P(CH) = 0
William Elliot
Let C = { A subset R | A countable },
F = { f:R -> C }
Thus for all x,y in R
P(x in f(y)) = 0 = P(y in f(x))
P(x not in f(y)) = 1 = P(y not in f(x))
So assume the probablity 1 proposition, for all f in F
some x,y in R with x not in f(y), y notin f(x).
Now assume ZFC and CH. Then R can be well ordered with
R = { r_xi | xi < omega_1 }.
Define f in F with f(r_eta) = { r_xi | xi <= eta }.
Now if psi <= chi, then r_psi in f(r_chi),
otherwise r_chi in f(r_psi).
Hence a probality one contradiction,
showing the probablity of CH is zero.
David C. Ullrich
<ma...@hevanet.remove.com> wrote:
>The probablity of the Continuum Hypothesis is zero.
>
>Let C = { A subset R | A countable },
> F = { f:R -> C }
>Thus for all x,y in R
> P(x in f(y)) = 0 = P(y in f(x))
> P(x not in f(y)) = 1 = P(y not in f(x))
This is a little imprecise - before you can talk about
the probability that an x in R does something you
need to specify what "distribution" you're talking
about. Statements about probability that do not
specify a distribution are usually taken to refer
to the "natural" distribution, but there is no such
thing on R.
That doesn't matter much - we can simply replace
R everywhere with the interval [0,1], where there
_is_ a natural distribution.
>So assume the probablity 1 proposition, for all f in F
> some x,y in R with x not in f(y), y notin f(x).
But _this_ is an error that's not going to be so easy
to fix. It's true that a countable intersection of sets
of full probability has full probability, but F is
uncountable, and that doesn't work for uncountable
intersections.
An example of the same technique that proves
something false:
For each x in [0,1], let P(x) be the statement
"a random y in [0,1] is not equal to x".
Each P(x) is true with probability 1.
Now we incorrectly jump to the conclusion
that, with probability 1, _all_ the P(x) are
true. It follows that almost every y in [0,1]
has the property that it is not equal to
any element of [0,1], in particular y is
not equal to y.
>Now assume ZFC and CH. Then R can be well ordered with
> R = { r_xi | xi < omega_1 }.
>
>Define f in F with f(r_eta) = { r_xi | xi <= eta }.
>
>Now if psi <= chi, then r_psi in f(r_chi),
> otherwise r_chi in f(r_psi).
>Hence a probality one contradiction,
>showing the probablity of CH is zero.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
William Elliot
> <ma...@hevanet.remove.com> wrote:
>
> >The probablity of the Continuum Hypothesis is zero.
> >
> >Let C = { A subset R | A countable },
> > F = { f:R -> C }
> >Thus for all x,y in R
> > P(x in f(y)) = 0 = P(y in f(x))
> > P(x not in f(y)) = 1 = P(y not in f(x))
> This is a little imprecise - before you can talk about the probability
> that an x in R does something you need to specify what "distribution"
> you're talking about. Statements about probability that do not specify a
> distribution are usually taken to refer to the "natural" distribution,
> but there is no such thing on R.
>
Oh? Doesn't P(0 < x) = 1/2 ?
P(x in \/{ [2j, 2j+1] : j in Z } = 1/2 ?
P(x in A) = lim(n->oo) P(x in [-n,n] /\ A) ?
> That doesn't matter much - we can simply replace
> R everywhere with the interval [0,1], where there
> _is_ a natural distribution.
>
P(x in A) = measure A ?
> >So assume the probablity 1 proposition, for all f in F
> > some x,y in R with x not in f(y), y notin f(x).
>
> But _this_ is an error that's not going to be so easy
> to fix. It's true that a countable intersection of sets
> of full probability has full probability, but F is
> uncountable, and that doesn't work for uncountable
> intersections.
>
That I dispute by using your counter example
below using P(n /= 1) = 1 instead of P(x /= a) = 1.
> An example of the same technique that proves something false:
>
> For each x in [0,1], let P(x) be the statement
> "a random y in [0,1] is not equal to x".
> Each P(x) is true with probability 1.
>
> Now we incorrectly jump to the conclusion that, with probability 1,
> _all_ the P(x) are true. It follows that almost every y in [0,1] has the
> property that it is not equal to any element of [0,1], in particular y
> is not equal to y.
>
Of course, not (P(A) = 1 implies A), or equivalently,
not (P(A) = 0 implies not A). This is shown using
the same examples P(n = 0) = 0 = P(x = a).
> >Now assume ZFC and CH. Then R can be well ordered with
> > R = { r_xi | xi < omega_1 }.
> >
> >Define f in F with f(r_eta) = { r_xi | xi <= eta }.
> >
> >Now if psi <= chi, then r_psi in f(r_chi),
> > otherwise r_chi in f(r_psi).
