## 1 Introduction

The famous Banach–Mazur game was invented by Mazur in 1935. For the history of game theory and facts about game theory, the reader is referred to the survey . Let X be a topological space and $$X=A\cup B$$ be any given decomposition of X into two disjoint sets. The game BM(XAB) is played as follows: Two players, named $$\alpha$$ and $$\beta$$, alternately choose open nonempty sets with $$U_0\supseteq V_0\supseteq U_1\supseteq V_1\supseteq \cdots$$.

$$\begin{array}{lccccr} \alpha &{}U_0&{} &{}U_1&{} &{}\\ &{} &{} &{} &{} &{}\cdots \\ \beta &{} &{}V_0&{} &{}V_1&{} \end{array}$$

Player $$\alpha$$ wins this game if $$A\cap \bigcap _{n\in \omega }U_n\ne \emptyset$$. Otherwise $$\beta$$ wins.

We study a well-known modification of this game considered by Choquet in 1958, known as Banach–Mazur game or Choquet game. Player $$\alpha$$ and $$\beta$$ alternately choose open nonempty sets with $$U_0\supseteq V_0\supseteq U_1\supseteq V_1\cdots$$. In the first round, player $$\beta$$ starts by choosing a nonempty open set $$U_0.$$

$$\begin{array}{lccccr} \beta &{}U_0&{} &{}U_1&{} &{}\\ &{} &{} &{} &{} &{}\cdots \\ \alpha &{} &{}V_0&{} &{}V_1&{} \end{array}$$

Player $$\alpha$$ wins this play if $$\bigcap _{n\in \omega }V_n\ne \emptyset$$. Otherwise $$\beta$$ wins. Denote this game by BM(X). Every finite sequence of sets $$(U_0,\ldots ,U_n)$$, obtained by the first n steps in this game is called partial play of $$\beta .$$ A strategy for player $$\alpha$$ in the game BM(X) is a map s that assigns to each partial play $$(U_0,\ldots ,U_n)$$ of $$\beta$$ a nonempty open set $$V_n\subseteq U_n$$. The strategy s is called a winning strategy for player $$\alpha$$ if player $$\alpha$$ always wins the play of the game using this strategy. The space X is called weakly$$\alpha$$-favorable (see ) if X admits a winning strategy for player $$\alpha$$ in the game BM(X). We say that a partial play $$(W_0,\ldots ,W_k)$$ is stronger than $$(U_0,\ldots , U_m)$$ if $$m\le k$$ and $$U_0=W_0,\ldots ,U_m=W_m$$. Notice that if $$(W_0,\ldots ,W_k)$$ is stronger than $$(U_0,\ldots , U_m)$$, then $$s(W_0,\ldots ,W_k)\subseteq s(U_0,\ldots , U_m)$$, we denote this by $$(U_0,\ldots , U_m)\preceq (W_0,\ldots ,W_k)$$. We denote a sequence $$(U_0,\ldots , U_k)$$ by $$\overrightarrow{U}(k).$$

The strong Choquet game is defined as follows:

$$\begin{array}{lccccr} \beta &{}U_0\ni x_0&{} &{}U_1\ni x_1&{} &{}\\ &{} &{} &{} &{} &{}\cdots \\ \alpha &{} &{}V_0&{} &{}V_1&{} \end{array}$$

Player $$\beta$$ and $$\alpha$$ take turns in playing nonempty open subset, similar to the Banach–Mazur game. In the first round, player $$\beta$$ starts by choosing a point $$x_0$$ and an open set $$U_0$$ containing $$x_0$$, then player $$\alpha$$ responds with an open set $$V_0$$ such that $$x_0\in V_0\subseteq U_0$$. In the n-th round, player $$\beta$$ selects a point $$x_n$$ and an open set $$U_n$$ such that $$x_n\in U_n\subseteq V_{n-1}$$ and $$\alpha$$ responds with an open set $$V_n$$ such that $$x_n\in V_n\subseteq U_n$$. Player $$\alpha$$ wins if $$\bigcap _{n\in \omega }V_n\ne \emptyset$$. Otherwise $$\beta$$ wins. We say that a partial play $$(W_0,x_0,\ldots ,W_k,x_k)$$ is stronger than $$(U_0,y_0,\ldots , U_m,y_m)$$ if $$m\le k \text{ and } U_0{=}W_0,\ldots ,U_m=W_m \text{ and } x_0=y_0,\ldots ,x_m=y_m.$$ We denote this by $$(U_0,y_0,\ldots , U_m,y_m)\preceq (W_0,x_0,\ldots , W_k,x_k)$$. We denote a sequence $$(W_0,x_0,\ldots , W_k,x_k)$$ by $$(\overrightarrow{x}\circ \overrightarrow{W})(k)$$. A topological space X is called Choquet complete if player $$\alpha$$ has a winning strategy in the strong Choquet game, and we then write Ch(X).

