In particular, by (3)(a), for each $\beta<\alpha<(2^{\aleph_{0}})^{+}$,
$k_{\alpha}(\Gamma_{\iota_{\alpha}})\cap
k_{\beta}(\Gamma_{\iota_{\beta}})\neq\emptyset.$

Let $\iota_{0}$ be any finite iteration witnessing (1). Since we are assuming
that the lemma is false there is $X_{0}\in\mathcal{U}$ which is bad for
$\iota_{0}$. So, suppose that
$\langle(\iota_{\beta},X_{\beta})\mathrel{|}\allowbreak\beta<\alpha\rangle$
was defined. By $\kappa$-completeness,
$X^{0}_{\alpha}:=\bigcap_{\beta<\alpha}X_{\beta}\in\mathcal{U}$. Arguing as
before, there is a finite iteration $\iota_{\alpha}$ for which
$X^{0}_{\alpha}$ is good. Also, since we are assuming that the conclusion of
the lemma is false, there is $X_{\alpha}\subseteq X^{0}_{\alpha}$ that is bad
for $\iota_{\alpha}.$ Finally, $X_{\alpha}$ is bad for all $\iota_{\beta}$,
$\beta\leq\alpha$: Otherwise, there is $s\in
D_{\dot{X}_{\alpha},\iota_{\beta}}$ such that $k_{\beta}(s)\in H$. However,
$D_{\dot{X}_{\alpha},\iota_{\beta}}\subseteq
D_{\dot{X}_{\beta},\iota_{\beta}}$, which would contradicts that $X_{\beta}$
is bad for $\iota_{\beta}.$

Let us now look at the sets $\Gamma_{\alpha}:=\Gamma_{\iota_{\alpha}}$. By the
argument above, for every $\alpha<\beta$, every disagreement point between
(the evaluations of) $k_{\alpha}(r_{\alpha})$ and $k_{\beta}(r_{\beta})$ is a
member of $k_{\alpha}(\Gamma_{\alpha})\cap k_{\beta}(\Gamma_{\beta})$. Let us
apply the $\Delta$-system lemma to
$\\{k_{\alpha}(\Gamma_{\alpha})\mathrel{|}\allowbreak\alpha<(2^{\aleph_{0}})^{+}\\}$.
This way we obtain a set $I$ of cardinality $(2^{\aleph_{0}})^{+}$ and a root
$\Delta$ such that for all $\alpha,\beta\in I$,
$k_{\alpha}(\Gamma_{\alpha})\cap k_{\beta}(\Gamma_{\beta})=\Delta.$ Fix a
coloring $c\colon[(2^{\aleph_{0}})^{+}]^{2}\rightarrow\Delta$ sending
$\langle\alpha,\beta\rangle$ to the least coordinate $\zeta$ in the root
$\Delta$ which exhibits an incompatibility between the stems of
$k_{\alpha}(r_{\alpha})^{H\mathbin{\upharpoonright}\zeta}$ and
$k_{\beta}(r_{\beta})^{H\mathbin{\upharpoonright}\zeta}$. By the Erdős-Rado
theorem this coloring admits a homogeneous set $J\subseteq I$ of order type
$\omega_{1}+1$. So, let $\gamma\in\Delta$ be such that $c``J=\\{\gamma\\}$;
namely, $\gamma$ is the common least coordinate witnessing a disagreement
between the stems of
$k_{\alpha}(r_{\alpha})(\gamma)^{H\mathbin{\upharpoonright}\gamma}$ and
$k_{\beta}(r_{\beta})(\gamma)^{H\mathbin{\upharpoonright}\gamma}$, for all
$\langle\alpha,\beta\rangle\in[J]^{2}$. Since for each such $\alpha,\beta$ the
lengths of
$\operatorname{stem}(k_{\alpha}(r_{\alpha})(\gamma)^{H\mathbin{\upharpoonright}\gamma})$
and
$\operatorname{stem}(k_{\beta}(r_{\beta})(\gamma)^{H\mathbin{\upharpoonright}\gamma})$
are finite and different we derive a contradiction with our departing
assumption. ∎

Given Lemma 5.17 the proof of our coding lemma is easy:

###### Proof of Lemma 5.8.

Let $\mathcal{U}\in V[G]$ be a $\kappa$-complete ultrafilter on $\kappa$ and
let $j\colon V[G]\rightarrow M[H]$ be the induced ultrapower. By our setup
considerations in page 5.3 there is a finite iteration $\iota\colon
V\rightarrow N$ such that $k\circ\iota=j\mathbin{\upharpoonright}V$ and
$\epsilon:=[\operatorname{id}]_{\mathcal{U}}$ belongs to
$\operatorname{range}(k)$. Invoke Lemma 5.17 with respect to these inputs and
define $\mathcal{V}$ to be the set of all $X\in\mathcal{P}(\kappa)^{V[G]}$
such that there are $q\in k^{-1}(H)$ and $p\in G$ such that

$q\leq^{*}_{\Gamma_{\iota}}(\iota(p)\wedge r_{\iota})\;\text{ and
}\;q\Vdash_{\iota(\mathbb{P}_{\kappa})}\epsilon\in\iota(\dot{X}).$

Note that any two conditions $q,q^{\prime}$ witnessing the above admit an
explicit $\leq^{*}_{\Gamma_{\iota}}$-extension. In particular, $\mathcal{V}$
is a filter. Also, by the moreover part of Lemma 5.17, $\mathcal{V}$ contains
the $\kappa$-complete ultrafilter $\mathcal{U}$. Thus,
$\mathcal{V}=\mathcal{U}.$ ∎

Before proving the consistency of the $\omega$-gluing property (Theorem 5.19
below) let us pose an attractive conjecture, which is a strengthening of Lemma
5.16:

######  Conjecture 5.18.

Let $j\colon V[G]\to M[H]$ be an ultrapower embedding with critical point
$\kappa$, such that $j\mathbin{\upharpoonright}V$ is a normal iteration using
normal measures.

