Proof Assume first that Cut has a winning strategy $\sigma $ in the game ${\mathcal {U}}(\kappa ,{\le }\gamma )$ . The strategy $\sigma $ can be identified with a binary tree T of height $\gamma $ , where the root of the tree is labelled with $\kappa $ , and if a node of the tree is labelled with a set X which is not a singleton, then its immediate successor nodes are labelled with the sets from the partition that is the response of $\sigma $ to the sequence of cuts and choices leading up to the choice of X, and limit nodes are labelled with the intersection of the labels of their predecessors. $\sigma $ being a winning strategy means that the intersection of labels along any branch of T of length $\gamma $ is empty. Thus, for each ordinal $\alpha <\kappa $ there has to be a node labelled with $\{\alpha \}$ , for this is the only reason why $\alpha $ would not appear in an intersection of choices along some branch of T. However there are only at most $2^{<\gamma }$ -many nodes in this tree, hence $\kappa \le 2^{<\gamma }$ .

Now assume that $\kappa \le 2^{<\gamma }$ . Let $X\subseteq {}^{\gamma } 2$ be such that for $y\in {}^{\gamma } 2$ , we have $y\in X$ if and only if there is $\alpha <\gamma $ such that for all $\beta <\gamma $ ,

• • $y(2\cdot \beta )=1\,\leftrightarrow \,\beta \ge \alpha $ and

• • $\beta \ge \alpha \,\to \,y(2\cdot \beta +1)=1$ .

By our assumption on $\kappa $ , we may identify $\kappa $ with a subset Y of the space X. The winning strategy for Cut is to increasingly partition the space Y in $\gamma $ -many steps using sets of the form $[f]\cap Y$ for functions $f\colon \xi \to 2$ with increasingly large $\xi <\gamma $ . That is, during a run of ${\mathcal {U}}(\kappa ,{\le }\gamma )$ , Cut and Choose work towards constructing a function $F\colon \gamma \to 2$ , the only possible element of the intersection of all choices of Choose, fixing one digit in each round of the game. Now in any even round, Choose cannot possibly pick the digit $1$ , for this would correspond to picking a set $[f]\cap Y$ that is only a singleton, by the definition of the set X. But this means that either Cut already wins at some stage less than $\gamma $ , or that $F\colon \gamma \to 2$ is not an element of X. But this again means that Cut wins, for it implies that the intersection of all choices of Choose is in fact empty.

Fact 2.4. Assume that U is a nonprincipal V-ultrafilter on $\kappa $ in a P-generic extension of the universe, that U yields a wellfounded ultrapower of V, and that j is the generic embedding induced by U. Let $\delta \le \kappa $ be the V-completeness of U, that is the least $\bar \delta $ such that U is not V- $<\bar \delta ^+$ -complete. Then, $\operatorname {\mathrm {crit}} j=\delta $ .

Proof Pick a partition $\langle E_{\xi }\mid \xi <\delta \rangle \in V$ of $\kappa $ with each $E_{\xi }\not \in U$ . Define $h\colon \kappa \to \delta $ by letting $h(\alpha )=\xi $ if $\alpha \in E_{\xi }$ . Using that U is nonprincipal, and letting $c_{\alpha }$ denote the function with domain $\kappa $ and constant value $\alpha $ for any ordinal $\alpha $ , we have that $[c_{\alpha }]_U<[h]_U<[c_{\delta }]_U$ for every $\alpha <\kappa $ . It follows that $j(\delta )>\delta $ , and by the V- ${<}\delta $ -completeness of U, we obtain that $\operatorname {\mathrm {crit}} j=\delta $ , as desired.

