Proof Let B be as in the statement of the lemma. Pick $A'\subseteq j_{U^0}(\kappa )$ such that $k(A')=B$ . Then $A' \not \in \operatorname {\mathrm {rng}}(j_{U^0})$ , since otherwise its image B will be in the range of $j_U=k\circ j_{U^0}$ . Denote by

$$ \begin{align*}\{A^{\prime}_{\nu}\mid \nu<j_{U^0}(\kappa^+)\}=j_{U^0}(\{A_i\mid i<\kappa^+\}),\end{align*} $$

$$ \begin{align*}\{A^{\prime\prime}_{\nu}\mid \nu<j_{U}(\kappa^+)\}=j_{U}(\{A_i\mid i<\kappa^+\}).\end{align*} $$

Since $U^0$ is normal, there is $f:\kappa \rightarrow \kappa ^+$ such that $A'=A^{\prime }_{j_{U^0}(f)(\kappa )}$ and thus

$$ \begin{align*}B=k(A')=k(A^{\prime}_{{j_{U^0}(f)(\kappa)}})=A^{\prime\prime}_{j_{U}(f)(\kappa)}.\end{align*} $$

Since B is not in the range of k, f is not constant. Recall that $\{ A_\alpha \mid \alpha <\kappa ^+\}$ is a strong witness for U being non-Galvin ultrafilter over $\kappa $ . Apply this to the family $\{ A_{f(\nu )} \mid \nu <\kappa \}.$ It follows that $[id]_U\not \in A^{\prime \prime }_{j_{U}(f)(\kappa )}=B.$

Proof Note that $(\kappa ^+)^{V^*}=(\kappa ^+)^V$ . Just otherwise, $(\kappa ^{++})^V $ will be $\leq (\kappa ^+)^{V^*}$ , and then, $j^*(\kappa )>(\kappa ^{++})^V$ . This is impossible, since $j^*$ extends $j_U$ . The rest follows from the previous lemma and the fact that $[\kappa , [id]_U)\subseteq \operatorname {\mathrm {rng}}(k)\setminus \operatorname {\mathrm {rng}}(j_{U})$ since $crit(k)=j_{U^0}(\kappa )=[id]_U$ .

Proof Suppose now that ${\langle } A_{\alpha _\xi } \mid \xi <\kappa {\rangle }$ is a subfamily of $\{ A_\alpha \mid \alpha <\kappa ^+\}$ of size $\kappa $ in ${V^*}$ .

Work in V. Let be a name of $\alpha _\xi $ . By $\kappa $ -c.c., then for every $\xi <\kappa $ there will be $s_\xi \subseteq \kappa ^+$ of cardinality less than $\kappa $ , such that .

Let $S=\sup _{\xi <\kappa }s_\xi $ . Enumerate $S={\langle } \beta _i\mid i<\kappa {\rangle }$ such that we if $\beta _i\in s_\zeta $ and $\beta _j\in s_\mu $ where $\zeta <\mu $ then $i<j$ , i.e., enumerate first $s_0$ then $s_1$ and so on, such that the resulting enumeration of S is of order-type $\kappa $ . This is possible since each $s_\zeta $ has cardinality less than $\kappa $ . Define

$$ \begin{align*}C=\{\nu<\kappa \mid \forall \xi<\nu (\sup(\gamma\mid \beta_\gamma\in s_\xi)<\nu)\}.\end{align*} $$

Clearly, C is a club. Hence $[id]_U\in j_U(C)$ . Then, by elementarity, for every $\zeta <[id]_U$ , and every $\beta _i\in s^{\prime }_\zeta $ , $i<[id]_U$ .

Let us use the fact that the sequence ${\langle } A_\alpha \mid \alpha <\kappa ^+{\rangle }$ is a strong witness for U being non-Galvin, hence $[id]_U \not \in A^{\prime }_{\beta _\zeta }$ , for every $\kappa \leq \zeta < [id]_U$ . Fix any $\kappa \leq \xi <[id]_U$ , then by elementarity we have in $M_U$ . Therefore there is some $\gamma <\kappa $ such that $\alpha ^{\prime }_\xi =\beta _\gamma $ . Clearly, $\gamma \geq \kappa $ , and by the closure property of $[id]_U$ , we conclude that $\gamma <[id]_U$ . Hence, in $M^*$ , $[id]_U \not \in A^{\prime }_{\beta ^{\prime }_\gamma }=A^{\prime }_{\alpha ^{\prime }_\xi }$ , as wanted.

Proof The forcing is simply adding for each inaccessible $\alpha \leq \kappa $ , $\alpha ^+$ -many Cohen functions to $\alpha $ . Namely, consider the Easton support iteration

such that for $\alpha \leq \kappa $ ,

is trivial unless $\alpha $ is inaccessible, in which case it is a $\mathcal {P}_\alpha $ -name for $\operatorname {\mathrm {Cohen}}(\alpha ,\alpha ^+)$ .

Let $G:=G_\kappa *g_\kappa $ be V-generic for . Denote ${\langle } f_{\kappa ,\alpha }\mid \alpha <\kappa ^+{\rangle }$ be the enumeration of the $\kappa ^+$ Cohen functions added by $g_\kappa $ . The idea is that the sets which are going to be a strong witness for the failure of the Galvin property are ${\langle } A_\alpha \mid \alpha <\kappa ^+{\rangle }$ , where

$$ \begin{align*}A_\alpha=\{\beta<\kappa\mid f_{\kappa,\alpha}(\beta)=1\}.\end{align*} $$

The next step is to construct the measure for this witness by extending ground model embeddings to $V[G]$ . Let $U\in V$ be a normal measure over $\kappa $ and consider the second ultrapower by U and the corresponding commutative diagram

$$ \begin{align*}j_1:=j_{U}:V\rightarrow M_U=:M_1, \ j_2:=j_{U^2}:V\rightarrow M_{U^2}=:M_2\end{align*} $$

$$ \begin{align*}k:M_1\rightarrow M_{2}, \ j_{2}=k\circ j_1,\end{align*} $$

where k is simply the ultrapower embedding defined in $M_U$ using the ultrafilter $j_{1}(U)$ . Denote $\kappa _1=j_1(\kappa )$ and $\kappa _2=j_{2}(\kappa )$ , then $k(\kappa _1)=\kappa _2$ .

By Easton support and elementarity,

where

is the quotient forcing above $\kappa $ , which is forcing equivalent to the continuation of the iteration above $\kappa $ using the same recipe as $\mathcal {P}_\kappa $ .

In $V[G]$ , let us first construct an M-generic filter for . Take $G_{\kappa }*g_{\kappa }$ to be the generic up to $\kappa $ including $\kappa $ . Above $\kappa $ , from the point of view of $V[G]$ , we have $\kappa ^+$ -closure for $\mathcal {P}_{(\kappa ,\kappa _1)}$ . By $GCH$ , and since $j_1$ is an ultrapower by a measure, there are only $\kappa ^+$ -many dense open subsets of this forcing to meet. Therefore we can construct in $V[G]$ by standard construction an $M_1[G]$ -generic filter $G_{(\kappa ,\kappa _1)}$ for $\mathcal {P}_{(\kappa ,\kappa _1)}$ . By $\kappa _1^+-cc$ of $Q_{\kappa _1}$ , we can find $g^{\prime }_{\kappa _1}$ which is $M_1[G*G_{(\kappa ,\kappa _1)}]$ -generic for $Q_{\kappa _1}$ . We need to change the values of $g^{\prime }_{\kappa _1}={\langle } f^{\prime }_{\kappa _1,\alpha }\mid \alpha <\kappa _1^{+}{\rangle }$ to $g_{\kappa _1}={\langle } f_{\kappa _1,\alpha }\mid \alpha <\kappa _1^{+}{\rangle }$ such that for every $\alpha <\kappa ^{+}$ , $f_{\kappa _1,j_1(\alpha )}\restriction \kappa =f_{\kappa ,\alpha }$ . This will ensure that the Silver criterion to lift an elementary embedding holds, namely, $j_1^{\prime \prime }G_{\kappa }*g\subseteq G_{\kappa }*g*G_{(\kappa ,\kappa _1)}*g^{\prime }_{\kappa _1}$ . Also, we would like to tweak the values of $f_{\kappa _1,j_1(\alpha )}(\kappa )$ to ensure that the sets $A_\alpha $ are members of the ultrafilter generated by $\kappa $ . By the definition of $A_\alpha $ , the way to do this is to set $f_{\kappa _1,j_1(\alpha )}(\kappa )=1$ .

Formally, for each condition $p\in \operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^{M_1[G_{\kappa }*G*G_{(\kappa ,\kappa _1)}]}$ , define a function $p^*$ with $\operatorname {\mathrm {dom}}(p^*)=\operatorname {\mathrm {dom}}(p)$ and for every ${\langle } \gamma ,\alpha {\rangle }\in \operatorname {\mathrm {dom}}(p^*)$ ,

$$ \begin{align*}p^*({\langle}\gamma,\alpha{\rangle})=\begin{cases} f_{\kappa,\beta}(\gamma), & \gamma<\kappa\wedge j_1(\beta)=\alpha,\\ 1, & \gamma=\kappa\wedge j_1(\beta)=\alpha,\\ p({\langle}\gamma,\alpha{\rangle}), & \text{else}.\end{cases}\end{align*} $$

Let $g_{\kappa _1}:=\{p^*\mid p\in g^{\prime }_{\kappa _1}\}$ . Clearly, the functions ${\langle } f_{\kappa _1,\alpha }\mid \alpha <\kappa _1^+{\rangle }$ derived from $g_{\kappa _1}$ satisfy that $f_{\kappa _1,j_1(\beta )}\restriction \kappa =f_{\kappa ,\beta }$ and $f_{\kappa _1,j_1(\beta )}(\kappa )=1$ for every $\beta <\kappa ^+$ . It remains to show that $g_{\kappa _1}$ is generic:

