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Topic: A rational X irrational game
Replies: 10   Last Post: Mar 20, 2004 5:52 PM

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Keith Ramsay

Posts: 1,745
Registered: 12/6/04
Re: A rational X irrational game
Posted: Mar 20, 2004 5:52 PM
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In article <c3ggid$meho$>, Alexander Malkis
<> 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

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 two-step 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 Zermelo-Frankel
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

Keith Ramsay

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