> >Hence a probality one contradiction,
> >showing the probablity of CH is zero.
>
This proof by Freiling seems to show that
CH, f in F ==> P(some x,y with x not in f(y), y not in f(x)) = 1
and
CH ==> P(for all f in F, some x,y with x not in f(y), y not in f(x)) = 1
Is that a resolution to his intuitive disproof of CH?
----
David C. Ullrich
<ma...@hevanet.remove.com> wrote:
>On Sat, 2 Aug 2008, David C. Ullrich wrote:
>> <ma...@hevanet.remove.com> wrote:
>>
>> >The probablity of the Continuum Hypothesis is zero.
>> >
>> >Let C = { A subset R | A countable },
>> > F = { f:R -> C }
>> >Thus for all x,y in R
>> > P(x in f(y)) = 0 = P(y in f(x))
>> > P(x not in f(y)) = 1 = P(y not in f(x))
>
>> This is a little imprecise - before you can talk about the probability
>> that an x in R does something you need to specify what "distribution"
>> you're talking about. Statements about probability that do not specify a
>> distribution are usually taken to refer to the "natural" distribution,
>> but there is no such thing on R.
>>
>Oh? Doesn't P(0 < x) = 1/2 ?
No. Say A is a subset of R. What's the _definition_ of P(x in A)?
(The reason I ask you the definition instead of just telling
you what it is is that in fact there _is_ no such definition...)
>P(x in \/{ [2j, 2j+1] : j in Z } = 1/2 ?
>
>P(x in A) = lim(n->oo) P(x in [-n,n] /\ A) ?
>
>> That doesn't matter much - we can simply replace
>> R everywhere with the interval [0,1], where there
>> _is_ a natural distribution.
>>
>P(x in A) = measure A ?
Yes (for the "uniform" distribution.)
>> >So assume the probablity 1 proposition, for all f in F
>> > some x,y in R with x not in f(y), y notin f(x).
>>
>> But _this_ is an error that's not going to be so easy
>> to fix. It's true that a countable intersection of sets
>> of full probability has full probability, but F is
>> uncountable, and that doesn't work for uncountable
>> intersections.
>>
>That I dispute by using your counter example
>below using P(n /= 1) = 1 instead of P(x /= a) = 1.
Learn a little actual probability before "disputing"
things like this.
Or note that the integers are countable but the
reals are not, and note that my comment has
to do with _uncountable_ unions and intersections.
Yes, _if_ P is a probability on the integers such that
P({n}) > 0 for every n then what you say is so.
That has no relevance whatever to the counterexample
I gave.
>> An example of the same technique that proves something false:
>>
>> For each x in [0,1], let P(x) be the statement
>> "a random y in [0,1] is not equal to x".
>> Each P(x) is true with probability 1.
>>
>> Now we incorrectly jump to the conclusion that, with probability 1,
>> _all_ the P(x) are true. It follows that almost every y in [0,1] has the
>> property that it is not equal to any element of [0,1], in particular y
>> is not equal to y.
>>
>Of course, not (P(A) = 1 implies A),
For heaven's sake, learn some probability before lecturing
people about it. No, P(A) = 1 does not imply A.
>or equivalently,
>not (P(A) = 0 implies not A). This is shown using
>the same examples P(n = 0) = 0 = P(x = a).
>
>> >Now assume ZFC and CH. Then R can be well ordered with
>> > R = { r_xi | xi < omega_1 }.
>> >
>> >Define f in F with f(r_eta) = { r_xi | xi <= eta }.
>> >
>> >Now if psi <= chi, then r_psi in f(r_chi),
>> > otherwise r_chi in f(r_psi).
>> >Hence a probality one contradiction,
>> >showing the probablity of CH is zero.
>>
>This proof by Freiling seems to show that
>
>CH, f in F ==> P(some x,y with x not in f(y), y not in f(x)) = 1
>and
>CH ==> P(for all f in F, some x,y with x not in f(y), y not in f(x)) = 1
>
>Is that a resolution to his intuitive disproof of CH?
>
>----
David C. Ullrich
umu...@gmail.com
> The probablity of the Continuum Hypothesis is zero.
Sure. The probability of anything exact is zero. Not so ?
Han de Bruijn
William Elliot
> <ma...@hevanet.remove.com> wrote:
> >>
> >> >The probablity of the Continuum Hypothesis is zero.
> >> >
> >> >Let C = { A subset R | A countable }, F = { f:R -> C }
> >> >Thus for all x,y in R
> >> > P(x in f(y)) = 0 = P(y in f(x))
> >> > P(x not in f(y)) = 1 = P(y not in f(x))
> >
> >> This is a little imprecise - before you can talk about the
> >> probability that an x in R does something you need to specify what
> >> "distribution" you're talking about. Statements about probability
> >> that do not specify a distribution are usually taken to refer to the
> >> "natural" distribution, but there is no such thing on R.
