Proof. Fix n and enumerate $\Sigma _n$ injectively as $\{\nu _0,\ldots ,\nu _N\}$ . Since $[\nu _0]\cap K$ is perfect by $(e_f)$ , we can pick $E_0(\nu _0)\in ([\nu _0]\cap K)\setminus X$ . Fix a number $k<N$ and put $C:=\{\nu _i:i\leq k\}$ . Assume that we have already defined a coherent map $E_k:C\to K\setminus X$ . Then for every $m>n$ and $\nu _i\in C$ , there is $e^m_0(\nu _i)\in \Sigma _m$ such that $E_k(\nu _i)\in [e^m_0(\nu _i)]$ . In that way we define a map $e^m_0\colon C\to \Sigma _m$ . It is clear that this map is coherent. By $(e_f)$ there are $\mu _{\langle 0\rangle }\neq \mu _{\langle 1\rangle }\in \Sigma _{n+1}$ such that both $e^{n+1}_{\langle 0\rangle }=e^{n+1}_0\cup \{\langle \nu _{k+1}, \mu _{\langle 0\rangle }\rangle \}$ and $e^{n+1}_{\langle 1\rangle }=e^{n+1}_0\cup \{\langle \nu _{k+1}, \mu _{\langle 1\rangle }\rangle \}$ are coherent as maps from $C\cup \{\nu _{k+1}\}$ to $\Sigma _{n+1}$ .

Suppose that for some $m>n$ we have defined mutually different $\{\mu _s:s\in 2^{m-n}\}\subseteq \Sigma _m\setminus e_0^m[C]$ such that $e^{m}_{ s}=e^{m}_0\cup \{\langle \nu _{k+1}, \mu _{ s}\rangle \}$ is coherent as a map from $C\cup \{\nu _{k+1}\}$ to $\Sigma _{m}$ for all $s\in 2^{m-n}$ . By $(e_f)$ for every $s\in 2^{m-n}$ there are $\mu _{ s\hat {\ \ } 0}\neq \mu _{ s\hat {\ \ }1}\in \Sigma _{m+1}$ such that

$$ \begin{align*}\{\langle e^m_0(\nu_i), e^{m+1}_0(\nu_i)\rangle:i\leq k\}\cup \{\langle \mu_s, \mu_{ s\hat{\ \ }j} \rangle\}\end{align*} $$

is coherent as a map from $\Sigma _m$ to $\Sigma _{m+1}$ for all $j\in 2$ . It follows that

$$ \begin{align*}e^{m+1}_{s\hat{\ \ }j}:=e^{m+1}_0\cup\{\langle \nu_{k+1}, \mu_{ s\hat{\ \ }j}\rangle\}\end{align*} $$

is coherent as a map from $C\cup \{\nu _{k+1}\}\to \Sigma _{m+1}$ for all $s\in 2^{m-n}$ and $j\in s$ , which completes our construction of the maps $e^m_s$ and elements $\mu _s$ as above for all $m>n$ and $s\in 2^{m-n}$ .

For every $t\in 2^\omega $ let $z_t\in K$ be the unique element in $\bigcap _{m>n}[\mu _ {t\upharpoonright (m-n)}]$ and note that $z_t\neq z_{t'}$ for any $t\neq t'$ in $2^\omega $ . Since the set $\{\,z_t : t\in 2^\omega \,\}$ is perfect, there exists t with $z_t\not \in X$ . It suffices to observe that $E_{k+1}:=E_k\cup \{\langle \nu _{k+1},z_t\rangle \}$ is a coherent map whose range is disjoint from X, which allows us to complete our proof by induction on $k<N$ .

Proof. We shall describe a strategy $\S $ for Alice in the Menger game played on X such that each play lost by Alice gives rise to the objects whose existence we need to establish. For every $n\in \omega $ , let $E_n\colon \Sigma _n\to K$ be a coherent selection from Lemma 2.1. Put

$$\begin{align*}i_0:=0,\quad C_0:=\Sigma_{i_0},\quad Z_0:=E_{i_0}[\Sigma_{i_0}]\subseteq K\setminus X. \end{align*}$$

For every $j\geq i_0$ and $\nu \in \Sigma _{i_0}$ , let $ \sigma _{0,j}(\nu )$ be the unique element of $\Sigma _j$ such that $E_{i_0}(\nu )\in [\sigma _{0,j}(\nu )]$ . Then,

$$\begin{align*}{\big\{\,\bigcup_{\nu\in\Sigma_{i_0}}[\sigma_{0,j}(\nu)] : j\geq i_0\,\big\}} \end{align*}$$

is a decreasing family of clopen sets in $(2^\omega )^S$ , whose intersection is equal to the set $Z_0$ . Since $Z_0\subseteq K\setminus X$ , the family $\mathcal {U}_0$ of all sets

$$\begin{align*}U^0_j:=K\setminus \bigcup_{\nu\in\Sigma_{i_0}}[\sigma_{0,j}(\nu)], \end{align*}$$

where $j\geq i_0$ , is an increasing open cover of X. Now, $\S $ instructs Alice to start the play with $\mathcal {U}_0$ .

Suppose that Bob chooses $U^0_{j_0}$ for some $j_0\geq i_0$ . For each $\nu \in \Sigma _{i_0}$ , let $e_0(\nu ):=\sigma _{0,j_0}(\nu )$ , an element of $\Sigma _{j_0}$ . Since $E_{i_0}$ is a coherent selection, the map $e_0\colon C_0\to \Sigma _{j_0}$ is coherent.

Then we put

$$\begin{align*}i_1:=j_0+1,\quad C_{1}:=\bigcup_{\nu\in C_0}\{\,\sigma\in\Sigma_{i_{1}} : \sigma \succ e_0(\nu)\,\}. \end{align*}$$

Suppose that a natural number $i_n$ and a set $C_n\subseteq \Sigma _{i_n}$ have already been defined for some $n>0$ . Let $Z_n:=E_{i_n}[C_{n}]\subseteq K\setminus X$ . For every $j\geq i_n$ and $\nu \in C_n$ , let $ \sigma _{n,j}(\nu )$ be the unique element of $\Sigma _j$ such that $E_{i_n}(\nu )\in [\sigma _{n,j}(\nu )]$ . Then,

$$\begin{align*}{\big\{\,\bigcup_{\nu\in C_{n}}[\sigma_{n,j}(\nu)] : j\geq i_n\,\big\}} \end{align*}$$

is a decreasing family of clopen sets in $(2^\omega )^S$ , whose intersection is equal to the set $Z_n$ . Since $Z_n\subseteq K\setminus X$ , the family $\mathcal {U}_0$ of all sets

$$\begin{align*}U^n_j:=K\setminus \bigcup_{\nu\in C_{n}}[\sigma_{n,j}(\nu)], \end{align*}$$

where $j\geq i_n$ , is an increasing open cover of X. Now, $\S $ instructs Alice to play the family $\mathcal {U}_n$ .

