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

On 21.12.2024 14:19, Richard Damon wrote:

On 12/21/24 6:00 AM, WM wrote:

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.

 But you admit that it didn't because it ended up past it.

The reason is that after every visible unit fraction there are more visible unit fractions created. That is potential infinity. You can't understand that matter.

 The numbers didn't just show up, they were always there and we didn't do anything that prohibited it from stopping at any of them.

They were dark.

 The only thing that prevented it from stopping at the "first" unit fraction is the fact that "the first unit fraction" (from the 0 end) just doesn't exist.

Before any unit fraction where it stops there show up many more - but only afterwards.

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

You think circular definitions are acceptable.

It is not circular.

Sure it does, as NUF(x) doesn't exist

You can't grasp it. That's not my problem.

>
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.

 What Natural Number can't be used as an index?

The true infinite can be exhausted, but only collectively:
ℕ \ {1, 2, 3, ...} = { }. So it gets empty. The  getting empty happens collectively but cannot be observed individually. We cannot construct a list that assigns natural numbers to all natural numbers. Almost all cannot be manipulated individually. But
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
proves that 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, endsegments containing only 1, 2, 3, or n ∈ ℕ numbers, have been passed.
Regards, WM