For a topological space X, let $$\tau (X)$$ denote the topology on the set X and $$\tau ^*(X)=\tau (X){\setminus }\{\emptyset \}.$$ A family $${\mathcal {P}}$$ of open nonempty sets is called a $$\pi$$-base if for every open nonempty set U, there is $$P\in {\mathcal {P}}$$ such that $$P\subseteq U.$$

A dcpo (directed complete partial order) is a poset $$(P,\sqsubseteq )$$ in which every directed set has a supremum. If $$p,q\in P$$, then we say that “p is far below q” whenever for any directed set D with $$q \sqsubseteq \sup (D)$$, there is some $$d \in D$$ with $$p\sqsubseteq d.$$ A domain is a dcpo in which every element q is the supremum of the directed set $$\{p\in P: p \text{ is } \text{ far } \text{ below } q\hbox {''}\}$$. This notion has been introduced by D. Scott as a model for the $$\lambda$$-calculus, for more information see [1, 10]. Domain representable topological spaces were introduced by Bennett and Lutzer . We say that a topological space is domain representable if it is homeomorphic to the space of maximal elements of some domain topologized with the Scott topology. In 2013, Fleissner and Yengulalp  introduced an equivalent definition of a domain representable space for $$T_1$$ topological spaces. We do not assume the antisymmetry condition on the relation $$\ll$$. As Önal and Vural suggested in , if we need an additional antisymmetric property, let us consider the equivalent relation E on the set Q defined by “pEq if and only if ($$p\ll q$$ and $$q\ll p$$) or $$p=q$$”. We do not assume any separation axioms, if it is not explicitly stated.

We say that a topological space X is F-Y (Fleissner–Yengulalp) countably domain representable if there is a triple $$(Q,\ll , B)$$ such that

(D1):

$$B: Q\rightarrow \tau ^*(X)$$ and $$\{B(q):q\in Q\}$$ is a base for $$\tau (X)$$,

(D2):

$$\ll$$ is a transitive relation on Q,

(D3):

for all $$p,q\in Q$$, $$p\ll q$$ implies $$B(p)\supseteq B(q)$$,

(D4):

for all $$x\in X$$, the set $$\{q\in Q:x\in B(q)\}$$ is upward directed by $$\ll$$ (every pair of elements has an upper bound),

(D5$$_{\omega _1}$$):

if $$D\subseteq Q$$ and $$(D,\ll )$$ is countable and upward directed, then $$\bigcap \{B(q): q\in D\}\ne \emptyset$$.

If the conditions (D1)–(D4) and the condition

1. (D5)

if $$D\subseteq Q$$ and $$(D,\ll )$$ is upward directed, then $$\bigcap \{B(q): q\in D\}\ne \emptyset$$

are satisfied, we say that the space X is F-Y domain representable.

In , Fleissner and Yengulalp introduced the notion of a $$\pi$$-domain representable space, as this is analogous to the notion of a domain representable space.

We say that a topological space X is F-Y (Fleissner–Yengulalp) countably$$\pi$$-domain representable if there is a triple $$(Q,\ll , B)$$ such that

($$\pi$$D1):

$$B: Q\rightarrow \tau ^*(X)$$ and $$\{B(q):q\in Q\}$$ is a $$\pi$$-base for $$\tau (X)$$,

($$\pi$$D2):

$$\ll$$ is a transitive relation on Q,

($$\pi$$D3):

for all $$p,q\in Q$$, $$p\ll q$$ implies $$B(p)\supseteq B(q)$$,

($$\pi$$D4):

if $$q,p\in Q$$ satisfy $$B(q)\cap B(p)\ne \emptyset$$, there exists $$r\in Q$$ satisfying $$p,q\ll r$$,

($$\pi$$D5$$_{\omega _1}$$):

if $$D\subseteq Q$$ and $$(D,\ll )$$ is countable and upward directed, then $$\bigcap \{B(q): q\in D\}\ne \emptyset$$.