Proof Let us generically Lévy collapse $2^{\kappa }$ to become of size $\gamma $ , by the ${<}\gamma $ -closed notion of forcing ${\mathrm {Coll}}(\gamma ,2^{\kappa })$ . In the generic extension, we perform a run of the game ${\mathcal {U}}(\kappa ,{\le }\gamma )$ with Choose following their ground model winning strategy $\sigma $ , and with the moves of Cut following an enumeration of ${P}(\kappa )^V$ in order-type $\gamma $ . More precisely, let $\langle x_{\beta }\mid \beta <\gamma \rangle $ be an enumeration of ${P}(\kappa )^V$ in our generic extension. At any stage $\beta <\gamma $ , assume that $\langle D_{\alpha }\mid \alpha <\beta \rangle $ denotes the sequence of choices of Choose so far, and let $D_{\beta }'=\bigcap _{\alpha <\beta }D_{\alpha }$ . Let Cut play the partition $C_{\beta }=\langle D_{\beta }'\cap x_{\beta },D_{\beta }'\setminus x_{\beta }\rangle $ at stage $\beta $ , and let $D_{\beta }$ denote the response of Choose. Note that since the Lévy collapse is ${<}\gamma $ -closed, any proper initial segment of this run is in the ground model V, and therefore it is possible for Choose to apply their strategy $\sigma $ in each step. Having finished the above run of ${\mathcal {U}}(\kappa ,{\le }\gamma )$ , let U be the collection of all $x_{\beta }$ ’s such that $D_{\beta }=D_{\beta }'\cap x_{\beta }$ . Equivalently, for any $x\subseteq \kappa $ , $x\in U$ if and only if $D_{\beta }\subseteq x$ for all sufficiently large $\beta <\gamma $ .

Proof Assume for a contradiction that this is not the case. We may thus assume that $\gamma =\omega $ , for otherwise U is ${<}\omega _1$ -complete in a $\sigma $ -closed forcing extension of the universe V and therefore yields a well-founded ultrapower of V. Let T be the tree of tuples of the form $\langle \vec f,\vec n,t\rangle $ with the following properties:

1. (1) $t=\langle \langle A_i,B_i\rangle \mid i<k\rangle $ is a partial run of the game ${\mathcal {U}}(\kappa ,{\le }\omega )$ of length $k<\omega $ that is consistent with $\sigma $ , where $A_i$ denotes the partition played by Cut, and $B_i$ denotes the choice of Choose at stage i for every $i<k$ .

2. (2) $\vec n=\langle n_j\mid j<l\rangle $ is a strictly increasing sequence of natural numbers for some $l\le k$ , and if $l>0$ then $n_{l-1}=k-1$ .

3. (3) $\vec f=\langle f_j\mid j<l\rangle $ is such that $f_j\colon B_{n_j}\to \operatorname {\mathrm {Ord}}$ for each $j<l$ .

4. (4) $f_{j+1}(\alpha )<f_j(\alpha )$ for all $\alpha \in B_{n_{j+1}}$ whenever $j+1<l$ .

The ordering relation on T is componentwise extension of sequences, that is for $\langle \vec f,\vec n,t\rangle $ and $\langle \vec f',\vec n',t'\rangle $ both in T, we have $\langle \vec f,\vec n,t\rangle <_T\langle \vec f',\vec n',t'\rangle $ if $\vec f'$ extends $\vec f$ , $\vec n'$ extends $\vec n$ , and $t'$ extends t as a sequence.

Subclaim 2.8. T has a branch in $V[G]$ .

Proof Using our assumption of ill-foundedness, pick a decreasing $\omega $ -sequence of ordinals in the generic ultrapower of V by U, which are represented by functions $g_i\colon \kappa \to \operatorname {\mathrm {Ord}}$ in V. For $i<\omega $ , let $U_i=\{\alpha <\kappa \mid g_{i+1}(\alpha )<g_i(\alpha )\}$ . Consider the run $\langle \langle A_i,B_i\rangle \mid i<\omega \rangle $ of the game ${\mathcal {U}}(\kappa ,{\le }\omega )$ in $V[G]$ , in which Choose plays according to $\sigma $ , and in which Cut plays based on the enumeration $\langle x_i\mid i<\omega \rangle $ of ${P}(\kappa )^V$ . We define sequences $\langle n_j\mid j<\omega \rangle $ and $\langle f_j\mid j<\omega \rangle $ inductively. Let $n_0=0$ , and let $f_0=g_0\operatorname {\mathrm {\!\upharpoonright \!}} B_0$ . Given $n_j$ and $f_j$ , let $n_{j+1}$ be least above $n_j$ such that $B_{n_{j+1}}\subseteq U_j$ —note that such $n_{j+1}$ must exist for $U_j\in U$ . Let $f_{j+1}=g_{j+1}\operatorname {\mathrm {\!\upharpoonright \!}} B_{n_{j+1}}$ . It is now straightforward to check that the sequence

$$\begin{align*}\left\langle\langle\langle f_j\mid j<l\rangle,\langle n_j\mid j<l\rangle,\langle\langle A_i,B_i\rangle\mid i\le n_{l-1}\rangle\rangle\mid l<\omega\right\rangle\end{align*}$$

is a branch through T in $V[G]$ .

