Chris M. Thomasson schrieb am Montag, 10. Oktober 2022 um 08:59:46 UTC+2:
>
> is the point 0+1/2 dark in the interval [0...1]?
>
> Well, not dark since I named it.

Correct.

> Think of a number, is it dark now?

Everything easily accessible in a system is not dark in that system.

Regards, WM

Jim Burns schrieb am Montag, 10. Oktober 2022 um 13:46:05 UTC+2:

> The natural numbers are
> the least infinite initial segment.
> ℕ = 𝕃𝕀𝕀𝕊

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.

Regards, WM

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.
> 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 natnumbers and ω which cannot be defined 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, this is impossible. If you subtract each and every n then nothing remains.
|ℕ \ {1, 2, 3, ...}| = 0 .
{0, 1, 2, 3, ..., ω} \ ℕ = {0, ω},

Regards, WM

Fritz Feldhase schrieb am Montag, 10. Oktober 2022 um 15:22:48 UTC+2:
> On Sunday, October 9, 2022 at 11:39:23 AM UTC+2, WM wrote:

> And right, there are INFINITELY many FISONs. :-)

The sequence is potentially infinite.
>
> Hint: card({{0, ..., n} : n e IN} = aleph_0.

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

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

The first column contains ℵo elements.
==> The complete figure contains ℵo elements.
No set of lines contains more than one line of the set.
No line contains ℵo elements.

> > Cantor claims that from 0 all natural numbers can be reached such that none remains unreached.
> Oh, really? Where does *he* claim that. Could you give a quote?

Jede lineare isolirte Pm. ist abzählbar.

Every counted number is reached.

> > That is precondition for every bijection with ℕ.
> If you say so. Can you PROVE your claim?
>
> Hint: It's not clear what you mean be "reached" here.

A counted number has been reached by the counter

> It's unclear if your claim _means_:
>
> a) For every n in IN that can be reached one-by-one there is a FISON F, such that n is in F.
>
> or
>
> b) There is a FISION F such that for every n in IN that can be reached one-by-one n is in F.

(a) is correct. (b) fails because the definable numbers are potentially infinite. They can outperform every FISON.
>
> If all natural number "can be reached from 0" then a) is TRUE, but b) is FALSE. After all, IN is not a FISON, you know.
> > The infinite set with |ω| elements can never be reached step-by-step.
> Well, maybe IN "can never be reached step-by-step from 0", but it might be possible for each and every element in IN.
>
> It would be helpful, if you could give definition for
>
> x can be be reached step-by-step from 0
>
> => short: x can be reached
>
> => shorter: x is reachable
>
> (for any ordinal x).
>
> How about
>
> x is /reachable/ iff card({y e ORD : 0 < y < x}) e IN
>
> ?
>
> Meaning: there are FINITELY MANY consecutive ordinals "between" 0 and x.
>
> This way we would immediately get
>
> 0 is reachable, 1 is reachable, 2 is reachable, etc.
>
> Or more general:
> For all n e IN: n is reachable.
> (easy proof by induction).

All this is correct for definable numbers. They are subject to induction.
>
> And, for example,
> ω is NOT reachable.
>
> Btw. So if "Cantor" claimed, as you say, that "all natural nunbers can be reached", then he's right.
>
> That's quite a good thing after all, as you say, this "is [a] precondition for every bijection with ℕ".

No. The simplest proof is this: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
And the best proof is this:
XOOO...
XOOO...
XOOO...
XOOO...
....

Then the indices are taken from their initial positions and are distributed by Cantor's formula. The indexed fractions are denoted by X, the fractions without indices are denoted by O. Index 1 remains at 1/1. The next step takes the index 2 from 2/1 and attaches it to 1/2:

XXOO...
OOOO...
XOOO...
XOOO...
....

In the end, when all exchanges of X and O have been carried through, according to Cantor, we have

XXXX...
XXXX...
XXXX...
XXXX...
....

All positions occupied by O's are dark.

Regards, WM

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 where IN/omega is defined due to von Neumann, you silly crank.

FUCK OFF, YOU SILLY ASSHOLE!

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
....

The first column contains ℵo elements.
==> The complete figure contains ℵo elements.
No set of lines contains more than one line of the set.
No line contains ℵo elements.

Or would that argument become invalid for infinite sets of FISONs?
Chuckle. It holds for every finite set, i.e., for the potentially infinite set. And more is impossible by the fact that all FISONs are finite.