Proof First let us prove that $g_{\kappa _1}\subseteq \operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^{M_1[G_{\kappa }*G*G_{(\kappa ,\kappa _1)}]}$ . Indeed, $g^{\prime }_{\kappa _1}\subseteq \operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^{M_1[G_{\kappa }*G*G_{(\kappa ,\kappa _1)}]}$ and for any $p\in g^{\prime }_{\kappa _1}$ ,

$$ \begin{align*} M_1[G_{\kappa}*G*G_{(\kappa,\kappa_1)}]\models|p|<\kappa_1, \end{align*} $$

and hence $\operatorname {\mathrm {dom}}(p)_{\leq \kappa }:=\{\alpha \mid \exists {\langle }\gamma ,\alpha {\rangle }\in \operatorname {\mathrm {dom}}(p),\gamma \leq \kappa \}$ is bounded in $\kappa _1^+$ while $j_1^{\prime \prime }\kappa ^+$ is unbounded. It follows that there is $\theta <\kappa ^+$ such that

$$ \begin{align*} \operatorname{\mathrm{dom}}(p)_{\leq\kappa}\cap j_1^{\prime\prime}\kappa^+\subseteq j_1^{\prime\prime}\theta. \end{align*} $$

Hence from the V-perspective, $|\operatorname {\mathrm {dom}}(p)_{\leq \kappa }\cap j_1^{\prime \prime }\kappa ^+|\leq \kappa $ . The difference between p and $p^*$ is only on the coordinates of $\operatorname {\mathrm {dom}}(p)_{\leq \kappa }\cap j_1^{\prime \prime }\kappa ^+$ and by closure of $M_1[G_{\kappa }*g*G_{(\kappa ,\kappa _1)}]$ to $\kappa $ -sequences it follows that

$$ \begin{align*}p^*\in \operatorname{\mathrm{Cohen}}(\kappa_1,\kappa_1^+)^{M_1[G_{\kappa}*G*G_{(\kappa,\kappa_1)}]}, \ g_{\kappa_1}\subseteq\operatorname{\mathrm{Cohen}}(\kappa_1,\kappa_1^+)^{M_1[G_{\kappa}*G*G_{(\kappa,\kappa_1)}]}.\end{align*} $$

To see that $g_{\kappa _1}$ is generic over $M_1[G_{\kappa }*G*G_{(\kappa ,\kappa _1)}]$ , let $D\in M_1[G_{\kappa }*G*G_{(\kappa ,\kappa _1)}]$ be dense open. In $M_1[G_{\kappa }*G*G_{(\kappa ,\kappa _1)}]$ , define $D^*$ to consist of all conditions $p\in \operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)$ . Such that

$$ \begin{align*}\forall q. \operatorname{\mathrm{dom}}(q)=\operatorname{\mathrm{dom}}(p)\wedge |\{x\mid p(x)\neq q(x)\}|\leq \kappa\rightarrow q\in D\end{align*} $$

then $D^*$ is dense open. To see this, pick any $p\in \operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^{M_1[G_{\kappa }*G*G_{(\kappa ,\kappa _1)}]}$ and enumerate by ${\langle } q_r\mid r<\theta {\rangle }$ all the conditions q such that

$$ \begin{align*}\operatorname{\mathrm{dom}}(q)=\operatorname{\mathrm{dom}}(p)\wedge |\{x\mid p(x)\neq q(x)\}|\leq\kappa.\end{align*} $$

Note $\theta <\kappa _1$ since $\kappa _1$ is inaccessible in $ M_U[G_{\kappa }*g*G_{(\kappa ,\kappa _1)}]$ . We define inductively and increasing sequence ${\langle } p_r\mid r<\theta {\rangle }$ , and exploit the $\kappa _1$ -closure of $\operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)$ to take care of limit stages. Define $p_0=p$ , and suppose that $p_r$ is defined, let $p_{r+1}^{\prime }:=q_r\cup p_r\restriction (\operatorname {\mathrm {dom}}(p_r)\setminus \operatorname {\mathrm {dom}}(p))$ , find $p_{r+1}^{\prime }\leq t_{r+1}\in D$ which exists by density and set

$$ \begin{align*}p_{r+1}=p_r\restriction \operatorname{\mathrm{dom}}(p)\cup t_{r+1}\restriction(\operatorname{\mathrm{dom}}(t_{r+1})\setminus \operatorname{\mathrm{dom}}(p)).\end{align*} $$

Then $p_r\leq p_{r+1}$ . Let

$$ \begin{align*}p^*:=\cup_{r<\theta} p_{r}\end{align*} $$

then $p^{*}$ has the property that for $\kappa $ many changes of $p^*$ from the domain of p stays inside D. Namely any q with $\operatorname {\mathrm {dom}}(q)=\operatorname {\mathrm {dom}}(p^*)$ ,

$$ \begin{align*}q\restriction (\operatorname{\mathrm{dom}}(p^*)\setminus \operatorname{\mathrm{dom}}(p))=p^*\restriction(\operatorname{\mathrm{dom}}(p^*)\setminus \operatorname{\mathrm{dom}}(p))\end{align*} $$

and $|\{x\in \operatorname {\mathrm {dom}}(p)\mid p(x)\neq q(x)\}|\leq \kappa $ , $q\restriction \operatorname {\mathrm {dom}}(p)=q_r$ for some r, therefore $q\geq t_{r+1}\in D$ . Now we define inductively ${\langle } p^{(r)}\mid r<\kappa ^+{\rangle }$ , $p^{(0)}=p$ at limit we take union, and at successor step we take $p^{(r+1)}=(p^{(r)})^*$ . We claim that $p_*:=\cup _{r<\kappa ^+}p^{(r)}\in D^*$ . First note that $\kappa ^+<\kappa _1$ , hence $|p_*|<\kappa _1$ (all the definition is inside $M_U[G_{\kappa }*g_{\kappa }*G_{(\kappa ,\kappa _1)}]$ ). Let q be any condition with $\operatorname {\mathrm {dom}}(q)=\operatorname {\mathrm {dom}}(p^*)$ and denote by

$$ \begin{align*}I=\{x\in \operatorname{\mathrm{dom}}(p_*)\mid q(x)\neq p_*(x)\}\end{align*} $$

and suppose that $|I|\leq \kappa $ . Since $\operatorname {\mathrm {dom}}(p_*)=\cup _{r<\kappa ^+}\operatorname {\mathrm {dom}}(p^{(r)})$ and $\operatorname {\mathrm {dom}}(p^{(r)})$ is $\subseteq $ -increasing, there is $j<\kappa ^+$ such that $I\subseteq \operatorname {\mathrm {dom}}(p^{(j)})$ . The condition $q\restriction I$ is enumerated in the construction of $p^{(j+1)}$ , hence $q\restriction \operatorname {\mathrm {dom}}(p^{(j+1)})\in D$ and since D is open, $q\in D$ . This means that $p_*\in D^*$ .

Finally, by genericity of $g^{\prime }_{\kappa _1}$ , we can find $p\in D^*\cap g^{\prime }_{\kappa _1}$ . By definition, $p^*\in g_{\kappa _1}$ and since $\operatorname {\mathrm {dom}}(p^*)=\operatorname {\mathrm {dom}}(p)$ and $|\{x\mid p(x)\neq p^*(x)\}|\leq \kappa $ it follows that $p^*\in D$ .

Denote by $H=G_\kappa *g_{\kappa }*G_{(\kappa ,\kappa _1)}*g_{\kappa _1}$ , then $j_1^{\prime \prime }G\subseteq H$ . Let

$$ \begin{align*}j_1^*:V[G]\rightarrow M_1[H]\end{align*} $$

be the extended ultrapower and derive the normal ultrafilter over $\kappa $ ,

$$ \begin{align*}U_1:=\{X\subseteq \kappa\mid \kappa\in j^*_1(X)\}\end{align*} $$

then $U\subseteq U_1$ and $j^*_1=j_{U_1}$ . Indeed let $k_1:M_{U_1}\rightarrow M_1[H]$ be the usual factor map $k_1(j_{U_1}(f)(\kappa ))=j^*_1(f)(\kappa )$ . We will prove that $k_1$ is onto and therefore $k_1=id$ . For every $A\in M_1[H]$ , there is a name

such that

. $M_U$ is the ultrapower by U, hence there is $f\in V$ such that

. By elementarity for every $\alpha <\kappa $ , $f(\alpha )$ is a name. In $V[G]$ define $f^*(\alpha )=(f(\alpha ))_G$ , then by elementarity

Denote by $M_1^*=M_1[H]$ and consider $j^*_1(U_1)\in M^*_1$ . Let us now define inside $M^*_1$ an $M_{2}$ -generic filter for

in a similar fashion as H was defined. First we take H to be the generic for

. Note that $M_{2}$ is closed under $\kappa _1$ -sequences with respect to $M_1$ . Therefore, from the $M^*_1$ -point of view,

is $\kappa _1^+$ -closed, and we can construct an $M_{2}[H]$ -generic filter $G_{(\kappa _1,\kappa _2)}*g^{\prime }_{\kappa _2}\in M_1^*$ for it. We change the values of $g^{\prime }_{\kappa _2}$ a bit differently from the way we changed the values of $g^{\prime }_{\kappa _1}$ . If $\alpha <\kappa _1^+$ is of the form $j_1(\beta )$ let $f_{\kappa _2,k(\alpha )}(\kappa _1)=1$ (to guarantee that $A_\alpha $ ’s belong to the ultrafilter generated by $\kappa _1$ ) and if $\alpha \in \kappa _1^+\setminus j_1^{\prime \prime }\kappa ^+$ letFootnote ² $f_{\kappa _2,k(\alpha )}(\kappa _1)=0$ . Also, we would like that $f_{\kappa _2,\kappa _1}(0)=\kappa $ . Formally, for every $p\in \operatorname {\mathrm {Cohen}}(\kappa _2,\kappa _2^+)^{M_2[H*G_{(\kappa _1,\kappa _2)}]}$ , define $p^*$ to be a function with $\operatorname {\mathrm {dom}}(p)=\operatorname {\mathrm {dom}}(p^*)$ and for every ${\langle }\gamma ,\alpha {\rangle }\in \operatorname {\mathrm {dom}}(p^*)$ ,

$$ \begin{align*}p^*({\langle}\gamma,\alpha{\rangle})=\begin{cases} f_{\kappa_1,\beta}(\gamma),& \gamma<\kappa_1\wedge \alpha=k(\beta), \\ 1, & \gamma=\kappa_1\wedge \alpha=k(j_1(\beta)),\\ 0, & \gamma=\kappa_1\wedge \alpha=k(\beta),\beta\notin j_1^{\prime\prime}\kappa^+,\\ \kappa, & \gamma=0\wedge \alpha=\kappa_1,\\ p({\langle}\gamma,\alpha{\rangle}), & \text{else}.\end{cases}\end{align*} $$

