FOM: CH
Henry Cohn
cohn at abel.MATH.HARVARD.EDU
Wed Sep 16 14:10:14 EDT 1998
Soren Riis writes:
>Assuming CH, we have demonstrated that it indeed matter who moves
>first (compare A and C). Yet, you acknowledge (correctly of course)
>that it is completely irrelevant who moves first (as I have defined
>the game).
>
>Thus you fell into the trap!!
>Your attempt to show the game is not well defined (because the order
>shouldn't matter) backfired. You statement that the order does
>not matter is in direct contradiction with provable facts
>(compare A, and C).
>Well, I should certainly stipulate: provable under the assumption
>of CH. The contradiction (you seems to acknowledge) demonstrates
>not-CH (using an argument which has not (yet?!) been incorporated
>in naive set theory).
and also:
>The contradiction we arrive at is of course somewhat different,
>but again the essential point is that in some sense both experts
>has a winning strategy. This is indeed a contradiction (though
>not a traditional one).
>
>Thus we have to (and the mathematicians I discussed this with
>indeed did) accept non-CH.
What if we phrase things without using CH at all? Let X be an
uncountable well-ordered set with all initial segments countable.
Define a probability measure by P(A) = 0 if A is countable and
P(A)=1 if X-A is countable. Two players independently choose
elements x and y of X, and the player choosing y wins iff y>x.
If x is chosen first, then choosing y according to P wins with
probability 1, since {y : y<=x} is countable and thus has
probability 0. (This does not depend on how x was chosen.)
Similarly, if y is chosen first, then choosing x according
to P wins with probability 1.
This seems to me to be exactly the same situation that arose
in the discussion of CH, so I don't find the earlier argument
convincing evidence for not-CH.
It's definitely counterintuitive that if both players choose
their elements according to P, the winner depends on who chooses
first. However, I think Schlottmann is right in attributing
this weirdness to the fact that {(x,y) : x<y} is not measurable
with respect to PxP - it doesn't even make sense to talk about
the probability that one player or the other wins (without
converting the double integral to one of several inequivalent
iterated integrals), since the probability that y wins should be
(PxP){(x,y) : x < y}.
Of course, one might object that P is not defined on a very large
or interesting sigma-algebra, but the same argument applies to any
probability measure that gives countable sets probability 0.
Henry Cohn
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