Re: The non-existence of "dark numbers"

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

On 16.03.2025 13:17, Alan Mackenzie wrote:

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

On 15.03.2025 12:57, Alan Mackenzie wrote:

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

[ .... ]

All natural numbers "have" a FISON

Then all natural numbers would be in FISONs. But because of
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo
all FISONs fail to contain all natural numbers.

If you really think there is a non-empty set of natural numbers which
don't "have" FISONs,

Of course there is such a set. It contains almost all natural numbers.
This has been proven in the OP: All separated definable natural numbers
can be removed from the harmonic series. When only terms containing all
definable numbers together remain, then the series diverges. All its
terms are dark.

then please say what the least natural number in that set is, or at
the very least, how you'd go about finding it.

The definable numbers are  potentially infinite sequence. With n also
n+1 and n^n^n belong to it.

So, no answer.

But let's take the implications of your second sentence there, ".... with
n, also n+1 ....".  Since I think we'd agree that 0 is a definable
number, then we've just defined the "definable numbers" as the inductive
set.

Thus the set of "definable numbers", N_def is N.  The complement of N_def
in N (what you've called "dark numbers") is thus the empty set.

This is another proof that "dark numbers" don't exist, one you cannot
disagree with without contradicting what you've previously written.

I doubt I can be bothered any more to address the other falsehoods,
contradictions, and delusions in your last post.  But it's clear that
what you write about such topics is wrong in general, if not totally.

[ .... ]

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).