Denote by $g_{\kappa _2}=\{p^*\mid p\in g^{\prime }_{\kappa _2}\}\in V[G]$ the resulting filter. It is important that for each $p\in g^{\prime }_2$ , the set

$$ \begin{align*}X_1:=j_{2}^{\prime\prime}\kappa^+\cap \operatorname{\mathrm{dom}}(f)_{\leq\kappa_1}=\{j_2(\alpha)\mid \exists {\langle}\gamma,j_2(\alpha){\rangle}\in\operatorname{\mathrm{dom}}(f), \gamma\leq\kappa_1\}\end{align*} $$

has size at most $\kappa $ . This ensured that $X_1\in M_1^*$ . Also, $k"\kappa _1^+$ is unbounded in $\kappa _2^+$ and conditions in $\operatorname {\mathrm {Cohen}}(\kappa _2,\kappa _2^+)^{M_2[H*G_{(\kappa _1,\kappa _2)}]}$ have $M_{2}[H*G_{(\kappa _1,\kappa _2)}]$ -cardinality less than $\kappa _2$ , which guarantees that for each $p\in \operatorname {\mathrm {Cohen}}(\kappa _2,\kappa _2^+)$ ,

$$ \begin{align*}X_2:= k"\kappa_1^+\cap \operatorname{\mathrm{dom}}(p)_{\leq\kappa_1}\end{align*} $$

has size at most $\kappa _1$ . Note that $p^*$ is definable in $M_1^*$ from the parameters $p,X_1,X_2\in M_1^*$ , and $p^*$ differs from p at most on $\kappa _1$ -many values. By the closure of $M_{2}[H*G_{(\kappa _1,\kappa _2)}]$ to $\kappa _1$ -sequences from $M_1^*$ ,

$$ \begin{align*}p^*\in M_{2}[H*G_{(\kappa_1,\kappa_2)}]\text{ and }g_{\kappa_2}\subseteq \operatorname{\mathrm{Cohen}}(\kappa_2,\kappa_2^+)^{M_{2}[H*G_{(\kappa_1,\kappa_2)}]}.\end{align*} $$

The genericity argument of Lemma 2.7 extends to the models $M_1$ and $M_{2}[H*G_{(\kappa _1,\kappa _2)}]$ , hence $g_{\kappa _2}$ is $M_{2}[H*G_{(\kappa _1,\kappa _2)}]$ -generic. Denote by $M_2^*=M_{2}[H*G_{(\kappa _1,\kappa _2)}*g_{\kappa _2}]$ . It follows that k can be extended (in $V[G]$ ) to $k^*$ and also $j_{2}$ to $j^*_2=k^*\circ j^*_1:V[G]\rightarrow M^*_2$ . Finally, let

$$ \begin{align*}W:=\{X\in P^{V[G]}(\kappa)\mid \kappa_1\in j^*_2(X)\}\in V[G].\end{align*} $$

Let us prove that W witnesses the theorem:

Proof To see (1), let us denote by $j_W:V[G]\rightarrow M_W$ the ultrapower embedding by W and $k_W:M_W\rightarrow M^*_2$ defined by $k_W([f]_W)=j^*_2(f)(\kappa _1)$ the factor map satisfying $k_W\circ j_W=j^*_2$ . Let us argue that $k_W$ is onto and therefore $k_W=id$ and $[id]_W=\kappa _1$ . Indeed, let $A\in M_2^*$ then there is

such that

. Since $j_2=j_{U^2}$ there is $h\in V$ such that

. Note that $\kappa =j^*_2(f_\kappa )_{\kappa _1}(0)$ , hence define in $V[G]$ , $h^*(\alpha )=(h(f_{\kappa ,\alpha }(0),\alpha ))_G$ . We have that

To see $(2)$ , for every club $C\in Cub_\kappa $ , $j^*_2(C)$ is closed and $j^*_1(C)$ is unbounded in $\kappa _1$ . Since $crit(k^*)=\kappa _1$ and $j^*_{2}(C)=k^*(j^*_1(C))$ it follows that $j^*_2(C)\cap \kappa _1=j^*_1(C)$ , hence $j^*_2(C)\cap \kappa _1$ is unbounded in $\kappa _1$ which implies that $\kappa _1\in j^*_2(C)$ .

For $(3)$ , since $M_2^*\models cf(\kappa _1)=\kappa _1$ , it follows that $\{\alpha \mid cf(\alpha )=\alpha \}\in W$ . Finally, for every $\alpha <\kappa ^+$ ,

$$ \begin{align*}j^*_2(A_\alpha)=\{\beta<\kappa_2\mid f_{\kappa_2,j_2(\alpha)}(\beta)=1\}.\end{align*} $$

Since $j_2(\alpha )=k(j_1(\alpha ))$ , by the definition of $g_{\kappa _2}$ , $f_{\kappa _2,j_2(\alpha )}(\kappa _1)=1$ , thus $\kappa _1\in j^*_2(A_\alpha )$ , and by definition of W, $A_\alpha \in W$ .

For (3), let $\{A_{\alpha _i}\mid i<\kappa \}$ be any subfamily of length $\kappa $ and $\kappa \leq \eta <[id]_W=\kappa _1$ . Denote

$$ \begin{align*}j_2^*({\langle} A_{\alpha_i}\mid i<\kappa{\rangle})={\langle} A^{(2)}_{\alpha^{(2)}_i}\mid i<\kappa_2{\rangle}, \ j_1^*({\langle} A_{\alpha_i}\mid i<\kappa{\rangle})={\langle} A^{(1)}_{\alpha^{(1)}_i}\mid i<\kappa_1{\rangle}.\end{align*} $$

Since $\kappa \leq \eta <\kappa _1$ , then $\eta \notin j_1^{\prime \prime }\kappa ^+$ and thus $\alpha ^{(1)}_\eta \notin j_1^{\prime \prime }\kappa ^+$ . Also, $k(\alpha ^{(1)}_\eta )=\alpha ^{(2)}_{k(\eta )}=\alpha ^{(2)}_\eta $ . Hence by definition, $f_{\kappa _2,\alpha ^{(2)}_\eta }(\kappa _1)=0$ , hence $\kappa _1\notin A^{\prime }_{\alpha ^{(2)}_\eta }.$

Proof Suppose that $Gal(U,\kappa ,\kappa ^+)$ holds and let $G\subseteq \operatorname {\mathrm {Prikry}}(U)$ be V-generic. By [Reference Gitik18, Proposition 1.3] every set $A\in V[G]$ of size $\kappa ^+$ contains a set $B\in V$ of cardinality $\kappa $ . Toward a contradiction suppose that $H\in V[G]$ is a V-generic filter for $\operatorname {\mathrm {Cohen}}(\kappa ,\kappa ^+)$ . Code $H:\kappa \times \kappa ^+\rightarrow 2$ as $X\subseteq \kappa ^+$ , just pick a bijection $\phi $ from $\kappa ^+$ to $\kappa ^+\times \kappa $ , and let $X=\{\alpha <\kappa ^+\mid H(\phi (\alpha ))=1\}$ . The set X does not contain an old subset of cardinality $\kappa $ ; this is a contradiction. To see this, let $Y\in V$ such that $|Y|=\kappa $ , proceed with a density argument: any condition $p\in \operatorname {\mathrm {Cohen}}(\kappa ,\kappa ^+)$ has size $<\kappa $ and therefore can be extended to a condition $p'$ such that for some $y\in Y$ , $\phi (y)\in \operatorname {\mathrm {dom}}(p')$ and $p'(\phi (y))=0$ .

Proof The model $V^*$ is obtained by iterating with Easton support the lottery sum of Cohen forcings for adding $\alpha ^+$ -Cohen functions ${\langle } f_{\alpha \gamma }\mid \gamma <\alpha ^+{\rangle }$ over $\alpha $ , and Cohen ${}^2$ for adding two blocks of $\alpha ^+$ -Cohen functions

$$ \begin{align*}{\langle} f_{\alpha \gamma}\mid \gamma<\alpha^+{\rangle},{\langle} h_{\alpha\gamma}\mid \gamma<\alpha^+{\rangle}.\end{align*} $$

More specifically, let

denotes the Easton support iteration, such that for each $\alpha <\kappa $ ,

is the trivial forcing unless $\alpha $ is inaccessible in which case

is a ${\mathcal P}_\alpha $ -name for the lottery sum

$$ \begin{align*}\operatorname{\mathrm{LOTT}}(\operatorname{\mathrm{Cohen}}(\alpha,\alpha^+),\operatorname{\mathrm{Cohen}}(\alpha,\alpha^+)\times \operatorname{\mathrm{Cohen}}(\alpha,\alpha^+)).\end{align*} $$

At $\kappa $ itself we let

. Let $G_\kappa *F_\kappa $ be a V-generic subset of

and let $V^*=V[G_\kappa *F_\kappa ]$ . We denote by $F_\alpha :={\langle } f_{\alpha \gamma }\mid \gamma <\alpha ^+{\rangle }$ the generic Cohen function if $\operatorname {\mathrm {Cohen}}(\alpha ,\alpha ^+)$ was forced in $G_\kappa $ and by

$$ \begin{align*}F_\alpha:={\langle} f_{\alpha \gamma}\mid \gamma<\alpha^+{\rangle},\ H_\alpha:={\langle} h_{\alpha,\gamma}\mid \gamma<\alpha^+{\rangle}\end{align*} $$

if $\operatorname {\mathrm {Cohen}}(\alpha ,\alpha ^+)\times \operatorname {\mathrm {Cohen}}(\alpha ,\alpha ^+)$ was.

