Re: The non-existence of "dark numbers"

On 15.03.2025 00:11, Alan Mackenzie wrote:

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

ℕ_def contains all numbers the subtraction of which from ℕ does not
result in the empty set.

 What does "which" refer to?

It refers to the numbers. An Englishman should comprehend that.

To N_def or to a
member of the "all numbers"?

That is one and the same.

 Assuming the former, then if X is any proper subset of N, N \ X is
non-empty.  So by this "definition", N_def is any proper subset of N.

No, ℕ_def contains only definable numbers.

 

Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ. Therefore |ℕ \ ℕ_def| = ℵo. Do you agree?

 Of course not.

Then you cannot think logically.

It all depends on the X from which N_def is formed.  If
X is N \ {1},

Then its elements are mostly undefined as individuals.

Yes, of course. But ℕ_def is not completed by its definition.

 You haven't defined N_def - what appears above is not a coherent
definition.

It is coherent enough. Every element has a finite FISON. ℕ is infinite. Therefore it cannot be emptied by the elements of ℕ_def and also not by ℕ_def.

The tending takes place, but not in a "place".

 

No? Tending means that hitherto undefined natural numbers become
defined. That takes place on the ordinal line.

 "Hitherto" ("bis jetzt" in German) is purely a time based adverb.  The
natural numbers are not defined in a time based sequence.  They are
defined all together.

Not the defined numbers.
Regards, WM