Then, there is a finite sub-iteration iteration $\iota\colon V\to N$ factoring
$j\mathbin{\upharpoonright}V$, with critical points $\vec{\mu}$, and a
condition $r\in\iota(\mathbb{P}_{\kappa})$ such that
$\operatorname{supp}r=\vec{\mu}$ and for all $s\leq^{*}r$ in $N$ with
$\operatorname{supp}s=\vec{\mu}$ then $k(s)\in H$.

The meaning of this conjecture is the following: There is a finite sub-
iteration $\iota$ of $j$ for which there is a
$\iota(\mathbb{P}_{\kappa})$-generic filter that can be defined from $G$
together with finitely-many additional information, and all its conditions are
sent to $H$ by $k$. Moreover, this filter is $\leq^{*}$-generic.

If our conjecture holds then, in particular, the above-mentioned filter will
decide all statements of the form $``\epsilon\in\iota(\dot{X})$” for a
$\mathbb{P}_{\kappa}$-name $\dot{X}$ for a subset of $\kappa$. Thus the filter
will determine the measure $\mathcal{U}$ defining $j$.

Unfortunately we do not know how to prove the conjecture, but we got pretty
close to it, by showing that its corollary holds; to wit, there is a finite
iteration that determines $\mathcal{U}$ (Lemma 5.17 above).

###  5.4. A model for the $\omega$-gluing property from $o(\kappa)=\omega_{1}$

We are finally in conditions to derive the main result of Section 5.

######  Theorem 5.19 ($\mathrm{V}=\mathcal{K}$).

Let $\kappa$ be a measurable cardinal with $o(\kappa)=\omega_{1}$ and assume
that there are no other measurables of Mitchell order $\geq\omega_{1}$.

Then, there is a forcing extension in which the $\omega$-gluing property
holds.

###### Proof.

Let
$\vec{\mathcal{U}}=\langle\mathcal{U}(\mu,\zeta)\mathrel{|}\allowbreak\mu\leq\kappa,\,\zeta<o^{\vec{\mathcal{U}}}(\mu)\rangle$
be the coherent sequence of normal measures in $\mathcal{K}$. Force with
$\mathbb{S}$ and, working in $\mathcal{K}[S]$, let $F=\bigcup S$ be the
obtained generic function. Let us define a non-stationary support iteration of
Prikry type forcings. The forthcoming definition will ensure that
$\mathbb{Q}_{\alpha}$ exists in the generic extension by
$\mathbb{S}\mathbin{\upharpoonright}\alpha\times\mathbb{P}\mathbin{\upharpoonright}\alpha$.

We say that a tuple $\langle\vec{\rho},\vec{\zeta},f,g\rangle$ _codes a
measure_ on $\alpha$ if:

  1. (1)

$\vec{\rho}=\langle\rho_{0},\dots,\rho_{n-1}\rangle$ where
$\rho_{i}\colon\alpha^{i}\to(\alpha+1)$ and $\rho_{0}(\langle\rangle)=\alpha$,

  2. (2)

$\vec{\zeta}=\langle\zeta_{0},\dots,\zeta_{n-1}\rangle$ is a sequence of
countable ordinals,

  3. (3)

$f\colon\alpha^{n}\to\mathbb{S}\mathbin{\upharpoonright}\alpha\ast\mathbb{P}_{\alpha}$,

  4. (4)

$g\colon\alpha^{n}\to\alpha$,

  5. (5)

For each $i<n$, $\rho_{i}(\langle\mu_{j}\mathrel{|}\allowbreak
j<i\rangle)=\mu_{i}$ and $\zeta_{i}<o^{N_{i}}(\mu_{i})$, where
$\iota_{i,i+1}\colon N_{i}\to
N_{i+1}\cong\mathrm{Ult}(N_{i},\iota_{i}(\vec{\mathcal{U}})(\mu_{i},\zeta_{i}))$,
being $\iota_{i+1}=\iota_{i,i+1}\circ\iota_{i}$, $N_{0}=V$ and
$\iota_{0}=\operatorname{id}$,

and moreover setting $r=\iota_{n}(f)(\langle\mu_{i}\mathrel{|}\allowbreak
i<n\rangle)$, $\epsilon=\iota_{n}(g)(\langle\mu_{i}\mathrel{|}\allowbreak
i<n\rangle)$, and

