Jim Burns schrieb am Donnerstag, 13. Oktober 2022 um 05:19:27 UTC+2:
> On 10/12/2022 5:11 PM, WM wrote:
>
> >>> The dark numbers have no discernible well-order.
> >>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> >>> The least discernible infinite segment
> >>> therefore is the complete set ℕ.
> >>
> >> The least infinite initial segment
> >> is an infinite initial segment, and
> >> is less than (⊆) each infinite initial segment.
> >
> > There is only one in ℕ.
> It's a curious thing how you manage to
> make all these claims at the same time.

What is curious? All FISONs are finite. The only infinite set is creted by including the dark numbers. But they cannot be subdivided. Therefore there is only one infinite set.
> Like:
> <WM>
> > Nevertheless it is impossible to distinguish
> > different infinite initial segments.
> </WM>

> > There is only one in ℕ.
> Let's say that
> ℕ is the only infinite initial segment in ℕ
> and
> an infinite initial segment
> contains 0 and contains all its successors j⁺⁺

Most of which cannot be distinguished.

Regards, WM

FromTheRafters schrieb am Mittwoch, 12. Oktober 2022 um 23:40:42 UTC+2:
> WM has brought this to us :
> > Fritz Feldhase schrieb am Mittwoch, 12. Oktober 2022 um 11:13:52 UTC+2:
> >
> >> {{0, ..., n} : n e IN} = {{0}, {0, 1}, {0, 1, 2}, ...} = (due to von
> >> Neumann) = {1, 2, 3, ...} = IN \ {0}.
> >>
> >> Now card( IN \ {0} ) = aleph_0, since card( IN ) = aleph_0 BY DEFINITION.
> >
> > Contradicted by counting FISONs by elements in FISONs:
> >
> > 1
> > 2, 1
> > 3, 2, 1
> > ...
> >
> > If the first column contains ℵo elements.
> The natural numbers. The second column also lists the natural numbers,
> so does the third and so on.

Each column has one number less than the first column. But that does not matter in potential infinity.

> > then the complete figure contains ℵo elements.
> You got it, the natural numbers are still countable in two dimensional
> tables.

Here all columns contain the same. No two-dimensional counting necessary.

> > But no set of lines contains more than one line of the set.
> > And no line contains ℵo elements.

That is important!

Regards, WM

On 10/13/2022 5:32 AM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 13. Oktober 2022 um 05:19:27 UTC+2:
>> On 10/12/2022 5:11 PM, WM wrote:

>>> There is only one in ℕ.
>>
>> Let's say that
>> ℕ is the only infinite initial segment in ℕ
>> and
>> an infinite initial segment
>> contains 0 and contains all its successors j⁺⁺
>
> Most of which cannot be distinguished.

However,
you (WM) agree that
ℕ is the only infinite initial segment in ℕ

That's enough for us to know that
ℕ = ⋃𝓕 the union of FISONs

What does "cannot be distinguished" mean?
Never mind. We don't need it.
It doesn't matter.

⋃𝓕 subset ℕ
[1]

⋃𝓕 is an infinite initial segment.
[2]

There is only one subset of ℕ which is
an infinite initial segment,
and it is ℕ

Therefore,
⋃𝓕 = ℕ

----
[1]
⋃𝓕 subset ℕ

| Assume otherwise.
| Assume ⋃𝓕 is not subset ℕ
| Assume something in ⋃𝓕 is not in ℕ
| | Let m ∈ ⋃𝓕 and m ∉ ℕ
| m is in some FISON 𝐹 in 𝓕
| | {i: ~∃i₂=<i, i₂∉ℕ} {i: ∃i₂=<i, i₂∉ℕ}
| is a split of 𝐹
| | Because 𝐹 is a FISON,
| {i: ~∃i₂=<i, i₂∉ℕ} contains a last, j
| {i: ∃i₂=<i, i₂∉ℕ} contains a first, k
| and k = j⁺⁺
| | From the definition of the split,
| j ∈ ℕ and k ∉ ℕ
| | However,
| ℕ is an infinite initial segment
| ∀j ∈ ℕ, j⁺⁺ ∈ ℕ
| and k = j⁺⁺
| Contradiction.

Therefore
⋃𝓕 subset ℕ

----
[2]
⋃𝓕 is an infinite initial segment.

An infinite initial segment
contains 0 and contains all its successors j⁺⁺

| ⟨0⟩ is a FISON
| ⟨0⟩ ∈ 𝓕

0 ∈ ⋃𝓕

| Assume j ∈ ⋃𝓕
| | j is in some FISON 𝐹 in 𝓕
| Either
| j⁺⁺ ∈ 𝐹
| or
| j⁺⁺ ∉ 𝐹 and j⁺⁺ ∈ 𝐹⊕⟨j⁺⁺⟩
| | 𝐹⊕⟨j⁺⁺⟩ is 𝐹 with j⁺⁺ appended at its end.
| 𝐹 is a FISON. 𝐹⊕⟨j⁺⁺⟩ is a FISON too.
| j⁺⁺ is in a FISON, 𝐹 or 𝐹⊕⟨j⁺⁺⟩
| | j⁺⁺ ∈ ⋃𝓕

if j ∈ ⋃𝓕 then j⁺⁺ ∈ ⋃𝓕

⋃𝓕 contains all its successors. And 0.

Therefore,
⋃𝓕 is an infinite initial segment.

On 10/11/2022 11:57 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 11. Oktober 2022 um 18:29:04 UTC+2:
>> On Tuesday, October 11, 2022 at 5:16:23 PM UTC+2, WM wrote:
>>
>>> ω is the first infinite ordinal number.
>>> ℕ is the set of all finite ordinal numbers.
>>> There is an obvious difference.
>> Except that there is no difference in the context of ZFC
>
> ZFC is wrong.
>
>> where IN/omega is defined due to von Neumann,
>
> The name v. Neumann cannot beat logic.
>
> Counting FISONs by elements in FISONs we have to count ℵo elements in ℵo FISONs.
>
> 1
> 2, 1
> 3, 2, 1
> ...
>

> Regards, WM

WMs math usefulness is only in counting rocks, or sheeps.

the rest is made up bunk, lies, fiction, imagination, drivelings, and always => without math.