If the conditions ($$\pi$$D1)–($$\pi$$D4) and the condition

($$\pi$$D5):

if $$D\subseteq Q$$ and $$(D,\ll )$$ is upward directed, then $$\bigcap \{B(q): q\in D\}\ne \emptyset$$

are satisfied, we say that the space X is F-Y$$\pi$$-domain representable.

## 2 $$\pi$$-domain representable spaces

In , Kenderov and Revalski have shown that the set $$E=\{f\in C(X):f\text { attains its minimum in }X\}$$ contains a $$G_\delta$$ dense subset of C(X) is equivalent to the existence of a winning strategy for player $$\alpha$$ in the Banach–Mazur game. Oxtoby  showed that if X is a metrizable space, then player $$\alpha$$ has a winning strategy in BM(X) if and only if X contains a dense completely metrizable subspace. Krawczyk and Kubiś  have characterized the existence of winning strategies for player $$\alpha$$ in the abstract Banach–Mazur game played with finitely generated structures instead of open sets. In , there has been presented a version of the Banach–Mazur game played on a partially ordered set. We give a characterization of the existence of a winning strategy for player $$\alpha$$ in the Banach–Mazur game using the notion “$$\pi$$-domain representable space” introduced by W. Fleissner and L. Yengulalp.

### Theorem 1

A topological space X is weakly $$\alpha$$-favorable if and only if X is F-Y countably $$\pi$$-domain representable.

### Proof

If X is F-Y countably $$\pi$$-domain representable, then it is easy to show that X is weakly $$\alpha$$-favorable.

Assume that X is weakly $$\alpha$$-favorable. We shall show that X is F-Y countably $$\pi$$-domain representable. Let s be a winning strategy for player $$\alpha$$ in BM(X). We consider a family Q consisting of all finite sequences $$\Big (\overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{U}_i(j_i)\Big )$$, where $$\overrightarrow{U}_m(j_m)=(U^m_0,\ldots , U^m_{j_m})$$ is a partial play and $$m\le i,$$ i.e.,

\begin{aligned} U^m_0\supseteq s(U^m_0)\supseteq U^m_1\supseteq s(U^m_0,U^m_1)\supseteq \ldots \supseteq U^m_{j_m} \supseteq s(U^m_0,\ldots ,U^m_{j_m}) \end{aligned}

and $$s(\overrightarrow{U}_0(j_0))\supseteq \ldots \supseteq s(\overrightarrow{U}_i(j_i))$$.

Let us define a relation $$\ll$$ on the family Q:

\begin{aligned}&\left( \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{U}_i(j_i)\right) \ll \left( \overrightarrow{W}_0(l_0),\ldots ,\overrightarrow{W}_k(l_k)\right) \text { iff }\\&s(\overrightarrow{U}_i(j_i))\supseteq s(\overrightarrow{W}_0(l_0))\\&\quad \& \;i\le k \; \& \; \forall \,{t\le i}\;\exists \,{r\le k}\; \overrightarrow{U}_t(j_t)\preceq \overrightarrow{W}_r(l_r). \end{aligned}

Since $$\preceq$$ is transitive, $$\ll$$ is transitive.

Let us define a map $$B:Q\rightarrow \tau ^*(X)$$ by the formula

\begin{aligned} B\left( \left( \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{U}_i(j_i)\right) \right) =s(\overrightarrow{U}_i(j_i)) \end{aligned}

for $$\left( \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{U}_i(j_i)\right) \in Q$$.

Since $$\{s(V):V\in \tau ^*(X)\}$$ is a $$\pi$$-base, $$\{B(q):q\in Q\}$$ is a $$\pi$$-base for $$\tau$$. It is easy to see that the map B satisfies the condition $$(\pi \mathrm{D}3)$$.

