subgroup of $G$ and for each $x\in G_n$ there is a $V\in\mathcal{U}$ such that $x* U(\mathcal{U},n) \subseteq V$. Let $V(x,\mathcal{U},n)$ be such a $V$.

Define a strategy $\sigma$ for TWO of ${{\sf G}_1}({\mathcal{O}},{\mathcal{O}})$ as follows: When ONE plays the open cover $O_1$ in the first inning, TWO simulates a move for ONE in ${{\sf G}_1}({{\mathcal{O}}_{nbd}},{\mathcal{O}})$ as ${\mathcal{O}}(U(O_1,1))$, applies the winning strategy $F$ to this move to obtain $x_1*U(O_1,1) = F({\mathcal{O}}(U(O_1,1)))$. Then if $G_1\cap x_1*U(O_1,1)\neq \emptyset$ we find for an $x\in G_1$ that $x*U(O_1,1) = x_1*U(O_1,1)$, a left coset of $U(O_1,1)$ in $G$. Then TWO fixes such an $x$ and plays $$\sigma(O_1) = V(x,O_1,1) \in O_1.$$ If $G_1\cap x_1*U(O_1,1) = \emptyset$ then TWO chooses an arbitrary element $x\in G_1$ and plays $$\sigma(O_1) = V(x,O_1,1) \in O_1.$$ When ONE next moves $O_2$, TWO simulates a move for ONE in ${{\sf G}_1}({{\mathcal{O}}_{nbd}},{\mathcal{O}})$ as follows: Define ${\mathcal{O}}= \{U\cap V:U\in O_1 \mbox{ and }V\in O_1\}\setminus\{\emptyset\}$ and let ONE’s move be ${\mathcal{O}}(U({\mathcal{O}},2))$. TWO’s response using $F$ is $x_2*U({\mathcal{O}},2)=F({\mathcal{O}}(U(O_1,1)),{\mathcal{O}}(U({\mathcal{O}},2))$. Consider $x_2*U({\mathcal{O}},2)\cap G_2$. If this is nonempty select any $x$ in this intersection. as before we have $x*U({\mathcal{O}},2)=x_2*U({\mathcal{O}},2)$, and now TWO responds with $$\sigma(O_1,O_2) = V(x,{\mathcal{O}},2)\in O_2.$$ If on the other hand the intersection is empty then TWO 