Dan, I'll bite: What do you mean when you claim that "assuming AC one can pick a random integer.
From which some interesting consequences follow"?
Jim Propp On Friday, November 14, 2014, Dan Asimov <dasimov@earthlink.net> wrote:
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 <javascript:;>> 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
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