-Woodin cardinal. On the other hand, we show thatassuming there is no transitive model of KP with a Woodin cardinal theconjecture holds. In the course of this we will also show that if
1 Introduction
In the book ‘The Core Model Iterability Problem’ [Reference Steel10, p. 28], John Steel conjectured the following:
Conjecture 1. Let
$\mathcal {W}$
and
$\mathcal {R}$
be
$1$
-small weasels which are
$\Omega +1$
-iterable. Then the following are equivalent
$:$
-
1.
$\mathcal {W}\leq ^{*}\mathcal {R}$
, and -
2. there is a club
$C \subset \Omega $
such that
$(\alpha ^{+})^{\mathcal {W}}\leq (\alpha ^{+})^{\mathcal {R}}$
for all regular cardinals
$\alpha \in C$
.
In the terminology of the book a weasel is premouse of ordinal height
$\Omega $
, where
$\Omega $
is a fixed measurable cardinal. The relation
$\leq ^{*}$
is the mouse order, i.e., if
$\mathcal {W}$
and
$\mathcal {R}$
are weasels which are sufficiently iterable to successfully
coiterate (by results of the book
$\Omega +1$
-iterability suffices), then
$\mathcal {W}\leq ^{*}\mathcal {R}$
if and only if
$\mathcal {R}$
wins the coiteration, i.e. if
$(\mathcal {T},\mathcal {U})$
is the successful coiteration of
$(\mathcal {W},\mathcal {R})$
and
$\operatorname {\mathrm {lh}}(T)=\theta +1$
and
$\operatorname {\mathrm {lh}}(\mathcal {U})=\gamma +1$
, then
$\mathcal {M}_{\theta }^{\mathcal {T}}\unlhd \mathcal {M}_{\gamma }^{\mathcal {U}}$
.
Steel showed in [Reference Steel10] that Conjecture 1 holds for weasels small enough that linear iterations suffice for comparison. His proof is based on universal linear iterations.
In the following we will prove Conjecture 1 under the assumption that neither
$\mathcal {W}$
nor
$\mathcal {R}$
have an initial segment which models the theory KP + “there
is a Woodin cardinal”, see Theorem 26. In particular, Conjecture 1 holds if there is no transitive model of KP +
“there is a Woodin cardinal”.
On the other hand, assuming the existence of an iterable admissible passive premouse
M with a 
-Woodin
cardinal,Footnote
1
we will show that there is an inner model that thinks there is a
counterexample to Conjecture 1,
see Section 4.2.
For the construction of the counterexample from the assumption just described we will
need to investigate the extender algebra over admissible mice. In the course of
this, we will show that if M is an iterable premouse modelling KP
with a largest, regular, uncountable cardinal
$\delta $
,
$\mathbb {P} \in M$
is a forcing poset such that
$M\models "\mathbb {P}\text { has the }\delta \text {-c.c.}"$
, and
$g \subseteq \mathbb {P}$
is M-generic,
$M[g] \models \text {KP}$
, see Theorem 10.
In Section 5 we will answer another question
implicit in remarks from [Reference Steel10, p. 32] about
the S-hull property in
$1$
-small weasels. We will show that for any
$\Omega +1$
-iterable weasel
$\mathcal {W}$
such that
$\Omega $
is S-thick in
$\mathcal {W}$
the set of points that have the S-hull property
is almost closed, see Theorem 37
and Definition 36.
We will use the notation from [Reference Schlutzenberg9]. A k-maximal iteration tree is as in Definition 3.4 of [Reference Steel11].
2 Admissible premice
Definition 2. A passiveFootnote
2
premouse
$M=(\lfloor M \rfloor ,\in ,\mathbb {E}^{M},\emptyset )$
is called admissible if
$M\models \text {KP}$
.Footnote
3
Remark. Note that an active premouse M, i.e. so that
$F^M \neq \emptyset $
, cannot model KP. We leave this as an exercise.
For
$n\leq \omega $
, we say an n-sound premouse
M is n-countably iterable if every
countable elementary substructure of M is
$(n,\omega _{1},\omega _{1}+1)^{*}$
-iterable. If M is
$\omega $
-countably iterable we also say that M is
countably iterable.
The following is a nice criterion for the admissibility of a passive premouse, whose proof we leave to the reader.
Lemma 3. Let M be a passive premouse. Then
$M\models \text {KP}$
if and only if for all
$f\in \Sigma _{1}^{M}(M)$
such that f is a function with
$\operatorname {\mathrm {dom}}(f)\in M$
,
$f\in M$
.
Moreover, if M has a largest cardinal
$\delta $
, then
$M\models \text {KP}$
if and only if for all
${f\in \Sigma _{1}^{M}(M)}$
such that f is a function with
$\operatorname {\mathrm {dom}}(f)=\delta $
,
$f\in M$
.
Remark. If M is a
$0$
-countably iterable premouse without a largest cardinal,
then M is admissible. The proof of this uses the
Condensation Lemma.
Lemma 4. Suppose M is an admissible passive premouse. If
$\rho _{1}^{M}<\operatorname {\mathrm {OR}}^{M}$
, then
$\rho _{1}^{M}$
is the largest cardinal of M.
Proof. Let
$\rho :=\rho _{1}^{M}$
. Assume for the sake of contradiction that
$\rho ^{+M}$
exists, i.e.,
$\rho ^{+M}<\operatorname {\mathrm {OR}}^{M}$
. Let
$$\begin{align*}H^{\prime}:=\text{Hull}_{1}^{M}(\rho\cup\{p_{1}^{M},\rho\}). \end{align*}$$
Let
$\xi :=\sup (H^{\prime }\cap \rho ^{+M})$
. Suppose first that
$\xi =\rho ^{+M}$
. In this case by the upwards absoluteness of
$\Sigma _{1}$
formulas
$$\begin{align*}M\models\forall\beta<\xi \exists\alpha\exists\zeta(\zeta\in\text{Hull}_{1}^{M\vert\alpha}(\rho\cup\{p_{1}^{M},\rho\})\land \beta < \zeta<\xi). \end{align*}$$
Note that the part of the formula in parentheses is
$\Sigma _{1}$
. Thus, by
$\Delta _{0}$
-collection, there is some
$\gamma <\operatorname {\mathrm {OR}}^{M}$
such that
$$\begin{align*}M\models\forall\beta<\xi \exists\zeta(\zeta\in\text{Hull}_{1}^{M\vert\gamma}(\rho\cup\{p_{1}^{M},\rho\})\land\beta < \zeta<\xi). \end{align*}$$
Thus, there is
$f\in J(M\vert \gamma )\unlhd M$
such that
$f:\rho \rightarrow \rho ^{+M}$
is cofinal. But then there is also
$f^{\prime }:\rho \rightarrow \rho ^{+M}$
onto and
$f^{\prime }\in J(M\vert \gamma )$
. Contradiction!
Let us suppose now that
$\xi <\rho ^{+M}$
. Note that since
$\text {Hull}_{1}^{M}(\rho \cup \{p_{1}^{M}\})\subset H^{\prime }$
and
$\text {Hull}_{1}^{M}(\rho \cup \{p_{1}^{M}\})$
is unbounded in
$\operatorname {\mathrm {OR}}^{M}$
,
$H^{\prime }$
is unbounded in
$\operatorname {\mathrm {OR}}^{M}$
. Again we have that
$$\begin{align*}M\models\forall\beta<\xi\exists\alpha\exists\zeta(\zeta\in\text{Hull}_{1}^{M\vert\alpha}(\rho\cup\{p_{1}^{M},\rho\})\land\beta < \zeta<\xi). \end{align*}$$
Thus, by
$\Delta _{0}$
-collection there is some
$\gamma <\operatorname {\mathrm {OR}}^{M}$
such that
$$\begin{align*}M\models\forall\beta<\xi\exists\zeta(\zeta\in\text{Hull}_{1}^{M\vert\gamma}(\rho\cup\{p_{1}^{M},\rho\})\land\beta < \zeta<\xi). \end{align*}$$
Since
$H^{\prime }$
is unbounded in
$\operatorname {\mathrm {OR}}^{M}$
, there is some such
$\gamma \in H^{\prime }$
. But then since
$\rho \in H^{\prime }$
there is
$f\in H^{\prime }$
such that
$f:\rho \rightarrow \xi $
is cofinal and thus also some
$f^{\prime }:\rho \rightarrow \xi $
which is surjective and
$f^{\prime }\in H^{\prime }$
. But this means that
$\xi \in H^{\prime }$
. Contradiction!
2.1 Forcing over admissible premice
In this subsection, we will prove that if M is an admissible
passive premouse with a largest cardinal
$\delta $
which is Woodin in M,
$\mathbb {B}$
is the extender algebra as defined inside M,
and
$g\subset \mathbb {B}$
is M-generic, then
$M[g]\models \text {KP}$
. This will be Corollary 11 which is an instance of the more general Theorem
10.
Note that if M is an admissible passive premouse with a largest,
regular, and uncountable cardinal
$\delta $
and
$\mathbb {P}\in M$
is a forcing poset, then we may assume without loss of
generality that
$\mathbb {P}\subset \delta $
, as there is a surjection
$f:\delta \rightarrow \mathbb {P}$
in M. We will do so throughout without
further mention.
Lemma 5. Let M be an admissible passive premouse with a largest,
regular, and uncountable cardinal
$\delta $
and let
$\mathbb {P}\in M$
be a forcing poset such that
${M\models "\mathbb {P}\text { has the }\delta \text {-c.c.}"}$
Then there is no
$A\in \Sigma _{1}^{M}(M)$
such that
$A\subset \mathbb {P}$
is an antichain in
$\mathbb {P}$
which is unbounded in
$\delta $
.
Proof. Suppose there is
$A\in \Sigma _{1}^{M}(M)$
is such that
$A\subset \mathbb {P}$
is an antichain in
$\mathbb {P}$
which is unbounded in
$\delta $
. Since
$\mathbb {P}$
has the
$\delta $
-c.c. in M, and
$\delta $
is regular in M, this means that
$A\notin M$
. Let
$\varphi $
be a
$\Sigma _{1}$
-formula such that for some
$p\in M$
,
$$\begin{align*}x\in A\iff M\models\varphi(x,p). \end{align*}$$
Define for
$\alpha <\operatorname {\mathrm {OR}}^{M}\setminus \delta $
such that
$p\in M\vert \alpha $
,
$$\begin{align*}A_{\alpha}:=\{\xi\in\delta:M\vert\alpha\models\varphi(\xi,p)\}. \end{align*}$$
Note that
$A=\bigcup _{\alpha <\operatorname {\mathrm {OR}}^{M}}A_{\alpha }$
and
$A_{\alpha }\in M$
for all
$\alpha $
such that
$\alpha \in \operatorname {\mathrm {OR}}^{M}\setminus \delta $
and
${p\in M\vert \alpha} $
. In particular, since
$A_{\alpha }\subset A$
is an antichain,
$\mathbb {P}$
has the
$\delta $
-c.c. in M, and
$\delta $
is regular in M,
$A_{\alpha }$
is bounded in
$\delta $
for all such
$\alpha $
.
Note that since
$A=\bigcup _{\alpha <\operatorname {\mathrm {OR}}^{M}}A_{\alpha }$
is by assumption unbounded in
$\delta $
,
$$\begin{align*}M\models\forall\beta<\delta\exists\alpha(\exists\gamma(\gamma>\beta\land\gamma\in A_{\alpha})). \end{align*}$$
Since the last part of this formula is
$\Sigma _{1}$
, we have by
$\Delta _{0}$
-collection that there is some
$\alpha ^{\prime }\in M$
such that
$\alpha ^{\prime }$
works for all
$\beta <\delta $
uniformly, i.e.
$$\begin{align*}M\models\exists\alpha^{\prime}\forall\beta<\delta(\exists\gamma(\gamma>\beta\land\gamma\in A_{\alpha^{\prime}})). \end{align*}$$
This contradicts the fact that every
$A_\alpha $
is bounded in
$\delta $
!
Lemma 6. Suppose that M is an admissible passive premouse with a
largest, regular, and uncountable cardinal
$\delta $
. Then
$\delta $
is a
$\Sigma _{1}^{M}(M)$
-regular cardinal, i.e. for all
$\eta <\delta $
and
$f\in \Sigma _{1}^{M}(M)$
such that
$f:\eta \rightarrow \delta $
,
$\operatorname {\mathrm {ran}}(f)$
is bounded in
$\delta $
, i.e.,
$\sup (\operatorname {\mathrm {ran}}(f))<\delta $
.
The proof of the Lemma is very similar to the proof of Lemma 5. Thus, we leave it as an exercise for the reader.
Definition 7. Let M be an admissible passive premouse,
$\mathbb {P}\in M$
be a forcing poset, and
$g\subset \mathbb {P}$
be M-generic. Let
$\xi < \operatorname {\mathrm {OR}}^M$
such that
$\mathbb {P} \in M \vert \xi +1$
. For
$\beta \geq \omega \cdot \xi $
it is a standard fact that
$\Vdash _{M\vert \vert \beta }^{\mathbb {P}}\in M$
, where
$\Vdash _{M\vert \vert \beta }^{\mathbb {P}}$
is the syntactical forcing relation with
$M\vert \vert \beta $
as a ground model using
$\mathbb {P}$
-names in
$M \vert \vert \beta $
. Moreover, the function F is
$\Delta _{1}^{M}(\{\mathbb {P}\})$
, where
$F(\beta )= {\Vdash }_{M\vert \vert \beta }^{\mathbb {P}}$
for
$\beta \geq \omega \cdot \xi $
. We let
$M[g] \vert \vert \beta = (L_\beta [\mathbb {E}^M, g], \in , \mathbb {E}^M \!\upharpoonright \! \beta ,g)$
and have that
$\lfloor M[g] \vert \vert \beta \rfloor $
is the collection of the evaluations of the
$\mathbb {P}$
-names in
$M \vert \vert \beta $
. We let
$M \vert \vert \beta [g] = M[g]\vert \vert \beta $
.Footnote
4
Suppose that
$g\subset \mathbb {P}$
is M-generic and that for a
$\Sigma _{0}$
-formula
$\psi $
,
$M[g]\models \exists x\psi (x,z_{1},...,z_{n})$
, where
$\{z_{1},...,z_{n}\}\subset M[g]$
. Let
$\varphi \equiv \exists x\psi $
be the corresponding
$\Sigma _{1}$
formula and let
$\dot {z}_{1},...,\dot {z}_{n}$
be
$\mathbb {P}$
-names for
$z_{1},...,z_{n}$
. We let
$$\begin{align*}W_{\varphi,\dot{z}_{1},...,\dot{z}_{n}}^{M}:=\{p\in\mathbb{P}:\exists\beta<\operatorname{\mathrm{OR}}^{M}\exists\dot{x}\in (M\vert \beta)^{\mathbb{P}}(p\Vdash_{M\vert\beta}^{\mathbb{P}}\psi(\dot{x},\dot{z}_{1},...,\dot{z}_{n}))\}. \end{align*}$$
We call the set
$W_{\varphi ,\dot {z}_{1},...,\dot {z}_{n}}^{M}$
the set of conditions strongly
forcing
$\varphi $
(with parameters
$\dot {z}_{1},...,\dot {z}_{n}$
).Footnote
5
Remark. Since
$M\models "\text {Pairing}"$
it follows easily (using
$\mathbb {P}$
-names) that
$M[g]\models "\text {Pairing}"$
. Therefore, we may arrange that n has
any specific value which we want it to have. In particular,
$1$
.
We will write
$W_{\varphi ,\dot {z}_{1},...,\dot {z}_{n}}$
for
$W_{\varphi ,\dot {z}_{1},...,\dot {z}_{n}}^{M}$
if it is clear from the context what
M is. Since
$F\in \Delta _{1}^{M}(\{\mathbb {P}\})$
, it follows easily that
$W_{\varphi ,\dot {z}}\in \Sigma _{1}^{M}(\{\dot {z},\varphi ,\mathbb {P}\})$
. Moreover, since
$\mathbb {P}\in M$
, there is
$\beta <\operatorname {\mathrm {OR}}^{M}$
such that
$\mathbb {P}\in M\vert \beta $
so that
$W_{\varphi ,\dot {z}}$
is not trivially empty. However, in general we cannot
assume that
$W_{\varphi ,\dot {z}}\in M$
.
We also write
$p\Vdash _{M}^{w.\mathbb {P}}\varphi (\dot {z})$
for
$p\in W_{\varphi ,\dot {z}}$
.
For
$\beta < \operatorname {\mathrm {OR}}^M$
we might interpret the
$\Sigma _1$
formula defining
$W_{\varphi ,\dot {z}}$
over the structure
$M \vert \beta $
. We will denote the set defined over
$M \vert \beta $
in this way by
$W_{\varphi ,\dot {z}}^{M \vert \beta }$
.
We have the following forcing theorem for
$\Sigma _{1}$
statements for admissible premice.
Theorem 8. Suppose that M is an admissible passive premouse. Let
$\varphi (y)$
be the formula
$\exists x\psi (x,y)$
, where
$\psi $
is a
$\Sigma _{0}$
formula. Let
$\mathbb {P}\in M$
be a forcing poset,
$\dot {z} \in M^{\mathbb {P}}$
, and suppose that
$g\subset \mathbb {P}$
is M-generic. Then the following are
equivalent:
-
•
$M[g]\models \varphi (\dot {z}^{g})$
, -
• there is
$p\in W_{\varphi ,\dot {z}}\cap g$
, and -
• there is
$p\in g$
such that
$p\Vdash _{M}^{\mathbb {P}}\varphi (\dot {z})$
.
Remark. Here
$p\Vdash _{M}^{\mathbb {P}}\varphi (\dot {z})$
refers to the classical notion of forcing an
existential statement, i.e., for every
$q\leq p$
there is some
$r\leq q$
such that there is
$\dot {x}\in M^{\mathbb {P}}$
such that
$r\Vdash _{M}^{\mathbb {P}}\psi (\dot {x},\dot {z})$
. In order to show that this is equivalent to the
existence of some
$p\in W_{\varphi ,\dot {z}}\cap g$
, we must use the admissibility of M.
The rest of the proof is standard.