Suppose that Bob chooses $U^n_{j_n}$ for some $j_n\geq i_n$ . Since $E_{i_n}$ is a coherent selection, the map $e_n\colon C_n\to \Sigma _{j_n}$ is coherent. Then we put

$$\begin{align*}i_{n+1}:=j_n+1,\quad C_{n+1}:=\bigcup_{\nu\in C_n}\{\,\sigma\in\Sigma_{i_{n+1}} : \sigma \succ e_n(\nu)\,\}. \end{align*}$$

This completes our definition of the strategy $\S $ for Alice such that each infinite play in the Menger game on X in which Alice uses $\S $ gives rise to a sequence

$$\begin{align*}{\langle\, \langle i_n,j_n,C_n,e_n,\mathcal U_n\rangle : n\in\omega\,\rangle} \end{align*}$$

as described above. In particular, conditions $(1)$ – $(4)$ are satisfied by the construction. By Theorem 1.4, there is a play, where Alice uses the strategy $\S $ and the play is won by Bob. Then $X\subseteq \bigcup _{n\in \omega }U^n_{j_n}$ , i.e.,

$$\begin{align*}\emptyset=X\cap \bigcap_{n\in\omega}\bigcup_{\nu\in C_n}[\sigma_{n,j_n}(\nu)]. \end{align*}$$

For each n, we have

$$\begin{align*}\bigcup_{\nu\in C_n}[\sigma_{n,j_n}(\nu)]=\bigcup_{\nu\in C_n}[e_n(\nu)]=\bigcup_{\nu\in C_n}\bigcup\{\,[\sigma] : \sigma\in\Sigma_{i_{n+1}},\sigma \succ e_n(\nu)\,\}=\bigcup_{\sigma\in C_{n+1}}[\sigma], \end{align*}$$

and thus

$$\begin{align*}\emptyset=X\cap \bigcap_{n\in\omega}\bigcup_{\nu\in C_n}[\sigma_{n,j_n}(\nu)]=X\cap \bigcap_{n\in\omega}\bigcup_{\nu\in C_{n+1}}[\nu]. \end{align*}$$

It follows that condition $(5)$ is also satisfied.

Observation 3.1. In the notation above, if $\sigma $ is consistent with p, then $\sigma \upharpoonright \beta $ is consistent with both $p\upharpoonright \beta $ and p, and $\big (p|(\sigma \upharpoonright \beta )\big )\upharpoonright \beta =(p\upharpoonright \beta )|(\sigma \upharpoonright \beta )$ .

Observation 3.2. Let $p\in \mathbb S_\alpha $ be an $(F,n)$ -determined condition, $\Sigma $ the set of all maps $\nu \colon F\to 2^n$ consistent with p and $\beta <\alpha $ . Then the following assertions hold.

1. (1) If $(q,n)\geq _F (p,n)$ , then q is also $(F,n)$ -determined;

2. (2) If $\sigma \in \Sigma $ , then $p|(\sigma \upharpoonright \beta )$ is $(F\setminus \beta , n)$ -determined, and $\nu \in (2^n)^{F\setminus \beta }$ is consistent with $p|(\sigma \upharpoonright \beta )$ iff $(\sigma \upharpoonright \beta )\cup \nu \in \Sigma $ ;

3. (3) The set $\{\,p|\sigma : \sigma \in \Sigma \,\}$ is a maximal antichain above p;

4. (4) p is $(F\cap \beta ,n)$ -determined and $\{\sigma \upharpoonright (F\cap \beta ):\sigma \in \Sigma \}$ is the family of all functions from $F\cap \beta $ to $2^n$ consistent with p;

5. (5) If $\sigma \in \Sigma $ and $r\geq p|\sigma $ , then there exists $q\in \mathbb S_\alpha $ with $(q,n)\geq _F(p,n)$ and ${q|\sigma =r;}$

6. (6) If $D\subseteq \mathbb S_\alpha $ is open and dense, then there exists $q\in \mathbb S_\alpha $ with $(q,n)\geq _F(p,n)$ and $q|\sigma \in D$ for all $\sigma \in \Sigma $ ; and

7. (7) If $\tau $ is an $\mathbb S_\alpha $ -name for a real, then for each natural number l there is a condition $q\in \mathbb S_\alpha $ and a family $\{y_\sigma :\sigma \in \Sigma \}\subseteq 2^l$ such that $(q,n)\geq _F (p,n)$ and $q|\sigma \Vdash \tau \upharpoonright l= y_\sigma $ for all $\sigma \in \Sigma $ .

Observation 3.4. Suppose that p is $(F,n)$ -determined, $\beta \leq \alpha $ , and $p\upharpoonright \beta \in G_\beta $ . Then in $V[G_\beta ]$ , $p\upharpoonright [\beta ,\alpha )^{G_\beta }\in \mathbb S_{\beta ,\alpha }^{G_\beta }$ is $(F\setminus \beta ,n)$ -determined.

Moreover, if $\sigma \in (2^n)^F$ is consistent with p and $(p|(\sigma \upharpoonright \beta ))\upharpoonright \beta \in G_\beta $ , then $\sigma \upharpoonright (F\setminus \beta )$ is consistent with $p\upharpoonright [\beta ,\alpha )^{G_\beta }$ in $V[G_\beta ]$ ; and if $(p|(\sigma \upharpoonright \beta ))\upharpoonright \beta \in G_\beta $ and $\nu \in (2^n)^{F\setminus \beta }$ is consistent with $p\upharpoonright [\beta ,\alpha )^{G_\beta }$ in $V[G_\beta ]$ , then $(\sigma \upharpoonright \beta )\cup \nu \in (2^n)^F$ is consistent with p.

Proof. Let $\xi =\min (F)$ and note that the fact that p is $(F,n)$ -determined yields $N\leq 2^n$ and $\{s_i:i<N\}\subseteq 2^n$ such that $p\upharpoonright \xi $ forces $p(\xi )\cap 2^n=\{s_i:i<N\}$ . For every $i<N$ let $\mu _i$ be the map $\{\langle \xi ,s_i\rangle \}$ and note that $\mu _i$ is consistent with p.

By induction on $i<N$ , using Lemma 3.3 and Observation 3.2(2), we can find mutually disjoint finite sets $X_i\subseteq {2^\omega }\cap V$ , $i<N$ , such that $|X_i|\leq 2^{n\cdot (|F|-1)}$ , and for each natural number l there is a condition $u^l_i\in \mathbb S_\alpha $ with $u^l_j\upharpoonright \xi \geq u^l_i\upharpoonright \xi $ for all $i<j\leq N$ , $u^l_0\upharpoonright \xi \geq p\upharpoonright \xi $ , $(u^l_i,n)\geq _{F\setminus \{\xi \}} (p|\mu _i,n)$ and

$$\begin{align*}u^l_{i}\Vdash \big(\operatorname{\mathrm{\exists}} x\in X_i\big)\big(\tau\upharpoonright l= x\upharpoonright l\big). \end{align*}$$

Pick a natural number $l_{*}>k$ such that $x\upharpoonright l_{*}\neq x'\upharpoonright l_{*}$ for any distinct $x,x'$ in $\bigcup _{i<N} X_i$ . As a result, the elements of the family $\{\,X_i\upharpoonright l_{*} : i<N\,\}$ are mutually disjoint.

