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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Mon, 11 Nov 2024 09:41:41 +0100
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On 10.11.2024 18:49, Jim Burns wrote:
> On 11/10/2024 4:35 AM, WM wrote:

> The set
>    {3,4,5}
> does not _change_ to the set
>    {6,7,8}
> because
> our sets do not change.

But points or intervals in geometry can be translated on the real axis.

> Our sets do not change.

But points or intervals in geometry can be translated on the real axis.
> 
>>>
>>> In the first case, with the not.changing sets,
>>> a finite ๐˜€๐—ฒ๐—พ๐˜‚๐—ฒ๐—ป๐—ฐ๐—ฒ of ๐—ฐ๐—น๐—ฎ๐—ถ๐—บ๐˜€ which
>>>   has only true.or.not.first.false ๐—ฐ๐—น๐—ฎ๐—ถ๐—บ๐˜€
>>> has only true ๐—ฐ๐—น๐—ฎ๐—ถ๐—บ๐˜€.
>>
>> But it
> 
> "It" refers to who or what?

To that finite ๐˜€๐—ฒ๐—พ๐˜‚๐—ฒ๐—ป๐—ฐ๐—ฒ.

>> But it will never complete
>> an infinite set of claims.
> 
> We do not need an infinite ๐˜€๐—ฒ๐—พ๐˜‚๐—ฒ๐—ป๐—ฐ๐—ฒ of ๐—ฐ๐—น๐—ฎ๐—ถ๐—บ๐˜€ completed.
> We do not want an infinite ๐˜€๐—ฒ๐—พ๐˜‚๐—ฒ๐—ป๐—ฐ๐—ฒ of ๐—ฐ๐—น๐—ฎ๐—ถ๐—บ๐˜€ completed.

But you claim that _all_ fractions are in bijection with all natural 
numbers, don't you?
>> It will forever remain in the status nascendi.
>> Therefore
>> irrelevant for actual or completed infinity.
> 
> A finite ๐˜€๐—ฒ๐—พ๐˜‚๐—ฒ๐—ป๐—ฐ๐—ฒ of ๐—ฐ๐—น๐—ฎ๐—ถ๐—บ๐˜€, each of which
> is true.or.not.first.false,
> will forever remain
> a finite ๐˜€๐—ฒ๐—พ๐˜‚๐—ฒ๐—ป๐—ฐ๐—ฒ of ๐—ฐ๐—น๐—ฎ๐—ถ๐—บ๐˜€, each of which
> is true.or.not.first.false.

Therefore such a sequence does not entitle you to claim infinite mappings.
> 
>>> Infinite sets can correspond to
>>> other infinite sets which,
>>> without much thought about infinity,
>>> would seem to be a different "size".
>>
>> But they cannot become such sets.
> 
> Our sets do not change.

My intervals I(n) = [n - 1/10, n + 1/10] must be translated to all the 
midpoints 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 
3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... if you want to 
contradict my claim.
> 
>> But they cannot be completely transformed
> < into each other.
> 
> Our sets do not change.

But intervals can be shifted.
> Consider again the two sets of midpoints
> โŸจ 1, 2, 3, 4, 5, ... โŸฉ and
> โŸจ 1/1, 1/2, 2/1, 1/3, 2/2, ... โŸฉ
> 
> They both _are_
> And their points correspond
> by i/j โ†ฆ n = (i+j-1)(i+j-2)/2+i

The first few terms do correspond or can be made c orresponding. That 
can be proven by translating the due intervals. But the full claim is 
nonsense because it is impossible to satisfy.

Regards, WM