It is not in general the case that if N is admissible and
$g\subset \mathbb {P}$
is N-generic for some forcing poset
$\mathbb {P}\in N$
, then
$N[g]$
is admissible. See Proposition 13 and the remark preceding it for examples of
admissible structures where the generic extension fails to be admissible. For
$N[g]$
to be admissible, it is sufficient that g
meets all dense open subsets of
$\mathbb {P}$
that are unions of a
$\Sigma _{1}^N(N)$
and a
$\Pi _{1}^N (N)$
class over N. See [Reference Mathias5] for more on this. With Theorem 10 we will give another criterion
for ensuring that the forcing extension of an admissible premouse is a model of
KP.
Lemma 9. Let M be an admissible passive premouse with a largest,
regular, and uncountable cardinal
$\delta $
. Let
$\mathbb {P}\in M$
be a forcing such that
$M\models "\mathbb {P}\text { has the }\delta \text {-c.c.}"$
. Let
$g\subset \mathbb {P}$
be M-generic. Let
$\lambda <\delta $
and suppose that
$M[g]\models \forall \alpha <\lambda \exists x\psi (x,\alpha ,z)$
, where
$\psi $
is a
$\Sigma _{0}$
-formula and
$z\in M[g]$
. Let
$\varphi \equiv \exists x\psi $
and
$\dot {z}\in M^{\mathbb {P}}$
be a name for z.
Then there is
$A\in M$
such that
$A\subset \lambda \times \delta $
and if for
$\alpha <\lambda $
,
$A_{\alpha }:=\{\beta :(\alpha ,\beta )\in A\}$
, then
$A_{\alpha }\subset W_{\varphi ,\check {\alpha },\dot {z}}$
is a maximal antichain in
$W_{\varphi ,\check {\alpha },\dot {z}}$
, in particular, for every
$p\in W_{\varphi ,\check {\alpha },\dot {z}}$
there is some
$q\in A_{\alpha }$
such that
$q\mid \mid p$
.
Proof. We construct the set A recursively along the ordinals of
M. More specifically, we will define via a
$\Sigma _{1}$
-recursion a sequence
$\langle (A^{i},\beta _{i}):i<\operatorname {\mathrm {OR}}^{M}\rangle $
with each
$A^i \in M$
and
$\beta _i \in \operatorname {\mathrm {OR}}^M$
, and
$\langle (A^i,\beta _i) : i< \lambda \rangle \in M$
for each
$\lambda <\operatorname {\mathrm {OR}}^M$
. We will then set
$A=\bigcup _{i<\operatorname {\mathrm {OR}}^{M}}A^{i}$
. Let
$\beta _{0}$
be the least
$\beta $
such that
$$\begin{align*}M\vert\beta\models\exists\alpha<\lambda\exists p\in\mathbb{P}(\exists\dot{x}\in M^{\mathbb{P}}(p\Vdash_{M\vert\beta}^{\mathbb{P}}\psi(\dot{x},\check{\alpha},\dot{z}))), \end{align*}$$
i.e.,
$\beta $
is such that there is some
$\alpha <\lambda $
such that
$W_{\varphi ,\check {\alpha },\dot {z}}^{ M\vert \beta }\neq \emptyset $
. Let
$A^{0}$
be the set of all
$(\alpha ,p)\in A^{0}$
such that
$\alpha <\lambda $
,
$W_{\varphi ,\check {\alpha },\dot {z}}^{M\vert \beta _0} \neq \emptyset $
, and p is the
$<_{M}$
-least element in
$W_{\varphi ,\check {\alpha },\dot {z}}^{M\vert \beta _0}$
. Here,
$<_{M}$
denotes the canonical
$\Sigma _{1}$
-definable well-order of M. Note that
$A^{0}\in M$
, since it is definable over
$M \vert \beta _0$
. Moreover,
$A^{0}$
is a bounded subset of
$\delta $
. (Actually, of
$\delta ^{2}$
, but via coding we may assume that
$A^{0}\subset \delta $
). Before we continue the construction let us introduce
the following notation: For
$\alpha <\lambda $
and
$i<\operatorname {\mathrm {OR}}^{M}$
let
$A_{\alpha }^{i}:=\{q\in \mathbb {P}:(\alpha ,q)\in A^{i}\}$
, where
$A^{i}$
is to be defined.
Let
$\gamma +1<\operatorname {\mathrm {OR}}^{M}$
and suppose that
$\langle (A^{i},\beta _{i}):i\leq \gamma \rangle $
is defined such that
$\langle (A^{i},\beta _{i}):i\leq \gamma \rangle \in \Sigma _{1}^{M}(\{\lambda ,\psi ,\dot {z},\mathbb {P}\})$
. We define
$\beta _{\gamma +1}$
as the least
$\beta $
such that
$$\begin{align*}M\vert\beta\models\exists\alpha<\lambda\exists p\in\mathbb{P}((\forall q\in A_{\alpha}^{\gamma}(p\perp q))\land p\in W_{\varphi,\check{\alpha},\dot{z}}), \end{align*}$$
if there is such
$\beta $
. In other words
$\beta $
is such that for some
$\alpha $
there is an element
$p\in W_{\varphi ,\check {\alpha },\dot {z}}^{M\vert \beta }$
incompatible with the elements of
$A_{\alpha }^{\gamma }$
. In case
$\beta _{\gamma +1}$
is undefined we stop the recursion. Note that in this
case
$A^{\gamma }\in M$
.
In case
$\beta _{\gamma +1}$
is defined, we set
$A^{\gamma +1}$
to be the union of
$A^{\gamma }$
with the set of all
$(\alpha ,p)$
such that p is the
$<_{M}$
-least q such that
$$\begin{align*}M\vert\beta_{\gamma+1}\models q\in W_{\varphi,\check{\alpha},\dot{z}}\land\forall r\in A_{\alpha}^{\gamma}(r\perp q), \end{align*}$$
if there is such q. Since
$A^{\gamma +1}$
is definable over
$M\vert \beta _{\gamma +1}$
,
$A^{\gamma +1}\in M$
.
Now suppose that
$\gamma <\operatorname {\mathrm {OR}}^{M}$
is a limit ordinal and
$\langle A^{i}:i<\gamma \rangle $
is defined. We set
$\beta _{\gamma }=0$
and
$A^{\gamma }=\bigcup _{i<\gamma }A^{i}$
. Note that since
$\langle A_{i}:i<\gamma \rangle \in \Sigma _{1}^{M}(M)$
by KP,
$A^{\gamma }\in M$
.
We aim to see that the recursive definition of
$\langle (A^{i},\beta _{i}):i<\operatorname {\mathrm {OR}}^{M}\rangle $
stops before the ordinal height of
$\operatorname {\mathrm {OR}}^{M}$
, i.e. there is some
$\gamma <\operatorname {\mathrm {OR}}^{M}$
such that
$\beta _{\gamma }$
is not defined. Suppose that this is not the case. We
claim that
$\bar {A}:=\bigcup _{i<\operatorname {\mathrm {OR}}^{M}}A^{i}\in \Sigma _{1}^{M}(M)$
is an unbounded subset of
$\delta $
. For suppose not. Then, since
$\rho _{1}^{M}=\delta $
,
$\bar {A}\in M$
. But the definition of
$\bar {A}$
gives a cofinal and total
$f:\bar {A}\rightarrow \operatorname {\mathrm {OR}}^{M}$
such that
$f\in \Sigma _{1}^{M}(M)$
. Contradiction!
In particular, the following holds in M,
$$\begin{align*}\forall\beta<\delta\exists(\alpha,\gamma)(\alpha<\lambda\land\gamma\in(\beta,\delta)\land\exists\beta_{i}("\gamma\text{ is added to }A_{\alpha}\text{ at stage }\beta_{i}")). \end{align*}$$
But by
$\Delta _{0}$
-collection, there is some
$\beta _{i}<\operatorname {\mathrm {OR}}^{M}$
such that for all
$\beta <\delta $
, there is some
$\gamma \in (\beta ,\delta )$
added to
$A_{\alpha }^{i}$
for some
$\alpha <\lambda $
. But this is inside
$M\vert \beta _{i}+\omega $
. Thus, as
$\delta $
is a regular cardinal of M, there is
some
$\alpha $
to which unboundedly many
$\gamma <\delta $
are added. Contradiction, since
$\mathbb {P}$
has the
$\delta $
-c.c. in M.
Let
$\gamma <\operatorname {\mathrm {OR}}^{M}$
be least such that
$\beta _{\gamma }$
is undefined. Note that
$\gamma $
is by definition a successor ordinal, i.e.
$\gamma =\eta +1$
for some
$\eta <\operatorname {\mathrm {OR}}^{M}$
. Set
$A:=A^{\eta }$
. By construction
$A\in M$
. For
$\alpha <\lambda $
, set
$A_{\alpha }:=\{p\in \mathbb {P}:(\alpha ,p)\in A\}$
. Since
$\beta _{\eta +1}$
is undefined, we have that for all
$\alpha <\lambda $
,
$A_{\alpha }\subset W_{\varphi ,\check {\alpha },\dot {z}}$
is a maximal antichain in
$W_{\varphi ,\check {\alpha },\dot {z}}$
.
Theorem 10. Let M be an admissible passive premouse with a largest,
regular, and uncountable cardinal
$\delta $
and
$\mathbb {P}\in M$
be such that
$M\models "\mathbb {P}\text { has the }\delta \text {-c.c.}"$
Then
$M[g]\models \text {KP}$
for any M-generic
$g\subset \mathbb {P}$
.
Proof. We will verify that
$M[g]$
satisfies
$\Delta _{0}$
-collection and leave the remaining axioms of KP as an
exercise.
Let
$g\subset \mathbb {P}$
be M-generic and suppose for the sake
of contradiction that
$M[g]\not \models "\Delta _{0}\text {-collection}"$
. Let
$\psi $
be a
$\Sigma _{0}$
-formula and
$y\in M[g]$
such that they constitute a counterexample to
$\Delta _{0}$
-collection, i.e.
$$\begin{align*}M[g]\models\forall\alpha<\delta\exists x(\psi(\alpha,x,y)), \end{align*}$$
but there exists no
$z\in M[g]$
such that
$$\begin{align*}M[g]\models\forall\alpha<\delta\exists x\in z(\psi(\alpha,x,y)). \end{align*}$$
We distinguish two cases. First, suppose that there is some
$\lambda <\delta $
such that there exists no
$z\in M[g]$
such that
$$\begin{align*}M[g]\models\forall\alpha<\lambda\exists x\in z(\psi(\alpha,x,y)). \end{align*}$$
Then, by the previous Lemma 9 there is
$A\in M$
such that for
$\alpha <\lambda $
,
$A_{\alpha }:=\{p\in \mathbb {P}:(\alpha ,p)\in A\}$
is a maximal antichain in
$W_{\varphi ,\check {\alpha },\dot {y}}$
, where
$\varphi \equiv \exists x\psi $
and
$\dot {y}\in M^{\mathbb {P}}$
is a
$\mathbb {P}$
-name for y.
Claim 1. For
$\alpha <\lambda $
,
$A_{\alpha }\cap g\neq \emptyset $
.
Proof. Fix some
$\alpha <\lambda $
. Note that since
$M[g]\models \exists x\psi (\alpha ,x,y)$
, there is some
${b\in M[g]}$
such that
$M[g]\models \psi (\alpha ,b,y)$
. By Theorem 8 we can fix some
$p\in g\cap W_{\varphi ,\check {\alpha },\dot {y}}$
. We claim that this implies that
$A_{\alpha }\cap g\neq \emptyset $
. To this end let
$D_{\alpha }:=\{r\in \mathbb {P}:\forall q\in A_{\alpha }(r\perp q)\}$
. Note that since
$A_{\alpha }\in M$
,
$D_{\alpha }\in M$
. Moreover,
$A_{\alpha }\cup D_{\alpha }\in M$
is a pre-dense subset of
$\mathbb {P}$
in M and thus,
$g\cap (A_{\alpha }\cup D_{\alpha })\neq \emptyset $
by the M-genericity of
g. Suppose for the sake of contradiction
that
$g\cap D_{\alpha }\neq \emptyset $
and let
$q\in g\cap D_{\alpha }$
. Since
$q,p\in g$
, there is
$r\in g$
such that
$r\leq q,p$
. However, this means that
$r\in W_{\varphi ,\check {\alpha },\dot {y}}$
, since
$r\leq p$
. But r is incompatible with
every element of
$A_{\alpha }$
, so
$A_{\alpha }$
is not a maximal antichain in
$W_{\varphi ,\check {\alpha },\dot {y}}$
. Contradiction! Thus,
$A_{\alpha }\cap g\neq \emptyset $
.
We have established that for all
$\alpha <\lambda $
,
$A_{\alpha }\cap g\neq \emptyset $
. Moreover, by Theorem 8, we have that
$$\begin{align*}M\models\forall a\in A\exists\dot{x}(\exists p\in\mathbb{P}\exists\alpha<\lambda(a=(\alpha,p)\land p\Vdash_{M}^{\mathbb{P}}\psi(\check{\alpha},\dot{x},\dot{y}))). \end{align*}$$
This is the antecedence of an instance of the
$\Delta _{0}$
-collection scheme since the part in parentheses is
$\Sigma _{1}$
(note that we are using here the fact that
$\Vdash _{M}^{\mathbb {P}}$
for
$\Sigma _{1}$
statements is
$\Sigma _{1}$
definable over M which is true by
Theorem 8). Thus, there
is
$z\in M$
such that
$$\begin{align*}M\models\forall a\in A\exists\dot{x}\in z(\exists p\in\mathbb{P}\exists\alpha<\lambda(a=(\alpha,p)\land p\Vdash_{M}^{\mathbb{P}}\psi(\check{\alpha},\dot{x},\dot{y}))). \end{align*}$$
Let
$z^{g}:=\{\dot {x}^{g}:x\in z\cap M^{\mathbb {P}}\}$
. Note that
$z^{g}\in M[g]$
, as
$z\cap M^{\mathbb {P}}\in M\cap M^{\mathbb {P}}$
. It follows that
$$\begin{align*}M[g]\models\forall\alpha<\lambda\exists x\in z^{g}(\psi(\alpha,x,y)), \end{align*}$$
a contradiction!
Let us now turn towards the second case, i.e., we assume that for all
$\lambda <\delta $
there is
$z\in M[g]$
such that
$$\begin{align*}M[g]\models\forall\alpha<\lambda\exists x\in z(\psi(\alpha,x,y)). \end{align*}$$
Let us associate to
$\psi $
a function
$f\in \Sigma _{1}^{M[g]}(M[g])$
such that
$f:\delta \rightarrow \operatorname {\mathrm {OR}}^{M[g]}$
and for
$\alpha <\delta $
,
$f(\alpha )$
is the least
$\beta $
such that there is
$x\in M\vert \vert \beta [g]$
such that
$M[g]\models \psi (\alpha ,x,y)$
. By our assumption
$f\!\upharpoonright \!\alpha \in M[g]$
for every
$\alpha <\delta $
, but
$f\notin M[g]$
. Let us define an auxiliary function
F with domain
$\delta $
such that
$F(\alpha )=f\!\upharpoonright \!\alpha $
. Note that
$F\in \Sigma _{1}^{M[g]}(M[g])$
. Let
$\varphi ^\prime _{F}$
be a
$\Sigma _1$
formula and
$z\in M[g]$
such that
$$\begin{align*}(c,d)\in F\iff M[g]\models\varphi^\prime_{F}(c,d,z), \end{align*}$$
and let
$\varphi _F(a,b,z)$
be the formula
$$ \begin{align} \exists c \exists d (\varphi^\prime_F (c,d,z) \land a \leq c \land b = d \!\upharpoonright\! a). \end{align} $$
Let
$$\begin{align*}X:=\{p\in\mathbb{P}:\exists\beta<\operatorname{\mathrm{OR}}^{M}(p\Vdash_{M\vert\beta}^{\mathbb{P}}\forall\alpha<\check{\delta}\exists y\varphi_{F}(\alpha,y,\dot{z}))\}. \end{align*}$$
Note that
$X\in \Sigma _{1}^{M}(M)$
and let
$\varphi _{X}$
be the defining formula and
$a\in M$
the corresponding parameter.
We now aim to construct in a similar way as in the proof of Lemma 9 a maximal antichain
A in X. If
$X \neq \emptyset $
, let
$A_{0}:=\{p\}$
for some
$p\in X$
. Otherwise, set
$A_0 = \emptyset $
. Suppose that
$\langle A_{i}:i<\gamma \rangle $
is defined via a
$\Sigma _{1}$
-recursion for some
$\gamma <\operatorname {\mathrm {OR}}^{M}$
. If
$\gamma $
is a limit ordinal, let
$A_{\gamma }=\bigcup _{i<\gamma }A_{i}$
. Then,
$A_{\gamma }\in M$
and
$\langle A_{i}:i\leq \gamma \rangle \in \Sigma _{1}^{M}(M)$
. If
$\gamma $
is not a limit, let
$\beta _{\gamma }<\operatorname {\mathrm {OR}}^{M}$
be the least
$\beta $
such that
$$\begin{align*}M\vert\beta\models\exists p\in\mathbb{P} (\varphi_{X}(p,a) \land \forall q\in A_{\gamma-1}(q \perp p)), \end{align*}$$
if there exists such
$\beta $
. In the case that there is no such
$\beta $
, stop the recursion. In case
$\beta _{\gamma }$
is defined, let
$p_{\gamma }$
be the
$<_{M}$
-least p such that
$$\begin{align*}M\vert\beta_{\gamma}\models \varphi_{X}(p,a) \land \forall q\in A_{\gamma-1} (q\perp p). \end{align*}$$
Let
$A_{\gamma }:=A_{\gamma -1}\cup \{p_{\gamma }\}$
. It is not hard to verify that
$\langle A_{i}:i\leq \gamma \rangle \in \Sigma _{1}^{M}(M)$
and
${A_{\gamma }\in M}$
.
Similar to the proof of the Lemma 9 we aim to see that there is a least
$\gamma <\operatorname {\mathrm {OR}}^{M}$
such that
$\beta _{\gamma }$
is undefined, which will show that
$A:=A_{\gamma -1}$
is a maximal antichain in X and
$A\in M$
.
