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On the following page, Taranovsky is talking about his "Determinacy Maximum" axiom: http://web.mit.edu/dmytro/www/DeterminacyMaximum.htm

He also justifies the choice of the name, by pointing out that many simple extensions of the theorem lead to inconsistency.

One of the results is theorem 3, giving an example of a game which has length $\omega_1+\omega$ and ordinal definable payoff and is undetermined. However, the proof uses dichotomy "$\omega_1=\Bbb R$ or $\omega_1<\Bbb R$" which need not hold when not assuming AC (e.g. it fails if we assume AD).

My question thus is

Can Taranovsky's game be modified so that it is undetermined without assuming axioms other than ZF?

Thanks in advance.

(PS. Is game-theory tag fitting for this kind of question? If not, what tag is to be used in the future?)

Added: I have noticed that Wikipedia claims existence of such long undetermined game on Determinacy article, but it has no references nor claim about background system.

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    $\begingroup$ Look at the tag wiki of the descriptive set theory tag, it's right on the money! :-) $\endgroup$ – Asaf Karagila Jun 3 '15 at 13:49
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Yes. Depending on the extent that choice holds, there are two cases. In both, the payoff is ordinal definable.
Case A: There is a sequence of $ω_1$ distinct reals. In the game, player 1 must play such a sequence X, and the final ω moves will be an undetermined game based on X. Either X does not have a perfect subset (so the perfect subset game corresponding to X is undetermined), or a well-ordering of ℝ (and hence an undetermined game) can be defined from X.
Case B: There is no such X. The following game of length $ω_1$ is undetermined. Player 1 plays a countable ordinal α, and player 2 must respond with a sequence of α distinct reals. A winning strategy for player 2 would give $ω_1$ distinct reals.

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  • $\begingroup$ Thank you very much! I really like the trick used in case A: no matter how X ends up looking like, we can define some undetermined game from it. $\endgroup$ – Wojowu Nov 4 '16 at 15:14

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