Axiom wA2. There is a quasi-ordering on $\mathcal {AR}$ such that:

• (w1) is countable for all $b \in \mathcal {AR}$ .

• (2) iff .

• (3) .

Proof. (1) $\implies $ (2) follows from the abstract Ellentuck theorem. For the converse, let $A \in \mathcal {R}$ , $a \in \mathcal {AR}\mathord {\upharpoonright } A,$ and . Define

Since $\mathcal {X}$ is clopen, it is Ramsey. Therefore, there exists some $B \in [a,A]$ such that $[a,B] \subseteq \mathcal {X}$ or $[a,B] \subseteq \mathcal {X}^c$ . If $[a,B] \subseteq \mathcal {X}$ , then , and if $[a,B] \subseteq \mathcal {X}^c$ , then . By A3, we may let such that $[a,B'] \subseteq [a,B]$ , so $B'$ witnesses that A4 holds for .

Proof. Let be a countable family of subsets of $\mathcal {R}$ , and let . Fix any $A \in \mathcal {R}$ and $a \in \mathcal {AR}\mathord {\upharpoonright } A$ , and assume that (2) fails for all $B \in [a,A]$ . In particular, applying $n = 0$ to the negation of (2), I has a strategy in $K[a,B]$ to reach $\mathcal {X}_0$ for all $B \in [a,A]$ .

If $B \in [a,A]$ and $b \in \mathcal {AR}\mathord {\upharpoonright }[a,B]$ , say that “(2) holds for $(b,B)$ ” if II has a strategy $\tau $ in $K[b,B]$ such that, for any total state s following $\tau $ , there exists some such that if and , then I has no strategy in $K[a_n,B_n]$ to reach $\mathcal {X}_n$ . Note that this would also imply that I has no strategy in $K[a_n,C]$ to reach $\mathcal {X}_n$ for all $C \in [a_n,B_n]$ .

Let denote the set of all such that for all $B \in [a',A]$ , (2) does not hold for $(a',B)$ . Since we assumed that (2) fails for all $B \in [a,A]$ , by the previous claim there exists some $A_\emptyset \in [a,A]$ such that . As stated in the first paragraph, I has a strategy $\sigma _\emptyset $ in $K[a,A_\emptyset ]$ to reach $\mathcal {X}_0$ . To finish the proof, we shall construct a strategy $\sigma $ for I in $K[a,A_\emptyset ]$ to reach $\mathcal {X}$ .

For the rest of this proof, all states are assumed to be for II. Let $\sigma (\emptyset ) := \sigma _\emptyset (\emptyset )$ . Now suppose for each state s of $K[a,A_\emptyset ]$ following $\sigma $ of rank n, we define the following:

1. (1) If $n> 0$ and $s' = s\mathord {\upharpoonright }(n - 1)$ , then $A_s \in [a(s),A_{s'}]$ .

2. (2) $\sigma _s$ is a strategy for I in $K[a(s),A_s]$ to reach $\mathcal {X}_n$ .

3. (3) If , then $t_s^i$ is a state of $K[a(s\mathord {\upharpoonright } i),A_{s\mathord {\upharpoonright } i}]$ following $\sigma _{s\mathord {\upharpoonright } i}$ of rank $n - i$ , and $a(t_s^i) = a(s\mathord {\upharpoonright } i)$ .

4. (4) $\sigma _{s\mathord {\upharpoonright }(i+1)}(t_s^{i+1}) \in [a_s,\sigma _{s\mathord {\upharpoonright } i}(t_s^i)]$ , and $\sigma (s) \in [a(s),\sigma _s(t_s^n)]$ .

5. (5) For all and $B \in [b,A_s]$ , (2) does not hold in $(b,B)$ .