Jim Burns schrieb am Donnerstag, 13. Oktober 2022 um 17:47:04 UTC+2:
> On 10/13/2022 5:32 AM, WM wrote:
> > Jim Burns schrieb am Donnerstag,
> > 13. Oktober 2022 um 05:19:27 UTC+2:
> >> On 10/12/2022 5:11 PM, WM wrote:
>
> >>> There is only one in ℕ.
> >>
> >> Let's say that
> >> ℕ is the only infinite initial segment in ℕ
> >> and
> >> an infinite initial segment
> >> contains 0 and contains all its successors j⁺⁺
> >
> > Most of which cannot be distinguished.
> However,
> you (WM) agree that
> ℕ is the only infinite initial segment in ℕ

Yes. The reason is that the set of dark numbers cannot be divided.
Without dark numbers there were two alternatives:
Either there is a gap before omega or the natural numbers reach till omega.
If omega exists, then all smaller numbers can be subtracted. If all were definable, then a last one could be subtracted.

> That's enough for us to know that
> ℕ = ⋃𝓕 the union of FISONs
>

>
> What does "cannot be distinguished" mean?

Dark.
>
> ⋃𝓕 subset ℕ

Of course.

> [1]

> ⋃𝓕 is an infinite initial segment.

No. ℵo > n. Counting ℵo FISONs by elements in FISONs we have to count ℵo elements in ℵo FISONs.

1
2, 1
3, 2, 1
....

Assume that the first column contains ℵo elements.
Then the complete figure contains ℵo elements.
No line contains ℵo elements.
By inclusion-monotony: no set of lines contains ℵo elements.
The figure does not contain ℵo lines.

Regards, WM

On 10/13/2022 4:35 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 12. Oktober 2022 um 23:40:42 UTC+2:
>> WM has brought this to us :
>>> Fritz Feldhase schrieb am Mittwoch, 12. Oktober 2022 um 11:13:52 UTC+2:
>>>
>>>> {{0, ..., n} : n e IN} = {{0}, {0, 1}, {0, 1, 2}, ...} = (due to von
>>>> Neumann) = {1, 2, 3, ...} = IN \ {0}.
>>>>
>>>> Now card( IN \ {0} ) = aleph_0, since card( IN ) = aleph_0 BY DEFINITION.
>>>
>>> Contradicted by counting FISONs by elements in FISONs:
>>>
>>> 1
>>> 2, 1
>>> 3, 2, 1
>>> ...
>>>
>>> If the first column contains ℵo elements.
>> The natural numbers. The second column also lists the natural numbers,
>> so does the third and so on.
>
> Each column has one number less than the first column.

?#!? All you did is *put spaces in the top of the columns*, dummy.

>
>>> then the complete figure contains ℵo elements.
>> You got it, the natural numbers are still countable in two dimensional
>> tables.
>
> Here all columns contain the same. No two-dimensional counting necessary.

you are lost again, out of math, each column has extra spaces on top.......

>
>>> But no set of lines contains more than one line of the set.
>>> And no line contains ℵo elements.
>
> That is important!

nope, it's all bullshit.

>
> Regards, WM

On 10/9/2022 4:02 AM, WM wrote:
> Chris M. Thomasson schrieb am Freitag, 7. Oktober 2022 um 22:35:00 UTC+2:
>> On 10/7/2022 12:09 AM, WM wrote:
>
>>> below omega there is no gap. No leap can end in a gap. The result of this is the existence of dark nunbers.
>> With regard to artificially restricting ones self to the natural
>> numbers, there is no gap between say, [41...42].
>
> Yes, there is no gap. And if ω exists, then there is no gap between ω and ℕ. But every leap from ω into ℕ covers infinitely many natural numbers.
>
> Reagrds, WM
the gap(s) is(are) expanding in your gray matter.

bunny leap?, or leaping lizard ?

On 10/11/2022 10:16 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 9. Oktober 2022 um 15:58:35 UTC+2:
>> On Sunday, 9 October 2022 at 06:02:36 UTC-3, WM wrote:
>> [...]
>>> Yes, there is no gap. And if ω exists, then there is no gap between ω and ℕ. But every leap from ω into ℕ covers infinitely many natural numbers.
>> WTF is a "leap from ω into ℕ"?
>
> It means to enter ℕ from above.

like second floor to first floor ?

>
>> What even do *YOU* mean by a "gap between ω and ℕ"?
>
> Nothing existing.

so it is not real. Is this part of your delusion?

>
>> More basically, is ω =/= ℕ?
>
> ω is the first infinite ordinal number.
> ℕ is the set of all finite ordinal numbers.
> There is an obvious difference.

of course, they are spelt differently for a reason.

>
> Regards, WM

On 10/11/2022 10:45 AM, WM wrote:
> Fritz Feldhase schrieb am Montag, 10. Oktober 2022 um 14:48:55 UTC+2:
>> On Sunday, October 9, 2022 at 11:02:36 AM UTC+2, WM wrote:
>>
>>> there is no gap between ω and ℕ.

QUACK!!

>> I'll interprete this to mean: There is no ordinal o such that for all n in IN: n < o < ω.
>
> Yes. But there are many (ℵo) ordinals between all definable

QUACK!! natnumbers and ω which cannot be defined

QUACK!! because

> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>
>> Hint: For all n e IN: card({k e IN: n < k} = aleph_0.
>>
>> Meaning: For each and every n in IN there are countably-infintely many elements in IN which are larger than n.
>
> If ℕ is completed

QUACK!!
>, this is impossible. If you subtract each and every n then nothing remains.

QUACK!!

> |ℕ \ {1, 2, 3, ...}| = 0 .
> {0, 1, 2, 3, ..., ω} \ ℕ = {0, ω},

QUACK!!