Suppose for the sake of contradiction that for all
$\gamma <\operatorname {\mathrm {OR}}^{M}$
,
$\beta _{\gamma }$
is defined. Let
$\bar {A}:=\bigcup _{\gamma <\operatorname {\mathrm {OR}}^{M}}A_{\gamma }$
. Clearly,
$\bar {A}\in \Sigma _{1}^{M}(M)$
. Since
$\bar {A}$
is by construction an antichain, by Lemma 5
$\bar {A}$
is bounded in
$\delta $
. In particular,
$\bar {A}\in M$
. However, as in the previous proof, the definition of
$\bar {A}$
give rise to a function
$f\in \Sigma _{1}^{M}(M)$
such that
$\operatorname {\mathrm {dom}}(f)=\bar {A}$
and f is cofinal in
$\operatorname {\mathrm {OR}}^{M}$
. But this is a contradiction!
Let
$D:=\{p\in \mathbb {P}:\forall q\in A(q\perp p)\}$
. As
$A\in M$
,
$D\in M$
and thus,
$A\cup D\in M$
.
$A\cup M$
is pre-dense and therefore,
$g\cap (A\cup D)\neq \emptyset $
. Note that
$g\cap A=\emptyset $
, so
${g\cap D\neq \emptyset} $
.
Let
$\tilde {p}\in D\cap g$
. Note that this means that there is no extension
p of
$\tilde {p}$
such that
$$ \begin{align} p\Vdash_{M}^{w.\mathbb{P}} \exists \beta ((M \vert \beta) [\dot{g}]\models \forall\alpha<\check{\delta}\exists y\varphi_{F}(\alpha,y,\dot{z})). \end{align} $$
We claim that there is
$\vec {p} = \langle p_{i}:i<\delta \rangle \in M$
such that for all
$i < \delta $
,
$p_i \leq \tilde {p}$
and
$M \models \psi (i,\dot {z},p_i)$
, where
$\psi (i,\dot {z},p)$
is the statement
$$ \begin{align} p\Vdash_{M}^{w.\mathbb{P}}\exists y\varphi_{F}(i,y,\dot{z}). \end{align} $$
We may find such
$\langle p_{i}:i<\delta \rangle $
in M, as
$\Delta _{0}$
-collection holds in M and
$$\begin{align*}M\models\forall i<\delta\exists p((p\Vdash_{M}^{w.\mathbb{P}}\exists y\varphi_{F}(i,y,\dot{z}))\land p\leq\tilde{p}). \end{align*}$$
For every
$i<\delta $
there exists such
$p\in \mathbb {P}$
, since
$\tilde {p}\in g$
and
$M[g]\models \forall i<\delta \exists y\varphi _{F}(i,y,z)$
and so by Theorem 8 the existence of p
follows.
Claim 2. There is
$i_0 <\delta $
such that for all
$i \in [i_0,\delta )$
and all
$q\in \mathbb {P}$
with
$q\leq p_i$
and for all
$j \in (i,\delta )$
, there is
$k\in [j,\delta )$
such that
$q \mid \mid p_k$
.
Proof. This follows routinely from the
$\delta $
-c.c. of
$\mathbb {P}$
in M (otherwise, working in
M, construct by recursion on
$\beta $
an antichain
$\langle q_i \rangle _{i<\delta }$
).
Now since
$p_{i_0}\leq \tilde {p}$
, the following claim gives a contradiction, completing
the proof:
Claim 3. Replacing p with
$p_{i_0}$
, line (2.2) holds.
Proof. Let h be
$(M,\mathbb {P})$
-generic with
$p_{i_0} \in h$
. By Claim 2, for every
$\beta \in (i_0,\delta )$
,
$h \cap \{ p_j \mid j \in (i, \delta ) \} \neq \emptyset $
. So there are cofinally many
$i < \delta $
such that
$p_i \in h$
. By KP in M, we can fix
$\xi <\operatorname {\mathrm {OR}}^M$
such that
$M \vert \xi $
satisfies
$\psi (i,\dot {z},p_i)$
for all
$i<\delta $
. Therefore,
$M \vert \vert \xi [h]$
satisfies
$\exists y \varphi _F (i,y,z)$
for cofinally many
$i < \delta $
. Note that for all
$j_0<j_1<\delta $
and all
$p\in \mathbb {P}$
, if
$M \vert \xi $
satisfies
$\psi (j_1,\dot {z},p)$
then
$M \vert \xi $
satisfies
$\psi (j_0,\dot {z},p)$
.Footnote
6
Thus,
$M \vert \vert \xi [h]$
satisfies
$\exists y \varphi _F(i,y,z)$
for all
$i < \delta $
.
As stated prior the claim, this completes the proof.
From Theorem 10 we immediately get the following corollary.
Corollary 11. Suppose that M is an admissible passive premouse with a
largest, regular, and uncountable cardinal
$\delta $
which is Woodin in M and let
$\mathbb {B}\in M$
be the extender algebra with
$\delta $
-many generators as defined inside M.
Let
$g\subset \mathbb {B}$
be M-generic. Then
$M[g]\models \text {KP}$
.
If we replaced
$M\models "\mathbb {P}\text { has the }\delta \text {-c.c.}$
” in the statement of Theorem 10 with
${M\models "\mathbb {P}\text { is }<\delta \text {-closed,}"}$
Theorem 10 is
false as Proposition 13
shows.
Lemma 12. Suppose that M,
$M_{1}$
, and
$M_{2}$
are sound premice such that
$\operatorname {\mathrm {OR}}^{M}$
is a regular, uncountable cardinal (in
V),
$\rho _{\omega }^{M_{1}}=\rho _{\omega }^{M_{2}}=\operatorname {\mathrm {OR}}^{M}$
,
$M \unlhd M_1$
,
$M \unlhd M_2$
, and Condensation holds of
$M_{1}$
and
$M_{2}$
. Then, either
$M_{1}\unlhd M_{2}$
or
$M_{2}\unlhd M_{1}$
.
The Lemma follows from the proof of Lemma 3.1 in [Reference Jensen, Schimmerling, Schindler and Steel4].
Let
$M = L_{\omega _1^{\operatorname {\mathrm {CK}}}}$
and
$\mathbb {P}$
be Cohen forcing. By [Reference Greenberg and Monin2], there are
$(M,\mathbb {P})$
-generics g such that
$M[g] \not \models \text {KP}$
.Footnote
7
The following proposition establishes a variant of this fact.
Proposition 13. Let
$M = (\lfloor M \rfloor , \in , \mathbb {E}^M, \emptyset )$
be a
$1$
-sound admissible passive premouse with a largest,
regular cardinal
$\delta $
such that
$\rho _{1}^{M}<\operatorname {\mathrm {OR}}^{M}$
and suppose that Condensation holds in
M. Let
$\mathbb {C}_{\delta }:=(\delta ^{<\delta })^{M}$
. Then there is an M-generic
$g\subset \mathbb {C}_{\delta }$
such that
$(\lfloor M[g] \rfloor , \in ) \not \models \text {KP}$
.
Proof. Note that
$\mathbb {\mathbb {C}_{\delta }}\in M$
is such that
$$\begin{align*}M\models "\mathbb{C}_{\delta}\text{ is a }<\delta\text{-closed, separable, and atom-less forcing}". \end{align*}$$
By Lemma 4
and
$1$
-soundness,
$\delta =\rho _{1}^{M}$
and
$M=H_{1}^{M}(\delta \cup \{p_{1}^{M}\})$
. Thus, there is a partial surjective function
$h:\delta \rightarrow M$
such that
$h\in \Sigma _{1}^{M}(\{p_{1}^{M}\})$
. Let
$$\begin{align*}\tilde{D}:=\{\xi<\delta:M\models "h(\xi)\subset\mathbb{C}_{\delta}\text{ is dense in }\mathbb{C}_{\delta}"\}. \end{align*}$$
Note that
$\tilde {D}\notin M$
, as otherwise M could construct an
M-generic. Let
$$\begin{align*}D:=\langle\xi_{i}:i<\delta\rangle, \end{align*}$$
be the monotone enumeration of
$\tilde {D}$
. Note that
$\tilde {D}$
has ordertype
$\delta $
. Since
$h\in \Sigma _{1}^{M}(M)$
,
$\tilde {D}\in \Sigma _{1}^{M}(M)$
. However,
$D\notin \Sigma _{1}^{M}(M)$
, as otherwise by
$\Delta _{0}$
-collection,
$D\in M$
.
The idea is now to construct an M-generic
g which codes in a
$\Sigma _{1}$
-fashion the set D so that
$D\in M[g]$
if
$(\lfloor M[g] \rfloor , \in ) \models \text {KP}$
.
Let us define
$\tilde {g} = \langle p_{i}:i<\delta \rangle $
via a recursion on i such that for
all
$\alpha < \delta $
,
$\langle p_{i}:i<\alpha \rangle $
is uniformly
$\Sigma _1$
over M in the parameters
$\{D\!\upharpoonright \! \alpha , \mathbb {C}_\delta \}$
. We will have that
$p_{i}\leq p_{j}$
for
$j<i$
. Set
$p_{0}:=\langle \xi _{0}\rangle $
and
$p_{1}$
to be the
$<_{M}$
-least
$p\in h(\xi _{0})$
such that
$p\leq p_{0}$
. Note that there is such p, since
$h(\xi _{0})$
is dense in
$\mathbb {C}_{\delta }$
.
Suppose that
$\langle p_{i}:i<\lambda \rangle $
is defined, where
$\lambda <\delta $
is a limit ordinal, such that
$p_{i}\leq p_{j}$
for
$j\leq i<\lambda $
. Note that
$D\!\upharpoonright \!\lambda \in M$
, since
$\tilde {D}\in \Sigma _{1}^{M}(M)$
and
$\delta $
is
$\Sigma _{1}^M (M)$
-regular by Lemma 6. By our induction hypothesis the construction
so far is uniformly
$\Sigma _{1}$
in the parameters
$D\!\upharpoonright \!\lambda $
and
$\mathbb {C}_{\delta }$
. This implies that
$\langle p_{i}:i<\lambda \rangle \in M$
. Thus, we can set
$p_{\lambda }=(\bigcup _{i<\lambda }p_{i})^{\frown }\langle \xi _{\lambda }\rangle $
. Let
$p_{\lambda +1}$
be the
$<_{M}$
-least
$p\in h(\xi _{\lambda })$
such that
$p\leq p_{\lambda }$
.
We now turn towards the successor case. Suppose we have defined
$\langle p_\beta \rangle _{\beta \leq \gamma }$
where
$\gamma $
is an odd successor (that is,
$\gamma =\lambda +2n+1$
for some
$n<\omega $
and
$\lambda $
is a limit less than
$\gamma $
or equal to
$0$
). Set
$p_{\gamma + 1} := p_{\gamma }^{\frown }\langle \xi _{\lambda +n}\rangle $
and let
$p_{\gamma + 2}$
be the
$<_{M}$
-least
$p\in h(\xi _{\lambda +n})$
such that
$p\leq p_{\gamma +1}$
.
Let g be the upwards-closure of
$\tilde {g}$
in
$\mathbb {\mathbb {C}_{\delta }}$
. By definition of
$\tilde {g}$
, g is M-generic.
Note that since
$\mathbb {E}^{M}\!\upharpoonright \!\delta \in M[g]$
by Lemma 12 we can define
$\mathbb {E}^{M}$
over
$M[g]$
in a
$\Sigma _{1}$
fashion as the extender sequence of the stack of sound
premice extending
$M\vert \delta $
which project to
$\delta $
and for which condensation holds. Thus, it follows
that
${(M, \in , \mathbb {E}^M) \in \Sigma _{1}^{(\lfloor M[g] \rfloor , \in )}(\{ M \vert \delta \})}$
. Note that therefore
$\tilde {g}$
is definable over
$(M[g], \in )$
from
$\{\mathbb {C}_{\delta },\delta ,g, M \vert \delta \}$
via a
$\Sigma _{1}$
recursion of length
$\delta $
. Thus, if
$(\lfloor M[g] \rfloor , \in ) \models \text {KP}$
, then
$\tilde {g} \in M[g]$
and so
$D \in M[g]$
.
Let
$h^{\prime }:\delta \rightarrow \operatorname {\mathrm {OR}}^{M}$
be the partial function such that
$h^{\prime }(\alpha )$
is the least
${\beta <\operatorname {\mathrm {OR}}^{M}}$
such that
$h(\alpha )\in M \vert \beta $
, if
$h(\alpha )$
is defined. Clearly,
$h^{\prime }\in \Sigma _{1}^{M}(\{p_{1}^{M}\})$
. We have that
$h^{\prime }\in \Sigma _{1}^{(\lfloor M[g] \rfloor , \in )}(\lfloor M[g] \rfloor )$
. Moreover, as
$\mathbb {C}_{\delta }$
is
$<\delta $
-closed and atom-less,
${\operatorname {\mathrm {ran}}(h^{\prime }\!\upharpoonright \! D)}$
is cofinal in M. Since
$\operatorname {\mathrm {OR}}^{M}=\operatorname {\mathrm {OR}}^{M[g]}$
,
$h^\prime \!\upharpoonright \! D$
is cofinal in
$M[g]$
. But now if
$(\lfloor M[g] \rfloor , \,{\in}) \models \text {KP,}$
since
$D\in M[g]$
,
$\operatorname {\mathrm {OR}}^{M[g]}\in M[g]$
. Contradiction! Thus,
${(\lfloor M[g] \rfloor , \in ) \not \models \text {KP}}$
.
2.2 Preservation of admissibility between iterates of premice
Next we deal with the preservation of admissibility between iterates of premice. Parts of this will be used in the proof of Lemma 25.
Lemma 14. Suppose that M and N are premice and
let
$i:M\rightarrow N$
be
$r\Sigma _{3}$
-elementary. Then
$M\models \text {KP}$
iff
$N\models \text {KP}$
.
The lemma follows directly from the fact that the part of KP excluding the
induction axioms has an
$r\Pi _{3}$
axiomatization and is therefore preserved under
$r\Sigma _{3}$
-elementary embeddings between premice.
Thus, by Lemma 14, if
$\text {Ult}_{n}(M,E)$
is wellfounded,
$\text {Ult}_{n}(M,E) \models \text {KP}$
if
$n\geq 2$
and M is an admissible passive
premouse.
Lemma 15. Let
$k\leq \omega $
and let N be a
k-sound premouse. Let
$\mathcal {T}$
be a k-maximal iteration tree on
N such that
$\operatorname {\mathrm {lh}}(\mathcal {T})=\theta +1$
. Let
$b:=[0,\theta ]^{\mathcal {T}}$
be the main branch of
$\mathcal {T}$
and
$\alpha $
be least such that
$\alpha + 1 \in b$
and
$(\alpha +1,\theta ]^{\mathcal {T}}$
does not drop in model. Let
$\eta = \alpha +1$
. Then
$\mathcal {M}_{\eta }^{*\mathcal {T}}$
is an admissible passive premouse with a largest,
regular, and uncountable cardinal if and only if
$\mathcal {M}_{\theta }^{\mathcal {T}}$
is an admissible passive premouse with a largest,
regular, and uncountable cardinal.
Proof. We prove the two directions separately.
Claim 1. Suppose
$\mathcal {M}_{\eta }^{*\mathcal {T}}$
is an admissible passive premouse with a
largest, regular, and uncountable cardinal. Then so is
$\mathcal {M}_{\theta }^{\mathcal {T}}$
.
Proof. Note that in the case that
$\deg ^{\mathcal {T}}(\theta )\geq 2$
, the lemma holds by Lemma 14. Thus, we may
assume that
$\deg ^{\mathcal {T}}(\theta ) \leq 1$
.
By standard facts
$\mathcal {M}_{\theta }^{\mathcal {T}}$
is a passive premouse with a largest, regular,
and uncountable cardinal. We have to verify that
$\mathcal {M}_{\theta }^{\mathcal {T}}$
is admissible. By Lemma 3, it suffices to
show that
$\Delta _{0}$
-collection holds in
$\mathcal {M}_{\theta }^{\mathcal {T}}$
.
Case 1.
$\theta = \zeta + 1$
is a successor ordinal.
We inductively assume that
$\mathcal {M}^{*\mathcal {T}}_{\theta }$
is an admissible passive premouse with
a largest, regular, and uncountable cardinal.
Let
$n := \deg ^{\mathcal {T}} (\theta )$
,
$M:= \mathcal {M}^{*\mathcal {T}}_\theta $
,
$E:=E_{\zeta }^{\mathcal {T}}$
, and
$M^\prime := \text {Ult}_{n}(\mathcal {M}_\theta ^{*\mathcal {T}},E)$
. By our initial remarks
$n \leq 1$
. By standard arguments
$\kappa := \operatorname {\mathrm {crit}}(E)$
is strictly less than the largest
cardinal of M. In particular, by Lemma
4,
$\kappa < \rho _1^M$
.
Since M is admissible we have by
$\Delta _{0}$
-collection that for any
$m<\omega $
,
$f\in \Sigma _{1}^{M}(M)\cap {^{[\kappa ]^{m}}M}$
implies that
$f\in M$
. This means that the canonical factor
map
$\pi \colon \text {Ult}_{0}(M,E) \to \text {Ult}_{1}(M,E)$
is the identity. Thus,
$\text {Ult}_{0}(M,E)=\text {Ult}_{1}(M,E)$
and the associated ultrapower maps are
the same. In particular,
$i^{*\mathcal {T}}_{\theta }\colon M\to M^{\prime }$
is a
$1$
-embedding and
$i^{*\mathcal {T}}_{\theta }$
is cofinal in
$\operatorname {\mathrm {OR}}^{M^{\prime }}$
.
Subclaim 1. For
$a \in [\operatorname {\mathrm {lh}}(E)]^{<\omega }$
,
$\text {Ult}_{0}(M,E_a) \models \text {KP}$
.
Proof. Let
$a \in [\operatorname {\mathrm {lh}}(E)]^{<\omega }$
and suppose for contradiction
that
${\text {Ult}_{0}(M,E_a) \not \models \text {KP}}$
. By our initial remarks this
means that
$\Delta _{0}$
-collection fails in
$\text {Ult}_{0}(M,E_a)$
, i.e., there is a
$\Sigma _{0}$
-formula
$\varphi $
,
$\xi < \operatorname {\mathrm {OR}}^{\text {Ult}_{0}(M,E_a)}$
, and
$p\in \text {Ult}_{0}(M,E_a)$
such that
$$ \begin{align} \text{Ult}_{0}(M,E_a) \models\forall\alpha<\xi\exists x\varphi(\alpha,x,p), \end{align} $$
but there is no
$z\in \text {Ult}_{0}(M,E_a)$
such that
$$ \begin{align} \text{Ult}_{0}(M,E_a) \models\forall\alpha<\xi\exists x\in z\varphi(\alpha,x,p). \end{align} $$
Let
$i \colon M \to \text {Ult}_{0}(M,E_a)$
be the ultrapower embedding. By
our previous remarks we have that
i is a
$1$
-embedding and cofinal in
$\operatorname {\mathrm {OR}}^{\text {Ult}_{0}(M,E_a)}$
.