Let $U\in V$ be a normal ultrafilter, $j_1:=j_U:V\rightarrow M_U$ the corresponding elementary embedding, $\kappa _1=j_1(\kappa )$ , $k:=j_{j_1(U)}:M_U\rightarrow M_{U^2}$ , $j_2=k\circ j_1$ , and $\kappa _2=j_2(\kappa )$ . Let us extend $j_1, k,j_2$ in $V[G_{\kappa }*F_\kappa ]$ :

We first extend $j_{1}: V \to M_{{U}}$ to $j_1^{*}: V[G_\kappa *F_\kappa ]\to M_{U}[G_{\kappa _1}*F_{\kappa _1}]$ . Do this by taking first $G_{\kappa _1}\cap P_{\kappa }=G_\kappa $ , at $\kappa $ we force with the lottery sum so we can choose to force only one block of Cohens and take $F_\kappa $ as a generic. Then defining a master condition sequence, using the closure of the forcing above $\kappa $ in $M_{U}$ exploiting $GCH$ to ensure that there are only $\kappa ^+$ -many dense sets to meet. This defines $G_{\kappa _1}$ . As for $F_{\kappa _1}$ , we first find an $M_U[G_{\kappa _1}]$ -generic $F^{\prime }_{\kappa _1}\times H^{\prime }_{\kappa _1}\in V[G_{\kappa }*F_\kappa ]$ again using $GCH$ , closure of $M_U[G_{\kappa _1}]$ under $\kappa $ -sequences and the closure of the forcing $(\operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^2)^{M_U[G_{\kappa _1}]}$ . Let us alter some values of $F^{\prime }_{\kappa _1}$ and $H^{\prime }_{\kappa _1}$ to define $F_{\kappa _1}={\langle } f_{\kappa _1,\gamma }\mid \gamma <\kappa _1^+{\rangle }$ and $H_{\kappa _1}={\langle } h_{\kappa _1,\gamma }\mid \gamma <\kappa _1^+{\rangle }$ such that for every $\alpha <\kappa _1^+$ :

1. (1) $f_{\kappa _1,j_1(\alpha )}\restriction \kappa =h_{\kappa _1,j_1(\alpha )}\restriction \kappa =f_{\kappa ,\alpha }$ .

2. (2) $f_{\kappa _1,j_1(\alpha )}(\kappa )=\alpha $ .

Formally, we change every pair of partial functions $p={\langle } p_0,p_1{\rangle } \in F^{\prime }_{\kappa _1}\times H^{\prime }_{\kappa _1}$ to the pair of partial functions $p_*={\langle } p_0^*,p_1^*{\rangle }$ such that $\operatorname {\mathrm {dom}}(p_0^*)=\operatorname {\mathrm {dom}}(p_0)$ , $\operatorname {\mathrm {dom}}(p_1^*)=\operatorname {\mathrm {dom}}(p_1)$ and for every ${\langle } \alpha ,\delta {\rangle }\in \operatorname {\mathrm {dom}}(p_0)$ :

$$ \begin{align*}p^*_0({\langle}\alpha,\delta{\rangle})=\begin{cases} f_{\kappa,\alpha_0}(\delta), &\exists \alpha_0<\kappa^+.\alpha=j_1(\alpha_0)\text{ and }\delta<\kappa,\\ \alpha_0, & \exists \alpha_0<\kappa^+.\alpha=j_1(\alpha_0)\text{ and }\delta=\kappa,\\ p_0({\langle}\alpha,\delta{\rangle}), & \text{else}.\end{cases}\end{align*} $$

$$ \begin{align*}p^*_1({\langle}\alpha,\delta{\rangle})=\begin{cases} f_{\kappa,\alpha_0}(\delta), &\exists \alpha_0<\kappa^+.\alpha=j_1(\alpha_0)\text{ and }\delta<\kappa,\\ p_1({\langle}\alpha,\delta{\rangle}), & \text{else}.\end{cases}\end{align*} $$

Note that for every $p_0,p_1\subseteq \operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^{M_U[G_{\kappa _1}]}$ we only change $\kappa $ -many values as $M_U[G_{\kappa _1}]\models |\operatorname {\mathrm {dom}}(p_0)|,|\operatorname {\mathrm {dom}}(p_1)|<\kappa _1$ , hence

$$ \begin{align*}|j_1^{\prime\prime}\kappa^+\cap\{\alpha\mid\exists\delta.{\langle}\alpha,\delta{\rangle}\in \operatorname{\mathrm{dom}}(p_0)\}|\leq\kappa\end{align*} $$

since $j_1(\kappa ^+)=\bigcup j_1^{\prime \prime }\kappa ^+$ , the same holds for $p_1$ . It follows that

$$ \begin{align*}p^*\in (\operatorname{\mathrm{Cohen}}(\kappa_1,\kappa_1^+)^2)^{M_U[G_{\kappa_1}]}.\end{align*} $$

Changing less than $\kappa _1$ -many values of a generic for $\operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^2$ does not impact the genericity. Hence $F_{\kappa _1}\times H_{\kappa _1}:=\{p^*\mid p\in F^{\prime }_{\kappa _1}\times H^{\prime }_{\kappa _1}\}\in V[G_{\kappa }*F_\kappa ]$ is still $M_U[G_{\kappa _1}]$ -generic.

Since at $\kappa $ we only force $\operatorname {\mathrm {Cohen}}(\kappa ,\kappa ^+)$ , in order to extend $j_1$ we only need a generic for $\operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)$ in the $M_U$ -side. We constructed $F_{\kappa _1}$ so that $j_1^{\prime \prime }F_\kappa \subseteq F_{\kappa _1}$ , hence $j_1^{\prime \prime }G_\kappa *F_\kappa \subseteq G_{\kappa _1}*F_{\kappa _1}$ ( $H_{\kappa _1}$ will be used later). Thus in $V[G_{\kappa }*F_{\kappa }]$ , we have extended $j_1\subseteq j_1^*:V[G_{\kappa }*F_{\kappa }]\rightarrow M_U[G_{\kappa _1}*F_{\kappa _1}]$ . Let us note that $j_1^*$ is actually the elementary embedding derived from the normal measure $U\subseteq U^0:=\{X\in P^{V[G_\kappa *F_\kappa ]}(\kappa )\mid \kappa \in j_1^*(X)\}$ :

Clearly the function $k_0:M_{U^0}\rightarrow M_U[G_{\kappa _1}*F_{\kappa _1}]$ defined by $k_0([f]_{U^0})=j^*_1(f)(\kappa )$ is elementary. To see the $k_0=id$ let us prove that $k_0$ is onto. Fix

and let $f\in V$ be such that

and define in $V[G_{\kappa }*F_\kappa ]$ the function $f^*(x)=(f(f_{\kappa ,\kappa }(x)))_{G_{\kappa }*F_{\kappa }}$ . Then

$$ \begin{align*}k_0(j_{U^0}(f^*)(\kappa))=j^*_1(f^*)(\kappa)=(j^*_1(f)(j^*_1(f_{\kappa,\kappa})(\kappa)))_{G_{\kappa_1}*F_{\kappa_1}}=\end{align*} $$

Recall that we have constructed the function $H_{\kappa _1}\in V[G_{\kappa }*F_{\kappa }]$ such that $F_{\kappa _1}\times H_{\kappa _1}$ is $M_U[G_{\kappa _1}]$ -generic for $\operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)^2$ . Now we wish to extend $k:M_U\rightarrow M_{U^2}$ to $k^*:M_U[G_{\kappa _1}*F_{\kappa _1}]\rightarrow M_{U^2}[G_{\kappa _2}*F_{\kappa _2}]$ in $V[G_{\kappa }*F_{\kappa }]$ . We do this by taking $G_{\kappa _2}\cap \kappa _1= G_{\kappa _1}$ , at $\kappa _1$ we force $\operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)\times \operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)$ putting the generic $F_{\kappa _1}\times H_{\kappa _1}$ , then exploiting the closure and $GCH$ to complete to a generic $G_{\kappa _2}*F^{\prime }_{\kappa _2}\in V[G_\kappa *F_\kappa ]$ . Finally, we wish to modify some values of $F^{\prime }_{\kappa _2}$ to a generic $F_{\kappa _2}={\langle } f_{\kappa _2,\gamma }\mid \gamma <\kappa _2^+{\rangle }$ so that for every $\alpha <\kappa _1^+$ :

1. (1) $f_{\kappa _2,k(\alpha )}\restriction \kappa _1=f_{\kappa _1,\alpha }$ .

2. (2) For $\alpha \in j_1^{\prime \prime }\kappa ^+$ , $f_{\kappa _2,k(\alpha )}(\kappa _1)=1$ .

3. (3) For $\alpha \in \kappa _1^+\setminus j_1^{\prime \prime }\kappa ^+$ , $f_{\kappa _2,k(\alpha )}(\kappa _1)=0$ .

4. (4) $f_{\kappa _2,\kappa _1}(\kappa _1)=\kappa $ .

Again, this is possible since we do not change too many values of $F^{\prime }_{\kappa _2}$ . At this point, let us emphasize that we do not use $H_{\kappa _1}$ in the generic we have in the $M_U$ -side Footnote ³ . The generic $H_{\kappa _1}$ is used in the construction of the generic on the $M_{U^2}$ -side where we can choose (due to the lottery sum) to force at $\kappa _1$ two copies of $\operatorname {\mathrm {Cohen}}(\kappa _1,\kappa _1^+)$ , of course, that at $\kappa _2=j_{2}(\kappa )$ we are still obligated to force one copy of $\operatorname {\mathrm {Cohen}}(\kappa _2\kappa _2^+)$ which contains the point-wise image of $F_{\kappa _1}$ under the factor map k.

Hence we extended in $V[G_{\kappa }*F_\kappa ]$ , $k\subseteq k^*:M_U[G_{\kappa _1}*F_{\kappa _1}]\rightarrow M_{U^2}[G_{\kappa _2}*F_{\kappa _2}]$ .

Let $j_2^*=k^*\circ j_1^*$ , $V^*=V[G_{\kappa }*F_\kappa ]$ , $M_1^*=M_U[G_{\kappa _1}*F_{\kappa _1}]$ and $M_2^*=M_{U^2}[G_{\kappa _2}*F_{\kappa _2}]$ .

In $V^*$ , define

$$ \begin{align*}U^*=\{X\subseteq\kappa\mid \kappa\in j_2^*(U)\},\end{align*} $$

$$ \begin{align*}W=\{X\subseteq \kappa\mid \kappa_1\in j^*_2(X)\},\end{align*} $$

and for every $\alpha <\kappa ^+$ ,

$$ \begin{align*}A_\alpha=\{\nu<\kappa\mid f_{\kappa,\alpha}(\nu)=1\}.\end{align*} $$

Then as in Claim 2.8, we have that W is a $\kappa $ -complete ultrafilter over $\kappa $ such that:

1. (1) $j_1^*=j_U^*$ , $j^*_2=j_W$ and $[id]_W=\kappa _1$ .

