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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).

Martin Sleziak
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Noah Schweber
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  • Can you phrase it in combinatorial terms, rather than games? – Asaf Karagila Jan 20 '20 at 18:04
  • @AsafKaragila Not off the top of my head (although maybe I'm being overly restrictive in my interpretation of "combinatorial" here) - do you have something in mind? (I could easily be missing something obvious here ...) – Noah Schweber Jan 20 '20 at 18:05
  • I've had a long week, so I cannot parse the question in terms of "does ZF prove the existence of a topological space such that every open set contains such and such chain of sets, but also possibly such and such chain of sets". – Asaf Karagila Jan 20 '20 at 18:07
  • So, you're asking, essentially, is it consistent that all spaces are universally Baire. – Asaf Karagila Jan 20 '20 at 18:08
  • @AsafKaragila Not sure what "universally Baire" means, but probably ... let me check to make sure. – Noah Schweber Jan 20 '20 at 18:09
  • @AsafKaragila Actually crud, my prior comment (now deleted) wasn't right since "property of Baire" is not the same as "has determined BM-game." So I don't think universally Baire works (also it just seems to be about sets of reals). So I guess the inverse phrasing of the problem is simply: "Is there a model of ZF where every topological space's Banach-Mazur game is determined?" Not much better ... – Noah Schweber Jan 20 '20 at 18:11
  • And do you have a combinatorial condition equivalent to BM being determined, that is my question. – Asaf Karagila Jan 20 '20 at 18:16
  • @AsafKaragila Nope. But in the absence of choice, I wouldn't expect determinacy principles in general to be very combinatorial unless we significantly stretch the meaning of "combinatorial" ... at which point games themselves are combinatorial. – Noah Schweber Jan 20 '20 at 18:22
  • Okay, let's think this together. For the game to be determined, that means that in any open set you can either find a decreasing sequence whose intersection is empty, or a decreasing sequence whose intersection is not empty. Or something like that? – Asaf Karagila Jan 20 '20 at 19:16
  • @AsafKaragila I don't think that takes into account the "back-and-forth" flavor of the game sufficiently. – Noah Schweber Jan 20 '20 at 21:22
  • We can alternate picking $A_i$'s that are the first term of a decreasing sequence contained in $B_i$ with empty intersection, and $B_i$'s that are the first term of a decreasing sequence in $A_{i-1}$ with nonempty intersection. It's not clear to me what happens in that case even if such sequences always exist – Alessandro Codenotti Jan 20 '20 at 21:31
  • @AlessandroCodenotti Nothing good - consistently every open set has both those forms! Take e.g. $\mathbb{Q}$ with the usual topology: starting from any open set we can "zoom in" on some rational or on some irrational. – Noah Schweber Jan 20 '20 at 21:33
  • Yes, I was hoping you can find a way to make this more... correct. – Asaf Karagila Jan 20 '20 at 21:37
  • Of course, good point @Noah – Alessandro Codenotti Jan 20 '20 at 21:38
  • By this question Baire spaces which are not productively Baire are undetermined. Let $\Bbb P_S$ be the usual forcing that adds to $\omega_1$ a club contained in the stationary subset $S$. And let $S_1$ and $S_2$ be two disjoint stationary subsets of $\omega_1$. Then $\Bbb P_{S_1}$ and $\Bbb P_{S_2}$ are Baire ($\sigma$-distributive) but their product is not otherwise forcing with it would add disjoint clubs to $\omega_1$. (continues) – Alessandro Codenotti Jan 20 '20 at 21:52
  • Now does the above argument uses choice at some point? I'm afraid so, though I haven't checked carefully. I still suggest that Baire spaces which are not productively Baire might be a good place to look for examples. – Alessandro Codenotti Jan 20 '20 at 21:52
  • @AlessandroCodenotti Without choice there need not be disjoint stationary sets. – Noah Schweber Jan 20 '20 at 21:58
  • Oh, wow, I didn't expect things to go that wrong. That was just the easiest example I know of, my real question is whether ZF+the product of Baire spaces is Baire is consistent relative to ZF, which seems suspicious but I learned not to be surprised by ZF's weirdness – Alessandro Codenotti Jan 20 '20 at 22:00

1 Answers1

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This has now been partially answered at Mathoverflow by James Hanson, with the remaining case separately asked; I'm posting this answer to move this question off the unanswered queue (and I've made it CW so I don't get reputation for his work). The answer is affirmative under DC, and currently wide open otherwise.

Noah Schweber
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