Let
$\delta $
be the largest cardinal of
M and
$\delta ^{\prime }=i(\delta )$
be the largest cardinal of
$\text {Ult}_{0}(M,E_a)$
. By Lemma 3 we may
assume that
$\xi $
in (2.4) is equal to
$\delta ^{\prime }$
. Let
$j=\lvert a\rvert $
. Let
$g^{\prime }\colon \delta ^\prime \to \text {Ult}_{0}(M,E_a)$
be the canonical function
derived from
$\varphi $
by taking the
$<_{\text {Ult}_{0}(M,E_a)}$
-least witness. Note that
$g^{\prime }\in \Sigma _{1}^{\text {Ult}_{0}(M,E_a)}(\{p\})$
as
$\varphi $
is
$\Sigma _{0}$
. We will now show that
$g^\prime $
is bounded in
$\text {Ult}_{0}(M,E_a)$
. This will contradict the
failure of
$\Delta _0$
-collection, completing the
proof.
By the definition of
$\text {Ult}_{0}(M,E_a)$
we have for
$\alpha <\delta ^{\prime }$
some
$f_{\!\alpha } \in {^{[\kappa ]^{j}}}\delta \cap M$
such that
$\alpha =[a,f_{\!\alpha }]_{E_a}^{M}$
. Also, there is
$f_{\!\!p}\in ({^{[\kappa ]^{j}}}M) \cap M$
such that
$p=[a,f_{\!\!p}]_{E}^{M}$
. Since we assumed that
$\delta $
is regular in M
and
$\kappa < \delta $
we may assume that for
$\alpha <\delta ^\prime $
,
$f_{\!\alpha }\in {^{[\kappa ]^{j}}}\delta \cap (M\vert \delta )$
. Moreover, since
Łoś’s Theorem holds for
$\Sigma _{1}$
-formulae, we have by (2.4) for
all
$\alpha <\delta ^{\prime }$
,
$$\begin{align*}A_{\alpha}:=\{b\in[\kappa]^{j}:M\models\exists x\varphi(f_{\!\alpha}(b),x,f_{\!\!p}(b))\}\in E_a. \end{align*}$$
Note that for any
$h\in {^{[\kappa ]^{j}}} \delta \cap (M\vert \delta )$
there is some
$\alpha <\delta ^{\prime }$
such that
$\alpha =[a,h]_{E}^{M}$
. Thus,
$$\begin{align*}M\models\forall h\in {^{[\kappa]^{j}}}\delta\cap(M\vert\delta)\exists A(A\in E_a \land\forall b\in A\exists x\varphi(h(b),x,f_{\!\!p}(b))). \end{align*}$$
Note that this does make sense
since
$E_a \in M$
: By standard facts about
k-maximal iteration trees
E is close to M
and therefore in particular, for every
$b \in [\operatorname {\mathrm {lh}}(E)]^{<\omega }$
,
$E_{b} \in \Sigma _{1}^{M}(M)$
. But since M is
admissible and
$\kappa < \delta $
, this implies that
$E_a \in M$
.
Using
$\Delta _{0}$
-collection, we have for every
$h\in {^{[\kappa ]^{j}}} M \cap (M\vert \delta )$
and A such that
$$\begin{align*}M \models A \in E_a \land\forall b\in A\exists x\varphi(h(b),x,f_{\!\!p}(b)), \end{align*}$$
there is some
$Y\in M$
such that
$$\begin{align*}M\models A \in E_a \land\forall b\in A\exists x\in Y\varphi(h(b),x,f_{\!\!p}(b)). \end{align*}$$
Thus,
$$\begin{align*}M\models\forall h\in {^{[\kappa]^{j}}}\delta\cap(M\vert\delta)\exists A\exists Y(A\in E_a \land\forall b\in A\exists x\in Y\varphi(h(b),x,f_{\!\!p}(b))). \end{align*}$$
By another application of
$\Delta _{0}$
-collection this gives us
$\beta < \operatorname {\mathrm {OR}}^M$
such that
$$\begin{align*}M\models\forall h\in {^{[\kappa]^{j}}}\delta\cap(M\vert\delta)\exists A\in E_a (\forall b\in A\exists x\in (M \vert \beta)\varphi(h(b),x,f_{\!\!p}(b))). \end{align*}$$
It now easily follows that
$g^{\prime }$
is bounded by
$M^\prime \vert i(\beta )$
.
In the case that E is finitely generated
this argument shows that
$M^\prime \models \text {KP}$
. Thus, we may assume that
E is not finitely generated. In
this case,
$M^{\prime }$
is the direct limit of
$$\begin{align*}(\text{Ult}_{0}(M,E_{a}),\pi_{ab}:a,b\in[\nu(E)]^{<\omega}\land a\subset b), \end{align*}$$
where
$\pi _{ab}$
is the canonical factor embedding. For
$a\in [\nu (E)]^{<\omega }$
let
$X_{a}:=\pi _{a\infty }[\text {Ult}_{0}(M,E_{a})]$
. By Łoś’s Theorem
it follows that
$\pi _{a\infty }:\text {Ult}_{0}(M,E_{a})\rightarrow M^{\prime }$
is
$\Sigma _{1}$
-elementary.
Suppose now that
$f ^\prime \in \Sigma _{1}^{M^{\prime }} (M^{\prime })$
. There is some
$a \in [\nu (E)]^{<\omega }$
such that
${f^\prime \in \Sigma _1^{M^\prime } (\{ p^\prime \})}$
for some
$p^\prime \in X_{a}$
. Let
$p = \pi _{a \infty }^{-1}(p^\prime )$
and let f be the
function, which is defined over
$\text {Ult}_{0}(M,E_a)$
via the parameter p,
as
$f^\prime $
is defined over
$M^\prime $
via the parameter
$p^\prime $
. By the claim we have that
$f \in \text {Ult}_{0}(M,E_a)$
. But this means that
$f^\prime \in M^\prime $
. Thus, by Lemma 3,
$M^\prime $
is admissible.
Case 2.
$\operatorname {\mathrm {lh}}(\mathcal {T}) = \theta $
is a limit ordinal.
This is proven much as in the case that
$\theta = \zeta +1$
is a successor ordinal and
$E_\zeta $
is not finitely generated.
This finishes the proof of Claim 1.
Claim 2. Suppose
$\mathcal {M}_{\theta }^{\mathcal {T}}$
is an admissible passive premouse with a
largest, regular, and uncountable cardinal. Then so is
$\mathcal {M}_{\eta }^{*\mathcal {T}}$
.
Proof. Again we argue by induction on
$\theta $
. Suppose first that
$\theta =\zeta +1$
and let
$\mathcal {M}^{\mathcal {T}}_\theta = M^\prime $
and
$M = \mathcal {M}_\theta ^{*\mathcal {T}}$
. Again, by standard arguments we may assume
that M is a premouse with a largest, regular,
and uncountable cardinal
$\delta $
. It suffices to see that M is
admissible for which in turn it suffices to see that
$\Delta _0$
-collection holds in M.
Suppose for the sake of contradiction that this is not true and
let
$\eta < \operatorname {\mathrm {OR}}^M$
witness the least failure of
$\Delta _{0}$
-collection, i.e.,
-
1. there is a
$\Sigma _{1}$
-formula
$\varphi $
and
$p\in M$
such that
$M\models \forall \alpha <\eta \exists x\varphi (x,\alpha ,p)$
, but there is no
$z\in M$
such that
$M\models \forall \alpha <\eta \exists x\in z\varphi (x,\alpha ,p)$
, and -
2. for all
$\Sigma _1$
-formulas
$\psi $
and
$q \in M$
we have that if
$\gamma < \eta $
and
$M \models \forall \alpha < \gamma \exists x \psi (x,q)$
, then there is
$z \in M$
such that
$M \models \forall \alpha < \gamma \exists x \in z \psi (x,q)$
.
Let
$n = \deg _\theta ^{\mathcal {T}}$
so that
$M^\prime = \text {Ult}_{n}(M,E)$
, where
$E = E_{\zeta }^{\mathcal {T}}$
. As before we may assume that
$n\leq 1$
. Let
$i:=i_{\theta }^{*\mathcal {T}}:M\rightarrow M^{\prime }$
be the ultrapower map and let
$\delta ^{\prime }=i(\delta )$
be the largest cardinal of
$M^{\prime }$
. Note that we can no longer assume that
i is both a
$0$
- and a
$1$
-embedding, unlike in Claim 1.
Case 1.
$n=1$
.
Note that it suffices to see that
$M^{\prime }\models \forall \alpha <i(\eta )\exists x \varphi (x,\alpha ,i(p))$
because then by the admissibility of
$M^{\prime }$
,
$M^{\prime }\models \exists z^{\prime }\forall \alpha <i(\eta )\exists x\in z^{\prime }\varphi (x,\alpha ,i(p))$
and so by the elementarity of
i there is a bound in
M. Note that i is
a
$1$
-embedding and therefore
$r\Sigma _2$
-elementary. But this means that
$M^{\prime }\models \forall \alpha <i(\eta )\exists x \varphi (x,\alpha ,i(p))$
.
Case 2.
$n=0$
.
Let
$\eta ^{\prime }:=\sup (i[\eta ])$
and note that
$\eta ^{\prime }$
is a limit ordinal. We claim that for
every
$\alpha < \eta ^\prime $
,
$M^\prime \models \exists x \varphi (x,\alpha ,i(p))$
. Suppose otherwise and let
$\alpha ^{\prime }<\eta ^{\prime }$
be a counterexample, i.e.
$$\begin{align*}M^{\prime}\models\forall x\neg\varphi(x,\alpha^{\prime},i(p)). \end{align*}$$
Let
$\alpha <\eta $
be such that
$i(\alpha )>\alpha ^{\prime }$
. Note that since
$\eta $
is the minimal failure of
$\Delta _{0}$
-collection in M,
$$\begin{align*}M\models\exists z\forall\beta<\alpha\exists x\in z\varphi(x,\beta,p). \end{align*}$$
But this is a
$\Sigma _{1}$
-statement, so that
$$\begin{align*}M^{\prime}\models\exists z\forall\beta<i(\alpha)\exists x\in z\varphi(x,\beta,i(p)). \end{align*}$$
But then in particular,
$$\begin{align*}M^{\prime}\models\exists x\varphi(x,\alpha^{\prime},i(p)). \end{align*}$$
Contradiction! Therefore, by
$\Delta _{0}$
-collection in
$M^\prime $
, there is
$\beta ^{\prime }<\operatorname {\mathrm {OR}}^{M^{\prime }}$
such that
$M^{\prime }\models \forall \alpha <\eta ^{\prime }\exists x\in (M^{\prime }\vert \beta ^{\prime })\varphi (x,\alpha ,i(p)).$
Since i is cofinal,
we may assume without loss of generality that
$\beta ^{\prime }\in \operatorname {\mathrm {ran}}(i)$
. Let
$\beta <\operatorname {\mathrm {OR}}^{M}$
be such that
$i(\beta )=\beta ^{\prime }$
. We claim that
$$\begin{align*}M\models\forall\alpha<\eta \exists x\in(M\vert\beta)\varphi(x,\alpha,p), \end{align*}$$
which would be a contradiction! So
suppose that there is
$\alpha <\eta $
such that
$$\begin{align*}M\models\forall x\in(M\vert\beta)\neg\varphi(x,\alpha,p). \end{align*}$$
This is a
$\Delta _0$
statement, so that
$$\begin{align*}M^{\prime}\models\forall x\in(M^{\prime}\vert\beta^{\prime})\neg\varphi(x,\alpha,i(p)). \end{align*}$$
Contradiction!
Now let us suppose that
$\theta $
is a limit ordinal. In the case that there is
some
$\gamma \in b\cap \theta $
such that for all
$\xi \in (\gamma ,\theta )\cap b$
,
$\deg _\xi ^{\mathcal {T}} \geq 1$
, we can argue as in Case 1 of the successor
case. If otherwise we can use the argument from Case 2 of the successor
case, since
$i_{\gamma \theta }^{\mathcal {T}}$
will be cofinal.
This completes the proof of the lemma.
If we would not require
$\delta $
to be regular in the statement of Lemma 15, the lemma would be provably
false as the following example shows.
Example 16. Let M be a
$1$
-sound premouse such that
$M\models \text {KP}$
. Suppose that M has a largest
cardinal
$\delta>\omega $
such that for some
$\kappa <\delta $
,
$\operatorname {\mathrm {cof}}^{M}(\delta )=\kappa $
and there is an M-total
$E\in \mathbb {E}^{M}$
such that
$\operatorname {\mathrm {crit}}(E)=\kappa $
. Let
$\pi :M\rightarrow \text {Ult}_{1}(M,E)$
be the ultrapower embedding.
Then
$\pi $
is discontinuous at
$\delta $
so that
$\pi (\delta )>\sup (\pi [\delta ])$
. However, since
$\pi $
is a
$1$
-embedding,
$\rho _{1}^{M}=\sup (\pi [\delta ])$
. But
$\pi (\delta )$
is the largest cardinal of
$\text {Ult}_{1}(M,E)$
. So by Lemma 4,
$\text {Ult}_{1}(M,E) \not \models \text {KP}$
.
In this subsection we have established the following (cf. [Reference Steel11, Chapter 7.2]).
Theorem 17. Let
$k\leq \omega $
and suppose that M is a
k-sound,
$(k,\vert M\vert ^{+}+1)$
-iterable admissible passive premouse with a largest,
regular, and uncountable cardinal
$\delta $
. Suppose that
$\delta $
is Woodin in M and let
$X\subset \vert M \vert $
. Then there is a k-maximal iteration
tree
$\mathcal {T}$
on M with last model
$\mathcal {M}_\infty ^{\mathcal {T}}$
, which does not drop in model anywhere, such that
X is generic over
$\mathcal {M}_{\infty }^{\mathcal {T}}$
for the extender algebra of
$\mathcal {M}_\infty ^{\mathcal {T}}$
and
$\mathcal {M}_{\infty }^{\mathcal {T}}[X]\models \text {KP}$
.
2.3 A version of the truncation Lemma
For the proof of Lemma 24, we
need a version of the Truncation Lemma for premice. Recall the following coarse
definition. If M is a possibly ill-founded structure in some
signature extending
$\mathcal {L}_{\dot {\in }}$
, we call
$$\begin{align*}\text{wfp}(M) := \{ x \in \lfloor M \rfloor \mid \in^M \!\upharpoonright\! (\text{trc}_{\in^M} (\{x\}))^2 \text{ is wellfounded} \} \end{align*}$$
the wellfounded part of M. By [Reference Barwise1] and Problem 5.27 of [Reference Schindler8], if
$M \models \text {KP}$
, then
$\text {wfp}(M) \models \text {KP}$
. We aim to show something similar in the case that
M is an ill-founded premouse.
Definition 18. Let
$M = (\lfloor M \rfloor , \in ^M, \mathbb {E}^M)$
be an
$\mathcal {L}_{\dot {\in }, \dot {E}}$
-structure. We say that
${M \models "V=L[E]"}$
, if M models the Axiom of
Extensionality, the Axiom of Foundation, and
$$ \begin{align} \forall x \exists \alpha (x \in S_\alpha [\dot{E}]) \land \forall \alpha \exists x (x \notin S_\alpha [\dot{E}]), \end{align} $$
where
$S_\alpha [\dot {E}]$
is the refinement of the
$J[\dot {E}]$
-hierarchy as described in Chapter 5 of [Reference Schindler8].
Note that the second conjunct of 2.6 makes sure that the model is an actual instance of its internal J-hierarchy.
Definition 19. Let
$M = (\lfloor M \rfloor , \in ^M, \mathbb {E}^M)$
be an
$\mathcal {L}_{\in , \dot {E}}$
-structure such that
$M \models "V = L[E]"$
and
$\text {wfp} (M)$
is transitive. Letting
$\text {wfo}(M) = \operatorname {\mathrm {OR}} \cap \text {wfp}(M)$
, we call
$$\begin{align*}\text{wfc} (M) := (S_{\text{wfo}(M)}^{\mathbb{E}^M}, \in, \mathbb{E}^M \!\upharpoonright\! \text{wfo}(M)) \end{align*}$$
the wellfounded cut of M.
The next lemma is the version of the Truncation Lemma we need for the proof of Lemma 24. The proof is essentially identical to the proof of Proposition 2.4 in [Reference Hachtman3].
Lemma 20. Let
$M = (\lfloor M \rfloor , \in ^M, \mathbb {E}^M)$
be an ill-founded
$\mathcal {L}_{\in , \dot {E}}$
-structure such that
${M \models "V = L[E]"}$
and
$\text {wfp} (M)$
is transitive. Then
$\text {wfc}(M) \models \text {KP}$
.
Proof. We may assume that
$\omega \in \text {wfc}(M)$
, as otherwise
$\text {wfc}(M) = L_\omega $
which is clearly admissible. It suffices to see that
$\text {wfc}(M) \models \Delta _0 \text {-Collection}$
. By induction, it easily follows that for
$\alpha < \text {wfo}(M)$
,
$S_\alpha ^{\mathbb {E}^M} = (S_\alpha )^M \in \text {wfp}(M)$
, so that
$\text {wfc} (M) \subseteq \text {wfp}(M)$
. Let
$\varphi $
be a
$\Sigma _0$
formula and
$a,p \in \text {wfc} (M)$
such that
$$\begin{align*}\text{wfc} (M) \models \forall x \in a \exists y \varphi(x,y,p). \end{align*}$$
Note that by
$\Sigma _1$
upwards absoluteness this holds in M.
Let
$\gamma $
be a non-standard ordinal of M. In
$S_\gamma ^M$
, we may define a function F with
$\operatorname {\mathrm {dom}}(F) = a$
such that for
$x \in a$
,
$$ \begin{align*} F(x) = \eta \iff S_\gamma^M \models x \in a \land \exists y \in S_{\eta+1} \varphi(x,y,p) \land \forall y \in S_\eta \neg \varphi(x,y,p). \end{align*} $$
Since
$\text {wfc} (M) \subseteq S_\gamma ^M$
, it follows by
$\Sigma _1$
upwards absoluteness that
$F(x) < \text {wfo}(M)$
. However, this means that
$\eta := \bigcup _{x \in a} F(x) \subset \text {wfo}(M)$
. Since F is definable over
M, we must have that
$\eta < \text {wfo}(M)$
. This means that
$$\begin{align*}\text{wfc} (M) \models \forall x \in a \exists y \in S_{\eta} \varphi(x,y,p). \end{align*}$$
Thus,
$\text {wfc} (M) \models \text {KP}$
.
