximal tree with stem
$\dot{\eta}_{\alpha}$ and $A_{\iota}$ being as in Lemma 5.13.

The following gives better control on the support of $r_{\iota}$:

######  Lemma 5.15.

Let $\iota\colon V\rightarrow N$ be a finite iteration factoring
$j\mathbin{\upharpoonright}V.$

Using the notations of Lemmas 5.11 and 5.13,
$A_{\iota}\subseteq\Gamma_{\iota}$.

###### Proof.

Assume otherwise and let $\alpha<\iota(\kappa)$ be such that $\alpha\in
A_{\iota}$ but $\alpha\notin\Gamma_{\iota}$. Let $\tau$ be a
$\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}\alpha$-name for the
ordinal $\max\dot{\eta}_{\alpha}$. We may assume that this name is forced by
the trivial condition to be strictly below $\alpha$. Let $D_{0}$ be the set of
all conditions $q\in\iota(\mathbb{P}_{\kappa})$ with
$\alpha\in\operatorname{supp}q$ such that

$\mathop{1\hskip-3.0pt{\rm
l}}\Vdash_{\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}\alpha}\min(\mathrm{Lev}_{1}(T^{q(\alpha)}))>\tau.$

An easy application of the mixing lemma shows that $D_{0}$ is $\leq^{*}$-dense
open in $\iota(\mathbb{P}_{\kappa})$. The goal is to find a condition $p\in G$
with $\iota(p)\in D_{0}$.

Recall that for $q,q^{\prime}\in\iota(\mathbb{P}_{\kappa})$ we write
$q^{\prime}\leq^{*}_{\Gamma_{\iota}}q$ if $q^{\prime}\leq^{*}q$ and

$\\{\beta<\iota(\kappa)\mathrel{|}\allowbreak
q^{\prime}\mathbin{\upharpoonright}\beta\nVdash_{\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}\beta}q^{\prime}(\beta)=q(\beta)\\}\subseteq{\Gamma_{\iota}}.$

Define $D:=\\{q\in\iota(\mathbb{P}_{\kappa})\mathrel{|}\allowbreak\forall
q^{\prime}\leq^{*}_{\Gamma_{\iota}}q\,(q^{\prime}\in D_{0})\\}$. It follows
that $D$ is $\leq^{*}$-dense open in $\iota(\mathbb{P}_{\kappa})$. Also, if
$q\in\iota(\mathbb{P}_{\kappa})$, $\alpha\in\operatorname{supp}q$ and $D\ni
q^{\prime}\leq^{*}_{\Gamma_{\iota}}q$ then $q\in D_{0}$.171717Because we are
assuming that $\alpha\notin\Gamma_{\iota}.$

Using Lemma 5.11, for each each $r\in\iota(\mathbb{P}_{\kappa})$ with
$\operatorname{supp}r\subseteq{\Gamma_{\iota}}$, there is a $\leq^{*}$-dense
open set $E\subseteq\mathbb{P}_{\kappa}$ with the following property: for
every $p\in E$ there is $D\ni
q^{\prime}\leq^{*}_{\Gamma_{\iota}}\iota(p)\wedge r$. Thus, if
$\alpha\in\operatorname{supp}\iota(p)$ then $\iota(p)\in D_{0}$.

We note that $\alpha\in\operatorname{supp}p$ for $\leq^{*}$-densely many $p\in
E$: by Claim 5.3.1, as $\alpha\notin{\Gamma_{\iota}}$ (and in particular
$\alpha$ is not one of the critical points of the iteration) there is a
nowhere dense set $T$ such that $\alpha\in\iota(T)$. Thus, if
$\operatorname{supp}p\supseteq T$, $\alpha\in\operatorname{dom}\iota(p)$.
Clearly, the collection of conditions $p$ with support $\operatorname{supp}p$
containing a given nowhere stationary set $T$ is dense open.

By density, there is $p\in G\cap E$ with $\iota(p)\in D_{0}$. Now use that
$\alpha\in A_{\iota}$ to find $r\in k^{-1}(H)$ such that
$k(r(\alpha))_{H\mathbin{\upharpoonright}k(\alpha)}<j(p)(k(\alpha))_{H\mathbin{\upharpoonright}k(\alpha)}$.
On one hand,

$r\mathbin{\upharpoonright}\alpha\Vdash_{\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}\alpha}\max(\mathrm{stem}(r(\alpha)))\leq\tau.$

On the other hand, since $\iota(p)\in D_{0}$,

$r\mathbin{\upharpoonright}\alpha\Vdash_{\iota(\mathbb{P}_{\kappa})\mathbin{\upharpoonright}\alpha}\max(\mathrm{stem}(r(\alpha)))>\tau.$

This is a contradiction. ∎

The previous lemma acknowledges that the condition $r_{\iota}$ of Definition
5.14 complies with the requirements of Lemma 5.11.