By the absoluteness of well-foundedness, T thus has a branch in V. Such a branch yields a run $\langle \langle A_i,B_i\rangle \mid i<\omega \rangle $ of the game ${\mathcal {U}}(\kappa ,{\le }\omega )$ in which Choose follows their winning strategy, hence there is $\beta \in \bigcap _{n\in \omega }B_n\ne \emptyset $ . This branch also yields a strictly increasing sequence $\langle n_i\mid i<\omega \rangle $ of natural numbers, and a sequence of functions $\langle f_i\mid i<\omega \rangle $ so that for each $i<\omega $ , $f_i\colon B_{n_i}\to \operatorname {\mathrm {Ord}}$ , and $f_{i+1}(\alpha )<f_i(\alpha )$ whenever $\alpha \in B_{n_{i+1}}$ . But then, our choice of $\beta $ yields a decreasing $\omega $ -sequence $\langle f_{n_i}(\beta )\mid i<\omega \rangle $ of ordinals, which is a contradiction, as desired.

By Fact 2.4, it follows that $\gamma \le \operatorname {\mathrm {crit}} j\le \kappa $ , and hence by the weak homogeneity of the Lévy collapse, it follows that $\operatorname {\mathrm {crit}} j\le \kappa $ is generically measurable, as witnessed by ${<}\gamma $ -closed forcing.Footnote ⁴

Proof Let U be a normal measurable ultrafilter on $\kappa $ and work in $L[U]$ . Choose has a winning strategy in the game ${\mathcal {U}}(\kappa ,\gamma )$ whenever $\gamma <\kappa $ . Assume for a contradiction that there were some $\lambda <\kappa $ for which Choose had a winning strategy in the game ${\mathcal {U}}(\lambda ,\omega )$ . By Theorem 2.5, there is a generically measurable cardinal $\nu \le \lambda $ . But then, by standard inner model theory results (see the discussion before this observation), $\nu <\kappa $ would be measurable in some inner model of our universe $L[U]$ . Let u be a normal measurable ultrafilter on $\nu $ in that model, and consider the model $L[u]$ . By classical results of Kunen (see [Reference Kanamori17, Theorem 20.12]), $L[U]$ can be obtained by iterating the measure u over the model $L[u]$ . But then, $u\in L[U]$ contradicts the fact that the ultrafilter u could not be an element of its induced ultrapower of $L[u]$ (see [Reference Kanamori17, Proposition 5.7(e)]).

Proof Fix $X\subseteq {}^{\gamma } 2$ of size $\kappa $ that does not contain a continuous and injective image of ${}^{\gamma } 2$ .Footnote ⁶ Let ${\mathcal {U}}(X,{\le }\gamma )$ be the variant of ${\mathcal {U}}(\kappa ,{\le }\gamma )$ where we play on the underlying set X rather than $\kappa $ . Noting that these two games are equivalent, assume for a contradiction that Choose had a winning strategy $\sigma $ for the game ${\mathcal {U}}(X,{\le }\gamma )$ . We consider the following quasistrategy $\tau $ for Cut:Footnote ⁷

• • In each even round $2i$ , given a set $A\subseteq {}^{\gamma } 2$ , Cut splits it into the sets $A_0=\{x\in A\mid x(i)=0\}$ and $A_1=\{x\in A\mid x(i)=1\}$ .

• • In each odd round $2i+1$ , Cut splits off some singleton $\{x_i\}$ , i.e., presents a partition of the form $\langle \{x_i\},X_i\rangle $ .

Note that if Choose wins a run of the game ${\mathcal {U}}(X,{\le }\gamma )$ in which Cut plays according to their quasistrategy $\tau $ , then by the definition of $\tau $ at even stages, if $x\in {}^{\gamma } 2$ is in the intersection of choices made by Choose in such a run, $x(i)$ has been fixed for every $i<\gamma $ , that is the intersection of these choices will only have a single element.

Using the above claim, since $\sigma $ is a winning strategy for Choose and $\gamma $ is regular, we can construct a full binary tree T of height $\gamma $ of partial plays t such that partial plays on the same level of T have the same length, and such that the final choices made by Choose in any two such partial plays of successor length which are on the same level of T will be disjoint. Let $\pi $ be an order-preserving isomorphism from ${}^{<\gamma }2\to T$ , and for $a\in {}^{\gamma } 2$ , let $\pi (a)=\bigcup \{\pi (a\operatorname {\mathrm {\!\upharpoonright \!}}\alpha )\mid \alpha <\gamma \}$ . Since $\sigma $ is a winning strategy for Choose, the intersection of choices from any run of ${\mathcal {U}}(X,{\le }\gamma )$ is nonempty, and thus, using the way the quasistrategy $\tau $ was defined at even stages, this yields a continuous and injective map $f\colon {}^{\gamma } 2\rightarrow X$ , letting $f(a)=x$ whenever $a\in {}^{\gamma } 2$ , $b=\pi (a)$ is a branch through T, and x is the unique element of the intersection of choices of Choose in the run $\bigcup b$ . This shows that X contains a continuous and injective image of ${}^{\gamma } 2$ , contradicting our choice of X.