Now we proceed by induction on the cardinality of F. If $F=\{\xi \}$ , then $|X_i|=1$ for all $i<N$ (because $ 2^{n\cdot (|F|-1)}=1$ ), i.e., $X_i=\{x_i\}$ for some $x_i\in 2^\omega $ . Put $y_{\mu _i}:=x_i\upharpoonright l_{*}$ for all $i<N$ and let $r\in \mathbb S_\alpha $ be a condition such that $r\upharpoonright \xi =u^{l_{*}}_{N-1}\upharpoonright \xi $ , $r\upharpoonright \xi $ forces $r(\xi )$ to be $\bigcup \{\,u^{l_{*}}_i(\xi ) : i<N\,\}$ , and $(r|\mu _i)\upharpoonright \beta $ forces $r(\beta )=u^{l_{*}}_i(\beta )$ for all $\beta>\xi $ . It follows from the above that $r|\mu _i\geq u^{l_{*}}_i$ and hence $r|\mu _i$ forces $\tau \upharpoonright l_{*}=x_i\upharpoonright l_{*}=y_{\mu _i}$ , hence $q:=r$ and $l:=l_{*}$ are as required.

Now assume that $\left |F\right |>1$ and the statement holds for each set of cardinality smaller than $\left |F\right |$ . Let $\xi ,l_{*},r$ be as above and note that by the construction we have

$$ \begin{align*}(r|\mu_i,n)\geq_{F\setminus\{\xi\}}(u^{l_{*}}_i,n) \geq_{F\setminus\{\xi\}}(p|\mu_i,n)\end{align*} $$

for all $i<N$ , and hence $(r,n)\geq _F (p,n)$ . Fix $i<N$ and let G be an $\mathbb {S}_{\xi +1}$ -generic filter containing $(r|\mu _i)\upharpoonright (\xi +1)$ . Work in $V[G]$ . Then

$$ \begin{align*}r^{\prime}_i:=\big((r|\mu_i)\upharpoonright[\xi+1,\alpha)\big)^G \in \mathbb{S}_{\xi+1,\alpha} ^G\end{align*} $$

is $(F\setminus \{\xi \},n)$ -determined by Observation 3.2(1) because

$$ \begin{align*}(r^{\prime}_i,n) \geq_{F\setminus\{\xi\}} \Big(\big(u^{l_{*}}_i\upharpoonright[\xi+1,\alpha)\big)^{G},n\Big)\end{align*} $$

by the construction, and $\big (u^{l_{*}}_i\upharpoonright [\xi +1,\alpha )\big )^{G}$ is $(F\setminus \{\xi \},n)$ -determined by Observation 3.4. Note that

$$\begin{align*}r_i^{\prime}\Vdash \tau\in ({2^\omega}\cap V[\dot{G}_{\xi+1,\alpha}])\setminus\bigcup_{\xi<\beta<\alpha}({2^\omega}\cap V[\dot{G}_{\xi+1,\beta}]). \end{align*}$$

because $r^{\prime }_i\geq p\upharpoonright [\xi +1,\alpha )^G$ and $p\restriction (\xi +1)\in G$ . By the inductive assumption, there exist a condition $r^{\prime \prime }_i\in \mathbb {S}_{\xi +1,\alpha }^{G} $ , a natural number $l_i>l_{*}$ and pairwise different elements $t_{\sigma '}\in 2^{l_i}$ for all maps $\sigma '\colon F\setminus \{\xi \}\to 2^n$ consistent with $r_i^{\prime \prime }$ , such that $(r_i^{\prime \prime },n)\geq _{F\setminus \{\xi \}}(r_i^{\prime },n)$ and $r_i^{\prime \prime }|\sigma '\Vdash \tau \upharpoonright l_i = t_{\sigma '}$ . Let $\Sigma ^{\prime }_i$ be the set of all maps $\sigma '\colon F\setminus \{\xi \}\to 2^n$ consistent with $r_i^{\prime \prime }$ .

Now we work in V. Let $\Sigma $ be the set of all maps $\sigma \colon F\to 2^n$ consistent with p. Let , , and be $\mathbb {S}_{\xi +1}$ -names for the condition $r_i^{\prime \prime }$ , the set $\Sigma ^{\prime }_i$ , natural number $l_i$ and finite sequences $t_{\sigma '}$ , respectively. Note that

by the second part of Observation 3.4. By induction on $i<N$ pick a condition $r_i\in \mathbb S_{\xi +1}$ stronger than $(r|\mu _i)\upharpoonright (\xi +1)$ and such that $r_j\upharpoonright \xi \geq r_i\upharpoonright \xi $ for all $i<j<N$ , $r_0\upharpoonright \xi \geq r\upharpoonright \xi $ , and which forces all the above properties, and also decides all the names mentioned in the previous sentences. More precisely, there exist $l_i>l_{*}$ , $t^i_\sigma \in 2^{l_i}$ for all $\sigma \in \Sigma $ with $\sigma (\xi )=s_i$ , and $r_i^{\prime \prime }\in \mathbb S_{\xi +1,\alpha }$ such that $r_i$ forces that , and for all maps $\sigma \in \Sigma $ with $\sigma (\xi )=s_i$ . Thus the elements $t^i_{\sigma }$ are pairwise different for all maps $\sigma \in \Sigma $ with $\sigma (\xi )=s_i$ .

