Proof. For every $\theta < \delta $ let $S_\theta = \{ \sup (X) : X \in B \text { and } |\mathcal {A}_X| < \theta \}$ . We want to show that there is some $\theta $ such that $\sup S_\theta = \lambda $ . Since $M_U \models |\mathcal {A}| < j_U(\delta )$ , by Łoś we have $Y := \{X \in B : |\mathcal {A}_X| < \delta \} \in U$ . Note that

$$ \begin{align*}Y = \bigcup_{\theta \leq \delta} \{X \in B : |\mathcal{A}_X| < \theta \}.\end{align*} $$

Suppose now towards a contradiction that $\sup S_\theta < \lambda $ for all $\theta < \mathrm {cf}(\lambda )$ . It follows that

$$ \begin{align*}\zeta := \sup\{\sup(X) : X \in Y\} < \lambda\ \ \text{ (recall }\mathrm{cf}(\lambda)> \delta).\end{align*} $$

But then U concentrates on $P_\kappa (\zeta )$ , contradicting our assumption.

Proof. Let $\mathcal {A} = \{j_U(Y) \cap (\sup j_U[\lambda ]) : Y \subseteq \lambda \}$ . Note that $M_U$ is only closed under $\lambda $ -sequences and therefore it is not immediate that $\mathcal {A}\in M_U$ . However, since $2^{<\lambda }=\lambda $ , by Lemma $4.3.4$ in [Reference Goldberg22], we do get $\mathcal {A} \in M_U$ . For completeness, we will sketch the proof of this lemma. Since $2^{<\lambda } = \lambda $ and $M_U$ is closed under $\lambda $ sequences, $j_U[P_\lambda (\lambda )] \in M_U$ . Also $j_U \upharpoonright P_\lambda (\lambda ) \in M_U$ since this is the inverse of the transitive collapse of $j_U[P_\lambda (\lambda )]$ . By supercompactness, $P(\lambda ) \subseteq M_U$ , so $j_U \upharpoonright P_\lambda (\lambda )^M = j_U \upharpoonright P_\lambda (\lambda )$ . Since $j_U(X) \cap \sup j_U[\lambda ] = \bigcup _{\xi < \lambda } j_U(X \cap \xi )$ we can define $\mathcal {A}$ from $j_U \upharpoonright P_\lambda (\lambda )$ .

Let $f \colon P_\kappa (\lambda ) \to P(P(\lambda ))$ represent $\mathcal {A}$ in $M_U$ , so that $j_U(f)(j_U[\lambda ]) = \mathcal {A}$ . For each $X \in P_\kappa (\lambda )$ let $\mathcal {A}_X = f(X)$ . We may assume that for every X, $f(X) \subseteq P(\sup (X))$ . For every $Y \subseteq \lambda $ , define

(4) $$ \begin{align} B_Y := \{X \in P_\kappa(\lambda) : Y \cap \sup(X) \in \mathcal{A}_X\}, \end{align} $$

then $B_Y \in U$ as $j_U[\lambda ]\in j_U(B_Y)$ by definition of $\mathcal {A}$ . We claim that $\{B_Y : Y \subseteq \lambda \}$ witnesses $\lnot \operatorname {\mathrm {Gal}}(U,\kappa ,2^{\lambda })$ . Otherwise, there is a sequence of distinct $\langle Y_i \subseteq \lambda : i < \kappa \rangle $ so that $B = \bigcap _{i<\kappa } B_{Y_i} \in U$ .

Note that since $\lambda <j_U(\kappa )$ , and $j_U(\kappa )$ is measurable in $M_U$ , $M_U\models |\mathcal {A}|=2^\lambda <j_U(\kappa )$ so by Lemma 3.4, we can fix $\theta _0<\kappa $ such that $S_{\theta _0}$ is unbounded in $\lambda $ . For $i\neq j < \theta _0^+$ let $\beta _{i,j}$ be least such that $Y_i \setminus \beta _{i,j} \neq Y_j \setminus \beta _{i,j}$ . Since $\mathrm {cf}(\lambda )>\kappa >\theta _0$ , $ \beta ^* := \sup _{i\neq j<\theta _0^+}(\beta _{i,j})<\lambda $ , and by our choice of $\theta _0$ , there is an $X^* \in B$ with $|\mathcal {A}_{X^*}| < \theta _0$ and $\sup (X^*) \geq \beta ^*$ . Then the map $i \mapsto Y_i \cap \sup (X^*)$ is now a one-to-one map from $\theta _0^+$ into $\mathcal {A}_{X^*}$ , a contradiction.

Proof. For each $\alpha $ with $\kappa \leq \alpha < \lambda $ , let $f_\alpha \colon \lambda \to P(\alpha ) $ be a surjection (here we use $2^{<\lambda } = \lambda $ ). Since U is fine, $\operatorname {\mathrm {id}}_U \supseteq j_U[\lambda ]$ . It follows that $j_U(f_\alpha )[\operatorname {\mathrm {id}}_U ]\supseteq j_U[P(\alpha )]$ .