Proof They simply win by picking their choices according to U.

Proof The $2^{\kappa }$ -strong compactness of $\kappa $ allows us to extend I to a ${<}\kappa $ -complete prime ideal, the complement of which thus is a measurable ultrafilter containing X as an element. The result then follows from Observation 3.2.

Proof The proof of (1) is analogous to the proof of Observation 1.2, and the proof of (2) is analogous to the proof of Observation 2.2, making use of the fact that $\kappa $ cannot be written as a ${<}\kappa $ -union of elements of I in the forward directions. For (3), assume for a contradiction that $\kappa>2^{<\gamma }$ , however Cut has a winning strategy $\sigma $ for the game ${\mathcal {U}}(\kappa ,I,{<}\gamma )$ . $\sigma $ can be identified with a binary tree T of height at most $\gamma $ , where the root of the tree is labelled with $\kappa $ , and if a node of the tree is labelled with $y\in I^+$ , then its immediate successor nodes are labelled with the sets from the partition that is the response of $\sigma $ to the sequence of cuts and choices leading up to the choice of y, and limit nodes are labelled with the intersection of the labels of their predecessors. If a node is labelled with a set in I, then it does not have any successors, and it means that Choose has lost at such a point. $\sigma $ being a winning strategy means that T has no branch of length $\gamma $ . Thus, the union of all the labels of the leaves of T has to be $\kappa $ , which clearly contradicts our assumption on I, for the number of leaves of T is at most $2^{<\gamma }<\kappa $ .

Regarding (4), assume first that $\kappa $ is weakly compact, however Cut has a winning strategy $\sigma $ in the game ${\mathcal {U}}(\kappa ,I,{<}\kappa )$ . Let $\theta $ be a sufficiently large regular cardinal, and let $M\supseteq (\kappa +1)$ be an elementary substructure of $H(\theta )$ of size $\kappa $ that is closed under ${<}\kappa $ -sequences and with $\sigma \in M$ . Using that $\kappa $ is weakly compact, let U be a uniform ${<}\kappa $ -complete M-ultrafilter on $\kappa $ . Let us consider a run of the game ${\mathcal {U}}(\kappa ,I,{<}\kappa )$ in which Cut follows their winning strategy $\sigma $ , and Choose responds according to U. This is possible for proper initial segments of such a run will be elements of M by the ${<}\kappa $ -closure of M, and hence can be used as input for $\sigma $ in M, yielding the individual moves of Cut to be in M as well. But since U is uniform and ${<}\kappa $ -complete, all choices of Choose will be in U and therefore I-positive. This means that Choose wins against $\sigma $ , which is our desired contradiction.

In the other direction, assume that Cut does not have a winning strategy in the game ${\mathcal {U}}(\kappa ,I,{<}\kappa )$ . We verify that under the assumptions of our observation, $\kappa $ has the filter property (as in [Reference Holy and Schlicht11, Definition 2.3]) and is thus weakly compact. Recall that the filter property states that for every subset X of ${\mathcal {P}}(\kappa )$ of size at most $\kappa $ , there is a ${<}\kappa $ -complete filter F on $\kappa $ which measures X. Note that the definition of ${<}\kappa $ -complete filters in [Reference Holy and Schlicht11, Definition 2.2] is nonstandard, but every such filter induces a ${<}\kappa $ -complete filter in the usual sense as its upwards closure, and every ${<}\kappa $ -complete filter in the usual sense is ${<}\kappa $ -complete in the sense of [Reference Holy and Schlicht11, Definition 2.2]. Therefore the filter property in [Reference Holy and Schlicht11, Definition 2.3] is equivalent to the standard one. A proof of the equivalence of weak compactness and the filter property can be found in [Reference Abramson, Harrington, Kleinberg and Zwicker1, Theorem 1.1.3]. To verify the filter property for $\kappa $ , let $\mathcal A=\langle A_i\mid i<\kappa \rangle $ be a collection of subsets of $\kappa $ . We need to find a ${<}\kappa $ -complete filter $\mathcal F$ on $\mathcal A$ , that is, ${<}\kappa $ -sized subsets of $\mathcal F$ need to have $\kappa $ -sized intersections. At any stage $i<\kappa $ , let Cut play the partition $\langle A_i,\kappa \setminus A_i\rangle $ of $\kappa $ . Since Cut does not have a winning strategy in the game ${\mathcal {U}}(\kappa ,I,{<}\kappa )$ , there is a sequence $\langle B_i\mid i<\kappa \rangle $ of choices of Choose in such a run such that for every $\lambda <\kappa $ , $\bigcap _{i<\lambda }B_i\in I^+$ . But now we may clearly define our desired filter $\mathcal F$ by letting $A_i\in \mathcal F$ if $B_i=A_i$ .