Now let s be a state of $K[a,A_\emptyset ]$ following $\sigma $ of rank $n + 1$ . We may write $s = (s\mathord {\upharpoonright } n)^\frown (a(s),B_s)$ . Since $B_s \in [a(s),\sigma (s\mathord {\upharpoonright } n)] \subseteq [a(s),\sigma _\emptyset (t_{s\mathord {\upharpoonright } n}^0)]$ , we may define $t_s^0 := {t_{s\mathord {\upharpoonright } n}^0}^\frown (a(s),B_s)$ , which is a legal state in $K[a,A_\emptyset ]$ following $\sigma _\emptyset $ . We have $\sigma _\emptyset (t_s^0) \in [a(s),B_s] \subseteq [a(s),\sigma _{s\mathord {\upharpoonright } 1}(t_{s\mathord {\upharpoonright } n}^1)]$ , so we may define $t_s^1 := {t_{s\mathord {\upharpoonright } n}^1}^\frown (a(s),\sigma _\emptyset (t_s^0))$ . This again, gives us a legal state in $K[a(s\mathord {\upharpoonright } 1),A_{s\mathord {\upharpoonright } 1}]$ following $\sigma _{s\mathord {\upharpoonright } 1}$ . We may repeat this process to give us states $t_s^i$ for . We let $A_s' := \sigma _{s\mathord {\upharpoonright } n}(t_s^n) \in [a(s\mathord {\upharpoonright } n),A_{s\mathord {\upharpoonright } n}]$ .

Let be the set of all such that there exists some $B \in [a(s),A_s']$ in which (2) holds for $(a',B)$ . By (5) of the induction hypothesis and the claim, we may obtain $A_s \in [a(s),A_s']$ such that , and I has a winning strategy $\sigma _s$ in $K[a(s),A_s]$ to reach $\mathcal {X}_{n+1}$ . Define $\sigma (s) := \sigma _s(\emptyset )$ . This is indeed a legal move, as

$$ \begin{align*} \sigma(s) \in [a(s),A_s] \subseteq [a(s),A_s'] \subseteq [a(s),A_{s\mathord{\upharpoonright} n}] \subseteq [a(s),B_s]. \end{align*} $$

This completes the induction. We see that $\sigma $ is indeed a strategy for I in $K[a,A_\emptyset ]$ to reach $\mathcal {X}$ —if s is a total state of $K[a,A_\emptyset ]$ following $\sigma $ , then $A(s) = A(t_s^n) \in \mathcal {X}_n$ for all n, so . This completes the proof.

Proof. Let be a countable family of subsets of $\mathcal {R}$ , and let . Fix any $A \in \mathcal {R}$ and $a \in \mathcal {AR}\mathord {\upharpoonright } A$ . If there exists some $B \in [a,A]$ such that II has a strategy in $K[a,B]$ to reach $\mathcal {X}$ , then we’re done, so assume otherwise. Consider applying Lemma 3.5 to . We claim that (2) fails for all $(a,B),$ where $B \in [a,A]$ , so by the same lemma, I has a strategy in $K[a,A]$ to reach $\mathcal {X}^c$ . Indeed, otherwise let $\tau $ be a strategy in $K[a,B]$ witnessing that (2) holds for $(a,B)$ . Player II shall follow $\tau $ until they reach some turn n, ending with II playing $(a_n,B_n')$ , such that I has no strategy in $K[a_n,B_n']$ to reach $\mathcal {X}_n^c$ . Since $\mathcal {X}_n$ is Kastanas Ramsey, II may instead play $(a_n,B_n)$ in the last turn, where $B_n \in [a_n,B_n']$ , such that II has a strategy in $K[a_n,B_n]$ to reach $\mathcal {X}_n$ . Afterwards, II follows this strategy to reach $\mathcal {X}_n$ . Since $\mathcal {X}_n \subseteq \mathcal {X}$ , this constitutes a strategy for II in $K[a,B]$ to reach $\mathcal {X}$ , contradicting our assumption.