Towards item ($$\pi$$D4), let $$p,q\in Q$$ be such that $$B(q)\cap B(p)\ne \emptyset$$ and $$p=\left( \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{U}_i(j_i)\right)$$, $$q=\left( \overrightarrow{W}_0(l_0),\ldots ,\overrightarrow{W}_k(l_k)\right)$$. Since $$V=B(p)\cap B(q)\subseteq s(\overrightarrow{U}_0(j_0))$$ and s is a winning strategy, we find an element $$\overrightarrow{U}'_0(j'_0)$$ stronger than $$\overrightarrow{U}_0(j_0)$$ such that $$s(\overrightarrow{U}'_0(j'_0))\subseteq V$$. Step by step we find a partial play $$\overrightarrow{U}'_t(j'_t)$$ such that $$\overrightarrow{U}_t(j_t)\preceq \overrightarrow{U}'_t(j'_t)$$ and $$s(\overrightarrow{U}'_t(j'_t))\subseteq s(\overrightarrow{U}'_{t-1}(j'_{t-1}))$$ for $$t\le i$$. Since $$s(\overrightarrow{U}'_i(j'_i))\subseteq s(\overrightarrow{W}_{0}(l_0))$$, we find a partial play $$\overrightarrow{W}'_0(l'_0)$$ such that $$\overrightarrow{W}_0(l_0) \preceq \overrightarrow{W}'_0(l'_0)$$ and $$s(\overrightarrow{W}'_0(l'_0))\subseteq s(\overrightarrow{U}'_i(j'_i))$$. Similarly, as for the sequence p, for the sequence q, we define $$\overrightarrow{W}'_t(l'_t)$$ with $$\overrightarrow{W}_t(l_t)\preceq \overrightarrow{W}'_t(l'_t)$$ and $$s(\overrightarrow{W}'_t(l'_t))\subseteq s(\overrightarrow{W}'_{t-1}(l'_{t-1}))$$ for all $$t\le k$$.

Continuing in this way, we get an element $$r=\Big (\overrightarrow{U}'_0(j'_0),\ldots ,\overrightarrow{U}'_i(j'_i), \overrightarrow{W}'_0(l'_0), \ldots ,\overrightarrow{W}'_k(l'_k)\Big )$$ such that $$p,q\ll r$$ and $$r\in Q.$$

Next we show the condition ($$\pi$$D5$$_{\omega _1}$$). Let $$D\subseteq Q$$ be a countable upward directed set and let $$D=\{p_n: n\in \omega \}$$. We define a chain $$\{q_n: n\in \omega \}\subseteq D\subseteq Q$$ such that $$p_n\ll q_n$$ for $$n\in \omega$$. By the condition ($$\pi$$D3), we get $$\bigcap \{B(q_n):n\in \omega \}\subseteq \bigcap \{B(p):p\in D\}$$. Each $$q_n\in Q$$ is of the form $$q_n=\left( \overrightarrow{W}^n_0(l^n_0),\ldots , \overrightarrow{W}^n_{k_n}(l^n_{k_n})\right) .$$

Since $$q_0\ll q_1$$, there is $$j_1\le k_1$$ such that $$\overrightarrow{W}^0_0(l^0_0)\preceq \overrightarrow{W}^1_{j_1}(l^1_{j_1})$$. We have

\begin{aligned} s(\overrightarrow{W}^0_0(l^0_0))\supseteq B(q_0)=s(\overrightarrow{W}^0_{k_0}(l^0_{k_0}))\supseteq s(\overrightarrow{W}^1_{j_1}(l^1_{j_1}))\supseteq B(q_1)=s(\overrightarrow{W}^1_{k_1}(l^1_{k_1})) . \end{aligned}

Let $$\overrightarrow{U}'_0(l^0_0)=\overrightarrow{W}^0_0(l^0_0)$$ and $$\overrightarrow{U}'_1(l^1_{j_1})=\overrightarrow{W}^1_{j_1}(l^1_{j_1})$$. Inductively, we can choose a sequence $$\{s(\overrightarrow{U}'_n(l^n_{j_n})):n\in \omega \}$$ such that $$\overrightarrow{U}_n'(l^n_{j_n}) \preceq \overrightarrow{U}'_{n+1}(l^{n+1}_{j_{n+1}})$$ and

\begin{aligned} B(q_{n}) \supseteq s(\overrightarrow{U}'_{n+1}(l^{n+1}_{j_{n+1}}))\supseteq B(q_{n+1}). \end{aligned}

Since s is a winning strategy for player $$\alpha$$, we have

\begin{aligned} \emptyset \ne \bigcap \{s(\overrightarrow{U}'_n(l^n_{j_n})):n\in \omega \}=\bigcap \{B(q_n): n\in \omega \}\subseteq \bigcap \{B(p):p\in D\}. \end{aligned}

$$\square$$

We give an example of a space, which is F-Y countably domain representable, but which is not F-Y $$\pi$$-domain representable. Note that this space is F-Y countably $$\pi$$-domain representable and not F-Y domain representable.