Proof It is straightforward to check that $I\supseteq \operatorname {\mathrm {bd}}_{\kappa }$ is normal, using that $\dot U$ is forced to be uniform and V-normal. We will describe a winning strategy for Choose in the game ${\mathcal {U}}(X,I,{<}\gamma )$ . At each stage $\alpha $ , Choose not only decides for a set $C_{\alpha }$ to actually respond with, but they also pick a condition $p_{\alpha }\in P$ forcing that $\check C_{\alpha }\in \dot U$ , such that these conditions form a decreasing sequence of conditions.

At stage $0$ , assume that Cut presents the partition $\langle A_0,B_0\rangle $ of X. Since some condition forces that $\check X\in \dot U$ , Choose may pick $C_0$ to either be $A_0$ or $B_0$ , and pick a condition $p_0$ forcing that $\check C_0\in \dot U$ . At successor stages $\alpha +1$ , we proceed essentially in the same way. Assume that Cut presents the partition $\langle A_{\alpha +1},B_{\alpha +1}\rangle $ of $C_{\alpha }$ . Since $p_{\alpha }\operatorname {\mathrm {\Vdash }}\check C_{\alpha }\in \dot U$ , Choose may pick $C_{\alpha +1}$ to either be $A_{\alpha +1}$ or $B_{\alpha +1}$ , and pick a condition $p_{\alpha +1}\le p_{\alpha }$ forcing that $\check C_{\alpha +1}\in \dot U$ .

At limit stages $\alpha <\gamma $ , Cut presents a partition $\langle A_{\alpha },B_{\alpha }\rangle $ of $\bigcap _{\beta <\alpha }C_{\beta }$ . Since the forcing notion P is ${<}\gamma $ -closed, we may let $p^{\prime }_{\alpha }$ be a lower bound of $\langle p_{\beta }\mid \beta <\alpha \rangle $ . Then, $p^{\prime }_{\alpha }$ forces that $\bigcap _{\beta <\check \alpha }\check C_{\beta }\in \dot U$ , and Choose may pick $C_{\alpha }$ to either be $A_{\alpha }$ or $B_{\alpha }$ , and pick a condition $p_{\alpha }\le p^{\prime }_{\alpha }$ forcing that $\check C_{\alpha }\in \dot U$ .

Proof Apply the Lévy collapse forcing to make $\theta $ become $\kappa ^+$ , which is ${<}\kappa $ -closed. Work in a generic extension for this forcing. As we argued above, $\kappa ^+$ is generically measurable as witnessed by ${<}\kappa $ -closed forcing. Thus by Observation 3.5, Choose has a winning strategy in the game ${\mathcal {U}}(\theta ,\gamma )$ whenever $\gamma <\kappa $ . Assume for a contradiction that there were some $\lambda $ with $\gamma <\lambda <\theta $ for which Choose had a winning strategy in the game ${\mathcal {U}}(\lambda ,{\le }\gamma )$ . By Theorem 2.5, we obtain a generically measurable cardinal $\nu \le \lambda $ . But then clearly, $\nu $ is also generically measurable in our ground model, contradicting our assumption.

Proof We first force with a reverse Easton iteration, adding a Cohen subset to every inaccessible cardinal below $\kappa $ . Let us consider the generic extension thus obtained as our ground model in the following. By well-known standard arguments (similar to those in the proof of Silver’s theorem about violating the $\operatorname {\mathrm {GCH}}$ at a measurable cardinal (see [Reference Jech14, Theorem 21.4])), adding a Cohen subset of $\kappa $ to that model will resurrect the measurability of $\kappa $ in the extension. We will force to add a Suslin tree T to $\kappa $ that is closed under ascending ${<}\gamma ^+$ -sequences, and show that $\kappa $ is generically measurable in that extension, as witnessed by forcing with that Suslin tree (with its reversed ordering), which now is a ${<}\gamma ^+$ -closed notion of forcing.