>
> Regards, WM

each QUACK!! indicates mistake that needs correcting.

WM, please post in alt.delusional.fake.math

On 10/13/2022 2:05 PM, WM wrote:
> Jim Burns schrieb am Donnerstag, 13. Oktober 2022 um 17:47:04 UTC+2:
>> On 10/13/2022 5:32 AM, WM wrote:
>>> Jim Burns schrieb am Donnerstag,
>>> 13. Oktober 2022 um 05:19:27 UTC+2:
>>>> On 10/12/2022 5:11 PM, WM wrote:
>>
>>>>> There is only one in ℕ.
>>>>
>>>> Let's say that
>>>> ℕ is the only infinite initial segment in ℕ
>>>> and
>>>> an infinite initial segment
>>>> contains 0 and contains all its successors j⁺⁺
>>>
>>> Most of which cannot be

[QUACK!!]distinguished.
>> However,
>> you (WM) agree that
>> ℕ is the only infinite initial segment in ℕ
>
> Yes. The reason is that the set of dark

[QUACK!!] numbers cannot be divided.
> Without dark

[QUACK!!] numbers there were two alternatives:
> Either there is a gap

[QUACK!!] before omega or the natural numbers reach

[QUACK!!] till omega.
> If omega exists, then all smaller numbers can be subtracted. If all were

[QUACK!!] definable, then a last

[QUACK!!] one could be subtracted.
>
>> That's enough for us to know that
>> ℕ = ⋃𝓕 the union of FISONs
>>
>
>>
>> What does "cannot be distinguished" mean?
>
> Dark. [QUACK!!]
>>
>> ⋃𝓕 subset ℕ
>
> Of course.
>
>> [1]
>
>> ⋃𝓕 is an infinite initial segment.
>
>
> No. ℵo > n. Counting ℵo FISONs by elements in FISONs we have to count ℵo elements in ℵo FISONs.

No. No FISON is ℵo.

>
> 1
> 2, 1
> 3, 2, 1
> ...
>
> Assume that the first column contains ℵo elements.
> Then the complete figure contains ℵo elements.
> No line contains ℵo elements.
> By inclusion [QUACK!!]-monotony: no set of lines contains ℵo elements.
> The figure does not contain ℵo lines.

because you stop at ...

>
> Regards, WM

On 10/13/2022 4:28 PM, Sergi o wrote:
> On 10/11/2022 10:45 AM, WM wrote:

>> |ℕ \ {1, 2, 3, ...}| = 0 .
>> {0, 1, 2, 3, ..., ω} \ ℕ = {0, ω},
>
> QUACK!!
>
>>
>> Regards, WM
>
> each QUACK!! indicates mistake that needs correcting.

You offer multiple QUorrACKtions.

On 10/13/2022 3:05 PM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 13. Oktober 2022 um 17:47:04 UTC+2:
>> On 10/13/2022 5:32 AM, WM wrote:
>>> Jim Burns schrieb am Donnerstag,
>>> 13. Oktober 2022 um 05:19:27 UTC+2:
>>>> On 10/12/2022 5:11 PM, WM wrote:

>>>>> There is only one in ℕ.
>>>>
>>>> Let's say that
>>>> ℕ is the only infinite initial segment in ℕ
>>>> and
>>>> an infinite initial segment
>>>> contains 0 and contains all its successors j⁺⁺
>>>
>>> Most of which cannot be distinguished.
>>
>> However,
>> you (WM) agree that
>> ℕ is the only infinite initial segment in ℕ
>
> Yes.

And I quote: "Yes".

You claim you have reasons.
I am not in the mood for looking a gift-horse
in the mouth.

We agree that
ℕ is an infinite initial segment
(ℕ contains all its successors j⁺⁺ and 0)
and
each non-ℕ subset B of N is
NOT an infinite initial segment
(B does NOT contain all its successors j⁺⁺ and 0)

> The reason is that
> the set of dark numbers cannot be divided.
> Without dark numbers there were two alternatives:
> Either there is a gap before omega
> or the natural numbers reach till omega.
> If omega exists,
> then all smaller numbers can be subtracted.
> If all were definable,
> then a last one could be subtracted.
>
>> That's enough for us to know that
>> ℕ = ⋃𝓕 the union of FISONs

>> ⋃𝓕 subset ℕ
>
> Of course.

>> ⋃𝓕 is an infinite initial segment.
>
> No.

Yes.
An infinite initial segment
contains all its successors j⁺⁺ and 0

⋃𝓕 contains all its successors j⁺⁺ and 0

⋃𝓕 is an infinite initial segment.

----
ℕ does not contain dark numbers
because
ℕ = ⋃𝓕
because
⋃𝓕 is an infinite initial segment in ℕ
because
⋃𝓕 contains all its successors j⁺⁺
because
𝐹ⱼ⊕⟨j⁺⁺⟩ is a FISON too

<WM<JB>>
>
>> That's because 𝐹ₘ₋₁⊕⟨m⟩ is a FISON too.
>
> Boring.
>
</WM<JB>>
Date: Fri, 7 Oct 2022 14:34:33 -0400

You were correct. It is boring.

Thrills and chills are not an essential feature
of mathematical facts.