2. (2) ${\langle } A_\alpha \mid \alpha <\kappa ^+{\rangle }$ is a strong witness for W being non-Galvin.

3. (3) $Cub_\kappa \subseteq W$ .

4. (4) $L_0=\{\alpha <\kappa \mid \operatorname {\mathrm {Cohen}}(\alpha ,\alpha ^+)\times \operatorname {\mathrm {Cohen}}(\alpha ,\alpha ^+)\text { was forced in }G_\kappa \}\in W$ .

Also, recall that $j_2:V\rightarrow M_2$ is also the ultrapower by $U\times U$ under the identification(isomorphism):

$$ \begin{align*}j_{U^2}(f)(\kappa,\kappa_1)=j_{2,1}(j_1(\nu\mapsto f(\nu,*))(\kappa))(\kappa_1).\end{align*} $$

Clearly, the projections $\pi _{1},\pi _2:\kappa \times \kappa \rightarrow \kappa $ on the first and second coordinates (resp. Rudin–Keisler) project $U^2$ on U. Also, $W\cap V=U^*\cap V=U$ and $U^*\leq _{R-K} W$ and the projection map is denoted by $\nu \mapsto \pi _{nor}(\nu )$ .Footnote ⁴

Let us prove that W witnesses the theorem:

Proof of Theorem 2.11

Let ${\langle } c_n \mid n<\omega {\rangle } $ be the W-Prikry sequence corresponding to H. Suppose without loss of generality that for every $n<\omega $ , $c_n\in L_0$ , this will hold from a certain point and the proof can be adjusted in a straightforward way. This guarantees that the generic $H_{c_n}={\langle } h_{c_n,\gamma }\mid \gamma <\alpha ^+{\rangle }$ for the second component of the generic we have in $G_\kappa $ for $\operatorname {\mathrm {Cohen}}(c_n,c_n^+)\times \operatorname {\mathrm {Cohen}}(c_n,c_n^+)$ is defined for every $n<\omega $ . The functions $h_{c_n,\gamma }$ will be used below to define the Cohen generic functions.

Define, for every $n<\omega $ , the set

$$ \begin{align*}Z_n=\{\alpha<\kappa^+\mid \{c_m \mid n\leq m<\omega\} \subseteq A_\alpha \text{ and } n \text{ is least possible}\}.\end{align*} $$

For every $\alpha <\kappa ^+$ , let $n_\alpha $ be the unique n such that $\alpha \in Z_n$ . Let $\alpha <\kappa ^+$ , and define $f^*_\alpha :\kappa \to \kappa $ as follows:

Fix a sequence ${\langle } s_\alpha \mid \alpha <\kappa ^+ {\rangle }\in V^*$ of canonical functions in $\prod _{\nu <\kappa }\nu ^+$ :

$$ \begin{align*}f^*_\alpha\restriction c_{n_\alpha}= h_{c_{n_\alpha} s_\alpha(c_{n_\alpha})},\end{align*} $$

$$ \begin{align*}f^*_\alpha\restriction [c_{m-1}, c_m)= h_{ c_m, s_\alpha(c_m)}\restriction [c_{m-1}, c_m), \text{for }m, n_\alpha< m<\omega.\end{align*} $$

Let us argue that $F={\langle } f^*_\alpha \mid \alpha <\kappa ^+{\rangle } $ induces a $\operatorname {\mathrm {Cohen}}(\kappa ,\kappa ^+)^{V^*}$ generic filter over $V^*$ .

Claim 2.12. Let $G^*=\{p\in \operatorname {\mathrm {Cohen}}(\kappa ,\kappa ^+)^{V^*}\mid p\subseteq F\}$ , then $G^*$ is a $V^*$ -generic filter.

Let $\mathcal {A}\in V^*$ be a maximal antichain in the forcing $\operatorname {\mathrm {Cohen}}(\kappa , \kappa ^+)^{V^*}$ . Note that since $\operatorname {\mathrm {Cohen}}(\kappa ,\kappa ^{+})^{V^*}$ is $\kappa $ -closed then

$$ \begin{align*}Cohen(\kappa,\kappa^{+})^{V[G_\kappa]}=Cohen(\kappa,\kappa^{+})^{V^*}\!\!.\end{align*} $$

By $\kappa ^+$ -cc of the forcing $\mathcal {P}_{\kappa +1}$ , there is $Y\subseteq \kappa ^{+}$ , $Y\in V$ such that $|Y|=\kappa $ and $\mathcal {A}\subseteq \operatorname {\mathrm {Cohen}}(\kappa ,Y)^{V^*}$ . Also, since $|\mathcal {A}|=\kappa $ , $\mathcal {A}\in V[G_\kappa *F_\kappa ]$ , there is $Z\subseteq \kappa ^{+}$ such that $|Z|=\kappa $ such that $\mathcal {A}\in V[G_\kappa *F_\kappa \restriction Z]$ . Without loss of generality assume that $Z=Y\in V$ (Otherwise just take the union). Let $V\ni \phi :\kappa \rightarrow Y$ be a bijection.

Claim 2.13. There is an $\in $ -increasing continuous chain ${\langle } N_\beta \mid \beta <\kappa {\rangle }$ of elementary submodels of $H_\chi $ such that:

1. (1) $|N_\beta |<\kappa $ .

2. (2) $G_\kappa ,F_\kappa ,\mathcal {A},\phi ,{\langle } s_\alpha \mid \alpha <\kappa ^+{\rangle }\in N_0$ .

3. (3) $N_\beta \cap \kappa =\gamma _\beta $ is a cardinal $<\kappa $ , $\gamma _{\beta +1}$ is regular.

4. (4) For every $\rho ,\delta \in \phi "\gamma _\beta .\rho <\delta \rightarrow \forall \gamma _\beta \leq \mu <\kappa , s_\rho (\mu )<s_\delta (\mu )$ .

5. (5) If $\gamma _\beta $ is regular, then $N_\beta ^{<\gamma _\beta }\subseteq N_\beta $ . In particular $\operatorname {\mathrm {Cohen}}(\gamma _\beta ,\phi "\gamma _\beta )=\operatorname {\mathrm {Cohen}}(\kappa ,Y)\cap N_\beta $ .

Proof of Claim 2.13

Let us construct such a sequence inductively. Note that $(4)$ follows from elementarity and $(2)$ . Requirements $(1)$ – $(5)$ are preserved at limit stages due to continuity. At successor stages, suppose we have constructed $N_\beta $ , find an elementary submodel $N^0_{\beta +1}$ such that $N_\beta \subseteq N^0_{\beta +1}, \ {\langle } N_\alpha \mid \alpha <\beta {\rangle }\in N^0_{\beta +1}$ , then we construct an auxiliary $\in $ -increasing and continuous chain of elementary submodels ${\langle } N^\alpha _{\beta +1}\mid \alpha <\kappa {\rangle }$ as follows: $N^0_{\beta +1}$ is already defined. At limits we take the union and at successor let us take care of requirements 3 and 5. Let $\gamma ^{\prime }_\alpha =\sup (N^{\alpha }_{\beta +1}\cap \kappa )<\kappa $ . Let $N^{\alpha +1}_{\beta +1}$ be an elementary submodel such that $N^{\alpha }_\beta ,^{<\gamma ^{\prime }_\alpha },\subseteq N^{\alpha +1}_{\beta +1}$ and $|N^{\alpha +1}_{\beta +1}|<\kappa $ . Note that the sets

$$ \begin{align*}C_1=\{\alpha<\kappa\mid N^{\alpha}_{\beta+1}\cap\kappa=\gamma^{\prime}_\alpha\in\kappa\},\end{align*} $$

$$ \begin{align*}C_2=\{\alpha\in C_1\mid \text{if }\gamma_\alpha\text{ is regular then } N_{\alpha}^{<\gamma_\alpha}\subseteq N_\alpha\}\end{align*} $$

are clubs and also $\bar C=C_1\cap C_2$ is. It follows that $\{\gamma ^{\prime }_\alpha \mid \alpha \in \bar C\}$ is a club and since $\kappa $ is measurable, there is a $\alpha ^*\in \bar C$ limit such that $\gamma ^{\prime }_{\alpha ^*}$ is regular. Let $N_{\beta +1}=N^{\alpha ^*}_{\beta +1}$ , to conclude $2$ since $\gamma _{\beta +1}=\gamma ^{\prime }_{\alpha ^*}$ is regular.

Set

$$ \begin{align*}C=\{ \beta<\kappa \mid \gamma_\beta=\beta \}.\end{align*} $$

This is club in $\kappa $ since the sequence $\gamma _\beta $ is continuous and since the set $\{\beta \mid \gamma _\beta =\beta \}$ is a club.

Claim 2.14. Let

$$ \begin{align*}E:=\{\beta<\kappa\mid\forall\gamma\in\phi"\beta.\exists\ \delta<\beta^+.f_{\kappa,\gamma}\restriction\beta=f_{\beta,\delta}\}.\end{align*} $$

Then $E\in W$ .

Proof of Claim 2.14

By construction, for every $\alpha <\kappa _1^{+}$ , $f_{\kappa _2,k(\alpha )}\restriction \kappa _1=f_{\kappa _1,\alpha }$ and therefore for every $\alpha \in j^*_2(\phi )"\kappa _1$ , there is $\nu <\kappa _1^+$ such that $\alpha =k^*(j_1^*(\phi ))(\nu )=k^*(j_1^*(\phi )(\nu ))$ and $j_1^*(\phi )(\nu )<\kappa _1^{+}$ . Hence $f_{\kappa _2,\alpha }\restriction \kappa _1=f_{\kappa _1,\beta }$ for some $\beta <\kappa _1^{+}$ . Reflecting this we obtain the set $E\in W$ .

To see that $G^*\cap \mathcal {A}\neq \emptyset $ , we will need to catch a piece of $\mathcal {A}$ in the elementary submodels constructed and pick the Prikry points in the club C prepared:

Claim 2.15. For every $\nu _0\in C\cap E$ , there is $d=d^{\nu _0}\in N_{\nu _0}\cap \mathcal {A}$ such that d is extended by ${\langle } h_{\nu _0,s_{\tau }(\nu _0)}\mid \tau \in \phi "\nu _0{\rangle }$ .