> >>
> >Oh? Doesn't P(0 < x) = 1/2 ?
>
> No. Say A is a subset of R. What's the _definition_ of P(x in A)?
P(x in A) = lim(n->oo) P(x in [-n,n] /\ A)
> >> That doesn't matter much - we can simply replace R everywhere with
> >> the interval [0,1], where there _is_ a natural distribution.
> >>
> >P(x in A) = measure A ?
> Yes (for the "uniform" distribution.)
>
> >> >So assume the probablity 1 proposition, for all f in F
> >> > some x,y in R with x not in f(y), y notin f(x).
> >>
> >> But _this_ is an error that's not going to be so easy to fix. It's
> >> true that a countable intersection of sets of full probability has
> >> full probability, but F is uncountable, and that doesn't work for
> >> uncountable intersections.
> Or note that the integers are countable but the
> reals are not, and note that my comment has
> to do with _uncountable_ unions and intersections.
Does P(n /= 1) = 1, P(n = 1) = 0?
> Yes, _if_ P is a probability on the integers such that
> P({n}) > 0 for every n then what you say is so.
>
> >> An example of the same technique that proves something false:
> >>
> >> For each x in [0,1], let P(x) be the statement
> >> "a random y in [0,1] is not equal to x".
> >> Each P(x) is true with probability 1.
> >>
> >> Now we incorrectly jump to the conclusion that, with probability 1,
> >> _all_ the P(x) are true. It follows that almost every y in [0,1] has the
> >> property that it is not equal to any element of [0,1], in particular y
> >> is not equal to y.
> >>
> >Of course, not (P(A) = 1 implies A),
>
> For heaven's sake, learn some probability before lecturing
> people about it. No, P(A) = 1 does not imply A.
>
Notice that I negated 'P(A) = 1 implies A'.
David C. Ullrich
<ma...@hevanet.remove.com> wrote:
>On Sun, 3 Aug 2008, David C. Ullrich wrote:
>> <ma...@hevanet.remove.com> wrote:
>> >>
>> >> >The probablity of the Continuum Hypothesis is zero.
>> >> >
>> >> >Let C = { A subset R | A countable }, F = { f:R -> C }
>> >> >Thus for all x,y in R
>> >> > P(x in f(y)) = 0 = P(y in f(x))
>> >> > P(x not in f(y)) = 1 = P(y not in f(x))
>> >
>> >> This is a little imprecise - before you can talk about the
>> >> probability that an x in R does something you need to specify what
>> >> "distribution" you're talking about. Statements about probability
>> >> that do not specify a distribution are usually taken to refer to the
>> >> "natural" distribution, but there is no such thing on R.
>> >>
>> >Oh? Doesn't P(0 < x) = 1/2 ?
>>
>> No. Say A is a subset of R. What's the _definition_ of P(x in A)?
>
>P(x in A) = lim(n->oo) P(x in [-n,n] /\ A)
First, you just made that up - that's not "the" definition.
Second, what's the definition of P(x in [-n,n] /\ A) ? If the
"P"'s on both sides of the equation refer to the same thing
then this "definition" is circular. If otoh on the right
side you're referring to the uniform distribution on
[-n,n] then (i) you need to specify this - maybe call
that distribution P_n and define
P(x in A) = P_n(x in [-n,n] /\ A)
(ii) _if_ we clarify the "definition" as in (i), which
is what I think you must have meant, then the P
you define is _not_ a probability measure.
>> >> That doesn't matter much - we can simply replace R everywhere with
>> >> the interval [0,1], where there _is_ a natural distribution.
>> >>
>> >P(x in A) = measure A ?
>> Yes (for the "uniform" distribution.)
>>
>> >> >So assume the probablity 1 proposition, for all f in F
>> >> > some x,y in R with x not in f(y), y notin f(x).
>> >>
>> >> But _this_ is an error that's not going to be so easy to fix. It's
>> >> true that a countable intersection of sets of full probability has
>> >> full probability, but F is uncountable, and that doesn't work for
>> >> uncountable intersections.
>
>> Or note that the integers are countable but the
>> reals are not, and note that my comment has
>> to do with _uncountable_ unions and intersections.
>
>Does P(n /= 1) = 1, P(n = 1) = 0?
It's impossible to say, since I have no idea what P
you're talking about.
Lemme try to explain it this way. You know some topology.
You know that there's no definition of "open set" in
topology - the collection of open sets is _given_, a
collection T forms a topology on X if and only if [etc].
So if someone asked you whether a certain set was open
you'd answer that it was impossible to say unless he told
you what topology he was talking about (except in some
cases where there's a standard "default" topology).
