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Noah Schweber
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This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; however, the situation with ZF+$\neg$DC is still open (there was an error in a claimed answer, and this question has been modified from its original form to accommodate that).

I'd like to ask what the ZF-situation is. However, in ZF alone things are rather weird so there are multiple plausible "right" versions of this question (which are all equivalent in ZF+DC). Unfortunately this does make the question rather subjective (presumably based on the answers we somehow pin down which version was "right," but how?), but I think it's within $\epsilon$ of precise for a sufficiently small $\epsilon$.

The most obvious question to ask is whether the result, phrased exactly as previously, continues to hold:

Version 1: Does ZF prove "There is an undetermined Banach-Mazur game"?

On second thought, however, it's not obvious to me that this is the right question. Without DC determinacy is a bit weird, and it's arguably better to talk about quasistrategies. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

A game is quasidetermined if one player or the other has a winning quasistrategy. In ZF+DC determinacy is the same as quasideterminacy, but in ZF they are distinct. So we can separately ask:

Version 2: Does ZF prove "There is an un-quasidetermined Banach-Mazur game?"

Yet on third thought, even that isn't obviously the right question: it's consistent with ZF that both players have winning quasistrategies in a given game! (Consider "Alternate choosing new elements of an amorphous set.") So we might want to land somewhere in the middle - say that a game is weakly determined if exactly one player has a winning quasistrategy.

Version 3: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?

This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; however, the situation with ZF+$\neg$DC is still open (there was an error in a claimed answer, and this question has been modified from its original form to accommodate that).

I'd like to ask what the ZF-situation is. However, in ZF alone things are rather weird so there are multiple plausible "right" versions of this question (which are all equivalent in ZF+DC). Unfortunately this does make the question rather subjective (presumably based on the answers we somehow pin down which version was "right," but how?), but I think it's within $\epsilon$ of precise for a sufficiently small $\epsilon$.

The most obvious question to ask is whether the result, phrased exactly as previously, continues to hold:

Version 1: Does ZF prove "There is an undetermined Banach-Mazur game"?

On second thought, however, it's not obvious to me that this is the right question. Without DC determinacy is a bit weird, and it's arguably better to talk about quasistrategies. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

A game is quasidetermined if one player or the other has a winning quasistrategy. In ZF+DC determinacy is the same as quasideterminacy, but in ZF they are distinct. So we can separately ask:

Version 2: Does ZF prove "There is an un-quasidetermined Banach-Mazur game?"

Yet on third thought, even that isn't obviously the right question: it's consistent with ZF that both players have winning quasistrategies in a given game! (Consider "Alternate choosing new elements of an amorphous set.") So we might want to land somewhere in the middle - say that a game is weakly determined if exactly one player has a winning quasistrategy.

Version 3: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?

This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; however, the situation with ZF+$\neg$DC is still open (there was an error in a claimed answer, and this question has been modified from its original form to accommodate that).

I'd like to ask what the ZF-situation is. However, in ZF alone things are rather weird so there are multiple plausible "right" versions of this question. Unfortunately this does make the question rather subjective (presumably based on the answers we somehow pin down which version was "right," but how?), but I think it's within $\epsilon$ of precise for a sufficiently small $\epsilon$.

The most obvious question to ask is whether the result, phrased exactly as previously, continues to hold:

Version 1: Does ZF prove "There is an undetermined Banach-Mazur game"?

On second thought, however, it's not obvious to me that this is the right question. Without DC determinacy is a bit weird, and it's arguably better to talk about quasistrategies. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

A game is quasidetermined if one player or the other has a winning quasistrategy. In ZF+DC determinacy is the same as quasideterminacy, but in ZF they are distinct. So we can separately ask:

Version 2: Does ZF prove "There is an un-quasidetermined Banach-Mazur game?"

Yet on third thought, even that isn't obviously the right question: it's consistent with ZF that both players have winning quasistrategies in a given game! (Consider "Alternate choosing new elements of an amorphous set.") So we might want to land somewhere in the middle - say that a game is weakly determined if exactly one player has a winning quasistrategy.

Version 3: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?

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*Really* undetermined Undetermined Banach-Mazur games?: beyond DC

This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; however, the situation with ZF+$\neg$DC is still open This question is(there was an error in a follow-upclaimed answer, and this question has been modified from its original form to this one; seeaccommodate that question for the definition of Banach-Mazur games.).

I'd like to ask what the ZF-situation is. However, in ZF alone things are rather weird so there are multiple plausible "right" versions of this question (which are all equivalent in ZF+DC). Unfortunately this does make the question rather subjective (presumably based on the answers we somehow pin down which version was "right," but how?), but I think it's within $\epsilon$ of precise for a sufficiently small $\epsilon$.

