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

On 12/15/24 6:33 AM, WM wrote:

On 15.12.2024 12:03, Mikko wrote:

On 2024-12-14 09:50:52 +0000, WM said:
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On 14.12.2024 09:52, Mikko wrote:

On 2024-12-12 22:06:58 +0000, WM said:

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In mathematics, a set A is Dedekind-infinite (named after the German mathematician Richard Dedekind) if some proper subset B of A is equinumerous to A. [Wikipedia].

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Do you happen to know any set that is Dedekind-infinite?
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No, there is no such set.

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The set of natural numbers, if there is any such set,

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If ℕ is a set, i.e. if it is complete such that all numbers can be used for indexing sequences or in other mappings, then it can also be exhausted such that no element remains. Then the sequence of intersections of endsegments
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content. Then, by the law
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
the content must become finite.
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is Dedekind-infinte:
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.

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This "bijection" appears possible but it is not.

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So you say that there is a natural number that does not have a next
natural number. What number is that?

 We cannot name dark numbers as individuals. All numbers which can be used a individuals belong to a potentially infinite collection ℕ_def. There is no firm end. When n belongs to ℕ_def, then also n+1 and 2n and n^n^n belong to ℕ_def. The only common property is that all the numbers belong to a finite set and have an infinite set of dark successors.
 ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
 This is the only way to explain that the intersection of endegments
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content in a sequences which allow the loss of only one number per step:
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
 Regards, WM
 

You can't "name" your dark numbers, because they don't exist in the set you claim they are part of,
You already admitted that the set of definable natural numbers was in fact the set of all natural numbers, so there is nothing left to be dark except the dark hole created by the explosion of your logic by its nconsistancies when msused the way you do.