Comments by "EebstertheGreat" (@EebstertheGreat) on "Conway Checkers (proof) - Numberphile" video.

@Bodyknock No. Supertasks are defined in such a way that they have a countably infinite number of states that cannot be numbered. For instance, there may be an ultimate state, but no penultimate state. Or there may be an ultimate and penultimate state, but no antepenultimate state. And there is always an initial state. But this cannot happen with this sort of game. Every state in this game has a predecessor (except the initial state) and successor (except the final state). Playing this game with infinitely many steps and a defined initial state is equivalent to counting down from infinity to one. It is nonsensical. What we could do is define a sequence of "unmoves" from the solved state toward an initial state. This could be well-defined, and although I haven't actually tried it, perhaps such a sequence could in some sense have a limit which is a completely filled bottom half of the board. This is a meaningful idea that can be defined mathematically. However, it will have no final "unmove." Therefore the corresponding "solution" will have no initial move. So you can't even start.
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@Bodyknock I think you mean that in a supertask, there are countably infinitely many states. They can still "be numbered" in the sense that we can put them into one-to-one correspondence with the natural numbers. For example, when you say, "the well-ordered set which is the natural numbers plus W which is an ordinal greater than every natural number is still a countable set; the function f(W) = 0 and f(n) = n+1 is an example of a bijection between that set and the naturals.," I assume you mean ω by W, but it doesn't make any difference here. The fact is that no "supertask" as defined in this video solves Conway's Soldiers. By this I mean that there does not exist any identification s: N→N such that every n encodes a legal move to s(n) and there is a natural number m such that s(m) is the final state. To be clear, the task is to create a sequence of moves which starts with some arrangement of checkers below the line and which ends with a single checker five squares above the line. (The final state may have any number of additional checkers, but it must have at least that one.) There does not exist such a sequence. If there did, that sequence would be equivalent (specifically, homeomorphic) to one that immediately solves the problem, that is, in which s(1) was the final state. There is no requirement of finiteness in the working out; such a sequence simply does not exist.
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@Bodyknock I never implied such a thing. There is no countable set in question here. That's the whole point of the proof. Every attempt at enumeration fails, so the set is not denumerable.
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