$\begin{matrix}U&=&\\{\dot{X}^{S\mathbin{\upharpoonright}\alpha\ast
G\mathbin{\upharpoonright}\alpha}\subseteq\alpha\mathrel{|}\allowbreak&\exists
p\in S\mathbin{\upharpoonright}\alpha\ast G\mathbin{\upharpoonright}\alpha,\\\
&&&s\leq^{*}_{\Gamma_{\iota_{n}}}\iota_{n}(p)\wedge r_{\iota},s\in N_{n},\\\
&&&N_{n}\models
s\Vdash_{\iota_{n}(\mathbb{S}\mathbin{\upharpoonright}\alpha\ast\mathbb{P}_{\alpha})}\epsilon\in\iota_{n}(\dot{X})\\}\end{matrix},$

it is the case that $U$ is an $\alpha$-complete ultrafilter on $\alpha$. In
this case, we will say that the measure $U$ is _coded_ by
$\langle\vec{\rho},\vec{\zeta},f,g\rangle$. Note that a code
$\langle\vec{\rho},\vec{\zeta},f,g\rangle$ for a measure on $\alpha$ belongs
to $H(\alpha^{+})$. In particular, $\langle\vec{\rho},\vec{\zeta},f,g\rangle$
is a legitimate value for $s(\alpha)$ provided $s\in\mathbb{S}$ (see our
definition in page 5.1).

Let us define $\mathbb{Q}_{\alpha}$.202020Recall that this definition is in
the generic extension by
$\mathbb{S}\mathbin{\upharpoonright}(\alpha+1)\ast\mathbb{P}\mathbin{\upharpoonright}\alpha$.
If $F(\alpha)$ is not a countable sequence of codes for measures on $\alpha$,
we let $\mathbb{Q}_{\alpha}$ to be trivial. Otherwise, let
$\mathbb{Q}_{\alpha}$ be the tree Prikry forcing of Section 4 defined with
respect to the measures coded by $F(\alpha)$.

Let us show that in $V[S\times G]$ the $\omega$-gluing property holds.

Indeed, by Lemma 5.8, in the generic extension, every $\kappa$-complete
ultrafilter on $\kappa$ has a code. Fix a countable sequence of measures in
the generic extension, $\langle U_{n}\mathrel{|}\allowbreak n<\omega\rangle$,
and let
$c_{n}=\langle\vec{\rho}_{n},\vec{\zeta}_{n},f_{n},g_{n}\rangle\mathrel{|}\allowbreak
n<\omega\rangle$ be a sequence of codes for them. Let $\zeta_{*}$ be a
countable ordinal larger than all the countable ordinals which are mentioned
in all $\vec{\zeta}_{n}$’s.

Let $\iota\colon V\to\mathrm{Ult}(V,\mathcal{U}(\kappa,\zeta_{*}))=M$. By
Lemma 5.3 regarded with respect to $\mathbb{S}$, one can lift $\iota$ to
$\tilde{\iota}\colon V[S]\to M[\tilde{S}]$, in a way that

$\left(\bigcup\tilde{S}\right)(\kappa)=\langle c_{n}\mathrel{|}\allowbreak
n<\omega\rangle.$

Let us verify that, in $M$, each
$c_{n}=\langle\vec{\rho}_{n},\vec{\zeta}_{n},f_{n},g_{n}\rangle$ is still a
code for $U_{n}$. Fix a code $c_{*}=c_{n}$. By our choice of $\zeta_{*}$
(i.e., above $\sup_{n<\omega}\vec{\zeta}_{n}$), each measure which is
mentioned in the process of decoding $c_{*}$ exists in $M$ and in its
iterations. Since $M$ is closed under $\kappa$ sequences, the computations of
the models $N_{i}$ and the embeddings $\iota_{i}$ are the same. Given the
embedding and the models, the definition of $U_{n}$ is clearly absolute.

Finally, we argue as in Section 4. Let $\mathcal{V}$ be the measure on
$\kappa^{\omega}$ defined by $\dot{X}\in\mathcal{V}$ if and only if there is a
direct extension of $\iota(p)$ forcing the canonical name of the generic of
$\tilde{\iota}(\mathbb{Q})_{\kappa}$ to be in $\tilde{\iota}(\dot{X})$. ∎

##  6\. A lower bound for the $\omega$-gluing property

######  Theorem 6.1.

Assume that $\kappa$ is a measurable cardinal having the $\omega$-gluing
property. If there is no inner model of $``\exists\alpha\,(o(\alpha)=\alpha)$”
then $o^{\mathcal{\mathcal{K}}}(\kappa)\geq\omega_{1}$.

###### Proof.

Assume that there is no inner model of $``\exists\alpha\,(o(\alpha)=\alpha)$”.
Through the proof
$\langle\mathcal{U}_{\kappa,\xi}\mathrel{|}\allowbreak\xi<o^{\mathcal{\mathcal{K}}}(\kappa)\rangle$
will denote an enumeration of the normal measures in $\mathcal{K}$ indexed by
their corresponding Mitchell orders. We aim to produce –by induction on
$\beta<\omega_{1}$– a sequence
$\langle\mathcal{V}_{\beta}\mathrel{|}\allowbreak\beta<\omega_{1}\rangle$ of
$\kappa$-complete $\mathcal{K}$-normal measures on $\kappa$ such that
$\langle\mathcal{V}_{\beta}\cap\mathcal{K}\mathrel{|}\allowbreak\beta<\omega_{1}\rangle$
is Mitchell-order increasing. By maximality of the core model, if
$\mathcal{V}_{\beta}$ is $\mathcal{K}$-normal and $\kappa$-complete then
$\mathcal{V}_{\beta}\cap\mathcal{K}\in\mathcal{K}$. In particular, every
$\mathcal{V}_{\beta}$ will admit an index
$\xi_{\beta}<o^{\mathcal{K}}(\kappa)$ such that
$\mathcal{V}_{\beta}\cap\mathcal{K}=\mathcal{U}_{\kappa,\xi_{\beta}}$. The
sequence
$\langle\mathcal{V}_{\beta}\cap\mathcal{K}\mathrel{|}\allowbreak\beta<\omega_{1}\rangle$
will thus produce an evidence for $o^{\mathcal{K}}(\kappa)\geq\omega_{1}$.

