Re: The non-existence of "dark numbers"

On 3/16/2025 2:40 PM, WM wrote:

On 16.03.2025 18:19, Jim Burns wrote:

On 3/16/2025 12:17 PM, WM wrote:

On 16.03.2025 13:17, Alan Mackenzie wrote:

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

ℵo numbers have no FISONs.

>
You haven't said what you mean by F.

>
I did in the discussion with JB:
F is the set of FISONs.

>
And a FISON is
a Finite Initial Segment Of Naturals.
However,
if you (WM) have _non.circularly_ said
what 'finite' means, I have missed it.

>
Finite means a natural number.

Circular.

If you don't know
try to learn what they are.
1 and if n then n+1.

Non.circular!
But incomplete.
Add 'emptiest' to it, and you're there.
The _emptiest_ set with 1 and if n then n+1
does not hold dark numbers,
has all its end.segments same.sized,
is its.own.only.inductive.subset,
and
a subset can be proved to be the whole set
by proving it is an inductive subset.

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.

>
You (WM) use an unreliable quantifier swap
to arrive at that conclusion.

>
No
it is very reliable to say that
when all FISONs have infinite distance from ω,
then no FISON comes closer.

Your unreliable quantifier swap contradicts that.
Your unreliable quantifier swap is unreliable.
For _our_ ℕ
the _emptiest_ set with 0 and if n then n+1,
all its end.segments are same.sized.
----
For emptiest inductive ℕ
each element is followed.
∀ᴺj ∃ᴺk: j < k
Equally true  and
contradicting your unreliable quantifier swap,
no element follows all.
¬∃ᴺk ∀ᴺj: j ≤ k