It thus follows that after forcing with $\mathbb Q*\dot T$ , $\kappa $ is measurable, and thus $\mathbb Q$ forces that $\kappa $ is generically measurable as witnessed by the notion of forcing $\dot T$ , which is ${<}\gamma ^+$ -closed, as desired. It is also clear that $\kappa $ is Mahlo after forcing with $\mathbb Q$ , for otherwise it could not be measurable in the further $\dot T$ -generic extension.

Proof The proofs of (1)–(3) and (5) are analogous to those in Observation 3.4. The argument for (4) is a minor adaption of that for (3).

Proof Write X as a disjoint union of I-positive sets $X_i$ for $i<\omega $ . At any stage $n<\omega $ , let Cut play a disjoint partition $\langle A^n_{\alpha }\mid \alpha <\kappa \rangle $ of X into I-positive sets such that each $A^n_{\alpha }$ contains exactly one element of $X_n$ , and such that $A^n_{\alpha }\cap X_m\in I^+$ whenever $m>n$ .Footnote ¹⁴ Choose has to pick some $B^n=A^n_{\alpha }$ . Let $A^n_{\alpha }\cap X_n=\{\alpha _n\}$ . Note that the above defines a strategy for Cut which ensures that for any $i<\omega $ , $X_i\cap \bigcap _{n<\omega }B^n$ contains at most one element, and hence the intersection $\bigcap _{n<\omega }B^n$ of choices of Choose is countable, showing this strategy to be a winning strategy for Cut, as desired.Footnote ¹⁵

Proof We only treat the equivalence between games of the form ${\mathcal {G}}_{\nu }(X,I,\gamma )$ and ${\mathcal {U}}_{\nu }(X,I,\gamma )$ when $\nu <\kappa $ is a cardinal and $\gamma <\kappa $ is a limit ordinal, for the other equivalences are analogous. Making use of the comments after Definition 5.1, if Cut wins ${\mathcal {U}}_{\nu }(X,I,\gamma )$ , then Cut wins ${\mathcal {G}}_{\nu }(X,I,\gamma )$ , because every disjoint partition of an I-positive set X into less than $\kappa $ -many I-positive sets is an I-partition of X. Analogously, if Choose wins ${\mathcal {G}}_{\nu }(X,I,\gamma )$ , then Choose wins ${\mathcal {U}}_{\nu }(X,I,\gamma )$ .

Given an I-partition $W=\{w_{\alpha }\mid \alpha <\theta \}$ of some set $X\in I^+$ , with $\theta \le \kappa $ , we call $W'=\{w^{\prime }_{\alpha }\mid \alpha <\theta \}$ a full disjointification of W in case $w^{\prime }_0=w_0\cup (X\setminus \bigcup W)$ and $w^{\prime }_{\alpha }=w_{\alpha }\setminus \bigcup _{\bar \alpha <\alpha }w_{\bar \alpha }$ for every nonzero $\alpha <\theta $ .Footnote ¹⁶ Then $W'$ is a partition of X and moreover, $w^{\prime }_{\alpha }\subseteq w_{\alpha }$ for every nonzero $\alpha <\theta $ .

Suppose that $\sigma $ is a winning strategy for Cut in ${\mathcal {G}}_{\nu }(X,I,\gamma )$ . To define a winning strategy $\tau $ for Cut in ${\mathcal {U}}_{\nu }(X,I,\gamma )$ , we use an auxiliary run of ${\mathcal {G}}_{\nu }(X,I,\gamma )$ in which Cut plays according to $\sigma $ . Given a move $W_{\alpha }$ of Cut in round $\alpha <\gamma $ in ${\mathcal {G}}_{\nu }(X,I,\gamma )$ , we let $\tau $ perform two consecutive moves in ${\mathcal {U}}_{\nu }(X,I,\gamma )$ . The first one is the full disjointification $W_{\alpha }'$ of $W_{\alpha }$ . The second one splits X into $\bigcup W_{\alpha }$ and $X\setminus \bigcup W_{\alpha }$ . Choose first picks an element $Y_{\alpha }$ of $W_{\alpha }'$ and then they have to pick $\bigcup W_{\alpha }$ . By the definition of full disjointifications, there is some $X_{\alpha } \in W_{\alpha }$ with $Y_{\alpha }\cap \bigcup W_{\alpha }\subseteq X_{\alpha }$ . We let Choose play such an $X_{\alpha }$ in ${\mathcal {G}}_{\nu }(X,I,\gamma )$ , and Cut again responds in the next round by using $\sigma $ . Since $\sigma $ is a winning strategy for Cut, it follows that $\bigcap _{\alpha <\gamma }(Y_{\alpha }\cap \bigcup W_{\alpha }) \subseteq \bigcap _{\alpha <\gamma }X_{\alpha }\in I$ , and therefore that $\tau $ is a winning strategy for Cut, as desired.