3 Where the conjecture holds
In this section, we will show under the assumption that there is no transitive model of KP with a Woodin cardinal that Conjecture 1 holds. This will be a consequence of Theorem 26. Lemma 24 is the key insight for proving Theorem 26. First, let us recall some well-known basic properties of weasels and their coiterations.
The following theorem from [Reference Steel10] guarantees
that the coiteration of two
$\Omega +1$
-iterable weasels of height
$\Omega $
is successful.
Theorem 21. Let
$\kappa $
be an inaccessible cardinal. Let
$\mathcal {M}$
and
$\mathcal {N}$
be premice such that
$\operatorname {\mathrm {OR}}^{\mathcal {M}}=\operatorname {\mathrm {OR}}^{\mathcal {N}}=\kappa $
. Let
$(\mathcal {T},\mathcal {U})$
be a successful coiteration of
$(\mathcal {M},\mathcal {N})$
. Then
$\max \{\operatorname {\mathrm {lh}}(\mathcal {T}), \operatorname {\mathrm {lh}}(\mathcal {U})\}\leq \kappa +1$
. Moreover, setting
$\operatorname {\mathrm {lh}}(\mathcal {T})=\theta +1$
and
$\operatorname {\mathrm {lh}}(\mathcal {U})=\gamma +1$
, either
-
1.
$D^{\mathcal {T}}\cap [0,\theta ]^{\mathcal {T}}=\emptyset $
,
$i_{0,\theta }^{\mathcal {T}}[\kappa ]\subset \kappa $
,
$\mathcal {M}_{\theta }^{\mathcal {T}}\unlhd \mathcal {M}_{\gamma }^{\mathcal {U}}$
, and
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\theta }^{\mathcal {T}}}=\kappa $
, or -
2.
$D^{\mathcal {U}}\cap [0,\gamma ]^{\mathcal {U}}=\emptyset $
,
$i_{0,\gamma }^{\mathcal {U}}[\kappa ]\subset \kappa $
,
$\mathcal {M}_{\gamma }^{\mathcal {U}}\unlhd \mathcal {M}_{\theta }^{\mathcal {T}}$
, and
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\gamma }^{\mathcal {U}}}=\kappa $
.
The next lemma collects basic facts about iteration trees whose proof is well-known.
Lemma 22. Let
$\kappa $
be a regular and uncountable cardinal. Let
$\mathcal {M}$
be a premouse such that
$\operatorname {\mathrm {OR}}^{\mathcal {M}}=\kappa $
and let
$\mathcal {T}$
be a
$0$
-maximal iteration tree on
$\mathcal {M}$
such that
$\operatorname {\mathrm {lh}}(\mathcal {T})=\kappa +1$
. Let
$b := [0,\kappa ]^{\mathcal {T}}$
. Then the following hold:
-
1. if
$b \cap D^{\mathcal {T}} = \emptyset $
and
$i_{0\kappa }^{\mathcal {T}}[\kappa ]\subset \kappa $
, then there is a club
$C_1 \subset b$
such that for all
$\alpha \in C_1 \cap b$
,
$i_{0\alpha }^{\mathcal {T}}[\alpha ]\subset \alpha $
, -
2. if
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\alpha }^{\mathcal {T}}} \leq \kappa $
for all
$\alpha <\kappa $
, then there is a club
$C_2 \subset b$
such that for all
$\alpha \in C_2$
,
$\alpha =\sup \{\operatorname {\mathrm {lh}}(E_{\beta }^{\mathcal {T}}):\beta <\alpha \}$
, and -
3. if
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\alpha }^{\mathcal {T}}} \leq \kappa $
for all
$\alpha <\kappa $
, and
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\kappa }^{\mathcal {T}}}>\kappa $
, then there is a club
$C_3 \subset b$
such that for all
$\alpha \in C_3$
,
$\alpha> \sup (D^{\mathcal {U}} \cap b)$
,
$i_{\alpha \kappa }^{\mathcal {T}}(\alpha )=\kappa $
and
$\operatorname {\mathrm {crit}}(i_{\alpha \kappa }^{\mathcal {T}})=\alpha $
.
We need the following version of the
$\Sigma _{1}^{1}$
-Bounding Theorem in order to prove Lemma 24.
Theorem 23. Let
$x \in {{^\omega }\omega }$
and suppose that
$A\subset \text {WO}$
is
$\Sigma _{1}^{1}(x)$
-definable, where
$\text {WO} \subset {{^\omega }\omega }$
is the set of reals coding well orderings of
$\omega $
. Then
$$\begin{align*}\sup\{\operatorname{\mathrm{otp}}(R_{y}):y\in A\}<\omega_{1}^{\operatorname{\mathrm{CK}}}(x), \end{align*}$$
where
$R_{y}\subset \omega \times \omega $
is the relation coded by y and
$\omega _{1}^{\operatorname {\mathrm {CK}}}(x)$
is the least ordinal
$\alpha $
such that
$L_\alpha [x]\models \text {KP}$
.Footnote
8
The proof of Theorem 23 is basically a refinement of the Kunen–Martin Theorem, which may be found in [Reference Moschovakis6, p. 75].
Lemma 24. Let M be a premouse such that
$\delta $
is a regular, uncountable cardinal of M
and
$M \models "\delta ^{+} \text { exists}"$
. Suppose that M is
$(0,\delta +1)$
-iterable, and there is no initial segment of
M which models KP + “there is a Woodin
cardinal”. Let
$\mathcal {U}$
be a
$0$
-maximal iteration tree on M such that
-
•
$\operatorname {\mathrm {lh}}(\mathcal {U})=\delta +1$
, -
•
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {U}}):\alpha <\delta \}=\delta $
, -
•
$b\cap D^{U}=\emptyset $
, where
$b:=[0,\delta ]^{\mathcal {U}}$
, i.e.,
$\mathcal {U}$
is non-dropping on its main branch, and -
•
$i_{0\delta }^{\mathcal {U}}(\delta )=\delta $
.
Then
$i_{0\delta }^{\mathcal {U}}(\delta ^{+M})=\delta ^{+M}$
.
Proof. Suppose otherwise, and let
$\mathcal {U}$
be a counterexample. Let
$\xi <\delta ^{+M}$
be such that
$i_{0\delta }^{\mathcal {U}}(\xi )>\delta ^{+M}$
and
$\rho _{\omega }^{M\vert \xi }=\delta $
. Note that such
$\xi $
exists since
$i_{0\delta }^{\mathcal {U}}$
is continuous at
$\delta ^{+M}$
, as
$\delta ^{+M}$
and its images are not of measurable cofinality in any
$\mathcal {M}_\alpha ^{\mathcal {U}}$
for
$\alpha \in [0,\delta )^{\mathcal {U}}$
and
$\mathcal {U}$
is
$0$
-maximal. Moreover, it is a standard fact that the ordinals
$\eta $
such that
$\rho _{\omega }^{M\vert \eta }=\delta $
are cofinal in
$\delta ^{+M}$
. We also assume that
$\xi>\operatorname {\mathrm {OR}}^{Q}$
, where
$Q\lhd M$
is least such that
$J(Q)\models "\delta \text { is not Woodin}"$
. Note that Q exists and
$\operatorname {\mathrm {OR}}^{Q}<\delta ^{+M}$
, since there is no initial segment of M
which models KP and has
$\delta $
as a Woodin.
Let
$g\subset \operatorname {\mathrm {Col}}(\omega ,\delta )$
be V-generic. In particular,
g is also M-generic. Let
$x\in M[g]\cap {^{\omega }\omega }$
code
$M\vert \xi +\omega $
.
Let us define in
$V[g]$
the set
$\bar {A}$
such that
$\mathcal {T}\in \bar {A}$
if and only if
-
1.
$\mathcal {T}$
is a
$0$
-maximal putative iteration tree on
$M\vert \xi +\omega $
such that
$\operatorname {\mathrm {lh}}(\mathcal {T})=\delta +1$
-
2.
$\mathcal {T}$
is non-dropping on its main branch, -
3.
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {T}}):\alpha <\delta \}=\delta $
, and -
4.
$i_{0\delta }^{\mathcal {T}}(\delta )=\delta $
, thus in particular,
$\mathcal {M}_{\delta }^{\mathcal {T}}\vert \delta \in \operatorname {\mathrm {wfp}}(\mathcal {M}_{\delta }^{\mathcal {T}})$
.
Note that if
$\mathcal {T} \in \bar {A}$
, then
$\mathcal {T} \!\upharpoonright \! \delta $
is guided by Q-structures.
Claim 1. If
$\mathcal {T}\in \bar {A}$
, then there is a Q-structure for
$\mathcal {T}\!\upharpoonright \!\delta $
in
$V[g]$
, i.e., there is
$\alpha <\operatorname {\mathrm {OR}}$
such that
$J_{\alpha }(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta ))\models "\delta =\delta (\mathcal {T}\!\upharpoonright \!\delta )\text { is not Woodin}"$
.
Proof. Suppose that there is no Q-structure in
$V[g]$
. This means that for all
${\alpha <\operatorname {\mathrm {OR}}}$
,
$J_{\alpha }(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta ))\models "\delta \text { is Woodin}"$
. Let
$\alpha $
be least such that
$J_{\alpha }(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta ))\models \text {KP}$
. Since
$\delta $
is regular in
$L(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta ))$
, there is
$X\prec J_{\alpha }(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta ))$
such that
${X\cap \delta \in \delta} $
. Let N be the transitive collapse
of X; then by condensation,
${N \lhd \mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta )}$
. Note that since
$\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta )=\bigcup _{\alpha \in [0,\delta )^{\mathcal {T}}}\mathcal {M}_{\alpha }^{\mathcal {T}}\vert \operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {T}})$
and by elementarity there is no initial segment of
$\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta )$
which models KP and has a Woodin. However,
N models KP and has a Woodin.
Contradiction!
Since for
$\mathcal {T}\in \bar {A}$
there is a Q-structure for
$\mathcal {T}\!\upharpoonright \!\delta $
and
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {T}}):\alpha <\delta \}=\delta $
(and thus Q-structures for
$\mathcal {T}\!\upharpoonright \!\alpha $
with
$\alpha <\delta $
are of ordinal height less than
$\delta $
) it follows by the usual absoluteness arguments that there
is a unique cofinal well-founded branch b for
$\mathcal {T}\!\upharpoonright \!\delta $
in
$V[g]$
.
However, at this point we do not know whether for
$\mathcal {T}\in \bar {A}$
, the unique cofinal wellfounded branch of
$\mathcal {T}\!\upharpoonright \!\delta $
was actually chosen, i.e. whether
$\mathcal {T}$
is an iteration tree. This is verified in the next
claim.
Claim 2. If
$\mathcal {T}\in \bar {A}$
, then
$\mathcal {M}_{\delta }^{\mathcal {T}}$
is wellfounded.
Proof. Suppose otherwise. Let
$\mathcal {T}\in \bar {A}$
be such that
$\mathcal {M}_{\delta }^{\mathcal {T}}$
is ill-founded. By 4 we have that
$\mathcal {M}_{\delta }^{\mathcal {T}}\vert \delta \in \operatorname {\mathrm {wfp}}(\mathcal {M}_{\delta }^{\mathcal {T}})$
. As we are assuming that
$\mathcal {M}_{\delta }^{\mathcal {T}}$
is ill-founded, we may apply Lemma 20 and get that
$\text {wfc}(\mathcal {M}_{\delta }^{\mathcal {T}}) \models \text {KP}$
. To arrive at a contradiction, we distinguish two
cases.
The first case is that
$Q(c,\mathcal {T}\!\upharpoonright \!\delta )\lhd \text {wfc}(\mathcal {M}_{\delta }^{\mathcal {T}})$
, where
$c=[0,\delta ]^{\mathcal {T}}$
. Let b be the unique, cofinal,
and wellfounded branch b for
$\mathcal {T}\!\upharpoonright \!\delta $
in
$V[g]$
. By
$1$
-smallness, we have that
$Q(c,\mathcal {T}\!\upharpoonright \!\delta )=Q(b,\mathcal {T}\!\upharpoonright \!\delta )$
. Note that
$Q(b,\mathcal {T}\!\upharpoonright \!\delta ) \neq \mathcal {M}^{\mathcal {T}}_b$
, since otherwise there is a drop on
b and thus
$Q(c,\mathcal {T}\!\upharpoonright \!\delta )$
is not sound, but
$Q(c,\mathcal {T}\!\upharpoonright \!\delta ) \lhd \text {wfc}(\mathcal {M}_{\delta }^{\mathcal {T}})$
, which would be a contradiction! But then by the
Zipper Lemma, we have that
$\delta $
is Woodin in
$J(Q(b,\mathcal {T}\!\upharpoonright \!\delta ))$
. Contradiction!
The second case is that the Q-structure is not an element in
$\text {wfc}(\mathcal {M}_{\delta }^{\mathcal {T}})$
. In this case we have that for every proper
initial segment of
$\text {wfc}(\mathcal {M}_{\delta }^{\mathcal {T}})$
,
$\mathcal {M}_{\delta }^{\mathcal {T}}$
thinks that
$\delta $
is Woodin in that segment. Moreover, by
$1$
-smallness we have that
$\text {wfc}(\mathcal {M}_{\delta }^{\mathcal {T}}) = J_{\beta }(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta ))$
for some
$\beta $
. But this means that
$\delta $
is Woodin in
$\text {wfc}(\mathcal {M}_{\delta }^{\mathcal {T}})$
.
Let b be unique cofinal and wellfounded branch
b for
$\mathcal {T}\!\upharpoonright \!\delta $
in
$V[g]$
. By
$1$
-smallness,
$Q(b,\mathcal {T}\!\upharpoonright \!\delta )= J_{\alpha }(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta ))$
for some
$\alpha \in \operatorname {\mathrm {OR}}$
. Since
$J_{\beta }(\mathcal {M}(\mathcal {T}\!\upharpoonright \!\delta )) \models "\delta \text { is Woodin}"$
, we must have that
$\beta \leq \alpha $
and
$J_\alpha (\mathcal {M}(\mathcal {T}\!\upharpoonright \! \delta )) \unlhd \mathcal {M}^{\mathcal {T}}_b$
. But since
$J_\beta (\mathcal {M}(\mathcal {T} \!\upharpoonright \! \delta ))$
satisfies KP + “
$\delta (\mathcal {T})$
is Woodin” and M has no
segment modelling this theory, we must have
$\beta = \alpha = \operatorname {\mathrm {OR}}(\mathcal {M}^{\mathcal {T}}_b)$
; that is,
$J_\beta (\mathcal {M}(\mathcal {T} \!\upharpoonright \! \delta )) = \mathcal {Q}(\mathcal {T} \!\upharpoonright \! \delta ,b) = \mathcal {M}^{\mathcal {T}}_b$
. Since
$M \vert (\xi +\omega )$
satisfies “there is no Woodin cardinal
$\leq \delta $
”, b must drop. But since
$\mathcal {M}^{\mathcal {T}}_b \models \text {KP} + "\text {there is a Woodin}"$
, by Lemma 15, the last drop in model along b is
to a segment modelling this same theory, contradicting the fact that
M has no segment modelling this theory.
The set
$\bar {A}$
gives rise to the set
$A=\{\operatorname {\mathrm {OR}}^{\mathcal {M}_{\infty }^{\mathcal {T}}}:\mathcal {T}\in \bar {A}\}$
. Note that A is a set of countable
ordinals in
$V[g]$
, since the trees are countable in
$V[g]$
and also
$M\vert \xi +\omega $
is countable in
$V[g]$
. Thus, we may code A as a set of reals
$A^{*}$
by setting
$A^{*}:=\{y \in \text {WO}:\operatorname {\mathrm {otp}}(R_{y})\in A\}$
.
Claim 3.
$A^{*}$
is
$\Sigma _{1}^{1}(x)$
.
Proof. The main difficulty is in expressing in a
$\Sigma _{1}^{1}(x)$
-fashion that for
$\alpha <\delta $
,
$\mathcal {M}_{\alpha }^{\mathcal {T}}$
is wellfounded. Fix some
$\alpha <\delta $
. Note that if
$\alpha $
is a successor, then
$\mathcal {M}_{\alpha }^{\mathcal {T}}$
is always wellfounded, since M is
$(0,\delta +1)$
-iterable in
$V[g]$
as can be seen by standard arguments, and we may
assume inductively that
$\mathcal {U}\!\upharpoonright \!\alpha $
is guided by Q-structures. Suppose that
$\alpha <\delta $
is a limit. Since
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {T}}):\alpha <\delta \}=\delta $
, we have that
$\delta (\mathcal {T}\!\upharpoonright \!\alpha )<\delta $
. Note that
$\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {T}})>\delta (\mathcal {T}\!\upharpoonright \!\alpha )$
and
$\mathcal {M}_{\alpha }^{\mathcal {T}}$
is
$1$
-small below
$\delta $
. Thus,
$Q(b,\mathcal {T}\!\upharpoonright \!\alpha )$
, where
$b=[0,\alpha ]^{\mathcal {T}}$
, must exist and
$Q(b,\mathcal {T}\!\upharpoonright \!\alpha )\lhd \mathcal {M}_{\alpha }^{\mathcal {T}}\vert \operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {T}})$
. Thus, the wellfoundedness of
$Q(b,\mathcal {T}\!\upharpoonright \!\alpha )$
can be expressed via the existence of an
isomorphism between the ordinals of
$Q(b,\mathcal {T}\!\upharpoonright \!\alpha )$
and an initial segment of the ordinals of the
model coded by x. This is
$\Sigma _{1}^{1}(x)$
.
Claim 4.
$\sup (A) < \omega _{1}^{\operatorname {\mathrm {CK}}}(x)$
.
Proof. This follows immediately from Theorem 23 and the previous claim.