Now let $\langle f^*_\alpha : \alpha < j_U(\lambda ) \rangle = j_U(\langle f_\alpha : \alpha < \lambda \rangle ),$ and set $X_\alpha = f^*_\alpha {} [\operatorname {\mathrm {id}}_U].$ Notice ${\langle } X_\alpha : \alpha < j_U(\lambda ) {\rangle }$ is in $M_U$ : $X_\alpha \subseteq P(\alpha )^{M_U} \in M_U$ and $\operatorname {\mathrm {id}}_U \in M_U$ , and the sequence is definable from $\operatorname {\mathrm {id}}_U $ and $\langle f^*_\alpha : \alpha < j_U(\lambda ) \rangle $ . Again, since $\operatorname {\mathrm {id}}_U \supseteq j_U[\lambda ]$ it follows that $j_U[P(\alpha )] \subseteq X_{j_U(\alpha )}$ .

Let $\lambda ^* = \sup j_U[\lambda ]$ . In $M_U$ , define the following set:

(5) $$ \begin{align} \mathcal{A} = \{ Y \subseteq \lambda^* : Y \cap \alpha \in X_\alpha \text{ for unboundedly many } \alpha \in \operatorname{\mathrm{id}}_U\cap \lambda^*\}. \end{align} $$

Working in $M_U$ , we have that $|\operatorname {\mathrm {id}}_U|<j_U(\kappa )$ and $j_U(\kappa )$ is strong limit. Therefore,

$$ \begin{align*}|\mathcal{A}| \leq |\prod_{\alpha\in \operatorname{\mathrm{id}}_U\cap \lambda^*} X_\alpha|\leq |\operatorname{\mathrm{id}}_U|^{|\operatorname{\mathrm{id}}_U\cap\lambda^*|}< j_U(\kappa).\end{align*} $$

By definition of $\mathcal {A}$ and elementarity we have $j_U(Y) \cap \lambda ^* \in \mathcal {A}$ for all $Y \subseteq \lambda $ . To see this, we will prove that for all $\alpha \in j_U[\lambda ]$ we have $j_U(Y) \cap \lambda ^* \cap \alpha \in X_\alpha $ . Note that if $\alpha = j_U(\beta )$ for some $\beta < \lambda $ then

$$ \begin{align*}j_U(Y) \cap j_U(\beta) = j_U(Y \cap \beta) \in j_U[P(\beta)] \subseteq X_\beta.\end{align*} $$

Now let $f \colon P_\kappa (\lambda ) \to P(P(\lambda ^*))$ represent $\mathcal {A}$ in the ultrapower so that

$$ \begin{align*}j_U(f)(\operatorname{\mathrm{id}}_U) = \mathcal{A}.\end{align*} $$

For each $X \in P_\kappa (\lambda )$ let $\mathcal {A}_X = f(X)$ . Let g represent $\lambda ^*$ in $M_U$ , so $j_U(g)(\operatorname {\mathrm {id}}) = \lambda ^*$ . By the previous paragraph, $j_U(Y) \cap \sup (\operatorname {\mathrm {id}}_U) \in \mathcal {A} = j_U(f)(\operatorname {\mathrm {id}})$ . Reflecting this we have

(6) $$ \begin{align} \forall Y \subseteq \lambda, \, B_Y := \{X \in P_\kappa(\lambda) : Y \cap g(X) \in \mathcal{A}_X\} \in U. \end{align} $$

This is true as $\operatorname {\mathrm {id}}_U \in j_U(B_Y)$ by elementarity and the definition of $\mathcal {A}$ . We claim that $\{B_Y : Y \subseteq \lambda \}$ witnesses $\lnot \operatorname {\mathrm {Gal}}(U,\kappa ,2^{\lambda })$ . Otherwise there is $\langle Y_i \subseteq \lambda : i < \kappa \rangle $ so that $B = \bigcap _{i<\kappa } B_{Y_i} \in U$ .

By Lemma 3.4, fix $\theta _0<\kappa $ such that $\{\sup (X) : X\in B \text { and } |\mathcal {A}_X| \leq \theta _0\}$ is unbounded in $\lambda $ . Now for $i\neq j < \theta _0^+$ let $\beta _{i,j}$ be least such that

$$ \begin{align*}Y_i \setminus \beta_{i,j} \neq Y_j \setminus \beta_{i,j}.\end{align*} $$

Now take $\alpha ^*> \sup _{i\neq j<\theta _0^+}(\beta _{i,j})$ below $\lambda $ such that there is an $X^* \in B$ with $|\mathcal {A}_{X^*}| \leq \theta _0$ and $g(X^*) = \alpha ^*$ . Then the map $i \mapsto Y_i \cap g(X^*)$ is a 1–1 map from $\theta _0^+$ into $\mathcal {A}_{X^*}$ , a contradiction.

