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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions? (infinitary)
Date: Thu, 10 Oct 2024 19:56:26 +0200
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On 09.10.2024 19:08, joes wrote:
> Am Wed, 09 Oct 2024 14:48:17 +0200 schrieb WM:

>> Not all infinite sets can be compared by size, but we can establish some
>> useful rules.

> That is a weakness of your notion of cardinality.

The weaker weakness of Cantors Cardinality is that it is complete nonsense.

> How do you compare finite sets?

By their numbers of elements.

>> Nonsense. Only potential infinity is used. Never the main body is
>> applied.
> What "main body"?

The actually infinite numbers of dark elements.
> 
>>>> For Cantor's enumeration of all fractions I have given a simple
>>>> disproof.
>>> Your "proofs" tend to be nonsense.
>> Theorem: If every endsegment has infinitely many numbers, then
>> infinitely many numbers are in all endsegments.
>> Proof: If not, then there would be at least one endsegment with less
>> numbers.
> I struggle to follow this illogic. Why should one segment have less
> numbers?

All have the same numbers, namely ℕ. Some of the first numbers are 
transformed from contents to indices and than lost. But almost all 
numbers, namely ℵo, remain (because after every definable natnumber n ℵo 
numbers follow). If the intersection is less than ℵo, at least one 
endsegment must have fewer than ℵo numbers.
> 
>> Note: The shrinking endsegments cannot acquire new numbers.
> Not necessary, they already contain as many as needed.

It would be nessessary if all are infinite but their intersection is 
empty. Then the infinitely many numbers cannot be the same in all 
endsegments. Consider this finite example:

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .

Satz: Jede Menge, die mindestens drei Elemente enthält (nennen wir diese 
Mengen DE), enthält drei Elemente gemeinsam mit allen DE-Mengen. Es gibt 
also drei Elemente, die in allen DE-Mengen enthalten sind. Das ist ein 
erlaubter Quantorentausch.

Regards, WM