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

On 2024-12-15 11:33:15 +0000, WM said:

On 15.12.2024 12:03, Mikko wrote:

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

On 14.12.2024 09:52, Mikko wrote:

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

 

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].

 Do you happen to know any set that is Dedekind-infinite?
 

No, there is no such set.

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

 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.
 

is Dedekind-infinte:
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.

 This "bijection" appears possible but it is not.

 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.

We needn't. The axioms of natural numbers ensure that every natural number
has a successor, no natural number is its own successor, and no two natural
numbers has the same successor. If that is not possible then there are no
natural numbers.
--
Mikko