Let $\mathcal{V}_{0}$ be a normal $\kappa$-complete measure in $V$. Clearly,
$\mathcal{V}_{0}\cap\mathcal{K}$ is $\mathcal{K}$-normal so that
$\mathcal{V}_{0}\cap\mathcal{K}\in\mathcal{K}$. Next, let us assume that we
have built
$\langle\mathcal{V}_{\beta}\mathrel{|}\allowbreak\beta<\alpha\rangle$.

If $\alpha$ takes the form $\alpha=\gamma+1$ then we apply the $\omega$-gluing
property to the constant sequence
$\vec{\mathcal{V}}=\langle\mathcal{V}_{\gamma}\mathrel{|}\allowbreak
n<\omega\rangle$. Otherwise, we fix $\langle\beta_{n}\mathrel{|}\allowbreak
n<\omega\rangle$ a cofinal sequence in $\alpha$ and apply the $\omega$-gluing
property to the sequence
$\vec{\mathcal{V}}=\langle\mathcal{V}_{\beta_{r(m)}}\mathrel{|}\allowbreak
m<\omega\rangle$, where $r\colon\omega\to\omega$ is a function such that
$|r^{-1}(\\{n\\})|=\aleph_{0}$ for all $n<\omega$.

Let $\bar{\mathcal{V}}_{\alpha}$ be the obtained gluing measure and
$j=j_{\alpha}\colon V\to M$ be its ultrapower map. By virtue of our anti-large
cardinal hypothesis, a classical theorem of Mitchell [Mit84] implies that
$j\mathbin{\upharpoonright}\mathcal{K}\colon\mathcal{K}\to\mathcal{K}^{M}$ is
a normal iteration of measures in $\mathcal{K}$. Denote this latter by
$\langle\iota_{\gamma,\delta}\mathrel{|}\allowbreak\gamma\leq\delta<\delta_{*}\rangle$,
being
$\iota_{\gamma,\gamma+1}\colon\mathcal{K}_{\gamma}\rightarrow\mathcal{K}_{\gamma+1}$.
In addition, we put
$\mu_{\gamma}:=\operatorname{crit}(\iota_{\gamma,\gamma+1})$.

Since $\bar{\mathcal{V}}_{\alpha}$ glues the sequence $\vec{\mathcal{V}}$
there is, by Definition 1.1, an increasing sequence of ordinals
$\langle\eta_{n}\mathrel{|}\allowbreak n<\omega\rangle$ such that

$\vec{\mathcal{V}}(n)=\\{X\subseteq\kappa\mathrel{|}\allowbreak\eta_{n}\in
j(X)\\}.$

Let $\eta_{\omega}:=\sup_{n<\omega}\eta_{n}$ and define

$\mathcal{V}_{\alpha}:=\\{X\subseteq\kappa\mathrel{|}\allowbreak\eta_{\omega}\in
j(X)\\}.$

By our induction hypothesis $\vec{\mathcal{V}}(n)$ is a normal measure, hence
$\vec{\mathcal{V}}(n)\cap\mathcal{K}$ contains the club filter
$\mathrm{Club}_{\kappa}^{\mathcal{K}}$, and thus $\eta_{n}\in\bigcap
j``\mathrm{Club}_{\kappa}^{\mathcal{K}}$ for all $n<\omega$.

For $m<\omega$ and a function $f\colon\kappa^{m}\to\kappa$ in $\mathcal{K}$,
the set of its closure points $\\{\alpha<\kappa\mathrel{|}\allowbreak
f``\alpha^{m}\subseteq\alpha\\}$ belongs to
$\mathrm{Club}^{\mathcal{K}}_{\kappa}$. In particular, all the $\eta_{n}$’s
and $\eta_{\omega}$ are closure points of $j(f)$, for every such function
$f\in\mathcal{K}$.

######  Claim 6.1.1.

For $n<\omega$, there is $\gamma_{n}<\delta_{*}$ with
$\iota_{\gamma_{n}}(\kappa)=\operatorname{crit}(\iota_{\gamma_{n},\gamma_{n+1}})=\eta_{n}$.

###### Proof.

We may assume without loss of generality that $\mu_{\gamma}\leq
i_{\gamma}(\kappa)$ for $\gamma\leq\delta_{*}$: If for some index $\gamma$,
$\mu_{\gamma}>\iota_{\gamma}(\kappa)$ then the latter must be $j(\kappa)$,
hence above $\eta_{n}$ for all $n<\omega$. So, if that happens, we truncate
the iteration at that stage.

