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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Feb 11, 2012 at 1:06 answer added Patrick Reardon timeline score: 1
Feb 7, 2012 at 8:42 vote accept Dávid Natingga
Feb 6, 2012 at 9:43 answer added Jirka Hanika timeline score: 1
Feb 5, 2012 at 16:39 comment added François G. Dorais @Ramiro: pN still can't win because she will exhaust all of N by herself in some countable ordinal number of steps. If pN plays first on limit rounds, then pR can just wait the game out copying pN's moves using an injection from pN's countable set into $\mathbb{R}$. If pR plays first on limit rounds, then it's less clear what happens. It's probable that a winning strategy for pR gives an injection from $\omega_1$ into $\mathbb{R}$, but that's not entirely clear to me right now.
Feb 5, 2012 at 16:29 comment added Ramiro de la Vega What if pN plays with a countable set not containing any real and they are playing in a universe where there is no subset of the reals of size $\aleph_1$?
Feb 5, 2012 at 15:19 comment added Dávid Natingga Yes, they are reals, thank you for pointing out I can use $\LaTeX$. So if I understand, in general p$\aleph_N$ with a set A can be defeated by another player whose set is a superset of A in $\omega_N+1$ steps?
Feb 5, 2012 at 14:58 history edited Dávid Natingga CC BY-SA 3.0
added 279 characters in body; added 56 characters in body
Feb 5, 2012 at 14:40 comment added François G. Dorais None of the players can win in $\omega$ steps or fewer. With $\omega+1$ steps, pN's opponent can always win by choosing the first available natural number at each round, exhausting N in $\omega$ steps.
Feb 5, 2012 at 14:36 comment added François G. Dorais What is R, the reals? How long is the game? Longer than $\omega$ steps, I suppose? (You can use $\LaTeX$ expressions by enclosing in dollar signs.)
Feb 5, 2012 at 14:31 history edited Dávid Natingga CC BY-SA 3.0
edited tags; edited title
Feb 5, 2012 at 14:23 history asked Dávid Natingga CC BY-SA 3.0