Regards, WM

On 10/11/2022 11:22 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 9. Oktober 2022 um 20:23:03 UTC+2:
>> On 10/9/2022 5:39 AM, WM wrote:

>>> But every FISON contains less than half of
>>> the infinite initial segment,
>>> i.e., of all natural numbers {1, 2, 3, ...}.
>>> Proof:
>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>> or
>>> ∀n ∈ ℕ_def: 2n < ℵo ==> n < ℵo/2.
>>
>> The 0-side of the 0-side,ω-side split
>> contains _all_ of the 0-side of the
>> 0-side,ω-side split, without exception.
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>
> Your discussion deleted since
> it does not refer to dark numbers.

What's to say? "They're dark."

But, what I could say, I said.

>>> Nevertheless it is impossible to distinguish
>>> different infinite initial segments.

>>> Why? Because the additional numbers are dark.
>>
>> The additional numbers, if dark, if existing,
>> are on the ω-side of the 0-side,ω-side split.

----
>>> Every FISON-end can be reached one-by-one.
>
>> Nothing on the 0-side of the 0-side,ω-side
>> split cannot be reached one-by-one.
>
> Not clear what you mean.

A _split_ partitions an ordered set into two non-∅
classes, each of one before each of the other.

There is a split between all which can be
reached one-by-one from 0 and everything else.

We must define everything else to be after
everything reachable, but I don't think you
object to that. Define it that way, and
anything reachable is before anything not
reachable. Hence, a split.

We can describe being reachable one-by-one
from 0

k is _reachable one-by-one from 0_ ⟺
an ordered set 𝐹ₖ exists
("set 𝐹ₖ" == "permanent collection 𝐹ₖ")
such that
0 begins 𝐹ₖ and k ends 𝐹ₖ and,
for each split of 𝐹ₖ
a last-before i and first-after j exist
and j = i⁺⁺

We CAN'T describe being NOT reachable one-by-one
from 0
-- not very much, beyond saying "otherwise".

That is the 0-side,ω-side which I refer to.

Let 𝓕 be the set of all sets 𝐹ₖ as described.

j is on the 0-side ⟺
𝐹 ∈ 𝓕 exists, j ∈ 𝐹 and j⁺⁺ ∉ 𝐹

j is on the ω-side ⟺
for each 𝐹 ∈ 𝓕 j ∉ 𝐹

<WM>
>
>> A leap which can't be reached one-by-one
>> is _not finite_
>
> Correct.
> It covers dark numbers which cannot be
> distingusihed.

Everything on the 0-side of the split
is not-dark.

A leap from the ω-side to the 0-side
is not-finite _because_
one of the splits does not have a last-before

Recall that, in a finite leap, each split
must have a last-before and a first-after.

Suppose ω exists and is first-after the split
(which is how ω is defined).

A leap from ω across the split is not-finite,
because there is no last-before for that split,
but
only not-dark numbers are covered by the leap,
because only 0-side numbers and ω are in
the leap.

The leaps are infinite _without_ dark numbers.
The leaps are infinite because one split
has no last-before. Ultimately, because
j⁺⁺ the end of 𝐹ⱼ⊕⟨j⁺⁺⟩ is 0-side too.

----
>>> Nevertheless it is impossible to distinguish
>>> different infinite initial segments.

Yes,
the elements on the ω-side have been
very vaguely described.

However,
one infinite initial segment is distinguished
from all the others, even so.

Define an _infinite initial segment_
to contain 0 and each of its successors j⁺⁺

A unique infinite initial segment exists
which is subset all infinite initial segments.
It is the _least infinite initial segment_
It is what everyone (non-WM) defines as ℕ

We can say a lot about ℕ

On 10/11/2022 8:32 AM, WM wrote:
> Chris M. Thomasson schrieb am Montag, 10. Oktober 2022 um 08:59:46 UTC+2:
>>
>> is the point 0+1/2 dark in the interval [0...1]?
>>
>> Well, not dark since I named it.
>
> Correct.
>
>> Think of a number, is it dark now?
>
> 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? Is that what you mean?

On 10/11/2022 8:45 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.
>> 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 natnumbers and ω which cannot be defined 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, this is impossible. If you subtract each and every n then nothing remains.
> |ℕ \ {1, 2, 3, ...}| = 0 .
> {0, 1, 2, 3, ..., ω} \ ℕ = {0, ω},

An infinite process is never completed for it has no end.

On 10/11/2022 11:37 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 10. Oktober 2022 um 13:46:05 UTC+2:

>> The natural numbers are
>> the least infinite initial segment.
>> ℕ = 𝕃𝕀𝕀𝕊
>
> 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.

If
your dark numbers manage to exist somehow in
some infinite initial segment,
then
the least infinite initial segment is subset
that partly-dark infinite initial segment too.

