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

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. The fact that there are element with the same "value" in N is irrelevent.

You just don' understand what it means to PAIR elements of two 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.
All you are doing is proving your logic doesn't want to actually try to handle the infinite sets because it can't.
The problem is that you can't GET to "beyond all n" in the pairing, as there are always more n to get to.
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.
Your logic woud say that Aleph_0/10 would be some value between (possible dark) Natural Numbers, but those are all finite, and thus 10x the number above that break would also be finite so we can say that for your logic, Aleph_0 is a finite number, and thus you logic doesn't HAVE i

Regards, WM