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Re: A rational X irrational game
Posted:
Mar 20, 2004 5:52 PM


In article <c3ggid$meho$1@hades.rz.unisaarland.de>, Alexander Malkis <alexloeschediesmalk@line.cs.unisb.de> writes: Your game doesn't terminate in a finite time. And at each time point the intersection contains intinitely many rational and irrational points. So the game is not well formulated.
There is a considerable literature about infinite games, and the family of infinite games like the one described here, but with the set of rationals replaced with any arbitrary set of real numbers, is famous. The description given in this thread is a little informal, but is pretty much the way that the game is described in the literature.
It's possible to give a rigorous definition of what a strategy for one of the players is, and for what it means for a strategy to be a winning strategy. These are the essential questions about such games, and they can be defined without supposing that it's possible to play out such a game to its completion. A strategy is a function assigning a choice of play to each position in the game, and it's a winning strategy if every (infinite) sequence of play which adheres to it counts as a win for that player.
The axiom of determinacy (AD) in set theory is the statement that for infinite games in which the two players take turns choosing bits, generating a countable sequence b0,b1,..., and player 1 wins if the sequence lies in some set A of sequences of bits, there always is a winning strategy for one of the players. AD has a lot of specific consequences, but is inconsistent with the axiom of choice, so people don't really believe it's true. One has to be a little careful with defining what it means to have a winning strategy in a more general game, because in general one can't prove the existence of a strategy, whether winning or not, without assuming the axiom of choice. If U is a family of sets, we can define the twostep game where player 1 chooses a set S in U, and player 2 in order to win has to choose an element of S. If U is a family of nonempty sets without a choice function, then neither player has a winning strategy in the sense I defined above. Each move of layer 1 can be refuted because S is never empty. On the other hand, a winning strategy for player 2 is a choice function. (So perhaps in the absense of choice we want to define winning strategy more loosely. We could allow a winning strategy merely to be an assignment of a nonempty subset of the available strategies, with the property that staying inside the subset is sufficient to guarantee a win. I don't know whether this looser notion of winning strategy is considered interesting or not.)
There are weakenings of AD consistent with AC. Set theorists seem mostly to believe that a class of sets known as L[R] satisfy AD. L[R] is enough like the usual set theoretic universe to satisfy the ZermeloFrankel axioms for set theory (ZF). L[R] is the smallest model of ZF containing each real number (R stands for the reals) and all the ordinals. I find it interesting that such a natural property as determinacy fails to be satisfied by every game of this kind, but that the exceptions are so very elusive.
Keith Ramsay



