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

On 1/24/25 4:46 AM, WM wrote:

On 23.01.2025 13:01, Richard Damon wrote:

On 1/23/25 3:32 AM, WM wrote:

 

Numbers do not change. Only their state of being known.

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So, why do you claim they changed?

 I did not.

YOu did, but you don't undetstand your words. When ypu said that the ... was the "dark numbers" you

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You seem to think that you can't use "visible" numbers collectively, they become dark when you do, nor can your "dark" numbers be used individually,

 You have not understood anything. All numbers can be used collectively but visible numbers can be used as individuals.

But All of the numbers can also be used individually too. You just can't use them all at once with your logic, since your logic can't actually handle an infinite set.
Your seem to think that the need to handle an infinite set of numbers, that each could be used individually, collectively means these numbers couldn't be used individually,
It is just YOUR LOGIC that is "dark" and can't do what you need it to do.

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No. Every number that is defined in a system by its FISON is visible. Many numbers smaller than 10^99 are defined on the pocket calculator. No greater number can be defined in that system.

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Numbers are not definied by its "FISON", its FISON is defined by the number.

 Both is correct for visible numbers.

And all Natural Numbers meet your requirement for being a "visible" number.

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ALL Natural Numbers are defined, and thus visible, and not "dark"

 Up to every FISON |ℕ \ {1, 2, 3, ..., n}| = ℵo. Since every FISON is the union of all its predecessors we get
F(n): |ℕ \ UF(n)| = ℵo.
If you don't believe in the union of all F(n), find the first exception.

But your conclusion, that some Natual Numbers can't be used individually doesn't follow.
There is no execption, but there doesn't need to be, as it is an accepted fact that all finite sets of numbers are not infinite, so no finite set of numbers is teh full set of the Natural Numbers.
All you are doing is proving your logic can't handle actually (or even potentially) infinite sets.

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There are an infinite set of FISONs, one for every Natural Number.

 The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem would require the existence of a first necessary FISON.

Sure it does, but you need to use an INFINITE set of them, and thus can not individually name all of them at once. That doesn't mean that there are any that can't be named individually.

 Regards, WM