$X_k$. However, if $x_k*N_k\cap X_k\neq\emptyset$, then let $y$ be an element of this intersection. We claim that $x_k*N_k\cap X_k\subseteq y*N_k^2$. For let $z\in x_k*N_k\cap X_k$ be given. Choose $v$ in $N_k$ with $z = x_k*v$. Since we also have $y\in x_k*N_k$, choose $w\in N_k$ with $y=x_k*w$. Now $x_k = y*w^{-1} \in N_k*N_k$ and so $z = y*(w^{-1}*v)\in y*N_k^2$.

Thus, we may choose for each $k\in S_m$ an $x_k\in X_k$ such that $(x_k*U_k^2:k\in S_m)$ covers $X_k$. Finally, recursively choose a sequence $(T_k:k<\omega)$ as follows: Choose $T_0\in \mathcal{U}(\emptyset)$ with $x_0*N_0^2 \subseteq T_0$. With $T_0,\, \cdots,\, T_m$ chosen, choose $T_{m+1}\in\mathcal{U}(T_0,\cdots,T_m)$ with $x_{m+1}*N_{m+1}^2\subseteq T_{m+1}$. Then the sequence $$F(\emptyset),\, T_0,\, F(T_0),\, \cdots,\, T_k,\, F(T_0,\cdots,T_k),\, T_{k+1},\, \cdots$$ is an $F$-play lost by ONE.

In the proof of (5) implies (1) we use that ${{\sf S}_1}({\mathcal{O}},{\mathcal{O}}_X)$ is equivalent to ${{\sf S}_1}(\Omega,{\mathcal{O}}_X)$. $\Box$

Note that $(4)$ also implies that ONE has no winning strategy in the game ${{\sf G}_1}({\mathcal{O}}_{nbd},{\mathcal{O}}_X)$ on $(G,*)$, which in turn implies $(2)$. In particular we have the following Corollary:

\[rothbbdedgp\] Let $(H,*)$ be a subgroup of a $\sigma$-compact group $(G,*)$. The following are equivalent:

1.  [$(G,*)$ satisfies ${{\sf S}_1}({\mathcal{O}},{\mathcal{O}}_H)$.]{}

2.  [$(H,*)$ is Rothberger bounded.]{}

3.  [ONE has no winning strategy in the game ${{\sf 