Also,
the least infinite initial segment is subset
any infinite initial segment NOT partly-dark.
No dark numbers are in _it_

A key step to that result is
to show that the intersection 𝔸∩𝔹 of
two infinite initial segments 𝔸 and 𝔹
is also an infinite initial segment.

So
the least infinite initial segment of 𝔸
is subset 𝔹
and
the least infinite initial segment of 𝔹
is subset
the least infinite initial segment of 𝔸

And also the other way around.
Each least infinite initial segment
is subset all
least infinite initial segments.

Therefore,
the least infinite initial segment
is unique.

On Tuesday, 11 October 2022 at 13:10:00 UTC-3, WM wrote:
> Fritz Feldhase schrieb am Montag, 10. Oktober 2022 um 15:22:48 UTC+2:
[...]
> > Hint: card({{0, ..., n} : n e IN} = aleph_0.
> No. ℵo > n. Counting FISONs by elements in FISONs we have to count ℵo elements in ℵo FISONs.

Learn to read, idiot! And after that, learn at least a *little* bit of mathematics.

tisdag 11 oktober 2022 kl. 17:27:48 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 10. Oktober 2022 um 07:32:27 UTC+2:
> > torsdag 6 oktober 2022 kl. 21:25:22 UTC+2 skrev WM:
> > > Fritz Feldhase schrieb am Mittwoch, 5. Oktober 2022 um 18:36:21 UTC+2:
>
> > > > Because for each and every natural number n there are infinitely many natural numbers (finite ordinals) between n and omega. That's why.
> > > >
> > > > Hope this helps.
> > > It helps to understand why they are dark. They cannot be removed. They cannot be the ends of leaps. They are undefinable.
> > No, it helps nothign because what you say has no meaning!
> So it appears to you. Don't worry.
>
> Regards, WM
So it is because your shit is not mathematics!

On Tuesday, October 11, 2022 at 6:10:00 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 10. Oktober 2022 um 15:22:48 UTC+2:
> >
> > Hint: card( {{0, ..., n} : n e IN} ) = aleph_0.
> >
> No.

Yes, you silly idiot.

{{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.

HUCK OFF, YOU SILLY ASSHOLE FULL OF SHIt!

On Tuesday, October 11, 2022 at 6:10:00 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 10. Oktober 2022 um 15:22:48 UTC+2:
> >
> > Hint: card( {{0, ..., n} : n e IN} ) = aleph_0.
> >
> No.

Yes, you silly idiot.

{{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.

HUCK OFF, YOU SILLY ASSHOLE FULL OF SHIt!

On Tuesday, October 11, 2022 at 6:10:00 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 10. Oktober 2022 um 15:22:48 UTC+2:
> >
> > Hint: card( {{0, ..., n} : n e IN} ) = aleph_0.
> >
> No.

Yes, you psychotic idiot.

{{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.

FUCK OFF, YOU SILLY ASSHOLE FULL OF SHIT!

Jim Burns schrieb am Dienstag, 11. Oktober 2022 um 20:46:25 UTC+2:
> On 10/11/2022 11:22 AM, WM wrote:

> There is a split between all which can be
> reached one-by-one from 0 and everything else.

That means there is a split behind the end of every FISON.

> We can describe being reachable one-by-one
> from 0

Yes, this is the potentially infinite union of FISONs.
>
> k is _reachable one-by-one from 0_ ⟺
> an ordered set 𝐹ₖ exists
> ("set 𝐹ₖ" == "permanent collection 𝐹ₖ")
> such that
> 0 begins 𝐹ₖ and k ends 𝐹ₖ and,
> for each split of 𝐹ₖ
> a last-before i and first-after j exist
> and j = i⁺⁺
>
Yes.

> We CAN'T describe being NOT reachable one-by-one
> from 0

That is the dark part.

>
> That is the 0-side,ω-side which I refer to.
>
> Let 𝓕 be the set of all sets 𝐹ₖ as described.

There is no fixed set 𝓕 but only a collection

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
∀n ∈ U𝓕: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>
> j is on the 0-side ⟺
> 𝐹 ∈ 𝓕 exists, j ∈ 𝐹 and j⁺⁺ ∉ 𝐹
>
> j is on the ω-side ⟺
> for each 𝐹 ∈ 𝓕 j ∉ 𝐹

There are ℵo dark = undefinable natnumbers . Therefore no such j can be defined.
>
> <WM>
> >
> >> A leap which can't be reached one-by-one
> >> is _not finite_
> >
> > Correct.
> > It covers dark numbers which cannot be
> > distingusihed.
>
> Everything on the 0-side of the split
> is not-dark.

and is in finite distance from 0 while an infinite dark realm has been covered.

