Counterintuitions and the well-ordering theorem
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From: George Greene on 20 Nov 2009 08:31 On Nov 19, 1:03 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Sorry. Here's the whole gory detail: This was extremely kind of you. I would've settled for a link; you were not obligated to tutor so intensively -- I am willing to do my own homework occasionally! The link I chose was http://en.wikipedia.org/wiki/Gale-Stewart_game > 1. A "game" (of the type we are discussing) is defined by > a set G of infinite sequences of naturals (we identify > an infinite sequence of naturals with a function from > omega to omega). They call the set of *all* w-sequences-of-elements-of-w "Baire space". So we are just calling G a subset of that. Since "beat" below will be defined in terms of membership in G, G is (apparently) the sequences where the first player f wins. > 5. f is a winning strategy for the first player if: > forall g, beats(f,g,G) > > 6. g is a winning strategy for the second player if: > forall f, ~beats(f,g,G) > > 7. There is no winning strategy for the first player if: > forall f, exists g, beats(f,g,G) > > 8. There is a winning strategy for the second player if: > exists g, forall f, beats(f,g,G) > > Because of the order of the quantifiers, the lack of > a winning strategy for the first player does not > logically imply there is a winning strategy for the > second player. But some sub-contexts DO imply that. The actual TITLE of the article I linked to is NOT "Gale-Stewart game" but DETERMINACY. Some of the time, some of these things ARE determined. In particular, for checkers, chess, and Go, they are determined.
From: George Greene on 20 Nov 2009 08:34 On Nov 19, 4:31 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > On the other hand, > we can by fiat declare some new object (not in ZF, but in Morse Kelly > set theory or in Quine's NF set theory) to be the complete collection > of all sets, and we can consistently deal with such a completed collection. Only in the narrow technical sense of consistency. Remember, it is YOU who were alleging that the set/class distinction was a lame dodge "forced on us by consistency"! It is NOT a REAL distinction! > So what exactly, is the sense in which the collection of sets cannot > be completed? We cannot "consistently" refuse to apply the concept of a powerclass to this class, is the problem. The refusal is arbitrary and ad hoc. The collection of all collections *IS* a collection, whether you theory WANTS TO CALL it one OR NOT! It therefore HAS a powerclass, whether your theory wants to admit OR NOT!
From: George Greene on 20 Nov 2009 08:36 On Nov 19, 4:31 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > There is a sense > in which sets cannot be thought of as a "completed totality", which > is the reason that there cannot be a set of all sets. This underestimates the plasticity of "thought", especially of what "can be" thought, as opposed to what can be justified as consistent. If there can be a class of all sets but not a set of all sets, then is the set/class distinction real or fake? If sets can have powersets, why can't classes have powerclasses? And if they can, why isn't the limit of the power-class process simply a hyper-class?? AD NAUSEAM???
From: George Greene on 20 Nov 2009 08:46 Gale, Stewart, and wikipedia are laughing at you. I, on the other hand, am not. On Nov 19, 1:03 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > 5. f is a winning strategy for the first player if: > forall g, beats(f,g,G) > > 6. g is a winning strategy for the second player if: > forall f, ~beats(f,g,G) > > 7. There is no winning strategy for the first player if: > forall f, exists g, beats(f,g,G) > > 8. There is a winning strategy for the second player if: > exists g, forall f, beats(f,g,G) > > Because of the order of the quantifiers, the lack of > a winning strategy for the first player does not > logically imply there is a winning strategy for the > second player. wikipedia rebuts: > The proof that such games are determined is rather simple: > Player I simply plays not to lose; that is, he plays to make > sure that player II does not have a winning strategy after I's move. > If player I cannot do this, then it means player II had a winning strategy from the beginning. > On the other hand, if player I can play in this way, then he must win, because the game > will be over after some finite number of moves, and he can't have lost at that point. The missing assumption here is basically that the game has a finite board and a finite number of pieces and is therefore (like chess or go) in some sense finite after all, despite the fact that we were defining everything in terms of infinite sequences. If, at the beginning, as you were saying, neither player had a winning strategy, then all player 1 has to do is choose a move that allows that fact to continue to be the case; there must exist such a safe move; otherwise, player II already has a winning strategy. But player 1's repetition of these "safe" choices will eventually run up against draw-by-repetition or some other aspect of the inherent finitude of the game.
From: Daryl McCullough on 20 Nov 2009 09:06
Herman Jurjus says... >Can you prove "if ZF minus powerset is consistent, then so is ZF >including powerset", without presuming the latter? No, definitely not. I'm talking about the case in which one grants the *consistency* of the power set axiom. I can certainly understand the doubts that the power set axiom is consistent. -- Daryl McCullough Ithaca, NY |