Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : noreply (at) *nospam* example.org (joes)
Groupes : sci.mathDate : 15. Dec 2024, 12:15:17
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <e11a34c507a23732d83e3d0fcde7b609cdaf3ade@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
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
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.
…