Let $\gamma_{n}:=\gamma\leq\delta_{*}$ be the maximal ordinal such that
$\mu_{\bar{\gamma}}<\eta_{n}$ for $\bar{\gamma}<\gamma$.

Recall that every member of $\mathcal{K}_{\gamma}$ is represented as
$\iota_{\gamma}(f)(\mu_{\gamma_{0}},\dots,\mu_{\gamma_{m-1}})$, where
$m<\omega$, $f\colon\kappa^{m}\to\mathcal{K}$ is in $\mathcal{K}$ and
$\gamma_{0},\dots,\gamma_{m-1}<\gamma$. In particular, any ordinal below
$\iota_{\gamma}(\kappa)$ is represented by a function
$f\colon\kappa^{m}\rightarrow\kappa$ in $\mathcal{K}$.

Let us show that $\eta_{n}=\iota_{\gamma}(\kappa)=\mu_{\gamma+1}$: For each
$\alpha<\iota_{\gamma}(\kappa)$ there is a function
$f\colon\kappa^{m}\rightarrow\kappa$ with
$\alpha=\iota_{\gamma}(f)(\mu_{\gamma_{0}},\dots,\mu_{\gamma_{m-1}})$.
Clearly, $\iota_{\gamma}(f)(\mu_{\gamma_{0}},\dots,\mu_{\gamma_{m-1}})\leq
j(f)(\mu_{\gamma_{0}},\dots,\mu_{\gamma_{m-1}})$. Also, in that $\eta_{n}$ is
an accumulation point of $j(f)$ and $\mu_{\gamma_{m-1}}<\eta_{n}$, we have
$\alpha\leq j(f)(\mu_{\gamma_{0}},\dots,\mu_{\gamma_{m-1}})<\eta_{n}.$ All in
all, $\iota_{\gamma}(\kappa)\leq\eta_{n}$.

By maximality of $\gamma$, $\eta_{n}\leq\mu_{\gamma}$. From altogether,
$\iota_{\gamma_{n}}(\kappa)=\mu_{\gamma_{n}}=\eta_{n}.$ ∎

Let $\gamma_{\omega}:=\sup_{n<\omega}\gamma_{n}$.

######  Claim 6.1.2.

$\eta_{\omega}=\iota_{\gamma_{\omega}}(\kappa)=\operatorname{crit}(\iota_{\gamma_{\omega},\gamma_{\omega}+1})$.

###### Proof.

On one hand, since $\iota_{\gamma_{n}}(\kappa)=\eta_{n}$ and this latter is
the critical point of $\iota_{\gamma_{n},\gamma_{n+1}}$,
$\iota_{\gamma_{\omega}}(\kappa)=\eta_{\omega}$. On the other hand,
$\gamma_{\omega}<\delta^{*}$ and
$\operatorname{crit}(\iota_{\gamma_{\omega},\gamma_{\omega+1}})=\eta_{\omega}$.
In effect, the latter assertion will follow from normality of the iteration
once we show that $\gamma_{\omega}<\delta^{*}$. Towards a contradiction,
suppose that $\delta_{*}=\gamma_{\omega}$. Then,
$j(\kappa)=\iota_{\gamma_{\omega}}(\kappa)=\eta_{\omega}$. However, $M$ is
closed under $\omega$-sequences and so $\mathrm{cof}^{V}(j(\kappa))>\omega$.
Thus, in particular, $j(\kappa)$ cannot be $\eta_{\omega}$. ∎

Let us analyse the measures from the iteration of $\mathcal{K}$ which are used
during the steps $\gamma_{n}$ and $\gamma_{\omega}$. By our initial
hypothesis, $o^{\mathcal{K}}(\kappa)<\kappa$ and in particular,
$<(\kappa^{+})^{\mathcal{K}}$. Therefore, the normal measures on $\kappa$ in
$\mathcal{K}$ are discrete –there is a sequence of pairwise disjoint sets
$\langle A_{\xi}\mathrel{|}\allowbreak\xi<o^{\mathcal{K}}(\kappa)\rangle$,such
that

$A_{\xi}\in U_{\kappa,\xi}\setminus\bigcup_{\zeta\neq\xi}U_{\kappa,\zeta}.$

Let us look at the measure which is applied at step $\gamma_{n}$ of the
iteration. As
$\operatorname{crit}(\iota_{\gamma_{n},\gamma_{n+1}})=\iota_{\gamma_{n}}(\kappa)$
this is one of the measures on $\iota_{\gamma_{n}}(\kappa)$ lying in
$\mathcal{K}_{\gamma_{n}}$. Recall that $\langle
U_{\kappa,\xi}\mathrel{|}\allowbreak\xi<o^{\mathcal{K}}(\kappa)\rangle$
denotes the sequence of normal measures on $\kappa$ in $\mathcal{K}$ and that
they are indexed according to their Mitchell order.

Since $\operatorname{crit}(\iota_{\gamma})=\kappa>o^{\mathcal{K}}(\kappa)$ we
conclude that

$\iota_{\gamma}(\langle
U_{\kappa,\xi}\mathrel{|}\allowbreak\xi<o^{\mathcal{K}}(\kappa)\rangle)=\langle\iota_{\gamma}(U_{\kappa,\xi})\mathrel{|}\allowbreak\xi<o^{\mathcal{K}}(\kappa)\rangle)$

are the only normal measures on $\iota_{\gamma}(\kappa)$ lying in
$\mathcal{K}_{\gamma}$. In particular, the measure that is iterated at tage
$\gamma_{n}$ must have the form $\iota_{\gamma_{n}}(U_{\kappa,\zeta_{n}})$ for
$\zeta_{n}<o^{\mathcal{K}}(\kappa)$.