> ℵo > n.
> Counting ℵo FISONs by elements in FISONs
> we have to count ℵo elements in ℵo FISONs.
>
> 1
> 2, 1
> 3, 2, 1
> ...
>
> Assume that the first column contains ℵo elements.
> Then the complete figure contains ℵo elements.
> No line contains ℵo elements.
> By inclusion-monotony: no set of lines contains ℵo elements.
> The figure does not contain ℵo lines.

torsdag 13 oktober 2022 kl. 21:05:51 UTC+2 skrev WM:
> Jim Burns schrieb am Donnerstag, 13. Oktober 2022 um 17:47:04 UTC+2:
> > On 10/13/2022 5:32 AM, WM wrote:
> > > Jim Burns schrieb am Donnerstag,
> > > 13. Oktober 2022 um 05:19:27 UTC+2:
> > >> On 10/12/2022 5:11 PM, WM wrote:
> >
> > >>> There is only one in ℕ.
> > >>
> > >> Let's say that
> > >> ℕ is the only infinite initial segment in ℕ
> > >> and
> > >> an infinite initial segment
> > >> contains 0 and contains all its successors j⁺⁺
> > >
> > > Most of which cannot be distinguished.
> > However,
> > you (WM) agree that
> > ℕ is the only infinite initial segment in ℕ
> Yes. The reason is that the set of dark numbers cannot be divided.
> Without dark numbers there were two alternatives:
> Either there is a gap before omega or the natural numbers reach till omega.
> If omega exists, then all smaller numbers can be subtracted. If all were definable, then a last one could be subtracted.
> > That's enough for us to know that
> > ℕ = ⋃𝓕 the union of FISONs
> >
>
> >
> > What does "cannot be distinguished" mean?
> Dark.
> >
> > ⋃𝓕 subset ℕ
>
> Of course.
> > [1]
>
> > ⋃𝓕 is an infinite initial segment.
> No. ℵo > n. Counting ℵo FISONs by elements in FISONs we have to count ℵo elements in ℵo FISONs.
> 1
> 2, 1
> 3, 2, 1
> ...
> Assume that the first column contains ℵo elements.
> Then the complete figure contains ℵo elements.
> No line contains ℵo elements.
> By inclusion-monotony: no set of lines contains ℵo elements.
> The figure does not contain ℵo lines.
>
> Regards, WM
"dark" is meaningless!

On Thursday, October 13, 2022 at 9:05:51 PM UTC+2, WM wrote:

> The figure does not contain ℵo lines.

Look, dumbo, if you consider the infinite sequence (F_n)_(n e IN) with F_n = {1, ..., n}, then "it contains" ℵo (different) terms BY DEFINITION.

You see: card((F_n)_(N e IN)) = card(IN) = ℵo, since An,m e IN: n =/= m -> F_n =/= F_m.

AND NOW FUCK OFF, YOU SILLY ASSHOLE FULL OF SHIT!

Fritz Feldhase schrieb am Freitag, 14. Oktober 2022 um 09:13:07 UTC+2:
> On Thursday, October 13, 2022 at 9:05:51 PM UTC+2, WM wrote:
>
> > The figure does not contain ℵo lines.
> if you consider the infinite sequence (F_n)_(n e IN) with F_n = {1, ..., n}, then "it contains" ℵo (different) terms BY DEFINITION.

1
2, 1
3, 2, 1
....

By mathematics / symmetry there cannot be more lines than columns.
By definition there are not ℵo columns.
Any questions?

Regards, WM

Jim Burns schrieb am Freitag, 14. Oktober 2022 um 05:26:20 UTC+2:
> On 10/13/2022 3:05 PM, WM wrote:
> > Jim Burns schrieb am Donnerstag,
> > 13. Oktober 2022 um 17:47:04 UTC+2:
> >> On 10/13/2022 5:32 AM, WM wrote:
> >>> Jim Burns schrieb am Donnerstag,
> >>> 13. Oktober 2022 um 05:19:27 UTC+2:
> >>>> On 10/12/2022 5:11 PM, WM wrote:
>
> >>>>> There is only one in ℕ.
> >>>>
> >>>> Let's say that
> >>>> ℕ is the only infinite initial segment in ℕ
> >>>> and
> >>>> an infinite initial segment
> >>>> contains 0 and contains all its successors j⁺⁺
> >>>
> >>> Most of which cannot be distinguished.
> >>
> >> However,
> >> you (WM) agree that
> >> ℕ is the only infinite initial segment in ℕ
> >
> > Yes.
> And I quote: "Yes".

> We agree that
> ℕ is an infinite initial segment

> > the set of dark numbers cannot be divided.
> > Without dark numbers there were two alternatives:
> > Either there is a gap before omega
> > or the natural numbers reach till omega.
> > If omega exists,
> > then all smaller numbers can be subtracted.
> > If all were definable,
> > then a last one could be subtracted.

> >> ⋃𝓕 subset ℕ
> >
> > Of course.
> >> ⋃𝓕 is an infinite initial segment.
> >
> > No.
> Yes.
> An infinite initial segment
> contains all its successors j⁺⁺ and 0

You have to distinguish potential infinity and actual infinity.
> ℕ does not contain dark numbers
> because
> ℕ = ⋃𝓕

Wrong.

> > 1
> > 2, 1
> > 3, 2, 1
> > ...
> >
By mathematics / symmetry there cannot be more lines than columns.
By definition there are not ℵo columns.

> > Assume that the first column contains ℵo elements.
> > Then the complete figure contains ℵo elements.
> > No line contains ℵo elements.
> > By inclusion-monotony: no set of lines contains ℵo elements.
> > The figure does not contain ℵo lines.

Regards, WM

On 10/14/2022 6:46 AM, WM wrote:
> Fritz Feldhase schrieb am Freitag, 14. Oktober 2022 um 09:13:07 UTC+2:
>> On Thursday, October 13, 2022 at 9:05:51 PM UTC+2, WM wrote:
>>
>>> The figure does not contain ℵo lines.
>> if you consider the infinite sequence (F_n)_(n e IN) with F_n = {1, ..., n}, then "it contains" ℵo (different) terms BY DEFINITION.
>
> 1
> 2, 1
> 3, 2, 1
> ...
>
> By mathematics / symmetry there cannot be more lines than columns.
> By definition there are not ℵo columns.
> Any questions?
>
> Regards, WM

sure, WTF are you talking about ?

you are only adding spaces at the beginning of each column, so what is that suppose to show ?

your three dots mean the columns go to infinity, and the rows go to infinity. You are known to have problems with common math nomenclature, so try again.

