for which TWO has a winning strategy in the game ${{\sf G}_1}({\mathcal{O}},{\mathcal{O}})$.]{}

Rothberger groups not embedding into any $\sigma$-compact space.
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Not every ${\sf T}_0$ Rothberger bounded group is a subgroup of a $\sigma$-compact group, as illustrated by the following example of Comfort and Ross ([@CR], Example 3.2). We precede the example with a few general remarks about $P$-spaces. A topological space is said to be a $P$-space if each ${\sf G}_{\delta}$ set is open. Evidently, every subspace of a $P$-space is a $P$-space. Every countably infinite subspace of a ${\sf T}_2$ $P$-space is closed and discrete. It follows that a compact $P$-space is finite, and thus a $\sigma$-compact $P$-space is countable. Thus, no uncountable Lindelöf $P$-space is a subspace of a $\sigma$-compact ${\sf T}_2$-space. If a topological group $(G,*)$ is a Lindelöf $P$-space then it is Rothberger bounded in a strong sense: Let $(U_n:n<\omega)$ be a sequence of neighborhoods for the identity. Then $U = \cap_{n<\omega}U_n$ is a neighborhood for the identity. Since the group is $\aleph_0$ bounded fix a sequence $(x_n:n<\omega)$ of elements of the group such that $x_n*U,\, n<\omega$ covers the group. Then the sequence $(x_n*U_n:n<\omega)$ witnesses that the group is Rothberger bounded.

We now define the 