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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: The non-existence of "dark numbers"
Date: Fri, 14 Mar 2025 16:27:23 -0000 (UTC)
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Am Fri, 14 Mar 2025 13:09:24 +0100 schrieb WM:
> On 13.03.2025 18:53, Alan Mackenzie wrote:
>> WM <wolfgang.mueckenheim@tha.de> wrote:
> 
>> "Definable number" has not been defined by you, except in a
>> sociological sense.
> Then use numbers defined by induction:
> |ℕ \ {1}| = ℵo.
> If |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
> Here the numbers n belonging to a potentially infinite set are defined.
> This set is called ℕ_def. It strives for ℕ but never reaches it
This set *is* N.

>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo infinitely many numbers
>>> remain. That is the difference between dark and definable numbers.
>> Rubbish!  It's just that the set difference between an infinite set and
>> a one of its finite subsets remains infinite.
> Yes, just that is the dark part. All definable numbers belong to finite
> sets.
All naturals do.

>>> ℕ_def is a subset of ℕ. If ℕ_def had a last element, the successor
>>> would be the first dark number.
>> If, if, if, ....  "N_def" remains undefined, so it is not sensible to
>> make assertions about it.
> See above. Every inductive set (Zermelo, Peano, v. Neumann) is
> definable.
As is N.

>>>> But I can agree with you that there is no first "dark number".  That
>>>> is what I have proven.  There is a theorem that every non-empty
>>>> subset of the natural numbers has a least member.
>>> That theorem is wrong in case of dark numbers.
>> That's a very bold claim.  Without further evidence, I think it's fair
>> to say you are simply mistaken here.
> The potebtially infinite inductive set has no last element. Therefore
> its complement has no first element.
Because it is empty.

>>>>> When |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| =
>>>>> ℵo. How do the ℵo dark numbers get visible?
>> There are no such things as "dark numbers", so talking about their
>> visibility is not sensible.
> But there are ℵo numbers following upon all numbers of ℕ_def.
No.

>>>> There is no such thing as a "dark number".  It's a figment of your
>>>> imagination and faulty intuition.
>>> Above we have an inductive definition of all elements which have
>>> infinitely many dark successors.
>> "Dark number" remains undefined, except in a sociological sense.  "Dark
>> successor" is likewise undefined.
> "Es ist sogar erlaubt, sich die neugeschaffene Zahl ω als Grenze zu
> denken, welcher die Zahlen ν zustreben, wenn darunter nichts anderes
> verstanden wird, als daß ω die erste ganze Zahl sein soll, welche auf
> alle Zahlen ν folgt, d. h. größer zu nennen ist als jede der Zahlen ν."
> Between the striving numbers ν and ω lie the dark numbers.
Rebutted elsewhere.

>>> The set ℕ_def defined by induction does not include ℵo undefined
>>> numbers.
>> The set N doesn't include ANY undefined numbers.
>   ℵo
Neither undefined nor in N.

>>>> Quite aside from the fact that there is no
>>>> mathematical definition of a "defined" number.  The "definition" you
>>>> gave a few posts back was sociological (talking about how people
>>>> interacted with eachother) not mathematical.
>>> Mathematics is social, even when talking to oneself. Things which
>>> cannot be represented in any mind cannot be treated.
>> 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,
> ...} = { }.
Maybe not in your mind.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.