Re: The non-existence of "dark numbers"

Am Fri, 14 Mar 2025 15:25:22 +0100 schrieb WM:

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You're confusing yourself with the outdated notion "potentially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

That's pretty obvious.

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.

Whatever. They are infinite in any case, which you continually fail to
comprehend.

ℕ \ {1, 2, 3, ...} = { }.

Yes, N = {1, 2, 3, ...}.

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

The complement (wrt what?) is empty.

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.

Come again?

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

N_def remains undefined

"Dark number" remains undefined, except in a sociological sense.
"Dark successor" is likewise undefined.

[ "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.

Wrong. There are no other numbers than the "striving" ones.

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.

And it is filled with nu.

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

There is no number that "makes N empty", wtf.

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.... ?

And what?

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

I see no dark numbers here.

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.

It doesn't. Why do you think there is a number k such that N = {1, ...,
k}?

--
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.