Re: The non-existence of "dark numbers"

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You're confusing yourself with the outdated notion "potentially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"?

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, whereby elements are "created" one per second, or
whatever.  That is a wrong and misleading way of thinking about it.  The
elements of N are defined and proven to exist.  There is no process
involved in this.

∀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.

Gibberish.  What does it mean for a number to "belong to" a finite set?
If you just mean "is an element of", then it's trivially true, since any
number n is a member of the singleton set {n}.

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

Du siehst den Wald vor Bäumen nicht.
[ You can't see the wood for the trees. ]

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.

"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?

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 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.

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.

N_def remains undefined, so talk about numbers following it is not
sensible.

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.
[ "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".

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

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

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).