We now argue that a winning strategy $\sigma $ for Choose in ${\mathcal {U}}_{\nu }(X,I,\gamma )$ yields a winning strategy $\tau $ for Choose in the game ${\mathcal {G}}_{\nu }(X,I,\gamma )$ , making use of an auxiliary run of ${\mathcal {U}}_{\nu }(X,I,\gamma )$ in which Choose plays according to $\sigma $ . Given a move $W_{\alpha }$ of Cut in the game ${\mathcal {G}}_{\nu }(X,I,\gamma )$ , we let Cut perform two consecutive moves in the game ${\mathcal {U}}_{\nu }(X,I,\gamma )$ : The first one is the full disjointification $W_{\alpha }'$ of $W_{\alpha }$ , and the second one splits X into $\bigcup W_{\alpha }$ and $X\setminus \bigcup W_{\alpha }$ . The strategy $\sigma $ will pick some element $Y_{\alpha }\in W_{\alpha }'$ , and then decides for $\bigcup W_{\alpha }$ . We let the next move of Choose according to $\tau $ be some $X_{\alpha }\in W_{\alpha }$ for which $Y_{\alpha }\cap \bigcup W_{\alpha }\subseteq X_{\alpha }$ . Since $\sigma $ is a winning strategy for Choose, it follows that $\bigcap _{\alpha <\gamma }(Y_{\alpha }\cap \bigcup W_{\alpha }) \subseteq \bigcap _{\alpha <\gamma }X_{\alpha }\in I^+$ , and therefore that $\tau $ is a winning strategy for Choose, as desired.

Proof The idea of the arguments for the above is that we may simulate a single move of Cut, in the games where they are allowed to play larger antichains, by less than $\gamma $ -many moves in the corresponding games where they are only allowed to play antichains of size at most $\nu $ in each of their moves. Let us go through some of the details of one of those equivalences in somewhat more detail. For example, let us assume that in (1), Cut has a winning strategy in the game ${\mathcal {G}}_{\nu ^{\beta }}(X,{\mathbb {Q}},\gamma )$ . Assume that in one of their moves, they play a maximal antichain of ${\mathbb {Q}}$ below X of the form $W=\{x_r\mid r\in {}^{\beta }\nu \}$ . Let Cut make $\beta $ -many moves in the game ${\mathcal {G}}_{\nu }(X,{\mathbb {Q}},\gamma )$ , playing maximal antichains $W_i$ below X for $i<\beta $ , with $W_i=\{w_i^j\mid j<\nu \}$ such that $w_i^j=\sup \{x_r\mid r(i)=j\}$ for all $j<\nu $ . Let $r\colon \beta \to \nu $ be such that Choose picks $w_i^{r(i)}$ in their ith move, for each $i<\beta $ . Now clearly $x_r\le w_i^{r(i)}$ for each $i<\beta $ , hence $x_r\le \inf \{w_i^{r(i)}\mid i<\beta \}$ . Note that for $r\ne r'\in {}^{\beta }\nu $ , $\inf \{w_i^{r(i)}\mid i<\beta \}$ and $\inf \{w_i^{r'(i)}\mid i<\beta \}$ are incompatible, and hence the collection of these infima for different $r'\in {}^{\beta }\nu $ forms a maximal antichain of ${\mathbb {Q}}$ below X. Thus, by the separativity of ${\mathbb {Q}}$ , it follows that in fact $x_r=\inf \{w_i^{r(i)}\mid i<\beta \}$ . Let Choose respond to W by picking $x_r\in W$ . Cut will win this run of the game ${\mathcal {G}}_{\nu ^{\beta }}(X,{\mathbb {Q}},\gamma )$ for they are using their winning strategy, but then they will also win the above run of ${\mathcal {G}}_{\nu }(X,{\mathbb {Q}},\gamma )$ , for the responses of Choose in this run will be cofinal in the sequence of their corresponding responses in the run of ${\mathcal {G}}_{\nu ^{\beta }}(X,{\mathbb {Q}},\gamma )$ , and thus the set of responses of Choose in either game will not have a lower bound. We have thus produced a winning strategy for Cut in the game ${\mathcal {G}}_{\nu }(X,I,\gamma )$ in this way, as desired.