We need a further technical lemma. The following lemma gives us almost a
complete description of the conditions in a finite iteration that are going to
be sent to $H$.

######  Lemma 5.16.

Let $\iota\colon V\rightarrow N$ be a finite iteration factoring
$j\mathbin{\upharpoonright}V.$

For every $q\in H$ with $q\in\operatorname{range}(k)$, there is $p\in G$ and
$q^{\prime}\in k^{-1}(H)$ with $k(q^{\prime})\leq q$ such that
$q^{\prime}\leq^{*}_{\Gamma_{\iota}}\iota(p)\wedge r_{\iota}$.

###### Proof.

Set $\bar{q}:=k^{-1}(q)$. Since $\operatorname{supp}\bar{q}$ is a nowhere
stationary set, there is $p\in G$ such that $\iota(\operatorname{supp}p)$
covers $\operatorname{supp}\bar{q}\setminus\Gamma_{\iota}$. In addition, we
may assume that $p\leq q\mathbin{\upharpoonright}\kappa$ and that
$p\Vdash_{\mathbb{P}}``\forall
p^{\prime}\in\dot{G}\;\text{$(\iota(p^{\prime})$ is compatible with
$r_{\iota})$''}$.

We would like to use Lemma 5.11: First, the set
$D\subseteq\iota(\mathbb{P}_{\kappa})$ of all conditions that are either
$\leq^{*}$-incompatible with $\bar{q}$ or $\leq^{*}$-stronger than $\bar{q}$
is $\leq^{*}$-dense open. Second, for every $q\leq p$, $\iota(q)$ is
compatible with $r_{\iota}$ because, by our choice of $p$,
$q\Vdash_{\mathbb{P}}\text{$``\iota(q)$ is compatible with $r_{\iota}$''}$.
Thus, we may appeal to Lemma 5.11 with respect to $\langle
D,r_{\iota},p\rangle$ and get a $\leq^{*}$-dense open set
$E\subseteq\mathbb{P}$ below $p$ such that, for each $p^{\prime}\in E$, there
is a $\leq^{*}_{\Gamma_{\iota}}$-extension of $\iota(p^{\prime})\wedge
r_{\iota}$ in $D$. Extending $p$ we assume that there is a
$\leq^{*}_{\Gamma_{\iota}}$ extension of $\iota(p)\wedge r_{\iota}$ in $D$.

Let us define an auxiliary condition $q^{\prime\prime}$ as follows:
$q^{\prime\prime}(\alpha):=\iota(p)(\alpha)\wedge\bar{q}(\alpha)$ for
$\alpha\notin\Gamma_{\iota}$ and otherwise
$q^{\prime\prime}(\alpha):=r_{\iota}(\alpha)\wedge\bar{q}(\alpha)$.181818More
precisely, we define $q^{\prime\prime}(\alpha)$ this way assuming that the
outcome is a condition. Otherwise, we stipulate $q^{\prime\prime}(\alpha)$ to
be the trivial condition. Let

$q^{\prime}:=\iota(p)\mathbin{\upharpoonright}\left(\iota(\kappa)\setminus\Gamma_{\iota}\right)\cup
q^{\prime\prime}\mathbin{\upharpoonright}\Gamma_{\iota}.$

By elementarity, for each $\alpha$, $k(q^{\prime\prime})(\alpha)$ is forced to
be the maximal condition in $j(\mathbb{Q})_{\alpha}$ stronger than
$j(p)(\alpha)\wedge k(r_{\iota})(\alpha)$ and $q(\alpha)$, if such a condition
exists. Let $s$ be a condition in $H$ stronger than $j(p),k(r_{\iota})$ and
$q$. Then, clearly, $s$ forces that $k(q^{\prime\prime})\leq j(p)\wedge
k(r_{\iota}),q$ and that $k(q^{\prime\prime})\in H$. We would like to show
that $s\Vdash_{j(\mathbb{P}_{\kappa})}k(q^{\prime})=k(q^{\prime\prime})$. To
this end, fix $u\leq^{*}_{\Gamma_{\iota}}\iota(p)\wedge r_{\iota}.$

######  Claim 5.16.1.

$u\leq^{*}\bar{q}$.