Let $w\in \mathbb S_\alpha $ be a condition such that $w\upharpoonright \xi =r_{N-1}\upharpoonright \xi $ ,

$$\begin{align*}w(\xi)=\bigcup\{\,r_i(\xi) : i<N\,\}, \end{align*}$$

and for every ordinal number $\beta $ with $\xi <\beta <\alpha $ and natural number $i<N$ , we have

$$\begin{align*}(w|\mu_i)\upharpoonright\beta \Vdash w(\beta)=r_i^{\prime\prime}(\beta). \end{align*}$$

It follows that

$$ \begin{align*}w|\sigma\Vdash \tau\upharpoonright l_i = t^i_\sigma\end{align*} $$

for all $\sigma \in \Sigma $ with $\sigma (\xi )=s_i$ because

$$ \begin{align*}w|\sigma\geq r_i\cup r_i^{\prime\prime}|\big(\sigma\upharpoonright(F\setminus\{\xi\})\big).\end{align*} $$

Let $\sigma ,\nu \in \Sigma $ be different maps. Assume that $\sigma (\xi )\neq \nu (\xi )$ . Then there are different natural numbers $i,j$ such that $\sigma (\xi )=s_i$ and $\nu (\xi )=s_j$ . Since $w|\sigma \geq w|\mu _i\geq u^{l_{*}}_i$ and $w|\nu \geq w|\mu _j\geq u^{l_{*}}_j$ , we have

$$\begin{align*}w|\sigma\Vdash\tau\upharpoonright l_i=t^i_\sigma,\quad\tau\upharpoonright l_{*}\in \{\,x\upharpoonright l_{*} : x\in X_i\,\} \end{align*}$$

and

$$\begin{align*}w|\nu\Vdash\tau\upharpoonright l_j=t^j_\nu\quad\tau\upharpoonright l_{*}\in \{\,x\upharpoonright l_{*} : x\in X_j\,\}. \end{align*}$$

The sets $\{\,x\upharpoonright l_{*} : x\in X_i\,\}$ and $\{\,x\upharpoonright l_{*} : x\in X_j\,\}$ are disjoint, and thus $t^i_\sigma \upharpoonright l_{*}\neq t^j_\nu \upharpoonright l_{*}$ . Now assume that there is a natural number i such that $\sigma (\xi )=\nu (\xi )=s_i$ . Then $\sigma \upharpoonright F\setminus \{\xi \}\neq \nu \upharpoonright F\setminus \{\xi \}$ and the condition $(w|\mu _i)\upharpoonright (\xi +1)$ forces that and because $(w|\mu _i)\upharpoonright (\xi +1)\geq r_i$ , while $t^i_\sigma \neq t^i_\nu $ . Summarizing, if $\nu (\xi )=\sigma (\xi )=s_i$ , then $t^i_\sigma \neq t^i_\nu $ , and if $\nu (\xi )=s_j\neq s_i=\sigma (\xi )$ , then $t^i_\sigma \upharpoonright l_{*}\neq t^j_\nu \upharpoonright l_{*}$ .

Finally, applying Observation 3.2(7) to $l=\max _{i<N}l_i$ and w which is $(F,n)$ -determined (because $(w,n)\geq _F(p,n)$ by the construction), we get a condition q such that $(q,n)\geq _F (w,n)$ , and for every $\sigma \in \Sigma $ a sequence $y_\sigma \in 2^l$ such that $q|\sigma \Vdash \tau \upharpoonright l=y_\sigma $ . Since $q|\sigma \geq w|\sigma ,$ we have $y_\sigma \upharpoonright l_i=t^i_\sigma $ for all $\sigma \in \Sigma $ with $\sigma (\xi )=s_i$ . It follows from the above that the $y_\sigma $ ’s are mutually different: if $\nu (\xi )=\sigma (\xi )=s_i$ , then $y_\sigma \upharpoonright l_i =t^i_\sigma \neq t^i_\nu =y_\nu \upharpoonright l_i$ ; and if $\nu (\xi )=s_j\neq s_i= \sigma (\xi )$ , then $y_\sigma \upharpoonright l_{*}=t^i_\sigma \upharpoonright l_{*}\neq t^j_\nu \upharpoonright l_{*}=y_\nu \upharpoonright l_{*}$ .

Proof. We proceed by induction on $|F|$ . Suppose that $F=\{\beta \}$ for some $\beta <\alpha $ and pick $r\in \mathbb S_\beta $ , $r\geq p\upharpoonright \beta $ which decides $k>n$ and $p(\beta )\cap 2^k$ (and hence also decides $p(\beta )\cap 2^n$ ), so that each element $s\in p(\beta )\cap 2^n$ has at least 2 extensions in $p(\beta )\cap 2^k$ . Then $q:=r\cup p\upharpoonright [\beta ,\alpha )$ is as required.

Now assume that $|F|>1$ and let $\beta =\max (F)$ . By the inductive assumption there exists $r\in \mathbb S_\beta $ and $k_0>n$ such that r is $(F\cap \beta ,k_0)$ -determined and $(r,k_0)\geq _{F\cap \beta } (p\upharpoonright \beta ,n)$ . Let $\Sigma $ be the family of all $\sigma :F\cap \beta \to 2^{k_0}$ consistent with r. Let $N=|\Sigma |$ and write $\Sigma $ in the form $\{\sigma _i:i<N\}$ . By induction on $i<N$ let us construct a sequence

$$ \begin{align*}(r^0_{N-1},k_0)\geq_{F\cap\beta}\cdots \geq_{F\cap\beta}(r^0_{0},k_0)\geq_{F\cap\beta} (r^0_{-1},k_0),\end{align*} $$

where $r^0_{-1}=r$ and $r^0_i\in \mathbb S_\beta $ for all $i<N$ , as follows: Given $r^0_{i-1}$ for some $i<N$ , let $\mathbb S_\beta \ni u^0_i\geq r^0_{i-1}|\sigma _i$ be a condition such that there exists $T^0_i\subseteq 2^{l_i}$ for some $l_i\geq k_0$ such that $u^0_i$ forces $p(\beta )\cap 2^{l_i}=T^0_i$ , and for every $s\in T^0_i\upharpoonright n$ there exists at least two $t\in T^0_i$ extending s. Now, let $r^0_i\in \mathbb S_\beta $ be such that $(r^0_i,k_0)\geq _{F\cap \beta }(r^0_{i-1},k_0)$ and $r^0_i|\sigma _i=u^0_i$ , by Observation 3.2(5).

Let $k=\max _{i<N}l_i$ and set $r^1_{-1}=r^0_{N-1}$ . By induction on $i<N$ let us construct a sequence

$$ \begin{align*}(r^1_{n-1},k_0)\geq_{F\cap\beta}\cdots \geq_{F\cap\beta}(r^1_{0},k_0)\geq_{F\cap\beta} (r^1_{-1},k_0),\end{align*} $$

where $r^1_i\in \mathbb S_\beta $ for all $i<N$ , as follows: Given $r^1_{i-1}$ for some $i<N$ , let $\mathbb S_\beta \ni u^1_i\geq r^1_{i-1}|\sigma _i$ be a condition such that there exists $T^1_i\subseteq 2^{k}$ for which $u^1_i$ forces $p(\beta )\cap 2^{k}=T^1_i$ . Now, let $r^1_i\in \mathbb S_\beta $ be such that $(r^1_i,k_0)\geq _{F\cap \beta }(r^1_{i-1},k_0)$ and $r^1_i|\sigma _i=u^1_i$ . Note that $T^0_i=\{t\upharpoonright l_i:t\in T^1_i\}$ .

