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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Sat, 7 Dec 2024 12:09:19 +0100
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On 06.12.2024 19:17, Jim Burns wrote:
> On 12/6/2024 3:19 AM, WM wrote:

>>> ⎜ With {} NOT as an end.segment,
>>
>> all endsegments hold content.
> 
> But no common.to.all finite.cardinals.

Show two endsegments which do not hold common content.
> 
>>> ⎜ there STILL are
>>> ⎜ more.than.finite.many end.segments,
>>
>> Not actually infinitely many however.
> 
> More.than.finitely.many are enough to
> break the rules we devise for finitely.many.

More than finitely many are finitely many, unless they are actually 
infinitely many. Therefore they are no enough.
> 
> For each finite.cardinal,
> up.to.that.cardinal are finitely.many.
> A rule for finitely.many holds.
> 
> All.the.finite.cardinals are more.than.finitely.many.
> A rule for more.than.finitely.many holds.

All the fite cardinals are actually infinitely many. That is impossible 
as long as an upper bound rests in the contentents of endsegments.
> 
>> If all endsegments have content,
>> then not all natnumbers are indices,
> 
> That seems to be based on the idea that
> no finite.cardinal is both index and content.

By an unfortunate definition (made by myself) there is always one 
cardinal content and index: E(2) = {2, 3, 4, ...}. But that is not 
really a problem.
> 
> Elsewhere, considering one set, that's true.
> No element is both
> index(minimum) and content(non.minimum).
> 
> However,
> here, we're considering all the end segments.
> Each content is index in a later set.
> Each non.zero index is content in an earlier set.

"All at once" is the seductive attempt of tricksters. All that happens 
in a sequence can be investigated at every desired step.

> Each content is index in a later set.

Only if all content is lost. That is not possible for visible 
endsegments. They all are infinite and therefore are finitely many.
> 
>>> ⎜ And therefore,
>>> ⎜ the intersection of all
>>> ⎝ STILL holds no finite cardinal.
>>
>> No definable finite cardinal.
> 
> Wasn't there a time when you (WM)
> thought 'undefinable finite.cardinal'
> was contradictory?

Yes, until about six years ago.

	Preface
This book contains the collection of my writings on dark numbers most of 
which have been published already in lectures, at conferences, and here 
and there in the internet. Although I was a strong opponent of Cantor's 
actual infinity, an internet discussion in 2018 [1] has changed my mind 
in that without actual infinity the real axis would have gaps. That is 
my reason for accepting it and investigating its consequences.
[W. Mückenheim: "Evidence for Dark Numbers", ELIVA Press, Chisinau 2024. 
ISBN 978-99993-2-218-8, in press]

> Round up the usual suspects
> and label them 'definable'.

∀k ∈ ℕ : E(k+1) = E(k) \ {k} cannot come down to the empty set in 
definable numbers. No other way however is accessible.
> 
> The intersection of all non.empty.end.segments
> of the definable finite.cardinals,
> which are each infinite non.empty.end.segments,
> is empty.

A clear selfcontradiction because of inclusion monotony.
> 
> Generalizing,
> the intersection of all non.empty end.segments is empty.
> 
> It is an argument considering finites,
> of which there are more.than.finitely.many.

It is violating mathematics and logic. Like Bob.

Regards, WM