Re: The non-existence of "dark numbers"

On 16.03.2025 18:23, Alan Mackenzie wrote:

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

The definable numbers are  potentially infinite sequence. With n also
n+1 and n^n^n belong to it.

 So, no answer.
 But let's take the implications of your second sentence there, ".... with
n, also n+1 ....".  Since I think we'd agree that 0 is a definable
number, then we've just defined the "definable numbers" as the inductive
set.

Yes.

 Thus the set of "definable numbers", N_def is N.

No. The definable natural numbers strive to the smallest infinite number ω. But they are never there.

This is another proof that "dark numbers" don't exist, one you cannot
disagree with without contradicting what you've previously written.

A proof exists that they are there. ℕ_def contains all numbers the subtraction of which from ℕ does not result in the empty set. Obviously the subtraction of all numbers which cannot empty ℕ cannot empty ℕ.

 I doubt I can be bothered any more to address the other falsehoods,
contradictions, and delusions in your last post.  But it's clear that
what you write about such topics is wrong in general, if not totally.

Obviously both of you are cranks. JB who claims that lossless exchange causes losses but refuses to explain this in detail, and you who claims that the subtraction of all numbers which cannot empty ℕ can empty ℕ
Regards, WM