Proof. Let be an $(a,A)$ -biasymptotic set for some $A \in \mathcal {R}$ and $a \in \mathcal {AR}\mathord {\upharpoonright } A$ . Define

Since is $(a,A)$ -biasymptotic, for any $B \in [a,A]$ , there exists some $C \in [a,B]$ such that , and some $C' \in [a,B]$ such that . Consequently, $[a,B] \cap \mathcal {X} \neq \emptyset $ and $[a,B] \cap \mathcal {X}^c \neq \emptyset $ , so $\mathcal {X}$ is not Ramsey as B is arbitrary. However, $\mathcal {X}$ is a countable union of clopen sets, so it is Borel (under the metrisable topology). By the Borel determinacy for , the game $K[a,A]$ to reach $\mathcal {X}$ or $\mathcal {X}^c$ is always determined, so .

Proof. Fix a biasymptotic set , and let $\mathcal {X} \subseteq \mathcal {R}$ be not Kastanas Ramsey. Define two sets as follows:

Observe that both $\mathcal {X}_0^c$ and $\mathcal {X}_1^c$ are Kastanas Ramsey: Indeed, for any $A \in \mathcal {R}$ and $a \in \mathcal {AR}\mathord {\upharpoonright } A$ , II has a winning strategy in $K[a,A]$ to reach $\mathcal {X}_0^c$ by playing for all n, and II also has a winning strategy in $K[a,A]$ to reach $\mathcal {X}_1^c$ by playing for all n. On the other hand, we have that $\mathcal {X}_0 \cup \mathcal {X}_1 = \mathcal {X}$ , so if both $\mathcal {X}_0$ and $\mathcal {X}_1$ are Kastanas Ramsey, then so is $\mathcal {X}$ by Proposition 3.6, a contradiction. Thus, at least one of $\mathcal {X}_0$ or $\mathcal {X}_1$ witnesses that $\boldsymbol {\mathcal {KR}}$ is not closed under complements.

Proof. Fix a biasymptotic set , and let $\mathcal {X} \subseteq \mathcal {R}$ be a Kastanas Ramsey set in which $\mathcal {X}^c$ is not Kastanas Ramsey. Define two sets as follows:

By the same argument as in Proposition 3.11, $\mathcal {X}_0^c$ and $\mathcal {X}_1^c$ are Kastanas Ramsey. However, $\mathcal {X}^c = \mathcal {X}_0^c \cap \mathcal {X}_1^c$ is not.

Proof. Fix some $A \in \mathcal {R}$ and $a \in \mathcal {AR}\mathord {\upharpoonright } A$ . Suppose that $\vec {\mathcal {D}} = \{\mathcal {D}_b\}_{b \in \mathcal {AR}\mathord {\upharpoonright }[a,A]}$ is dense open below $[a,A]$ . We shall define a fusion sequence in $[a,A]$ , with , such that $A_{n+1} \in [a_{n+1},A_n]$ : Let $A_0 := A$ , and suppose that $A_n$ has been defined. Let $\{b_i : i < N\}$ enumerate the set of all $b \in \mathcal {AR}\mathord {\upharpoonright } A_n$ such that $a \sqsubseteq b$ and . Let $A_{n+1}^0 := A_n$ . If $A_{n+1}^i \in [a_{n+1},A_n]$ has been defined, let $B_{n+1} \in \mathcal {D}_{b_i}$ be such that $B_{n+1} \in [b_i,A_{n+1}^i]$ , which exists as $\mathcal {D}_{b_i}$ is dense open in $[b_i,A]$ . By A3, we then let $A_{n+1}^{i+1} \in [a_{n+1},A_n^i]$ such that $[b_i,A_{n+1}^{i+1}] \subseteq [b_i,B_{n+1}]$ . We complete the induction by letting $A_{n+1} := A_{n+1}^N$ . Let B be the limit of the fusion sequence , and we have that B diagonalises $\vec {\mathcal {D}}$ .