It's the same here: what P are you talking about?
David C. Ullrich
William Elliot
> <ma...@hevanet.remove.com> wrote:
>
> >> No. Say A is a subset of R. What's the _definition_ of P(x in A)?
m(A) = measure of A
for all n in N, P_n(x in [-n,n] /\ A) = m([-n,n] /\ A)/m([-n,n])
> If otoh on the right side you're referring to the uniform distribution
> on [-n,n] then (i) you need to specify this - maybe call that
> distribution P_n and define
>
> P(x in A) = P_n(x in [-n,n] /\ A)
lim(n->oo) P_n(x in [-n,n] /\ A)
> (ii) _if_ we clarify the "definition" as in (i), which
> is what I think you must have meant, then the P
> you define is _not_ a probability measure.
>
Why not? P(x in A) in [0,1] for all measurable A.
> >Does P(n /= 1) = 1, P(n = 1) = 0?
>
> It's impossible to say, since I have no idea what P
> you're talking about.
>
P defined like above using the counting measure.
> >> Yes, _if_ P is a probability on the integers such that
> >> P({n}) > 0 for every n then what you say is so.
> >>
> >> >> An example of the same technique that proves something false:
> >> >>
> >> >> For each x in [0,1], let P(x) be the statement
> >> >> "a random y in [0,1] is not equal to x".
> >> >> Each P(x) is true with probability 1.
P(x /= a) = 1.
> >> >> Now we incorrectly jump to the conclusion that, with probability 1,
> >> >> _all_ the P(x) are true. It follows that almost every y in [0,1] has the
> >> >> property that it is not equal to any element of [0,1], in particular y
> >> >> is not equal to y.
> >> >or equivalently,
> >> >not (P(A) = 0 implies not A). This is shown using
> >> >the same examples P(n = 0) = 0 = P(x = a).
----
David C. Ullrich
<ma...@hevanet.remove.com> wrote:
>On Mon, 4 Aug 2008, David C. Ullrich wrote:
>> <ma...@hevanet.remove.com> wrote:
>>
>> >> No. Say A is a subset of R. What's the _definition_ of P(x in A)?
>
>m(A) = measure of A
>
>for all n in N, P_n(x in [-n,n] /\ A) = m([-n,n] /\ A)/m([-n,n])
>
>> If otoh on the right side you're referring to the uniform distribution
>> on [-n,n] then (i) you need to specify this - maybe call that
>> distribution P_n and define
>>
>> P(x in A) = P_n(x in [-n,n] /\ A)
>
>lim(n->oo) P_n(x in [-n,n] /\ A)
>
>> (ii) _if_ we clarify the "definition" as in (i), which
>> is what I think you must have meant, then the P
>> you define is _not_ a probability measure.
>>
>Why not? P(x in A) in [0,1] for all measurable A.
This is getting silly. You simply have no idea what
you're talking about here - you need to _learn_
some probability theory...
Ok. First, it's not true that that limit _exists_ for
every measurable A. That's not so important, we
could just consider the sets A for which the
limit does exist.
More important is that the P you have defined
is not countable additive (hence none of the
results in standard probability theory apply).
For example, let A_n = [n, n+1]. Then
P(A_n) = 0 for every n, but the union of the
A_n is R, and P(R) = 1.
(Does countable additivity matter? Yes,
if you want to apply probability theory.
But what if you don't want to do that?
Well, in the original post you were applyiug
the much _stronger_ property of _uncountable_
additivity.
Not to mention the fact that in your original
post you never did specify what "P" you
were talking about.)
>> >Does P(n /= 1) = 1, P(n = 1) = 0?
>>
>> It's impossible to say, since I have no idea what P
>> you're talking about.
>>
>P defined like above using the counting measure.
Exactly the same problems arise as with the P
you defined for subsets of R above (take
A_n = {n} this time).
>> >> Yes, _if_ P is a probability on the integers such that
>> >> P({n}) > 0 for every n then what you say is so.
>> >>
>> >> >> An example of the same technique that proves something false:
>> >> >>
>> >> >> For each x in [0,1], let P(x) be the statement
>> >> >> "a random y in [0,1] is not equal to x".
>> >> >> Each P(x) is true with probability 1.
>
>P(x /= a) = 1.
>
>> >> >> Now we incorrectly jump to the conclusion that, with probability 1,
>> >> >> _all_ the P(x) are true. It follows that almost every y in [0,1] has the
>> >> >> property that it is not equal to any element of [0,1], in particular y
>> >> >> is not equal to y.
>
>> >> >or equivalently,
>> >> >not (P(A) = 0 implies not A). This is shown using
>> >> >the same examples P(n = 0) = 0 = P(x = a).
>
>----
David C. Ullrich