######  Claim 6.1.3.

$\mathcal{V}_{\alpha}$ is $\mathcal{K}$-normal.

###### Proof.

Let $f\colon\kappa\rightarrow\kappa$ be a function in $\mathcal{K}$ with
$\\{\alpha<\kappa\mathrel{|}\allowbreak
f(\alpha)<\alpha\\}\in\mathcal{V}_{\alpha}\cap\mathcal{K}$. By definition of
$\mathcal{V}_{\alpha}$ and since $f\in\mathcal{K}$,
$\iota_{\delta_{*}}(f)(\eta_{\omega})<\eta_{\omega}$. Also, Claim 6.1.2 yields
$\iota_{\gamma_{\omega}+1}(f)(\eta_{\omega})<\eta_{\omega}$, so that
$\\{\alpha<\eta_{\omega}\mathrel{|}\allowbreak
i_{\gamma_{\omega}}(f)(\alpha)<\alpha\\}$ belongs to
$\iota_{\gamma_{\omega}}(U_{\kappa,\zeta_{\omega}})$, the normal measure
iterated at stage $\gamma_{\omega}$. Thus, there is
$X\in\iota_{\gamma_{\omega}}(U_{\kappa,\zeta_{\omega}})$ and
$\theta<\eta_{\omega}=\iota_{\gamma_{\omega}}(\kappa)$ such that
$\iota_{\gamma_{\omega}}(f)``X=\\{\theta\\}$. By elementarity, there is $X\in
U_{\kappa,\zeta_{\omega}}$ and $\theta<\kappa$ such that $f``X=\\{\theta\\}$.
Moreover, $X\in\mathcal{V}_{\alpha}\cap\mathcal{K}$:

$X\in
U_{\kappa,\zeta_{\omega}}\Leftrightarrow\iota_{\gamma_{\omega}}(X)\in\iota_{\gamma_{\omega}}(U_{\kappa,\zeta_{\omega}})\Leftrightarrow\eta_{\omega}\in\iota_{\gamma_{\omega},\gamma_{\omega}+1}(X)\Leftrightarrow\eta_{\omega}\in\iota_{\delta_{*}}(X).\qed$

######  Claim 6.1.4.

Let $\zeta_{n}$ be as above. Then,

$\zeta_{n}=\begin{cases}\xi_{\gamma},&\text{if $\alpha=\gamma+1$;}\\\
\xi_{\beta_{r(n)}},&\text{if $\alpha\in\mathrm{acc}(\omega_{1})$.}\end{cases}$

###### Proof.

By discreteness of the measures $U_{\kappa,\xi}$’s, $\zeta_{n}$ is the unique
ordinal $\zeta$ such that
$\iota_{\gamma_{n}}(A_{\zeta})\in\iota_{\gamma_{n}}(U_{\kappa,\zeta_{n}})$. By
Claim 6.1.1 and normality of the iteration this is equivalent to say that

$\eta_{n}\in\iota_{\gamma_{n},\gamma_{n+1}}(\iota_{\gamma_{n}}(A_{\zeta}))\;\Leftrightarrow\;\eta_{n}\in
j(A_{\zeta}).$

Finally, this is equivalent to
$A_{\zeta}\in\vec{\mathcal{V}}(n)\cap\mathcal{K}$. But
$\vec{\mathcal{V}}(n)\cap\mathcal{K}$ is exactly $U_{\kappa,\xi_{\gamma}}$ in
the successor case and $U_{\kappa,\xi_{\beta_{r(n)}}}$ in the limit case, so
the claim follows. ∎

Let us look at $\mathcal{V}_{\alpha}\cap\mathcal{K}$. Since
$\mathcal{V}_{\alpha}$ is $\kappa$-complete and $\mathcal{K}$-normal there is
$\zeta<o^{\mathcal{K}}(\kappa)$ such that
$\mathcal{V}_{\alpha}\cap\mathcal{K}=U_{\kappa,\zeta}$. We next show that
$\zeta>\zeta_{n}$ for all $n<\omega$. From this we shall be able to infer that
$\mathcal{V}_{\beta}\cap\mathcal{K}\unlhd\mathcal{V}_{\alpha}\cap\mathcal{K}$.

######  Claim 6.1.5.

$o^{\mathcal{K}^{M}}(\eta_{\omega})=\zeta$.

###### Proof of claim.

Arguing as before, the measure used at stage $\gamma_{\omega}$ of the
iteration is of the form $\iota_{\gamma_{\omega}}(U_{\kappa,\zeta_{\omega}})$
for some $\zeta_{\omega}<o^{\mathcal{K}}(\kappa)$. In particular,
$o^{\mathcal{K}_{\gamma_{\omega}}}(\eta_{\omega})=\zeta_{\omega}$ and so,
since
$\eta_{\omega}<\operatorname{crit}(\iota_{{\gamma_{\omega}}+1,\delta_{*}})$,
$o^{\mathcal{K}^{M}}(\eta_{\omega})=\zeta_{\omega}.$ Indeed,

$o^{\mathcal{K}^{M}}(\eta_{\omega})=o^{\mathcal{K}^{M}}(\iota_{\gamma_{\omega},\delta_{*}}(\eta_{\omega}))=\iota_{\gamma_{\omega},\delta_{*}}(o^{\mathcal{K}_{\gamma_{\omega}}}(\eta_{\omega}))$

, since $\iota_{\gamma_{\omega},\delta_{*}}(\eta_{\omega})=\eta_{\omega}$ and
$\iota_{\gamma_{\omega},\delta_{*}}(\zeta_{\omega})=\zeta_{\omega}$.

