FOM: Re: Elaboration on not-CH proof:
Soren Moller Riis
smriis at daimi.aau.dk
Tue Sep 15 18:13:17 EDT 1998
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Re: Elaboration on not-CH proof:
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Martin Schlottmann writes:
> To my understanding the analysis of the game is not
> correct unless we also stipulate a strategy for player I.
> Given the other rules of the game, it is completely
> irrelevant that I has the first pick.
>
>
> Let P_I be the probability measure of player I, P_II
> the probability measure of player II. If we call
> A(r):={r'|r'<r} and B(s):={s'|s'<=s},
> then the event that II wins is:
> {II wins}={(r,s)|r\in B(s)}={(r,s)|s\not\in A(r)}
> which occurs with probability:
> (P_I x P_II)( {(r,s)|r\in B(s)} )
> = (P_I x P_II)( {(r,s)|s\not\in A(r)} ),
> but only in the case that {II wins} is a measurable
> set (then we could use Fubini in order to calculate
> this probability to be 1, given certain properties of
> P_I and P_II).
>
> Therefore, the only thing shown is that certain sets
> in the product space are not measurable, a rather
> non-surprise.
It seems you are addressing Chris Freiling original
argument rather than my version of his argument.
Assuming CH, Player II has a winning strategy in the game
in the following sense:
----------------------------------------------------------
A: Consequence of CH
For ANY r \in R, r \in B(s):={s'|s'<=s} with probability 1,
provided s is chosen randomly according to a non-singular
probability distribution.
----------------------------------------------------------
This is a significantly stronger statement than the one you
consider and which I think is closer to Freilings argument
where the question really is about swapping the order of
integration of a function with a non-measurable graph.
You (and Freiling) seems to consider a somewhat weaker
consequence of CH:
-----------------------------------------------------------
A': Weaker consequence of CH
If r \in R is chosen randomly (first) and then player II
randomly choose s \in R, then r \in B(s):={s'|s'<=s} with
probability 1.
-----------------------------------------------------------
You (and Freiling) in effect compare this statement with
the statement:
-----------------------------------------------------------
C': Letting player II move first:
If s \in R is chosen randomly (first) (by player II) and
player I then randomly choose r \in R, then
r \in B(s):={s'|s'<=s} with probability 0.
-----------------------------------------------------------
However in my version of the argument I contrast A with:
------------------------------------------------------------
C: Letting player II move first:
For ANY countable set B, we have r \in B with probability 0,
provided r is chosen randomly according to a non-singular
probability distribution.
------------------------------------------------------------
You (and essentially also Freiling as he put his argument) seems
to contrast A' and C'. The skeptics then argue that there
is no contradiction here. Put in your words:
"Therefore, the only thing shown is that certain sets
in the product space are not measurable, a rather non-surprise".
What you do not seems to recognise is that I am contrasting
A and C (rather than A' and C'). If we do this and combine
this with the fact that (USING YOUR OWN WORDS):
"Given the other rules of the game, it is completely
irrelevant that I has the first pick" we must accept not-CH.
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).
Soren Riis
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