# FOM: Further elaboration on not-CH:

Soren Moller Riis smriis at daimi.aau.dk
Wed Sep 16 05:57:08 EDT 1998

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Further elaboration on not-CH:
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In my latest elaboration [Riis,  Wed, 16 Sep 1998]
I got carried away when I wrote:

> using an argument which has not (yet?!) been incorporated
> in naive set theory.

This is of course debatable and I forgot to add a smiley ;-)

I certainly accepts that the proof of non-CH. So
do all the mathematicians I have spoken to.

The point I am making is that I do not think Chris Freiling
argument necessarily needs to depend on swapping the order
of integration (in a case where the function involved is
non-measurable).
The probabilities involved in the variant I have suggested
ARE well defined.

Let me give some examples which might clarify my point:

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Example 1:

Fix some predicate A(x,y,z,w) and consider the game
in which player I choose x, player II choose y, Player I
then choose z, and Player II chose w. Let us say player
II wins if A(x,y,z,w) holds. Otherwise player I wins.

It is plain that player II has a winning strategy in
this game iff and only if

\forall x \exists y \forall z \exists w A(x,y,z,w)

is valid.
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Thus if we play the game between two experts (player I
and player II) either player I or player II has a
winning strategy. This is "tertium non datur"

The game I just defined (in the example) is well defined
though we did not specify any strategy for the loosing
player. All we did was to present the winning player with
a strategy which works against ANY defence.

Now not all games between experts are well defined in the
sense that one has a successful winning strategy. Consider
for example:

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Example 2:

Player I selects a natural number.

Player II selects a natural number.

The one who choose the highest number win the game.

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None of the players has a winning strategy and it clearly
makes no sense to assign a probability (say 1/2) that
player I (player II) wins.

We can easily prove (in ZFC) that none of the players has a
strategy which guarantee victory with any probability $p>0$.
More specifically there is no probability space U, and map
f: U-> N, such that for ANY n \in N, the probability
f(u)>N is at least p.

Thus we can show (in ZFC) that none of the players have
any strategy which guarantee victory with some non-zero
probability.

Now consider the game:

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Example 3:

Player I selects r \in R.

Player II selects B \subseteq R, B countable.

Player II wins if and only if r \in B. Otherwise
player I wins.

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This game is well defined if we require player I "moves"
first.  In this case (assuming CH) player II has a winning
strategy in the sense that there exists a probability
space U and a map f from U into the collection of countable
subsets of R, such that for ANY r \in R, the probability
r \in f(u) is 1.

Thus player II has a winning strategy and this winning strategy
only involves well defined measurable functions.

Now consider the game:

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Example 4:

Player II selects B \subseteq R, B countable.

Player I  selects r \in R.

Player II wins if and only if r \in B. Otherwise
player I wins.

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This game is also well defined.  In this case player I has a winning
strategy in the sense that there exists a probability
space U and a map g from U into R such that for ANY countable
B \subseteq R  the probability g(u) \not\in B is 1.

The games (presented in example 3 and example 4) can be played
in the same sense as we could play the game in Example 1.
And the winning strategies are completely well defined (nothing
involving non-measurable functions).
Yet, the outcome depends (if we assume CH) on whether player I or
player II moves first. If the players moves simultaneously
the game is not even well defined (again provided we assume CH).

of two experts in a game like the one in example 1, has a winning
strategy.

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

Thus we have to (and the mathematicians I discussed this with
indeed did) accept non-CH.

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Should we consider the above argument as part of naive set theory?
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I do NOT think it should. Rather it seems to belong naturally
to some meta-theory of set-theory. Since Godel we know that it
sometimes might be necessary to step outside a given system and
have a look at things from a higher perspective. I think this is
what is happening in the variant of Chris Freiling proof which
I have presented.

Soren Riis