###### Proof.

Otherwise, $u$ is $\leq^{*}$-incompatible with $\bar{q}$ and thus so are
$k(u)$ and $q$. Without loss of generality, this is forced by $s.$ Let us now
look at $k(q^{\prime\prime})$. The only coordinates in which we don’t have
$k(q^{\prime\prime})(\alpha)^{H\mathbin{\upharpoonright}\alpha}\leq
k(u)(\alpha)^{H\mathbin{\upharpoonright}\alpha}$ are the coordinates from
$k(\Gamma_{\iota})$. However, in those coordinates
$k(q^{\prime\prime})(\alpha)$ and $k(u)(\alpha)$ are forced by $s$ to be
direct extensions of $k(r_{\iota})(\alpha)$, hence they are compatible. ∎

Since $u$ agrees with $\iota(p)$ on all coordinates except those in
$\Gamma_{\iota}$, $s\mathbin{\upharpoonright}\alpha$ forces that
$\iota(p)(\alpha)$ is stronger than $\bar{q}(\alpha)$ for all
$\alpha\notin\Gamma_{\iota}$. Therefore, $s$ forces that
$k(q^{\prime})=k(q^{\prime\prime})$, so that $k(q^{\prime})\in H$ and
$k(q^{\prime})\leq q$, as needed.∎

The next lemma culminates the discussion of this section. Indeed, from it we
will be able to infer our key coding lemma (Lemma 5.8).

######  Lemma 5.17.

For each $\epsilon<j(\kappa)$ there is a finite iteration $\iota\colon V\to N$
factoring $j\mathbin{\upharpoonright}V$ such that
$\epsilon\in\operatorname{range}(k)$ and for which the following holds:

For every $\mathbb{P}_{\kappa}$-nice name $\dot{X}$ for a subset of $\kappa$,

$\text{$M[H]\models\epsilon\in j(\dot{X})_{H}\;\Longleftrightarrow\;\exists
q\in\iota(\mathbb{P}_{\kappa})\,(k(q)\in
H\,\wedge\,q\Vdash_{\iota(\mathbb{P_{\kappa}})}k^{-1}(\epsilon)\in\iota(\dot{X}))$}.$

Moreover, letting $r_{\iota}$ be the condition of Definition 5.14, we may take
$q$ so that, for some $p\in G$, $q\leq^{*}_{\Gamma_{\iota}}\iota(p)\wedge
r_{\iota}$ and $\operatorname{supp}q=\operatorname{supp}(\iota(p)\wedge
r_{\iota})$.

###### Proof.

Suppose that the lemma is false. Fix a finite iteration $\iota\colon
V\rightarrow N$ with $\epsilon\in\operatorname{Im}(k)$, and set
$\bar{\epsilon}=k^{-1}(\epsilon)$. For each $\mathbb{P}_{\kappa}$-name
$\dot{X}$ for a subset of $\kappa$,

$D_{\dot{X},\iota}:=\\{s\in\iota(\mathbb{P}_{\kappa})\mathrel{|}\allowbreak
s\Vdash_{\iota(\mathbb{P}_{\kappa})}\bar{\epsilon}\in\iota(\dot{X})\vee
s\Vdash_{\iota(\mathbb{P}_{\kappa})}\bar{\epsilon}\notin i(\dot{X})\\}.$

By the Prikry lemma this is a $\leq^{*}$-dense open set and
$D_{\dot{X},\iota}\in N$.

Denote by $\Gamma_{\iota}$ the set from Lemma 5.11. Likewise, $r_{\iota}$
denotes the condition arising from Definition 5.14; namely, $r_{\iota}$ is the
weakest condition with support $A_{\iota}$ and stem $\dot{\eta}_{\alpha}$ at
each $\alpha\in A_{\iota}$. By Lemma 5.15, $A_{\iota}\subseteq\Gamma_{\iota}$,
hence $\operatorname{supp}r_{\iota}\subseteq\Gamma_{\iota}$.

