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

On 31.12.2024 16:37, Richard Damon wrote:

On 12/30/24 4:38 PM, WM wrote:

On 30.12.2024 17:08, Richard Damon wrote:

On 12/30/24 3:44 AM, WM wrote:

On 30.12.2024 01:36, Richard Damon wrote:

On 12/29/24 5:31 PM, WM wrote:

>

My theorem: Every union of FISONs which stay below a certain
threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.

But what does that prove?

That proves the existence of dark numbers.

How?
It shows that there are numbers you didn't look at, but you admit to
not looking at all the numbers.

Neither can you look at more numbers.

Of course I can look at "more" numbers, as I can look at the one after
where you stopped.

I don't stop but know that every FISON is finite as its name says and
covers at most 1 % of ℕ.

If ℕ is an actually infinite set then it is not made by FISONs.

But there is no Natural Number that isn't in a FISON, which is NOT what
you have been saying.

You cannot find a natural number that isn't in a FISON. But there must
be more because FISONs cover less than 1 % of ℕ.

The problem is your logic can't have "Every" FISON, because "every
FISON" is an infinite set of them,