From: Han de Bruijn on
Mike Kelly wrote:

> Han.deBruijn(a)DTO.TUDelft.NL wrote:
>
>>Mike Kelly wrote:
>>
>>>The limit of a sequence need not have a property that each of its
>>>elements has.
>>
>>The limit of the sequence 1,1,1,1,1,1,1,1, ... ,1, ... is 1.
>
> True. How does that refute anything I said?
>
> We have a sequence of sets..
>
> {0}, {0,1}, {0,1,2}, {0,1,2,3} ....
>
> and for each we can define a uniform probability distribution to choose
> one element of the set. The limit of this sequence of sets is N.
>
> We construct a sequence of corresponding "Kolmogorov sums", the
> summations of the elementary probabilities.
>
> 1/1, 2/2, 3/3, 4/4, ... =
> 1, 1, 1, 1...
>
> The limit of this sequence of real numbers is 1.
>
> Now you want to conclude that there is a uniform distribution on N.
> However, you offer no justification for this.

No. I just want to conclude that this is a sufficient base for inferring
that the probability of a natural being divisible by three is 1/3 .

Han de Bruijn

From: Mike Kelly on

Han de Bruijn wrote:
> Mike Kelly wrote:
>
> > Han.deBruijn(a)DTO.TUDelft.NL wrote:
> >
> >>Mike Kelly wrote:
> >>
> >>>The limit of a sequence need not have a property that each of its
> >>>elements has.
> >>
> >>The limit of the sequence 1,1,1,1,1,1,1,1, ... ,1, ... is 1.
> >
> > True. How does that refute anything I said?
> >
> > We have a sequence of sets..
> >
> > {0}, {0,1}, {0,1,2}, {0,1,2,3} ....
> >
> > and for each we can define a uniform probability distribution to choose
> > one element of the set. The limit of this sequence of sets is N.
> >
> > We construct a sequence of corresponding "Kolmogorov sums", the
> > summations of the elementary probabilities.
> >
> > 1/1, 2/2, 3/3, 4/4, ... =
> > 1, 1, 1, 1...
> >
> > The limit of this sequence of real numbers is 1.
> >
> > Now you want to conclude that there is a uniform distribution on N.
> > However, you offer no justification for this.
>
> No. I just want to conclude that this is a sufficient base for inferring
> that the probability of a natural being divisible by three is 1/3 .

It isn't.

--
mike.

From: Han de Bruijn on
Virgil wrote:

> In article <3a6c6$4510f00a$82a1e228$27505(a)news2.tudelft.nl>,
> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote:
>
>>David R Tribble wrote:
>>
>>>mueckenh wrote:
>>>
>>>>>Nothing has changed. There is no complete set of natural numbers. Any
>>>>>set that can be established is a finite set. Hence, the probability to
>>>>>select a number divisible by 3 is 1/3 or very very close to 1/3.
>>>
>>>Virgil wrote:
>>>
>>>>That presumes that the allegedly finite set of naturals that can be
>>>>constructed is nearly uniform with respect to divisibility by 3 at
>>>>least, and probably by other numbers as well. What is the justification
>>>>for this assumption?
>>
>>Wolfgang says litteraly: "_or_ very very close to 1/3".
>
> Which requires "_nearly uniform_ with respect to divisibility by 3".

Sorry, Virgil. I don't get it.

Han de Bruijn

From: Han de Bruijn on
stephen(a)nomail.com wrote:

> Assuming that Tony's definition of infinitesimal has anything
> to do with any standard definition of infinitesimal. :)

What is "any standard definition of infinitesimal" ?

Han de Bruijn

From: Han de Bruijn on
Tony Orlow wrote:

> Han de Bruijn wrote:
>
>> Tony Orlow wrote:
>>
>>> Han, if I prove inductively, say, that 2^x>2*x for all x>2, do you
>>> find it objectionable to say that this also applies to any infinite
>>> value, if such a thing existed, given that any infinite value would
>>> be greater than any finite value, and therefore greater than 2?
>>
>> Give me one reason, Tony, why I would find such a theorem interesting
>> in the first place. I'd prefer the ultimate terseness in mathematics,
>> especially if it comes to infinities.
>
> Okay, but it's not that particular theorem that is of interest, but the
> system of theorems this extension of induction makes possible. If we can
> say that such inductive arguments hold in the infinite case, and within
> any range up to n one set has size 2n (multiples of 1/2) and the other
> x^2 (logs base 2 of naturals), then proving that n>2 -> n^2>2n proves
> the second to be bigger than the first, for infinite n, since they are
> larger than 2. This gives us a nice exact way of comparing infinite sets
> over an infinite range, free from "limit ordinals" and other nonsense.
> This allows us to distinguish between the sizes of the naturals vs. the
> evens, and even detect the addition or removal of a single element from
> an infinite set.
>
> Hope that made sense.

Thought I've made my point. When I said that your infinities are not
worse than their standard alephs, then I didn't say that I find your
infinities "better" :-( Sorry ...

Han de Bruijn