Re: The non-existence of "dark numbers"

WM wrote on 3/14/2025 :

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 because
>

∀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.
>

That doesn't shed any light
on "dark" or "defi[n]able" numbers.

>
Du siehst den Wald vor Bäumen nicht.
>

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.
>

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.

 

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.

 

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." E. Zermelo (ed.): "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 195.
>
Between the striving numbers ν and ω lie the dark numbers.
>

The set ℕ_def defined by induction does not include ℵo undefined numbers.

 The set N doesn't include ANY undefined numbers.

>
  ℵo
>

  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, ...} = { }.

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. There are no undefined elements in a set.