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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Sun, 15 Dec 2024 11:15:17 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:
> On 14.12.2024 15:08, Richard Damon wrote:
>> On 12/14/24 3:38 AM, WM wrote:
>>> On 14.12.2024 01:03, Richard Damon wrote:
>>>> On 12/13/24 12:00 PM, WM wrote:
>>>>> On 13.12.2024 13:11, Richard Damon wrote:
>>>>>
>>>>>> Note, the pairing is not between some elements of N that are also
>>>>>> in D, with other elements in N, but the elements of D and the
>>>>>> elements on N.
>>>>> Yes all elements of D, as black hats attached to the elements 10n of
>>>>> ℕ, have to get attached to all elements of ℕ. There the simple shift
>>>>> from 10n to n (division by 10) is applied.
>>>> No, the black hats are attached to the element of D, not N.
>>> They are elements of D and become attached to elements of ℕ.
>> No, they are PAIR with elements of N.
>> There is no operatation to "Attach" sets.
> To put a hat on n is to attach a hat to n.
Oh, you mean including the pair (n, 10n) in the bijection. Note that
the larger number is on the right and the pair (10n, 100n) is
unaffected.

>>>>> That pairs the elements of D with the elements of ℕ. Alas, it can be
>>>>> proved that for every interval [1, n] the deficit of hats amounts to
>>>>> at least 90 %. And beyond all n, there are no further hats.
>>>> But we aren't dealing with intervals of [1, n] but of the full set.
>>> Those who try to forbid the detailed analysis are dishonest swindlers
>>> and tricksters and not worth to participate in scientific discussion.
>> No, we are not forbiding "detailed" analysis
> Then deal with all infinitely many intervals [1, n].
??? The bijection is not finite. 

>>>> The problem is that you can't GET to "beyond all n" in the pairing,
>>>> as there are always more n to get to.
>>> If this is impossible, then also Cantor cannot use all n.
>> Why can't he? The problem is in the space of the full set, not the
>> finite sub sets.
> The intervals [1, n] cover the full set.
Only in the limit.

>>>> Yes, there are only 1/10th as many Black Hats as White Hats, but
>>>> since that number is Aleph_0/10, which just happens to also equal
>>>> Aleph_0, there is no "deficit" in the set of Natual Numbers.
>>> This example proves that aleph_0 is nonsense.
>> Nope, it proves it is incompatible with finite logic.
> There is no other logic.
There is the logic of the infinite.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.