Re: The non-existence of "dark numbers"

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.