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I'm working through some lecture notes on the axiom of determinacy, and have run into some trouble with the proof of the incompatibility of the axiom of determinacy with the axiom of choice. Specifically, the theorem takes the following form,

Assume that $\omega^\omega$ can be well ordered. There is a set $A\subseteq\omega^\omega$ which is not determined.

The proof in the notes is a bit confusing, and I can't quite follow it, although I do get the basic idea that we use a diagonal argument by recursively defining a set for which there can be no winning strategy. Could anybody either post a proof or direct me to one?

Thanks in advance. Ben.

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    Without knowing much about the location in the notes, the fact that the AoD implies all sets of real numbers are Lebesgue measurable would be indication that it is incompatible with Choice.2012-07-11
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    (Addendum: Since choice is used to prove the existence of sets of reals which are not Lebesgue measurable.) The Wikipedia page for the AofD also has an outline of a proof of the incompatibility: http://en.wikipedia.org/wiki/Axiom_of_determinacy2012-07-11

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