Proof. By Corollary 2.8, there is fine ultrafilter W on $P_\kappa (\lambda )$ such that $W\leq _{RK} U$ . Also, since U is $\sigma $ -complete so is W. By Theorem 3.8, $\neg \operatorname {\mathrm {Gal}}(W,\kappa ,2^\lambda )$ . By Lemma 3.3, also $\neg \operatorname {\mathrm {Gal}}(U,\kappa ,2^\lambda )$ .

Proof. Applying Theorem 3.13, U is Rudin–Keisler equivalent to an isonormal ultrafilter W on $|\theta | \geq \lambda $ , where $\theta $ is as in the previous remark. Note that $M_U$ and $M_W$ are in fact closed under $\lambda ^{<\kappa }$ -sequences and in this case $\lambda ^{<\kappa } \geq \lambda ^+$ . Hence we may derive a normal, fine ultrafilter Z on $P_\kappa (\lambda ^+)$ from $j_W$ using $j_W[\lambda ^+]$ . Now Theorem 3.6 applies to Z and $Z \leq _{RK} W \cong _{RK}U$ , so by Lemma 3.3 we may conclude $\lnot \operatorname {\mathrm {Gal}}(U,\kappa ,2^{\lambda ^+})$ .

Proof. By Theorem $4.4.37$ in [Reference Goldberg22], there is a unique weakly normal ultrafilter W on $\lambda $ such that $W \cong _{RK} U$ . Furthermore, $M_W$ is closed under $\lambda $ -sequences. We will show that $\operatorname {\mathrm {Gal}}(W, \kappa , \lambda ^+)$ holds. Let ${\langle } X_i : i < \lambda ^+ {\rangle } \subseteq W$ . For $\alpha < \lambda ^+$ and $\xi < \lambda $ let

(7) $$ \begin{align} H_{\alpha, \xi} = \{ i < \lambda^+ : X_i \cap \xi = X_\alpha \cap \xi \}. \end{align} $$

Let $\alpha ^*$ be as in the claim. Fix $f : \lambda \to \kappa $ so that $[f]_W = \kappa $ . Note that we may take f to be monotone. To see this, let $g \colon \kappa \to \lambda $ be increasing, continuous, and cofinal, and assume the range of g consists only of cardinals. For every $\beta < \lambda $ let $f(\beta ) = \alpha $ where $\alpha $ is least such that $g(\alpha )> \beta $ . Then f is clearly weakly monotone. Now we check that $[f]_W = \kappa $ . As mentioned in Remark 3.14, there is some ordinal $\theta $ strictly between $\sup j_U[\lambda ]$ and $j(\lambda )$ such that

(8) $$ \begin{align} X \in W \iff \theta \in j_U(X). \end{align} $$

So we shall show that $j_W(f)([\operatorname {\mathrm {id}}]_W) = j_U(f)(\theta ) = \kappa $ . If $\alpha < \kappa $ then $j_U(g)(\alpha ) = j_U(g(\alpha )) < \sup j_U[\lambda ] < \theta $ . Since g is continuous, $\sup j_U[\lambda ] = j_U(g)(\kappa )$ . Since $g(\alpha )$ is always a cardinal, $\theta < |\sup j_U[\lambda ]|^{+M_U} \leq j(g)(\kappa +1)$ . Hence $j_U(f)(\theta ) = \kappa $ , as desired.

Now for all $i < \kappa $ choose $\beta _i \in H_{\alpha ^*,g(i)+1} \setminus \{\beta _j \mid j < i\}$ for $i < \kappa $ . We define the f-diagonal intersection:

(9) $$ \begin{align} D = \Delta^f_{i < \kappa} X_{\beta_i} = \{ \eta < \lambda \mid \forall i < f(\eta) \, (\eta \in X_{\beta_i})\}. \end{align} $$

Note that $j_W(\vec {X})_i= j_W(X_{\beta _i})$ for all $i < \kappa $ and $j_W(f)([\operatorname {\mathrm {id}}]_W) = \kappa $ , so it follows that W is closed under f-diagonal intersections. Now it suffices to show that

$$ \begin{align*}X_{\alpha^*} \cap D \subseteq \bigcap_{i < \kappa} X_{\beta_i}.\end{align*} $$

Towards this end suppose $\zeta \in X_{\alpha ^*} \cap D$ . Fix $i < \kappa $ . If $i < f(\zeta )$ then $\zeta \in X_{\beta _i}$ by definition of D. If $i \geq f(\zeta )$ , since $\zeta \in X_{\alpha ^*}$ then $X_{\beta _i} \cap \lambda _i+1 = X_{\alpha ^*} \cap \lambda _i+1$ , so $\zeta \in X_{\beta _i}$ .

