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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: The non-existence of "dark numbers"
Date: Sat, 15 Mar 2025 13:42:35 -0700
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On 3/15/2025 1:56 AM, WM wrote:
> 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.

When you see { 1, 2, 3 ... }

4 is dark to you?