On 10/14/2022 4:53 AM, WM wrote:
> Jim Burns schrieb am Freitag, 14. Oktober 2022 um 05:26:20 UTC+2:
>> On 10/13/2022 3:05 PM, WM wrote:
>>> Jim Burns schrieb am Donnerstag,
>>> 13. Oktober 2022 um 17:47:04 UTC+2:
>>>> On 10/13/2022 5:32 AM, WM wrote:
>>>>> Jim Burns schrieb am Donnerstag,
>>>>> 13. Oktober 2022 um 05:19:27 UTC+2:
>>>>>> On 10/12/2022 5:11 PM, WM wrote:
>>
>>>>>>> There is only one in ℕ.
>>>>>>
>>>>>> Let's say that
>>>>>> ℕ is the only infinite initial segment in ℕ
>>>>>> and
>>>>>> an infinite initial segment
>>>>>> contains 0 and contains all its successors j⁺⁺
>>>>>
>>>>> Most of which cannot be distinguished.
>>>>
>>>> However,
>>>> you (WM) agree that
>>>> ℕ is the only infinite initial segment in ℕ
>>>
>>> Yes.
>> And I quote: "Yes".
>
>> We agree that
>> ℕ is an infinite initial segment
>
>>> the set of dark numbers cannot be divided.
>>> Without dark numbers there were two alternatives:
>>> Either there is a gap before omega
>>> or the natural numbers reach till omega.
>>> If omega exists,
>>> then all smaller numbers can be subtracted.
>>> If all were definable,
>>> then a last one could be subtracted.
>
>>>> ⋃𝓕 subset ℕ
>>>
>>> Of course.
>>>> ⋃𝓕 is an infinite initial segment.
>>>
>>> No.
>> Yes.
>> An infinite initial segment
>> contains all its successors j⁺⁺ and 0
>
> You have to distinguish potential infinity and actual infinity.
[...]

Are you infinitely stupid, or potentially stupid?

Well, stupid is as stupid does, right? ;^)

lol.

On 10/14/2022 4:28 PM, Chris M. Thomasson wrote:
> On 10/14/2022 4:53 AM, WM wrote:
>> Jim Burns schrieb am Freitag, 14. Oktober 2022 um 05:26:20 UTC+2:
>>> On 10/13/2022 3:05 PM, WM wrote:
>>>> Jim Burns schrieb am Donnerstag,
>>>> 13. Oktober 2022 um 17:47:04 UTC+2:
>>>>> On 10/13/2022 5:32 AM, WM wrote:
>>>>>> Jim Burns schrieb am Donnerstag,
>>>>>> 13. Oktober 2022 um 05:19:27 UTC+2:
>>>>>>> On 10/12/2022 5:11 PM, WM wrote:
>>>
>>>>>>>> There is only one in ℕ.
>>>>>>>
>>>>>>> Let's say that
>>>>>>> ℕ is the only infinite initial segment in ℕ
>>>>>>> and
>>>>>>> an infinite initial segment
>>>>>>> contains 0 and contains all its successors j⁺⁺
>>>>>>
>>>>>> Most of which cannot be distinguished.
>>>>>
>>>>> However,
>>>>> you (WM) agree that
>>>>> ℕ is the only infinite initial segment in ℕ
>>>>
>>>> Yes.
>>> And I quote: "Yes".
>>
>>> We agree that
>>> ℕ is an infinite initial segment
>>
>>>> the set of dark numbers cannot be divided.
>>>> Without dark numbers there were two alternatives:
>>>> Either there is a gap before omega
>>>> or the natural numbers reach till omega.
>>>> If omega exists,
>>>> then all smaller numbers can be subtracted.
>>>> If all were definable,
>>>> then a last one could be subtracted.
>>
>>>>> ⋃𝓕 subset ℕ
>>>>
>>>> Of course.
>>>>> ⋃𝓕 is an infinite initial segment.
>>>>
>>>> No.
>>> Yes.
>>> An infinite initial segment
>>> contains all its successors j⁺⁺ and 0
>>
>> You have to distinguish potential infinity and actual infinity.
> [...]
>
> Are you infinitely stupid, or potentially stupid?
>
> Well, stupid is as stupid does, right? ;^)
>
> lol.

WM streams his consciousness into the (actual) ANT LIST

(this short section is from 9/18/2020 ANT LIST 4.0- so, is WM a bot or not?)

154 degenerate Ants
155 Delusional Ants
156 Demasking Ants
157 Deodor Ant *
158 despicable bastard Ants
159 destroy every converging sequence Ants
160 Differing Ants
161 Dimentionless Ants
162 discontinuity Ants
163 dishonest cheating Ants
164 distinguish between sense and nonsense Ants
165 Distracted Ants
166 Dodgy Shtick Ants
167 Dumb and Dumber Ants
168 Dumbo Ants
169 Each and every Ant
170 Eggsegmented Ants
171 Eggsegments Ants
172 Element Ants
173 Emirp Ants *
174 Empty set Ants
175 Empty Ants
176 Empty limit Ants
177 Endless Ants
178 Endpointless Ants
179 Endsegment Infesting Ants

On 10/14/2022 7:53 AM, WM wrote:
> Jim Burns schrieb am Freitag,
> 14. Oktober 2022 um 05:26:20 UTC+2:
>> On 10/13/2022 3:05 PM, WM wrote:
>>> Jim Burns schrieb am Donnerstag,
>>> 13. Oktober 2022 um 17:47:04 UTC+2:
>>>> On 10/13/2022 5:32 AM, WM wrote:
>>>>> Jim Burns schrieb am Donnerstag,
>>>>> 13. Oktober 2022 um 05:19:27 UTC+2:
>>>>>> On 10/12/2022 5:11 PM, WM wrote:

>>>>>>> There is only one in ℕ.
>>>>>>
>>>>>> Let's say that
>>>>>> ℕ is the only infinite initial segment in ℕ
>>>>>> and
>>>>>> an infinite initial segment
>>>>>> contains 0 and contains all its successors j⁺⁺
>>>>>
>>>>> Most of which cannot be distinguished.
>>>>
>>>> However,
>>>> you (WM) agree that
>>>> ℕ is the only infinite initial segment in ℕ
>>>
>>> Yes.
>>
>> And I quote: "Yes".
>
>> We agree that
>> ℕ is an infinite initial segment

Should I read that as you (WM) NOT agreeing to
the part that you deleted?