Proof. Let W be the isonormal ultrafilter on $\lambda $ that is RK-equivalent to U. Let $\lambda ^* := \sup (j_U[\lambda ])$ and $\mathcal {A} = {\langle } j_U(S) \cap \lambda ^* \mid S \subseteq \lambda {\rangle }$ . As in Theorem 3.6, $\mathcal {A}\in M_W$ so we may let $\mathcal {A} = [\beta \mapsto \mathcal {A}_\beta ]_W$ . We may assume that for every $\beta <\lambda $ , $|\mathcal {A}_\beta |<\kappa $ . Let ${\langle } X_\alpha \mid \alpha < 2^\lambda {\rangle } $ be a one-to-one enumeration of subsets of $\lambda $ . For $\alpha < \lambda $ let

(10) $$ \begin{align} B_\alpha = \{ \beta < \lambda : X_\alpha \cap \beta \in \mathcal{A}_\beta\}. \end{align} $$

Suppose towards a contradiction that there is some $I \subseteq 2^\lambda $ such that $|I| = (2^\kappa )^+$ and $B := \bigcap _{\alpha \in I} B_\alpha \in U$ . Let $C \subseteq B$ be cofinal in $\lambda $ with $ot(C) = \kappa $ . Now consider

$$ \begin{align*} f \colon \,& I \to \prod_{\beta \in C} \mathcal{A}_\beta\\ &\alpha \mapsto {\langle} X_\alpha \cap \beta \mid \beta \in C {\rangle.} \end{align*} $$

Then f is a well-defined injection by the definition of B but $|I| = (2^\kappa )^+$ and the above product has size $2^\kappa $ , a contradiction.

Question 3.19. Let U be a normal, fine ultrafilter on $P_\kappa (\lambda )$ , where $\mathrm {cf}(\lambda ) = \kappa $ . Must $\operatorname {\mathrm {Gal}}(U,2^\kappa ,2^\lambda )$ hold? Must $\operatorname {\mathrm {Gal}}(U,2^\kappa ,2^\lambda )$ fail?

Question 3.23. Is it consistent to have a normal $\kappa ^+$ -dense ideal on $\kappa ^+$ where $\kappa $ is $\kappa ^+$ -supercompact?

Proof. Let U be a $\sigma $ -complete ultrafilter on $\kappa ^+$ . By Theorem $8.2.24$ in [Reference Goldberg22] there is an ordinal $\lambda < \kappa ^+$ and an ultrafilter D on $\lambda $ with $D \leq _{RK} U$ such that $j_D(U) = W$ is a $j_D(\kappa ^+)$ -irreducible ultrafilter and $j_U = j_W^{M_D} \circ j_D$ . By Theorem $8.2.22$ in [Reference Goldberg22] $j_W^{M_D}$ is a $j_D(\kappa ^+)$ supercompactness embedding from the perspective of $M_D$ . Hence $j_W^{M_D}[j_D(\kappa ^+)] \in M_U$ . Now define an ultrafilter F on $P_\kappa (\kappa ^+)$ by

(10) $$ \begin{align} X \in F \iff j_W^{M_D}[j_D(\kappa^+)] \in j_U(X) \end{align} $$

then F is $\sigma $ -complete. To see that F is fine, notice $j_D(\kappa ^+)> \kappa ^+$ we have $j_W^{M_D}[j_D(\kappa ^+)] \supseteq j_U[\kappa ^+]$ . Also $j_U(\kappa ) = j_W(j_D(\kappa ))> j_D(\kappa ^+)$ since $j_W(\kappa )> \kappa ^+$ . Hence

$$ \begin{align*}|j_W^{M_D}[j_D(\kappa^+)]|^{M_U} = |j_D(\kappa^+)|^{M_U} < j_U(\kappa).\end{align*} $$

Since $j_W^{M_D}[j_D(\kappa ^+)]$ covers $j_U[\kappa ^+]$ and has size less than $j_U(\kappa )$ we conclude that F is fine. Furthermore $F \leq _{RK} U$ since F was derived using $j_U$ .

Proof. By Theorem 3.6, in V there is ${\langle } A_\alpha : \alpha < 2^\lambda {\rangle } \subseteq U$ such that no intersection of $\kappa $ many $A_\alpha $ ’s lies in U. An easy density argument ensures that for every $\alpha <2^\lambda $ there is $n_\alpha <\omega $ such that for every $m>n_\alpha $ , $X_m\in A_\alpha $ . Since $\mathrm {cf}(2^\lambda )>\omega $ , there is $S\in [2^\lambda ]^{2^\lambda }$ , and $n^*<\omega $ such that for every $\alpha \in S$ , $n_\alpha =n^*$ . This means that $\{X_m\mid m<n^*\}\subseteq \bigcap _{\alpha \in S}A_\alpha $ . We will show that S contains no ground model set of size $\kappa $ . Otherwise $T \in [S]^\kappa $ in V, then by the failure of Galvin’s property, $\bigcap _{\alpha \in T} A_\alpha \notin U$ . On the other hand, $\{X_m \mid m>n^*\}\subseteq \bigcap _{\alpha \in T} A_\alpha $ . However, since $\kappa \setminus (\bigcap _{\alpha \in T} A_\alpha ) \in U$ , the same density argument as before yield that for some $n'<\omega $ , for every $m>n'$ , $X_m\notin \bigcap _{\alpha \in T} A_\alpha $ , this is a contradiction.