### Example 1

We consider the space

\begin{aligned} X=\sigma \big (\{0,1\}^{\omega _1}\big )=\big \{x\in \{0,1\}^{\omega _1}: |\hbox {supp}\; x|\le \omega \big \}, \end{aligned}

where $$\hbox {supp}\;x=\{\alpha \in \omega _1:x(\alpha )=1\}$$ for $$x\in \{0,1\}^{\omega _1}$$, with the topology ($$\omega _1$$-box topology) generated by the base

\begin{aligned} {\mathcal {B}}=\big \{\hbox {pr}_A^{-1}(x): A\in [\omega _1]^{\le \omega }, x\in \{0,1\}^A\big \} , \end{aligned}

where $$\hbox {pr}_A: \sigma (\{0,1\}^{\omega _1})\rightarrow \{0,1\}^A$$ is a projection.

We shall define a triple $$(Q,\ll , B)$$. Let $$Q={\mathcal {B}}$$, and the map $$B:Q\rightarrow Q$$ be the identity. Define a relation $$\ll$$ in the following way:

\begin{aligned} \hbox {pr}^{-1}_A(x_A)\ll \hbox {pr}^{-1}_B(x_B) \Leftrightarrow \hbox {pr}^{-1}_A(x_A)\supseteq \hbox {pr}^{-1}_B(x_B) \end{aligned}

for any $$\hbox {pr}^{-1}_A(x_A), \hbox {pr}^{-1}_B(x_B)\in {\mathcal {B}}$$. It is easy to see that the relation $$\ll$$ is transitive and that it satisfies the condition (D3). Now, we prove the condition (D4). Let $$x\in X$$ and $$\hbox {pr}^{-1}_{A_1}(x_{A_1}), \hbox {pr}^{-1}_{A_2}(x_{A_2}) \in \{\hbox {pr}^{-1}_A(x_A)\in {\mathcal {B}}:x\in \hbox {pr}^{-1}_A(x_A)\}$$. Since $$x\in \hbox {pr}^{-1}_{A_1}(x_{A_1})\cap \hbox {pr}^{-1}_{A_2}(x_{A_2})$$, we get $$x_{A_1}\upharpoonright A_2=x_{A_2}\upharpoonright A_1$$. Set $$A_3=A_1\cup A_2$$ and let $$x_{A_3}\in \{0,1\}^{A_3}$$ be such that $$x_{A_3}\upharpoonright A_2=x_{A_2}$$ and $$x_{A_3}\upharpoonright A_1=x_{A_1}$$. We have $$x_{A_3}\in \{0,1\}^{A_3}$$ such that $$x\in \hbox {pr}^{-1}_{A_3}(x_{A_3})\subseteq \hbox {pr}^{-1}_{A_1}(x_{A_1})\cap \hbox {pr}^{-1}_{A_2}(x_{A_2})$$. Hence $$\hbox {pr}^{-1}_{A_1}(x_{A_1}),\hbox {pr}^{-1}_{A_2}(x_{A_2}) \ll \hbox {pr}^{-1}_{A_3}(x_{A_3})$$.

We prove the condition $$(D5_{\omega _1})$$. Let $$D\subseteq {\mathcal {B}}$$ be a countable upward directed family. We can construct a chain $$\{\hbox {pr}^{-1}_{A_{n}}(x_{A_{n}}): n\in \omega \}\subseteq D$$ such that for each set $$\hbox {pr}^{-1}_{A}(x_{A})\in D$$, there exists $$n\in \omega$$ such that $$\hbox {pr}^{-1}_{A}(x_{A})\ll \hbox {pr}^{-1}_{A_{n}}(x_{A_{n}})$$.

Let $$B=\bigcup \{A_n:n\in \omega \}.$$ Since $$\{\hbox {pr}^{-1}_{A_n}(x_{A_n}): n\in \omega \}$$ is a chain, there is $$x_B\in \{0,1\}^B$$ such that $$x_B\upharpoonright A_n=x_{A_n}$$ for $$n\in \omega .$$ Then

\begin{aligned} \bigcap \{\hbox {pr}^{-1}_{A_n}(x_{A_n}): n\in \omega \}=\hbox {pr}^{-1}_B(x_B)\in {\mathcal {B}}, \end{aligned}

and $$\hbox {pr}^{-1}_B(x_B)\subseteq \bigcap D.$$ This completes the proof that the space $$\sigma \big (\{0,1\}^{\omega _1}\big )$$ is F-Y countably domain representable.