Set $r^2_{-1}=r^1_{N-1}$ . By induction on $i<N$ let us construct a sequence

$$ \begin{align*}(r^2_{n-1},k_0)\geq_{F\cap\beta}\cdots \geq_{F\cap\beta}(r^2_{0},k_0)\geq_{F\cap\beta} (r^2_{-1},k_0),\end{align*} $$

where $r^2_i\in \mathbb S_\beta $ for all $i<N$ , as follows: Given $r^2_{i-1}$ for some $i<N$ , let $\mathbb S_\beta \ni u^2_i\geq r^2_{i-1}|\sigma _i$ be a condition such that for every $\gamma \in F\cap \beta $ there exists $\nu _i(\gamma )\in 2^k$ such that $\sigma _i(\gamma )=\nu _i(\gamma )\upharpoonright k_0$ and $u^2_i\upharpoonright \gamma $ forces that $\nu _i(\gamma )$ is an initial segment of the stem of $u^2_i(\gamma )$ . Such an $u^2_i$ can be constructed recursively over $\gamma \in F\cap \beta $ , moving from the bigger to smaller elements. Now, let $r^2_i\in \mathbb S_\beta $ be such that $(r^2_i,k_0)\geq _{F\cap \beta }(r^2_{i-1},k_0)$ and $r^1_i|\sigma _i=u^2_i$ .

We claim that $q=r^2_{N-1}\cup p\upharpoonright [\beta ,\alpha )$ and k are as required. Indeed, we have that $(q\upharpoonright \beta ,k)\geq _{F\cap \beta } (p\upharpoonright \beta ,n)$ because $q\upharpoonright \beta =r^2_{N-1}$ , $k\geq k_0$ , and $(r^2_{N-1},k_0)\geq _{F\cap \beta } (r,k_0)\geq _{F\cap \beta } (p\upharpoonright \beta ,n)$ . Moreover, since $r^2_{N-1}|\sigma _i\geq u^0_i$ , we have that $r^2_{N-1}|\sigma _i$ decides $p(\beta )\cap 2^k$ as $T^1_i$ , which has the property that any $s\in T^1_i\upharpoonright n$ has at least two extensions in $T^1_i$ (because $T^0_i=T^1_i\upharpoonright l_i$ has this property). It follows that $r^2_{N-1}|\sigma _i$ forces $q(\beta )=p(\beta ) $ and $(p(\beta ),k)\geq (p(\beta ),n)$ . Since $\{r^2_{N-1}|\sigma _i:i<N\}$ is dense above $r^2_{N-1}$ , we conclude that $r^2_{N-1}=q\upharpoonright \beta $ forces $(q(\beta ),k)\geq (p(\beta ),n)$ , and therefore $(q,k)\geq (p,n)$ .

Finally, by the construction of $r^2_{N-1}$ we have that q is $(F,k)$ -determined, with

$$ \begin{align*}\big\{\nu_i\cup\{\langle\beta,t\rangle\}\: :\: i<N,t\in T^1_i\big\}\end{align*} $$

being the family of those $\nu :F\to 2^k$ which are consistent with q.

Proof. The choice of the $F_n$ ’s is standard and thus will not be specified, except that we set $F_0=\{0\}$ . Set also $k_0=0$ . Trivially, $p_0$ is $(F_0,k_0)$ -determined since the unique map $\{\langle 0,\emptyset \rangle \} $ in $(2^{k_0})^{F_0}$ is consistent with $p_0$ . By Lemma 3.5, there are a condition $q_0\in \mathbb S_\alpha $ , a natural number $l_0$ and pairwise different elements $y_{\sigma }\in 2^{l_0}$ for all mapsFootnote ² $\sigma \colon F_0\to 2^{k_0}$ consistent with $p_0$ such that $(q_0,k_0)\geq _{F_0}(p_0,k_0)$ and $q_0|\sigma \Vdash \tau \upharpoonright l_0=y_{\sigma }$ .

Let $k_1>k_0$ be a natural number and $p_1\in \mathbb S_\alpha $ a condition from Lemma 3.6, applied to the set $F_1$ and the condition $q_0$ . Then $p_1$ is $(F_1,k_1)$ -determined and $(p_1,k_1)\geq _{F_0}(p_0,k_0)$ .

Fix a natural number $n\geq 1$ and assume that a set $F_n$ , natural number $k_n$ , and an $(F_n,k_n)$ -determined condition $p_n\in \mathbb S_\alpha $ with $(p_n,k_n)\geq _{F_{n-1}}(q_{n-1},k_{n-1})$ have already been defined. By Lemma 3.5, there are a condition $q_n\in \mathbb S_\alpha $ , a natural number $l_n>l_{n-1}$ and pairwise different elements $y_{\sigma }\in 2^{l_n}$ for all maps $\sigma \colon F_n\to 2^{k_n}$ consistent with $p_n$ such that $(q_n,k_n)\geq _{F_n}(p_n,k_n)$ and $q_n|\sigma \Vdash \tau \upharpoonright l_n=y_{\sigma }$ .

Let p be the fusion of the sequence ${\langle \, (p_n,k_n,F_n) : n\in \omega \,\rangle }$ [Reference Baumgartner and Laver3, Lemma 1.2] and note that it is as required.

Observation 3.8. In the notation above, if $F,H\subseteq \alpha $ , $p\in \mathbb S_\alpha $ , $\nu ,\sigma $ are consistent with p and $\nu \prec \sigma $ , then $p|\sigma \geq p|\nu $ .

Proof. We start with proving the first part. Proceed by induction on $|G|$ . If $G=\emptyset $ , then there is nothing to prove. Let $\gamma :=\max G$ and assume that the statement holds for $G':=G\setminus \{\gamma \}$ . Fix a map $\mu \colon G\to 2^k$ consistent with p and let $\mu ':=\mu \upharpoonright G'$ . By Observation 3.2(4), p is both $(F\cap \gamma ,n)$ - and $(G\cap \gamma ,k)$ -determined, and hence there exists $\nu ':(F\cap \gamma )\to 2^n$ consistent with p such that $\nu '\succ \mu '$ . Since $(p\upharpoonright \gamma )|\mu '\Vdash \mu (\gamma )\in p(\gamma ),$ the condition $(p\upharpoonright \gamma )|\nu '$ also forces this because $(p\upharpoonright \gamma )|\nu '\geq (p\upharpoonright \gamma )|\mu '$ by Observation 3.8. Strengthening the latter condition to some r if necessary, we may find $t\in 2^n$ extending $\mu (\gamma )$ such that $r\Vdash t\in p(\gamma )$ . Since p is $(F,n)$ -determined, we conclude that $(p\upharpoonright \gamma )|\nu '\Vdash t\in p(\gamma ).$ It follows that $\nu '\cup \{(\gamma ,t)\}\in (2^n)^F$ is consistent with p and extends $\mu $ .

The second part of the lemma is rather straightforward. Item $(2)$ is an equivalent reformulation of $(1)$ , by Observation 3.2(2). We shall present the proof of $(3)$ , because we find it the least obvious one.

