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

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:
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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.

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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.

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No, the black hats are attached to the element of D, not N.

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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.

 

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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.

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But we aren't dealing with intervals of [1, n] but of the full set.

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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 problem is that you can't GET to "beyond all n" in the pairing, as there are always more n to get to.

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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.

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.

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This example proves that aleph_0 is nonsense.

 Nope, it proves it is incompatible with finite logic.

There is no other logic.
Regards, WM