If so, it's mysterious what you meant by
<WM>
>>>>>>> There is only one in ℕ.

----
Still,
we at least agree that
ℕ is an infinite initial segment...
until further notice.

From that, it arises that
ω = 𝛚ℕ = ⋃𝓕

I define a _set operator_ '𝛚𝑆'
as the intersection of
all infinite initial segments 𝐵 in 𝑆
𝛚𝑆 = ⋂{𝐵 ⊆ 𝑆 : 𝐵⋮₀}

That will lead us to the unique _set_ ω
but, without more work, we might consider whether
𝛚𝑆 could be different for each 𝑆

I have introduced the notation '𝑆⋮₀' for
"𝑆 is an infinite initial segment"
⟺
"𝑆 contains all its successors j⁺⁺ and 0"

ℕ is an infinite initial segment.
ℕ contains all its successors j⁺⁺ and 0
ℕ⋮₀

Lemma 1.
The intersection of infinite initial segments
is an infinite initial segment.
(⋂{ 𝐵⋮₀ })⋮₀

| if
| j ∈ ⋂{ 𝐵⋮₀ }
| then
| j is in each 𝐵 and 𝐵⋮₀
| and
| j⁺⁺ is in each 𝐵
| and
| j⁺⁺ ∈ ⋂{ 𝐵⋮₀ }

If 𝑆 is an infinite initial segment,
then 𝛚𝑆 is the least (⊆) infinite initial
sub-segment in 𝑆
𝑆⋮₀ ⟹
(𝛚𝑆)⋮₀ ∧
∀𝐵⋮₀ ⊆ 𝑆⋮₀ : 𝛚𝑆 ⊆ 𝐵⋮₀ ⊆ 𝑆⋮₀

If 𝐴 and 𝐵 are infinite initial segments,
then their _least_ infinite initial sub-segments
are the same.
𝐴⋮₀ ∧ 𝐵⋮₀ ⟹ 𝛚𝐴 = 𝛚𝐵

| Assume 𝐴 and 𝐵 are infinite initial segments.
| 𝐴⋮₀ ∧ 𝐵⋮₀
| | The intersection 𝐴∩𝐵 of 𝐴 and 𝐵 is
| an infinite initial segment (lemma 1) and
| is subset both 𝐴 and 𝐵
| (𝐴∩𝐵)⋮₀
| 𝐴∩𝐵 ⊆ 𝐴 ∧ 𝐴∩𝐵 ⊆ 𝐵
| | The least infinite initial segment 𝛚𝐵 of 𝐵
| is subset 𝐴∩𝐵 subset 𝐵
| 𝛚𝐵 ⊆ (𝐴∩𝐵)⋮₀ ⊆ 𝐵⋮₀
| | Also,
| 𝛚𝐵 subset 𝐴∩𝐵 subset 𝐴
| thus,
| the least infinite initial segment 𝛚𝐴 of 𝐴
| is subset 𝛚𝐵 subset 𝐴∩𝐵 subset 𝐴
| 𝛚𝐴 ⊆ 𝛚𝐵⋮₀ ⊆ (𝐴∩𝐵)⋮₀ ⊆ 𝐴⋮₀
| | Thus, 𝛚𝐴 ⊆ 𝛚𝐵
| Similarly, 𝛚𝐵 ⊆ 𝛚𝐴
| Thus, 𝛚𝐴 = 𝛚𝐵

Therefore,
if 𝐴 and 𝐵 are infinite initial segments,
then their _least_ infinite initial sub-segments
are the same.
𝐴⋮₀ ∧ 𝐵⋮₀ ⟹ 𝛚𝐴 = 𝛚𝐵

<WM>
> Nevertheless it is impossible to distinguish
> different infinite initial segments.
>
Date: Sun, 9 Oct 2022 02:39:18 -0700 (PDT)

ω is distinguished as the unique
least infinite initial segment.

For any infinite initial segment 𝑆⋮₀
ω is the least infinite initial sub-segment in 𝑆
∀𝑆⋮₀ : ω⋮₀ ⊆ 𝑆

<WM>
> Why?
> Because the additional numbers are dark.
>
Date: Sun, 9 Oct 2022 02:39:18 -0700 (PDT)

ω = ⋃𝓕

∀j ∈ ω, ∃𝐹 ∈ 𝓕 : j ∈ 𝐹 ∈ 𝓕

ω does not contain dark numbers.

>>>> ⋃𝓕 is an infinite initial segment.
>>>
>>> No.
>>
>> Yes.
>> An infinite initial segment
>> contains all its successors j⁺⁺ and 0
>
> You have to distinguish potential infinity
> and actual infinity.

If j is in FISON 𝐹
then j⁺⁺ is in 𝐹 or in FISON 𝐹ⱼ⊕⟨j⁺⁺⟩

Jim Burns schrieb am Samstag, 15. Oktober 2022 um 04:48:47 UTC+2:
> On 10/14/2022 7:53 AM, WM wrote:
> > Jim Burns schrieb am Freitag,
> > 14. Oktober 2022 um 05:26:20 UTC+2:
> >> On 10/13/2022 3:05 PM, WM wrote:
> >>> Jim Burns schrieb am Donnerstag,
> >>> 13. Oktober 2022 um 17:47:04 UTC+2:
> >>>> On 10/13/2022 5:32 AM, WM wrote:
> >>>>> Jim Burns schrieb am Donnerstag,
> >>>>> 13. Oktober 2022 um 05:19:27 UTC+2:
> >>>>>> On 10/12/2022 5:11 PM, WM wrote:
>
> >>>>>>> There is only one in ℕ.
> >>>>>>
> >>>>>> Let's say that
> >>>>>> ℕ is the only infinite initial segment in ℕ
> >>>>>> and
> >>>>>> an infinite initial segment
> >>>>>> contains 0 and contains all its successors j⁺⁺
> >>>>>
> >>>>> Most of which cannot be distinguished.
> >>>>
> >>>> However,
> >>>> you (WM) agree that
> >>>> ℕ is the only infinite initial segment in ℕ
> >>>
> >>> Yes.
> >>
> >> And I quote: "Yes".
> >
> >> We agree that
> >> ℕ is an infinite initial segment
> Should I read that as you (WM) NOT agreeing to
> the part that you deleted?

