Talk:Banach–Mazur game

New article
Feel free to edit/correct mistakes, etc. I'm not an expert on this, so I'm sure others could add more insight. Revolver

I'm not sure I got the last part right, about Baire spaces. I'm pretty tired. Check that out. Revolver

I removed this:
 * The relationship of the Banach-Mazur game to Baire spaces follows immediately from the above result:


 * Let X be a topological space.  Then there exists a subset C of X such that Player II does not have a winning strategy if and only if X is a Baire space.

This statement seems to use the conjecture that a space X is Baire iff X has a non-meagre subset, i.e. iff X is of second category. But that conjecture is false, as was pointed out by an anonymous contributor on Baire space. AxelBoldt 18:39, 16 Dec 2003 (UTC)

Which player is trying to get into the set?
The article has II winning if the final point winds up in the set determining the Banach-Mazur game. Moschovakis has it the other way; I is trying to get into the set. Obviously it doesn't really matter mathematically, but I want to put a link to this article from Determinacy, and I think it's confusing if usual games are defined with I trying to get into the set, and Banach-Mazur games are defined the opposite way.

Would anyone object if I switched it, and changed other things to match? --Trovatore 05:21, 3 September 2005 (UTC)

It's also important to remember how the winning conditions change as the definition changes. For example, if player 2 is trying to get into the set, the set C must be comeager, but if player 1 is trying to get into the set, it need not be comeager, but rather there must be some Yi in Y such that the intersection of C with Yi is a comeager subset of Yi. Also, the definition certainly varies by author. Robert Soare discusses banach-mazur games in his Computability Theory book, and in his games, player 1 is trying to get into the set C rather than player 2, but Soare also requires that player 1's first move is the entire space C, so player 1 plays a role very similar to player 2 in other formulations. For Wikipedia's purposes, I definitely think it would be best to pick one formulation, and stick with it in various articles, but also mention that the phrasing varies by authors. Althai 16:02, 4 March 2007 (UTC)

Applications to Computability Theory
Banach-Mazur games show up in computability theory (they are related to finite extension construction, forcing, and other types of constructions which show up in computability theory, and sometimes model theory and other areas of logic). I would be happy to write about some of these applications if people would find it useful, but I'm not sure if maybe it would be beyond the scope of the article. Also, does anyone know of other places outside of topology where these games show up? Althai 16:05, 4 March 2007 (UTC)
 * I would encourage you to add this information.  You've got me curious now.  --Trovatore 21:27, 5 March 2007 (UTC)

Minor notation request
Can someone define the notation $$P_2 \uparrow MB(X,Y,W)$$? I guess that it means that $$P_2$$ wins, but I don't see where it's defined. --Dylan Thurston (talk) 22:43, 31 October 2011 (UTC)

Definition and properties
I could find no adequate explanation of siftable and strongly-siftable spaces nor of the Choquet game in the Wikipedia. Would someone knowledgeable please write an article on these? Howard McCay (talk) 07:58, 6 July 2013 (UTC)

Inconsistency in the article
The article says, But it also says where
 * Any winning strategy of $$P_2$$ [in $$MB(X,Y,W)$$] can be reduced to a stationary winning strategy.
 * A Markov winning strategy for $$P_2$$ in $$BM(X)$$ can be reduced to a stationary winning strategy. Furthermore, if $$P_2$$ has a winning strategy in $$BM(X)$$, then she has a winning strategy depending only on two preceding moves.  It is still an unsettled question whether a winning strategy for $$P_2$$ can be reduced to a winning strategy that depends only on the last two moves of $$P_1$$.
 * $$BM(X)$$ denote[s] a modification of $$MB(X,Y,W)$$ where $$X=Y$$, $$W$$ is the family of all nonempty open sets in $$X$$, and $$P_2$$ wins a play $$(W_0, W_1, \cdots)$$ if and only if $$\cap_{n<\omega} W_n \neq \emptyset$$.

But since BM(X) is a special case of MB(X,Y,W), it follows the first claim that if P2 has a winning strategy in BM(X), it has a stationary winning strategy. So one of these two bullet points must be false. My guess is it's the former, but I'm not sure, so I'm hoping someone else will fix the article. --skeptical scientist (talk) 06:12, 29 November 2015 (UTC)