> Recall that, in a finite leap, each split
> must have a last-before and a first-after.

Of course.
>
> Suppose ω exists and is first-after the split
> (which is how ω is defined).

No. If ω exists, then no definable number is next to it.

> The leaps are infinite _without_ dark numbers.

Then definable numbers or a gap would touch ω. Contradiction.

> A unique infinite initial segment exists
> which is subset all infinite initial segments.
> It is the _least infinite initial segment_
> It is what everyone (non-WM) defines as ℕ
>
> We can say a lot about ℕ

You have said a lot of nonsense already. Like the empty intersection of infinite inclusion- monotonic endsegments.

Regards, WM

On 10/12/2022 1:52 PM, WM wrote:
> Jim Burns schrieb am Dienstag, 11. Oktober 2022 um 20:46:25 UTC+2:
>> On 10/11/2022 11:22 AM, WM wrote:
>
>> There is a split between all which can be
>> reached one-by-one from 0 and everything else.
>
> That means there is a split behind the end of every FISON.
>
>> We can describe being reachable one-by-one
>> from 0
>
> Yes, this is the potentially infinite union of FISONs.
>>
>> k is _reachable one-by-one from 0_ ⟺
>> an ordered set 𝐹ₖ exists
>> ("set 𝐹ₖ" == "permanent collection 𝐹ₖ")
>> such that
>> 0 begins 𝐹ₖ and k ends 𝐹ₖ and,
>> for each split of 𝐹ₖ
>> a last-before i and first-after j exist
>> and j = i⁺⁺
>>
> Yes.
>
>> We CAN'T describe being NOT reachable one-by-one
>> from 0
>
> That is the dark part.
>
>>
>> That is the 0-side,ω-side which I refer to.
>>
>> Let 𝓕 be the set of all sets 𝐹ₖ as described.
>
> There is no fixed set 𝓕 but only a collection
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> ∀n ∈ U𝓕: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>
>> j is on the 0-side ⟺
>> 𝐹 ∈ 𝓕 exists, j ∈ 𝐹 and j⁺⁺ ∉ 𝐹
>>
>> j is on the ω-side ⟺
>> for each 𝐹 ∈ 𝓕 j ∉ 𝐹
>
> There are ℵo dark = undefinable natnumbers . Therefore no such j can be defined.
>>
>> <WM>
>>>
>>>> A leap which can't be reached one-by-one
>>>> is _not finite_
>>>
>>> Correct.
>>> It covers dark numbers which cannot be
>>> distingusihed.
>>
>> Everything on the 0-side of the split
>> is not-dark.
>
> and is in finite distance from 0 while an infinite dark realm has been covered.
>
>> Recall that, in a finite leap, each split
>> must have a last-before and a first-after.
>
> Of course.
>>
>> Suppose ω exists and is first-after the split
>> (which is how ω is defined).
>
> No. If ω exists, then no definable number is next to it.
>
>> The leaps are infinite _without_ dark numbers.
>
> Then definable numbers or a gap would touch ω. Contradiction.
>
>> A unique infinite initial segment exists
>> which is subset all infinite initial segments.
>> It is the _least infinite initial segment_
>> It is what everyone (non-WM) defines as ℕ
>>
>> We can say a lot about ℕ
>
> You have said a lot of nonsense already. Like the empty intersection of infinite inclusion- monotonic endsegments.

Are you some sort of Bot?

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.

> 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 .

Regards, WM

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.
then the complete figure contains ℵo elements.
But no set of lines contains more than one line of the set.
And no line contains ℵo elements.

Regards, WM

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.

>> 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.

Jim Burns schrieb am Dienstag, 11. Oktober 2022 um 22:05:16 UTC+2:
> On 10/11/2022 11:37 AM, WM wrote:
> > Jim Burns schrieb am Montag,
> > 10. Oktober 2022 um 13:46:05 UTC+2:
>
> >> The natural numbers are
> >> the least infinite initial segment.
> >> ℕ = 𝕃𝕀𝕀𝕊
> >
> > 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-

> Each least infinite initial segment
> is subset all
> least infinite initial segments.

Dark numbers cannot be distinguished or subdivided.

Up to every FISON-end all is finite. After the set of dark numbers (which cannot be subdivided) there is the infinite set ℕ.
>
> Therefore,
> the least infinite initial segment
> is unique.

It is unique because the set of dark numbers can only be used as whole.

Regards, WM

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.

> then the complete figure contains ℵo elements.