We agree that ℕ is an infinite initial segment
>
> If so, it's mysterious what you meant by
> <WM>
> >>>>>>> There is only one in ℕ.

There is only one infinite initial segment that can be discerned in ℕ because the infinite set of dark nunbers cannot be subdivided.
> ----
> Still,
> we at least agree that
> ℕ is an infinite initial segment...
> until further notice.

Yes. There are man more but not initial infinite segemnts:
{2, 3, 4, ...}
{3, 4, 5, ...}
and so on.
>
> From that, it arises that
> ω = 𝛚ℕ = ⋃𝓕
>
No.

> I define a _set operator_ '𝛚𝑆'
> as the intersection of
> all infinite initial segments 𝐵 in 𝑆
> 𝛚𝑆 = ⋂{𝐵 ⊆ 𝑆 : 𝐵⋮₀}

The intersection all infinite segments is the set of dark numbers.
>
> I have introduced the notation '𝑆⋮₀' for
> "𝑆 is an infinite initial segment"
> ⟺
> "𝑆 contains all its successors j⁺⁺ and 0"
>
> ℕ is an infinite initial segment.
> ℕ contains all its successors j⁺⁺ and 0
> ℕ⋮₀
>
> Lemma 1.
> The intersection of infinite initial segments
> is an infinite initial segment.

There is only one discernible infinite initial segment.

> ω = ⋃𝓕

No.

⋃𝓕 has not cardinality ℵo.

The figure
1 2, 1
3, 2, 1
....
does not contain ℵo lines.

Proof: By symmetry there cannot be more lines than columns.
By definition there are not ℵo columns.

Regards, WM

Gus Gassmann has brought this to us :
> On Monday, 10 October 2022 at 11:22:48 UTC-3, FromTheRafters wrote:
> [...]
>> I don't think a well ordered set has an infinite initial segment.
>
> I think that is wrong. Consider the set of natural numbers ordered by << in
> the following manner
>
> if a is odd and b is even, then a << b
> if a and b have the same parity, then A << b iff a < b in the conventional
> order if a is even and b is odd, then b << a.
>
> This well-order(!) can also be written as 1 << 3 << 5 << ... << 2 << 4 << 6
> << ... .
>
> I would think that for every /even/ natural number the initial segment is
> infinite.

That works as a sequence I guess, but I can't help seeing this as {{set
of odds},{set of evens}} order type two in sets. You're probably right
though, I retract my statement.

On 10/15/2022 5:46 AM, WM wrote:
> Jim Burns schrieb am Samstag, 15. Oktober 2022 um 04:48:47 UTC+2:
>> On 10/14/2022 7:53 AM, WM wrote:
>>> Jim Burns schrieb am Freitag,
>>> 14. Oktober 2022 um 05:26:20 UTC+2:
>>>> On 10/13/2022 3:05 PM, WM wrote:
>>>>> Jim Burns schrieb am Donnerstag,
>>>>> 13. Oktober 2022 um 17:47:04 UTC+2:
>>>>>> On 10/13/2022 5:32 AM, WM wrote:
>>>>>>> Jim Burns schrieb am Donnerstag,
>>>>>>> 13. Oktober 2022 um 05:19:27 UTC+2:
>>>>>>>> On 10/12/2022 5:11 PM, WM wrote:
>>
>>>>>>>>> There is only one in ℕ.
>>>>>>>>
>>>>>>>> Let's say that
>>>>>>>> ℕ is the only infinite initial segment in ℕ
>>>>>>>> and
>>>>>>>> an infinite initial segment
>>>>>>>> contains 0 and contains all its successors j⁺⁺
>>>>>>>
>>>>>>> Most of which cannot be distinguished.
>>>>>>
>>>>>> However,
>>>>>> you (WM) agree that
>>>>>> ℕ is the only infinite initial segment in ℕ
>>>>>
>>>>> Yes.
>>>>
>>>> And I quote: "Yes".
>>>
>>>> We agree that
>>>> ℕ is an infinite initial segment
>> Should I read that as you (WM) NOT agreeing to
>> the part that you deleted?
>
> We agree that ℕ is an infinite initial segment

no, ℕ is an infinite endsegment.

>>
>> If so, it's mysterious what you meant by
>> <WM>
>>>>>>>>> There is only one in ℕ.
>
> There is only one infinite initial segment that can be discerned in ℕ because the infinite set of dark nunbers cannot be subdivided.
>> ----
>> Still,
>> we at least agree that
>> ℕ is an infinite initial segment...
>> until further notice.
>
> Yes. There are man more but not initial infinite segemnts:
> {2, 3, 4, ...}
> {3, 4, 5, ...}
> and so on.
>>
>> From that, it arises that
>> ω = 𝛚ℕ = ⋃𝓕
>>
> No.

yes.

>
>> I define a _set operator_ '𝛚𝑆'
>> as the intersection of
>> all infinite initial segments 𝐵 in 𝑆
>> 𝛚𝑆 = ⋂{𝐵 ⊆ 𝑆 : 𝐵⋮₀}
>
> The intersection all infinite segments is the set of dark numbers.
>>
>> I have introduced the notation '𝑆⋮₀' for
>> "𝑆 is an infinite initial segment"
>> ⟺
>> "𝑆 contains all its successors j⁺⁺ and 0"
>>
>> ℕ is an infinite initial segment.
>> ℕ contains all its successors j⁺⁺ and 0
>> ℕ⋮₀
>>
>> Lemma 1.
>> The intersection of infinite initial segments
>> is an infinite initial segment.
>
> There is only one discernible infinite initial segment.
>
>> ω = ⋃𝓕
>
> No.
>
> ⋃𝓕 has not cardinality ℵo.
>
> The figure
> 1
> 2, 1
> 3, 2, 1
> ...
> does not contain ℵo lines.

wrong, if you want that, do not use the three dots, ...