Proceed by induction on $|F|$ . Assume that $F=\{\beta \}$ . Let $\nu \in C$ and $\sigma := e_0(\nu )$ . Choose some distinct $\mu ,\mu '\in \Sigma _m$ such that $\mu ,\mu '\succ \nu _{*}$ and $\mu \restriction (H \cap \beta ) = \sigma \restriction (H \cap \beta ) = \mu ' \restriction (H \cap \beta )$ , which is possible by (1). Then the maps $e=e_0\cup \{\langle \nu _{*},\mu \rangle \}$ and $e'=e_0\cup \{\langle \nu _{*},\mu '\rangle \}$ are easily seen to be as required.

Now assume that $|F|>1$ and $(3)$ holds for any finite subset of the support of p of cardinality smaller than $\left |F\right |$ . Set $\beta =\max F$ , $C^-=\{\nu \upharpoonright (F\cap \beta ):\nu \in C\}$ , $e^-_0(\nu \upharpoonright (F\cap \beta ))= e_0(\nu )\upharpoonright (H\cap \beta )$ for all $\nu \in C$ , and $\nu ^-_{*}=\nu _{*}\upharpoonright (F\cap \beta )$ . Let also $\Sigma _m^-$ be the family of all maps $\sigma :H\cap \beta \to 2^m$ consistent with p. By the inductive assumption applied to the objects defined above we can get a coherent $e^-:C^-\cup \{\nu ^-_{*}\}\to \Sigma ^-_m $ such that $e^-\upharpoonright C^-= e^-_0$ . (Actually, we could get even two different such $e^-$ if $\nu ^-_{*}\not \in C^-$ , but thus irrelevant here.)

By $(2)$ applied to $\nu _{*}$ and $\rho =e^-(\nu ^-_{*})$ we can find distinct $\sigma ,\sigma '\in \Sigma _m$ such that

$$ \begin{align*}\sigma\upharpoonright (H\cap\beta)=\rho=\sigma'\upharpoonright (H\cap\beta).\end{align*} $$

It suffices to show that $e=e_0\cup \{\langle \nu _{*},\sigma \rangle \}$ and $e'=e_0\cup \{\langle \nu _{*},\sigma '\rangle \}$ are both coherent. We shall check this for e, the case of $e'$ is analogous. Pick $\nu \in C$ and let $\gamma \in F$ be the minimal element with $\nu (\gamma )\neq \nu _{*}(\gamma )$ . Such an ordinal $\gamma $ must exist because $\nu \neq \nu _{*}$ as $\nu \in C\not \ni \nu _{*}$ . Fix $\beta \geq \gamma $ . In the assertions from the cases below we use the fact that

$$ \begin{align*}e^-(\nu_{*}\upharpoonright (F\cap\beta)) & =e^-(\nu^-_{*})=\rho =\sigma\upharpoonright (H\cap\beta)\end{align*} $$

and $e(\nu _{*})=\sigma $ .

If $\gamma <\beta $ , then

$$ \begin{align*} e(\nu)\upharpoonright (H\cap\gamma) &= \big(e(\nu)\upharpoonright (H\cap\beta)\big) \upharpoonright (H\cap\gamma)= \big(e^-(\nu\upharpoonright (F\cap\beta)\big) \upharpoonright (H\cap\gamma) \\ & = \big(e^-(\nu_{*}\upharpoonright (F\cap\beta)\big) \upharpoonright (H\cap\gamma)= \big(e(\nu_{*})\upharpoonright (H\cap\beta)\big) \upharpoonright (H\cap\gamma) \\ &= e(\nu_{*})\upharpoonright(H\cap\gamma). \end{align*} $$

If $\gamma =\beta $ , then

$$\begin{align*}e(\nu)\upharpoonright (H\cap\beta)= e^-(\nu\upharpoonright (F\cap\beta)) = e^-(\nu_{*}\upharpoonright (F\cap\beta)) =e(\nu_{*})\upharpoonright (H\cap\beta). \end{align*}$$

Proof. For each natural number n, let $q_n\geq p$ be such that for every $\beta <\alpha $ and $\nu \in C_n$ we have

$$\begin{align*} & (q_n\upharpoonright\beta)|(\nu\upharpoonright (F_{i_n}\cap \beta))\Vdash_\beta q_n(\beta) \\ &\quad = \bigcup {\big\{\,(p|\nu')(\beta) : \nu'\in C_n,\nu'\upharpoonright (F_{i_n}\cap \beta)= \nu\upharpoonright (F_{i_n}\cap \beta)\,\big\}}. \end{align*}$$

The correctness of this definition formally requires to prove by recursion over $\beta \leq \alpha $ , along with defining $q_n$ , that each $\nu \upharpoonright (F_{i_n}\cap \beta )$ is consistent with $q_n\upharpoonright \beta $ , where $\nu \in C_n$ , and the set

$$\begin{align*}{\big\{\,(q_n\upharpoonright\beta)|(\nu\upharpoonright (F_{i_n}\cap \beta)) : \nu\in C_n\,\big\}} \end{align*}$$

is a maximal antichain above $q_n\upharpoonright \beta $ , but this is rather easy and standard. As a result, the set $ \{\,q_n|\nu : \nu \in C_n\,\}$ is a maximal antichain above $q_n$ .

It remains to show that

$$\begin{align*}(q_{n+1},k_{i_{n+1}})\geq_{F_{i_n}}(q_n,k_{i_n}) \end{align*}$$

for all $n\in \omega $ , and then let q be the fusion of the $q_n$ ’s. Suppose that for some $\beta \in F_{i_n}$ we have already shown that

$$\begin{align*}(q_{n+1}\upharpoonright\beta,k_{i_{n+1}})\geq_{F_{i_n}\cap\beta}(q_n\upharpoonright\beta,k_{i_n}), \end{align*}$$

and we will prove that

$$\begin{align*}q_{n+1}\upharpoonright\beta\Vdash (q_{n+1}(\beta), k_{i_{n+1}})\geq (q_n(\beta), k_{i_n}). \end{align*}$$

This boils down to proving that for every $\sigma \in C_{n+1}$ , if $\nu (\gamma ):=\sigma (\gamma )\upharpoonright k_n$ for all $\gamma \in F_{i_n}$ , then

$$ \begin{align*} & (q_{n+1}\upharpoonright\beta)|(\sigma\upharpoonright (F_{i_{n+1}}\cap \beta))\Vdash_\beta \\ &\quad \Big(\bigcup {\big\{\,(p|\sigma')(\beta) : \sigma'\in C_{n+1},\sigma'\upharpoonright (F_{i_{n+1}}\cap \beta)= \sigma\upharpoonright (F_{i_{n+1}}\cap \beta)\,\big\}}, k_{n+1}\Big) \\ &\qquad \geq \Big(\bigcup{\big\{\,(p|\nu')(\beta) : \nu'\in C_n,\nu'\upharpoonright (F_{i_n}\cap \beta)= \nu\upharpoonright (F_{i_n}\cap \beta)\,\big\}},k_n\Big). \end{align*} $$

