Re: The non-existence of "dark numbers"

Am Tue, 18 Mar 2025 17:29:08 +0100 schrieb WM:

On 18.03.2025 17:14, Alan Mackenzie wrote:

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

On 18.03.2025 12:57, Alan Mackenzie wrote:

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

 

The set of FISONs is an inductive set. But it is not ℕ because ∀n ∈
UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

That "because" doesn't hold

Because you say so? That's not significant.

There's no connection between the bit before the because and the bit
afterwards.  You're the one who's lacking a degree in maths,

No.

Headline please?

not me, and it shows

that degrees are not guaranteeing the ability to think mathematically.

Indeed.

it is decisive: ℕ_def contains all numbers the subtraction of which
from ℕ does not result in the empty set. Obviously the subtraction of
all numbers which cannot empty ℕ cannot empty ℕ.

I've told you and explained to you, I don't know how many times, that
that is ill-conceived meaningless gibberish.

Can you imagine a natural number that when subtracted from ℕ does not
empty ℕ? Here are some examples: 2, 5, 17. More are available.

This is so stupid. OF COURSE N contains more than one element.

Can you imagine a set of natural numbers that when subtracted from ℕ
does not empty ℕ? Here are some examples: {4, 9}, {1, 2, 3, 4, 5}. More
are vailable.
Now imagine all FISONs. None empties ℕ when applied one after the other.

Hold on. No single one or no FISON of them? Not that it changes anything.

Could that change when all are applied together?

Yup. Why shouldn't it?
--
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.