Let
$\mathcal {U}^*$
be the
$0$
-maximal iteration tree on
$M\vert (\xi +\omega )$
which is otherwise equivalent to
$\mathcal {U}$
, i.e. it has the same length and tree order, and uses the
same extenders as
$\mathcal {U}$
. Since
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {U}}):\alpha <\delta \}=\delta $
and
$\delta $
is a cardinal of M, this makes sense,
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {U}^*}):\alpha <\delta \}=\delta $
,
$i^{\mathcal {U}^*}_{0\delta }$
exists, and
$i^{\mathcal {U}^*}_{0\delta }(\delta )=\delta $
. It follows that
$\mathcal {U}^*$
is in
$\bar {A}$
.
The following claim is a simpler version of Lemma 4.64 of [Reference Sargsyan, Schindler and Schlutzenberg7].
Claim 5.
$i_{0\delta }^{\mathcal {U}}(\xi )=i_{0\delta }^{\mathcal {U}^{*}}(\xi )$
.
Proof. Let
$\kappa < \delta $
. It suffices to see that every function
$f:\kappa \rightarrow \xi $
which is in M is also in
$M\vert \xi +\omega $
. To this end note that since
$\rho _{\omega }^{M\vert \xi }=\delta $
, there is a surjection
$g:\delta \rightarrow \xi $
such that
$g\in M\vert \xi +\omega =J(M\vert \xi )$
. Define a function
$h:\kappa \rightarrow \delta $
such that for
$\alpha <\kappa $
,
$h(\alpha )=\min (g^{-1}(f(\alpha )))$
. As
$\delta $
is regular and
$\kappa <\delta $
, h is essentially a bounded
subset of
$\delta $
and thus,
$h\in M\vert \xi $
. But since f is definable from
g and h this means that
$f\in M\vert \xi +\omega $
.
By Claims 4, 5, and the fact that
$\mathcal {U}^{*}\in \bar {A}$
,
$i^{\mathcal {U}^*}_{0\delta }(\xi )+\omega = i^{\mathcal {U}}_{0\delta }(\xi )+\omega = \operatorname {\mathrm {OR}}(\mathcal {M}^{\mathcal {U}^*}_\delta )<\omega _1^{\text {CK}(x)}$
. But x is in
$M[g]$
, so
$\omega _1^{\text {CK}(x)} < \omega _1^{M[g]} = \delta ^{+M[g]}$
. But we chose
$\xi $
with
$\delta ^{+M[g]}<i^{\mathcal {U}}_{0\delta }(\xi )$
, a contradiction!
We need a second lemma with a similar flavor.
Lemma 25. Let M be a premouse such that
$\delta $
is a regular, uncountable cardinal of M
and
$M \models "\delta ^{+} \text { exists}"$
. Suppose that M is
$(0,\delta +1)$
-iterable, and there is no initial segment of
M which models KP + “there is a Woodin
cardinal”. Let
$\mathcal {U}$
be a
$0$
-maximal iteration tree on M such that
$\operatorname {\mathrm {lh}}(\mathcal {U})=\delta +1$
and
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {U}}):\alpha <\delta \}=\delta $
. Let
$b^{\mathcal {U}}:=[0,\delta ]^{\mathcal {U}}$
. Suppose that there is
$\eta \in b^{\mathcal {U}} \cap (\delta \setminus \sup (D^{\mathcal {U}} \cap b^{\mathcal {U}}))$
such that
$i_{\eta \delta }^{\mathcal {U}}(\gamma )=\delta $
for some
$\gamma <\delta $
. Then
$\delta ^{+\mathcal {M}_{\delta }^{\mathcal {U}}}<\delta ^{+M}$
.
Proof. The proof follows closely the proof of Lemma 24. Again we work in
$V[g]$
, where
$g\subset \operatorname {\mathrm {Col}}(\omega ,\delta )$
is V-generic. We modify the definition of
$\bar {A}$
by omitting 2 and modifying 4 to “there exists an
$\eta \in b^{\mathcal {T}} \cap (\delta \setminus \sup (D^{\mathcal {T}} \cap b^{\mathcal {T}}))$
and a
$\gamma <\delta $
such that
$i_{\eta \delta }^{\mathcal {T}}(\gamma )=\delta $
, where
$b^{\mathcal {T}} := [0,\delta ]^{\mathcal {T}}$
”. Moreover, we alter 1 by requiring that
$\mathcal {T}$
is a putative iteration tree on
$M\vert \delta $
.
As before, for all
$\mathcal {T}\in \bar {A}$
,
$\mathcal {T}$
is an iteration tree. Similar to before
$\mathcal {U}^{*}$
will be the tree
$\mathcal {U}$
considered on
$M\vert \delta $
. Again,
$\mathcal {U}^{*}\in \bar {A}$
and the tree- and dropping-structure of
$\mathcal {U}$
and
$\mathcal {U}^{*}$
are the same. We have that
$\mathcal {M}_{\alpha }^{\mathcal {U}^{*}}\unlhd \mathcal {M}_{\alpha }^{\mathcal {U}}$
for all
$\alpha \leq \delta $
.
By the same argument as in the proof of Lemma 24, we will have that
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\delta }^{\mathcal {U}^{*}}}<\omega _{1}^{\operatorname {\mathrm {CK}}}(x)<\delta ^{+M}$
where x codes
$M\vert \delta $
. We claim that this implies that
$\delta ^{+\mathcal {M}_{\delta }^{\mathcal {U}}} < \delta ^{+M}$
. If
$D^{\mathcal {U}}\cap b^{\mathcal {U}} \neq \emptyset $
, i.e.
$\mathcal {U}$
is dropping on its main branch,
${\mathcal {M}_{\delta }^{\mathcal {U}}=\mathcal {M}_{\delta }^{\mathcal {U}^{*}}}$
, so that
$\delta ^{+\mathcal {M}_{\delta }^{\mathcal {U}}}\leq \operatorname {\mathrm {OR}}^{\mathcal {M}_{\delta }^{\mathcal {U}}}<\delta ^{+M}$
. So suppose that
$D^{\mathcal {U}}\cap b^{\mathcal {U}}=\emptyset $
, equivalently that
$D^{\mathcal {U}^{*}}\cap b^{\mathcal {U}}=\emptyset $
. Then
$i^{\mathcal {U}}:M\rightarrow \mathcal {M}_{\delta }^{\mathcal {U}}$
and
$i^{\mathcal {U}^{*}}:M\vert \delta \rightarrow \mathcal {M}_{\delta }^{\mathcal {U}^{*}}$
exist. Let
$\eta <\delta $
be least such that there is
$\bar {\delta }<\delta $
such that
$i_{\eta \delta }^{\mathcal {U}^{*}}(\bar {\delta })=\delta $
. If
$\eta =0$
, then since
$\delta $
is regular in M and
$\sup \{\operatorname {\mathrm {lh}}(E_{\alpha }^{\mathcal {U}}):\alpha <\delta \}=\delta $
,
$\delta ^{+\mathcal {M}_{\delta }^{\mathcal {U}}}=i^{\mathcal {U}}(\bar {\delta }^{+M})\leq i^{\mathcal {U}}(\delta )=\sup i^{\mathcal {U}^{*}}[\delta ]<\omega _{1}^{\operatorname {\mathrm {CK}}}(x)<\delta ^{+M}$
. So suppose that
$\eta>0$
. Note that
$\bar {\delta }^{+\mathcal {M}_{\eta }^{\mathcal {U}^{*}}}$
must exist, since otherwise
$\bar {\delta }$
is the largest cardinal of
$\mathcal {M}_{\eta }^{\mathcal {U}^{*}}$
which would imply that
$\bar {\delta }\in \operatorname {\mathrm {ran}}(i_{0\eta }^{\mathcal {U}^{*}})$
and so
$\eta =0$
, contradicting our assumption on
$\eta $
. Moreover, since there is no dropping on the main branch,
$i_{0\eta }^{\mathcal {U}^{*}}$
is cofinal in
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\eta }^{\mathcal {U}^{*}}}$
. But then,
$\delta ^{+\mathcal {M}_{\delta }^{\mathcal {U}}}=i_{\eta \delta }^{\mathcal {U}}(\bar {\delta }^{+\mathcal {M}_{\eta }^{\mathcal {U}}})=i_{\eta \delta }^{\mathcal {U}^{*}}(\bar {\delta }^{+\mathcal {M}_{\eta }^{\mathcal {U}^{*}}})\leq i_{\eta \delta }^{\mathcal {U}^{*}}(i_{0\eta }^{\mathcal {U}^{*}}(\gamma ))=i^{\mathcal {U}^{*}}(\gamma )<\omega _{1}^{\operatorname {\mathrm {CK}}}(x)$
for some
$\gamma <\delta $
where it is not the case that
$\delta = \gamma ^{+\mathcal {M}}$
and
$i^{\mathcal {U}^*}_{0,\eta }(\gamma ) = \bar {\delta }$
, since
$\eta> 0$
.
We are now ready to prove the main theorem of this section.
Theorem 26. Let
$\Omega $
be a measurable cardinal. Let
$\mathcal {W}$
and
$\mathcal {R}$
be premice such that
$\operatorname {\mathrm {OR}}^{\mathcal {W}}=\operatorname {\mathrm {OR}}^{\mathcal {R}}=\Omega $
and suppose that both are
$(0,\Omega +1)$
-iterable. Suppose that neither
$\mathcal {W}$
nor
$\mathcal {R}$
have an initial segment which models
$\text {KP} \land \exists \delta ("\delta \text { is Woodin}")$
. Then the following are equivalent:
-
1.
$\mathcal {W}\leq ^{*}\mathcal {R}$
, -
2. there is a club
$C\subset \Omega $
such that for all
$\alpha \in C$
, if
$\alpha $
is regular, then
$(\alpha ^{+})^{\mathcal {W}}\leq (\alpha ^{+})^{\mathcal {R}}$
, and -
3. there is a stationary set
$S\subset \Omega $
such that for all
$\alpha \in S$
,
$\alpha $
is regular and
${(\alpha ^{+})^{\mathcal {W}}\leq (\alpha ^{+})^{\mathcal {R}}}$
.
Proof. Let us first show that clause 1 implies clause 2, i.e. suppose that
$\mathcal {W}\leq ^{*}\mathcal {R}$
. We aim to show that there is a club
$C\subset \Omega $
such that for all
$\alpha \in C$
, if
$\alpha $
is regular, then
$(\alpha ^{+})^{\mathcal {W}}\leq (\alpha ^{+})^{\mathcal {R}}$
. Let
$(\mathcal {T},\mathcal {U})$
be the coiteration of
$(\mathcal {W},\mathcal {R})$
. By Theorem 21, the coiteration is successful. Since
$\mathcal {W}\leq ^{*}\mathcal {R}$
,
$\mathcal {M}_{\Omega }^{\mathcal {T}}\unlhd \mathcal {M}_{\Omega }^{\mathcal {U}}$
and
$\mathcal {T}$
does not drop on its main branch. In particular,
$i^{\mathcal {T}}:\mathcal {W}\rightarrow \mathcal {M}_{\Omega }^{\mathcal {T}}$
exists. We may assume by padding the trees if necessary
that
$\operatorname {\mathrm {lh}}(\mathcal {T})=\operatorname {\mathrm {lh}}(\mathcal {U})=\Omega +1$
. Note that by Theorem 21,
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\Omega }^{\mathcal {T}}}=\Omega $
. Thus, by Lemma 22 (1), there is a club
$C_{\mathcal {T}}\subset \Omega $
such that for all
$\alpha \in C_{\mathcal {T}}$
,
$i_{0\alpha }^{\mathcal {T}}[\alpha ]\subset \alpha $
and
$b^{\mathcal {T}} \supset C_{\mathcal {T}}$
. Let
$\tilde {C}_{\mathcal {T}}$
be the club given by Lemma 22 (2) for
$\mathcal {T}$
. Set
$B_{\mathcal {T}}:=C_{\mathcal {T}}\cap \tilde {C}_{\mathcal {T}}$
. Let
$\alpha \in B_{\mathcal {T}}$
and suppose that it is regular. Then
$i_{0\alpha }^{\mathcal {T}}(\alpha )=\alpha $
. Thus, by Lemma 24, we have that
$\alpha ^{+\mathcal {W}}=\alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {T}}}$
. But since
$\alpha \in \tilde {C}_{\mathcal {T}}$
,
$\operatorname {\mathrm {crit}}(i_{\alpha \Omega }^{\mathcal {T}})\geq \alpha $
. Therefore,
$\alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {T}}}=\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}$
. Since
$\mathcal {M}_{\Omega }^{\mathcal {T}}\unlhd \mathcal {M}_{\Omega }^{\mathcal {U}}$
and
$\Omega $
is a cardinal, we have that
$\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}=\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {U}}}$
for all
$\alpha \in B_{\mathcal {T}}$
.
Case 1.
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\Omega }^{\mathcal {U}}}>\Omega $
.
Let
$C_{\mathcal {U}}\subset \Omega $
be the club given by Lemma 22 (3) for
$\mathcal {U}$
intersected with the club given by Lemma 22 (2) for
$\mathcal {U}$
. Then for all
$\alpha ,\beta \in C_{\mathcal {U}}$
such that
$\alpha <\beta $
, we have that
$i_{\alpha \beta }^{\mathcal {U}}(\alpha )=\beta $
. Suppose that
$\alpha \in C_{\mathcal {U}}^{\prime }$
is regular, where
$C_{\mathcal {U}}^{\prime }$
is the club of limit points of
$C_{\mathcal {U}}$
. We have by Lemma 25, that
$\alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {U}}}<\alpha ^{+\mathcal {R}}$
. Moreover, since
$\operatorname {\mathrm {crit}}(i_{\alpha \Omega }^{\mathcal {U}})= \alpha $
(note we are beyond the drops of
$b^{\mathcal {U}}$
),
$\alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {U}}}=\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {U}}}$
. Thus,
$C:=C_{\mathcal {U}}^{\prime }\cap B_{\mathcal {T}}$
witnesses clause 2 of the theorem in Case 1.
Case 2.
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\Omega }^{\mathcal {U}}}=\Omega $
.
Then
$i_{0\Omega }^{\mathcal {U}}$
exists. In this case we can construct a club
$C_{\mathcal {U}}$
for
$\mathcal {U}$
as
$C_{\mathcal {T}}$
was constructed for
$\mathcal {T}$
and set
$C:=C_{\mathcal {T}}\cap C_{\mathcal {U}}$
. This C witnesses clause 2 of the
theorem in Case 2.
The argument for showing that clause 2 implies clause 3 is standard. We now
show that clause 3 implies clause 1. Suppose that there is a stationary set
of regular cardinals
$S\subset \Omega $
such that for all
$\alpha \in S$
,
$\alpha ^{+\mathcal {W}}\leq \alpha ^{+\mathcal {R}}$
. We aim to show that
$\mathcal {W}\leq ^{*}\mathcal {R}$
. Let
$(\mathcal {T},\mathcal {U})$
be the coiteration of
$(\mathcal {W},\mathcal {R})$
which again by Theorem 21 is successful. Suppose for the sake of
contradiction that
$\mathcal {M}_{\infty }^{\mathcal {T}}\rhd \mathcal {M}_{\infty }^{\mathcal {U}}$
. By the same arguments as before we can construct a club
$D\subset \Omega $
such that for all regular
$\alpha \in D$
,
$\alpha ^{+\mathcal {W}}>\alpha ^{+\mathcal {R}}$
. But since S is stationary and consists
of regular cardinals,
$S\cap D\neq \emptyset $
, contradiction!
Corollary 27. Let
$\Omega $
be a measurable cardinal. Let
$\mathcal {W}$
and
$\mathcal {R}$
be premice such that
$\operatorname {\mathrm {OR}}^{\mathcal {W}}=\operatorname {\mathrm {OR}}^{\mathcal {R}}=\Omega $
and suppose that both are
$(0,\Omega +1)$
-iterable. Suppose that there is no transitive model of
$\text {KP}\land \exists \delta ("\delta \text { is Woodin}")$
. Then the following are equivalent:
-
•
$\mathcal {W}\leq ^{*}\mathcal {R}$
, -
• there is a club
$C\subset \Omega $
such that for all
$\alpha \in C$
, if
$\alpha $
is regular, then
$(\alpha ^{+})^{\mathcal {W}}\leq (\alpha ^{+})^{\mathcal {R}}$
, and -
• there is a stationary set
$S\subset \Omega $
such that for all
$\alpha \in S$
,
$\alpha \in S$
is regular and
$(\alpha ^{+})^{\mathcal {W}}\leq (\alpha ^{+})^{\mathcal {R}}$
.
4 The counterexample
In this section we construct a counterexample to Conjecture 1 assuming large cardinals. In
order to construct the counterexample we need an iterable premouse
M which has an initial segment
$N\lhd M$
such that
$N\models $
KP, N has a largest cardinal
$\kappa $
which is Woodin in N,
$\kappa $
is a measurable cardinal in the larger premouse
M, and there exists some
$\Omega \in (\kappa ^{+M},\operatorname {\mathrm {OR}}^{M})$
which is measurable in M.
We want to show that such N exists if we assume that
M is the least iterable sound premouse, which models
$\text {KP}$
and has a largest cardinal
$\delta $
which is 
-Woodin in
M.Footnote
9
In Section 4.1 we will construct the above
N. Note that since M is the least mouse with a
-Woodin, we will have that
$\kappa $
, the largest cardinal of N, is not
-Woodin in N. For the construction of the counterexample it will
be important that if g codes a wellorder and is generic over
N for the extender algebra, then the ordertype of
g is less than
$\operatorname {\mathrm {OR}}^N$
. This will work since we showed in Section 2.1 that
$N[g]\models \text {KP}$
if
$g\subset \mathbb {P}\in N$
is N-generic,
$N\models \text {KP}$
, and
$N\models "\mathbb {P}\text { has the }\eta \text {-c.c.}"$
, where
$\eta $
is the largest cardinal of N. In Section 4.2 we will construct the counterexample.
4.1 The construction of N
Definition 28. Let M be a passive premouse and let
$\delta < \operatorname {\mathrm {OR}} \cap M$
. We say that
$\delta $
is a 
-Woodin cardinal (of M) if for all
$A \in \Sigma _1^M(M)$
there is some
$\kappa < \delta $
which is
$<\delta ,A$
-reflecting in M.
Definition 29. Let 
be the least
$(0,\omega _{1}+1)$
-iterable and sound premouse which models
.
Let 
. Note that M has a unique Woodin
cardinal
$\delta $
. Moreover,
$\delta $
is acutally 
-Woodin in
M and is the largest cardinal of M.