Now we show that $$X=\sigma \big (\{0,1\}^{\omega _1}\big )$$ is not F-Y $$\pi$$-domain representable. Suppose that there exists a triple $$(Q,\ll , B)$$ satisfying the conditions ($$\pi$$D1)–($$\pi$$D5). The family $${\mathcal {P}}=\{B(q):q\in Q\}$$ is a $$\pi$$-base. By induction, we define a sequence $$\{Q_\alpha : \alpha <\omega _1\}$$ such that the following conditions are satisfied:

1. (1)

$$Q_\alpha \in [Q]^{\le \omega }$$ and $$Q_\alpha$$ is upward directed, for $$\alpha <\omega _1$$,

2. (2)

$$\bigcap \{B(q):q\in Q_\alpha \}=\hbox {pr}^{-1}_{A_\alpha }(x_{A_\alpha }) \in {\mathcal {B}}$$ for some $$A_\alpha \in [\omega _1]^{\le \omega }$$ and some $$x_{A_\alpha }\in \{0,1\}^{A_\alpha }$$, for $$\alpha <\omega _1$$,

3. (3)

$$Q_\alpha \subseteq Q_\beta ,$$ for $$\alpha<\beta <\omega _1$$,

4. (4)

if $$\bigcap \{B(q):q\in Q_\alpha \}=\hbox {pr}^{-1}_{A_\alpha }(x_{A_\alpha })$$ and $$\bigcap \{B(q):q\in Q_\beta \}=\hbox {pr}^{-1}_{A_\beta }(x_{A_\beta })$$ for some $$A_\alpha ,A_\beta \in [\omega _1]^{\le \omega }$$ and $$x_{A_\alpha }\in \{0,1\}^{A_\alpha }$$ and $$x_{A_\beta }\in \{0,1\}^{A_\beta }$$, then $$\hbox {supp}\; x_{A_\alpha }=\{\alpha \in A_\alpha :x(\alpha )=1\}\varsubsetneq \{\alpha \in A_\beta :x(\alpha )=1\}=\hbox {supp}\; x_{A_\beta }$$, for $$\alpha<\beta <\omega _1$$.

We define a set $$Q_0$$. Take any $$r_0\in Q$$. There exist a set $$A_0\in [\omega _1]^{\le \omega }$$ and $$x_{A_0}\in \{0,1\}^{A_0}$$ such that $$\hbox {pr}_{A_0}^{-1}(x_{A_0})\subseteq B(r_0)$$. By conditions $$(\pi D1), (\pi D3),(\pi D4)$$, there exists $$r_1\in Q$$ such that $$r_0\ll r_1$$ and $$B(r_1)\subseteq \hbox {pr}_{A_0}^{-1}(x_{A_0}).$$ Assume that we have defined $$r_0\ll \ldots \ll r_n$$ and a chain $$\{A_i:i\le n\}\subseteq [\omega _1]^{\le \omega }$$ and $$x_{A_i}\in \{0,1\}^{A_i}$$ such that

\begin{aligned} \hbox {pr}^{-1}_{A_{i-1}}(x_{A_{i-1}})\supseteq B(r_i)\supseteq \hbox {pr}_{A_i}^{-1}(x_{A_i}) \text { for } i\le n. \end{aligned}

By conditions ($$\pi$$D1), ($$\pi$$D3), ($$\pi$$D4), there exists $$r_{n+1}\in Q$$ such that $$r_n\ll r_{n+1}$$ and $$B(r_{n+1})\subseteq \hbox {pr}_{A_{n}}^{-1}(x_{A_{n}}).$$ There exist a set $$A_{n+1}\in [\omega _1]^{\le \omega }$$ and $$x_{A_{n+1}}\in \{0,1\}^{A_{n+1}}$$ such that $$\hbox {pr}_{A_{n+1}}^{-1}(x_{A_{n+1}})\subseteq B(r_{n+1})$$. Let $$Q_0=\{r_n:n\in \omega \}$$. Then $$\bigcap \{B(q):q\in Q_0\}=\bigcap \{\hbox {pr}_{A_{n}}^{-1}(x_{A_{n}}):n\in \omega \}=\hbox {pr}_{A}^{-1}(x_{A})$$, where $$A=\bigcup \{A_n:n\in \omega \}$$ and $$x_A\in \{0,1\}^A$$ and $$x_A\upharpoonright A_n=x_{A_n}$$ for all $$n\in \omega .$$

Assume that we have defined $$\{Q_\alpha : \alpha < \beta \}$$ which satisfies the conditions (1)–(4).

