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

On 20.12.2024 16:33, Richard Damon wrote:

On 12/20/24 9:50 AM, WM wrote:

On 20.12.2024 03:52, Richard Damon wrote:

On 12/19/24 10:47 AM, WM wrote:

On 19.12.2024 11:41, Mikko wrote:
>

Not really. What is acceptable for applied mathematics depends on the
application area, which you didn't specify.

>
It was obvious when the argument was discussed: The cursor moves from 0 to 1 on the real axis. For every unit fractions 1/n which it hits there are smaller unit fractions which it had not hit before because they were dark at the first time and came into being only later.

>

No, it means you missed them because you moved too far, because you closed your eyes.

>
The cursor moves until it hits a unit fraction.

 Then why did it it not stop till after it has passed one?

Ask it! My answer is that it stops at the smallest unit fraction. But you deny its existence. Then it can only stop where you allow it. But then many smaller unit fractions show up. So your permission concerns only visible unit fractions.

 

This shows that you can't move to the "first" (smallest valued) 1/n because no such number actually exist,

>
But as soon the cursor has met a unit fraction, many smaller ones show up. They had not "actually" existed as visible unit fractions.

 No, they were always there, you just didn't look for them.

Use the function NUF(x). It shows the smallest unit fraction.

 You find "dark numbers" because you seem to have a blind spot

The function NUF(x) has none. Like the function of endsegments:
If the complete sequence of indices indexing the Cantor list is accepted, then it must be possible to construct it. When indexing the entries by 1, 2, 3, ..., then always infinitely many natnumbers remain. I call these sets endsegments
E(n) = {n+1, n+2, n+3, ...}
with
E(0) = ℕ
and
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}.
This means the sequence of endsegments can decrease only by one natnumber per step. Therefore the sequence of endsegments cannot become empty (i.e., not all natnumbers can be applied as indices) unless the empty endsegment is reached, and before finite endsegments have been passed. These however, if existing at all, cannot be seen. They are dark. Therefore it is impossible to introduce the corresponding entries in Cantor's list.
Regards, WM