Re: Cantorian pseudomathematics

Han.deBruijn@xxxxxxxxxxxxxx writes:

> Jesse F. Hughes wrote:
>>
>> I understand the probability a divides n when we have a uniform
>> distribution on initial segments of N (although my terminology may be
>> off).
>
> Your terminology is clear enough.
>
>> It seems you're right that the limit of these values is 1/a.
>>
>> But there is no uniform distribution on N, so I don't see what Han
>> means there and your explanation doesn't clarify it.
>
> Who the hell has decided that "there is no uniform distribution on N"
> !?
>
> Your logic and mine are mutually incompatible if you can't agree
> upon the fact that an initial segment (1..n) of N is just N as n -> oo
> .

Okay. We'll use your logic for determining P( {n} ) for a uniform
distribution. My approach will mirror your method for determining
P( {na | n in N} ) where a is fixed. I take it that we agree that
{na | n in N} is the right way to define the event "a divides n"?

Let's see, you did something like this: Let P_k be the uniform
distribution on the set {1,...,k} and calculate

lim_{k -> oo} P_k( {na | n in N} n {1,...,k} )

This came out to be 1/a, so you conclude P(a divides x) = 1/a.

Now let's try that again. P_k( {n} n {1,...,k} ) = 1/k if n <= k, 0
else. Thus,

P( {n} ) = lim_{k -> oo}P_k( {n} n {1,...,k} ) = lim_{k-> oo} 1/k = 0.

Uh oh. By definition of distribution, P is not a distribution since
it fails to satisfy countable additivity.

Now maybe you have some other definition of distribution in mind, but
you haven't told me what it is.

Note: I don't deny that lim_{k->oo} {1,...,k} = N (given an
appropriate topology on PN). But that's not at issue.

> Oh, and for the sake of "clarity", the limit of a segment (1..n) for
> n -> oo "means": if you have a finite segment (1..m), then replace
> it by a larger finite segment (1..n) where n > m. Recursively.

That's not clearer at all. That isn't what limit means at all.

You need to give a topology on PN that allows one to prove

lim_{k->oo} {1,...,k} = N.

It seems to me that the topology with basis {(n,oo) | n in N} would
do, so we needn't dispute this point. But it's apparent you don't
know what mathematical clarity means.

--
"Just because you're ... in a Ph.d program it does not mean that
you're up to the challenge of being a real mathematician. Only those
who have a purity of mind and dedication to the truth as the highest
ideal have a chance." --James Harris, as Sir Galahad the Pure.
.

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