Proof. For each , we define

$$ \begin{align*} \mathcal{D}_{b,0} &:= \{D \in [b,A] : \exists B \in [a,A] \text{ s.t. } f(B) = b \wedge D \in [b,g(B)]\}, \\ \mathcal{D}_{b,1} &:= \{D \in [b,A] : \forall C \in [a,A], \, g(C) \in [b,D] \to g(C) \notin \mathcal{D}_{b,0}\}. \end{align*} $$

Let $\mathcal {D}_b := \mathcal {D}_{b,0} \cup \mathcal {D}_{b,1}$ . Observe that $\mathcal {D}_b$ is dense open in $[b,A]$ : Clearly both $\mathcal {D}_{b,0}$ and $\mathcal {D}_{b,1}$ are open. If $D \in [b,A]$ and $D \notin \mathcal {D}_{b,1}$ , then there exists some $C \in [a,A]$ such that $g(C) \in [b,D]$ . Then and $g(C) \in \mathcal {D}_{b,0}$ , so $\mathcal {D}_b$ is dense.

By Lemma 3.16, there exists some $D \in [a,A]$ diagonalising $(\mathcal {D}_b)_{b \in \mathcal {AR}\mathord {\upharpoonright }[a,A]}$ . Now let

By A4, there exists some $E_{f,g} \in [a,D]$ such that or . However, we see that the latter case is not possible: In this case, we let $b := f(E_{f,g})$ . Then , so let $B \in \mathcal {D}_{b,1}$ such that $[b,D] \subseteq [b,B]$ . Then $g(E_{f,g}) \in [b,E_{f,g}] \subseteq [b,D] \subseteq [b,B]$ , so $g(E_{f,g}) \notin \mathcal {D}_{b,0}$ . But $g(E_{f,g}) \in [b,B]$ , so this implies that $f(E_{f,g}) \neq b$ , a contradiction.

We shall show that $E_{f,g}$ works. Let . Then , so there exists some $B \in \mathcal {D}_{b,0}$ such that $f(B) = b$ and $[b,E_{f,g}] \subseteq [b,D] \subseteq [b,g(B)]$ , as desired.

Proof of Proposition 3.14.

Let $\sigma $ be a strategy for II in $K[a,A]$ to reach $\mathcal {X}$ . We shall construct a strategy $\tau $ for I in $K[a,A]$ to reach $\mathcal {X}$ as follows: We shall assign a state s (for II) in $K[a,A]$ following $\tau $ , to a state $t_s$ (for II) in $K[a,A]$ , following $\sigma $ , such that:

1. (1) $a(s) = a(t_s)$ .

2. (2) .

3. (3) If $s' \sqsubseteq s$ , then $t_{s'} \sqsubseteq t_s$ .

We begin by defining $t_\emptyset := \emptyset $ . Now suppose that s is a state (for II) in $K[a,A]$ following $\tau $ so far. We define the functions $f_s,g_s$ by stipulating that for all , $(f_s(B),g_s(B)) := \sigma ({t_s}^\frown B)$ . Observe that $f_s,g_s$ are suitable in (when restricted to ), so by Lemma 3.17 there exists some such that for all , there exists some such that $f_s(B_{s,b}) = b$ and $[b,E_s] \subseteq [b,g(B_{s,b})]$ . Thus, we define , and for all and $C \in [b,E_s]$ , define

$$ \begin{align*} t_{s^\frown(b,C)} := {t_s}^\frown(b,g(B_{s,b})). \end{align*} $$

Clearly, (1) and (3) of the induction hypothesis are satisfied. (2) is also satisfied, as

This completes the induction. Since every total state following $\tau $ corresponds to a total state following $\sigma $ with the same outcome, $\tau $ is a strategy for I to reach $\mathcal {X}$ .

Proof. It is easy to verify that satisfies A1, wA2 (resp., A2), and A3. A biasymptotic set would be

Proof. If $\sigma $ is a strategy for II in $K[(a,p),(A,\vec {0})]$ to reach $\mathcal {C}$ , then the strategy $\tau $ for II in $K[a,A]$ defined by $\tau (A_0,\dots ,A_{n-1}) := (b,B)$ , where $\sigma ((A_0,\vec {0}),\dots ,(A_{n-1},\vec {0})) = ((b,p),(B,u))$ for some $p,u$ , is a strategy to reach $\pi _0[\mathcal {C}]$ .