Let us now show that $\zeta_{\omega}=\zeta$. Once again, by discreteness,
$\zeta_{\omega}$ is the unique ordinal such that $A_{\zeta_{\omega}}\in
U_{\kappa,\zeta_{\omega}}$. By elementarity this is equivalent to
$\iota_{\gamma_{\omega}}(A_{\zeta_{\omega}})\in\iota_{\gamma_{\omega}}(U_{\kappa,\zeta_{\omega}})$,
which is equivalent to $\eta_{\omega}\in j(A_{\zeta_{\omega}})$. All in all,
we have that
$A_{\zeta_{\omega}}\in\mathcal{V}_{\alpha}\cap\mathcal{K}=U_{\kappa,\zeta}$.
Thus, by our choice on $A_{\zeta_{\omega}}$, $\zeta=\zeta_{\omega}$. ∎

######  Claim 6.1.6.

For each $n<\omega$ there is a set $B_{n}\in[\omega]^{\omega}$ such that
$\zeta_{m}=\zeta_{n}$ for all $m\in B_{n}$.

###### Proof of claim.

Fix $n<\omega$. If $\alpha$ was a successor ordinal then we can let
$B_{n}=\omega$ for in that case $\zeta_{m}=\xi_{\gamma}$ for every $m<\omega$.
Otherwise, if $\alpha$ is limit, put $B_{n}:=r^{-1}\\{r(n)\\}$. By our choice
upon $r$, $B_{n}\in[\omega]^{\omega}$. Also, by Claim 6.1.4
$\zeta_{m}=\xi_{\beta_{r(m)}}=\xi_{\beta_{r(n)}}=\zeta_{n}$. ∎

######  Claim 6.1.7.

For each $n<\omega$, $o^{\mathcal{K}^{M}}(\eta_{\omega})>\zeta_{n}$.

###### Proof of claim.

Fix $n<\omega$ and let $B_{n}\in[\omega]^{\omega}$ be as in the previous
claim. Since $M$ is closed under $\omega$-sequences we have that
$\langle\eta_{m}\mathrel{|}\allowbreak m\in B_{n}\rangle\in B$. Thus, working
in $M$ we can define the following filter:

$\mathcal{F}_{n}:=\\{X\subseteq\eta_{\omega}\mathrel{|}\allowbreak\exists
k\,\forall m\in(B_{n}\setminus k)\;\eta_{m}\in X\\}.$

We claim that
$\mathcal{F}_{n}\cap\mathcal{K}^{M}=\iota_{\gamma_{\omega}}(\mathcal{U}_{\kappa,\zeta_{n}})$.
Note that if this is the case $\mathcal{F}_{n}\cap\mathcal{K}^{M}$ is a
measure on $\eta_{\omega}$ in $\mathcal{K}^{M}$ of Mitchell order $\zeta_{n}$
and thus $o^{\mathcal{K}^{M}}(\eta_{\omega})\geq\zeta_{n}+1$.

Let $X\in\iota_{\gamma_{\omega}}(\mathcal{U}_{\kappa,\zeta_{n}})$. Note that
$X$ takes the form $\iota_{\gamma_{k},\gamma_{\omega}}(\bar{X}_{k})$ for a
tail of $k<\omega$ and $\bar{X}_{k}\in\mathcal{K}_{k}$. Let $\bar{k}$ be the
least of such indices and $m\in B_{n}\setminus\bar{k}$. Since
$X=\iota_{\gamma_{m},\gamma_{\omega}}(\bar{X}_{m})$ then, by elementarity,
$\bar{X}_{m}\in\iota_{\gamma_{m}}(\mathcal{U}_{\kappa,\zeta_{n}})=\iota_{\gamma_{m}}(\mathcal{U}_{k,\zeta_{m}}).$212121Note
that this latter equality follows by our choice of $B_{n}$. Since this latter
is the measure in $\mathcal{K}_{\gamma_{m}}$ used at stage $\gamma_{m}$ we
have that $\eta_{m}\in\iota_{\gamma_{m},\gamma_{m+1}}(\bar{X}_{m})$, which is
equivalent to $\eta_{m}\in\iota_{\gamma_{m},\gamma_{\omega}}(\bar{X}_{m})=X.$

Conversely, let $X\in\mathcal{K}^{M}$, $X\subseteq\eta_{\omega}$ be such that
$X\notin\iota_{\gamma_{\omega}}(\mathcal{U}_{\kappa,\zeta_{n}})$. Since
$\operatorname{crit}(\iota_{\gamma_{\omega},\gamma_{\omega}+1})=\eta_{\omega}$
we have that $X\in\mathcal{K}_{\gamma_{\omega}}$. Fix $k<\omega$. Arguing as
before find $m\in B_{n}\setminus k$ such that
$\iota_{\gamma_{m},\gamma_{\omega}}(\bar{X}_{m})=X$. By elementarity,
$\bar{X}_{m}\notin\iota_{\gamma_{m}}(\mathcal{U}_{\kappa,\zeta_{m}})$, hence
$\eta_{m}\notin\iota_{\gamma_{m},\gamma_{m+1}}(\bar{X}_{m})$, and thus
$\eta_{m}\notin X$. Thereby, $X\notin\mathcal{F}_{n}$. ∎

Thus, the above arguments show that
$o^{\mathcal{K}^{M}}(\eta_{\omega})=\zeta>\zeta_{n}$ for all $n<\omega$.