Proof of Claim 2.15

Fix any $\nu _0\in C\cap E$ . Consider the transitive collapse of $\pi :N_{\nu _0}\rightarrow N_{\nu _0}^*$ . Then the critical point of $\pi ^{-1}:N_{\nu _0}^*\rightarrow N_{\nu _0}$ is $\nu _0$ and $\pi ^{-1}(\nu _0)=\kappa $ . Denote by $\overline {F}_\kappa =\pi (F_\kappa ), \overline {\phi }=\pi (\phi )$ . Denote $\overline {F}_\kappa ={\langle } \overline {f}_{\kappa ,\gamma }\mid \gamma < \pi (\kappa ^+){\rangle }$ . For every $\gamma \in \overline {\phi }"{\nu _0}$ , there is some $\delta <{\nu _0}$ such that

$$ \begin{align*}\gamma=\pi(\phi)(\delta)=\pi(\phi(\delta))\text{ and }\overline{f}_{\kappa,\gamma}=\pi(f_{\kappa,\phi(\delta)}).\end{align*} $$

Moreover, since ${\nu _0}\in E$ , $f_{\kappa ,\phi (\delta )}\restriction {\nu _0}=f_{{\nu _0},\rho }$ for some $\rho <{\nu _0}^{+}$ and therefore $\overline {f}_{\kappa ,\gamma }=f_{{\nu _0},\rho }$ . Recall that

, hence

. We conclude that for some subset $Z\subseteq {\nu _0}^{+}$ ,

Since ${\nu _0}\in L_0$ , in $V[G_{\kappa }*F_\kappa ]$ we also have $H_{\nu _0}={\langle } h_{\nu _0,\alpha }\mid \alpha <\nu _0^+{\rangle }$ which are mutually Cohen-generic over $V[G_{\nu _0}*F_{\nu _0}\restriction Z]$ .

By construction, $\forall \tau _1<\tau _2\in \phi "\nu _0$ , $s_{\tau _1}(\nu _0)<s_{\tau _2}(\nu _0)$ , hence ${\langle } h_{\nu _0,s_\tau (\nu _0)}\mid \tau \in \phi "\nu _0{\rangle }$ are Cohen functions over $\nu _0$ which are distinct mutually $V[G_{\nu _0}*F_{\nu _0}\restriction Z]$ -generic. Also, $\overline {\mathcal {A}}\subseteq \pi (\operatorname {\mathrm {Cohen}}(\kappa ,Y))= \operatorname {\mathrm {Cohen}}(\nu _0,\pi (\phi )"\nu _0)=\operatorname {\mathrm {Cohen}}(\nu _0,\pi "[\phi "\nu _0])$ is a maximal antichain. Since $|\pi "\phi "\nu _0|=\nu _0=|\phi "\nu _0|$ , we can change the enumeration of the functions ${\langle } h_{\nu _0,s_{\tau }(\nu _0)}\mid \tau \in \phi "\nu _0{\rangle }$ to $h^{\prime }_{\pi (\tau )}=h_{\nu _0,s_{\tau }(\nu _0)}$ so that ${\langle } h^{\prime }_{\rho }\mid \rho \in \pi "\phi "\nu _0{\rangle }$ is generic for $\operatorname {\mathrm {Cohen}}(\nu _0,\pi "\phi _0)$ . Thus pick $d_0 \in \overline {\mathcal {A}}$ such that $d_0$ is extended by ${\langle } h^{\prime }_{\rho }\mid \rho \in \pi "\phi "\nu _0{\rangle }$ . It follows that

$$ \begin{align*}d:=\pi^{-1}(d_0)\in \mathcal{A}\cap N_{\nu_0}\end{align*} $$

is a condition with $\operatorname {\mathrm {dom}}(d)=\pi ^{-1}(\operatorname {\mathrm {dom}}(d_0))$ . Since the critical point of $\pi $ is $\nu _0$ , for every ${\langle }\alpha ,\beta {\rangle }\in \operatorname {\mathrm {dom}}(d_0)$ , $\pi ^{-1}({\langle }\alpha ,\beta {\rangle }))={\langle }\alpha ,\pi ^{-1}(\beta ){\rangle }$ , hence

$$ \begin{align*}d({\langle}\alpha,\pi^{-1}(\beta){\rangle})=\pi^{-1}(d_0(\alpha,\beta))=d_0(\alpha,\beta).\end{align*} $$

In particular for every ${\langle }\gamma ,\alpha {\rangle }\in \operatorname {\mathrm {dom}}(d)$ ,

$$ \begin{align*}d(\gamma,\alpha)=d_0(\gamma,\pi(\alpha))=h^{\prime}_{\pi(\alpha)}(\gamma)=h_{\nu_0,s_{\alpha}(\nu_0)}(\gamma).\end{align*} $$

Thus d is extended by ${\langle } h_{\nu _0,s_\tau (\nu _0)}\mid \tau \in \phi "\nu _0 {\rangle }$ .

It suffices to show that any condition in $\operatorname {\mathrm {Prikry}}(W)$ has an extension which forces that $G^*$ meets a member of $\mathcal {A}$ .

Let $p={\langle } {\langle }{\rangle }, B{\rangle }$ be a condition (we assume for simplicity that its finite sequence is empty) and shrink B to $B\cap C\cap E$ . For any $\nu _0\in B\cap C\cap E$ , we split $\phi "\nu _0$ into two sets:

$$ \begin{align*}X^{\nu_0}_0:=\{\tau\in\phi"\nu_0\mid \nu_0\in A_\tau\}\text{ and }X^{\nu_0}_1=\phi"\nu_0\setminus X^{\nu_0}_0.\end{align*} $$

The condition $p_0={\langle } {\langle } \nu _0{\rangle }, B\cap C\cap E\cap X\cap (\bigcap _{\tau \in \phi "\nu _0}A_\tau ){\rangle }$ forces the following:

1. (1) The Prikry sequence is included in each $A_\tau $ , $\tau \in X^{\nu _0}_0$ , i.e., $n_\tau =0$ .

2. (2) $n_\tau =1$ , for every $\tau \in X^{\nu _0}_1$ .

In particular, this condition forces some information about the Cohen functions. Namely that:

1. (1) For $\tau \in X_0^{\nu _0}$ , $f^*_{\tau }\restriction \nu _0=h_{\nu _0,s_\tau (\nu _0)}.$

2. (2) For $\tau \in X_1^{\nu _0}$ , .

We would like to find a condition in $\mathcal {A}$ which is below these decided parts of the Cohen. By the previous proposition, there is $d\in N_{\nu _0}\cap \operatorname {\mathrm {Cohen}}(\kappa ,Y)=\operatorname {\mathrm {Cohen}}(\nu _0,\phi "\nu _0)$ , which is extended by ${\langle } h_{\nu _0,s_\tau (\nu _0)}\mid \tau \in \phi "\nu _0{\rangle }$ . However, by $(1)$ and $(2)$ we can only ensure that the generic $f^*_\tau $ to extend $d\restriction \nu _0\times X^{\nu _0}_0$ in $X^{\nu _0}_0$ . We are left to extend $d\restriction \nu _0\times X^{\nu _0}_1$ . Let us show that for many $\nu _0$ , $X^{\nu }_0$ is a relatively large subset of $\phi "\nu _0$ :

Claim 2.16. Let

$$ \begin{align*}R=\{\nu<\kappa\mid \forall\alpha\in\phi"\pi_{nor}(\nu), \nu\in A_\alpha\}.\end{align*} $$

Then $R\in W$ .

Proof Clearly, for every $\alpha \in j^*_2(\phi )"\kappa $ , $\alpha =j^*_2(\phi (\gamma ))$ , and $f_{\kappa _2,\alpha }(\kappa _1)=1$ , reflecting this, we can find a W-large set of $\nu $ ’s such that for every $\alpha \in \phi "\pi _{nor}(\nu )$ , $f_{\kappa ,\alpha }(\nu )=1$ . And by definition of $A_\alpha $ , $\nu \in A_\alpha $ .

Denote $B_0:=B\cap C\cap E\cap R$ . In order to extend $d\restriction \nu _0\times X_1$ , we will need to pick $\nu _0$ high enough in $B_0$ , but also the next point $\nu _1\in B_0\setminus \nu _0+1$ in the Prikry sequence such that it will belong to all $A_\tau $ with $\tau \in X_1$ and in addition the relevant Cohen functions over $\nu _1$ extend $d\restriction \nu _0\times X_1$ .

Let us look at $B_0$ more carefully. Let

be its name in V. We fix a condition $m_0\in G_\kappa *F_\kappa $ which forces that if

then the properties of Claims 2.15 and 2.16 hold, namely there is

which is extended by

, and

. Recall that by the construction of $G_{\kappa _2}$ , we have $m_0\in G_{\kappa _2}*F_{\kappa _2}$ . Let $ m_0\leq t\in G_{\kappa _2}*F_{\kappa _2}$ be a condition such that

By the construction of $G_{\kappa _2}*F_{\kappa _2}$ , t has the form:

$$ \begin{align*}t={\langle} t_{<\kappa}, t_{\kappa},t_{(\kappa,\kappa_1)},\underset{t_{\kappa_1}}{\underbrace{{\langle} t_{\kappa_1}^0,t_{\kappa_1}^1{\rangle}}},t_{(\kappa_1,\kappa_2)},t_{\kappa_2}{\rangle}.\end{align*} $$

