Re: The non-existence of "dark numbers"

Am Thu, 13 Mar 2025 17:18:34 +0100 schrieb WM:

On 13.03.2025 13:59, Alan Mackenzie wrote:

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

 

I know that it is self-contradictory because it cannot distinguish
potential and actual infinity.

It can, but doesn't need to.  Potential and actual infinity are
needless concepts which only serve to confuse and obfuscate.  If you
disagree, feel free to cite a standard result in standard mathematics
which depends on these notions.

*crickets*

When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, ....
.... then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. This holds for all
elements of the inductive set, i.e., all FISONs F(n) or numbers n
which have more successors than predecessors.

I.e. all natural numbers.

No. All numbers can be subtracted from ℕ such that none remains:
ℕ \ {1, 2, 3, ...} = { }, let alone ℵo.

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

Only those contribute to the inductive set!

The inductive set is all natural numbers.  Why must you make such a
song and dance about it?

Because when only definable numbers are subtracted from ℕ, then ∀n ∈
ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo infinitely many numbers remain. That
is the difference between dark and defiable numbers.

I have no idea what your "definable numbers" are, but there can only
be finitely many of them (if they are a contiguous subset).

Modern mathematics must claim that contrary to the definition ℵo
vanishes to 0

What does this mean?

because ℕ \ {1, 2, 3, ...} = { }.  That is blatantly
wrong and shows that modern mathematicians believe in miracles.
Matheology.

Modern mathematics need not and does not claim such a ridiculous thing.

ℕ \ {1, 2, 3, ...} = { } is wrong?
>

You didn't point out any mistake in 3.  I doubt you can.

I told you that potential infinity has no last element, therefore
there is no first dark number.

The second part of your sentence does not follow clearly from the
first, therefore the sentence is false.  And even if it were not false,
it has no bearing on my item 3.

Try to think better. ℕ_def is a subset of ℕ. If ℕ_def had a last
element, the successor would be the first dark number.
>

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.

Which is why they disjunct from the naturals.

Try to remove all numbers individually from the harmonic series
such that none remains. If you can't, find the first one which
resists.

Why should I want to do that?

In order to experience that dark numbers exist and can't be
manipulated.

Dark numbers don't exist, as Jim and I have proven.

When |ℕ \ {1, 2, 3, ..., n}| = ℵo, then |ℕ \ {1, 2, 3, ..., n+1}| =
ℵo. How do the ℵo dark numbers get visible?

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.

They are not dark.

Induction cannot cover all natural numbers but only less than remain
uncovered.

The second part of that sentence is gibberish.  Nobody has been
talking about "uncovering" numbers, whatever that might mean.
Induction encompasses all natural numbers.  Anything it doesn't cover
is not a natural number, by definition.

Every defined number leaves ℵo undefined numbers. Try to find a
counterexample. Fail.

What the heck are you talking about?  What does it even mean for a
number to "leave" a set of numbers?

The set ℕ_def defined by induction does not include ℵo undefined
numbers.

Why should 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.