Fix $s\in 2^{k_n}$ such that there exists $\nu '\in C_n$ with $\nu '(\beta )=s$ and $\nu '\upharpoonright (F_{i_n}\cap \beta )=\nu \upharpoonright (F_{i_n}\cap \beta )$ . By Lemma 2.2(4), we have $e(\nu )\upharpoonright (F_{j_n}\cap \beta )= e(\nu ')\upharpoonright (F_{j_n}\cap \beta )$ . By Lemma 3.9(1), there are $\sigma _1,\sigma _2\in \Sigma _{i_{n+1}}$ , both extending $e(\nu ')$ (and hence $\sigma _1,\sigma _2\in C_{n+1}$ ) such that $\sigma _1(\beta )\neq \sigma _2(\beta )$ and

$$\begin{align*}\sigma_1\upharpoonright(F_{i_{n+1}}\cap\beta)=\sigma_2\upharpoonright(F_{i_{n+1}}\cap\beta)=\sigma\upharpoonright(F_{i_{n+1}}\cap\beta). \end{align*}$$

It follows that

$$\begin{align*}(q_{n+1}\upharpoonright\beta)|(\sigma\upharpoonright(F_{i_{n+1}}\cap\beta))\Vdash_\beta\sigma_1(\beta),\sigma_2(\beta)\in q_{n+1}(\beta)\cap 2^{k_{i_{n+1}}} \end{align*}$$

and $\sigma _1(\beta ),\sigma _2(\beta )$ are distinct extensions of $e(\nu ')(\beta ),$ which in its turn extends $\nu '(\beta )=s$ .

Proof of Theorem 4.1.

Let $\alpha $ be such as in Lemma 4.2. Working in $V[G_{\omega _2}]$ , we claim that $X\subseteq V[G_\alpha ]$ . Since in $V[G_\alpha ]$ the remainder $\mathbb S_{\alpha ,\omega _2}$ is order-isomorphic to $\mathbb S_{\omega _2}$ , there is no loss of generality in assuming that $\alpha =0$ , i.e., that $V=V[G_\alpha ]$ .

Let us pick $z\in X\setminus V$ and let $\gamma $ be the minimal ordinal with $z\in V[G_\gamma ]$ . Next, we work in V. Let $p_0\in G_{\gamma }$ and $\tau \in V^{\mathbb S_\gamma }$ be such that $\tau ^{G_\gamma }=z$ and

$$ \begin{align*}p_0\Vdash_\gamma\tau \in (2^\omega \cap V[\dot{G}_\gamma])\setminus\bigcup_{\beta<\gamma}V[\dot{G}_\beta].\end{align*} $$

We shall find $q\geq p_0$ , $q \in \mathbb {S}_\gamma $ such that $q\Vdash _{\omega _2}\tau \not \in X$ . This would accomplish the proof: The genericity of $G_{\gamma }$ implies that there is q as above which lies in $G_{\gamma } \subseteq G_{\omega _2}$ , which would yield $z=\tau ^{G_\gamma }=\tau ^{G_{\omega _2}}\not \in X$ .

Take p and $F_n,k_n,l_n, y_{\sigma _n}$ from Lemma 3.7, applied to $p_0$ and $\tau $ . Note also that Lemma 3.9(3) ensures that $S=\mathrm {supp}(p)$ and the sequences ${\langle \, k_n : n\in \omega \,\rangle }, {\langle \, F_n : n\in \omega \,\rangle }, {\langle \, \Sigma _n : n\in \omega \,\rangle }$ satisfy $(e_f)$ from the first paragraph of Section 2. Let K and $[\sigma ]$ for $\sigma \in \bigcup _{n\in \omega }\Sigma _n$ be defined in the same way as in the first paragraph of Section 2.

Fix an element $x\in K$ and let $\sigma _n\in \Sigma _n$ be such that $\{x\}=\bigcap _{n\in \omega }[\sigma _n]$ . Fix a natural number n. We have $\sigma _n(\beta )\subseteq \sigma _{n+1}(\beta )$ for all $\beta \in F_n$ , i.e., $\sigma _n \prec \sigma _{n+1}$ . Then $p|\sigma _{n+1}\geq p|\sigma _n$ and it follows from Lemma 3.7(4) that

$$\begin{align*}p|\sigma_{n+1}\Vdash \tau\restriction l_n=y_{\sigma_n}\text{ and }p|\sigma_{n+1}\Vdash \tau \restriction l_{n+1}=y_{\sigma_{n+1}}, \end{align*}$$

which gives $y_{\sigma _n}\subseteq y_{\sigma _{n+1}}$ . Thus, the map $h\colon K\to {2^\omega }$ such that

$$\begin{align*}h(x):=\bigcup_{n\in\omega}y_{\sigma_n}, \end{align*}$$

for all $x\in K$ , is well defined. By Lemma 3.7(5), the map h is a continuous injection. Consequently, the map $h\colon K\to h[K]$ is a homeomorphism, and hence $h[K]$ is perfect.

Fix a natural number n. By Lemma 3.7(2) and Observation 3.2(3), the set $\{\,p|\sigma : \sigma \in \Sigma _n\,\}$ is a maximal antichain above p. Applying Lemma 3.7(4), we have that

$$\begin{align*}p\Vdash \tau\upharpoonright l_n\in \{\,y_{\sigma} : \sigma\in\Sigma_n\,\}. \end{align*}$$

It follows from the above and from the definition of the function h that $p\Vdash \tau \in h[K].$

By our assumption on $\alpha =0$ , the set $h[K]\cap X\cap V$ is an element of V and it is totally imperfect and Menger in V. Let ${\langle \, \langle i_n,j_n,C_n\rangle : n\in \omega \,\rangle }$ be a sequence from Lemma 2.2, applied to $S:=\mathrm {supp}(p)$ and $h^{-1}[X\cap V]\subseteq K$ . Lemma 2.2(5) yields

$$\begin{align*}K':=\bigcap_{n\in\omega}\bigcup_{\nu\in C_n}[\nu]\subseteq K\setminus h^{-1}[X\cap V], \end{align*}$$

and therefore, $h[K']\subseteq h[K]\setminus (X\cap V)$ . Since $K,K',h$ are all coded in V, we conclude that $h[K']\subseteq h[K]\setminus X$ holds in $V[G_{\omega _2}]$ by Lemma 4.2(2).

Let $q\geq p$ be a condition given by Lemma 3.10. Since the sets $\{p|\nu :\nu \in C_n\}$ are predense above q for all $n\in \omega $ , we have $q\Vdash \tau \restriction l_{i_n} \in \{\,y_{\nu } : \nu \in C_n\,\}$ , and thus $q\Vdash \tau \in h[K']$ . We conclude that $q\Vdash \tau \not \in X$ .