Let $${\mathcal {R}}_\beta = \bigcup \{Q_\alpha :\alpha < \beta \}.$$ The set $${\mathcal {R}}_\beta$$ is upward directed by conditions (3), (1). Let $${\mathcal {R}}_\beta =\{p_n:n\in \omega \}$$. By (2) and (3), we get $$\bigcap \big \{B(p_n): n\in \omega \big \}=\hbox {pr}^{-1}_{A_\beta }(x_{A_\beta })\in {\mathcal {B}}$$ for some set $$A_\beta \in [\omega _1]^{\le \omega }$$ and $$x_{A_\beta }\in \{0,1\}^{A_\beta }.$$ There exist a set $$A\in [\omega _1]^{\le \omega }$$ and $$x_{A}\in \{0,1\}^{A}$$ such that $$\hbox {pr}_{A}^{-1}(x_{A})\varsubsetneq \hbox {pr}_{A_\beta }^{-1}(x_{A_\beta })$$ and $$\hbox {supp}\; x_{A_\beta }\varsubsetneq \hbox {supp}\; x_{A}.$$ Since $${\mathcal {P}}$$ is a $$\pi$$-base, we can find $$r_\beta \in Q$$ such that $$B(r_\beta )\subseteq \hbox {pr}_{A}^{-1}(x_{A})$$. Inductively, we can define a sequence $$\{q_n:n\in \omega \}\subseteq Q$$, a chain $$\{A_n:n\in \omega \}\subseteq [\omega _1]^{\le \omega }$$, and a sequence $$\{x_{A_n}\in \{0,1\}^{A_n}:n\in \omega \}$$ such that $$r_\beta , p_0\ll q_0$$, $$q_{n-1},p_n\ll q_n$$, and

\begin{aligned} B(q_n)\supseteq \hbox {pr}_{A_n}^{-1}(x_{A_n})\supseteq B(q_{n+1}) \text { for } n\in \omega . \end{aligned}

Let $$Q_\beta ={\mathcal {R}}_\beta \cup \{q_n:n\in \omega \}.$$ The set $$Q_\beta$$ satisfies conditions (1)–(4), so we finish the induction. The set $$\bigcup \{Q_\alpha :\alpha < \omega _1\}$$ is upward directed.

By conditions (2), (3), we have

\begin{aligned}&\bigcap \{B(q):q\in \bigcup \{Q_\alpha :\alpha< \omega _1\}\big \}=\bigcap \{\hbox {pr}_{A_\alpha }^{-1}(x_{A_\alpha }): \alpha<\omega _1\}=\\&\quad =\pi _A^{-1}(x_A), \text { for } A=\bigcup \{A_\alpha :\alpha<\omega _1\} \text { and } x_A\in \{0,1\}^A \\&\quad \text { such that } x_A\upharpoonright A_\alpha =x_{A_\alpha } \text { for } \alpha <\omega _1, \end{aligned}

where $$\pi _A:\{0,1\}^{\omega _1}\rightarrow \{0,1\}^A$$ is the projection. By condition (4), we get $$|\hbox {supp}\; x_A|=\omega _1$$. Hence $$\pi _A^{-1}(x_A)\cap \sigma \big (\{0,1\}^{\omega _1}\big )=\emptyset$$, a contradiction. $$\square$$

Note that by the proof of [4, Proposition 8.3] it follows that if there exists a triple $$(Q, \ll , B)$$, which satisfies the conditions of the definition of F-Y countably $$\pi$$-domain representable and $$|\bigcap \{B(q): q\in D\}|=1$$ for every countable and upward directed set $$D\subseteq Q$$, then the space X is F-Y $$\pi$$-domain representable by this triple.

### Theorem 2

The Cartesian product of any family of F-Y countably $$\pi$$-domain representable spaces is F-Y countably $$\pi$$-domain representable.