Proof. Let ${\langle } A_\alpha \mid \alpha <\lambda {\rangle }\subseteq U$ witness that $\neg \operatorname {\mathrm {Gal}}(U,\lambda ,\lambda )$ . Towards a contradiction, if $\chi (U)<\lambda $ , let ${\langle } X_\beta \mid \beta <\theta {\rangle }\subseteq U$ be a $\subseteq $ -base for U for some $\theta <\lambda $ . For each $\alpha <\lambda $ , we find $\beta _\alpha <\theta $ such that $X_{\beta _\alpha }\subseteq A_\alpha $ . By the regularity of $\lambda $ , we can find a single $\beta ^*$ ans $S\in [\lambda ]^\lambda $ such that for every $\alpha \in S$ , $\beta _\alpha =\beta ^*$ . It follows that $X_{\beta ^*}\subseteq \bigcap _{\alpha \in S}A_{\alpha }$ , contradicting that ${\langle } A_\alpha \mid \alpha <\lambda {\rangle }$ witnesses $\neg \operatorname {\mathrm {Gal}}(U,\lambda ,\lambda )$ .

Proof. Let $\bar {U}$ be the filter generated by U in $V[G]$ . Since $\text {Add}(\kappa ,\lambda ^{+\kappa ^+})$ is $\kappa $ -closed, $\bar {U}$ is a $\kappa $ -complete filter, and since $\kappa $ is indestructible, in $V[G]$ , we can extend $\bar {U}$ to a $\kappa $ -complete ultrafilter $U^*$ on $P_\lambda (\lambda ^+)$ . To see that $U^*$ remains fine, it suffices to use the fineness of U and note that $\{A\in P_\lambda (\lambda ^+)\mid \alpha \in A\}^V\subseteq \{A\in P_\lambda (\lambda ^+)\mid \alpha \in A\}^{V[G]}$ . It remains to see that $\operatorname {\mathrm {Gal}}(U^*,2^{\lambda ^+},2^{\lambda ^+})$ holds, since U is $\lambda $ -complete, it is in particular $\kappa ^+$ -complete, so we may apply Theorem 4.12, to conclude that the extension $U^*$ of U is generated by $\lambda ^{+\kappa ^+}$ -many sets. Also note that in $V[G]$ , $2^{\lambda ^+}=\lambda ^{+\kappa ^++1}$ . By Lemma 4.10, we conclude that $\operatorname {\mathrm {Gal}}(U^*,2^{\lambda ^+},2^{\lambda ^+})$ holds, as desired.

Proof. Let ${\langle } X_i : i < \chi _{\mathcal {F}}(U) {\rangle }$ be a $\subseteq _{\mathcal {F}}$ -base for U. By Lemma 4.8, $\chi _{\mathcal {F}}(U)=\chi (U) \geq \lambda ^+$ . By thinning out the $\subseteq _{\mathcal {F}}$ -base, we may assume that for any $j < \chi (U)$ , $X_j\notin \mathcal {F}[{\langle } X_i : i < j{\rangle }]$ , namely $X_i\not \subseteq _{\mathcal {F}} X_j$ for any $i<j$ . Now let ${\langle } \alpha _i\mid i<\mathrm {cf}(\chi (U)){\rangle }$ be increasing and cofinal in $\chi (U)$ . We claim that ${\langle } X_{\alpha _i}\mid i<\mathrm {cf}(\chi (U)){\rangle }$ witnesses that $\lnot \mathrm {Gal}^{*}(U,\lambda ,\lambda )$ . Otherwise, there is I unbounded in $\mathrm {cf}(\chi (U))$ such that $\{X_{\alpha _i}\mid i\in I\}$ has a $\subseteq _{\mathcal {F}}$ -lower bound $X\in U$ . Then there is $j< \mathrm {cf}(\chi (U))$ such that $X_j\subseteq _{\mathcal {F}} X$ . Since ${\langle } \alpha _i\mid i<\mathrm {cf}(\chi (U)){\rangle }$ is cofinal and I is unbounded, there is $i\in I$ such that $j<\alpha _i$ . But this is impossible since this would mean that $X_j\subseteq _{\mathcal {F}} X\subseteq _{\mathcal {F}} X_{\alpha _i}$ , contradicting the choice that $X_{\alpha _i}\notin \mathcal {F}[{\langle } X_j\mid j<\alpha _i{\rangle }]$ .

