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

On 12/16/24 3:30 AM, WM wrote:

On 15.12.2024 21:21, joes wrote:

Am Sun, 15 Dec 2024 16:25:55 +0100 schrieb WM:

On 15.12.2024 12:15, joes wrote:

Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:

>

That pairs the elements of D with the elements of ℕ. Alas, it can
be proved that for every interval [1, n] the deficit of hats
amounts to at least 90 %. And beyond all n, there are no further
hats.

But we aren't dealing with intervals of [1, n] but of the full set.

Those who try to forbid the detailed analysis are dishonest
swindlers and tricksters and not worth to participate in scientific
discussion.

No, we are not forbiding "detailed" analysis

Then deal with all infinitely many intervals [1, n].

??? The bijection is not finite.

Therefore we use all [1, n].

Those are all finite.

 All n are finite.

But N isn't, so the sets [1, n] aren't what the bijection is defined on.

>

The problem is that you can't GET to "beyond all n" in the pairing,
as there are always more n to get to.

If this is impossible, then also Cantor cannot use all n.

Why can't he? The problem is in the space of the full set, not the
finite sub sets.

The intervals [1, n] cover the full set.

Only in the limit.

With and without limit.

Wonrg. There is no natural n that „covers N”.

 All intervals do it because there is no n outside of all intervals [1, n]. My proof applies all intervals.

And all the intervals are finite, and thus not the INFINITE set N, which is where the bijection occurs.
Thus your "proof" is just a LIE.

 Regards, WM