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

On 12/7/24 5:47 AM, WM wrote:

On 06.12.2024 16:16, Richard Damon wrote:

On 12/6/24 8:04 AM, WM wrote:

 

And each of them has infinitely many successors,

>
The sequence has no successors. Therefore all numbers including their successors are contained in the sequence.

>
A sequence doesn't have successors,

 Therefore it includes all terms. So it is possible to use all terms. But it is impossible to use them as individuals.
 Regards, WM

No, it says your definition is just nonsense.
You are just stuck in your stupidity of not knowing the meaning of your terms.
The problem is you confuse your qualifiers, yes, we can't use *ALL* terms, as there is a infinite number of them, and we are finite.
But we can use *ANY* of the terms, as they all can individually be used individually.
Your problem is the concept of a "last" one, or one with only a finite number of successors, just don't exist, as if it did, the set could not be infinite, as we would have a finite count of the memberhip.
Thus, your logic is just blown up into smithereens by its contradictions.