Since $f_{\kappa _2,j_2(\alpha )}(\kappa _1)=1$ for every $\alpha <\kappa ^+$ , this will hold for every $\alpha \in \phi "\kappa $ as well. Also, recall that $Y\in V$ , hence $\phi \in V$ . Thus $j_2(\phi )\in M_2$ and $j_2(\phi )"\kappa \in M_2$ . Also, for $(t_{\kappa _2})_{G_{\kappa _2}}\in M_2[G_{\kappa _2}]$ ,

$$ \begin{align*}j_2^{\prime\prime}\kappa^+\cap \operatorname{\mathrm{Supp}}((t_{\kappa_2})_{G_{\kappa_2}})\in M_2[G_{\kappa_2}]\end{align*} $$

and $(t_{\kappa _2})_{G_{\kappa _2}}\restriction \kappa \times \{j_2(\alpha )\}\subseteq f_{\kappa ,\alpha }$ . We also fix $\theta <\kappa ^+$ such that $\operatorname {\mathrm {Supp}}((t_{\kappa _2})_{G_{\kappa _2}})\subseteq j_2(\theta )$ , there is such $\theta $ since $j_2^{\prime \prime }\kappa ^+$ is unbounded in $j_2(\kappa ^+)$ . Therefore, we can extend if necessary t such that

$$ \begin{align*}(2a)\ \ t_{<\kappa_2}\Vdash (\kappa\cup\{\kappa_1\})\times j_2(\phi)"\kappa \subseteq \operatorname{\mathrm{dom}}(t_{\kappa_2})\wedge (0,\kappa_1)\in\operatorname{\mathrm{dom}}(t_{\kappa_2})\wedge \operatorname{\mathrm{Supp}}(t_{\kappa_2})\subseteq j_2(\theta),\end{align*} $$

$$ \begin{align*}(2b) \ \ t_{<\kappa_2}\Vdash \ t_{\kappa_2}(\kappa_1,\alpha)=1,\text{ for every }\alpha\in j_2(\phi)"\kappa\text{ and } t_{\kappa_2,\kappa_1}(0)=\kappa,\end{align*} $$

Next consider $t_{\kappa _1}={\langle } t_{\kappa _1}^0,t_{\kappa _1}^1{\rangle }$ ; it is a ${\mathcal P}_{\kappa _1}$ -name for a condition in $F_{\kappa _1}\times H_{\kappa _1}$ . By the construction of the generic $F_{\kappa _1}\times H_{\kappa _1}$ , for every $\alpha <\kappa ^+$ , we made sure that $h_{\kappa _1,j_1(\alpha )}\restriction \kappa =f_{\kappa ,\alpha }$ . Also, $j_1(\phi )"\kappa \in M_2$ . Let

$$ \begin{align*}\mu_1=(j_1\restriction \phi"\kappa)^{-1}\in M_1.\end{align*} $$

Note that for every $\beta <\kappa ^+$ , $j_1(s_\beta )=s_{j_1(\beta )}:\kappa _1\rightarrow \kappa _1$ is the canonical function for $j_1(\beta )$ defined in $M_U$ , hence $j_2(s_\beta )(\kappa _1)=k(s_{j_1(\beta )})(\kappa _1)=j_1(\beta )$ . Hence

$$ \begin{align*}\operatorname{\mathrm{dom}}(\mu_1)=j_1(\phi)"\kappa=\{s_\gamma(\kappa_1)\mid \gamma\in j_2(\phi)"\kappa\}, \ rg(\mu_1)=\phi"\kappa\subseteq\kappa^+.\end{align*} $$

Extend if necessary $t_{<\kappa _1}$ , and assume that

As for the lower part, due to the Easton support, we have

$$ \begin{align*}(4) \ \ t_{<\kappa}\in V_\kappa.\end{align*} $$

Fix functions $r,\Gamma _1$ which represents $t,\mu $ resp. in the ultrapower $M_{U^2}$ , namely $j_2(r)(\kappa ,\kappa _1)=t, \ j_2(\Gamma _1)(\kappa ,\kappa _1)=\mu $ . Without loss of generality, suppose that for every $(\nu ',\nu )$ , it takes the form

$$ \begin{align*}r(\nu',\nu)={\langle} r_{<\nu'} r_{\nu'},r_{(\nu',\nu)},{\langle} r_{\nu}^0,r_{\nu}^1{\rangle},r_{(\nu,\kappa)},r_\kappa{\rangle}.\end{align*} $$

Reflecting some of the properties of t we obtain a set $B'\in U^2$ such that for every $(\nu ',\nu )\in B'$ :

1. $(1)_{(\nu ',\nu )}$ .

2. $(2a)_{(\nu ',\nu )}$ $ r_{<\kappa }\Vdash (\nu '\cup \{\nu \})\times \phi "\nu '\hspace{-1pt} \subseteq\hspace{-1pt} \operatorname {\mathrm {dom}}(r_{\kappa })\wedge {\langle }0,\nu {\rangle }\in \operatorname {\mathrm {dom}}(r_{\kappa })\wedge \operatorname {\mathrm {Supp}}(r_{\kappa }) \subseteq \theta .$

3. $(2b)_{(\nu ',\nu )}$ $r_{<\kappa }\Vdash \forall \alpha \in \phi "\nu '.r_{\kappa ,\alpha }(\nu )=1$ and $r_{\kappa ,\nu }(0)=\nu '$ .

4. $(3)_{(\nu ',\nu )}$ $ r_{<\nu }\Vdash \nu '\times \operatorname {\mathrm {dom}}(\Gamma _1(\nu ',\nu ))\hspace{-1pt}\subseteq\hspace{-1pt} \operatorname {\mathrm {dom}}(r^1_\nu )$ and for every .

5. $(4)_{(\nu ',\nu )}$ $r_{<\nu '}=t_{<\kappa }\in V_{\nu '}$ .

Let

$$ \begin{align*}B"=\{\nu \mid\exists(\nu',\nu)\in B'. r(\nu',\nu)\in G_\kappa*F_\kappa\}.\end{align*} $$

Since $B'\in U^2$ we have that $(\kappa ,\kappa _1)\in j_2(B')$ and since $j_2(r)(\kappa ,\kappa _1)=t\in j^*_2(G_\kappa *F_\kappa )=G_{\kappa _2}*F_{\kappa _2}$ , we conclude that $B"\in W$ . Also, $B"\subseteq B_0$ by clause $(1)$ .

We proceed by a density argument, recalling that by the definition of $G_2$ , we have that ${\langle } t_{<\kappa },t_\kappa {\rangle }\in G_\kappa *F_\kappa $ .

Claim 2.17. Let D be the set of all conditions $q\in \mathcal {P}_{\kappa +1}$ , such that exists $(\nu ^{\prime }_0,\nu _0),(\nu ^{\prime }_1,\nu _1)\in B', \ \nu ^{\prime }_1>\nu _0$ and a $\mathcal {P}_{\nu _0}$ -name such that:

1. (a) $r(\nu ^{\prime }_0,\nu _0),r(\nu ^{\prime }_1,\nu _1)\leq q$ .

2. (b) .

3. (c)

Then D is dense (open) above ${\langle } t_{<\kappa },t_\kappa {\rangle }$ and thus $D\cap G_\kappa *F_\kappa \neq \emptyset .$

Proof Work in V, and let ${\langle } t_{<\kappa },t_\kappa {\rangle }\leq p:={\langle } p_{<\kappa },p_\kappa {\rangle } \in {\mathcal P}_{\kappa +1}$ . We will define two extensions $p\leq q\leq q^*$ which corresponds to the choice of $(\nu ^{\prime }_0,\nu _0),(\nu ^{\prime }_1,\nu _1)$ such that $q^*\in D$ . By definition of $\mathcal {P}_{\kappa +1}$ , $p_{<\kappa }\Vdash p_\kappa \in \operatorname {\mathrm {Cohen}}(\kappa ,\kappa ^+)$ , by $\kappa -$ cc of $\mathcal {P}_\kappa $ , for some $Z\subseteq \kappa ^+,\ Z\in V$ , $|Z|<\kappa $ and some $\gamma <\kappa $ , $p_{<\kappa }\Vdash \operatorname {\mathrm {dom}}(p_\kappa )\subseteq \gamma \times Z$ . Applying $j_2$ , we have that

$$ \begin{align*}j_2(p_{<\kappa})=p_{<\kappa}\Vdash\operatorname{\mathrm{dom}}(j_2(p_\kappa))\subseteq j_2(\gamma\times Z)= \gamma\times j_2^{\prime\prime}Z\text{ and } j_2(p_{\kappa})_{j_2(\alpha)}=p_{\kappa,\alpha}\geq t_{\kappa,\alpha}.\end{align*} $$

Combining with $(2c)$ , we have both

$$ \begin{align*}p_{<\kappa}\Vdash Z\supseteq \operatorname{\mathrm{Supp}}(t_\kappa)\wedge \forall \beta\in Z.j_2(p_{\kappa})_{j_2(\beta)}\geq t_{\kappa,\beta},\end{align*} $$

To reflect this, denote $\mu =(j_2\restriction (Z\cup \theta ))^{-1}\in M_2$ , then

$$ \begin{align*}\operatorname{\mathrm{dom}}(\mu)=j_2(Z)\cup j_2^{\prime\prime}\theta, \ \operatorname{\mathrm{rng}}(\mu)=Z\cup\theta, \ \mu\text{ is 1--1},\end{align*} $$

and we can reformulate

$$ \begin{align*}p_{<\kappa}\Vdash \mu"j_2(Z)\supseteq \operatorname{\mathrm{Supp}}(t_\kappa)\wedge \forall \beta\in j_2(Z).j_2(p_{\kappa})_{\beta}\geq t_{\kappa,\mu(\beta)},\end{align*} $$

Also, since we can find $\delta <\kappa $ such that $t_{<\kappa }\Vdash \phi "(\delta ,\kappa )\cap Z=\emptyset .$ There exists such $\delta $ since $|Z|<\kappa $ , $t_{<\kappa }\Vdash |\operatorname {\mathrm {Supp}}(t_{\kappa })|<\kappa $ and by $\kappa $ -cc of $\mathcal {P}_\kappa $ . Recall that by the definition of $\mu _1$ , $\phi "(\delta ,\kappa )=\mu _1^{\prime \prime }\{s_\gamma (\kappa _1)\mid \gamma \in j_2(\phi )"(\delta ,\kappa )\}$ and that $\mu "\operatorname {\mathrm {Supp}}(j_2(p_\kappa ))=Z$ . Therefore in $M_2$ we will have that

$$ \begin{align*}p_{<\kappa}\Vdash [\mu_1^{\prime\prime}\{s_\gamma(\kappa_1)\mid \gamma\in j_2(\phi)"(\delta,\kappa)\}]\cap[ \mu"\operatorname{\mathrm{Supp}}(j_2(p_\kappa))]=\emptyset.\end{align*} $$

