Re: Replacement of Cardinality

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Sujet : Re: Replacement of Cardinality
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.logic sci.math
Date : 02. Aug 2024, 18:08:42
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <1b5b8ca644d046287b98425370d9b63969f29b77@i2pn2.org>
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User-Agent : Mozilla Thunderbird
On 8/2/24 12:32 PM, WM wrote:
Le 02/08/2024 à 18:24, Richard Damon a écrit :
On 8/2/24 12:05 PM, WM wrote:
 
The claim is that there is a finite difference that seperates any two specific unit fractions,
 Of course.
>
Not that there is a single finite difference that seperates any two arbirtry unit fractions.
 Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points.
So? that number is just ℵo (you don't understand the mathematics of transfinite values)

>
Not understanding the order of the arguement
 Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points. No order necessary.
So? That just shows that between any two unit fractions there IS a distance, and thus they are seperated, as required.

>
For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that provides a unit fraction smaller than x,
 No, eps has to be chosen. x has to be chosen. My claim is that most are dark and cannot be chosen.
It has to be chosen for a given pair of unit fractions.

Note: For every 1/n there exists a smaller real number. But they cannot be chosen because most 1/n already cannot be chosen. Proof: For every chosen eps you fail to separate infinitely many unit fractions.
WHY CAN'T IT BW CHOSEN? That seems to be the flaw in your logic, you think there exist numbers that are defined mathematically, but can't be chosen.
If there WAS just a single eps that seperated ALL unit fractions, then the could be no more than 1/eps unit fractions, and thus that would be larger than the largest Natural Number.
But EVERY Natural Number has a successor, so there is no largest Natural Number.
This just shows that your logic is limited and doesn't handle the sets you are trying to us it one, and it breaks.
 
and thus NUF(x) can not be 1 for any finite x.
 But for x belonging to the first ℵ₀*2ℵ₀ points we have less than ℵ₀ unit fractions.
Nope, because ℵo-1 = ℵo, and thus there are ℵo points below (and above) every point. All values of ℵo are not ordered within themselves.
If you math can't handle that, it can't handle the infinities. Since it doesn't handle the unbounded numbers, that it doesn't handle infinities isn't surprizing.

 Regards, WM

Date Sujet#  Auteur
2 Jul 25 o 

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