You got it, the natural numbers are still countable in two dimensional
tables. Wanna try for three?

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

On 10/12/2022 5:11 PM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 11. Oktober 2022 um 22:05:16 UTC+2:
>> On 10/11/2022 11:37 AM, WM wrote:
>>> Jim Burns schrieb am Montag,
>>> 10. Oktober 2022 um 13:46:05 UTC+2:

>>>> The natural numbers are
>>>> the least infinite initial segment.
>>>> ℕ = 𝕃𝕀𝕀𝕊
>>>
>>> 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.
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⁺⁺

Consider the subset of ℕ of numbers which are
not their own successors
{i ∈ ℕ : i ≠ i⁺⁺}

{i ∈ ℕ : i ≠ i⁺⁺} is an infinite initial
segment.
[1]

{i ∈ ℕ : i ≠ i⁺⁺} is an infinite initial
segment in ℕ
and ℕ is the only one of those.

{i ∈ ℕ : i ≠ i⁺⁺} = ℕ
∀j ∈ {i ∈ ℕ : i ≠ i⁺⁺} : j ≠ j⁺⁺
∀j ∈ ℕ : j ≠ j⁺⁺

That is know as _induction_
A critical part of using induction is knowing that
ℕ is the only infinite initial segment in ℕ

Thank you for agreeing to that.

----
[1]
{i ∈ ℕ : i ≠ i⁺⁺} is an infinite initial
segment.

| 0 ∈ {i ∈ ℕ : i ≠ i⁺⁺}
| || 0 ≠ 0++ because
|| 0 ≠ i++ for all i in ℕ
| | {i ∈ ℕ : i ≠ i⁺⁺} contains all its successors.
| || Assume j ∈ {i ∈ ℕ : i ≠ i⁺⁺}
|| j ≠ j⁺⁺
||
|| For the successor '⁺⁺'
|| Am An : m ≠ n -> m⁺⁺ ≠ n⁺⁺
|| In particular
|| j ≠ j⁺⁺ -> j⁺⁺ ≠ j⁺⁺⁺⁺
||
|| j⁺⁺ ≠ j⁺⁺⁺⁺
|| j⁺⁺ ∈ {i ∈ ℕ : i ≠ i⁺⁺}
| | if j ∈ {i ∈ ℕ : i ≠ i⁺⁺}
| then j⁺⁺ ∈ {i ∈ ℕ : i ≠ i⁺⁺}

{i ∈ ℕ : i ≠ i⁺⁺}
contains 0 and contains all its successors j⁺⁺

Therefore,
{i ∈ ℕ : i ≠ i⁺⁺}
is an infinite initial segment.

{i ∈ ℕ : i ≠ i⁺⁺} subset ℕ

>> Each least infinite initial segment
>> is subset all
>> least infinite initial segments.
>
> Dark numbers cannot be distinguished or subdivided.

ℕ is the only infinite initial segment in ℕ

Therefore,
we know an enormous amount about ℕ and
your dark numbers do not keep us from knowing.

On 10/6/2022 2:25 PM, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 5. Oktober 2022 um 18:36:21 UTC+2:
>> On Wednesday, October 5, 2022 at 11:10:59 AM UTC+2, WM wrote:
>>
>>> We can start [our leap at] omega and [jump to] 17. That is a leap.
>>> why must [there] be infinitely many natnumbers between omega and the end of the leap.
>>
>> Because the end of the leap is a natural number, and there are infinitely many natural numbers (finite ordinals) between any natural number and omega. That's why.
>
> If all of them were definable then all of them could be subtracted and nothing below omega would remain.

can you put that into an EQUATION ?

>
>>> But the leap from omega cannot pass less than aleph_0 natnumbers. Why that?
>> Because for each and every natural number n there are infinitely many natural numbers (finite ordinals) between n and omega. That's why.
>>
>> Hope this helps.
>
> It helps to understand why they are dark.

no such thing.

> They cannot be removed.

They are not there.

> They cannot be the ends of leaps.

They cannot be anything, they do not exist.

> They are undefinable.

They do not say that at all.

>
> Regards, WM

On 10/6/2022 2:28 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 5. Oktober 2022 um 23:00:29 UTC+2:
>> On 10/5/2022 5:10 AM, WM wrote:
>
>>> No, I ask
>>> why must be infinitely many natnumbers
>>> between omega and the end of the leap.
>> Because _there is no step_ from ω into ℕ.
>
> Irrelevant. There is a leap.

You called that "leap" a "gap" before, still stupid.

> Why must be infinitely many natnumbers
> between omega and the end of every leap into ℕ?

Leap Ants

>
> Regards, WM