Observation 5.2. If $X\subseteq 2^\omega $ is Hurewicz, then Alice has no winning strategy in the grouped Menger game played on X.

Proof. Let $\sigma $ be a strategy for Alice in the grouped Menger game played on $X:={2^\omega }\setminus Y$ . By Remark 5.1, we may assume that each family given by Alice according to the strategy $\sigma $ is countable and increasing and Bob chooses one set from the families given by Alice. We shall define a strategy $\S $ for Alice in the game $\mathsf {G}_1(\mathcal {K},\mathcal {O})$ played on X such that each play in $\mathsf {G}_1(\mathcal {K},\mathcal {O})$ played according to $\S $ and lost by Alice, gives rise to a play in the grouped Menger game on X in which Alice uses $\sigma $ and loses.

Suppose that $\sigma $ instructs Alice to start round $0$ by selecting a natural number $l_0>0$ . Let $L_0:=0$ and $L_1:=l_0$ . Then $\S $ instructs Alice to play the family of all sets

$$\begin{align*}\bigcap_{i\in[L_0,L_1)}U_i, \end{align*}$$

where

$$\begin{align*}(l_0,\mathcal U_0,U_0,\ldots,\mathcal U_{l_0-1},U_{l_0-1}) \end{align*}$$

is a play, where Alice uses the strategy $\sigma $ . Note that the family of all the intersections as above is indeed an open k-cover of X. Suppose that Bob replies in $\mathsf {G}_1(\mathcal {K},\mathcal {O})$ by selecting $\bigcap _{i\in [L_0,L_1)}U_i$ for some sequence $(\mathcal U_0,U_0,\dotsc ,\mathcal U_{l_0-1},U_{l_0-1})$ as above. The strategy $\sigma $ instructs Alice to proceed in round $1$ by selecting a natural number $l_1>0$ . Let $L_2:=l_0+l_1$ . Then $\S $ instructs Alice to play the family of all sets

$$\begin{align*}\bigcap_{i\in[L_1,L_2)}U_i, \end{align*}$$

where

$$\begin{align*}(l_0,\mathcal U_0,U_0,\dotsc,\mathcal U_{l_0-1},U_{l_0-1};\: l_1, \mathcal U_{l_0},U_{l_0},\ldots,\mathcal U_{l_0+l_1-1},U_{l_0+l_1-1}) \end{align*}$$

is a play in which Alice uses $\sigma $ , an open k-cover of X. Suppose that Bob replies in $\mathsf {G}_1(\mathcal {K},\mathcal {O})$ by selecting $\bigcap _{i\in [L_1,L_2)}U_i$ for some sequence $(\mathcal U_0,\mathcal F_0,\dotsc ,\mathcal U_{l_0-1},U_{l_0-1};\: \mathcal U_{l_0},U_{l_0},\ldots , \mathcal U_{l_0+l_1-1},U_{l_0+l_1-1})$ as above.

In general, let $\sigma $ instruct Alice to start round n by selecting a natural number $ l_n>0$ . Let $L_{n+1}:=l_0+l_1+\dotsb +l_n$ . Then the next move of Alice in $\mathsf {G}_1(\mathcal {K},\mathcal {O})$ according to $\S $ is, by the definition, the family of all sets

$$\begin{align*}\bigcap_{i\in[L_n,L_{n+1})} U_i, \end{align*}$$

where

$$ \begin{align*} & ( l_0,\mathcal U_0,U_0,\ldots, \mathcal U_{ l_0-1},U_{ l_0-1};\ l_1, \mathcal U_{ l_0},U_{ l_0},\ldots,\mathcal U_{ l_0+ l_1-1},U_{ l_0+ l_1-1};\ldots ;\\ &\quad l_n, \mathcal U_{ l_0+ l_1+\cdots+ l_{n-1}},U_{ l_0+ l_1+\dotsb+ l_{n-1}},\ldots, \mathcal U_{ l_0+ l_1+\dotsb+ l_{n-1}+ l_n-1},U_{ l_0+ l_1+\dotsb+ l_{n-1}+ l_n -1}) \end{align*} $$

is a play in which Alice uses $\sigma $ , an open k-cover of X.

Since the strategy $\S $ is not winning, there is a play in $\mathsf {G}_1(\mathcal {K},\mathcal {O})$ in which Alice uses $\S $ and loses, which gives rise to an infinite play

$$ \begin{align*} & ( l_0,\mathcal U_0, U_0,\ldots, \mathcal U_{ l_0-1},U_{ l_0-1};\ l_1, \mathcal U_{ l_0},U_{ l_0},\ldots,\mathcal U_{ l_0+ l_1-1},U_{ l_0+ l_1-1}; \ldots;\\ &l_n, \mathcal U_{ l_0+ l_1+\cdots+ l_{n-1}},U_{ l_0+ l_1+\cdots+ l_{n-1}},\ldots, \mathcal U_{ l_0+ l_1+\cdots+ l_{n-1}+ l_n-1},U_{ l_0+ l_1+\cdots+ l_{n-1}+ l_n -1}; l_{n+1},\ldots) \end{align*} $$

in which Alice uses $\sigma $ and

$$\begin{align*}X=\bigcup_{n\in\omega}\bigcap_{i\in[L_n,L_{n+1})} U_i. \end{align*}$$

This means that the strategy $\sigma $ is not winning as well.

Proof of Theorem 6.1.

Let $\alpha $ be such as in Lemma 6.2. Working in $V[G_{\omega _2}]$ , we claim that

(6.2.1) $$ \begin{align} X=\bigcup\big\{L\subseteq 2^\omega\: :\: L\subseteq X, L \mbox{ is compact, and } L \mbox{ is coded in }V[G_\alpha]\big\} \mbox{ and} \end{align} $$

(6.2.2) $$ \begin{align} 2^\omega\setminus X=\bigcup\big\{L\subseteq 2^\omega\: :\: L\subseteq 2^\omega\setminus X, L \mbox{ is compact, and } L \mbox{ is coded in } V[G_\alpha]\big\}. \end{align} $$

Since in $V[G_\alpha ]$ the remainder $\mathbb S_{\alpha ,\omega _2}$ is order-isomorphic to $\mathbb S_{\omega _2}$ , there is no loss of generality in assuming that $\alpha =0$ , i.e., that $V=V[G_\alpha ]$ .

First we prove (6.2.1). Let us pick $z\in X$ and let $\gamma $ be the minimal ordinal with $z\in V[G_\gamma ]$ . From now on, whenever we do not specify that we work in some other model, we work in V. Let $r\in G_{\omega _2}$ and $\tau \in V^{\mathbb S_\gamma }$ be such that $\tau ^{G_\gamma }=z$ and $r\Vdash \tau \in X$ . Let $p_0 \geq r \restriction \gamma $ be such that $p_0\in G_{\gamma }$ and