Proof. Let $\rho < \lambda ^+$ and let ${\langle } X_i : i <\rho {\rangle } \subseteq U$ . Let ${\langle } X_{i_\nu }\mid \nu <\lambda {\rangle }$ be a re-enumeration of order type $\lambda $ . Let

$$ \begin{align*}A = \Delta_{\nu < \lambda}X_{i_\nu} = \{x \in P_{\kappa}(\lambda) : x \in \bigcap_{\nu\in x}X_{i_\nu}\}.\end{align*} $$

Then $A \in U$ by normality. Furthermore, for each $i<\rho $ we find $\nu <\lambda $ such that $i=i_\nu $ . Then,

$$ \begin{align*}A \setminus X_i=A\setminus X_{i_\nu} \subseteq \{x : i_\nu \notin x\} \in \mathcal{F}^*.\\[-32pt]\end{align*} $$

Proof. Assume U is a $P_{\lambda ^+}$ -point and let ${\langle } X_i\mid i<\lambda {\rangle }\subseteq P(P_\kappa (\lambda )$ be such that for every $i<\lambda $ , $X_i\notin U$ . Then ${\langle } X_i^c\mid i<\lambda {\rangle }\subseteq U$ . By $(1)$ , there is $X\in U$ such that $X\subseteq _{\mathcal {F}} X_i^c$ for each i. This means that $X\cap X_i\in \mathcal {F}$ for every $i<\lambda $ . The other direction is similar.

Proof. For each $x\in P_\kappa (\lambda )$ and $\alpha <\lambda ^+$ let

$$ \begin{align*}H_{\alpha,x}=\{\beta<\lambda^+\mid A_\beta\cap P(x)=A_\alpha\cap P(x)\}.\end{align*} $$

Fix $\alpha ^*$ as in the claim above and let $H_x=H_{\alpha ^*,x}$ . For every $x\in P_\kappa (\lambda )$ , choose $\alpha _x\in H_{\alpha ^*,x}$ and let $A_x=A_{\alpha _x}$ . To see that ${\unicode{x29D6}}_{x\in P_\kappa (\lambda )}A_x\in U$ , we prove that $\operatorname {\mathrm {id}}_U=j_U[\lambda ]\in j_U({\unicode{x29D6}}_{x\in P_\kappa (\lambda )}A_x)=({\unicode{x29D6}}_{x\in P_{j(\kappa )}(j(\lambda ))}A^{\prime }_x))^{M_U}$ , where $j_U({\langle } A_x\mid x\in P_\kappa (\lambda ){\rangle })={\langle } A^{\prime }_x\mid x\in P_{j(\kappa )}(j(\lambda ))^{M_U}{\rangle }$ . Towards this, following the definition of the cone intersection, let $x\prec j_U[\lambda ]$ , then there is $y\in P_\kappa (\lambda )$ such that $j_U(y)=j_U[y]=x$ . In particular, $A^{\prime }_x=j_U(A_y)$ and since $A_y\in U$ , $\operatorname {\mathrm {id}}_U\in j_U(A_y)$ . For the second part of the definition of $({\unicode{x29D6}}_{x\in P_\kappa (\lambda )} A^{\prime }_x)^{M_U}$ , let $y\in (P_{j(\kappa )}(j(\lambda )))^{M_U}$ such that $\operatorname {\mathrm {id}}_U\subseteq y$ . By elementarity and the choice of the sequence ${\langle } \alpha _x\mid x\in P_\kappa (\lambda ){\rangle }$ , $M_U\models j_{U}(A_{\alpha ^*})\cap P(y)=A_{y}\cap P(y)$ . Since $\operatorname {\mathrm {id}}_U\subseteq y$ we conclude that $\operatorname {\mathrm {id}}_U\in j_U(A^*)\cap P(y)$ and therefore in $A_y$ . It follows that $ \operatorname {\mathrm {id}}_U\in j_U({\unicode{x29D6}}_{x\in P_{\kappa }(\lambda )}A_x)$ , as wanted.

Proof. The direction $(2) \Rightarrow (1)$ is trivial, since given the set produced by a strategy for $G_2^\gamma $ the Judge can play the ultrafilter determined by that set in $G_1^\gamma $ .

Given a $G^\gamma _1$ -winning strategy $\sigma $ , we will define a $G^\gamma _2$ -winning strategy $\sigma '$ . For the first move, the Challenger plays an ordinal $\alpha _0$ and $\sigma $ will yield a normal $N_{\alpha _0}+1$ ultrafilter $U_0$ . To determine the move $Y_0$ given by $\sigma '$ , choose $\beta $ large enough that $U_0 \in N_\beta $ and choose a minimal enumeration of $U_0 = \{X_\xi : \xi < \lambda \}$ is the well-ordering of $H_\theta $ . By elementarity, and since $U_0\in N_\beta $ , this enumeration is in $N_\beta $ . Then set $Y_0 = \Delta _{\xi < \lambda } X_\xi $ . Note that any legal move $\alpha _1$ of the challenger will have to satisfy that $Y_0\in N_{\alpha _1}$ and also $U_0\in N_{\alpha _1}$ . Again by elementarity, we will have that the minimal enumeration $U_0 = \{X_\xi : \xi < \lambda \}$ will also be in $N_{\alpha _1}$ , so whenever $U_0\subseteq U_1$ is a normal $N_{\alpha _1}$ -ultrafilter, $Y_0\in U_1$ .

