Re: The non-existence of "dark numbers"

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.
 

When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, ....

 The difference operator \
applies to sets, not to cardinal numbers.

I know, but erroneously I had used the sets. I corrected that but without correcting the sign

.... 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.

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.

 

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

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.

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.

 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.
Regards, WM