Let $p_{0}$ be any condition in $G$ forcing that $\iota(q)$ is compatible with
$r_{\iota}$, for every $q\in G$. Invoking Lemma 5.11 for the direct extension
order $\leq^{*}$ of $\mathbb{P}_{\kappa}$, the $\leq^{*}$-dense open set
$D_{\dot{X},\iota}$ and the conditions $r_{\iota}$ and $p_{0}$ we obtain a
condition $p\in G$, $p\leq p_{0}$ such that $\iota(p)\wedge r_{\iota}$ admits
a $\leq^{*}_{\Gamma_{\iota}}$-extension $s\in D_{\dot{X},\iota}$ with
$\operatorname{supp}s=\operatorname{supp}(\iota(p)\wedge r_{\iota})$. Note
that any two conditions $s$’s as above are compatible, hence the decision on
$``\bar{\epsilon}\in\iota(\dot{X})$” is independent of their choice.

If for every $\dot{X}_{G}\in\mathcal{U}$ there was a condition $s\in
D_{\dot{X},\iota}$ with $s\leq^{*}_{\Gamma_{\iota}}\iota(p)\wedge r_{\iota}$
(for some $p\in G$) and $k(s)\in H$ then the conclusion of Lemma 5.17 would
hold, thus contradicting our departing assumption. Hence, there must be some
$\dot{X}_{G}\in\mathcal{U}$ such that if $s\in D_{\dot{X},\iota}$ and
$s\leq^{*}_{\Gamma_{\iota}}\iota(p)\wedge r_{\iota}$ then $k(s)\notin H.$

Fix one such condition $s$. Since $k(s)\notin H$, and

$\\{\bar{s}\in j(\mathbb{P}_{\kappa})\mathrel{|}\allowbreak\bar{s}\leq
k(s)\,\vee\,\bar{s}\perp k(s)\\}$

is dense, we can find $\bar{s}\in H$ such that $\bar{s}\perp k(s)$ and
$\bar{s}\Vdash_{j(\mathbb{P}_{\kappa})}\epsilon\in j(\dot{X}).$

Since $M$ is a direct limit of a normal iteration, applying Lemma 5.2, we have
that $\bar{s}$ originates from another finite iteration $\iota^{\prime}\colon
V\to N^{\prime}$; to wit, if $k^{\prime}\colon N^{\prime}\rightarrow M$ is the
factor embedding between $j$ and $\iota^{\prime}$ then
$\bar{s}\in\operatorname{range}(k^{\prime}).$ Finally, by Lemma 5.16, there is
$s^{\prime}\leq^{*}_{\Gamma_{\iota^{\prime}}}\iota^{\prime}(p^{\prime})\wedge
r_{\iota^{\prime}}$ for some $p^{\prime}\in G$, such that
$k^{\prime}(s^{\prime})\leq\bar{s}$ and $k^{\prime}(s^{\prime})\in H$.
Clearly, $k^{\prime}(s^{\prime})$ is incompatible with $k(s)$.

######  Claim 5.17.1.

The following are true:

  1. (1)

$\\{\beta\mathrel{|}\allowbreak
k^{\prime}(s^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}\perp
k(s)(\beta)^{H\mathbin{\upharpoonright}\beta}\\}\subseteq
k(\Gamma_{\iota})\cap k^{\prime}(\Gamma_{\iota^{\prime}}).$

  2. (2)

For each $\beta$ in the above set,

$\operatorname{stem}(k^{\prime}(r^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta})\neq\operatorname{stem}(k(r)(\beta)^{H\mathbin{\upharpoonright}\beta}).$

###### Proof.

Let us stipulate that $r:=r_{\iota},r^{\prime}=r_{\iota^{\prime}}$,
$\Gamma:=\Gamma_{\iota}$ and $\Gamma^{\prime}=\Gamma_{\iota^{\prime}}$.

(1) Let $\beta$ be such that
$k^{\prime}(s^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}\perp
k(s)(\beta)^{H\mathbin{\upharpoonright}\beta}$. Clearly, $\beta\geq\kappa$.

Since any two conditions in
$j(\mathbb{Q})^{H\mathbin{\upharpoonright}\beta}_{\beta}$ with the same stem
are compatible it must be the case that
$k(s)(\beta)^{H\mathbin{\upharpoonright}\beta}$ and
$k(s^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}$ have different stems.
To simplify notations, let us put
$q:=k(s)(\beta)^{H\mathbin{\upharpoonright}\beta}$ and
$q^{\prime}:=k^{\prime}(s^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}$.

