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

Am Sun, 15 Dec 2024 12:33:15 +0100 schrieb WM:

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:

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.

Shame.

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.

And thus all n e N do.

The only common property is that all the numbers
belong to a finite set and have an infinite set of dark successors.

If all successors belong to N_def, it can’t be finite and the
successors can’t be dark.

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.

The explanation is that the sequence is infinitely long.

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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.