# Undetermined Banach-Mazur games in ZF?

This question was previously asked and bountied on MSE, with no response. This MO question is related, but is also unanswered and the comments do not appear to address this question.

Given a topological space $$\mathcal{X}=(X,\tau)$$, the Banach-Mazur game on $$\mathcal{X}$$ is the (two-player, perfect information, length-$$\omega$$) game played as follows:

• Players $$1$$ and $$2$$ alternately play decreasing nonempty open sets $$A_1\supseteq B_1\supseteq A_2\supseteq B_2\supseteq ...$$.

• Player $$1$$ wins iff $$\bigcap_{i\in\mathbb{N}} A_i=\emptyset$$.

ZFC implies that there is a subspace of $$\mathbb{R}$$ with the usual topology whose Banach-Mazur game is undetermined; on the other hand, it's consistent with ZF+DC (and indeed adds no consistency strength!) that no subspace of $$\mathbb{R}$$ does this ("every set of reals has the Baire property").

However, when we leave $$\mathbb{R}$$ things get much weirder. My question is:

Does ZF alone prove that there is some space $$\mathcal{X}$$ whose Banach-Mazur game is undetermined?

Controlling the behavior of all possible topological spaces in a model of ZF is extremely hard for me, and I suspect the answer to the question is in fact yes. In fact, I recall seeing a pretty simple proof of this; however, I can't track it down or whip up a ZF-construction on my own (specifically, everything I try ultimately winds up being a recursive construction killed by having too many requirements to meet in the given number of steps).

• I removed my answer since I'm not sure how to fix the issues you guys brought up. I'm not sure that it's unfixable, just I can't do it now. – Asaf Karagila Mar 2 '20 at 23:12

This is only a partial answer. ZF + DC + 'every Banach-Mazur game is determined' is inconsistent.

Let $$X$$ be the set of all functions of the form $$f: \alpha \rightarrow \{0,1\}$$, with $$\alpha$$ an ordinal such that for any ordinals $$\beta < \gamma$$ with $$\omega \cdot \gamma + \omega \leq \alpha$$, the sets $$(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$$ and $$(n \in \mathbb{N} : f(\omega \cdot \gamma + n) = 1)$$ are distinct.

Let $$\tau$$ be the topology on $$X$$ generated by sets of the form $$U_f = (g \in X : g \supseteq f)$$, and let $$\mathcal{X} = (X,\tau)$$.

Claim: Player $$1$$ does not have a winning strategy for the Banach-Mazur game on $$\mathcal{X}$$.

Proof of claim: For any strategy $$S$$ for player $$1$$. Let $$T$$ be the tree of all initial segments of plays against $$S$$ such that each play by player $$2$$ is of the form $$U_f$$ for some $$f \in X$$ of length $$\omega \cdot \alpha$$ for some $$\alpha$$. If this tree fails to be pruned, then player $$1$$ has in some play of the game played a set $$V$$ such that for any $$U_f \subseteq V$$, $$f$$ enumerates every element of $$2^{\mathbb{N}}$$ extending $$\sigma$$ for some $$\sigma \in 2^{<\omega}$$. This would imply that $$2^{\mathbb{N}}$$ can be well-ordered, allowing us to construct a Bernstein set, contradicting our assumptions. Hence this must be a pruned tree of height $$\omega$$, so by dependent choice it has a path. Let $$g$$ be the union of all $$f$$ such that $$U_f$$ is on that path somewhere. By construction, $$g \in X$$, so we have that the strategy where player $$2$$ blindly plays the moves in this path wins against $$S$$.

So, it must be the case that player $$2$$ has a winning strategy $$S$$. For any $$f \in X$$ with length $$\omega \cdot \alpha$$ for some $$\alpha$$, let $$T_f$$ be the strategy for player $$1$$ that plays $$U_f$$ on player $$1$$'s zeroth move and on player $$1$$'s $$n+1$$st move, if player $$2$$ played $$V$$, then player $$1$$ plays $$V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 0}$$, if this is non-empty, and otherwise player $$1$$ plays $$V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 1}$$. (Note that since the union of these two is $$V$$, one of these must be non-empty.) Since $$S$$ is a winning strategy, for any $$f\in X$$ of length $$\omega \cdot \alpha$$ for some $$\alpha$$, the play of $$T_f$$ against $$S$$ must result in a nonempty intersection. By construction, for any $$g$$ and $$h$$ in that intersection, $$g(\omega \cdot \alpha + n) = h(\omega \cdot \alpha +n)$$ for all $$n<\omega$$, and the set $$(n \in \mathbb{N} : h(\omega \cdot \alpha + n) = 1)$$ must be distinct from $$(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$$ for any $$\beta < \alpha$$.

Therefore $$S$$ gives us a uniform procedure for picking a real not on a given well-ordered list of reals. By iterating this gives us a well-ordering of the reals. Therefore we can construct a Bernstein set from $$S$$ and we have that the Banach-Mazur game on that set is not determined, which contradicts our assumption. Therefore ZF + DC proves that there is an undetermined Banach-Mazur game.

Right now I don't see how to use a failure of DC to build an undetermined game.

• Bernstein, not Berenstein... :-) – Asaf Karagila Feb 7 '20 at 13:12
• Berenstain, not Berenstein... – Gabe Goldberg Feb 7 '20 at 16:16
• @Gabe Not all games are bears. You can define an analogous notion to a mouse, and then finally have Bernstein bears. But that has yet to happen. – Asaf Karagila Feb 7 '20 at 17:49
• @GabeGoldberg So would this question be about Berenstain Baires? (I'll put that on the list of math books for children, along with If you give a mouse a measure.) – Noah Schweber Feb 7 '20 at 18:12
• When I was looking around at papers related to this I found out that 'barely Baire spaces' are a thing. – James Hanson Feb 7 '20 at 18:26