Now suppose that ${\langle }(\alpha _j,Y_j)\mid j<i{\rangle }$ is a run in the game according to $\sigma '$ which has already been defined. Also assume that we maintained an auxiliary $G^\gamma _1$ -run ${\langle } (\alpha _j,U_j)\mid j<i{\rangle }$ , such that each $Y_j$ is the diagonal intersection of $U_j$ according to the minimal enumeration of $U_j$ in the $H_\theta $ well-ordering.

Now suppose that the challenger in the game $G^\gamma _2$ plays a legal move $\alpha _i$ . Note that ${\langle } (U_j,Y_j)\mid j<i{\rangle } \in N_{\alpha _i+1}$ , and that for each $j<i$ , the minimal enumeration of $U_j$ is a member of $N_{\alpha _i+1}$ . Next, we let the challenger in the game $G^\gamma _1$ play $\alpha _i$ , and the strategy $\sigma '$ produces an $N_{\alpha _i+1}$ -normal ultrafilter $U_i$ which extends all the $U_j$ ’s for $j<i$ . It follows that $Y_j\in U_i$ for each $j<i$ . Let $U_i=\{X^i_\xi \mid \xi <\lambda \}$ be the minimal enumeration of $U_i$ , we set $Y_i=\Delta _{\xi < \lambda }X^i_\xi $ . It is routine to check that $\sigma $ is a winning strategy in the game $G^\gamma _2$ .

Proof. Given a winning strategy $\sigma $ for the Judge, we define the hopeless ideal $I(\sigma )$ as the collection of all $X \subseteq P_\kappa (\lambda )$ such that no play of the game played according to $\sigma $ ends with a $Y \subseteq _{\mathcal {F}} X$ . In other words, no ultrafilter generated in any play according to $\sigma $ assigns measure one to X.

More precisely, for any $\ell \leq \gamma $ we recursively define a play of $G^\ell _2$ according to $\sigma $ as a sequence $P = {\langle } (\alpha _i, Y_i) : \sigma ((P \upharpoonright i)^{\smallfrown }\alpha _i) = Y_i \text { for } i < \ell {\rangle }$ . Now define the hopeless ideal:

(13) $$ \begin{align} I(\sigma) = \{ X \subseteq P_\kappa(\lambda) : \text{ for any play } P \text{ and any } \alpha, \sigma(P{}^\smallfrown \alpha) \not\subseteq_{\mathcal{F}} X \}. \end{align} $$

For two plays P and Q, we say $P \sqsubseteq Q$ iff there is a sequence of valid moves $\vec {M}$ such that $P{}^{\smallfrown } \vec {M} = Q$ . We also define the conditional hopeless ideal given some play P of length $\ell < \gamma $ :

(14) $$ \begin{align} I(\sigma,P) = \{ X \subseteq P_\kappa(\lambda) : \text{for any play } Q \sqsupseteq P \text{ and any } \alpha, \sigma(Q{}^\smallfrown\alpha) \not\subseteq_{\mathcal{F}} X \}. \end{align} $$

To construct the desired ideal, we will first take an arbitrary winning strategy $\sigma $ for the Judge in $G^\gamma _2$ and build the tree $T(\sigma )$ together with a correspondence taking sets X in $T(\sigma )$ to plays of the game $R_X$ . We will then use $T(\sigma )$ to construct a new strategy $\sigma '$ for which $T(\sigma )$ will witness the required property $(3)$ of $I(\sigma ')$ .

The construction of $T(\sigma )$ and the assignment $X\mapsto R_X$ will go by induction on the levels of $T(\sigma )$ . We wish to maintain the following in our induction:

1. 1. $R_X$ has successor length and ends with X as the last move played by the Judge.

2. 2. The tree order is $\subseteq _{\mathcal {F}}$ restricted to the nodes of the tree.

3. 3. If $X \supseteq _{\mathcal {F}}Y$ are nodes in $T(\sigma )$ then $R_X \sqsubseteq R_Y$ .

4. 4. If $X,Y$ are on the same level of the tree then $X \cap Y \in \mathcal {F}^*$ (here $\mathcal {F}^*$ is the ideal dual to $\mathcal {F})$ .

For $\delta $ limit we will take $T(\sigma ) \upharpoonright \delta = \bigcup _{\alpha < \delta } (T(\sigma ) \upharpoonright \alpha )$ and $R_X$ will be already be defined for every $X \in T(\sigma ) \upharpoonright \delta $ . The induction hypothesis is trivially maintained. So we shall focus on the successor stage of the construction.