Proof. We shall construct a strategy $\tau $ for I in $K[a,A]$ to reach $\pi _0[\mathcal {C}]^c$ . Let enumerate the set . We define a decreasing sequence as follows: Let $\sigma _{p_0}$ be the strategy for I in $K[(a,p_0),(A,\vec {0})]$ to reach $\mathcal {C}^c$ , and let $C_0 := \sigma _{p_0}(\emptyset )$ . We also define $B_{p_0} := A$ . Suppose that $A_k^0$ has been defined. By the hypothesis, there exists some $B_{p_{k+1}} \in [a,C_k]$ such that I has a strategy in $K[(a,p_{k+1}),(B_{p_{k+1}},\vec {0})]$ to reach $\mathcal {C}^c$ . Now let $C_{k+1} := \sigma _{p_{k+1}}(\emptyset )$ . Having constructed the above sequence, we now define .

Note that we may assume that for all partial states t of $K[(a,p),(B_p,\vec {0})]$ for II, $\sigma _p(t) = (B,\vec {0})$ for some $B \in \mathcal {R}$ . Now let s be a partial state of $K[a,A]$ for II following $\tau $ so far, and , and assume that $\tau (s)$ has been defined. Suppose for the induction hypothesis that we have a set such that:

1. (1) $t_{s,q}$ is a partial state of following with $a(t_{s,q}) = (a(s),q)$ .

2. (2) If , then $\tau (s) \in [a(s),A_{s,q}]$ .

Note that for the base case, for each we let $t_{\emptyset ,a}$ be the empty state of the game $K[(a,p),(B_p,\vec {0})]$ . For each and $B_n \in [a_{n+1},\tau (s)]$ , Let enumerate the set , and for each k we let and (which differs from the enumeration in the first paragraph, but it doesn’t matter). We define a decreasing sequence as follows: Let $D_0 := \sigma _{p_0}({t_{s,q_0'}}^\frown ((a_{n+1},q_0),(B_n,\vec {0})))$ . Assuming that $D_k$ has been defined, we let $D_{k+1} := \sigma _{p_k}({t_{s,q_k'}}^\frown ((a_{n+1},q_k),(D_k,\vec {0})))$ . Note that by the induction hypothesis, all the partial states listed above are legal. We conclude the construction of $\tau $ by asserting that .

We shall now show that $\tau $ is a strategy for I in $K[a,A]$ to reach $\pi _0[\mathcal {C}]^c$ . If not, then there exists some complete play $K[a,A]$ following $\tau $ such that II plays $(a_1,B_0,a_2,B_1,\dots )$ , and . In particular, we have that $(C,x) \in \mathcal {C}$ for some . For each n, let and let $p := q_0$ . By our construction of $\tau $ , there exists a complete play of the game $K[(a,p),(B_p,\vec {0})]$ following $\sigma _p$ such that II plays $((a_1,q_1),(B_0,\vec {0}),(a_2,q_2),(B_1,\vec {0}),\dots )$ . But since $\sigma _p$ is a strategy for I in $K[(a,p),(A,\vec {0})]$ to reach $\mathcal {C}^c$ , we have that $(C,x) \in \mathcal {C}^c$ , a contradiction.

Proof. The set of Ramsey subsets of $\mathcal {R}$ is closed under complements by definition. By Proposition 3.6, the set of Ramsey subsets of $\mathcal {R}$ is closed under countable intersections, so it forms a $\sigma $ -algebra. By Theorems 1.1 and 1.2, every analytic subset of $\mathcal {R}$ is contained in this $\sigma $ -algebra.

Proof. We first note that since $\mathcal {R}$ is deep, it is always possible for II to respond with a legal move in $Y[a,A]$ . The deepness of $\mathcal {R}$ also allows us to view $Y[a,A]$ as a “special case” of $K[a,A]$ : In $K[a,A]$ , if II responded with $(a_k,B_{k-1})$ , and I wants to restrict the next response by II such that for some , then I can respond to $(a_k,B_{k-1})$ by playing any $A_k \in [a_k,B_{k-1}]$ such that for all , . Therefore, a strategy for II in $K[a,A]$ to reach $\mathcal {X}$ may be passed to a strategy for II in $Y[a,A]$ to reach $\mathcal {X}$ .