The remaining arguments for (1) are very similar to the above. Item (2) follows directly from the argument for (1), and (3) is an immediate consequence of (2).

Proof (1): Suppose that $\langle W^j \mid j<\delta \rangle $ is a sequence of maximal antichains in ${\mathbb {Q}}$ , each of size ${\leq }\nu ^{\gamma }$ . For each $j<\delta $ , we define $\langle W^j_i \mid i<\gamma \rangle $ as in the proof of Theorem 6.2. Since ${\mathbb {Q}}$ is $(\delta ,\nu )$ -distributive, there exists a positive branch through $\langle W^j_i \mid {\prec } i,j {\succ } \in \gamma \rangle $ with a lower bound p, where ${\prec } i,j {\succ }$ denotes the standard pairing function applied to i and j. As in the proof of Theorem 6.2, p induces a positive branch through $\langle W^j \mid j<\delta \rangle $ .

(2): Suppose that $\langle W^j \mid j<\delta \rangle $ is a sequence of maximal antichains in ${\mathbb {Q}}$ , each of size ${\leq }\nu ^{<\gamma }$ . Fix a cofinal sequence $\langle \gamma _i \mid i<\operatorname {\mathrm {cof}}(\gamma ) \rangle $ in $\gamma $ . For each $j<\delta $ , we partition $W^j$ into subsets $\langle W^{j,i}\mid i<\operatorname {\mathrm {cof}}(\gamma )\rangle $ such that $W^{j,i}$ has size ${\leq }\nu ^{\gamma _i}$ for each $i<\operatorname {\mathrm {cof}}(\gamma )$ . We can extend each $W^{j,i}$ to a maximal antichain $\bar {W}^{j,i}$ by adding a single condition, namely $\mathrm{sup}(W^j\setminus W^{j,i})$ , since ${\mathbb {Q}}$ is ${<}(\nu ^{<\gamma })^+$ -complete. As in the proof of (1), we then replace each $\bar {W}^{j,i}$ by a sequence $\langle \bar {W}^{j,i}_k \mid k<\gamma _i\rangle $ such that $\bar {W}^{j,i}_k$ has size ${\leq }\nu $ . Let $\langle \tilde W_l\mid l<\delta \rangle $ enumerate all the $\bar W^{j,i}_k$ in order-type $\delta $ . Since ${\mathbb {Q}}$ is $(\delta ,\nu )$ -distributive, there exists a positive branch through $\langle \tilde W_l \mid l<\delta \rangle $ . This is easily seen to induce a positive branch through $\langle W^j \mid j<\delta \rangle $ , as required.

(3): We proceed as in the proof of (2), except when $W^{j,i}$ is extended to a maximal antichain $\bar {W}^{j,i}$ : Since $|W^{j,i}| \leq \nu ^{\gamma _i}<\nu ^{<\gamma }$ and ${\mathbb {Q}}$ is ${<}(\nu ^{<\gamma })$ -complete, $\mathrm{sup}(W^{j,i})$ exists, and thus, using that ${\mathbb {Q}}$ is a Boolean algebra, also $\mathrm{sup}(W^j\setminus W^{j,i})=\neg \sup (W^{j,i})$ exists.

Proof (1): Any sequence of maximal antichains of ${\mathbb {Q}}$ below X, each of size at most $\nu $ (or of arbitrary size in case $\nu =\infty $ ), witnessing ${\mathbb {Q}}$ to not be $(\gamma ,\nu )$ -distributive with respect to X can be identified with a strategy for Cut in the game ${\mathcal {G}}_{\nu }(X,{\mathbb {Q}},\gamma )$ , and the nonexistence of a positive branch through such a sequence corresponds to the fact that Choose cannot win against this strategy, which means that it is in fact a winning strategy, as desired.

(2): Assume that ${\mathbb {Q}}$ is $(\gamma ,\nu )$ -distributive with respect to X. Assume for a contradiction that Cut did have a winning strategy $\sigma $ in the game ${\mathcal {G}}_{\nu }(X,{\mathbb {Q}},\gamma )$ .