First, $\beta\in k(\Gamma)$: Otherwise, $\beta\notin k(\Gamma)$ and so
$q=j(p)(\beta)$, which belongs to the generic filter $H_{\beta}\subseteq
j(\mathbb{Q})_{\beta}$. Since $q^{\prime}\in H_{\beta}$ (because
$k^{\prime}(s^{\prime})\in H$) we infer that $q^{\prime}$ and $q$ are
compatible, hence obtaining a contradiction.

Second, $\beta\in k^{\prime}(\Gamma^{\prime})$: Otherwise,
$q^{\prime}=j(p^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}$. Since the
condition $\iota(p^{\prime})$ is compatible with $r$, $(j(p^{\prime})\wedge
k(r))(\beta)^{H\mathbin{\upharpoonright}\beta}$ is well-defined and extends
$j(p^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}$. Now, $q$ is a
$\leq^{*}$-extension of $(j(p)\wedge
k(r))(\beta)^{H\mathbin{\upharpoonright}\beta}$ so in
$M[H\mathbin{\upharpoonright}\beta]$,

$\operatorname{stem}q=\operatorname{stem}(j(p)\wedge
k(r))(\beta)=\operatorname{stem}(j(p^{\prime})\wedge k(r))(\beta),$

where the last equality follows from the maximality of the stem of $r$. Thus,
$q$ is compatible with $(j(p^{\prime})\wedge k(r))(\beta)$, which in turn
extends $q^{\prime}$. Thereby, $q$ and $q^{\prime}$ are compatible, hence
yielding the sought contradiction.

(2) By the previous arguments, the stems of
$k(s)(\beta)^{H\mathbin{\upharpoonright}\beta}$ and
$k^{\prime}(s^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}$ must be
different. Also, since $\beta\in k(\Gamma)\cap k^{\prime}(\Gamma^{\prime})$,
these stems respectively coincide with those of
$k(r)(\beta)^{H\mathbin{\upharpoonright}\beta}$ and
$k^{\prime}(r^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta},$ which shows
that the stems of $k(r)(\beta)^{H\mathbin{\upharpoonright}\beta}$ and
$k^{\prime}(r^{\prime})(\beta)^{H\mathbin{\upharpoonright}\beta}$ must be
different. ∎

For $X\in\mathcal{U}$ and $\iota\colon V\rightarrow N$ a finite iteration with
$\epsilon\in\operatorname{range}(k)$ we say that _$X$ is good for $\iota$_ if
there is $s\in D_{\dot{X},\iota}$ with
$s\leq^{*}_{\Gamma_{\iota}}\iota(p)\wedge r_{\iota}$ (some $p\in G$) such that
$k(s)\in H$.191919Note that if $s\in D_{\dot{X},\iota}$ and $k(s)\in H$ then
$k(s)\Vdash_{j(\mathbb{P})}\epsilon\in j(\dot{X}).$ Alternatively, _$X$ is
bad_ for $\iota$ if it is not good.

The above argument shows that if $X$ is bad for $\iota$ then there is another
finite iteration $\iota^{\prime}$ for which $X$ is good and this discrepancy
entails $k(\Gamma_{\iota})\cap
k^{\prime}(\Gamma_{\iota^{\prime}})\neq\emptyset$. Moreover, at least one of
the $\beta$’s in this intersection witnesses a disagreement between the length
of the stems of $k(r_{\iota})$ and $k^{\prime}(r_{\iota^{\prime}})$.

Working in $V[G]$ let us define
$\langle(\iota_{\alpha},X_{\alpha})\mathrel{|}\allowbreak\alpha<(2^{\aleph_{0}})^{+}\rangle$
as follows:

  1. (1)

$\iota_{\alpha}\colon V\rightarrow N_{\alpha}$ is a finite iteration with
$\epsilon\in\operatorname{range}(k_{\alpha})$;

  2. (2)

$\langle
X_{\alpha}\mathrel{|}\allowbreak\alpha<(2^{\aleph_{0}})^{+}\rangle\subseteq\mathcal{U}$
is a $\subseteq$-decreasing sequence;

  3. (3)

for all $\alpha<(2^{\aleph_{0}})^{+}$,

    1. (a)

$\bigcap_{\beta<\alpha}X_{\beta}$ is good for $\iota_{\alpha}$;

    2. (b)

$X_{\alpha}$ is bad for $\iota_{\beta}$, for all $\beta\leq\alpha$.