>
> Proof: By symmetry there cannot be more lines than columns.
> By definition there are not ℵo columns.

wrong, you use the three dots ... so there are infinite rows and columns.

if that is not what you mean, then go study Mathematical Nomenclature for Dummies.

>
> Regards, WM

On 10/12/2022 4:05 PM, Chris M. Thomasson wrote:
> On 10/12/2022 1:56 PM, WM wrote:
>> Chris M. Thomasson schrieb am Dienstag, 11. Oktober 2022 um 21:48:30 UTC+2:
>>> On 10/11/2022 8:32 AM, WM wrote:
>>
>>>> Everything easily accessible in a system is not dark in that system.
>>> Okay. I am thinking of a number. It is not dark to me, but its dark to
>>> you, right?
>>
>> That depends, but may be possible.
>
> The number I am thinking of is able to be constructed.

I used to use BAUDOT 5 (ITA2) and paper tape, got good results, AND it was recorded so all could see, they called me crazy, but I knew better.

>
>
>>> Is that what you mean?
>>
>> That is not so important since the border between visible and dark is blurred. Important is that for everybody the dark part is and remains infinite,
>> much larger than the always finite visible part:
>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>
> The border between visible and dark? Okay, for some reason this makes me think of the numbers that escape vs the ones that do not in the Mandelbrot set.
>

yes, those that escape are dark areas, no they are points actually.

On 10/15/2022 6:46 AM, WM wrote:

> There are man more but not initial
> infinite segemnts:
>
> {2, 3, 4, ...}
> {3, 4, 5, ...}
> and so on.

>> I define a _set operator_ '𝛚𝑆'
>> as the intersection of
>> all infinite initial segments 𝐵 in 𝑆
>> 𝛚𝑆 = ⋂{𝐵 ⊆ 𝑆 : 𝐵⋮₀}
>
> The intersection all infinite segments
> is the set of dark numbers.

𝐵⋮₀ is an infinite _initial_ segment.
𝐵⋮₀ contains all its successors j⁺⁺ and 0

>> I have introduced the notation '𝑆⋮₀' for
>> "𝑆 is an infinite initial segment"
>> ⟺
>> "𝑆 contains all its successors j⁺⁺ and 0"
>>
>> ℕ is an infinite initial segment.
>> ℕ contains all its successors j⁺⁺ and 0
>> ℕ⋮₀
>>
>> Lemma 1.
>> The intersection of infinite initial segments
>> is an infinite initial segment.
>
> There is only one discernible
> infinite initial segment.

Being discernible is not an essential feature
of an infinite _initial_ segment.
Containing all its successors j⁺⁺ and 0 are
essential features of an infinite _initial_ segment.

>> ω = ⋃𝓕
>
> No.

Yes.
⋃𝓕 ⊆ ω and ω ⊆ ⋃𝓕

----
(1)
⋃𝓕 ⊆ ω
∀j : j ∈ ⋃𝓕 ⟹ j ∈ ω

| Assume otherwise.
| Assume j ∈ ⋃𝓕 and j ∉ ω
| | j ∉ ω and j ∈ 𝐹 ∈ 𝓕
| 𝐹 is a FISON
| There is a first j₁ in 𝐹 such that
| j₁ ∉ ω and j₁⁻⁻ ∈ ω
| | However,
| ∀j ∈ ω : j++ ∈ ω
| j₁⁻⁻ ∈ ω and j₁⁻⁻⁺⁺ = j₁ ∈ ω
| Contradiction.

Therefore,
∀j : j ∈ ⋃𝓕 ⟹ j ∈ ω
⋃𝓕 ⊆ ω

----
(2)
ω ⊆ ⋃𝓕

ω = 𝛚𝐵
for any infinite initial segment 𝐵⋮₀

𝐴⋮₀ ⟹ ω ⊆ 𝐴
for any infinite initial segment 𝐴⋮₀

(⋃𝓕)⋮₀
∀j ∈ ⋃𝓕 : j⁺⁺ ∈ ⋃𝓕 ∧ 0 ∈ ⋃𝓕
[1]

(⋃𝓕)⋮₀ ⟹ ω ⊆ ⋃𝓕

ω ⊆ ⋃𝓕

[1]
| Assume j ∈ ⋃𝓕
| | j ∈ 𝐹 ∈ 𝓕
| and 𝐹 is a FISON
| Either j⁺⁺ ∈ 𝐹 or j⁺⁺ ∉ 𝐹
| | If j⁺⁺ ∈ 𝐹 ∈ 𝓕
| then j⁺⁺ ∈ ⋃𝓕
| | If j ∈ 𝐹 and j⁺⁺ ∉ 𝐹
| then
| 𝐹⊕⟨j⁺⁺⟩ is a FISON too
| and
| j⁺⁺ ∈ 𝐹⊕⟨j⁺⁺⟩ ∈ 𝓕
| and
| j⁺⁺ ∈ ⋃𝓕

Therefore,
∀j ∈ ⋃𝓕 : j⁺⁺ ∈ ⋃𝓕

----
>> ω = ⋃𝓕
>
> No.

Yes.
⋃𝓕 ⊆ ω and ω ⊆ ⋃𝓕
ω = ⋃𝓕

> ⋃𝓕 has not cardinality ℵo.
>
> The figure
> 1
> 2, 1
> 3, 2, 1
> ...
> does not contain ℵo lines.
>
> Proof:
> By symmetry there cannot be
> more lines than columns.
> By definition there are not ℵo columns.

𝐹ⱼ⊕⟨j⁺⁺⟩ is a FISON too.