### Proof

Let X be a product of a family $$\{X_a:a\in A\}$$ of F-Y countably $$\pi$$-domain representable spaces. Let $$(Q_a,\ll _a, B_a)$$ be a triple which satisfies conditions ($$\pi$$D1)–($$\pi$$D4) and ($$\pi$$D5$$_{\omega _1}$$) for the space $$X_a.$$ Any basic nonempty open subset U in X is of the form $$U=\prod \{U_a:a\in A\}$$, where $$U_a$$ is nonempty open subset of $$X_a$$ and $$U_a=X_a$$ for all but a finite number of $$a\in A$$. We may assume that $$0_a\in Q_a$$ is the least element in $$Q_a$$ and $$B_a(0_a)=X_a$$ for each $$a\in A$$. Put

\begin{aligned} Q=\left\{ p\in \prod \{Q_a:a\in A\}:|\{a\in A:p(a)\ne 0_a\}|<\omega \right\} . \end{aligned}

Define a relation $$\ll$$ on Q by the formula

\begin{aligned} p\ll q\Longleftrightarrow p(a)\ll _a q(a) \text { for all } a\in A, \end{aligned}

where $$p,q\in Q$$. Let us define a map $$B:Q\rightarrow \tau ^*(X)$$ by $$B(p)=\prod \{B_a(p(a)):a\in A\}$$, where $$p\in Q$$. It is easy to check that $$(Q,\ll ,B)$$ is a F-Y countably $$\pi$$-domain representing X. $$\square$$

In a similar way, one can prove the above theorem also for F-Y countably domain representable, F-Y $$\pi$$-domain representable, and F-Y domain representable.

## 3 Domain representable spaces

In 2003, Martin  showed that if a space is domain representable, then player $$\alpha$$ has a winning strategy in the strong Choquet game. In 2015, Fleissner and Yengulalp  showed that it is sufficient that a space is F-Y countably domain representable. Now, we shall show that the property of being F-Y countably domain representable is necessary. For this purpose, we can use a triple $$(Q,\ll , B)$$ defined in [4, Proposition 8.3] or we can use a similar triple to the triple defined in the Theorem 1. Namely, if s is a winning strategy for player $$\alpha$$, we consider a family Q consisting of all finite sequences $$(\overrightarrow{x_0}\circ \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{x_i}\circ \overrightarrow{U}_i(j_i))$$, where $$\overrightarrow{x_m}\circ \overrightarrow{U}_m(j_m)=(U^m_0,x^m_0,\ldots , U^m_{j_m},x^m_{j_m})$$ is a partial play in the strong Choquet game for all $$m\le i$$, i.e.,

\begin{aligned}&U^m_0\supseteq s(U^m_0,x^m_0)\supseteq U^m_1\supseteq s(U^m_0,x^m_0,U^m_1,x^m_1)\supseteq \ldots \supseteq U^m_{j_m}\\&\quad \supseteq s(U^m_0,x^m_0,\ldots ,U^m_{j_m},x^m_{j_m}) \end{aligned}

and $$s(\overrightarrow{x_0}\circ \overrightarrow{U}_0(j_0))\supseteq \ldots \supseteq s(\overrightarrow{x_i}\circ \overrightarrow{U}_i(j_i))$$.

Let us define a relation $$\ll$$ on the family Q:

\begin{aligned}&\left( \overrightarrow{x_0}\circ \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{x_i}\circ \overrightarrow{U}_i(j_i)\right) \ll \left( \overrightarrow{y_0}\circ \overrightarrow{W}_0(l_0),\ldots ,\overrightarrow{y_k}\circ \overrightarrow{W}_k(l_k)\right) \\&\text { iff } s\left( \overrightarrow{x_i}\circ \overrightarrow{U}_i(j_i)\right) \supseteq s\left( \overrightarrow{y_0}\circ \overrightarrow{W}_0(l_0)\right) \& \; i\le k\; \& \\&\forall \,{t\le i}\;\exists \,{r\le k}\; \overrightarrow{x_t}\circ \overrightarrow{U}_t(j_t)\preceq \overrightarrow{y_r}\circ \overrightarrow{W}_r(l_r). \end{aligned}

We define a map $$B:Q\rightarrow \tau ^*$$ by the formula

\begin{aligned} B \left( \left( \overrightarrow{x_0}\circ \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{x_i}\circ \overrightarrow{U}_i(j_i)\right) \right) =s\left( \overrightarrow{x_i}\circ \overrightarrow{U}_i(j_i)\right) \end{aligned}

for each $$\left( \overrightarrow{x_0}\circ \overrightarrow{U}_0(j_0),\ldots ,\overrightarrow{x_i}\circ \overrightarrow{U}_i(j_i)\right) \in Q$$.

As a consequence, we obtain:

### Theorem 3

A topological space X is Choquet complete if and only if it is F-Y countably domain representable.