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

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

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.

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The cursor moves until it hits a unit fraction.

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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.
Your problem is you presume that a "smallest unit fraction" exists, because you don't understand what unbounded means.
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.
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.

>

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

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

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

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

Nope. It shows you are just stupid and don't undertand what you are talking about.
You think circular definitions are acceptable.

>
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

Sure it does, as NUF(x) doesn't exist as a finite mapping of finite values to finite vaules. If is just a figment of your stupidity that proves your mind is just a black hole.

 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 aa an index?
Your problem is you can't tell the difference between "for any" and "for all" and your logic can't do "for all" with the natural numbers as it can't actually handle an infinite set, only "really big" sets.
All you are doing is showing that finite sets and infinite sets are different, and you can't build an infinite set with a finite number of finite operations, which is all your logic allows you to do.
Your "darkness" is just the blindness of you and your logic to your limitiations based on assuming finiteness.

  Regards, WM