FOM: A proof of not-CH
Soren Moller Riis
smriis at daimi.aau.dk
Sun Sep 13 12:24:49 EDT 1998
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A proof of not-CH:
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Joe Shipman writes [Shipman, September 12, 1998] that there was no
"proof" of The continuum Hypothesis (or of its negation) using ANY
axioms that had any plausibility. I think it is important that
Joe Shipman very carefully uses the past tense in his formulation.
On Fom there have been quite a lot of discussions of the status of CH.
In my oppinions these discussions have been somewhat irrelevant
as there is indeed a proof of not-CH. I have tried this proof on many
mathematicians (certainly at least 10) and I never found anyone
who did not accept the proof.
My proof is a variant of a related well known argument by Chris
Freiling. Freilings argument was published in JSL 1986
(See "Axioms of Symmetry: Throwing Darts at the Real Line",
Journal of Symbolic Logic, 51, pages 190-200).
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Suppose two players play the following game: First player I choose a
real number r. Player I sends this number to the referee. Then
player II choose (without knowing r) a countable set B of reals.
Player II wins the game if the number r happens to belong to the set B.
Most mathematicians would have no problems in putting their money
on player I. Thus they will give a negative answer to the following
question:
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Question:
Does player II have a strategy which will guarantee him/her
victory almost certainly? More specific: does Player II have a method
of picking the countable set B (using any kind of selection mechanism
he/she would like) which will win the game with probability 1?
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The mathematicians I spoke to either ought-right deny that player II
should have such a strategy, or suggest (in rare cases and with some
hesitation) that the question might not be well defined.
Now the mathematicians I spoke to certainly accepts naive set-theory
(including the axiom of choice) as foundation of mathematics. Thus
they are FORCED to accept not-CH. Indeed they all accepted the
validity of this argument, and thus they all accepted not-CH.
First we notice:
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Theorem (ZFC/naive set theory):
If CH holds, then player II can choose a measure space
and select a set B such that r belongs to B with probability 1.
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Thus it does not matter whether you think player I ought to win
with probability 1, OR you think the question not is well defined. In
both cases you are in effect denying the antecedent in the above
theorem. Thus in both cases you are in effect accepting not-CH.
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Proof of theorem:
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Assume CH is valid. Before the game begin player II fix a well-ordering
"<" of the reals which have the property that each real have only
countable many smaller reals (smaller in the well-ordering). This
can be done if CH is valid. Player II also select a non-singular
probability measure M on the reals. Thus any countable set have
measure 0 with respect to this measure.
Now we are ready to play the game: Player I select a real. This real
can be selected in any way we wish. For ANY real r chosen by player I,
the set A:={r': r' < r} is countable. Now player II choose s according
to the probability distribution. Player II then select the countable
set B:={s': s' <s} as the countable set B. With probability 1, s is
not in A (because the set A has measure 0). Thus with probability 1,
r belongs to B and player II wins the game. QED
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Why I think my twist of Freilings argument makes a stronger argument:
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My argument is (as I already pointed out) a variant of Chris
Freilings argument. I think however there is an important
difference between Freilings original argument and my modification.
In Freilings version people might discard the whole
argument as meaningless by appealing to the danger of
building arguments on probabilities. After all (one might claim)
the whole notion of probability is in a very strong sense
in contradiction with the axiom of choice (as part of naive set theory).
For this reason Chris Freilings argument might be viewed with some skepsis
(though I completely accepts it). Technically Freilings argument uses
a version of Fubinis Theorem which might be considered as invalid. It is
certainly not provable in ZFC.
The mathematicians who believe that the Question (above) is ill-posed
will of course feel no strong reason to accept Freilings argument.
The point by my formulation is that mathematicians who discards
the question as meaningless in effect are ACCEPTING not-CH. This is
the major point in my version of Freilings argument.
Thus the ONLY position which is consistent with naive set theory
and with CH, is to take it as a fact that the answer to the question
is positive and that Player II indeed has a method of picking a
countable set B such that any fixed but unknown r belongs to B with
probability 1!
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Godel though it was possible to decide CH on the basis of some obvious
principles. I think he was right and that a suitable version of
Freilings argument does the job. I leave it to future historians to
explain why Freilings argument (or perhaps even more the version I suggest)
have not been generally accepted as a valid proof of not-CH.
Question: Could some of you help me collecting a list of mathematical
statements (like CH) which are independent of set theory. I think
it could be interesting to try to see if some of these statements
could be settled by thought experiments involving some kind of
idealised dialog/discussion/game involving two players and a referee.
Soren Riis
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