Furthermore,
$\delta = \rho ^M_1$
,
$p_1^M = \{\delta \}$
, and
$\rho _2^M = \omega $
. Also, by the remarks in the proof of Lemma 15, M is actually
$(1,\omega _{1}+1)$
-iterable.
Definition 30. Let M be an admissible passive premouse such that
$\rho _{1}^{M}<\operatorname {\mathrm {OR}}^{M}$
. We let
$T^{M}\subset \rho _{1}^{M}$
be the set of ordinals coding
$\operatorname {\mathrm {Th}}_{\Sigma _{1}}^{M}(\rho _{1}^{M}\cup \{p_{1}^{M}\})$
.
Note that by Lemma 4,
$\rho _{1}^{M}$
is the largest cardinal of M. Moreover, by
definition
$T^{M}\notin M$
, but
$T^{M}\in \Sigma _{1}^{M}(M)$
. However, for every
$\alpha <\rho _{1}^{M}$
,
$T^{M}\cap \alpha \in M$
. In the proof of Lemma 31 below, we will consider degree
$1$
ultrapower embeddings
$i_E: M \to U = \text {Ult}_{1}(M,E)$
, where U is wellfounded. Note that in this
case,
${T^{U} = i_E(T^M)}$
, where
$i_E(T^M)$
denotes
$\bigcup _{\alpha <\rho _1^M} i_E(T^M\cap \alpha )$
.
Lemma 31. Let
$\delta $
be the largest cardinal of 
.
In 
, let
$\kappa $
be 
-reflecting and let
$H:=\text {Hull}_{1}^{M}(\kappa \cup \{\delta \})$
. Then
-
•
$H\cap \delta =\kappa $
, and -
• if
is the transitive collapse of
H, then
$N\models \text {KP}$
.
is the transitive collapse of
H, then 
Proof. Let 
. Recall that
$p_1^M = \{\delta \}$
and
$\rho _1^M = \delta $
. Let us first proof that
$H\cap \delta =\kappa $
. Suppose for the sake of contradiction that there is
$\xi \in H\cap [\kappa ,\delta )$
. Let
$\varphi _{\xi }$
be a
$\Sigma _{1}$
formula and
$\alpha \in \kappa $
be such that
$\xi $
is the unique
$\eta $
such that
$M\models \varphi _{\xi }(\eta ,\alpha ,p_1^M)$
. Note that
$\varphi _{\xi }(\xi ,\alpha ,p_1^M)\in \operatorname {\mathrm {Th}}_{1}^{M}(\delta \cup \{ {p_1^M} \})$
. Thus, there is some
$\lambda <\delta $
such that the code for
$\varphi _{\xi }(\xi ,\alpha ,p_1^M)$
is below
$\lambda $
, i.e.
$\varphi _{\xi }(\xi ,\alpha ,\dot {p})\in T^{M}\cap \lambda $
. Let
$E\in \mathbb {E}^{M}$
be an extender witnessing that
$\kappa $
is
$(\lambda ,T^{M})$
-strong. Let
$U := \text {Ult}_{1}(M,E)$
and let
$i_{E}:M\rightarrow U$
be the canonical ultrapower embedding. Note that we
are taking a
$1$
-ultrapower, which makes sense since
$\kappa = \operatorname {\mathrm {crit}}(E) < \delta = \rho _1^M$
. This means that
$i_E(T^M) = T^U$
(see the remarks following Definition 30). But since
E is
$(\lambda ,T^{M})$
-strong, we have
$T^{U}\cap \lambda =i_{E}(T^{M})\cap \lambda =T^{M}\cap \lambda $
. Thus,
$\varphi _{\xi }(\xi ,\alpha ,\dot {p})\in T^{U}\cap \lambda $
, which means that
$U\models \varphi _{\xi }(\xi ,\alpha ,p_1^{U})$
. Moreover, since
$M \models \varphi _\xi (\xi , \alpha ,p_1^M)$
,
$U \models \varphi _\xi (i_E(\xi ), i_E(\alpha ), i_E(p_1^M))$
. But
$i_E(\alpha ) = \alpha < \kappa $
and
$i_E(p_1^M) = p_1^U$
,Footnote
10
so
${U \models \varphi _\xi (i_E(\xi ), \alpha , p_1^U)}$
. But since
$\xi $
was the unique witness for
$\exists x\varphi _{\xi }(x,\alpha ,p_1^M)$
in M, we have
$$\begin{align*}M\models\forall x(x\neq\xi\rightarrow\neg\varphi_{\xi}(x,\alpha,p_1^M)). \end{align*}$$
However, this is
$r\Pi _{1}$
and thus,
$$\begin{align*}U\models\forall x(x\neq i_{E}(\xi)\rightarrow\neg\varphi_{\xi}(x,\alpha,p_1^U)). \end{align*}$$
This means
$i_{E}(\xi )=\xi $
. Contradiction!
Let us now verify that
$N\models \text {KP}$
. By what we have shown so far
$\pi (\kappa )=\delta $
. In particular,
$\kappa $
is the largest cardinal of N. So by
Lemma 3 it suffices to
check
$\Delta _{0}$
-collection. Suppose that
$$\begin{align*}N\models\forall\alpha<\kappa\exists y\varphi(\alpha,y,\bar{p}), \end{align*}$$
where
$\varphi $
is a
$\Sigma _{1}$
formula and
$\bar {p}\in N$
. Note that since
$H:=\text {Hull}_{1}^{M}(\kappa \cup \{\delta \})$
, we may replace
$\bar {p}$
with some tuple of ordinals
$\vec {\alpha }_p \in [\kappa ]^{<\omega }$
and
$\{\kappa \}$
. Letting
$p=\pi (\bar {p})$
, we have that
$p = \vec {\alpha }_p \cup \{ \{ \delta \}\}$
. Then
$$\begin{align*}M\models\forall\alpha<\kappa\exists y\varphi(\alpha,y,\vec{\alpha}_p,\{ \delta \}). \end{align*}$$
We aim to see that the bound
$\kappa $
can be replaced by
$\delta $
here, giving that
$$\begin{align*}M\models\forall\alpha<\delta\exists y\varphi(\alpha,y,\vec{\alpha}_p,\{ \delta \}). \end{align*}$$
Suppose for the sake of contradiction that there is some
$\xi \in [\kappa ,\delta )$
such that
${M\models \forall y\neg \varphi (\xi ,y,\vec {\alpha }_p,\{ \delta \})}$
. This means that
$\exists y\varphi (\xi ,y,\vec {\alpha }_p,\dot {p})\notin T^{M}$
. But then there is some
$\lambda \in (\xi ,\delta )$
such that this is witnessed by
$T^{M}\cap \lambda $
, i.e. the formula
$\exists y\varphi (\xi ,y,\vec {\alpha }_p,\dot {p})\notin T^{M}\cap \lambda $
even though
$\xi ,\vec {\alpha }_p \in M\vert \lambda $
. (Note that since we coded
$T^{M}$
as a theory with constant symbol
$\dot {p}$
it suffices if we pick
$\lambda>\sup \{\xi ,\vec {\alpha }_p\}$
.) Let
$E\in \mathbb {E}^{M}$
be
$(\lambda ,T^{M})$
-strong. Then as in the previous argument,
$\exists y\varphi (\xi ,y,\vec {\alpha }_p,\dot {p}) \notin T^{U}\cap \lambda $
. However, since
$i_{E}$
is
$r\Sigma _{2}$
-elementary, we have that
$$\begin{align*}U\models\forall\alpha<i_{E}(\kappa)\exists y\varphi(\alpha,y,i_E(\vec{\alpha}_p,\{ \delta \})). \end{align*}$$
Note that
$i_E (\vec {\alpha }_p,\{ \delta \}) = (\vec {\alpha }_p,\{\delta \})$
, since
$\vec {\alpha }_p \in [\kappa ]^{<\omega }$
and
$i_E(\{ \delta \}) = p_1^{U} = p_1^M = \{ \delta \}$
. Thus, since
$i_{E}(\kappa )>\xi $
,
$$\begin{align*}U\models\exists y\varphi(\xi,y,\vec{\alpha}_p,\{ \delta \}). \end{align*}$$
Contradiction!
Lemma 32. Let N be as in Lemma 31. Then
$N\lhd M$
and N is an admissible passive
premouse with largest cardinal
$\kappa $
which is Woodin in N.
Proof. Note that a failure of
$\kappa $
being Woodin in N is an
$r\Sigma _{1}$
fact about
$N\vert \kappa $
. Since
$H\prec _{\Sigma _{1}}M$
, this would imply by upwards-absoluteness that
$\delta $
fails to be Woodin in M. Thus,
$\kappa $
is Woodin in N. Moreover, as already
mentioned in the previous proof
$\kappa $
is the largest cardinal of N. But
then as
$N\vert \kappa =M\vert \kappa $
and so
$N=J_{\alpha }(M\vert \kappa )$
, we will have that
$N\lhd M$
, and we are done.
4.2 The construction of the counterexample
Let 
and let
$N\lhd M$
be as given by Lemma 32, i.e. the following hold:
-
• M has a largest cardinal
$\delta $
which is -Woodin in M, -
• N is an admissible mouse with a largest cardinal
$\kappa $
which is Woodin in N, but not
-Woodin in N, -
•
$\operatorname {\mathrm {OR}}^{N}<\kappa ^{+M}$
, and -
•
$\kappa $
is measurable in M, as witnessed
by an extender E from the extender sequence of
M.
-Woodin in M,
-Woodin in N,
The construction of the counterexample is now as follows: Fix
$\Omega \in (\kappa ^{++M},\delta )$
, which is a regular cardinal in M. Working
inside M we will linearly iterate
$M\vert \operatorname {\mathrm {lh}}(E)$
for
$\Omega $
many times via E and its images. Let
$\mathcal {T}$
be the corresponding iteration tree on
$M\vert \operatorname {\mathrm {lh}}(E)$
of length
$\operatorname {\mathrm {lh}}(\mathcal {T})=\Omega +1$
. Note that
$\mathcal {T}\in M$
. Let
$i^{\mathcal {T}}:M\vert \operatorname {\mathrm {lh}}(E)\rightarrow \mathcal {M}_{\Omega }^{\mathcal {T}}$
be the iteration map. We will have the following:
-
•
$i^{\mathcal {T}}(\kappa )=\Omega $
, -
•
$\sup (i^{\mathcal {T}}[\kappa ^{+M}])=i^{\mathcal {T}}(\kappa ^{+M})=\Omega ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}<\Omega ^{+M}$
, -
•
$W:=i^{\mathcal {T}}(N)\models \text {KP}$
and
$\operatorname {\mathrm {OR}}^{W}<\Omega ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}$
, and -
• W is an admissible mouse with a largest cardinal
$\Omega $
which is Woodin in W, but not
-Woodin in W.
-Woodin in W. Since
$\Omega ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}<\Omega ^{+M}$
, there is, inside M, some
$A\subset \Omega $
which codes a well order
$<_A$
such that
$\operatorname {\mathrm {otp}}(<_A)=\Omega ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}$
. Let
$\mathcal {U}$
be the A-genericity iteration of
W with respect to the
$\Omega $
-generator extender algebra of W at
$\Omega $
. Note that
$\mathcal {U}\in M$
, since by standard arguments there is for every limit
$\alpha \leq \Omega $
a Q-structure for
$\mathcal {U}\!\upharpoonright \!\alpha $
in M. Moreover, since
$\Omega $
is inaccessible in W, and a cardinal in
M, by the usual arguments
$\mathcal {U}$
is non-dropping,
$\operatorname {\mathrm {lh}}(\mathcal {U})=\Omega +1$
, and
$i_{0\Omega }^{\mathcal {U}}(\Omega )=\Omega $
.
Inside M we construct a club
$C\subset \Omega $
such that for all
$\alpha \in C$
:
-
• there exists some
$\pi _{\alpha }:M_{\alpha }\cong X_{\alpha }\prec _{1000}M\vert \vert \Omega ^{++}$
such that
$M\vert (\operatorname {\mathrm {lh}}(E)+\omega ) \cup \{\Omega ,A\}\subset \operatorname {\mathrm {ran}}(\pi _{\alpha })$
,
$\operatorname {\mathrm {crit}}(\pi _{\alpha })=\alpha $
, and
$\pi _{\alpha }(\alpha )=\Omega $
, -
•
$\alpha =\sup \{\operatorname {\mathrm {lh}}(E_{\beta }^{\mathcal {U}}):\beta <\alpha \}$
, -
•
$\alpha \in b:=[0,\Omega ]^{\mathcal {U}}$
, and -
•
$\alpha =\operatorname {\mathrm {crit}}(E_{\alpha }^{\mathcal {T}})$
.
The construction of such C is fairly standard, so we will omit
it. Note that for
$\alpha \in C$
,
$\operatorname {\mathrm {crit}}(i_{\alpha \Omega }^{\mathcal {U}})\geq \alpha $
, since
$\alpha =\sup \{\operatorname {\mathrm {lh}}(E_{\beta }^{\mathcal {U}}):\beta <\alpha \}$
. Thus, by the usual argument
$\alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {U}}}=\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {U}}}$
.
Claim. For
$\alpha \in C$
,
$\alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {U}}}>\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}$
.
Proof. Fix
$\alpha \in C$
. Let
$(\bar {\mathcal {T}},\bar {\mathcal {U}},\bar {A},\bar {W})\in M_{\alpha }$
such that
$\pi _{\alpha }((\bar {\mathcal {T}},\bar {\mathcal {U}},\bar {A},\bar {W}))=(\mathcal {T},\mathcal {U},A,W)$
. Note that there are such
$\bar {\mathcal {T}},\bar {\mathcal {U}},\bar {W}\in M_{\alpha }$
, since
$\mathcal {T},\mathcal {U},W$
are definable from
$M\vert \operatorname {\mathrm {lh}}(E)$
, A, and
$\Omega $
over
$M \vert \vert \Omega ^{++}$
.
By the properties of
$\pi _\alpha $
,
$\bar {A}$
is generic over
$\mathcal {M}_{\alpha }^{\bar {\mathcal {U}}}$
and
$\operatorname {\mathrm {otp}}(<_{\bar {A}}) = \alpha ^{+\mathcal {M}^{\bar {\mathcal {T}}}_\alpha } = \alpha ^{+\mathcal {M}^{\mathcal {T}}_\alpha }$
. By Corollary 11,
$\mathcal {M}_{\alpha }^{\bar {\mathcal {U}}}[\bar {A}]\models \text {KP}$
. Since
${<_{\bar {A}}} \in {\mathcal {M}_{\alpha }^{\bar {\mathcal {U}}}[\bar {A}]}$
and the transitive collapse of
$<_{\bar {A}}$
is a
$\Sigma _{1}$
-recursion,
$\operatorname {\mathrm {otp}}(<_{\bar {A}})=\alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {T}}}\in \mathcal {M}_{\alpha }^{\bar {\mathcal {U}}}[\bar {A}]$
. In particular,
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\alpha }^{\bar {\mathcal {U}}}[\bar {A}]} = \operatorname {\mathrm {OR}}^{\mathcal {M}_{\alpha }^{\bar {\mathcal {U}}}}> \alpha ^{+\mathcal {M}_{\alpha }^{\mathcal {T}}}$
. Note that since
$\operatorname {\mathrm {crit}}(E^{\mathcal {T}}_\alpha ) = \alpha $
it follows by the usual argument that
$\alpha ^{+\mathcal {M}^{\mathcal {T}}_\alpha } = \alpha ^{+\mathcal {M}^{\mathcal {T}}_\Omega }$
. So in order to finish the proof it suffices to see
that
$\operatorname {\mathrm {OR}}^{\mathcal {M}^{\bar {\mathcal {U}}}_\alpha } \leq \alpha ^{+\mathcal {M}^{\mathcal {U}}_\alpha }$
.
Since
$\operatorname {\mathrm {crit}}(\pi _\alpha ) = \alpha $
, we have that
$\mathcal {M}_\alpha ^{\bar {\mathcal {U}}} \vert \alpha = \mathcal {M}_\alpha ^{\mathcal {U}} \vert \alpha $
. Moreover, since
$i^{\mathcal {U}}_{0,\Omega } (\Omega ) = \Omega $
,
$i^{\bar {\mathcal {U}}}_{0\alpha }(\alpha ) = \alpha $
and so
$\mathcal {M}^{\bar {\mathcal {U}}}_\alpha \models "\alpha \text { is Woodin}"$
, since
$\pi _\alpha (\mathcal {M}^{\bar {\mathcal {U}}}_\alpha ) = \mathcal {M}^{\mathcal {U}}_\Omega $
and
$\mathcal {M}^{\mathcal {U}}_\Omega \models "\Omega \text { is Woodin}"$
. By the
$1$
-smallness of
$\mathcal {M}_\Omega ^{\mathcal {U}}$
it follows that
$\mathcal {M}^{\bar {\mathcal {U}}}_\alpha = J_\beta (\mathcal {M}^{\bar {\mathcal {U}}}_\alpha \vert \alpha )$
for some
$\beta < \Omega $
and so
$\mathcal {M}^{\bar {\mathcal {U}}}_\alpha \unlhd \mathcal {Q}(\mathcal {U} \!\upharpoonright \! \alpha )$
, so that
$\mathcal {M}^{\bar {\mathcal {U}}}_\alpha \lhd \mathcal {M}_\Omega ^{\mathcal {U}}$
. Since
$\mathcal {Q}(\mathcal {U} \!\upharpoonright \! \alpha ) \unlhd \mathcal {M}_\Omega ^{\mathcal {U}} \vert \alpha ^{+\mathcal {M}_\Omega ^{\mathcal {U}}}$
, we have
$\operatorname {\mathrm {OR}}^{\mathcal {M}^{\bar {\mathcal {U}}}_\alpha } \leq \alpha ^{+\mathcal {M}^{\mathcal {U}}_\alpha }$
.
We have shown that for every
$\alpha \in C$
,
$\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {T}}}<\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {U}}}$
. If we consider the tree
$\mathcal {U}^{*}$
which is just the tree
$\mathcal {U}$
considered not on W but on
$W\vert \Omega $
, we will have that for all
$\alpha \in C$
,
$\alpha ^{+W\vert \Omega }<\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {U}^{*}}}$
. Moreover, since
$i^{\mathcal {U}}(\Omega )=\Omega $
, we have
$\operatorname {\mathrm {OR}}^{\mathcal {M}_{\Omega }^{\mathcal {U}^{*}}}=\Omega $
, so
$\mathcal {M}_{\Omega }^{\mathcal {U}^{*}}$
is a weasel in the sense of M.
