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Let $X$ be a topological space. A Banach-Mazur game $BM(X)$ is played by two players $\alpha$ and $\beta$, who select nonempty open subsets of $X$. The player $\beta$ starts a game by selecting a nonempty open subset $V_1$ of $X$. In return, $\alpha$-player chooses a nonempty open subset $W_1$ of $V_1$. In general, at the $n$-th stage of the game, $n \geq 1$, the player $\beta$ chooses a nonempty open subset $V_n ⊂ W_{n−1}$ and $\alpha$ answers by a nonempty open subset $W_n$ of $V_n$. Proceeding in this fashion, the players generate a sequence $(V_n,W_n)_{n=1}^{\infty}$ which is called a play. The player $\alpha$ is said to be the winner of the play $(V_n,W_n)_{n=1}^{\infty}$ if $\bigcap_{n\geq 1}V_n =\bigcap_{n\geq 1}W_n\neq \emptyset$ otherwise the player $\beta$ wins this play. A partial play is a finite sequence of sets consisting of the first few moves of a play. A strategy for player $\alpha$ is a rule by means of which the player makes his/her choices.

Here is a more formal definition of the notion strategy. A strategy s for the player $\alpha$ is a sequence of mappings $s = \{ s_n\}$, which is inductively defined as follows: The domain of $s_1$ is the set of all open subsets of $X$ and $s_1$ assigns to each nonempty open set $V_1 \subset X$, a nonempty open subset $W_1 = s_1(V_1)$ of $V_1$. In general, if a partial play $(V_1, \ldots ,W_{n−1})$ has already been specified, where $W_i = s_i(V_1, \ldots , V_i)$, $1 \leq i \leq n − 1$. Then the domain of $s_n$ would be the set $\{(V_1,W_1, . . . ,W_{n−1}, V ):V \subset W_{n−1}\}$ can be the next move of $\beta$-player and it assigns to each choice $V_n \subset W_{n−1}$ some nonempty open subset $W_n = s_n(V_1,W_1, \ldots ,W_{n−1}, V_n)$ of $V_n$.

An $s$-play is a play in which $\alpha$ selects his/her moves according to the strategy $s$. The strategy $s$ for the player $\alpha$ is said to be a winning strategy if every $s$-play is won by $\alpha$. A space $X$ is called $\alpha$-favorable if there exists a winning strategy for $\alpha$ in $BM(X)$.

My question: Is there a topological space $X$ which is neither $\alpha$-favorable nor $\beta$-favorable?

M.Ramana
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  • Your syntax is ambiguous but I believe you're asking for a space $X$ which is neither $\alpha$-favorable nor $\beta$-favorable. Fleissner and Kunen wrote a paper about what they humorously(?) call "Barely Baire Spaces", meaning a space $X$ such that $X$ is a Baire space but $X\times X$ is not a Baire space. Such spaces exist (at least if the Axiom of Choice is assumed) and neither player has a winning strategy in $BM(X)$. – bof Nov 12 '22 at 04:53
  • @bof Thanks for the comment and the paper. I've just edited my question. Could you please tell me how I can conclude that in such spaces neither player has a winning strategy in BM(X)? – M.Ramana Nov 12 '22 at 05:33
  • @bof Are there any concrete examples of Baire spaces $X$ in which neither player has a winning strategy in BM(X)? – M.Ramana Nov 12 '22 at 06:21
  • @bof By a well-know theorem, $X$ is Baire iff $\beta$ player has no a winning strategy in BM(X). So when $X$ is a Barely Baire space, then $\beta$ player has no a winning strategy in $BM(X\times X)$. But how do we know that $\alpha$ player has no a winning strategy in $BM(X\times X)$? – M.Ramana Nov 12 '22 at 06:36
  • If $X$ is a "barely Baire" space, then $\beta$ does have a winning strategy in $BM(X\times X)$, since $X\times X$ is not a Baire space; but neither player has a winning strategy in $BM(X)$. Player $\beta$ has no winning strategy in $BM(X)$ because $X$ is a Baire space. On the other hand, a straightforward argument shows that the product of two $\alpha$-favorable spaces is $\alpha$-favorable; since $X\times X$ is not $\alpha$-favorable, neither is $X$. I don't know any "concrete" example of a (positional, win/lose) game where neither player has a winning strategy. – bof Nov 12 '22 at 08:15
  • @bof Thank you very much for your good explanation. I got it perfectly. Appreciate that. – M.Ramana Nov 12 '22 at 08:39

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