InThe most obvious question to ask is whether the absence of dependent choiceresult, phrased exactly as previously, continues to hold:

Version 1: Does ZF prove "There is an undetermined Banach-Mazur game"?

On second thought, however, it's not obvious to me that this is the right question. Without DC determinacy is a bit weird, and it's arguably less natural thanbetter to talk about quasideterminacyquasistrategies. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

At the question linked above, Asaf Karagila showed how from a failure of DC we can produce an undetermined Banach-Mazur game. However, the resultingA game is isquasidetermined still quasidetermined (indeed nicely -if one player $2$or the other has a winning quasistrategy and player $1$ does not). My questionIn ZF+DC determinacy is whetherthe same as quasideterminacy, but in ZF they are distinct. So we can whip up a more extreme counterexampleseparately ask:

Question 1Version 2: Is the statement "EveryDoes ZF prove "There is an un-quasidetermined Banach-Mazur game is quasidetermined" consistent with ZF?"

Arguably this is actuallyYet on third thought, even that isn't obviously the wrongright question, since quasideterminacy is in one important sense quite weak: it isit's consistent with ZF that there is a game in which both players have winning quasistrategies in a given game! ("Alternate choosing new elements of an amorphous set.") Say(Consider "Alternate choosing new elements of an amorphous set.") So we might want to land somewhere in the middle - say that a game is weakly determined if exactly one player has a winning quasistrategy.

Question 2Version 3: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?

*Really* undetermined Banach-Mazur games?

This question is a follow-up to this one; see that question for the definition of Banach-Mazur games.

In the absence of dependent choice, determinacy is arguably less natural than quasideterminacy. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

At the question linked above, Asaf Karagila showed how from a failure of DC we can produce an undetermined Banach-Mazur game. However, the resulting game is still quasidetermined (indeed nicely - player $2$ has a winning quasistrategy and player $1$ does not). My question is whether we can whip up a more extreme counterexample:

Question 1: Is the statement "Every Banach-Mazur game is quasidetermined" consistent with ZF?

Arguably this is actually the wrong question, since quasideterminacy is in one important sense quite weak: it is consistent with ZF that there is a game in which both players have winning quasistrategies! ("Alternate choosing new elements of an amorphous set.") Say that a game is weakly determined if exactly one player has a winning quasistrategy.

Question 2: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?

Undetermined Banach-Mazur games: beyond DC

This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; however, the situation with ZF+$\neg$DC is still open (there was an error in a claimed answer, and this question has been modified from its original form to accommodate that).

I'd like to ask what the ZF-situation is. However, in ZF alone things are rather weird so there are multiple plausible "right" versions of this question (which are all equivalent in ZF+DC). Unfortunately this does make the question rather subjective (presumably based on the answers we somehow pin down which version was "right," but how?), but I think it's within $\epsilon$ of precise for a sufficiently small $\epsilon$.

The most obvious question to ask is whether the result, phrased exactly as previously, continues to hold:

Version 1: Does ZF prove "There is an undetermined Banach-Mazur game"?

On second thought, however, it's not obvious to me that this is the right question. Without DC determinacy is a bit weird, and it's arguably better to talk about quasistrategies. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

A game is quasidetermined if one player or the other has a winning quasistrategy. In ZF+DC determinacy is the same as quasideterminacy, but in ZF they are distinct. So we can separately ask:

Version 2: Does ZF prove "There is an un-quasidetermined Banach-Mazur game?"

Yet on third thought, even that isn't obviously the right question: it's consistent with ZF that both players have winning quasistrategies in a given game! (Consider "Alternate choosing new elements of an amorphous set.") So we might want to land somewhere in the middle - say that a game is weakly determined if exactly one player has a winning quasistrategy.

Version 3: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?

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Noah Schweber
Noah Schweber
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*Really* undetermined Banach-Mazur games?

This question is a follow-up to this one; see that question for the definition of Banach-Mazur games.

In the absence of dependent choice, determinacy is arguably less natural than quasideterminacy. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

At the question linked above, Asaf Karagila showed how from a failure of DC we can produce an undetermined Banach-Mazur game. However, the resulting game is still quasidetermined (indeed nicely - player $2$ has a winning quasistrategy and player $1$ does not). My question is whether we can whip up a more extreme counterexample:

Question 1: Is the statement "Every Banach-Mazur game is quasidetermined" consistent with ZF?

Arguably this is actually the wrong question, since quasideterminacy is in one important sense quite weak: it is consistent with ZF that there is a game in which both players have winning quasistrategies! ("Alternate choosing new elements of an amorphous set.") Say that a game is weakly determined if exactly one player has a winning quasistrategy.

Question 2: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?