Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

Liste des GroupesRevenir à s math 
Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.logic
Date : 14. Dec 2024, 15:08:51
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <c31edc62508876748c8cf69f93ab80c0a7fd84ac@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
User-Agent : Mozilla Thunderbird
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.

 
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, just your INCORRECT analysis based on idea that can show that 0 is 1, and that the fast Achillies can't pass a tortoise if he gives it a head start.

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.
This is your problem, you think infinity must be just like the finite, but it isn't, so you can't have infinity in your logic.

>
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.
Since Aleph_0 is part of the logic of the infinite, it doesn't need to follow the rules of the finite.
All you are doing is showing that you don't understand how to work with the infinite.

>
Your logic woud say that Aleph_0/10 would be some value between (possible dark) Natural Numbers
 My logic says that nonsense cannot be defended by accepting just this nonsense.
And that 0 is 1, and that Achilles can't pass the tortoise.
It is YOUR logic that is the nonsense, and it is just its nonsense that can't understand the infinite.

 Regards, WM
Sorry, but all you are doing is proving that you just don't understand about the infinite, and since you try to apply the logic of the finite to it, that you don't even really understand how the finite works.
This is the problem of your "Naive" mathematics, it has nothing to ground it, so it just blows itself up into smithereens with contradictions.

Date Sujet#  Auteur
22 Dec 24 o 

Haut de la page

Les messages affichés proviennent d'usenet.

NewsPortal