However, if
$(\mathcal {T}^{\prime },\mathcal {U}^{\prime })$
is the coiteration of
$(W\vert \Omega ,\mathcal {M}_{\Omega }^{\mathcal {U}^{*}})$
, then
$\mathcal {T}^{\prime }=\mathcal {U}^{*}$
and
$\mathcal {U}^{\prime }$
is the trivial tree. Thus,
$W\vert \Omega =^{*}\mathcal {M}_{\Omega }^{\mathcal {U}^{*}}$
and there exists a club
$C\subset \Omega $
such that for all
$\alpha \in C$
,
$\alpha ^{+W\vert \Omega }<\alpha ^{+\mathcal {M}_{\Omega }^{\mathcal {U}^{*}}}$
. This contradicts Conjecture 1 inside M.
Remark. The construction above also works if we picked
$\Omega $
to be
$\delta $
. However, in this case, we have to modify the
construction slightly: Let
$\eta>\kappa $
be a measurable cardinal of M and let
$F\in \mathbb {E}^{M}$
be the measure witnessing this. Let
$\mathcal {T}^{\prime }$
the linear iteration of
$M\vert \eta ^{++M}$
via F and its images of length
$\delta +1$
. Note that by
$\Delta _{0}$
-collection
$\mathcal {T}^{\prime }\in M$
and that
$\delta ^{+}$
exists in
$\mathcal {M}_{\delta }^{\mathcal {T}^{\prime }}$
. Now as before we let
$\mathcal {T}$
be a linear iteration of
$M\vert \operatorname {\mathrm {lh}}(E)$
via E and its images of length
$\delta +1$
. However, note that
$\mathcal {T}\in \mathcal {M}_{\delta }^{\mathcal {T}^{\prime }}$
. Thus, there is
$A\in \mathcal {M}_{\delta }^{\mathcal {T}^{\prime }}$
such that
$\operatorname {\mathrm {otp}}(<_A)=\delta ^{+\mathcal {M}_{\delta }^{\mathcal {T}}}$
and
$A\subset \delta $
. Since
$\delta ^{+\mathcal {M}_{\delta }^{\mathcal {T}^{\prime }}}$
exists and there is no initial segment of
$\mathcal {M}_{\delta }^{\mathcal {T}^{\prime }}$
which models KP and has a 
-Woodin cardinal, we see by the same argument as before, that
Q-structures exist for
$\mathcal {U}$
, where
$\mathcal {U}$
is the genericity iteration of
$i_{0\delta }^{\mathcal {T}}(N)$
making A generic.
5 On another question from CMIP
In this last section we discuss another question remarked about in [Reference Steel10, prior to Lemma 4.6] concerning the
S-hull property. Throughout this section
$\Omega $
is a fixed measurable cardinal and
$\mu _{0}$
is a fixed normal measure on
$\Omega $
. We call a premouse of ordinal height
$\Omega $
a weasel.
Note that our definitions of thickness and the Hull property are different from the ones in [Reference Steel10], yet equivalent. We chose these different definitions in order to emphasize that thickness is a property independent of a specific weasel.
Definition 33. Let
$\mathcal {W}$
be a weasel and
$S\subset \Omega $
be stationary. We say that S is good for
$\mathcal {W}$
iff there is a club
$C\subset \Omega $
such that for all
$\alpha \in C\cap S$
-
•
$\alpha $
is inaccessible, -
•
$\alpha ^{+}=(\alpha ^{+})^{\mathcal {W}}$
, and -
•
$\alpha $
is not the critical point of a total-on-
$\mathcal {W}$
extender from the extender sequence of
$\mathcal {W}$
.
Definition 34. Let
$\Gamma \subset \Omega $
and S be stationary. We say that
$\Gamma $
is S-thick iff there is a club
$C\subset \Omega $
such that for all
$\alpha \in C\cap S$
,
$\Gamma \cap \alpha ^{+}$
contains an
$\alpha $
-club and
$\alpha \in \Gamma $
.
Definition 35. Let
$\mathcal {W}$
be a weasel and
$S\subset \Omega $
such that S is good for
$\mathcal {W}$
. Let
$\alpha < \Omega $
. We say that
$\mathcal {W}$
has the S-hull property at
$\alpha $
iff for all
$\Gamma \subset \Omega $
which are S-thick
$$\begin{align*}\mathcal{P}(\alpha)\cap\mathcal {W}\subset\text{transitive collapse of }\text{Hull}_{\omega}^{\mathcal {W}}(\alpha\cup\Gamma). \end{align*}$$
In Lemma 4.6 of [Reference Steel10, p. 32] it is proven that
for an
$\Omega +1$
-iterable weasel
$\mathcal {W}$
, if S is good for
$\mathcal {W}$
, then for
$\mu _{0}$
-a.e.
$\alpha <\Omega $
the S-hull property holds at
$\alpha $
.
However, in [Reference Steel10, p. 32, prior to Lemma 4.6],
Steel raises the issue whether the set
$\operatorname {\mathrm {HP}}_S^{\mathcal {W}}:=\{\alpha <\Omega :\mathcal {W}\text { has the }S\text {-hull property at }\alpha \}$
is closed. Note that
$\operatorname {\mathrm {HP}}_S^{\mathcal {W}}$
clearly cannot be closed in the usual sense, as the following
example from [Reference Steel10, p. 29] shows: Suppose that
M is an
$\Omega +1$
-iterable weasel which has the S-hull property at
all
$\alpha <\Omega $
and there is a total-on-
$M E\in \mathbb {E}^{M}$
with at least two generators. Then
$\text {Ult}_{0}(M,E)$
has the S-hull property at all
$\alpha <(\operatorname {\mathrm {crit}}(E)^{+})^{M}$
but not at
$(\operatorname {\mathrm {crit}}(E)^{+})^{M}$
. However,
$\text {Ult}_{0}(M,E)$
is still
$\Omega +1$
-iterable. Thus,
$\operatorname {\mathrm {HP}}_S^{\mathcal {W}}$
cannot be closed in the usual sense.
Definition 36. We say that
$X\subset \Omega $
is almost closed if for every
$\delta < \Omega $
, if
$\delta $
is the supremum of elements of X and
elements of
$\Omega \setminus X$
, then
$\delta \in X$
.
Given the preceding remarks, we will substitute almost closure for closure in the
issue raised in [Reference Steel10], and hence consider the
following question: Let
$\mathcal {W}$
be a
$(0,\Omega +1)$
-iterable weasel and
$S \subset \Omega $
such that S is good for
$\mathcal {W}$
, is the set
$\operatorname {\mathrm {HP}}_S^{\mathcal {W}}$
almost closed?
This question is answered positively by the following theorem.
Theorem 37. Let
$\mathcal {W}$
be an
$\Omega +1$
-iterable weasel and
$S\subset \Omega $
be such that S is good for
$\mathcal {W}$
. Then the set
$\{\alpha <\Omega :\mathcal {W}\text { has the }S\text {-hull property at }\alpha \}$
is almost closed.
Proof. By Lemma 4.5 of [Reference Steel10] there is an
$\Omega +1$
-iterable weasel M and an elementary
embedding
$\pi :M\rightarrow \mathcal {W}$
such that
$\operatorname {\mathrm {ran}}(\pi )$
is S-thick and M has the
S-hull property at all
$\alpha <\Omega $
, i.e.
$\operatorname {\mathrm {HP}}_S^{M}=\Omega $
. Let
$(\mathcal {T},\mathcal {U})$
be the coiteration of
$\mathcal {W}$
and M. We will prove the theorem assuming
that
$\operatorname {\mathrm {lh}}(\mathcal {T})=\operatorname {\mathrm {lh}}(\mathcal {U})=\Omega +1$
and leave the remaining cases as an exercise to the
reader. Since
$\mathcal {W}$
and M are universal,
$i^{\mathcal {T}}:\mathcal {W}\rightarrow \mathcal {M}_{\Omega }^{\mathcal {T}}$
and
$i^{\mathcal {U}}:M\rightarrow \mathcal {M}_{\Omega }^{\mathcal {U}}$
exist and
$\mathcal {M}_{\Omega }^{\mathcal {T}}=\mathcal {M}_{\Omega }^{\mathcal {U}}=:M_{\infty }$
.
Note that by the remark following Example 4.3 in [Reference Steel10, p. 29],
$M_{\infty }$
has the S-hull property at
$\xi $
iff for no
$\beta +1\in [0,\Omega ]^{\mathcal {U}}$
,
$\xi \in [\operatorname {\mathrm {crit}}(E_{\beta }^{\mathcal {U}})^{+\mathcal {M}_{\eta }^{\mathcal {U}}},\nu (E_{\beta }^{\mathcal {U}}))$
, where
${\eta = \operatorname {\mathrm {pred}}^{\mathcal {U}}(\beta +1)}$
. Thus, the set
$\operatorname {\mathrm {HP}}_S^{M_{\infty }}$
is almost closed. Moreover, by arguments from the proof of
Lemma 4.6 in [Reference Steel10] (using that the set
of fixed points of
$i^{\mathcal {T}}$
is S-thick) we have that for all
$\alpha <\Omega $
,
$\mathcal {W}$
has the S-hull property at
$\alpha $
iff
$M_{\infty }$
has the S-hull property at
$i^{\mathcal {T}}(\alpha )$
.
Suppose for the sake of contradiction that
$\operatorname {\mathrm {HP}}_S^{\mathcal {W}}$
is not almost closed. Let
$\delta <\Omega $
be a witness for this. Since
$\operatorname {\mathrm {HP}}_S^{M_{\infty }}$
is almost closed this means that
$i^{\mathcal {T}}(\delta )>\sup (i^{\mathcal {T}}[\delta ])$
. There are two ways this can happen.
Case 1.
There is
$\beta +1\in [0,\Omega ]^{\mathcal {T}}$
such that
$\operatorname {\mathrm {crit}}(E_{\beta }^{\mathcal {T}})=i_{0\gamma }^{\mathcal {T}}(\delta )$
, where
$\gamma = \operatorname {\mathrm {pred}}_{\mathcal {T}}(\beta +1)$
.
Let us assume that
$\beta $
is least such. Note that in this case we must have
that
$\sup (i_{0\gamma }^{\mathcal {T}}[\delta ])=i_{0\gamma }^{\mathcal {T}}(\delta )$
and
$\delta ^{\prime }:=i_{0\gamma }^{\mathcal {T}}(\delta )\in \operatorname {\mathrm {HP}}_S^{M_{\infty }}.$
Since
$\operatorname {\mathrm {crit}}(i_{\beta +1,\Omega }^{\mathcal {T}})>\delta ^{\prime }$
,
$\mathcal {M}_{\beta +1}^{\mathcal {T}}$
has the S-hull property at
$\delta ^{\prime }$
. But then, since
$\mathcal {M}_{\beta +1}^{\mathcal {T}}$
has the S-hull property at
$\delta ^{\prime }$
and
$\mathcal {P}(\delta ^{\prime })\cap \mathcal {M}_{\gamma }^{\mathcal {T}}=\mathcal {P}(\delta ^{\prime })\cap \mathcal {M}_{\beta +1}^{\mathcal {T}}$
,
$\mathcal {M}_{\gamma }^{\mathcal {T}}$
has the S-hull property at
$\delta ^{\prime }$
. Contradiction!
Case 2.
$ W \models "\delta \text { is singular with cofinality } \mu "$
and there is a minimal
$\alpha \in [0,\Omega ]^{\mathcal {T}}$
such that
$i^{\mathcal {T}}_{0\alpha } (\mu ) = \operatorname {\mathrm {crit}}(i^{\mathcal {T}}_{\alpha \Omega })$
.
Let
$\gamma + 1 \in [0,\Omega ]^{\mathcal {T}}$
such that
$\operatorname {\mathrm {pred}}_{\mathcal {T}} (\gamma + 1) = \alpha $
and set
$W^{\prime }:=\mathcal {M}_{\alpha }^{\mathcal {T}}$
,
$W^{\prime \prime }:=\mathcal {M}_{\gamma +1}^{\mathcal {T}}$
,
$\delta ^{\prime }:=i_{0\alpha }^{\mathcal {T}}(\delta )$
,
$j:=i_{\alpha ,\gamma +1}^{\mathcal {T}}$
, and
$\delta ^{\prime \prime }:=\sup (j[\delta ^{\prime }])$
. Note that
$\delta ^{\prime \prime }<j(\delta ^{\prime })$
. Moreover, letting
$i:=i_{0\alpha }^{\mathcal {T}}$
, we have that
$\delta ^{\prime }=i(\delta )=\sup (i[\delta ])$
.
Claim 1. The S-hull property holds at
$\delta ^{\prime \prime }$
in
$W^{\prime \prime }$
.
Proof. Note that
$\delta ^{\prime \prime }$
is a limit of
$\operatorname {\mathrm {HP}}_S^{W^{\prime \prime }}$
and
$\Omega \setminus \operatorname {\mathrm {HP}}_S^{W^{\prime \prime }}$
. So in order to see that the
S-hull property holds at
$\delta ^{\prime \prime }$
in
$W^{\prime \prime }$
it suffices to see that
$i_{\gamma +1,\Omega }^{\mathcal {T}}$
is continuous at
$\delta ^{\prime \prime }$
. To this end note that
$\operatorname {\mathrm {cof}}(\delta ^{\prime \prime })^{W^{\prime \prime }}=\kappa $
and that all extenders used along
$[0,\Omega ]^{\mathcal {T}}$
after
$E_{\gamma }^{\mathcal {T}}$
have critical points greater than
$\kappa $
. Thus,
$i_{\gamma +1,\Omega }^{\mathcal {T}}$
is continuous at
$\delta ^{\prime \prime }$
from which the claim follows.
We claim that this implies that the S-hull property
holds at
$\delta ^{\prime }$
in
$W^{\prime }$
, which would be a contradiction. Let
$A\subset \delta ^{\prime }$
such that
$A\in W^{\prime }$
and
$\Gamma $
be an S-thick set. We need to
show that there is a term
$\tau $
,
$\vec {\xi }\in [\Gamma ]^{<\omega }$
, and
$\vec {\beta }\in [\delta ^{\prime }]^{<\omega }$
such that
$$\begin{align*}A=\tau^{W^{\prime}}[\vec{\xi},\vec{\beta}]\cap\delta^{\prime}. \end{align*}$$
Note that
$j(A)\cap \delta ^{\prime \prime }\in W^{\prime \prime }$
. Thus, there is a term
$\sigma $
,
$\vec {\zeta }\in [\Gamma ]^{<\omega }$
, and
$\vec {\gamma }\in \delta ^{\prime \prime }$
such that
$$\begin{align*}j(A)\cap\delta^{\prime\prime}=\sigma^{W^{\prime\prime}}[\vec{\zeta},\vec{\gamma}]\cap\delta^{\prime\prime}. \end{align*}$$
We may assume that j fixes
$\vec {\zeta }$
. Let
$\vec {\gamma }=([a_{0},f_{\!0}]_{E}^{W^{\prime }},...,[a_{n},f_{\!n}]_{E}^{W^{\prime }})$
, where for
$k\leq n$
,
$a_{k}\in [\nu (E)]^{<\omega }$
and
$f_{\!k}\in {^{[\kappa ]^{<\omega }}W^{\prime }}\cap W^{\prime }$
. Note that by Łoś’s Theorem,
$$ \begin{align} j(\xi) & \in \sigma^{W^{\prime\prime}}[\vec{\zeta},\vec{\gamma}]\iff \nonumber\\ \{b\in[\kappa]^{<\omega}:W^{\prime} \models \xi & \in \sigma [\vec{\zeta},f_{0}^{a_{0},a}(b),...,f_{n}^{a_{n},a}(b)]\}\in E_{a}, \end{align} $$
where
$a=\bigcup _{i\leq n}a_{n}$
. Furthermore, for
$\xi <\delta ^{\prime }$
,
$$ \begin{align} j(\xi)\in\sigma^{W^{\prime\prime}}[\vec{\zeta},\vec{\gamma}]\iff j(\xi)\in j(A)\iff\xi\in A. \end{align} $$
Note that for every
$c\in [\nu (E)]^{<\omega }$
,
$E_{c}$
is close to
$W^{\prime }$
. In particular, since
$W^{\prime }\models \text {ZFC}$
,
$E_{c}\in W^{\prime }$
for every
$c\in [\nu (E)]^{<\omega }$
. Moreover,
$\delta ^{\prime }$
is a limit cardinal in
$W^{\prime }$
and
$\text {GCH}$
holds in
$W^{\prime }$
. Thus, for every
$c\in [\nu (E)]^{<\omega }$
the ordinal of the measure
$E_{c}$
in the
$W^{\prime }$
-order
$<_{W^{\prime }}$
is an ordinal less than
$\delta ^{\prime }$
. Furthermore, as the ordinals below
$\delta ^{\prime \prime }$
might be represented via functions bounded in
$\delta ^{\prime }$
, we may assume that for
$k\leq n$
,
$f_{\!k}$
is bounded in
$\delta ^{\prime }$
and thus also their ordinals in the
$W^{\prime }$
-order are less than
$\delta ^{\prime }$
. But this means that 5.1 and 5.2 give us a term
$\tau $
and
$\vec {\beta }\in [\delta ^{\prime }]^{<\omega }$
such that
$$\begin{align*}A=\tau^{W^{\prime}}[\vec{\zeta},\vec{\beta}]\cap\delta^{\prime}. \end{align*}$$
This finishes the proof of the theorem.
Acknowledgments
The first author would like to thank the organizers of the “5th Münster Conference on Inner Model Theory”, “Advances in Set Theory”, and the “Arctic Set Theory Workshop 6” for giving him the opportunity to present parts of this paper. Moreover, the authors would like to thank Tapio Saarinen, Philipp Schlicht, Shervin Sorouri, and the anonymous referee for their helpful feedback.
Funding
The first author was funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) – project number 445387776.
The second author was funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy EXC 2044-390685587, Mathematics Münster: Dynamics-Geometry-Structure.
The first author would like to thank the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy EXC 2044-390685587, Mathematics Münster: Dynamics-Geometry-Structure for financial support in order to attend conferences and present material from this work.
-Woodin cardinal.
-Woodin cardinal.
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