Re: Incompleteness of Cantor's enumeration of the rational numbers

On 25.11.2024 22:05, joes wrote:

Am Mon, 25 Nov 2024 13:18:28 +0100 schrieb WM:

But there is no finite set with ALL natural numbers.
Like usual, you mess up with your qualifiers.

ℕ is fixed, that means |ℕ| is fixed.

What does that have to do with it?

It is impossible to add or to delete an element.
It is impossible to change |ℕ| by 1 or more.

 

Limit theory only works if the limit actually exists

If limits exist at all, then the limit of the sequence 1/10, 1/10,
1/10, ... does exist.

But the concept of 1/10th of an infinte set does not exist..

It does.

It has the same cardinality.

Yes, it is much.

 

You can get things that APPEAR to reach a limit, but actually don't.

But if infinite sets do exist, then the set ℕ does exist, and all its
elements are members of finite intervals (0, n].

No, any given element is a member of a finite set, but you can't then
say that ALL are in such a set.

All are in the union of all finite sets.

Why not just directly take N, made up of finite numbers?

Why not? Do it. Consider the black hats at every 10 n and white hats at all other numbers n. It is possible to shift the black hats such that every interval (0, n] is completely covered by black hats. There is no first n discernible that cannot be covered by  black hat. But the origin of each used black hat larger than n is now covered by a white hat. Without deleting all white hats it is not possible to cover all n by black hats. But deleting white hats is prohibited by logic. Exchanging can never delete one of the exchanged elements. Therefore we have here, like in all Cantor-pairings, the same impediment and further disussion is futile: You must deny logic. I do not.
Regards, WM