Selective Separability Game

Let $G_{fin}(\mathcal D,\mathcal D)$ denote the selective separability game, and $SS^+$ denote $\mathcal{II}\uparrow G_{fin}(\mathcal D,\mathcal D)$.

• Barman/Dow gave an example of a space which is $SS$ but not $SS^+$.

Markov strategies

A Markov strategy for a game only considers the round number and latest move of the opponent.

Let $SS^{+mark}$ denote $\mathcal{II}\uparrow_{mark} G_{fin}(\mathcal D,\mathcal D)$.

• Gruenhage has asked if all $SS^+$ spaces are $SS^{+mark}$.
• Barman/Dow showed that all countable $SS^+$ spaces are $SS^{+mark}$.

Countable Fan Tightness Game

We may similarly consider $G_{fin}(\mathcal B_x,\mathcal B_x)$. If $\mathcal{II}\uparrow G_{fin}(\mathcal B_x,\mathcal B_x)$ then we write $C(D)FT^{+}$

Similarly, the following are equivalent:

• $X$ is $SS^+$
• $X$ is separable and $CDFT^{+}$
• $\mathcal{II}\uparrow G_{fin}(\mathcal B_x,\mathcal B_x)$ for each point in some countable dense subset of $X$

The above also holds for Markov strategies.

Note: $CF(D)T$-type properties are hereditary.

Idea of proof

There's a natural corespondance between $\omega$ covers of completely regular spaces $X$ and subsets of $C(X)$ containing $\mathbf 0$ as a limit point.

Given an $\omega$ cover $\mathcal U$, we may assume it consists of co-zero sets. For $U\in\mathcal U$, let $\mathbf x_U\in C(X)$ satisfy $U=\mathbf x_U^{-1}[(-1,1)]$. It follows $\mathbf 0 \in cl(\{\frac{1}{2^n}\mathbf x_U:U\in\mathcal U,n<\omega\})$.

Likewise, given $\mathbf 0\in cl(B)$, let $\mathcal U_B = \{ \mathbf x^{-1}[(-\frac{1}{2^n},\frac{1}{2^n})] : \mathbf x\in B, n<\omega \}$; it follows $\mathcal U_B$ is an $\omega$ cover.

Barman/Dow made a similar observation:

• If $X$ is $\sigma$-compact, then $C(X)$ is $CFT^{+mark}$.

Since $CFT^{+mark}$ implies $CDFT^{+mark}$, any separable subspace of such a $C(X)$ must be $SS^{+mark}$.

We get a direct comparison with Arhangel'skii's result by observing the following:

So if $X=\bigcup_{n<\omega}X^n$ for $X_n$ compact and increasing, let $\sigma(\mathcal U,n)$ be an $n$ cover of $\mathcal U$ for $X_n$. It follows that $\bigcup_{n<\omega}\sigma(\mathcal U,n)$ is an $\omega$ cover.

Alternately, if given a Markov strategy, let $X_n = \bigcap_{\mathcal U\in\Omega}\bigcup\sigma(\mathcal U,n)$. Since any open cover may be closed under finite unions to obtain an $\omega$ cover, it's not hard to see that $X_n$ is relatively compact, and therefore $\overline{X_n}$ is compact by regularity of $X$. Finally, it's not hard to show $\sigma$ isn't winning if $X\not=\bigcup_{n<\omega}X_n$. $\Box$

$\Omega M$ vs. $CFT$

As it turns out, the idea of Arhangel'skii's original result yields all of the following:

• $X$ is $\Omega M$ iff $C(X)$ is $CFT$
• $X$ is $\Omega M^{+}$ iff $C(X)$ is $CFT^{+}$
• $X$ is $\Omega M^{+mark}$ iff $C(X)$ is $CFT^{+mark}$

So it's easy to find an example of a space which is $CFT^+$ but not $CFT^{+mark}$.

• Take the one-point Lindelofication $\omega_1^\dagger$; this space isn't $\sigma$-compact and therefore not $\Omega M^{+mark}$, but it can be shown that it is $\Omega M^+$.
• It follows $C(\omega_1^\dagger)$ is $CFT^+$ but not $CFT^{+mark}$.

So if $C(\omega_1^\dagger)$ was separable, we'd have a $SS^+$, $\neg SS^{+mark}$ space... but it's not even ccc.

So searching for $\Omega M^+$, $\neg\Omega M^{+mark}$ spaces $X$ yields $CFT^+$, $\neg CFT^{+mark}$ spaces $C(X)$, but finding an example where $C(X)$ is separable may not be possible.