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

On 12/17/24 4:05 AM, WM wrote:

On 16.12.2024 18:25, joes wrote:

Am Mon, 16 Dec 2024 17:49:20 +0100 schrieb WM:

On 16.12.2024 16:40, joes wrote:

Am Mon, 16 Dec 2024 14:59:27 +0100 schrieb WM:

On 16.12.2024 12:55, joes wrote:

Am Mon, 16 Dec 2024 09:30:18 +0100 schrieb WM:

>

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

It does not. It applies to every single finite „interval”,

What element is not covered by all intervals that I use?

but not to the whole N.

You do not cover N, only finite parts.

What do I miss to cover?

Inf.many numbers for every n.

 But Cantor using every n does not miss to cover anything?
  > N is infinite.
 Every element is the last element of a FISON [1, n]. ℕ is the set of all FISONs. I use all FISONs. ∀n ∈ ℕ: f([1, n]) =< 1/10.
Ever heard of the effect of the universal quantifier?
 Regards, WM
 Regards, WM
 

But your logic can't deal with ALL Fisons.
Note, the mapping isn't in your [1, n] but in N.
Your logic that if it holds for all FISONs, it holds for N, is what shows that 0 == 1, so we see that logic is broken when it is applied to truely infinite things.