Let $\delta < \gamma $ and suppose we have constructed every level up to $\delta $ . Let b be a cofinal branch of the tree constructed thus far. Let $P_b=\bigcup _{X\in b}R_X$ be the play of the game following each of the plays encoded by b. Then by $(2)$ , $P_b$ is a play according to $\sigma $ . Since we assume that there is no saturated ideal, we can choose some antichain $\mathcal {A}_b \subseteq I(\sigma ,P_b)^+$ with $|\mathcal {A}| = \lambda ^+$ . For each $A \in \mathcal {A}_b$ let $Q_A$ be a play extending $P_b$ such that if $Y_A$ is the last move of the Judge in $Q_A$ then $Y_A \subseteq _{\mathcal {F}} A$ . Such a play exists by the definition of $I(\sigma ,P_b)$ .

Now we construct notes at the $\delta ^{\text {th}}$ -level, which will all have b as their set of predecessors in the tree. This is again by recursion. Suppose we have constructed the successors ${\langle } Y_\xi : \xi < \bar {\xi }{\rangle }$ . Then each $Y_\xi $ corresponds to a play $R_{Y_\xi }$ where the last move of the Judge is $Y_\xi $ . Let $\alpha (Y_\xi )$ denote the last move of the Challenger in $R_{Y_\xi }$ . Since each $N_\beta $ has size $\lambda $ we may choose some $A \in \mathcal {A}_b$ which is not an element of any $N_{\alpha (Y_\xi )+1}$ for any $\xi < \bar {\xi }$ . Now let $\alpha $ be least such that A and every $Y_\xi $ is in $N_{\alpha +1}$ . Then $\alpha $ is a legal move following $Q_A$ , so we may let $Y_{\bar {\xi }}$ be the Judge’s response to $Q_A {}^\frown {\langle } \alpha {\rangle }$ given by $\sigma $ . Notice that $Y_\xi \subseteq _{\mathcal {F}} A$ as A was given by a previous move of the Judge. Also, set $R_{Y_{\bar {\xi }}} = Q_A {}^\frown (\alpha , Y_{\bar {\xi }})$ .

Now we shall show that the three conditions of the inductive hypothesis are maintained. The first is easily seen to be satisfied, so towards verifying the second condition, let X be any node in $T(\sigma )$ and $Y_\xi $ any node on the newly constructed $\delta ^{\text {th}}$ level of the tree. If X is a predecessor of $Y_\xi $ is the tree order, then X lies on the branch b, and therefore $Y_\xi \subseteq _{\mathcal {F}} X$ as $Y_\xi $ was given by a move of the Judge to a run of the game where X was played. In the other direction, if X is not in b, then either $X=Y_{\bar {\xi }}$ in which case $Y_\xi \not \subseteq _{\mathcal {F}} X$ , or the predecessors of X and $Y_\xi $ in the tree are different. Applying $(4)$ of the induction hypothesis to the level where X and $Y_\xi $ split, together with $(2)$ , we see that $Y_\xi \not \subseteq _{\mathcal {F}} X$ .

The third condition now follows from the second. If $X\subseteq Y_\xi $ , then X must be on the branch b leading to $Y_\xi $ . But then by definition,

$$ \begin{align*}R_X\sqsubseteq b\sqsubseteq Q_A\sqsubseteq R_{Y_\xi},\end{align*} $$

where $A\in \mathcal {A}_b$ was the set used in the construction of $Y_\xi $ .

Finally, we verify the fourth condition. $Y,Z$ lie above distinct branches then this follows from the inductive hypothesis. So suppose now that $\xi < \zeta $ and $Y_\xi ,Y_\zeta $ are above the same branch b. Let $A_\xi ,A_\zeta $ be the corresponding members of $\mathcal {A}_b$ , respectively. Then $Y_\xi \subseteq _{\mathcal {F}} A_\xi $ and similar for $\zeta $ . But $A_\xi \cap A_\zeta \in \mathcal {F}^*$ since $\mathcal {A}_b$ is an antichain, so $Y_\xi \cap Y_\zeta \in \mathcal {F}^*$ . Hence the induction hypothesis is maintained and so the construction of $T(\sigma )$ is completed.

Now given $T(\sigma )$ we shall recursively define a new strategy $\sigma '$ . Suppose $R = {\langle } (\alpha _i, Y_j): i < j {\rangle }$ is a play of the game according to $\sigma ' \upharpoonright j$ , and assume further that every move given by $\sigma '$ is in $T(\sigma )$ . Let $b_R$ be the corresponding branch through $T(\sigma )$ . For each legal move $\beta $ for the Challenger, set

$$ \begin{align*} \sigma'(R^\frown {\langle}\beta{\rangle}) &= \text{ the unique immediate successor } Y \text{ of } b_R\ in \\ & T(\sigma) \text{ such that } \alpha(Y)>\beta \text{ is minimal.} \end{align*} $$

Since $\sigma $ is a winning strategy and $T(\sigma )$ has height $\gamma $ we see that $\sigma '$ is also a winning strategy. Also since $T(\sigma )$ has height $\gamma $ and no terminal nodes we see that $T(\sigma )$ is a $\gamma $ -closed subset of $I(\sigma ')^+$ .