Proof of Theorem 4.8.

It suffices to construct a $\boldsymbol {\Sigma }_2^1$ subset of $\mathcal {R}$ which is not pre-Kastanas Ramsey. Define the following two sets:

$$ \begin{align*} S_{\textbf{I}} &:= \{(a,A,\sigma) : a \in \mathcal{AR}\mathord{\upharpoonright} A \wedge \sigma \text{ is a strategy for } \textbf{I} \text{ in } Z^*[a,A]\}, \\ S_{\textbf{II}} &:= \{(a,A,\sigma) : a \in \mathcal{AR}\mathord{\upharpoonright} A \wedge \sigma \text{ is a strategy for } \textbf{II} \text{ in } Y[a,A]\}. \end{align*} $$

Note that a strategy in a game is a function from the set of partial states of the game to a play of the game. Since $\mathcal {AR}$ is countable, the sets of partial states of $Z^*[a,A]$ and of $Y[a,A]$ are also countable, and in both games all players play from a countable set. Therefore, we may naturally embed both sets $S_{\textbf {I}}$ and $S_{\textbf {II}}$ to the reals, giving us a $\Sigma _2^1$ -well ordering $\prec _s$ of $S_{\textbf {I}}$ and $S_{\textbf {II}}$ . Note that $\prec _s$ is of order-type , so every triple in $S_{\textbf {I}} \cup S_{\textbf {II}}$ has countably many $\prec _s$ -predecessors.

Given a triple $(a,A,\sigma ) \in S_{\textbf {I}} \cup S_{\textbf {II}}$ , we build $B_{a,A,\sigma } \in \mathcal {R}$ by an increasing sequence $a = b_0 \sqsubseteq b_1 \sqsubseteq \cdots $ with , and let . We consider two cases.

1. (1) If $(a,A,\sigma ) \in S_{\textbf {I}}$ , then we shall construct some $B_{a,A,\sigma } \in \mathcal {R}$ such that $B_{a,A,\sigma }$ is the outcome of some full play in $Z^*[a,A]$ , with I following $\sigma $ , and that $B_{a,A,\sigma } \neq B_{a',A',\sigma '}$ for all $(a',A',\sigma ') \prec _s (a,A,\sigma )$ . We do this as follows: We first enumerate the $\prec _s$ -predecessors by . Following a play in $Z^*[a,A]$ where I follows $\sigma $ , suppose that I started the $n^{\text {th}}$ turn with $B_n \in [b_n,A]$ . If or , then pick any . Otherwise, since $\mathcal {R}$ is deep we may pick some such that . This construction ensures that $B_{a,A,\sigma } \neq B_{a_n,A_n,\sigma _n}$ for all .

2. (2) If $(a,A,\sigma ) \in S_{\textbf {II}}$ , then we shall construct some $B_{a,A,\sigma } \in \mathcal {R}$ such that $B_{a,A,\sigma }$ is the outcome of some full play in $Y[a,A]$ , with II following $\sigma $ , and that $B_{a,A,\sigma } \neq B_{a',A',\sigma '}$ for all $(a',A',\sigma ') \prec _s (a,A,\sigma )$ . We do this as follows: Again, we enumerate the $\prec _s$ -predecessors by . Following a play in $Y[a,A]$ where II follows $\sigma $ , suppose that the sequence $b_n$ has been played so far. If , we then ask that I respond with , so that for any $b_{n+1}$ that II respond with next, . Otherwise, I may respond with any . This construction ensures that $B_{a,A,\sigma } \neq B_{a_n,A_n,\sigma _n}$ for all .

Now let

$$ \begin{align*} \mathcal{X} := \{B_{a,A,\sigma} : (a,A,\sigma) \in S_{\textbf{I}}\}. \end{align*} $$

$\mathcal {X}$ is $\boldsymbol {\Sigma }_2^1$ , as the well-ordering $\prec _s$ is $\Sigma _2^1$ and the construction of $\mathcal {X}$ is natural from $\prec _s$ . We then see that $\mathcal {X}$ is not pre-Kastanas Ramsey: Let $A \in \mathcal {R}$ and $a \in \mathcal {AR}$ .

