Yes. I realized as soon as I posted that that may statement that "the outcome does not depend on what either player does" couldn't be right. Adam's proof is very nice. I love transfinite induction. It's elegant that you don't need to say something alternately about A's and B's strategies, but can pool them together. Conway and Croft (1964) used a similar argument to prove that R^3 can be partitioned into congruent planar unit circles.) I've been totally down with AC ever since learning it was equivalent to "The cartesian product of a nonempty collection of nonempty sets is nonempty." (Proof: immediate.) By the way, I claim that assuming AC one can pick a random integer. From which some interesting consequences follow. --Dan
On Nov 14, 2014, at 7:19 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Although I doubt this is what Jim had in mind . . .
[...]
So according to AC, there is some W(A) + W(B) for which there exists no strategy for either player.
I suspect this means that the outcome does not depend on what either player does. But I'm not absolutely sure of that.
No, it just means that neither player can force a win, even though the game cannot end in a draw.
The way to `construct' such a game is by (uncountable transfinite) induction:
1. Well-order the set of all player-1-strategies and player-2-strategies by the initial ordinal of cardinality 2^aleph_null.
2. For each strategy in a list, insist that a particular sequence of moves `beats' it. Since at any stage we have fixed < 2^aleph_null outcomes, we can always do this.
By the end of all time, we have a game where for any strategy that player n makes, player 3-n can make a sequence of moves to win.
Sincerely,
Adam P. Goucher