######  Claim 6.1.8.

$\mathcal{V}_{\beta}\cap\mathcal{K}\unlhd\mathcal{V}_{\alpha}\cap\mathcal{K}$
for all $\beta<\alpha$.

###### Proof of claim.

If $\alpha=\gamma+1$ then
$\mathcal{V}_{\gamma}\cap\mathcal{K}=\mathcal{U}_{\kappa,\xi_{\gamma}}=\mathcal{U}_{\kappa,\zeta_{n}}\unlhd\mathcal{U}_{\kappa,\zeta}=\mathcal{V}_{\alpha}\cap\mathcal{K}$.

Suppose that $\alpha$ is a limit ordinal and let $\beta<\alpha$. Choose
$\beta_{n}\in(\beta,\alpha)$ and observe that $\beta_{n}=\beta_{r(m)}$ for
some (infinitely-many) $m<\omega$. In particular,
$\mathcal{V}_{\beta}\cap\mathcal{K}\unlhd\mathcal{V}_{\beta_{n}}\cap\mathcal{K}=\mathcal{V}_{\beta_{r(m)}}\cap\mathcal{K}=\mathcal{U}_{\kappa,\zeta_{m}}\unlhd\mathcal{U}_{\kappa,\zeta}=\mathcal{V}_{\alpha}\cap\mathcal{K}.$
∎

This completes the argument for the inductive step of the construction and as
result the proof of the theorem. ∎

##  7\. Open problems

We would like to conclude the paper mentioning a few open problems.

In §3.2 we gave a forcing argument showing that consistently a cardinal
$\kappa$ which is $\kappa^{+}$-$\Pi^{1}_{1}$-subcompact can fail to have the
$(2^{\kappa})^{+}$-gluing property. Intuitively, we expect a better result:

######  Question 1.

Can the least cardinal $\kappa$ which is $\kappa^{+}$-$\Pi^{1}_{1}$-subcompact
have the $2^{2^{\kappa}}$-gluing property?

While we suspect that the consistency strength of the gluing property is low,
it is rather unclear which large cardinals weaker than strongly compact
outright imply gluing. For example, a standard core model argument shows that
if there is no inner model with a Woodin cardinal then in the core model the
least strong cardinal does not have the $\omega$-gluing property.222222Indeed,
if $j\colon K\to K^{M}$ is an ultrapower embedding and $M$ is closed under
$\kappa$-sequences, then it must be a finite iteration of extenders from $K$.
In particular, there are at most finitely many $\alpha<j(\kappa)$ which belong
to $\bigcap\\{j(C)\mathrel{|}\allowbreak C\subseteq\kappa,\text{ club}\\}$.

######  Question 2.

What large cardinals do have the $\omega$-gluing property? More generally,
which of them do have the $\lambda$-gluing property for every cardinal
$\lambda$?

In light of the striking structural consequences of Goldberg’s _Ultrapower
Axiom_ (UA)[Gol22] the following speculation seems natural:

######  Question 3.

Is there a characterization, under $\mathrm{UA}$, of those measurable
cardinals having the $\lambda$-gluing property?

In §5 and §6 we show that the consistency strength of the $\omega$-gluing
property is exactly $``\exists\kappa\,(o(\kappa)=\omega_{1})$”. This raises an
obvious question:

######  Question 4.

What is the consistency strength of “There is a measurable cardinal with the
$\lambda$-gluing property” for $\lambda\geq\omega_{1}$?

Our conjecture is that starting with a strong cardinal one should be able to
produce a model with a measurable cardinal having the $\lambda$-gluing
property for all cardinals $\lambda$. This combined with Gitik’s theorem from
§1 would yield the exact consistency strength of this compactness principle.

Another interesting question from the technical perspective is:

######  Question 5.

Suppose that $\kappa$ has the $\lambda$-gluing property. Is there a forcing
poset destroying this property? In general, using forcing, can we tune the
exact amount of gluing that $\kappa$ carries?

Lastly, in §5, we gave a characterization of the $\kappa$-complete
ultrafilters on $\kappa$ in a certain forcing extension of $\mathcal{K}$. By
recent intriguing results of Gitik and Kaplan [GK23], we suspect that the
assumption $\mathrm{V}=\mathcal{K}$ can be dropped (or replaced by a suitable
$\mathrm{GCH}$ hypothesis).

######  Question 6.

Let $\mathbb{P}$ be a non-stationary support iteration of tree Prikry forcings
over an arbitrary ground model $V$. Is there a full characterization of the
$\kappa$-complete ultrafilters on $\kappa$ in the generic extension, similar
to Lemma 5.8?

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