The Banach–Mazur game and domain theory

Abstract

We prove that player $\alpha$ has a winning strategy in the Banach–Mazur game on a space X if and only if X is F-Y countably $\pi$-domain representable. We show that Choquet complete spaces are F-Y countably domain representable. We give an example of a space, which is F-Y countably domain representable, but which is not F-Y $\pi$-domain representable.

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 [12]. Let X be a topological space and $X=A\cup B$ be any given decomposition of X into two disjoint sets. The game BM(X, A, B) 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 [13]) 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 [2]. 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 [3] 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 [11], 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

• (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 [4], 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 [5], 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 [9] 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 [6] 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 [7], 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) $Q_\alpha \in [Q]^{\le \omega }$ and $Q_\alpha$ is upward directed, for $\alpha <\omega _1$,

• (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) $Q_\alpha \subseteq Q_\beta ,$ for $\alpha<\beta <\omega _1$,

• (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 [8] 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 [4] 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.