1. (1) If $\sigma $ is a strategy for I in $Z^*[a,A]$ , then $B_{a,A,\sigma }$ is the outcome of a run following $\sigma $ such that $B_{a,A,\sigma } \notin \mathcal {X}^c$ , so $\sigma $ is not a winning strategy for I.

2. (2) If $\sigma $ is a strategy for II in $Y[a,A]$ , then $B_{a,A,\sigma }$ is the outcome of a run following $\sigma $ such that $B_{a,A,\sigma } \notin \mathcal {X}$ , so $\sigma $ is not a winning strategy for II.

This completes the proof.

Proof. By Corollary 4.17, it suffices to show that every wA2-space above is deep.

1. (1) Let $A = \{n_0,n_1,\dots \} \in [\mathbb {N}]^\infty $ , and let $a \subseteq A$ be finite. For all N such that $n_N> \max (a)$ , we have that . Therefore, and .

2. (2) Let $A = (x_0,x_1,\dots ) \in \mathbf {FIN}_k^{[\infty ]}$ , and let $a \in \mathbf {FIN}_k^{[<\infty ]}\mathord {\upharpoonright } A$ . For all N such that $x_N> a$ , we have that . Therefore, and .

3. (3) The proof is identical to that of .

4. (4) Let $A = (x_0,x_1,\dots ) \in W_{Lv}^{[\infty ]}$ , and let $a = (y_i)_{i<n} \in W_{Lv}^{[<\infty ]}\mathord {\upharpoonright } A$ . Since A is rapidly increasing, for N large enough we have that $\sum _{i<n} |y_i| < |x_N|$ . Therefore, and .

5. (5) Let $A \subseteq T$ be a strong subtree. Given any $a \in \S _{<\infty }\mathord {\upharpoonright } A$ , we let $S_a$ be the set of terminal nodes in a. Since a is a strong subtree, for some N. We observe that

   

   Fix any $a \in \S _{<\infty }\mathord {\upharpoonright } A$ and be large enough. For each $s \in S_a$ , let be such that $s \sqsubseteq t_{s,0}$ , $s \sqsubseteq t_{s,1}$ and $t_{s,0} \neq t_{s,1}$ . We let

   $$ \begin{align*} b := \{u \in A : u \sqsubseteq t_{s,i} \text{ for some } s \in S_a \text{ and } i \in \{0,1\}\}. \end{align*} $$

   Observe that a is an initial segment of b, and every terminal node in a splits in b. Thus, and .

6. (6) Let $A \in \mathcal {E}_\infty $ , and let $a \in \mathcal {AE}_\infty \mathord {\upharpoonright } A$ . Let be the increasing enumeration of the minimal representatives of A. Given , we define the equivalence relation b on $\{0,1,\dots ,p_N(A)-1\}$ as follows: Given $i,j \in \mathbb {N}$ , we define

   

   We shall show that and .

   Given $i,j < p_N(A)$ such that $(i,j) \in A$ , if or then $(i,j) \in b$ . Otherwise, $(i,j) \in b$ as well. Therefore, b is an equivalence relation on which is coarser than A, so . To see that $a \sqsubseteq b$ —if and $(i,j) \in b$ , then $(i,j) \in A$ , so $(i,j) \in a$ as a is coarser than A. Finally, the equivalence classes in b are either of the form $[i]_b$ for some (of which $(i,j) \notin a$ implies that $[i]_b \neq [j]_b$ ), or $[i]_b$ for any (of which $[i]_b = [j]_b$ for all ). Therefore, b has many equivalence classes, i.e., .

7. (7) The proof is identical to that of .

