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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: The non-existence of "dark numbers"
Date: Sat, 15 Mar 2025 09:34:18 +0100
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On 15.03.2025 01:49, FromTheRafters wrote:
> on 3/14/2025, WM supposed :
>> On 14.03.2025 17:09, FromTheRafters wrote:
>>> After serious thinking WM wrote :
>>>> On 14.03.2025 14:11, FromTheRafters wrote:
>>>>> WM wrote on 3/14/2025 :
>>>>
>>>>>>> Natural numbers can be "represented in a mind", in fact in any
>>>>>>> mathematician's mind.
>>>>>>
>>>>>> Not those which make the set ℕ empty by subtracting them
>>>>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>>>> like the dark numbers can do
>>>>>> ℕ \ {1, 2, 3, ...} = { }.
>>>>>
>>>>> The set of natural numbers is never empty. What happens is that you 
>>>>> decide which of its elements complies with the definition of 
>>>>> elements of the new set, and remove the rest from consideration.
>>>>
>>>> The new set defined by individual elements cannot be equal to ℕ.
>>>
>>> Why not?
>>
>> Because |ℕ \ ℕ_def| = ℵo.
>> 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. Do you agree?
> 
> There you go with the subtraction idea again.

A great idea! However, it is standard set theory.

> The difference set between 
> the set of naturals and your imagined set of 'defined naturals' has 
> cardinality aleph_zero because your 'defined naturals' is a finite set.

Try to define a natural with more than finitely many predecessors. Fail.
However the defined naturals have no last element because with n also 
n+1 is defined.

Regards, WM