Let $\Gamma $ be such that $j_2(\Gamma )(\kappa ,\kappa _1)=\mu $ , and there is a set $\overline {B}_0\subseteq B'$ , $\overline {B}_0\in U^2$ such that for every $(\nu ',\nu )\in \overline {B}_0$ ,

$$ \begin{align*}(i) \ \ p_{<\kappa}\Vdash \Gamma(\nu',\nu)"Z\supseteq \operatorname{\mathrm{Supp}}(r_{\nu'})\wedge \forall\beta\in Z. p_{\kappa,\beta}\geq r_{\nu',\Gamma(\nu',\nu)(\beta)},\end{align*} $$

$$ \begin{align*}(iii) \ \ p_{<\kappa}\Vdash\Gamma_1(\nu',\nu)"\{s_\gamma(\nu)\mid \gamma\in \phi"(\delta,\nu')\}\cap [\Gamma(\nu',\nu)"\operatorname{\mathrm{Supp}}(p_\kappa)]=\emptyset.\end{align*} $$

Let us move to the choice of $(\nu ^{\prime }_0,\nu _0),(\nu ^{\prime }_1,\nu _1)$ . In $V[G_{\kappa }*F_{\kappa }]$ , there exists $(\nu ^0_0,\nu _0),(\nu ^0_1,\nu _1)\in \overline {B}_0$ such that $r(\nu ^0_0,\nu _0),r(\nu ^0_1,\nu _1)\in G_{\kappa }*F_{\kappa }$ (hence they are compatible) such that $\nu ^0_0>\delta ,\gamma ,\sup (\operatorname {\mathrm {Supp}}(p_{<\kappa }))$ and $\nu ^0_1>\nu _0,\operatorname {\mathrm {Supp}}(r_{<\kappa }(\nu ^0_0,\nu _0))$ . In particular, in V we can find $(\nu ^{\prime }_0,\nu _0),(\nu ^{\prime }_1,\nu _1)\in \overline {B}_0$ such that $r(\nu _0^{\prime },\nu _0),r(\nu _1^{\prime },\nu _1)$ are compatible, $\nu ^{\prime }_0>\delta ,\gamma ,\sup (\operatorname {\mathrm {Supp}}(p_{<\kappa }))$ , and $\nu ^{\prime }_1>\nu _0,\sup (\operatorname {\mathrm {Supp}}(r_{<\kappa }(\nu _0^{\prime },\nu _0)))$ . Denote

$$ \begin{align*}r^0:=r(\nu^{\prime}_0,\nu_0)={\langle} r^0_{<\nu^{\prime}_0},r^0_{\nu^{\prime}_0},r^0_{(\nu^{\prime}_0,\kappa)},r^0_\kappa{\rangle},\end{align*} $$

$$ \begin{align*}r^1:=r(\nu^{\prime}_1,\nu_1)=(r^1_{<\nu^{\prime}_1},r_{\nu^{\prime}_1},r_{(\nu^{\prime}_1,\nu_1)},{\langle} r^{0,1}_{\nu_1},r^{1,1}_{\nu_1}{\rangle},r^1_{(\nu_1,\kappa)},r^1_\kappa{\rangle}.\end{align*} $$

Let us define the first extension q, and it has the form:

$$ \begin{align*}q=p_{<\kappa}{}^{\smallfrown}q_{\nu^{\prime}_0}{}^{\smallfrown}r^0_{(\nu^{\prime}_0,\kappa)}{}^{\smallfrown}q_\kappa.\end{align*} $$

First, $q_{\nu ^{\prime }_0}$ is a $\mathcal {P}_{\nu ^{\prime }_0}$ -name for a condition with $\operatorname {\mathrm {Supp}}(q_{\nu ^{\prime }_0})= \Gamma (\nu ^{\prime }_0,\nu _0)"Z$ , by $(i) \ \operatorname {\mathrm {Supp}}(q_{\nu ^{\prime }_0})\supseteq \operatorname {\mathrm {Supp}}(r^0_{\nu ^{\prime }_0})$ . Set $q_{\nu ^{\prime }_0,\Gamma (\nu ^{\prime }_0,\nu _0)(\beta )}= p_{\kappa ,\beta }$ . As for $q_\kappa $ , we set it to be a $\mathcal {P}_\kappa $ -name for $r^0_\kappa \cup p_\kappa $ .

Once we will prove that $p_{<\kappa },r^0_{<\kappa }\leq q_{<\kappa }$ , from $(i)$ and $(ii)$ it will follow that $q_{<\kappa }$ forces $q_\kappa $ to be a partial function. Indeed, for every $\beta \in \operatorname {\mathrm {Supp}}(r^0_\kappa )\cap Z$ , $q_{<\kappa }$ will force

Clearly $p\leq q$ . To see that $r^0\leq q$ , up to $\nu ^{\prime }_0$ , we have that by $(4)_{(\nu ^{\prime }_0,\nu _0)}$ that

$$ \begin{align*}q_{<\nu^{\prime}_0}=p_{<\kappa}\geq t_{<\kappa}=r^0_{<\nu^{\prime}_0}.\end{align*} $$

At $\nu ^{\prime }_0$ , if $\alpha = \Gamma (\nu ^{\prime }_0,\nu _0)(\beta )$ , then $(i)$ insures that $q_{\nu ^{\prime }_0,\alpha }=p_{\kappa ,\beta }\geq r^0_{\nu ',\alpha }$ . Since in the interval $(\nu ^{\prime }_0,\kappa )$ , q and $r^0$ are the same, it follows that $q_{<\kappa }\geq r^0_{<\kappa }$ and at $\kappa $ it is clear that $q_{<\kappa }\Vdash r^0_\kappa \leq q_\kappa $ .

Next let us move to the choice of . Since $r^0\leq q$ and , use the maximality principal to find a $\mathcal {P}_{\nu _0}$ -name, such that q forces $(b)$ .Footnote ⁵

Define the final condition $q\leq q^*$ ,

$$ \begin{align*}q^*=q_{<\kappa}{}^{\smallfrown}q^*_{\nu^{\prime}_1}{}^{\smallfrown}r^1_{(\nu_1^{\prime},\kappa)}{}^{\smallfrown}q^*_\kappa.\end{align*} $$

The crucial point here is that by $(2b)_{(\nu ^{\prime }_1,\nu _1)}$ ,

and since

we have that $r^0\Vdash X_1^{\nu _0}\subseteq \phi "(\nu ^{\prime }_0,\nu _0)\subseteq \phi "(\nu ^{\prime }_0,\nu _1^{\prime })$ . By $(iii)$ we have that $q_{<\kappa }\Vdash [\Gamma _1(\nu ^{\prime }_1,\nu _1)"\{s_\gamma (\nu _1)\mid \gamma \in X^{\nu _0}_1\}]\cap [\Gamma (\nu _1^{\prime },\nu _1)"Z]=\emptyset $ . This will permit to code $d^{\nu _0}$ , let

$$ \begin{align*}\operatorname{\mathrm{Supp}}(q^*_{\nu^{\prime}_1})=[\Gamma_1(\nu^{\prime}_1,\nu_1)"\{s_\gamma(\nu_1)\mid \gamma\in X^{\nu_0}_1\}]\uplus [\Gamma(\nu_1^{\prime},\nu_1)"Z]\end{align*} $$

and

and $q^*_\kappa =q_\kappa \cup r^1_\kappa $ . Note that if $\tau \in \operatorname {\mathrm {Supp}}(q_{\kappa })\cap \operatorname {\mathrm {Supp}}(r^1_{\kappa })$ then either $\tau \in \operatorname {\mathrm {Supp}}(r^0_{\kappa })\cap \operatorname {\mathrm {Supp}}(r^1_{\kappa })$ , and $r^0_{\kappa },r^1_{\kappa }$ are forced to be compatible by $q_{<\kappa }$ and if $\tau \in Z\cap \operatorname {\mathrm {Supp}}(r^1_{\kappa })$ then the same argument as before works. We conclude that $r^0\leq q\leq q^*$ , $r^1\leq q^*$ , namely $(a)$ . Finally, for every $\tau \in X^{\nu _0}_1$ , $s_\tau (\nu _1)\in \operatorname {\mathrm {dom}}(\Gamma _1(\nu ^{\prime }_1,\nu _1))$ and by $(3)_{(\nu ^{\prime }_1,\nu _1)}$ we have that $q^*$ forces that

Then $p\leq q^*$ and $q^*\in D.$

By density, we can find such a condition $p^*\in G_\kappa *F_\kappa \cap D$ and points $(\nu ^{\prime }_0,\nu _0),(\nu _1^{\prime },\nu _1)\in B'$ witnessing $p^*\in D$ . It follows that $r(\nu ^{\prime }_0,\nu _0),r(\nu ^{\prime }_1,\nu _1)\in G_\kappa *F_\kappa $ , and by $(1)_{(\nu ^{\prime }_0,\nu _0)},(1)_{(\nu ^{\prime }_1,\nu _1)}$ , $\nu _0,\nu _1\in B_0$ . Extend ${\langle } {\langle }{\rangle },B{\rangle }$ by $p^*={\langle } \nu _0,\nu _1,B_0\cap (\cap _{\tau \in \phi "\nu _0}A_\tau {\rangle }\setminus \nu _1+1{\rangle }$ . By $(2b)_{(\nu ^{\prime }_1,\nu _1)}$ , for every $\tau \in \phi "\nu _0\subseteq \phi "\nu ^{\prime }_1$ , $f_{\kappa ,\tau }(\nu _1)=r_{\kappa ,\tau }(\nu _1)=1$ , hence $\nu _1\in \cap _{\tau \in \phi "\nu _0}A_\tau $ and $p^*\Vdash n_\tau =\begin {cases} 0, &\tau \in X_0,\\ 1, & \tau \in X_1.\end {cases}$ In other words, since $\nu _0\in B_0$ ,

Let

, it follows that

extends $d_\tau $ , and by $(c)$ of the definition of D,

extends $d_\tau $ . Thus

. This concludes the genericity proof.

Proof Let E be a $(\kappa ,\kappa ^{++})$ -extender. Let $j_1=j_E:V\to M_E=:M_1$ be its ultrapower embedding with $crit(j_E)=\kappa $ and ${}^{\kappa } M_E\subseteq M_E$ . Denote by $E_\alpha $ the ultrafilter

$$ \begin{align*}E_\alpha:=\{X\subseteq \kappa\mid \alpha\in j_E(X)\}.\end{align*} $$

Denote $U:=E_\kappa $ the normal ultrafilter and let $k:M_U \to M_E$ be the factor map defined by setting $k(j_U(f)(\kappa ))=j_E(f)(\kappa )$ such that $j_E=k\circ j_U$ . Define an Easton support iteration as follows: