holds for $X$.]{}

The equivalence of (1) and (2) is due to Pawlikowski [@JP], and the equivalence with (3) was proved in [@msrothbramsey]. Using ideas of [@CH] we now show:

\[twowinsrothberger\] For each uncountable cardinal $\kappa$ there is a ${\sf T}_0$ Rothberger bounded group of cardinality $\kappa$ such that TWO has a winning strategy in ${{\sf G}_1}({\mathcal{O}}_{nbd},{\mathcal{O}})$.

In the proof we will make use of the following elementary game of $\omega$ innings played on an infinite set $S$: In inning $n$ player ONE chooses a countable subset $W_n$ of $S$ and TWO responds by choosing a point $b_n\in  W_n$. ONE must further obey the rule that for each $n$, $W_n\subseteq W_{n+1}$. A play $(W_0,\, b_0,\, \cdots,\, W_n,\, b_n,\,\cdots)$ is won by TWO if for each $x\in\bigcup_{n<\omega}W_n$ there are infinitely may $n$ with $b_n = x$. A standard argument shows that TWO has a winning perfect information strategy in this game. Call this game the “countable - one game".

Let $\kappa$ be an uncountable cardinal number. Let $(G_{\alpha}:\alpha<\kappa)$ be a sequence of discrete countable groups and define $G$ to be the direct product $$\prod_{\alpha<\kappa}G_{\alpha}$$ endowed with the countable box topology.

Then the subset $G^* = \{f\in G:\vert \{\alpha: f(\alpha)\neq {\sf id}_{\alpha}\}\vert<\aleph_0\}$ 