Re: The non-existence of "dark numbers"

On 12.03.2025 18:42, Alan Mackenzie wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

If the numbers are definable.

 Meaningless.  Or are you admitting that your "dark numbers" aren't
natural numbers after all?

They

 

Learn what potential infinity is.

 I know what it is.  It's an outmoded notion of infinity, popular in the
1880s, but which is entirely unneeded in modern mathematics.

That makes "modern mathematics" worthless.

 

3. The least element of the set of dark numbers, by its very
     definition, has been "named", "addressed", "defined", and
     "instantiated".

 So you counter my proof by silently snipping elements 4, 5 and 6 of it?
That's not a nice thing to do.

They were based on the mistaken 3 and therefore useless.

 

Try to remove all numbers individually from the harmonic series such
that none remains. If you can't, find the first one which resists.

 Why should I want to do that?

In order to experience that dark numbers exist and can't be manipulated.

 

Jim has supplied at least one other proof.

 

He claims that lossless exchange can produce losses. He is in
contradiction with logic.

 Irrelevant to the current discussion.  He has supplied at least one other
proof of the non-existence of "dark numbers".

As invalid as yours. If you were able to learn, then you would have the chance here:
ℕ \ {1} = ℵo
and if ℕ \ {1, 2, 3, ..., n} = ℵo
then ℕ \ {1, 2, 3, ..., n+1} = ℵo.
Induction cannot cover all natural numbers but only less than remain uncovered.
Regards, WM