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

Am Sun, 15 Dec 2024 16:23:19 +0100 schrieb WM:

On 15.12.2024 13:39, joes wrote:

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

On 15.12.2024 12:03, Mikko wrote:

 

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.

Never an n can be named which is responsible for ℕ\ℕ = { }. But ℕ\ℕ =
{} can happen.

Omfg. N is infinite. All n are finite. There is no n such that N =
{1, …, n}.

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.

ℕ_def is potentially infinite. But it does not contain the numbers which
complete the set ℕ.

Yes, it does. They are equal. You think N is finite and made up „dark”
numbers to patch it up.

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.

And that means what? The set ℕ cannot be emptied? The set cannot be
emptied one by one?

Not in any finite number of steps.

Not all elements can be used as indices?
Either this is fact or ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1),
E(2), ..., E(k)} \ {k} empties the set.

Not for any k.

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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.