Re: The non-existence of "dark numbers"

On 14.03.2025 14:35, Alan Mackenzie wrote:

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

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.

 You're confusing yourself with the outdated notion "potentially
infinite".  The numbers n in an (?the) inductive set are N, not N_def.
Why do you denote the natural numbers by "N_def" when everybody else just
calls them "N"?

Perhaps everybody is unable to see that
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo?

It strives for ℕ but never reaches it because .....

 It doesn't "strive" for N.  You appear to be thinking about a process
taking place in time

Induction and counting are processes. It need not be in time. But it fails to complete ℕ.
ℕ \ {1, 2, 3, ...} = { }.
ℕ \ ℕ_def =/= { }.

"Definable" remains undefined, so there's no point to answer here.  Did
Zermelo, Peano, or von Neumann use "definable" the way you're trying to
use it, at all?

Zermelo claimed that without their construction/proof by induction we don't know whether infinite sets exist at all.
Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir noch des folgenden ... Axioms. [Zermelo: Untersuchungen über die Grundlagen der Mengenlehre I, S. 266] The elements are defined by induction in order to guarantee the existence of infinite sets.

The potentially infinite inductive set has no last element. Therefore
its complement has no first element.

 You're letting "potentially infinite" confuse you again.  The inductive
set indeed has no last element.  So "its complement" (undefined unless we
assume a base set to take the complement in), if somehow defined, is
empty.  The empty set has no first element.

The empty set has not ℵo elements.

But there are ℵo numbers following upon all numbers of ℕ_def.

 N_def remains undefined,

 >> |ℕ \ {1}| = ℵo.
 >> If |ℕ \ {1, 2, 3, ..., n}| = ℵo
 >> then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.

"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.
[ "It is even permissible to think of the newly created number as a
limit to which the numbers nu tend.  If nothing else is understood,
it's held to be the first integer which follows all numbers nu, that
is, is bigger than each of the numbers nu." ]

 

Between the striving numbers ν and ω lie the dark numbers.

 That contradicts the long excerpt from Cantor you've just cited.
According to that, omega is the _first_ number which follows the numbers
nu.  I.e., there is nothing between nu (which we can identify with N) and
omega.  There is no place for "dark numbers".

There is place to strive or tend.

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

 That nonsense has no bearing on the representability of natural numbers
in a mathematician's mind.  You're just saying that the complement in N
of a finite subset of N is of infinite size.  Yes, and.... ?
 

like the dark numbers can do
ℕ \ {1, 2, 3, ...} = { }.

 Dark numbers remain undefined.

Yes, they cannot be determines as individuals.

The above identity, more succinctly
written as N \ N = { } holds trivially, and has nothing to say about the
mythical "dark numbers".

n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo proves that definable numbers are not sufficient.
Regards, WM