Proof. The forward direction for both statements has been proven in Proposition 2.6 of [Reference de Rancourt11], so we only prove the converse for (1) (the proof for the converse for (2) is almost verbatim). Suppose $\sigma $ is a strategy for I in $AG_{\textbf {I}}(p)$ to reach $\mathcal {X}$ , and we define a strategy $\tau $ for I in $R(p)$ . For each state s for II of $R(p)$ following $\tau $ , we shall correspond it to a state $t_s$ for II of $AG_{\textbf {I}}(p)$ realising $a(s)$ . Start by letting $t_{(p_0)} := (p_0)$ for any . Now let s be a state of $R(p)$ for II following $\tau $ so far, with $s = {s'}^\frown (y_n,p_{n+1})$ . Suppose by the induction hypothesis that there exists a corresponding state $t_s$ of the game $AG_{\textbf {I}}(p)$ such that:

1. (1) $a(s') = a(t_{s'})$ ;

2. (2) $\sigma (t_{s'}) = (x_n,q_n')$ for some .

Now let $t_s := {t_{s'}}^\frown (y_n,p_{n+1})$ , and suppose $\sigma (t_s) = (x_{n+1},q_{n+1}')$ for some . Since , by Property (2) of Definition 5.1 there exists some such that . Then $\tau (s) := (x_{n+1},p_{n+1})$ is a legal continuation. This completes the inductive definition of $\tau $ , which is a winning strategy as every complete play following $\tau $ corresponds to a complete play following $\sigma $ realising the same sequence.

Proof. We let be a Gowers space, and assume WLOG that $0 \notin X$ . We then define a relation such that for all :

1. (1) If is odd, then for all $x \in X \cup \{0\}$ and $p \in P$ , $s^\frown x \blacktriangleleft P$ iff $x = 0$ .

2. (2) If is even, then for all $x \in X \cup \{0\}$ and $p \in P$ , $s^\frown x \blacktriangleleft P$ iff $x \neq 0$ and .

It is easy to verify that is a Gowers space. We now define an injective function by

$$ \begin{align*} f((x_0,\dots,x_{n-1})) := (x_0,0,x_1,0\dots,x_{n-1},0) \end{align*} $$

and naturally extend f to . Note that f is injective. For each $p \in P$ , we also define the functions $g,h$ by

$$ \begin{align*} g(p_0,0,p_1,0,p_2,\dots) &:= (p_0,p_1,p_2,\dots), \\ h(x_0,q_0,x_1,q_1,\dots) &:= (x_0,x_1,\dots). \end{align*} $$

We may now observe that:

1. (1) $\sigma $ is a strategy for I in $K(p)$ to reach $\mathcal {X}$ iff:

   $$ \begin{align*} s \mapsto \begin{cases} \sigma(s), &\text{if } s = \emptyset, \\ (0,\sigma(s)), &\text{if } s \neq \emptyset \\ \end{cases} \end{align*} $$
   is a strategy for II in $R(p)$ to reach $f[\mathcal {X}]$ .

2. (2) $\sigma $ is a strategy for II in $K(p)$ to reach $\mathcal {X}$ iff $\sigma \circ g$ is a strategy for I in $R(p)$ to reach $f[\mathcal {X}]$ .

3. (3) $\sigma $ is a strategy for I in $F(p)$ to reach $\mathcal {X}$ iff:

   $$ \begin{align*} s \mapsto \begin{cases} (\sigma \circ h)(s), &\text{if } s = \emptyset, \\ (0,(\sigma \circ h)(s)), &\text{if } s \neq \emptyset \\ \end{cases} \end{align*} $$
   is a strategy for II in $AG_{\textbf {II}}(p)$ to reach $f[\mathcal {X}]$ .

4. (4) $\sigma $ is a strategy for II in $G(p)$ to reach $\mathcal {X}$ iff $s \mapsto ((\sigma \circ g)(s),p)$ is a strategy for I in $AG_{\textbf {I}}(p)$ to reach $f[\mathcal {X}]$ .

Therefore, the proposition follows from Proposition 5.10.

Notation 5.14. Let be a wA2-space. Given $A,B \in \mathcal {R}$ and $a \in \mathcal {AR}$ , we write iff there exists some $b \in \mathcal {AR}\mathord {\upharpoonright }[a,A]$ such that $[b,A] \subseteq [b,B]$ .